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Symon-Mechanics

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Textbook by Keith R. Symon (University of Wisconsin), Addison-Wesley, second edition, 1960, kept in the archive's downloaded physics books. The preface describes a two-semester intermediate course: Newton's laws, one-dimensional motion and the harmonic oscillator, vectors, conservation laws, rigid bodies, gravitation, moving coordinates, then continuous media, Lagrange and Hamilton equations, tensors, rigid body rotation and small vibrations.

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J ’d‘____ _ IN‘_I Q__*__-‘ﬁllI ’‘J_‘_'___‘I___ OI __N OmDEﬁn DN‘I_I‘_I__F'M!__T_ __t H______J__I:I__-ll _-_ ‘ll;‘:_l“‘IN“Hi‘_.HqII|''EEl|'ft I‘_I C__II ___‘.M __ i¢M ar~--- __- .._ ___-- \_,.__1.;¢ I ‘ THE AUTHOR :£;_' '_"" ' """-FFF .-_-1» __________“.|____"_____"__d__|“|_“_"___d___H_I____“_"_r_I_______“_H_____r__"_"______H_____'__m_m__¢_'HI?_’__in____”______________H_______1:________________“_5H'___“__r'____|_“"_‘__2_'“__H_h_I______“_HH___’__J_uJ_J_r“_h;__h_'lm__1_b_____n_:______“__'_T__"___vm_"_m1_,1__"U____ _I__l||_|_‘llIIIIIIII_II_I________U____ ______U ___ _ ______'__|_|____________________H__________________'___m"_h______|____1___:_________'__£_=m__"n_~___h__"___5Z_____v_H___Hf?___~__H}_J_______”W____&1_E__F__ﬁ___ﬂ"_P_““:_Ej%_____H_\m_'_"_E$ﬁ_"__4_£_1m"_“H":___H_“m__?FHmum__n__wmEmﬁmﬁwlcﬁngnﬁﬂﬁJ_?________ __ MECHANICS b I I " RThis book isinthe ADDISON-WESLEY SERIES INPHYSICS MECHANICS by KEITH R.SYMON University ofWisconsin SECOND EDITION A 1YV A ADDISON-WESLEY. PUBLISHING COMPANY, INC. READING, MASSACHUSETTS, U.S.A. LONDON, ENGLANDv i 1 i \< J4 \ Copyright 1953, 1960 ADDISON-WESLEY PUBLISHING COMPANY, INC. __N Printed intheUnited States ofAmerica ALL RIGHTS RESERVED. THIS BOOK, OR PARTS THERE- OF, MAY NOT BE REPRODUCED IN ANY FORM WITH- OUT WRITTEN PERMISSION OFTHE PUBLISHER. Library ofCongress Catalog Card N0.60-5164 Second Edition TomyFather ' i . . | f PREFACE This textisintended asthebasis foranintermediate course inmechanics attheundergraduate level. Such acourse, asessential preparation for advanced work inphysics, hasseveral major objectives. Itmust develop inthestudent athorough understanding ofthefundamental principles of mechanics. Itshould treat indetail certain speciﬁc problems ofprimary importance inphysics, forexample, theharmonic oscillator, and the motion ofaparticle under acentral force. The problems suggested and those worked outinthetexthave been chosen with regard totheir in- terest and importance inphysics, aswell astotheir instructive value. This book contains suﬂicient material foratwo-semester course, andis arranged insuch away that, with appropriate omissions, itcanbeused forasingle three- orfour-hour course foronesemester. The author has used thematerial intheﬁrst seven chapters inathree-hour course in mechanics. The choice oftopics andtheir treatment throughout thebook arein- tended toemphasize themodem point ofview. Applications toatomic physics aremade wherever possible, withanindication astotheextent of thevalidity oftheresults ofclassical mechanics. Theinadequacies in classical mechanics arecarefully pointed out,andthepoints ofdeparture forquantum mechanics andforrelativistic mechanics areindicated. The development, except forthelastfour chapters, proceeds directly from Newton's laws ofmotion, which form asuitable basis from which toattack most mechanical problems. More advanced methods, using Lagrange’s equations andtensor algebra, areintroduced inthelastfour chapters. Animportant objective ofaﬁrst course inmechanics istotrain the student tothink about physical phenomena inmathematical terms. Most students have afairly good intuitive feeling formechanical phenomena in aqualitative way. The study ofmechanics should aimatdeveloping an almost equally intuitive feeling fortheprecise mathematical formulation ofphysical problems and forthephysical interpretation ofthemathe- matical solutions. The examples treated inthetext have been worked outsoastointegrate, asfaraspossible, themathematical treatment with thephysical interpretation. After working anassigned problem, the student should study ituntil hebissure heunderstands thephysical inter- pretation ofevery feature ofthemathematical treatment. Heshould de- cidewhether theresult agrees with hisphysical intuition about theprob- lem. Ifnot,then either hissolution orhisintuition should beappropriately corrected. Iftheanswer isfairly complicated, heshould trytoseewhether vn viii PREFACE itcanbesimpliﬁed incertain special orlimiting cases. Heshould tryto formulate andsolve similar problems onhisown. I Only aknowledge ofdifferential andintegral calculus hasbeen presup- posed. Mathematical concepts beyond those treated intheﬁrst year of calculus areintroduced andexplained asneeded. Aprevious course in elementary differential equations orvector analysis may behelpful, butit istheauthor’s experience that students with anadequate preparation in algebra andcalculus areabletohandle thevector analysis anddifferential equations needed forthis course with theexplanations provided herein. Aphysics student islikely togetmore outofhisadvanced courses in mathematics ifhehaspreviously encountered these concepts inphysics. The text hasbeen written soastoafford maximum ﬂexibility inthe selection andarrangement oftopics tobecovered. With certain obvious exceptions, many sections orgroups ofsections canbepostponed or omitted without prejudice totheunderstanding oftheremaining material. Where particular topics presented earlier areneeded inlater parts ofthe book, references tosection andequation numbers make iteasy tolocate theearlier material needed. Intheﬁrst chapter thebasic concepts ofmechanics arereviewed, and thelaws ofmechanics andofgravitation areformulated andapplied toa fewsimple examples. Thesecond chapter undertakes afairly thorough study oftheproblem ofone-dimensional motion. Thechapter concludes with astudy oftheharmonic oscillator asprobably themost important example ofone-dimensional motion. Useismade ofcomplex numbers to represent oscillating quantities. Thelastsection, ontheprinciple ofsuper- position, makes some useofFourier series, andprovides abasis forcertain parts ofChapters 8and12.Ifthese chapters arenottobecovered, Sec- tion2-11 may beomitted or,better, skimmed togetabrief indication of thesigniﬁcance oftheprinciple ofsuperposition andtheway inwhich Fourier series areused totreat theproblem ofanarbitrary applied force function. Chapter 3begins with adevelopment ofvector algebra anditsusein describing motions inaplane orinspace. Boldface letters areused for vectors. Section 3—6isabrief introduction tovector analysis, which is used very little inthisbook except inChapter 8,anditmay beomitted orskimmed ifChapter 8andafewproofs insome other chapters are omitted. The author feels there issome advantage inintroducing the student totheconcepts andnotation ofvector analysis atthisstage, where thelevel oftreatment isfairly easy; inlater courses where thephysical concepts andmathematical treatment become more diﬂicult, itwillbewell ifthenotations arealready familiar. Thetheorems stating thetime rates ofchange ofmomentum, energy, andangular momentum arederived for amoving particle, andseveral problems arediscussed, ofwhich motion PREFACE ix under central forces receives major attention. Examples aretaken from astronomical andfrom atomic problems. 1 InChapter 4theconservation laws ofenergy, momentum, andangular momentum arederived, with emphasis ontheir position ascornerstones of present-day physics. They arethen applied totypical problems, particu- larly collision problems. Thetwo-body problem issolved, andthemotion oftwocoupled harmonic oscillators isworked out. Thegeneral theory of coupled oscillations isbest treated bymeans oflinear transformations in vector spaces, asinChapter 12,butthebehavior ofcoupled oscillating systems istooimportant tobeomitted altogether from even aone-semester course. Thesection ontwocoupled oscillators canbeomitted orpostponed until Chapter 12.The rigid body isdiscussed inChapter 5asaspecial kind ofsystem ofparticles. Only rotation about aﬁxed axisistreated; themore general study ofthemotion ofarigid body islefttoalater chap- ter,where more advanced methods areused. Thesection onstatics treats theproblem ofthereduction ofasystem offorces toanequivalent simpler system. Elementary treatments oftheequilibrium ofbeams, ﬂexible strings, andofﬂuids aregiven inSections 5-9, 5-10, and5-11. The theory ofgravitation isstudied insome detail inChapter 6.The lastsection, onthegravitational ﬁeld equations, may beomitted without disturbing thecontinuity= oftheremaining material. The laws ofmotion inmoving coordinate systems areworked outinChapter 7,andapplied tomotion ontherotating earth andtothemotion ofasystem ofcharged particles inamagnetic ﬁeld. Particular attention ispaid tothestatus inNewtonian mechanics ofthe “ﬁctitious forces” which appear when moving coordinate systems areintroduced, andtotheroletobeplayed bysuch forces inthegeneral theory ofrelativity. The lastﬁvechapters cover more advanced material andaredesigned primarily tobeused inthesecond semester ofatwo-semester course in intermediate mechanics. Inashorter course, anyorallofthelastﬁve chapters may beomitted without destroying theunity ofthecourse, although theauthor hasfound itpossible toutilize parts ofChapter 8or 9even inaone-semester course. InChapter 8anintroductory treatment ofvibrating strings andofthemotion ofﬂuids ispresented, with emphasis onthefundamental concepts andmathematical methods used intreating themechanic ofcontinuous media. Chapter 9onLagrange's equations isintended asanintroduction tothemethods ofadvanced dynamics. Hamilton’s equations andtheconcept ofphase space arepresented, since they areprerequisite toanylater course inquantum mechanics orstatis- tical mechanics, butthetheory ofcanonical transformations andtheuse ofvariational principles arebeyond thescope ofthisbook. Chapter 10 develops thealgebra oftensors, including orthogonal coordinate trans- formations, which arerequired inthelasttwochapters. Theinertia tensor > l l1 l Fx PREFACE andthestress tensor aredescribed insome detail asexamples. Section 10-6 onthestress tensor willenable thereader toextend thediscussion ofideal ﬂuids inChapter 8toasolid orviscous medium. The methods developed inChapters 9and10areapplied inChapter 11tothegeneral rotation ofarigid body about apoint, andinChapter 12tothestudy of small vibrations ofaphysical system about astate ofequilibrium orof steady motion. The problems attheendofeach chapter arearranged intheorder inwhich thematerial iscovered inthechapter, forconvenience inas- signment. Anattempt hasbeen made toinclude asuiﬁcient variety of problems toguarantee that anyone who cansolve them hasmastered the material inthetext. The converse isnotnecessarily true, since most problems require more orlessphysical ingenuity inaddition toanunder- standing ofthetext. Many oftheproblems arefairly easy andshould be tractable foranyone who hasunderstood thematerial presented. Afew areprobably toodiﬂicult formost college juniors orseniors tosolve with- outsome assistance. Those problems which areparticularly diﬂicult or time-consuming aremarked with anasterisk. The lastthree chapters andthelastthree sections ofChapter 9have been added tothepresent edition inorder toprovide enough material for afulltwo-semester course inmechanics. Except forcorrections anda fewminor. changes andadditions, theﬁrsteight chapters andtheﬁrst eight sections ofChapter 9remain thesame asintheﬁrstedition ofthis text. Grateful acknowledgment ismade toProfessor Francis W.Sears of Dartmouth College andtoProfessor George H.Vineyard ofBrookhaven National Laboratory fortheir many helpful suggestions, andtoMr.Charles Vittitoe andMr.Donald Roiseland foracritical reading ofthelastfour chapters. The author isparticularly grateful tothemany teachers and students who have offered corrections and suggestions forimprovement which have been incorporated inthisrevised edition. While space does not permit mentioning individuals here, Ihope that each may ﬁndmythanks expressed inthechanges that have been made inthisedition. January, 1.960 K.R.S. CONTENTS CHAPTER 1.Enmmms orNEWTONIAN MECHANICS . . 1-1 1-2 I»-4|-4|—lP-‘I-1\IO>O1>i>0OMechanics, anexact science ...... . Kinematics, thedescription ofmotion . . Dynamics. Mass andforce ... . Newton’s laws ofmotion .... . Gravitation ......... . Units anddimensions ....... . Some elementary problems inmechanics . . CHAPTER 2.MOTION orAPARTICLE INONE DIMENSION .... 2-1 2-2 2-3 2-4 2-5 2-6 2-7 2-8 2-9Momentum andenergy theorems ......... Discussion ofthegeneral problem ofone-dimensional motion . Applied force depending onthetime. ........ Damping force depending onthevelocity ....... Conservative force depending onposition. Potential energy . Falling bodies ............... The simple harmonic oscillator ........ . Linear differential equations with constant coeﬁicients . . Thedamped harmonic oscillator .......... -2-10 Theforced harmonic oscillator .......... 2-11_Theprinciple ofsuperposition. Harmonic oscillator with arbitrary appliedforce............... CHAPTER 3.MOTION orAPARTICLE INTwo onTHREE DIMENSIONS 3-1 3-2 3-3 3-4 3-5 3-6 COOOQD<O®\IVector algebra ............... Applications toasetofforces acting onaparticle . . Diﬁerentiation andintegration ofvectors ... . Kinematics inaplane ...... . Kinematics inthree dimensions . . Elements ofvector analysis ...... . Momentum andenergy theorems ......... Plane andvector angular momentum theorems ..... Discussion ofthegeneral problem oftwo-andthree-dimensional motion................. 3-10 The harmonic oscillator intwoandthree dimensions .. . 3-11 Projectiles ............. . 3-12 Potential energy .............. 3-13 Motion under acentral force ........... 3-14 The central force inversely proportional tothesquare ofthe distance................. xiUlbil-"l-" 7 10 11 13 21 21 22 25 28 30 35 39 41 47 50 59 68 68 77 81 87 91 95 100 101 104 106 108 112 120 125 xii CONTENTS 3-15 Elliptic orbits. TheKepler problem ......... 3-16 Hyperbolic orbits. The Rutherford problem. Scattering cross section................. 3-17 Motion ofaparticle inanelectromagnetic ﬁeld ..... CHAPTER 4.THE MoT1oN orASYsTEM orPARTICLES .. . 4-1 Conservation oflinear momentum. Center ofmass . . — Conservation ofangular momentum .... . — Conservation ofenergy .... . — Critique oftheconservation laws . . - Rockets, conveyor belts, andplanets . . - Collision problems .............. — The two-body problem ............ - Center-of-mass coordinates. Rutherford scattering byacharged particle ofﬁnite mass ............. The N-body problem ............. 0Two coupled harmonic oscillators ....... .H>r¥>»-Pr-Pl-¥>rI>r-I>00U1:-l>C»Jl\DK10’-3 r-Pr-l>|—1<O CHAPTER 5.R1011)‘ Booms. RoTATIoN ABoUT ANAXIS. STATIcs .. 5-1 5-2 5-3 5-4 5-5 5-6 5-7Thedynamical problem ofthemotion ofarigid body . . Rotation about anaxis .........' . Thesimple pendulum .......... . The"compound pendulum .......... . Computation ofcenters ofmass andmoments ofinertia . . Statics ofrigid bodies ........... . Statics ofstructures .......... . 5-8 Stress andstrain ........ . 5-9.Equilibrium ofﬂexible strings andcables . . 5-10 Equilibrium ofsolid beams ..... . 5-11 Equilibrium ofﬂuids .. . CHAPTER 6.GRAv1TAT1oN. ....... . 6-1 Centers ofgravity forextended bodies ... . 6-2 Gravitational ﬁeld andgravitational potential .. . 6-3 Gravitational ﬁeld equations ... CHAPTER 7.’MOVING CooRnINATE SYsTEMs . . 7-1 Moving origin ofcoordinates ... . 7-2 Rotating coordinate systems .... . — Laws ofmotion ontherotating earth . . The Foucault pendulum .... . Larmor’s theorem ..... . The restricted three-body problem . . \l\l\‘|\IC5U1>-LOO132 135 139 155 155 158 162 165 168 17]. 178 181 185 188 203 203 206 208 212 215 225 231 232 235 239 245 257 257 259 262 269 269 271 278 280 283 285 CONTENTS CHAPTER 8.INTRODUCTION TOTHEMEcHAN1cs orCoNTINUoUs MEn1A 8-1 The equation ofmotion forthevibrating string ....~. - Normal modes ofvibration forthevibrating string . . -Wave propagation along astring ...... . -The string asalimiting case ofasystem ofparticles . . - General remarks onthepropagation ofwaves ... . -Kinematics ofmoving ﬂuids ....... . -_Equations ofmotion foranideal ﬂuid . . -Conservation lawsforﬂuid motion .. . -Steady ﬂow ........... . -0Sound waves .......... . 1Normal vibrations ofﬂuid inarectangular box . . 2Sound waves inpipes ........ . 3The Mach number .. . . 4Viscosity ... . 00OOOOOOOOOOOOOOOOOOOOOO»—*>—*>—*QOO0\10>UIrI>0Jl\') 00 r-1|-A CHAPTER 9.LAoRANGE’s EoUAT1oNs. . . 9-1 Generalized coordinates .. . 9-2 Lagrange’s equations ... . 9-3 Examples ...... . . 9-4 Systems subject toconstraints ...... . —Examples ofsystems subject toconstraints ... . — Constants ofthemotion andignorable coordinates . . -Further examples ............ . Electromagnetic forces andvelocity-dependent potentials .. Lagrange’s equations forthevibrating string .... . -0Hamilton’s equations ........ . —1Liouville’s theorem ....... . <D<OQOQO<O<O©P-*>—‘<O®\IO>Ul CHAPTER 10.TENsoR ALGEBRA. INERTIA ANDSTREss TENsoRs .. 10-1 Angular momentum ofarigid body .._..... . 10-2 Tensor algebra ....... . 10-3 Coordinate transformations .... . 10-4 Diagonalization ofasymmetric tensor . . 10-5 Theinertia tensor ....... . 10-6 Thestress tensor ...... . CHAPTER 11. THE RoTATIoN orARIGID BODY . . 11-1 Motion ofarigid body inspace .... . 11-2 Euler’s equations ofmotion forarigid body . . 11-3 Poinsot’s solution forafreely rotating body . . 11-4 Euler’s angles ......... . 11-5 Thesymmetrical top . .xiii 294 294 296 300 305 310 313 321 323 329 332 337 341 343 345 354 354 365 368 369 375 381 384 388 391 396 399 406 406 407 414 421 430 438 450 450 451 455 458 4611< I 1 xiv CONTENTS CHAPTER 12.THEORY orSMALL V1RRATIoNs 12-1 12-2 12-3 12-4 12-5 12-6 12-7 12-8Normal modes ofvibration ... Forced vibrations ...... Perturbation theory ..... Small vibrations about steady motion Betatron oscillations inanaccelerator Stability ofLagrange’s three bodies . BIBLIOGRAPHY ....... ANswERs ToODD-NUMBERED PROBLEMS . LIsT orSYMBOLS . INDEX .Condition forstability near anequilibrium conﬁguration .. Linearized equations ofmotion near anequilibrium conﬁguration473 473 475 477 .481 484 490 497 500 .515 .521 .533 .545 CHAPTER 1 ELEMENTS OFNEWTONIAN MECHANICS 1-1Mechanics, anexact science. When wesaythat physics isan exact science, wemean that itslaws areexpressed intheform ofmathe- matical equations which describe andpredict theresults ofprecise quanti- tative measurements. Theadvantage inaquantitative physical theory is notalone thepractical onethat itgives usthepower accurately topredict and tocontrol natural phenomena. Byacomparison oftheresults of accurate measurements with thenumerical predictions ofthetheory, we cangain considerable conﬁdence that thetheory iscorrect, andwecan determine inwhat respects itneeds tobemodiﬁed. Itisoften possible toexplain agiven phenomenon inseveral rough qualitative ways, andif wearecontent with that, itmay beimpossible todecide which theory is correct. Butifatheory canbegiven which predicts correctly theresults ofmeasurements tofour orﬁve(oreven twoorthree) signiﬁcant ﬁgures, thetheory canhardly bevery farwrong. Rough agreement might bea coincidence, butclose agreement isunlikely tobe.Furthermore, there have been many cases inthehistory ofscience when small butigniﬁcant discrepancies between theory andaccurate measurements have ledtothe development ofnew andmore “far-reaching theories. Such slight discrep- ancies would noteven have been detected ifwehadbeen content with a merely qualitative explanation ofthephenomena. Thesymbols which aretoappear intheequations that express thelaws ofascience must represent quantities which canbeexpressed innumerical terms. Hence theconcepts interms ofwhich anexact science istobe developed must begiven precise numerical meanings. Ifadeﬁnition ofa quantity (mass, forexample) istobegiven, thedeﬁnition must besuch astospecify precisely how thevalue ofthequantity istobedetermined inanygiven case. Aqualitative remark about itsmeaning may behelpful, butisnotsuﬁicient asadeﬁnition. Asamatter offact, itisprobably not possible togive anideally precise deﬁnition ofevery concept appearing in aphysical theory. Nevertheless, when wewrite down amathematical equation, thepresumption isthat thesymbols appearing inithave precise meanings, andweshould strive tomake ourideas asclear andprecise as possible, andtorecognize atwhat points there isalack ofprecision or clarity. Sometimes anewconcept canbedeﬁned interms ofothers whose meanings areknown, inwhich casethere isnoproblem. Forexample, momentum =mass Xvelocity 1 2 ELEMENTS orNEWTONIAN MECHANICS [cH.u>. 1 gives aperfectly precise deﬁnition of“momentum” provided “mass” and “velocity ”areassumed tobeprecisely deﬁned already. Butthiskind of deﬁnition willnotdoforallterms inatheory, since wemust start some- where with asetofbasic concepts or“primitive” terms whose meanings areassumed known. Theﬁrst concepts tobeintroduced inatheory can- notbedeﬁned intheabove way, since atﬁrstwehave nothing toputon theright side oftheequation. The meanings ofthese primitive terms must bemade clear bysome means that liesoutside ofthephysical theories being setup.Wemight, forexample, simply usetheterms over andover until their meanings become clear. This isthewaybabies learn alanguage, andprobably, tosome extent, freshman physics students learn thesame way. Wemight deﬁne allprimitive terms bystating their meaning in terms ofobservation andexperiment. Inparticular, nouns designating measurable quantities, likeforce, mass, etc., may bedeﬁned byspecifying theoperational process formeasuring them. Oneschool ofthought holds that allphysical terms should bedeﬁned inthisway. Orwemight simply state what theprimitive terms are,with arough indication oftheir physi- calmeaning, andthen letthemeaning bedetermined more precisely by thelaws andpostulates welaydown andtherules that wegive forinter- preting theoretical results interms ofexperimental situations. This isthe most convenient andﬂexible way, andistheway physical theories are usually setup.Ithasthedisadvantage thatwearenever surethatour concepts have been given aprecise meaning. Itislefttoexperience to decide notonly whether ourlaws arecorrect, buteven whether thecon- cepts weusehave aprecise meaning. The modern theories ofrelativity andquanta arise asmuch from fuzziness inclassical concepts asfrom in- accuracies inclassical laws. Historically, mechanics wastheearliest branch ofphysics tobedeveloped asanexact science. Thelaws oflevers andofﬂuids instatic equilibrium were known toGreek scientists inthethird century B.C. Thetremendous development ofphysics inthelastthree centuries began with thediscovery ofthelaws ofmechanics byGalileo andNewton. Thelaws ofmechanics asformulated byIsaac Newton inthemiddle oftheseventeenth century andthelaws ofelectricity andmagnetism asformulated byJames Clerk Maxwell about twohundred years later arethetwobasic theories ofclassi- calphysics. Relativistic physics, which began with thework ofEinstein in1905, and quantum physics, asbased upon thework ofHeisenberg andSchroedinger in1925-1926, require amodiﬁcation andreformulation ofmechanics and electrodynamics interms ofnew physical concepts. Nevertheless, modern physics builds onthefoundations laidbyclassical physics, andaclear understanding oftheprinciples ofclassical mechanics andelectrodynamics isstillessential inthestudy ofrelativistic andquan- tum physics. Furthermore, inthevast majority ofpractical applications ofmechanics tothevarious branches ofengineering andtoastronomy, the 0 1-1] MECHANICS, ANEXACT SCIENCE 3 laws ofclassical mechanics canstillbeapplied. Except when bodies travel atspeeds approaching thespeed oflight, orwhen enormous masses or enormous distances areinvolved, relativistic mechanics gives thesame re- sults asclassical mechanics; indeed, itmust, since weknow from experi- ence that classical mechanics gives correct results inordinary applications. Similarly, quantum mechanics should anddoes agree with classical mechan- icsexcept when applied tophysical systems ofmolecular sizeorsmaller. Indeed, oneofthechief guiding principles informulating new physical theories istherequirement that they must agree with theolder theories when applied tothose phenomena where theolder theories areknown to becorrect. Mechanics isthestudy ofthemotions ofmaterial bodies. Mechanics may bedivided intothree subdisciplines, kinematics, dynamics, andstatics. Kinematics isthestudy anddescription ofthepossible motions ofmate- rialbodies. Dynamics isthestudy ofthelaws which determine, among allpossible motions, which motion willactually take place inanygiven case. Indynamics weintroduce theconcept offorce. The central prob- lemofdynamics istodetermine foranyphysical system themotions which willtake place under theaction ofgiven forces. Statics isthestudy of forces andsystems offorces, with particular reference tosystems offorces which actonbodies atrest. Wemay alsosubdivide thestudy ofmechanics according tothekind of physical system tobestudied. This is,ingeneral, thebasis fortheoutline ofthepresent book. The simplest physical system, andtheoneweshall study ﬁrst, isasingle particle. Next weshall study themotion ofasys- temofparticles. Arigid body may betreated asaspecial kind ofsystem ofparticles. Finally, weshall study themotions ofcontinuous media, elastic andplastic substances, solids, liquids, andgases. Agreat many oftheapplications ofclassical mechanics may bebased directly onNewton’s laws ofmotion. Alloftheproblems studied inthis book, except inChapters 9-12, aretreated inthisway. There are,how- ever, anumber ofother ways offormulating theprinciples ofclassical mechanics. The equations ofLagrange and ofHamilton areexamples. They arenotnewphysical theories, forthey may bederived from Newton’s laws, butthey aredifferent ways ofexpressing thesame physical theory. They usemore advanced mathematical concepts, they areinsome respects more elegant than Newton’s formulation, andthey areinsome cases more powerful inthat they allow thesolutions ofsome problems whose solution based directly onNewton’s laws would bevery diﬂicult. Themore differ- entways weknow toformulate aphysical theory, thebetter chance we have oflearning how tomodify ittoﬁtnew kinds ofphenomena asthey arediscovered. This isoneofthemain reasons fortheimportance ofthe more advanced formulations ofmechanics. They areastarting point for thenewer theories ofrelativity andquanta. , 0H4 ELEMENTS orNEWTONIAN MECHANICS [cHAP. 1 1-2Kinematics, thedescription ofmotion. Mechanics isthescience which studies themotions ofphysical bodies. Wemust ﬁrstdescribe mo- tions. Easiest todescribe arethemotions ofaparticle, that is,anobject whose sizeandinternal structure arenegligible fortheproblem with which weareconcerned. Theearth, forexample, could beregarded asaparticle formost problems inplanetary motion, butcertainly notforterrestrial problems. Wecandescribe theposition ofaparticle byspecifying apoint inspace. This may bedone bygiving three coordinates. Usually, rec- tangular coordinates areused. Foraparticle moving along astraight line (Chapter 2)only onecoordinate need begiven. Todescribe themotion ofaparticle, wespecify thecoordinates asfunctions oftime: onedimension: x(t), 11 three dimensions: x(t), y(t), z(t). () Thebasic problem ofclassical mechanics istoﬁndways todetermine func- tions likethese which specify thepositions ofobjects asfunctions oftime, foranymechanical situation. The physical meaning ofthefunction x(t) iscontained intherules which tellushowtomeasure thecoordinate asofa particle atatime t.Assuming weknow themeaning ofa;(t), oratleast that ithasameaning (this assumption, which wemake inclassical me- chanics, isnotquite correct according toquantum mechanics), wecan deﬁne thea:-component ofvelocity 12,,attime tas* ~$ s \L-_----Q>-U~—-l1-“is 1»,=:i:=%, (1-2) and, similarly, .dy .dzvuiyigfy vziziwl 1/"axis Wenow deﬁne thecomponents of , acceleration a,,a,,,a,asthederiva- /4--A---n tives ofthevelocity components ‘__ 1"'_-' with respect totime (welistseveral equivalent notations which may be used): 0 P a 1, dv, 5 dzxz1 11 i 1" Z i 7_**'J_ x___, dt dt2 . . 2onedimension av=1)”=% =17=%t%, (1__3):0-axis three dimensions FIG. 1-1. Rectangular coordinates 2 specifying theposition ofaparticle Pa_._Qv_=_ _Q_z_ relative toanorigin O. ‘_U‘_dt_Z"'dtz *Weshall denote atime derivative either byd/dt orbyadot. Both notations aregiven inEq.(1-2). 1-3] DYNAMICS. MASS AND FORCE 5 Formany purposes some other system ofcoordinates may bemore con- venient forspecifying theposition ofaparticle. When other coordinate systems areused, appropriate formulas forcomponents ofvelocity and acceleration must beworked out. Spherical, cylindrical, andplane polar coordinates willbediscussed inChapter 3.Forproblems intwoandthree dimensions, theconcept ofavector isvery useful asameans ofrepresent- ingpositions, velocities, andaccelerations. Asystematic development of vector algebra willbegiven inSection 3-1. Todescribe asystem ofparticles, wemay specify thecoordinates of each particle inanyconvenient coordinate system. Orwemay introduce other kinds ofcoordinates, forexample, thecoordinates ofthecenter of mass, orthedistance between twoparticles. Iftheparticles form a"rigid body, thethree coordinates ofitscenter ofmass andthree angular coordi- nates specifying itsorientation inspace aresuﬂicient tospecify itsposition. Todescribe themotion ofcontinuous matter, forexample aﬂuid, wewould need tospecify thedensity p(:c,y,z,t)atanypoint (av,y,z)inspace ateach instant tintime, andthevelocity vector v(a:,y,z,t)with which thematter atthepoint (x,y,z)ismoving attime t.Appropriate devices fordescrib- ingthemotion ofphysical systems willbeintroduced asneeded. 1-3Dynamics. Mass andforce. Experience leads ustobelieve that themotions ofphysical bodies arecontrolled byinteractions between them andtheir surroundings. Observations ofthebehavior ofprojectiles and ofobjects sliding across smooth, well-lubricated surfaces suggest theidea that changes inthevelocity ofabody areproduced byinteraction with its surroundings. Abody isolated from allinteractions would have acon- stant velocity. Hence, informulating thelawsofdynamics, wefocus our attention onaccelerations. Letusimagine twobodies interacting with each other and otherwise isolated from interaction with their surroundings. Asarough approxima- tion tothissituation, imagine twoboys, notnecessarily ofequal size, en- gaged inatugofwarover arigid pole onsmooth ice. Although notwo actual bodies canever beisolated completely from interactions with all other bodies, thisisthesimplest kind ofsituation tothink about andone forwhich weexpect thesimplest mathematical laws. Careful experiments with actual bodies lead ustoconclusions astowhat weshould observe ifwecould achieve ideal isolation oftwobodies. Weshould observe that thetwobodies arealways accelerated inopposite directions, andthat the ratio oftheir accelerations isconstant foranyparticular pair ofbodies no matter how strongly they may bepushing orpulling each other. Ifwe measure thecoordinates ac;and1:2ofthetwobodies along thelineoftheir accelerations, then 131/5'32 =—k12, (1-4) 6 p ELEMENTS orNEWTONIAN MECHANICS [cHAP. 1 where I012isapositive constant characteristic ofthetwobodies concerned. Thenegative signexpresses thefactthat theaccelerations areinopposite directions. Furthermore, weﬁndthat ingeneral thelarger orheavier ormore mas- sivebody isaccelerated theleast. Weﬁnd, infact, thatthe ratio I012is proportional totheratio oftheweight ofbody 2tothat ofbody 1.The accelerations oftwointeracting bodies areinversely proportional totheir weights. This suggests thepossibility ofadynamical deﬁnition ofwhat weshall callthemasses ofbodies interms oftheir mutual accelerations. Wechoose astandard body asaunit mass. Themass ofanyother body isdeﬁned astheratio oftheacceleration oftheunit mass totheaccelera- tion oftheother body when thetwoareininteraction: mt=kn="551/55¢, (1-5) where m,-isthemass ofbody i,andbody 1isthestandard unit mass. Inorder that Eq.(1-5) may beauseful deﬁnition, theratio kmofthe mutual accelerations oftwobodies must satisfy certain requirements. If themass deﬁned byEq.(1-5) istobeameasure ofwhat wevaguely call theamount ofmatter inabody, then themass ofabody should bethesum ofthemasses ofitsparts, andthisturns outtobethecase toavery high degree ofprecision. Itisnotessential, inorder tobeuseful inscientiﬁc theories, that physical concepts forwhich wegiveprecise deﬁnitions should correspond closely toanypreviously held common-sense ideas. However, most precise physical concepts have originated from more orlessvague common-sense ideas, andmass isagood example. Later, inthetheory ofrelativity, theconcept ofmass issomewhat modiﬁed, anditisnolonger exactly true that themass ofabody isthesum ofthemasses ofitsparts. Onerequirement which iscertainly essential isthat theconcept ofmass beindependent oftheparticular body which happens tobechosen as having unit mass, inthesense that theratio oftwomasses willbethe same nomatter what unit ofmass may bechosen. This willbetrue be- cause ofthefollowing relation, which isfound experimentally, between themutual acceleration ratios deﬁned byEq.(1-4) ofanythree bodies: 761279237931 =1- (1-5) Suppose that body 1istheunit mass. Then ifbodies 2and3interact with each other, weﬁnd, using Eqs. (1-4), (1-6), and(1-5), 552/553 =—k23 ="1/(k12k31) (1-7) =-7613/7@12 =—m3/mg. 1-4] NEwToN’s LAWS orMOTION 7 The ﬁnal result contains noexplicit reference tobody 1,which wastaken tobethestandard unit mass. Thus theratio ofthemasses ofanytwo bodies isthenegative inverse oftheratio oftheir mutual accelerations, in- dependently oftheunit ofmass chosen. ByEq.(1-7), wehave, fortwointeracting bodies, ‘WI/2.’-5-2 = _’!7'l/1&1. ‘ This suggests that thequantity (mass ><acceleration) willbeimportant, andwecallthisquantity theforce acting onabody. Theacceleration of abody inspace hasthree components, andthethree components offorce acting onthebody are F,=mi, F,=mi], F,= (1-9) The forces which actonabody areofvarious kinds, electric, magnetic, gravitational, etc., and depend onthebehavior ofother bodies. In general, forces duetoseveral sources may actonagiven body, anditis found that thetotal force given byEqs. (1-9) isthevector sum ofthe forces which would bepresent ifeach source were present alone. The theory ofelectromagnetism isconcerned with theproblem ofde- termining theelectric andmagnetic forces exerted byelectrical charges andcurrents upon oneanother. The theory ofgravitation isconcerned with theproblem of3determining thegravitational forces exerted by masses upon oneanother. The fundamental problem ofmechanics isto determine themotions ofanymechanical system, given theforces acting onthebodies which make upthesystem. 1-4Newton’s laws ofmotion. Isaac Newton was theﬁrst togive a complete formulation ofthelaws ofmechanics. Newton stated hisfamous three laws asfollows:* (1)Every body continues initsstate ofrest orofuniform motion inastraight lineunless_it iscompelled tochange that state byforces impressed upon it. (2)Rate ofchange ofmomentum isproportional totheimpressed force, andisinthedirection inwhich theforce acts. (3)Toevery action there isalways opposed anequal reaction. Inthesecond law,momentum istobedeﬁned astheproduct ofthemass and thevelocity oftheparticle. Momentum, forwhich weusethesymbol *Isaac Newton, Mathematical Principles ofNatural Philosophy andhisSystem oftheWorld, tr.byF.Cajori (p.13). Berkeley: University ofCalifornia Press, 1934. 8 ELEMENTS orNEwToNIAN MECHANICS [cHAP. 1 p,hasthree components, deﬁned along ac-,y-,andz-axes bytheequations 11,,=mv,,, p,,=mvy, p,=mvz. (1-10) Theﬁrsttwolaws, together with thedeﬁnition ofmomentum, Eqs. (1-10), andthefact that themass isconstant byEq.(1—4),* areequivalent to Eqs. (1-9), which express them inmathematical form. The third law states that when twobodies interact, theforce exerted onbody 1bybody 2 isequal andopposite indirection tothat exerted onbody 2bybody 1. This lawexpresses theexperimental fact given byEq. (1-4), and can easily bederived from Eq.(1-4) andfrom Eqs. (1-5) and(1-9). Thestatus ofNewton’s ﬁrsttwolaws, orofEqs. (1-9), isoften thesub- jectofdispute. Wemay regard Eqs. (1-9) asdeﬁning force interms of mass andacceleration. Inthiscase, Newton’s ﬁrsttwolaws arenotlaws atallbutmerely deﬁnitions ofanewconcept tobeintroduced inthetheory. Thephysical laws arethen thelaws ofgravitation, electromagnetism, etc., which telluswhat theforces areinanyparticular situation. Newton’s discovery was notthat force equals mass times acceleration, for/this is merely adeﬁnition of“force.” What Newton discovered was that the laws ofphysics aremost easily expressed interms oftheconcept offorce deﬁned inthisway. Newton’s third lawisstillalegitimate physical law expressing theexperimental result given byEq.(1-4) interms ofthecon- ceptofforce. This point ofview toward Newton’s ﬁrsttwolawsiscon- venient formany purposes andisoften adopted. Itschief disadvantage is that Eqs. (1-9) deﬁne only thetotal force acting onabody, whereas we often wish tospeak ofthetotal force asa(vector) sumofcomponent forces ofvarious kinds duetovarious sources. The whole science ofstatics, which deals with theforces acting instructures atrest, would beunintelli- gible ifwetook Eqs. (1-9) asourdeﬁnition offorce, forallaccelerations are zero inastructure atrest. Wemay alsotake thelawsof electromagnetism, gravitation, etc., to- gether with theparallelogram lawofaddition, asdeﬁning “force.” Equa- tions (1-9) then become alawconnecting previously deﬁned quantities. This hasthedisadvantage that thedeﬁnition offorce changes whenever a newkind offorce (e.g., nuclear force) isdiscovered, orwhenever modiﬁca- tions aremade inelectromagnetism oringravitation. Probably thebest plan, themost ﬂexible atleast, istotake force asaprimitive concept of *Inthetheory ofrelativity, themass ofabody isnotconstant, butdepends on itsvelocity. Inthiscase, law(2)andEqs. (1-9) arenotequivalent, anditturns outthat law(2)isthecorrect formulation. Force should then beequated totime rateofchange ofmomentum. Thesimple deﬁnition (1-5) ofmass isnotcorrect according tothetheory ofrelativity unless theparticles being accelerated move atlowvelocities. 1-4] NEw'roN’s LAWS orMOTION 9 ourtheory, perhaps deﬁned operationally interms ofmeasurements with aspring balance. Then Newton’s laws arelaws, andsoarethelaws of theories ofspecial forces likegravitation andelectromagnetism. Aside from thequestion ofprocedure inregard tothedeﬁnition offorce, there areother difﬁculties inNewton’s mechanics. The third lawisnot always true. Itfails tohold forelectromagnetic forces, forexample, when theinteracting bodies arefarapart orrapidly accelerated and, infact, it fails foranyforces which propagate from onebody toanother with ﬁnite velocities. Fortunately, most ofourdevelopment isbased ontheﬁrst twolaws. Whenever thethird lawisused, itsusewillbeexplicitly noted andtheresults obtained willbevalid only totheextent that thethird law holds. Another difﬁculty isthat theconcepts ofNewtonian mechanics arenot perfectly clear andprecise, asindeed noconcepts canprobably ever befor anytheory, although wemust develop thetheory asifthey were. An outstanding example isthefactthat nospeciﬁcation ismade ofthecoordi- nate system with respect towhich theaccelerations mentioned intheﬁrst twolaws aretobemeasured. Newton himself recognized thisdiﬂiculty butfound novery satisfactory way ofspecifying thecorrect coordinate system touse. Perhaps thebest way toformulate these laws istosay thatthere isacoordinate system with respect towhich theyhold, leaving ittoexperiment todetermine thecorrect coordinate system. Itcanbe shown thatifthese lawsholdinanycoordinate system, they holdalsoin anycoordinate system moving uniformly with respect totheﬁrst. This iscalled theprinciple ofNewtonian relativity, and will beproved in Section 7-1, although thereader should ﬁndlittle difﬁculty inproving it forhimself. . Two assumptions which aremade throughout classical physics arethat thebehavior ofmeasuring instruments isunaffected bytheir state of motion solong asthey arenotrapidly accelerated, andthat itispossible, inprinciple atleast, todevise instruments tomeasure anyquantity with assmall anerror asweplease. These two assumptions failinextreme cases, theﬁrstatvery high velocities, thesecond when very small magni- tudes aretobemeasured. The failure ofthese assumptions forms the basis ofthetheory ofrelativity andthetheory ofquantum mechanics, respectively. However, foravery wide range ofphenomena, Newton’s mechanics iscorrect toavery high degree ofaccuracy, and forms the starting point atwhich themodern theories begin. Notonly thelaws but also theconcepts ofclassical physics must bemodiﬁed according tothe modern theories. However, anunderstanding oftheconcepts ofmodern physics ismade easier byaclear understanding oftheconcepts ofclassical physics. These difficulties arepointed outhere inorder that thereader may beprepared toaccept later modiﬁcations inthetheory. This isnotto 10 ELEMENTS orNEWTONIAN MECHANICS [CHAP. 1 saythat Newton himself (orthereader either atthisstage) ought tohave worried about these matters before setting uphislaws ofmotion. Had hedone so,heprobably never would have developed histheory atall.It wasnecessary tomake whatever assumptions seemed reasonable inorder togetstarted. Which assumptions needed tobealtered, andwhen, and inwhat Way, could only bedetermined later bythesuccesses andfailures ofthetheory inpredicting experimental results. 1-5Gravitation. Although there had been previous suggestions that themotions oftheplanets andoffalling bodies onearth might bedueto aproperty ofphysical bodies bywhich they attract oneanother, theﬁrst toformulate amathematical t-if-eory ofthisphenomenon wasIsaac Newton. Newton showed, bymethods obeconsidered later, that themotions ofthe planets could bequantitativfly accounted forifheassumed that with every pairofbodies isassocia .edaforce ofattraction proportional totheir masses andinversely proportional tothesquare ofthedistance between them. Insymbols, GmmF=iii, (1-11) where m1,mgarethemasses oftheattracting bodies, risthedistance be- tween them, andGisauniversal constant whose value according toex- periment is* _ G=(6.670 =|=0.005) X10_8 cm3-sec'2-gm_1. (1-12) Foraspherically symmetrical body, weshall show later (Section 6—2) that theforce canbecomputed asifallthemass were atthecenter. Fora small body ofmass matthesurface oftheearth, theforce ofgravitation istherefore F=mg, (1-13) where g=9,;-‘-5=980.2cm-sec_2, (1-14) andMisthemass oftheearth andRitsradius. The quantity ghasthe dimensions ofanacceleration, andwecanreadily show byEqs. (1-9) and (1-13) that anyfreely falling body atthesurface oftheearth isaccelerated downward with anacceleration g. The fact that thegravitational force onabody isproportional toits mass, rather than tosome other constant characterizing thebody (e.g., itselectric charge), ismore orlessaccidental from thepoint ofview of Newton’s theory. This factisfundamental inthegeneral theory ofrela- *Smithsonian Physical Tables, 9thed.,1954. 1—6] UNITS AND DIMENSIONS 11 tivity. Theproportionality between gravitational force andmass isproba- blythereason why thetheory ofgravitation isordinarily considered a branch ofmechanics, While theories ofother kinds offorce arenot. Equation (1-13) gives usamore convenient practical way ofmeasuring mass than that contemplated intheoriginal deﬁnition (1—5)'. Wemay measure amass bymeasuring thegravitational force onit,asinaspring balance, orbycomparing thegravitational force onitwith that onastand- ardmass, asinthebeam orplatform balance; inother words, byweigh- ingit. 1 1-6Units anddimensions. Insetting upasystem ofunits interms of which toexpress physical measurements, weﬁrstchoose arbitrary standard units foracertain setof“fundamental” physical quantities (e.g., mass, length, andtime) andthen deﬁne further derived units interms ofthe fundamental units (e.g., theunit ofvelocity isoneunit length perunit of time). Itiscustomary tochoose mass, length, andtime asthefunda- mental quantities inmechanics, although there isnothing sacred inthis choice. Wecould equally well choose some other three quantities, oreven more orfewer than three quantities, asfundamental. There arethree systems ofunits incommon use,thecentimeter-gram- second orcgssystem, themeter-kilogram-second ormks system, andthe foot-pound-second orEnglish system, thenames corresponding to’the names ofthethree fundamental units ineach system.* Units forother kinds ofphysical quantities areobtained from their deﬁning equations by substituting theunits forthefundamental quantities which occur. For example, velocity, byEq.(1-2), U"-Q‘?-1”—dt isdeﬁned asadistance divided byatime. Hence theunits ofvelocity are cm/sec, m/sec, andft/sec inthethree above-mentioned systems, respec- tively. Similarly, thereader canshow that theunits offorce inthethree sys- tems asgiven byEqs. (1-9) aregm-cm-sec"2, kgm-m-sec_2, lb-ft-sec_2. These units happen tohave thespecial names dyne, newton, andpoundal, respectively. Gravitational units offorce aresometimes deﬁned byre- placing Eqs. (1-9) bytheequations " Fa: =mi/gx F11 =my/gr F2 =mg/gr *Inthemkssystem, there isafourth fundamental unit, thecoulomb ofelectri- calcharge, which enters intothedeﬁnitions ofelectrical units. Electrical units in thecgssystem arealldeﬁned interms ofcentimeters, grams, andseconds. Elec- trical units intheEnglish system arepractically never used.1 4 11 J 12 ELEMENTS orNEWTONIAN MECHANICS [cn.u>. 1 where g=980.2 cm-sec_2 =9.802 m-sec'2 =32.16 ft-sec-2 isthestand- ardacceleration ofgravity attheearth’s surface. Unit force isthen that force exerted bythestandard gravitational ﬁeld onunit mass. Thenames gram-weight, kilogram-weight, pound-weight aregiven tothegravita- tional units offorce inthethree systems. Inthepresent text, weshall write thefundamental lawofmechanics intheform (1-9) rather than (1-15) ;hence weshall beusing theabsolute units for‘force andnotthe gravitational units. Henceforth thequestion ofunits willrarely arise, since nearly allour examples willbeworked outinalgebraic form. Itisassumed that the reader issuﬁiciently familiar with theunits ofmeasurement and their manipulation tobeable towork outnumerical examples inanysystem of units should theneed arise. Inanyphysical equation, thedimensions orunits ofalladditive terms onboth sides oftheequation must agree when reduced tofundamental units. Asanexample, wemay check that thedimensions ofthegravita- tional constant inEq.(1-11) arecorrectly given inthevalue quoted in Eq.(1-12): __Gmlmg _F_-7 (1-11) Wesubstitute foreach quantity theunits inwhich itisexpressed: (gm-cm-sec_2) =(cm3'Sec—2€$;)1)(gmxgm) =(gm-cm-sec_2). (1-16) Thecheck does notdepend onwhich system ofunits weusesolong asWe useabsolute units offorce, andwemay check dimensions without any reference tounits, using symbols Z,m,tforlength, mass, time: (mu-2) = -@ =(mlt"2). (1-17) When constant factors like Gareintroduced, wecan, ofcourse, always make thedimensions agree inanyparticular equation bychoosing appro- priate dimensions fortheconstant. Iftheunits intheterms ofanequa- tion donotagree, theequation iscertainly wrong. Ifthey doagree, this does notguarantee that theequation isright. However, acheck ondimen- sions inaresult willreveal most ofthemistakes that result from algebraic errors. Thereader should form thehabit ofmentally checking thedimen- sions ofhisformulas atevery step inaderivation. When constants are introduced inaproblem, their dimensions should beworked outfrom the ﬁrstequation inwhich they appear, andused inchecking subsequent steps. 1-7] som: ELEMENTARY PROBLEMS INMECHANICS 13 1-7Some elementary problems inmechanics. Before beginning asys- tematic development ofmechanics based onthelaws introduced inthis chapter, weshall review afewproblems from elementary mechanics in order toﬁxthese laws clearly inmind. Oneofthesimplest mechanical problems isthat ofﬁnding themotion of abody moving inastraight line, andacted upon byaconstant force. If themass ofthebody ismandtheforce isF,wehave, byNewton’s second law, F=ma. (1-18) The acceleration isthen constant: dv FG—H?—-E' Ifwemultiply Eq.(1-19) bydt,weobtain anexpression forthechange in velocity dvoccurring during theshort time dt: \1 Fdo_Tndt. (1-20) Integrating, weﬁndthetotal change invelocity during thetime t: f dv= vo v—vo=gt, (1-22)—dt, (1-21)¢$_3'1: where voisthevelocity att=0.Ifa:isthedistance ofthebody from a ﬁxed origin, measured along itslineoftravel, then d Fv=3;=vo+Tnt. (1-23) Weagain multiply bydtandintegrate toﬁnd2:: as t F /,0..-/0(....,.).., .1... w=$0+at+sgs. (1-25) where 1:0represents theposition ofthebody att=0.Wenow have a complete description ofthemotion. Wecancalculate from Eqs. (1-25) and(1-22) thevelocity ofthebody atanytime t,andthedistance ithasi 4 14 ELEMENTS orNEWTONIAN MECHANICS [CHAP. 1 traveled. Abody falling freely near thesurface oftheearth isacted upon byaconstant force given byEq.(1-13), andbynoother force ifairre- sistance isnegligible. Inthiscase, if:1:istheheight ofthebody above some reference point, wehave F=—-mg. (1-26) Thenegative signappears because theforce isdownward andthepositive direction ofasisupward. Substituting inEqs. (1-19), (1-22), and(1-25), wehave thefamiliar equations aZ ‘gr v=vo—gt, (1-28) $= Z0 +Uot * %gi2. l Inapplying Newton’s law ofmotion, Eq. (1-18), itisessential to decide ﬁrst towhat body thelawistobeapplied, then toinsert the mass mofthat body andthetotal force Facting onit.Failure tokeep inmind thisrather obvious point isthesource ofmany difficulties, oneof which isillustrated bythehorse-and-wagon dilemma. Ahorse pulls upon awagon, butaccording toNewton’s third lawthewagon pulls back with anequal andopposite force upon thehorse. How then caneither the wagon orthehorse move? Thereader whocansolve Problem 4atthe endofthischapter willhave nodifficulty answering thisquestion. Consider themotion ofthesystem illustrated inFig. 1-2. Two masses m1andm2hang from theends ofarope over apulley, andwewillsuppose that m2isgreater than m1. Wetake acasthedistance from thepulley Oii " at Amlg T V"L29 FIG. 1-2. Atwood’s machine. 1-7] SOME ELEMENTARY PROBLEMS INMECHANICS 15 tomg. Since thelength oftherope isconstant, thecoordinate acﬁxes thepositions ofboth mlandm2. Both move with thesame velocity V L1)=%. (1-30) thevelocity being positive when mlismoving upward andmgismoving downward. Ifweneglect friction and airresistance, theforces onml andm2are F1=—m19 +T, '(1-31) F2=mgg—1', (1-32) where -risthetension intherope. Theforces aretaken aspositive when they tend toproduce apositive velocity dx/dt. Note that theterms involv- ing-rinthese equations satisfy Newton’s third law. The equations of motion ofthetwomasses are . -mlg +1'=mla, (1-33) mzg —'r=mza, (1-34) where aistheacceleration dv/dt, andisthesame forboth masses. By adding Eqs. (1-33) and(1-34), wecaneliminate 1'andsolve fortheaccel- eration: _ __d2x_(m —m)“-W” (H5) The acceleration isconstant and thevelocity vand position :1:canbe found atanytime tasinthepreceding example. Wecansubstitute for afrom Eq.(1-35) ineither Eq.(1-33) or(1-34) andsolve forthetension: _2mlm2 _1'--———m1_,_m2g. (136) Asacheck, wenote that ifml=m2,then a=0and T="$19 ="129, (P37) asitshould ifthemasses areinstatic equilibrium. Asamatter ofinterest, notethatifmg>>ml,then <1i9, (1-33) 1'-_2mlg. (1-39) The reader should convince himself that these tworesults aretobeex- pected inthiscase. \ 16 ELEMENTS orNEWTONIAN MECHANICS [cn.u>. 1 I\\ / \\ / \\ // F ll, N / F/ / / / / //1 m______._mgsin0 mgcos0I / I I / 1/ 9 // 9 / r‘ I’ \\ // I \\ 7!ly// mg I \\ // / \ , I ‘\/ ’ FIG. 1-4. Resolution offorces into FIG. 1-3. Forces acting onabrick components parallel andperpendicular sliding down anincline. totheincline. When several force actonabody, itsacceleration isdetermined bythe vector sumoftheforces which act.Conversely, anyforce canberesolved inanyconvenient manner intovector components whose vector sumisthe given force, andthese components canbetreated asseparate forces acting onthebody.* Asanexample, weconsider abrick ofmass msliding down anincline, asshown inFig.1-3. Thetwoforces which actonthebrick are theweight mgandtheforce Fwith which theplane acts onthebrick. These twoforces areadded according totheparallelogram lawtogive a resultant Rwhich actsonthebrick: R=ma. (1-40) Since thebrick isaccelerated inthedirection oftheresultant force, itis evident that ifthebrick slides down theincline without jumping offor penetrating intotheinclined plane, theresultant force Rmust bedirected along theincline. Inorder toﬁnd R,weresolve each force into com- ponents parallel and perpendicular totheincline, asinFig. 1-4. The force. Fexerted onthebrick bytheplane isresolved inFig. 1-4intotwo components, aforce Nnormal totheplane preventing thebrick from penetrating theplane, andaforce fparallel totheplane, andopposed to *Asystematic development ofvector algebra willbegiven inChapter 3.Only anunderstanding oftheparallelogram lawforvector addition isneeded forthe present discussion. 1-7] soME ELEMENTARY PROBLEMS INMECHANICS 17 themotion ofthebrick, arising from thefriction between thebrick andthe plane. Adding parallel components, weobtain R=mgsin0—f, (1-41) and 0=N—mgcos0. (1-42) Ifthefrictional force fisproportional tothenormal force N,asisoften approximately true fordrysliding surfaces, then f=,uN=,umg cos0,. (1-43) where /.4isthecoefficient offriction. Using Eqs. (1-43), (1-41), and(1-40), wecancalculate theacceleration: a=g(sin0—p.cos0). (1-44) The velocity andposition cannow befound asfunctions ofthetime t, asintheﬁrst example. Equation (1-44) holds only when thebrick is sliding down theincline. Ifitissliding uptheincline, theforce fwill oppose themotion, andthesecond term inEq. (1-44) willbepositive. This could only happen ifthebrick were given aninitial velocity upthe incline. Ifthebrick isatrest,thefrictional force fmayhave anyvalue uptoamaximum p,N: l l . fSMN, (1-45) where pl,thecoefficient ofstatic friction, isusually greater than _u.In thiscaseRiszero, and » f=mgsin05nsmg cos0. (1-46) According toEq.(1-46), theangle 0.oftheincline must notbegreater than alimiting value 6,,theangle ofrepose: tan0_§tan0,=11,. (1-47) If0isgreater than 0,,thebrick cannot remain atrest. Ifabody moves with constant speed 11around acircle ofradius r,its acceleration istoward thecenter ofthecircle, asweshall prove inChapter 3,andisofmagnitude 02 a=—- (1-48) T Such abody must beacted onbyaconstant force toward thecenter. This centripetal force isgiven by 2F=ma= (1-49)il l l 1 1 1 18 ELEMENTS orNEWTONIAN MECHANICS [cnA1>. 1 Note that mv2/r isnota“centrifugal force ”directed away from thecenter, butismass times acceleration andisdirected toward thecenter, asisthe centripetal force F.Asanexample, themoon’s orbit around theearth is nearly circular, andifweassume thattheearth isatrestatthecenter, then, byEq.(1-11), theforce onthemoon is GM FZ 71'”! where Misthemass oftheearth andmthat ofthemoon. Wecanex- press thisforce interms oftheradius Roftheearth andtheacceleration gofgravity attheearth’s surface bysubstituting forGMfrom Eq.(1-14): _"FR? _ F_,2 (151) Thespeed vofthemoon is A v= . (1-52) where Tistheperiod ofrevolution. Substituting Eqs. (1-51) and(1-52) inEq.(1-49), wecanﬁndr: 22gRT T3= This equation wasﬁrstworked outbyIsaac Newton inorder tocheck his inverse square lawofgravitation.* Itwillnotbequite accurate because themoon’s orbit isnotquite circular, andalsobecause theearth does not remain atrest atthecenter ofthemoon’s orbit, butinstead wobbles slightly duetotheattraction ofthemoon. ByNewton’s third law, this attractive force isalso given byEq. (1-51). Since theearth ismuch heavier than themoon, itsacceleration ismuch smaller, andEq.(1-53) willnotbefarwrong. The exact treatment ofthisproblem isgiven in Section 4-7. Another small error isintroduced bythefactthat g,asde- termined experimentally, includes asmall effect duetotheearth’s rota- tion. (SeeSection 7-3.) Ifweinsert themeasured values, g=980.2 cm-sec_2, R=6,368 kilometers, T=27%days, l weobtain, from Eq.(1-53), r=383,000 kilometers. *Isaac Newton, op.cit.,p.407. PROBLEMS 19 Themean distance tothemoon according tomodern measurements is r=385,000 kilometers. Thevalues ofrandRavailable toNewton would nothave given such close agreement. PROBLEMS 1.Compute thegravitational force ofattraction between anelectron and a protonat aseparation of0.5A(1A=10-8 cm). Compare with theelectrostatic force ofattraction atthesame distance. 2.Thecoefficient ofviscosity 11isdeﬁned bytheequation Z"__ £2A'”ds’ where Fisthefrictional force acting across anarea Ainamoving ﬂuid, anddvis thedifference invelocity parallel toAbetween twolayers ofﬂuid adistance ds apart, dsbeing measured perpendicular toA.Find theunits inwhich thevis- cosity 11would beexpressed inthefoot-pound-second, cgs,andmkssystems. Find thethree conversion factors forconverting coeﬂicients ofviscosity from oneof these systems toanother. 3.Amotorist isapproaching agreen trafﬁc light with speed vo,when the light turns toamber. (a)Ifhisreaction time isr,during which hemakes his decision tostop andapplies hisfoot tothebrake, andifhismaximum braking deceleration isa,what istheminimum distance smlllfrom theintersection atthe moment thelight turns toamber inwhich hecanbring hiscartoastop? (b)If theamber light remains onforatime tbefore turning red,what isthemaximum distance sm, from theintersection atthemoment thelight turns toamber such that hecancontinue into theintersection atspeed v0without running thered light? (c)Show thatifhisinitial speed voisgreater than "Ohm: =2a(t —7'): there willbearange ofdistances from theintersection such that hecanneither stop intime norcontinue through without running the_redlight. (d)Make some reasonable estimates of1,t,anda,andcalculate vomuinmiles perhour. Ifvo= §vomax,calculate smlnandsmllx. 4.Aboyofmass mpulls (horizontally) asled ofmass M.The coeﬁicient of friction between sledandsnow is/1.(a)Draw adiagram showing allforces acting ontheboy and onthesled. (b)Find thehorizontal and vertical components of», each force atamoment when boyandsled each have anacceleration a.(c)If' thecoefficient ofstatic friction between theboy’s feetandtheground isp,,what isthemaximum acceleration hecangive tohimself andthesled, assuming trac- tiontobethelimiting factor? 20 ELEMENTS orNEWTONIAN MECHANICS [CHAP. 1 5.Aﬂoor mop ofmass mispushed with aforce Fdirected along thehandle, which makes anangle 6with thevertical. Thecoefficient offriction with theﬂoor isp.(a)Draw adiagram showing allforces acting onthemop. (b)Forgiven 0,p,ﬁndtheforce Frequired toslide themop with uniform velocity across the ﬂoor. (c)Show that if0islessthan theangle ofrepose, themop cannot bestarted across theﬂoor bypushing along thehandle. Neglect themass ofthemop handle. 6.Aboxofmass mslides across ahorizontal table with coefficient offriction p.Theboxisconnected byarope which passes over apulley toabody ofmass M hanging alongside thetable. Find theacceleration ofthesystem andthetension intherope. 7.The brick shown inFigs. 1-3and1-4isgiven aninitial velocity v0upthe incline. The angle 0isgreater than theangle ofrepose. Find thedistance the brick moves uptheincline, andthetime required forittoslide upandback toits original position. 8.Acurve inahighway ofradius ofcurvature risbanked atanangle 0with thehorizontal. Ifthecoefficient offriction is#8,what isthemaximum speed with which acarcanround thecurve without skidding? 9.Assuming theearth moves inacircle ofradius 93,000,000 miles, with a period ofrevolution ofoneyear, ﬁndthemass ofthesunintons. 10.(a)Compute themass oftheearth from itsradius and thevalues ofg andG.(b)Look upthemasses anddistances ofthesunandmoon andcompute theforce ofattraction between earth and sunand between earth and moon. Check your results bymaking arough estimate oftheratio ofthese twoforces from aconsideration ofthefactthattheformer causes theearth torevolve about thesunonce ayear, whereas thelatter causes theearth towobble inasmall circle, approximately once amonth, about thecommon center ofgravity ofthe earth-moon system. 11.The sunisabout 25,000 light years from thecenter ofthegalaxy, and travels approximately inacircle ataspeed of175mi/sec. Find theapproximate mass ofthegalaxy byassuming that thegravitational force onthesuncanbe calculated asifallthemass ofthegalaxy were atitscenter. Express theresult as aratio ofthegalactic mass tothesun’s mass. (You donotneed tolook upeither Gorthesun's mass todothisproblem ifyou compare therevolution ofthesun around thegalactic center with therevolution oftheearth about ‘thesun.) vi. CHAPTER 2’ MOTION OFAPARTICLE INONE DIMENSION 2-1Momentum andenergy theorems. Inthischapter, westudy the motion ofaparticle ofmass malong astraight line, which wewilltake to betheas-axis, under theaction ofaforce Fdirected along the:1:-axis. The discussion willbeapplicable, asweshall see,toother cases where the motion ofamechanical system depends ononly onecoordinate, orwhere allbutonecoordinate canbeeliminated from theproblem. The motion oftheparticle isgoverned, according toEqs. (1—9), bythe equation d2mﬁg=F. (2-1) Before considering thesolution ofEq.(2~1), weshall deﬁne some concepts which areuseful indiscussing mechanical problems and prove some simple general theorems about one-dimensional motion. The linear mo- mentum p,according toEq.(1-10), isdeﬁned as p=mv=m%- (2-2) From Eq.(2—1), using Eq.(2-2) andthefactthat misconstant, weobtain dz»_E-F. (2—3) This equation states that thetime rate ofchange ofmomentum isequal totheapplied force, andis,ofcourse, justNewton’s second law. Wemay callitthe(diﬁerential) momentum theorem. Ifwemultiply Eq. (2-3) bydtandintegrate from t1tot2,weobtain anintegrated form ofthe momentum theorem: P2-P1=ft"Fdt <2-4) 1 Equation (2—4) gives thechange inmomentum duetotheaction ofthe force Fbetween thetimes t1andt2.The integral ontheright iscalled theimpulse delivered bytheforce Fduring thistime; Fmust beknown asafunction oftalone inorder toevaluate theintegral. IfFisgiven asF(a;, v,t),then theimpulse canbecomputed foranyparticular given motion :v(t), v(t).s ~ 21 22 MOTION orAPARTICLE INONEDIMENSION [cn.u>. 2 Aquantity which willturn outtobeofconsiderable importance isthe kinetic energy, deﬁned (inclassical mechanics) bytheequation T=%mv2. (2—5) Ifwemultiply Eq.(2—1) byv,weobtain dvmv-‘E_Fv, or i <1 2_Q_ E(5-mv )_dt-Fv. (2—6) Equation (2—6) gives therate ofchange ofkinetic energy, andmay be called the(differential) energy theorem. Ifwemultiply bydtandinte- grate from t1tot2,weobtain theintegrated form oftheenergy theorem: T2-T1=ft"Fm. (2-7) 1 Equation (2—7) gives thechange inenergy duetotheaction oftheforce F between thetimes t1andt2.The integral ontheright iscalled thework done bytheforce dining thistime. Theintegrand Fvontheright isthe time rateofdoing work, andiscalled thepower supplied bytheforce F. Ingeneral, when Fisgiven asF(x,v,t),thework canonlybecomputed foraparticular speciﬁed motion x(t), v(t). Since v=dx/dt, wecanre- write thework integral inaform which isconvenient when Fisknown asafunction ofac: T2—T1= Fdx. (2-s) $1 2-2Discussion ofthegeneral problem ofone-dimensional motion. If theforce Fisknown, theequation ofmotion (2-1) becomes asecond-order ordinary differential equation fortheunknown function x(t). Theforce F may beknown asafunction ofanyorallofthevariables t,x,andv.For anygiven motion ofadynamical system, alldynamical variables (ac,v,F, p,T,etc.) associated with thesystem are, ofcourse, functions ofthe time t,that is,each hasadeﬁnite value atanyparticular time t.However, inmany cases adynamical variable such astheforce may beknown to bear acertain functional relationship toac,ortov,ortoanycombination ofx,v,andt.Asanexample, thegravitational force acting onabody falling from agreat height above theearth isknown asafunction ofthe height above theearth. Thefrictional drag onsuch abody would depend onitsspeed andonthedensity oftheairandhence ontheheight above theearth; ifatmospheric conditions arechanging, itwould also depend ont.IfFisgiven asF(x, v,t),then when a:(t)andv(t)areknown, these 2—2] THE GENERAL PROBLEM 23 functions canbesubstituted togive Fasafunction ofthetime talone; however, ingeneral, thiscannot bedone until after Eq.(2—1) hasbeen solved, and even then thefunction F(t) may bedifferent fordiﬂerent possible motions oftheparticle. Inany case, ifFisgiven asF(x, v,t) (where Fmay depend onanyorallofthese variables), then Eq.(2-1) becomes adeﬁnite differential equation tobesolved: if 1 .1-J;=-T;F(x,:0,t). (2-9) This isthemost general type ofsecond-order ordinary diﬁerential equa- tion, andweshall beconcerned inthischapter with studying itssolutions andtheir applications tomechanical problems. Equation (2-9) applies toallpossible motions oftheparticle under the action ofthespeciﬁed force. Ingeneral, there willbemany such motions, forEq.(2—9) prescribes only theacceleration oftheparticle atevery in- stant interms ofitsposition andvelocity atthat instant. Ifweknow theposition andvelocity ofaparticle atacertain time, wecandetermine itsposition ashort time later (orearlier). Knowing alsoitsacceleration, wecanﬁnditsvelocity ashort time later. Equation (2-9) then gives the acceleration ashort time later. Inthismanner, Wecantrace outthepast orsubsequent positions andvelocities ofaparticle ifitsposition xoand velocity v0areknown atanyonetime to.Any pairofvalues ofmoandvo willlead toapossible motion oftheparticle. Wecalltotheinitial instant, although itmay beanymoment inthehistory oftheparticle, andthe values ofnoand120attowecalltheinitial conditions. Instead ofspecifying initial values for:1;andv,wecould specify initial values ofanytwoquan- tities from which acandvcanbedetermined; forexample, wemay specify moandtheinitial momentum p0=moo. These initial conditions, together with Eq. (2-9), then represent aperfectly deﬁnite problem Whose solu- tion should beaunique function :c(t) representing themotion ofthe particle under thespeciﬁed conditions. The mathematical theory ofsecond-order ordinary differential equa- tions leads toresults inagreement with what Weexpect from thenature ofthephysical problem inwhich theequation arises. Thetheory asserts that, ordinarily, anequation oftheform (2-9) hasaunique continuous solution x(t)which takes ongiven values IE0andvoofxandatatanychosen initial value tooft.“Ordinarily” here means, asfarasthebeginning mechanics student isconcerned, “inallcases ofphysical interest.”* The properties ofdifferential equations like Eq. (2-9) are derived inmost *Forarigorous mathematical statement oftheconditions fortheexistence of asolution ofEq.(2-9), seeW.Leighton, AnIntroduction totheTheory ofDiﬁ'eren- tialEquations. New York: McGraw—Hill, 1952. (Appendix 1.)1 l 4 24 MOTION orAPARTICLE INom:DIMENSION [cnxn 2 treatises ondifferential equations. Weknow that anyphysical problem must always have aunique solution, andtherefore anyforce function F(x,at,t)which canoccur inaphysical problem willnecessarily satisfy the required conditions forthose values ofcc,5:,tofphysical interest. Thus ordinarily wedonotneed toworry about whether asolution exists. How- ever, most mechanical problems involve some simpliﬁcation oftheactual physical situation, anditispossible tooversimplify orotherwise distort aphysical problem insuch aWay that theresulting mathematical problem nolonger possesses aunique solution. The general practice ofphysicists inmechanics and elsewhere istoproceed, ignoring questions ofmathe- matical rigor. Onthose fortunately rare occasions when werunintodiffi- culty, wethen consult ourphysical intuition, orcheckour lapses ofrigor, until thesource ofthedifficulty isdiscovered. Such aprocedure may bring shudders tothemathematician, butitisthemost convenient and rapid way toapply mathematics tothesolution ofphysical problems. Thephysicist, while hemay proceed inanonrigorous fashion, should never- theless beacquainted with therigorous treatment ofthemathematical methods which heuses. l The existence theorem forEq.(2—9) guarantees that there isaunique mathematical solution tothis equation forallcases which willarise in practice. 1Insome cases theexact solution canbefound byelementary methods. Most oftheproblems considered inthistextwillbeofthis nature. Fortunately, many ofthemost important mechanical problems inphysics canbesolved Without toomuch difficulty. Infact, oneofthe reasons why certain problems areconsidered important isthat they can beeasily solved. The physicist isconcerned with discovering andverify- ingthelaws ofphysics. Inchecking these laws experimentally, heisfree, toalarge extent, tochoose those cases where themathematical analysis isnottoodiﬁicult tocarry out. The engineer isnotsofortunate, since hisproblems areselected notbecause they areeasy tosolve, butbecause they areofpractical importance. Inengineering, andoften alsoinphysics, many cases arise where theexact solution ofEq.(2—9) isdiﬂicult orim- possible toobtain. Insuch cases various methods areavailable forobtain- ingatleast approximate answers. The reader isreferred tocourses and texts ondifferential equations foradiscussion ofsuch methods.* From thepoint ofview oftheoretical mechanics, theimportant point isthat asolution always does exist andcanbefound, asaccurately asdesired. Weshall restrict ourattention toexamples which canbetreated by simple methods. *W.E.Milne, Numerical Calculus. Princeton: Princeton University Press, 1949. (Chapter 5.) H.Levy and E.A.Baggott, Numerical Solutions ofDiﬁerential Equations. New York: Dover Publications, 1950. 2—3] APPLIED FORCE DEPENDING ONTHE TIME 25 2-3Applied force depending onthetime. Iftheforce Fisgiven asa function ofthetime, then theequation ofmotion (2—9) canbesolved in thefollowing manner. Multiplying Eq.(2—9) bydtandintegrating from aninitial instant totoanylater (orearlier) instant t,weobtain Eq.(2-4), which inthiscase wewrite intheform mv-ma,=/‘F(t)dt. (2-10) lo Since F(t) isaknown function oft,theintegral ontheright can, atleast inprinciple, beevaluated andtheright member isthen afunction oft (and to).Wesolve forv: dz 1tv_-,2_to+EL)F(t)dt. (2-11) Now multiply bydtandintegrate again from totot:' 1 t t Z—119=1)0(t '-'to) + lit] dt. (2-12) 29 to Toavoid confusion, wemay rewrite thevariable ofintegration ast’in theﬁrstintegral andt”inthesecond: ‘ E ll’ 2=xo+v0(t-:0)+%/dt”IF(t’)at’. (2-13) to to This gives therequired solution :v(t)interms oftwointegrals which can beevaluated when F(t)isgiven. Adeﬁnite integral canalways beevalu- ated. Ifanexplicit formula fortheintegral cannot befound, then at least itcanalways becomputed asaccurately asweplease bynumerical methods. Forthisreason, inthediscussion ofageneral type ofproblem such astheone above, weordinarily consider theproblem solved when thesolution hasbeen expressed interms ofoneormore deﬁnite integrals. Inapractical problem, theintegrals would have tobeevaluated toobtain theﬁnal solution inusable form.* *The reader who hasstudied diﬂerential equations may bedisturbed bythe appearance ofthree constants, to,vo,andxo,inthesolution (2-13), whereas the general solution ofasecond-order differential equation should contain only two arbitrary constants. Mathematically, there areonlytwoindependent constants in.Eq.(2-13), anadditive constant containing theterms :00——voteplus aterm from thelower limit ofthelastintegral, andaconstant multiplying tcontaining theterm voplus aterm from thelower limit oftheﬁrstintegral. Physically, we cantake any initial instant to,andthen just two parameters xoandcoarere- quired tospecify oneoutofallpossible motions subject tothegiven force. 26 MOTION orAPARTICLE INONE DIMENSION [CHAP. 2 Problems inwhich Fisgiven asafunction oftusually arise when we seektoﬁndthebehavior ofamechanical system under theaction ofsome external inﬂuence. Asanexample, weconsider themotion ofafreeelec- tron ofcharge -ewhen subject toanoscillating electric ﬁeld along the 2:-axis: E,=E0cos(wt—|—0). (2-14) Theforce ontheelectron is F=—eE, =-eE0 cos(wt+0). (2-15) Theequation ofmotion is dvm82=—eE0 cos(wt+0). (2-16) Wemultiply bydtandintegrate, taking to=0: v= %= v0+e—?—E°Sin0—@9sin(wt+6). W W (2-17) Integrating again, weobtain E 0 E '0 E . .=a-5-Pm-,%+(..+L%IL)i+,;-Wg...(..¢+@>. (2-18) Iftheelectron isinitially atrestatno=0,thisbecomes eEcos0eEsin0 eE:1:=— SW2 —|— gm” t+-mwoz cos(wt+0). (2-19) Itislefttothereader toexplain physically theorigin oftheconstant term andtheterm linear intinEq.(2-19) interms ofthephase oftheelectric ﬁeld attheinitial instant. How dotheterms inEq.(2-19) depend on e,m,E0,andw?Explain physically. Why does theoscillatory term turn outtobeoutofphase with theapplied force? Theproblem considered here isofinterest inconnection with thepropa- gation ofradio waves through theionosphere, which contains ahigh density offreeelectrons. Associated with aradio Wave ofangular frequency wis anelectric ﬁeld which may begiven byEq.(2-14). Theoscillating term in Eq. (2-18) hasthesame frequency wandisindependent oftheinitial conditions. This coherent oscillation ofthefree electrons modiﬁes the propagation ofthewave. Thenonoscillating terms inEq.(2-18) depend ontheinitial conditions, andhence onthedetailed motion ofeach electron asthewave arrives. These terms cannot contribute tothepropagation characteristics ofthewave, since they donotoscillate with thefrequency ofthewave, although theymayaffect theleading edgeofthewave which 2-3] APPLIED FORCE DEPENDING ONTHE TIME 27 arrives ﬁrst. Weseethat theoscillatory part ofthedisplacement zvis180° outofphase with theapplied force duetotheelectric ﬁeld. Since theelec- tron hasanegative charge, theresulting electric polarization is180° out ofphase with theelectric ﬁeld. The result isthat thedielectric constant oftheionosphere islessthan one. (Inanordinary dielectric atlowfre- quencies, thecharges aredisplaced inthedirection oftheelectric force onthem, andthedielectric constant isgreater than one.) Since theveloc- ityoflight is v=¢(#@)_1/2, (2-20) where c=3><101° cm/sec andeand/1arethedielectric constant and magnetic permeability respectively, and since ju=1here, the(phase) velocity vofradio waves intheionosphere isgreater than thevelocity c ofelectromagnetic waves inempty space. Thus waves entering the ionosphere atanangle arebent back toward theearth. Theeﬁ'ect isseen tobeinversely proportional to0:2,sothat forhigh enough frequencies, thewaves donotreturn totheearth butpass outthrough theionosphere. Only aslight knowledge ofelectromagnetic theory isrequired tocarry thisdis- cussion through mathematically.* The dipole moment oftheelectron displaced from itsequilibrium position is 2 2—e:c=--5”-5E0cos(wt+0)=-kE, (2-21) ifweconsider only theoscillating term. Ifthere areNelectrons percm3, the total dipole moment perunitvolume is 2NeP,-—W E,,. (2-22) The electric displacement is 41rNe2 Dz =.-E; +41TP, = —- E1. Since thedielectric constant isdeﬁned by D,=eE,, (2-24) weconclude that 2_1 41rNee— ——-2-» 7"/(0 and since ,u=1, ( 4orNe2>_1/2(2-25) U=C 1—-—-Tn“? ' *See, e.g., G.P.Harnwell, Principles ofElectricity andElectromagnetism, 2nd ed.New York: McGraw-Hill, 1949. (Section 2.4.) 28 MOTION orAPARTICLE INONEDIMENSION [cnA1=. 2 2-4Damping force depending onthevelocity. Another type offorce which allows aneasy solution ofEq.(2—9) isthecasewhen Fisafunction ofvalone: m$3;=F(t). (2-27) Tosolve, wemultiply by[mF(v)] 1dtandintegrate from totot: %';_)= (2-2s) Theintegral ontheleftcanbeevaluated, inprinciple atleast, when F(v) isgiven, andanequation containing theunknown vresults. Ifthisequa- tionissolved forv(weassume ingeneral discussions that thiscanalways bedone), wewillhave anequation oftheform . d t-t1)=7:;=<0<00,_1;,—‘0) ' (2-29) Thesolution forxisthen ' z-tox=mo+'/to<p(120,T) dt. (2-30) Inthecase ofone-dimensional motion, theonly important kinds offorces which depend onthevelocity arefrictional forces. Theforce ofsliding or rolling friction between drysolid surfaces isnearly constant foragiven pair ofsurfaces with agiven normal force between them, anddepends on thevelocity only inthat itsdirection isalways opposed tothevelocity. Theforce offriction between lubricated surfaces orbetween asolid body andaliquid orgaseous medium depends onthevelocity inacomplicated way, andthefunction F(v) canusually begiven only intheform ofa tabulated summary ofexperimental data. Incertain cases and over certain ranges ofvelocity, thefrictional force isproportional tosome ﬁxed power ofthevelocity: F=(=F)bv". (2-31) Ifnisanoddinteger, thenegative sign should bechosen intheabove equation. Otherwise thesign must bechosen sothat theforce hasthe opposite sign tothevelocity v.The frictional force isalways opposed to thevelocity, andtherefore does negative work, i.e.,absorbs energy from themoving body. Avelocity-dependent force inthesame direction as thevelocity would represent asource ofenergy; such cases donotoften occur. 2-4] DAMPING FORCE DEPENDING ONTHE VELOCITY 29 Asanexample, weconsider theproblem ofaboat traveling with initial velocity vo,which shuts offitsengines atto=0when itisattheposition xo=0.Weassume theforce offriction given byEq.(2-31) withn=1: m217;’=—bv. (2-32) Wesolve Eq. (2-32), following thesteps outlined above [Eqs. (2-27) through (2-30)]: /”<2__i, ,,ov— m’ v b llla)" —-Et, v=v0e'b”"‘. (2-33) Weseethat ast——>oo,v—>0,asitshould, butthat theboat never comes completely torestinanyﬁnite time. Thesolution for:1:is 8 :6=fv0e"l’”"‘ dt o =_£‘5% (1-e_b”"‘). (2-34) Ast—>co,xapproaches thelimiting value 11,= (2-35) Thus wecanspecify adeﬁnite distance that theboat travels instopping. Although according totheabove result, Eq. (2-33), thevelocity never becomes exactly zero, when tissufficiently large thevelocity becomes so small that theboat ispractically stopped. Letuschoose some small velocity v,such that when v<v,wearewilling toregard theboat as stopped (say, forexample, theaverage random speed given toananchored boat bythewaves passing byit).Then wecandeﬁne thetime t,required fortheboat tostop by v,=v0e_b"/m, ts=1;'ln%9- (2-36) 3 Since thelogarithm isaslowly changing function, thestopping time t,will notdepend toanygreat extent onprecisely what value ofv,wechoose so long asitismuch smaller than vo.Itisoften instructive toexpand solu- tions inaTaylor seriesin t.Ifweexpand theright sideofEqs. (2-33) 30 MOTION orAPARTICLE INONEDIMENSION [CHAP. 2 and(2-34) inpower series int,weobtain* i=1),-ll”fi+---, (2-37) lb $=U0t*§—:T0t2-l-"‘ Note that theﬁrsttwoterms intheseries forvandatarejusttheformulas foraparticle acted onbyaconstant force —bv0, which istheinitial value ofthefrictional force inEq.(2-32). This istobeexpected, andaffords a fairly good check onthealgebra which ledtothesolution (2-34). Series expansions areavery useful means ofobtaining simple approximate for- mulas valid forashort range oftime. The characteristics ofthemotion ofabody under theaction ofafric- tional force asgiven byEq.(2-31) depend ontheexponent n.Ingeneral, alarge exponent nwillresult inrapid initial slowing butslow ﬁnal stopping, andvice versa, asonecanseebysketching graphs ofFvs.11forvarious values ofn.Forsmall enough values ofn,thevelocity comes tozero in aﬁnite time. Forlarge values ofn,thebody notonly requires aninﬁnite time, buttravels aninﬁnite distance before stopping. This disagrees with ordinary experience, anindication thatwhile theexponent nmaybe large athigh velocities, itmust become smaller atlowvelocities. The exponent n=1isoften assumed inproblems involving friction, particu- larly when friction isonly asmall effect tobetaken intoaccount approxi- mately. The reason fortaking n=1isthat this gives easy equations tosolve, and isoften afairly good approximation when thefrictional force issmall, provided bisproperly chosen. .\ 2-5Conservative force depending onposition. Potential energy. One ofthemost important types ofmotion occurs when theforce Fisafunc- *The reader who hasnotalready done soshould memorize theTaylor series forafewsimple functions like 2 3 4:2: ac re¢;,z-1is+ 9z= 1+1-l'*§+§j3+ 2 3 4mu+o=x-%+%-§+~» (1+.x)n =1+nx+n(n2—l)x2+n(n These three series areextremely useful inobtaining approximations tocompli- cated formulas, valid when xissmall. 2-5] CONSERVATIVE FORCE DEPENDING ONPOSITION 31 tion ofthecoordinate zcalone: m%=F(a:). (2-39) Wehave then, bytheenergy theorem (2-8), émvz —%mv§ =frF(x) dx. (2-40) 10 Theintegral ontheright isthework done bytheforce when theparticle goes from xotox.Wenow deﬁne thepotential energy V(.r) asthework done bytheforce when theparticle goes from a:tosome chosen standard point 90,: V(1:) =i/:'F(x) dx=—I:F(x) dx. (2-41) The reason forcalling thisquantity potential energy willappear shortly. Interms ofV(x), wecanwrite theintegral inEq.(2-40) asfollows: f”Fa)dx=—V(x) +V(x0). (2-42) Io With thehelpofEq.(2-42), Eq.(2-40) canbewritten tmvz +V(w)=tmvg +V(=vo)- (2-43) The quantity ontheright depends only ontheinitial conditions andis therefore constant during themotion. Itiscalled thetotal energy E,and wehave thelawofconservation ofkinetic plus potential energy, which holds, aswecansee,only when theforce isafunction ofposition alone: %mv2+V(w)=T+V=E. (2-44) Solving forv,weobtain . v=9%=£112 -v(2)11'2. (2-45) Thefunction x(t)istobefound bysolving forattheequation ,, . 4 [E-V(@)]-"ax =t-:0. (2-16) Inthiscase, theinitial conditions areexpressed interms oftheconstants Eandmo. Inapplying Eq.(2-46), andintaking theindicated square root inthe integrand, caremust betaken tousetheproper sign, depending onwhether 32 MOTION orAPARTICLE INONE DIMENSION [CHAP. 2 thevelocity vgiven byEq.(2-45) ispositive ornegative. Incases where vispositive during some parts ofthemotion andnegative during other parts, itmay benecessary tocarry outtheintegration inEq. (2-46) separately foreach part ofthemotion. From thedeﬁnition (2-41) wecanexpress theforce interms ofthe potential energy. dv F=—75- (2-47) This equation canbetaken asexpressing thephysical meaning ofthepo- tential energy. The potential energy isafunction whose negative derivative gives theforce. The effect ofchanging thecoordinate ofthe standard point 2:,istoaddaconstant toV(a;). Since itisthederivative ofVwhich enters into thedynamical equations astheforce, thechoice ofstandard point at,isimmaterial. Aconstant canalways beadded to thepotential V(:c) without affecting thephysical results. (The same constant must, ofcourse, beadded toE.) Asanexample, weconsider theproblem ofaparticle subject toa linear restoring force,ifor example, amass fastened toaspring: F=-191 (2-48) Thepotential energy, ifwetakeav,=0,is IV =— —kd (x) A atat =419152. (2-19) Equation (2-46) becomes, forthiscase, with to=0, gI(E-=}lcx2)"1/2d:c =t. (2-50) 10 Now make thesubstitutions sin0=aval%1 (2-51) O)=- : sothat 4 :1: 0/4 _--1/2 _1f _1_ 2/mo(E ﬁler) dx-w 0°d0-w(0 00), and, byEq.(2-50), 0=wt+00. 2-5] CONSERVATIVE FORCE DEPENDING ONPOSITION 33 Wecannow solve foracinEq.(2-51): 2=,/%sin9=Aan(wt+00), (2-53) A= 3-... Thus thecoordinate :1;oscillates harmonically intime, with amplitude A andfrequency w/2-1r. The initial conditions arehere determined bythe constants Aand 00,which arerelated toEandmobywhere E=215212, (2-55) I0 Z ASID 00. Notice that inthis example wemeet thesign difficulty intaking the square root inEq.(2-50) byreplacing (1—sing0)_1/2 by(cos0)_1, a quantity which canbemade either positive ornegative asrequired by choosing 0intheproper quadrant. Afunction ofthedependent variable anditsﬁrst derivative which isconstant forallsolutions ofasecond-order differential equation, iscalled aﬁrstintegral oftheequation. Thefunction frn.:i:2 +V(x) iscalled the energy integral ofEq.(2-39). Anintegral oftheequations ofmotion of amechanical system isalso called aconstant ofthemotion. Ingeneral, anymechanical problem canbesolved ifwecanﬁnd enough ﬁrst inte- grals, orconstants ofthemotion. , Even incases where theintegral inEq.(2-46) cannot easily beevalu- ated ortheresulting equation solved togive anexplicit solution for:i:(t), theenergy integral, Eq. (2-44), gives ususeful information about the solution. Foragiven energy E,weseefrom Eq.(2-45) that theparticle isconﬁned tothose regions ontheac-axis where V(a:) §E.Furthermore, thevelocity isproportional tothesquare root ofthedifference between EandV(x). Hence, ifweplot V(:c) versus x,wecangive a.good qualita- tivedescription ofthekinds ofmotion that arepossible. Forthepotential- energy function shown inFig. 2-1wenote that theleast energy possible isE0. Atthis energy, theparticle canonly beatrestat2:0. With a slightly higher energy E1,theparticle canmove between :01and:02;its velocity decreases asitapproaches x1orx2,anditstops andreverses its direction when itreaches either 2:1or2:2,which arecalled turning points ofthemotion. With energy E2,theparticle may oscillate between turn- ingpoints 2:3and2:4,orremain atrestat2:5.With energy E2,there are four turning points and theparticle may oscillate ineither ofthetwo 34 MOTION orAPARTICLE INONEDIMENSION [CHA.P. 2 Va) __i._.i i; E4---—- ----- -—-—- - :1;-1:~IIi!i"!!! __.!._____i|!ii--l____-‘I.E3 _____ E2 —-—-— E1 - -— _1_ii: -4-|1 _.|__F. - - -— cl?___.‘£§[-Q_..I-1ﬁ__MH__>5 §__R Zt FIG. 2-1. Apotential-energy function forone-dimensional motion. potential valleys. With energy E4,there isonly oneturning point; if theparticle isinitially traveling totheleft, itwillturn at2:6andreturn totheright, speeding upover thevalleys atavgand:05,andslowing down over thehillbetween. Atenergies above E5,there arenoturning points andtheparticle willmove inonedirection only, varying itsspeed accord- ingtothedepth ofthepotential ateach point. Apoint where V(x) hasaminimum iscalled apoint ofstable equilibrium. Aparticle atrestatsuch apoint willremain atrest. Ifdisplaced aslight distance, itwillexperience arestoring force tending toreturn it,andit willoscillate about theequilibrium point. Apoint where V(x) hasamaxi- mum iscalled apoint ofunstable equilibrium. Intheory, aparticle at restthere canremain atrest, since theforce iszero, butifitisdisplaced theslightest distance, theforce acting onitwillpush itfarther away from theunstable equilibrium position. Aregion where V(x) isconstant is called aregion ofneutral equilibrium, since aparticle canbedisplaced slightly without suffering either arestoring orarepelling force. This kind ofqualitative discussion, based ontheenergy integral, is simple andvery useful. Study thisexample until youunderstand itwell enough tobeable toseeataglance, foranypotential energy curve, the types ofmotion that arepossible. Itmay bethat only part oftheforce onaparticle isderivable from a potential function V(x). LetF’betheremainder oftheforce: F=-%+F’. (2-57) Inthiscase theenergy (T-1-V)isnolonger constant. Ifwesubstitute F 2-6] FALLING BODIES 35 from Eq. (2-57) inEq. (2-1), and multiply bydx/dt, wehave, after rearranging terms, 7 %(T +V)=F'v. (2-58) The time rate ofchange ofkinetic plus potential energy isequal tothe power delivered bytheadditional force F’. 2-6Falling bodies. Oneofthesimplest andmost commonly occurring types ofone-dimensional motion isthat offalling bodies. Wetake up thistype ofmotion here asanillustration oftheprinciples discussed in thepreceding sections. Abody falling near thesurface oftheearth, ifweneglect airresistance, issubject toaconstant force F=—-mg, ' (2-59) where wehave taken thepositive direction asupward. The equation of motion is d2m8%=-—mg. (2440) The solution may beobtained byany ofthethree methods discussed inSections 2-3, 2-4, and2-5, since aconstant force may beconsidered asafunction ofeither t,v,or11:.The reader willﬁnd itinstructive to solve theproblem byallthree methods. Wehave already obtained the result inChapter 1[Eqs. (1-28) and (1—29)]. Inorder toinclude theeffect ofairresistance, wemay assume afric- tional force proportional tov,sothat thetotal force is F=—mg —bv. (2-61) The constant bwilldepend onthesizeand shape ofthefalling body, aswell asontheviscosity oftheair. The problem must now betreated asacase ofF(v): mgig=—mg —bv. (2-62) Taking vo=0att=0,weproceed asinSection 2-4[Eq. (2-28)]: "do bt/..m -“5 ‘H3’ Weintegrate andsolve forv: 5=_%(1-e-"”'"). ’ (2-54) 36 MOTION orAPARTICLE INONE DIMENSION [cIIAP. 2 Wemay obtain aformula useful forshort times offallbyexpanding the exponential function inapower series: v=—u+2%ﬁ+~- (aw Thus forashort time (t<<m/b), v=—gt, approximately, andtheeffect ofairresistance canbeneglected. After along time, weseefrom Eq.(2-64) that 5--lnb-9, ifz>>3”,;- Thevelocity mg/biscalled theterminal velocity ofthefalling body inques- tion. The body reaches within 1/eofitsterminal velocity inatime t=m/b. Wecould usetheexperimentally determined terminal velocity toﬁnd theconstant b.Wenow integrate Eq.(2-64), taking xo=0: m2g bt _m xi—-&—(1iEiebt/)° Byexpanding theexponential function inapower series, Weobtain b2--aF+4iP+~» one Ift<<m/b, xi—-%gt2, asinEq.(1-29). When t>>m/b, _'..L22_T4at—(b2 bt This result iseasily interpreted interms ofterminal velocity. Why is thepositive constant present? Forsmall heavy bodies with large terminal velocities, abetter approxi- mation may be F=552. (2-cs) The reader should beable toshow that with thefrictional force given by Eq.(2-68), theresult (taking no=vo=0atto=0)is 5=-,1.2511(,/1%t) (2-69) —gt, ift<<, __ [E, ' (Q, b If t>> by 2-6] FALLING BODIES 37 x=—Z-Zlncosh<4 t) ‘(2-70) -24):”, ift<<,lg, Z’! _H - E. 7 bln2 ‘lb t, If t>>‘lbg Again there isaterminal velocity, given thistime by(mg/b)1/2. The ter- minal velocity canalways befound asthevelocity atwhich thefrictional force equals thegravitational force, andwillexist whenever thefrictional force becomes sufﬁciently large athigh velocities. Inthecase ofbodies falling from agreat height, thevariation ofthe gravitational force with height should betaken intoaccount. Inthiscase, weneglect airresistance, and measure xfrom thecenter oftheearth. Then ifMisthemass oftheearth andmthemass ofthefalling body, mMGF_--7, (2-71) and I 1/(5)=-LF55=- (2-72) where wehave taken ac,=ooinorder toavoid aconstant term inV(7c). Equation (2-45) becomes . 51 2 MG1'2~=5’§=-h(E+"‘T) -<2-73> The plus sign refers toascending motion, theminus sign todescending motion. The function V(:c) isplotted in Fig.2-2. Weseethat there aretwo 5 types ofmotion, depending on whether Eispositive ornegative. When Eispositive, there isnoturn- ingpoint, andifthebody isinitially moving upward, itwillcontinue to move upward forever, with decreas- ingvelocity, approaching thelimit- ingvelocity 2E’”=\/7;'(2-74) Fm.2-2.P15951V(x)=—(mMG/at).V(x)1 1 i 38 MOTION orAPARTICLE INONE DIMENSION [onAP. 2 When Eisnegative, there isaturning point ataheight _ 25= - (2-75) Ifthebody isinitially moving upward, itwillcome toastop at$71,and fallback totheearth. The dividing case between these two types of motion occurs when theinitial position andvelocity aresuch that E=0. Theturning point isthen atinﬁnity, andthebody moves upward forever, approaching thelimiting velocity vl=O.IfE=0,then atanyheight 2:, thevelocity willbe - 5.=,12-Li? (2-75) This is,called theescape velocity forabody atdistance atfrom thecenter oftheearth, because abody moving upward atheight xwith velocity ve willjust have sufficient energy totravel upward indeﬁnitely (ifthere is noairresistance). Toﬁndx(t), wemust evaluate theintegral Z dx ___ 2 d: ___.__ 10 2? where xoistheheight att=0.Tosolve forthecasewhen Eisnegative, wesubstitute 1—Ex COS 0= W6 ' Equation (2-77) then becomes 0 2 _’\/2 _—-——(_E)3,2 Lo2cos 0d0_ mt. (279) (We choose apositive signfortheintegrand sothat 0willincrease when t increases.) Wecan,without lossofgenerality, take notobeattheturning point T1,since thebody willatsome time initspast orfuture career pass through myifnoforce except gravity actsupon it,provided E<0.Then 00=0,and MG . 12-—-(_1:LE)3,2 (0-1-S1110cos0)= -7;t, or 0-1-21-sin20=4 t, (2-80) 517T 2-7] THE SIMPLE HARMONIC OSCILLATOR 39 and x=mycos20. (2-81) This pair ofequations cannot besolved explicitly for:e(t). Anumerical solution canbeobtained bychoosing asequence ofvalues of0andﬁnding thecorresponding values ofxandtfrom Eqs. (2-80) and (2-81). That part ofthemotion forwhich asislessthan theradius oftheearth will, of course, notbecorrectly given, since Eq.(2-71) assumes allthemass ofthe earth concentrated atat=0(nottomention thefactthat wehave omitted from ourequation ofmotion theforces which would actonthebody when itcollides with theearth). The solution canbeobtained inasimilar way forthecases when Eis positive orzero. .. 2-7The simple harmonic oscillator. The most important problem in one-dimensional motion, andfortunately oneoftheeasiest tosolve, isthe harmonic orlinear oscillator. The simplest example isthat ofamass m fastened toaspring whose constant isIc.Ifwemeasure acfrom there- laxed position ofthespring, then thespring exerts arestoring force F=-I55. (2-32) The potential energy associated '2 ,6 with thisforce is - V(x)=21552. (2-s3) The equation ofmotion, ifwe I-1-1 assume noother force acts, is2 dz Fro. 2-_3. Model ofasimple har- mﬁg+kw:0_(2_84) monic oscillator. Equation (2-84) describes thefreeharmonic oscillator. Itssolution was obtained inSection 2-5. The motion isasimple sinusoidal oscillation about thepoint ofequilibrium. Inallphysical cases there willbesome frictional force acting, though itmay often bevery small. Asagood approximation inmost cases, particularly when thefriction issmall, we canassume that thefrictional force isproportional tothevelocity. Since thisistheonly kind offrictional force forwhich theproblem caneasily besolved, weshall restrict ourattention tothiscase. IfweuseEq.(2-31) forthefrictional force with n=1,theequation ofmotion then becomes 2 ..m%§+z>%,"5+155=o. (2-s5) This equation describes thedamped harmonic oscillator. Itsmotion, at least forsmall damping, consists ofasinusoidal oscillation ofgradually 40 MOTION orAPARTICLE INONE DIMENSION [cmua 2 decreasing amplitude, asweshall show later. Iftheoscillator issubject toanadditional impressed force F(t), itsmotion willbegiven by dzx dx IfF(t)isasinusoidally varying force, Eq.(2-86) leads tothephenomenon ofresonance, where theamplitude ofoscillation becomes very large when thefrequency oftheimpressed force equals thenatural frequency ofthe freeoscillator. Theimportance oftheharmonic oscillator problem liesinthefactthat equations ofthesame form asEqs. (2-84)—(2—86) turn upinawide variety ofphysical problems. Inalmost every caseofone-dimensional motion where thepotential energy function V(a:) hasoneormore mjnima, the motion oftheparticle forsmall oscillations about theminimum point’fol- lows Eq.(2—84). Toshow this, letV(x) have aminimum atav=mo,and expand thefunction V(x) inaTaylor series about thispoint: 2v<»>=V($o)+ (x—we+a (w—M +t(%)m(w —¢vo)a+----(2-87) The constant V(:c0) canbedropped without affecting thephysical results. Since xoisaminimum point, dV _ d2V)<75)“ _0,(dag, toZ0. (2-ss) Making theabbreidations 2k=(ill) , (2-89)d@x2 0 ac’=:1:-—x0, (2-90) wecanWrite thepotential function intheform V(x’) =%kx’2 +-~-. (2-91) Forsufﬁciently small values ofas’,provided k750,wecanneglect theterms represented bydots, andEq.(2-91) becomes identical with Eq.(2-83). Hence, forsmall oscillations about anypotential minimum, except inthe exceptional case k=0,themotion isthat ofaharmonic oscillator. When asolid isdeformed, itresists thedeformation with aforce propor- tional totheamount ofdeformation, provided thedeformation isnottoo 2-8] LINEAR DIFFERENTIAL EQUATIONS 41 great. This statement iscalled Ho0ke’s law. Itfollows from thefactthat theundeformed solid isatapotential-energy minimum and that the potential energy may beexpanded inaTaylor series inthecoordinate describing thedeformation. Ifasolid isdeformed beyond acertain point, called itselastic limit, itwillremain permanently deformed; that is,its structure isaltered sothat itsundeformed shape forminimum potential energy ischanged. Itturns outinmost cases that thehigher-order terms intheseries (2-91) arenegligible almost uptotheelastic limit, sothat Hooke’s lawholds almost uptotheelastic limit. When theelastic limit isexceeded andplastic ﬂowtakes place, theforces depend inacomplicated Way notonly ontheshape ofthematerial, butalsoonthevelocity ofde- formation and even onitsprevious history, sothat theforces canno longer bespeciﬁed interms ofapotential-energy function. Thus practically anyproblem involving mechanical vibrations reduces tothat oftheharmonic oscillator atsmall amplitudes ofvibration, that is, solong astheelastic limits ofthematerials involved arenotexceeded. Themotions ofstretched strings andmembranes, andofsound vibrations inanenclosed gasorinasolid, result inanumber ofso-called normal modes ofvibration, each mode behaving inmany ways likeanindependent harmonic oscillator. Anelectric circuit containing inductance L,resist- ance R,and capacitance C’inseries, and subject toanapplied electro- motive force E(t), satisﬁes theequation d2q dqq__LE5+RE+6—-E(t)1 (2-92) where qisthecharge onthecondenser anddq/dt isthecurrent. This equation isidentical inform with Eq.(2-86). Early work onelectrical circuits wasoften carried outbyanalogy with thecorresponding mechani- calproblem. Today thesituation isoften reversed, andthemechanical andacoustical engineers areable tomake useofthesimple andeffective methods developed byelectrical engineers forhandling vibration prob- lems. The theory ofelectrical oscillations inatransmission lineorina cavity issimilar mathematically totheproblem of-thevibrating string or resonating aircavity. Thequantum-mechanical theory ofanatom can beputinaform which isidentical mathematically with thetheory ofa system ofharmonic oscillators. 2-8Linear diﬁerential equations with constant coeﬁcients. Equa- tions (2—84)—(2—86) areexamples ofsecond-order linear differential equa- tions. The order ofadifferential equation istheorder ofthehighest derivative that occurs init.Most equations ofmechanics areofsecond order. (Why?) Alimzar differential equation isoneinwhich there are11 l l l 4 1 42 MOTION orAPARTICLE INONE DIMENSION [CHAP. 2 noterms ofhigher than ﬁrst degree inthedependent variable (inthis case x)anditsderivatives. Thus themost general type oflinear differ- ential equation oforder nwould be a..c> +@._.c> +---+am‘g+aocn=bu).<2-93> Ifb(t)=0,theequation issaidtobehomogeneous; otherwise itisinhomo- geneous. Linear equations areimportant because there aresimple general methods forsolving them, particularly when thecoefficients no,a1,...,an areconstants, asinEqs. (2-84)-(2-86). Inthepresent section, weshall solve theproblem ofthefreeharmonic oscillator [Eq. (2—84)], andatthe same time develop ageneral method ofsolving anylinear homogeneous differential equation with constant coefficients. This method isapplied inSection 2-9tothedamped harmonic oscillator equation (2-85). In Section 2-10 weshall study thebehavior ofaharmonic oscillator under a sinusoidally oscillating impressed force. InSection 2-11 atheorem is developed which forms thebasis forattacking Eq. (2-86) with anyim- pressed force F(t), andthemethods ofattack arediscussed brieﬂy. The solution ofEq.(2-84), which Weobtained inSection 2-5, Wenow write intheform 2:=Asin(wot+0), we=Vic/m. (2-94) This solution depends ontwo “arbitrary” constants Aand 0.They are called arbitrary because nomatter what values aregiven tothem, the solution (2-94) willsatisfy Eq.(2-84). They arenotarbitrary inaphys- icalproblem, butdepend ontheinitial conditions. Itcanbeshown that thegeneral solution ofanysecond-order differential equation depends on twoarbitrary constants. Bythiswemean that wecanwrite thesolution intheform w=W;C1,C2), (2-95) such that forevery value ofC1andC2,orevery value within acertain range, x(t;C1,C2) satisﬁes theequation and, furthermore, practically every solution oftheequation isincluded inthefunction x(t;C1,C2)for some value ofC1andC2.* IfWecanﬁndasolution containing twoarbi- trary constants which satisﬁes asecond-order differential equation, then wecanbesure that practically every solution willbeincluded init.The methods ofsolution ofthedifferential equations studied inprevious sec- tions have allbeen such astolead directly toasolution corresponding to *The only exceptions arecertain “singular” solutions which may occur in regions where themathematical conditions foraunique solution (Section 2-2) arenotsatisﬁed. 2-8] LINEAR DIFFERENTIAL EQUATIONS 43 theinitial conditions ofthephysical problem. Inthepresent andsubse- quent sections ofthischapter, weshall consider methods which lead to ageneral solution containing two arbitrary constants. These constants must then begiven theproper values toﬁttheinitial conditions ofthe physical problem; thefact that asolution with two arbitrary constants isthegeneral solution guarantees that wecanalways satisfy theinitial conditions byproper choice oftheconstants. Wenowstate twotheorems regarding linear homogeneous differential equations: THEOREM I.Ifx=x1(t) isanysolution ofalinear homogeneous diﬁ'er- ential equation, andCisanyconstant, thenx=Ca;1(t) isalsoasolution. THEOREM II.Ifas=x1(t) anda:=ac2(t) aresolutions ofalinear homo- geneous diﬂerential equation, then x=x1(t) +x2(t) isalso asolution. Weprove these theorems only forthecase ofasecond-order equation, since mechanical equations aregenerally ofthis type: . d2am3,;+11105)§‘+¢l0(t)5$=0. <2-96> Assume thatx=x1(t) satisﬁes Eq.(2-96). Then am +a1(t)9‘-2,511+a0c><Cw.> = 2clam “E,,—,’§‘+tic)%+a@<¢>x1] =0. Hence x=Cx1(t) alsosatisﬁes Eq.(2-96). Ifa:1(t) andx2(t) both satisfy Eq.(2-96), then _ 2 02(3) +Gift) giﬂrigi) +¢lo(t)(-T1 +1'2) ___ (Z2271 dZl?1—[a2(i) F +¢l1(t) W+<1o(t)11:i +l:l12(t) % -i"111(5) %+<10(t)$2] =0- Hence as=2:1(t)+x2(t) alsosatisﬁes Eq.(2-96). Theproblem ofﬁnding thegeneral solution ofEq. (2-96) thus reduces tothat ofﬁnding any twoindependent “particular” solutions x1(t)and$2(t),forthen Theorems I andIIguarantee that 53=C1fB1(t) +029520) (2-97) 44 MOTION orAPARTICLE INONEDIMENSION [CHAP. 2 isalso asolution. Since thissolution contains two arbitrary constants, itmust bethegeneral solution. The requirement that x1(t) and x2(t) beindependent means inthiscase that oneisnotamultiple oftheother. If:e1(t) were aconstant multiple of:v2(t), then Eq.(2-97) would really contain only onearbitrary constant. The right member ofEq. (2-97) iscalled alinear combination ofx1and902. Inthecase ofequations like(2-84) and (2-85), where thecoefficients areconstant, asolution oftheform x=emalways exists. Toshow this, assume that ao,a1,anda2areallconstant inEq.(2-96) andsubstitute dx 01%x=em, E=pep‘, -it;=pzept. (2-98) Wethenhave (a2p2 —|—alp+a0)e"t =0. (2-99) Canceling outep‘,wehave analgebraic equation ofsecond degree inp. Such anequation has, ingeneral, tworoots. Ifthey aredifferent, this gives twoindependent functions ep‘satisfying Eq.(2-96) andourprob- lemissolved. Ifthetworoots forpshould beequal, wehave found only onesolution, butthen, asweshall show inthenext section, thefunction :2:=M‘ 2 I(2-100) alsosatisﬁes thedifferential equation. Thelinear homogeneous equation ofnthorder with constant coefficients canalsobesolved bythismethod. Letusapply themethod toEq.(2-84). Making thesubstitution (2-98), wehave mp2 +lc=0, (2—101) whose solution is it . Itp=4,/-E==|;iw0, cog= (2-102) This gives, asthegeneral solution, x=C1e"‘*’°‘ +C2e_""’°‘. (2-103) Inorder tointerpret thisresult, weremember that e“=cos0+isin0. (2—104) Ifweallow complex numbers acassolutions ofthedifferential equation, then thearbitrary constants C1andC2must alsobecomplex inorder for Eq. (2—103) tobethegeneral solution. The solution ofthephysical problem must bereal, hence wemust choose C1andC2sothat xturns out 2-8] LINEAR DIFFERENTIAL EQUATIONS 45 tobereal. Thesumoftwocomplex numbers isrealifoneisthecomplex conjugate oftheother. If C=a+ib, (2—105) and C*=a—ib, (2—106) then C+C*=2a, C’—C*=2ib. (2-107) Now e""’°‘ isthecomplex conjugate ofe_"‘°°‘, sothat ifWesetC1=C, C2=C*,then a:willbereal: ’ ~2=cow+o*@-M. (2-108) Wecould evaluate xbyusing Eqs. (2—104), (2-105), and (2—106), but thealgebra issimpler ifwemake useofthe(polar representation ofa complex number: C=a—i—ib=re“, (2-109) C*=a—ib=re“"’, (2—110) where r=of+b*)"”, tan0= (2-111) a=rcos0, b=rsin0. (2—112) The reader should verify that these equations follow algebraically from Eq.(2—104). Ifwerepresent Casapoint inthecomplex plane, then a andbareitsrectangular coordinates, andrand0areitspolar coordinates. Using thepolar representation ofC,Eq.(2-108) becomes (wesetr=%A) 2=-5-Ae1'(wo¢+0) _|_%.Ae_'i(woi+0) =Acos(wot+0). (2—l13) This isthegeneral realsolution ofEq.(2-84). Itdiffers from thesolu- tion (2-94) only byashift of1r/2inthephase constant 0. Setting B1=Acos0,B2=-Asin0,wecanwrite oursolution in another form: " a:=B1coswot—|—B2sinwot. (2-114) The constants A,0,orB1,B2,aretobeobtained interms oftheinitial values zoo,voatt=10 bysetting l mo=Acos0=B1, (2-115) U9 = —_(DOA Sill 0= (IJQB2. Il l 4 1 l I 1 46 MOTION oFAPARTICLE INONE DIMENSION I [cIIAI>. 2 Thesolutions areeasily obtained: 2 0% 1/2 A= $0+C? 1 0 tan0=--"9-, (2-118)$0190 OI‘ B1 =I130, B2= (2-120) Another way ofhandling Eq. (2-103) would betonotice that, since Eq.(2-84) contains only realcoeﬁicients, acomplex function cansatisfy itonly ifboth realandimaginary parts satisfy itseparately. (The proof ofthis statement isamatter ofsubstituting x=u+iwand carrying outalittle algebra.) Hence ifasolution is(wesetr=A) x=Ceiwot :Aei(wot+6) K =Acos(wot-I—0)+iAsin(wot+0), (2—121) thenboth therealandimaginary parts ofthissolution must separately be solutions, andwehave either solution (2—113) or(2-94). Wecancarry through thesolutions oflinear equations likethis, andperform anyalge- braic operations weplease onthem intheir complex form ‘(solong aswe donotmultiply twocomplex numbers together), with theunderstanding that ateach step what wearereally concerned with isonly therealpart oronly theimaginary part. This procedure isoften useful inthetreat- ment ofproblems involving harmonic oscillations, andweshall useitin Section 2-10. Itisoften very convenient torepresent asinusoidal function asacom- plex exponential:i0 _n cos0=realpartof6*”= (2-122) _ 1.0__ -10 sin0=imaginary part ofe”=e——§;-—- (2—123) Exponential functions areeasier tohandle algebraically than sines and cosines. The reader willﬁndtherelations (2-122), (2-123), and (2—104) useful inderiving trigonometric formulas. The power series forthesine andcosine functions arereadily obtained byexpanding cl’inapower series andseparating therealandimaginary parts. The trigonometric rulefor sin(A—|—B)andcos(A+B)canbeeasily obtained from thealgebraic ruleforadding exponents. Many other examples could becited. 2-9] THEDAMPED HARMONIC osoILLAToR 47 2-9Thedamped harmonic‘ oscillator. Theequation ofmotion fora particle subject toalinear restoring force andafrictional force proportional toitsvelocity is[Eq. (2-85)] 9 mi?—I—bi:+lea:=O, (2—l24) where thedots stand fortime derivatives. Applying themethod ofSec- tion2-8, wemake thesubstitution (2-98) andobtain 2 mp2 +bp—i—k=0. (2—125) Thesolution is u b b212]“P——57.* "m' (H26) Wedistinguish three cases: (a)lc/m >(b/2m)2, (b)k/m <(b/2m)2, and (c)It/m =(b/2m)2. Incase (a),wemake thesubstitutions (.00 ,=' J5 ) b 7—5-1;: ‘-91=(wt—I2)”. <2—129> where ‘Yiscalled thedamping coefficient and(wo/21r) isthenatural fre- quency oftheundamped oscillator. There arenow twosolutions forp: p=—'Y:l':iw1. (2—130) Thegeneral solution ofthedifferential equation istherefore I x=C1e_”+i°"t +C2e_"_“"‘t. A(2—131) Setting 01-aw". 02=ale-"’, (H32)wehave Q?=Ar"cos(<21:+0). - (2-133) This corresponds toanoscillation offrequency (w1/21r) with anamplitude Ae—" which decreases exponentially with time (Fig. 2-4). The constants Aand0depend upon theinitial conditions. The frequency ofoscillation islessthan without damping. Thesolution (2—133) canalsobewritten 2=e-“(B1 cosw1t +B2sin0.11:). (2-134) 48 MOTION orAPARTICLE INONE DIMENSION [CHAP. 2 IAs oi t —A FIG. 2-4. Motion ofdamped harmonic oscillator. Heavy curve: x= AF" coswt,’Y =w/8. Light curve: :11:=:i=Ae‘"". Interms oftheconstants woand‘Y,Eq.(2—124) canbewritten it+2m+wﬁrv=0. _(2-135) This form oftheequation isoften used indiscussing mechanical oscilla- tions. Thetotal energy oftheoscillator is E=%rn:i:2 —l—-§ka:2. (2—136) Intheimportant case ofsmall damping, 7<<wo,wecansetw1éwoand neglect 'Ycompared with wo,andwehave fortheenergy corresponding to thesolution (2—133), approximately, Ea15,12,426-2*‘ =Eoe-2". (2-137) Thus theenergy fallsoffexponentially attwice therateatwhich theampli- tude decays. Thefractional rateofdecline orlogarithmic derivative ofEis 1dE dlnE _ F"E —T ——2'Y. Wenow consider case (b),(wo<7). Inthiscase, thetwo solutions forpare 2-9] THEDAMPED HARMONIC oscILLAToR 49 1»=—v.--v—<12-wt)“, (2—139) p=-rs=-Y+(12—wt)”- Thegeneral solution is x=C1e_“t +C2e““t. (2—140) These twoterms both decline exponentially with time, oneatafaster rate than theother. The constants C1andC2may bechosen toﬁttheinitial conditions. The reader should determine them fortwoimportant cases: mo¢0,vo=0andxo=0,vosé0,and draw curves x(t)forthetwo cases. Incase (c),(wo='Y),wehave only onesolution forp: ' ' p=-v. (2-141) Thecorresponding solution forxis 2=e-"". (2-142) Wenow show that, inthiscase, another solution is 2=tr". - (2-143) Toprove this,wecompute zt=e_" —7te_"‘, (2—144) :25=—2'Ye_“ +1/2te_"t. TheleftsideofEq.(2—135) is,forthis:0, _ e+2n+wgx=(<23-'Y2)te_". (2-145) This iszero ifwo=7.Hence thegeneral solution incasewo='Yis x=(C1+C2t)e"”. (2—146) This function declines exponentially with time atarate intermediate be- tween that ofthetwoexponential terms inEq.(2-140): 'Y1>1>‘Y2. (2-147) Hence thesolution (2-146) falls tozerofaster after asufﬁciently long time than thesolution (2-140), except inthecaseC2=0inEq.(2—140). Cases (a),(b),and (c)areimportant inproblems involving mechanisms which approach anequilibrium position under theaction ofafrictional dampingI l 1 l 1 l 50 MOTION orAPARTICLE INONE DIMENSION [CHAP. 2 I (b) (C) (a) I» l Fro. 2-5. Return ofharmonic oscillator toequilibrium. (a)Underdamped. (b)Overdamped. (c)Critically damped. force, e.g., pointer reading meters, hydraulic andpneumatic spring returns fordoors, etc. Inmost cases, itisdesired that themechanism move quickly andsmoothly toitsequilibrium position. Foragiven damping coefﬁcient ’Y,orforagiven wo,thisisaccomplished in"theshortest time without overshoot ifwo=‘Y[case (c)]. This caseiscalled critical damping. Ifwo<‘Y,thesystem issaidtobeoverdamped; itbehaves sluggishly and does notreturn asquickly to2:=0asforcritical damping. Ifwo>‘Y, thesystem issaidtobeunderdamped; thecoordinate xthenovershoots the value ac=0andoscillates. Note that atcritical damping, w1=0,so thattheperiod ofoscillation becomes inﬁnite. Thebehavior isshown in Fig.2-5forthecase ofasystem displaced from equilibrium andreleased (wo;é0,vo=0). The reader should draw similar curves forthecase where thesystem isgiven asharp blow att=O(i.e., xo=0,vo¢0). 2-10 The forced harmonic oscillator. The harmonic oscillator subject toanexternal applied force isgoverned byEq.(2-86). Inorder tosim- plify theproblem ofsolving thisequation, westate thefollowing theorem: THEoREM III. Ifac,-(t) isasolution ofaninhomogeneous linear equation [e.g., Eq.(2-86)], andx;,(t) isasolution ofthecorresponding homogeneous equation [e.g., Eq.(2-85)], then x(t)=ac,-(t) +a:;,(t) isalsoasolution of theinhomogeneous equation. This theorem applies whether thecoefﬁcients intheequation areconstants orfunctions oft.The proof isamatter ofstraightforward substitution, andislefttothereader. Inconsequence ofTheorem III,ifweknow the general solution xhofthehomogeneous equation (2-85) (wefound thisin Section 2-9), then weneed ﬁndonly oneparticular solution ac,ofthein- homogeneous equation (2-86). Forwecanadd20,-to90;,andobtain asolu- tionofEq.(2-86) which contains twoarbitrary constants andistherefore thegeneral solution. 2-10] THE FoRoED HARMONIC osoILLAToR 51 The most important case isthat ofasinusoidally oscillating applied force. Iftheapplied force oscillates with angular frequency wandampli- tude Fo,theequation ofmotion is dzrc ola: I612=FQC0S((a.>l-1~ 00), (2-I48) where 0oisaconstant specifying thephase oftheapplied force. There are, ofcourse, many solutions ofEq.(2-148), ofwhich weneed ﬁndonly one. From physical considerations, weexpect that onesolution willbeasteady oscillation ofthecoordinate xatthesame frequency astheapplied force: ac=A,cos(wt+0,). (2—149) Theamplitude A,andphase 6,oftheoscillations inxwillhave tobede- termined bysubstituting Eq.(2—149) inEq.(2-148). This procedure is straightforward andleads tothecorrect answer. The algebra issimpler, however, ifwewrite theforce astherealpart ofacomplex function:* F(t)=Re(Foe’-M), (2-150) F0=FM". (2-151) Thus ifwecanﬁndasolution x(t)of 4’ .1 ,,,,m-dt—f+bdlt‘+lcx=Foe‘, (2452) then, bysplitting theequation intorealandimaginary parts, wecanshow’ that therealpart ofx(t)willsatisfy Eq.(2—148). Weassume asolution of theform X=Xoeiwt 7sothat . . '5 .. 2 '¢x=iwxoew , x=—wxoew . (2-153) Substituting inEq.(2-152), wesolve forxo: X0= (2-154) wo—w-1-2i'Yw The solution ofEq.(2-152) istherefore _ iwtX=X06..."= (2-155, wo-—w-1-21/Yw *Note theuseofroman type (F,x)todistinguish complex quantities from thecorresponding realquantities (F,2:).1 1 1 1 1 1 1 52 MOTION oFAPARTICLE INONE DIMENsIoN 1CHAP. 2 Weareoften more interested inthevelocity - iwt2=3°22 (2-156) mwo—w+2i’Yw Thesimplest Way towrite Eq.(2-156) istoexpress allcomplex factors in polar form [Eq. (2-109)]: ‘ _ 1'=e“"2, (2-157) wg—w2-1-2i‘Yw =[(w§ -w2)2 -1-4'Y2w2]1/2 exp itan_1 -5%: - (.00 '-(.0 (2—158) Ifweusethese expressions, Eq.(2-156) becomes ~ *= W’) mwo—w w where 2__2 B=1-—tan_1 -% =tan_1 all 1 (2—160) 2 wo—w 2'Yw sin19=—-—"-‘-’?’—-"i?———, (2-161)Kw?) __w2)2 +4,720,211/2 2'YwB= ‘W’ ByEq.(2-159), ri;=Re(x) =F0 2 2)2@+ 472 211/2 cos(wt+0o-1-13), (2-163) m (D0 —OJ (.0 and :1:=Re(x) =Re(x/iw) Fo 1 .= .s1n(wt-1-0+/3). (2—164)m _ w2)2 +47203211/2 O This isaparticular solution ofEq.(2-148) containing noarbitrary con- stants. ByTheorem IIIandEq. (2-133), thegeneral solution (forthe underdamped oscillator) is —7t F0/m:1:=Ae cos(w1t-1-0)-1-—-ii-sin (wt-1-0o-1-/3).not-w2>2+4v%»21"’ . (2—165) 2-10] THE FORCED HARMONIC osoILLAToR 53 This solution contains twoarbitrary constants A,0,whose values arede- termined bytheinitial values zvo,voatt=0.Theﬁrstterm diesoutex- ponentially intime andiscalled thetransient. The second term iscalled thesteady state, andoscillates with constant amplitude. Thetransient de- pends ontheinitial conditions. The steady state which remains after the transient diesaway isindependent oftheinitial conditions. Inthesteady state, therateatwhich Work isdone ontheoscillator by theapplied force is 2 ;,i;F(¢) =F7:[(w2 _w%)2‘-°+ 472w2]1l2 cos(wt+0o)cos(wt-1-0o-1-/3) _F3,_wcos13cos2 (wt+0o), _F3 wsinBsin2(wt +Ho) _ m [(w2 ___ w%)2 +4,Y2w2]l/2 2m [(w2 _ w2))2 +4,y2w2]l/2 (2—166) Thelastterm ontheright iszero ontheaverage, while theaverage value ofcos2 (wt+0o)over acomplete cycle is5-.Hence theaverage power de- livered bytheapplied force is , F2cos18 w Pay = <1EF(t)>av = 02m -[(0)2 __w5)2 +472w2:|1/2 ! or 1 P8,.=%Fo0t,,, cos19, (2—168) where aimisthemaximum value of:i:.Asimilar relation holds forpower delivered toanelectrical circuit. The factor cosﬁ iscalled thepower factor. Intheelectrical case, 13isthephase angle between thecurrent and theapplied emf. Using formula (2-162) forcosB,wecanrewrite Eq. (2-167): F5 'Yw2 PW m(1.02—w§)2 -1-4’Y2w2 _ (2169) Itiseasy toshow that inthesteady state power issupplied totheoscillator atthesame average rate that power is‘being dissipated byfriction, asof course itmust be.The power Pa‘,hasamaximum forw=wo. InFig. 2-6, thepower P1,,(inarbitrary units) andthephase of5ofsteady-state forced oscillations areplotted against wfortwovalues of'Y.The heavy curves areforsmall damping; thelight curves areforgreater damping. Formula (2-169) canbesimpliﬁed somewhat incase ‘Y<<wo. Inthis case, P8,,islarge only near theresonant frequency wo,andweshall deduce a formula valid near w=wo.Deﬁning I ~ Aw=w—wo, (2—170) 54 MOTION onAPARTICLE INONEDIMENSION [CHAP- 2 Pl": (7= I ; ’ T/2 ﬁr(7=%°\’0) '3,(1=am) Pnv1('Y =wo) L . L w wo —1r/2 FIG. 2-6. Power andphase offorced harmonic oscillations. andassuming Aw<<wo,wehave ((4)2-wﬁ)=(co+coo)Awé2w0Aw, (2-171) ti’so3. (2-172) t Hence -°-£3 Y. _ Pm,_4m(M2+72 (2173) This simple formula gives agood approximation toPm,near resonance. Thecorresponding formula for/3is COS 65 sin §éKiA; '(2-174) When w<<wo,Hi1r/2, andEq.(2—164) becomes acé-17% cos(wt+00)=Q- (2-175) t _ wom This result iseasily interpreted physically; when theforce varies slowly, theparticle moves insuch away that theapplied force isjustbalanced by therestoring force. When w>>wo,Bi—1r/2, andEq.(2-164) becomes .F F(t)ac=—ﬁtcos(wt+00)=-(m- (2-176) 2-10] THEFORCED HARMONIC OSCILLATOR 55 Themotion now depends only onthemass oftheparticle andonthefre- quency oftheapplied force, andisindependent ofthefriction andthe restoring force. This result is,infact, identical with that obtained inSec- tion 2—3[seeEqs. (2-15) and (2——19)] forafree particle subject toan oscillating force. Wecanapply theresult (2—165) tothecase ofanelectron bound toan equilibrium position a:=0byanelastic restoring force, andsubject toan oscillating electric ﬁeld: E,=E0coswt, (2—177) F=—eE0 coswt. (2—178) Themotion willbegiven by _ _ eE sin(wt+B)x—Ac7‘cos(wlt+0)-—mo[(0)2 _way +4V2w2]1/2 -(2—179) The term ofinterest here isthesecond one, which isindependent ofthe initial conditions and oscillates with thefrequency oftheelectric ﬁeld. Expanding thesecond term, weget m=_eEo sinBcos wt _eEo cosﬁsinwt_ m [(w2 __wg)2 +4.y2w2]1/2 m [(6)2 __wg)2 +4,Y2w2]1/2 _—eE0 coswt cog—-wz 5 m Kw”—w§)2+4Y2w2] _eE0sinQt 2 37;.» 22_ (2_18O) m [(w -1.00) +4’Yw] Theﬁrstterm represents anoscillation ofacinphase with theapplied force atlowfrequencies, 180° outofphase athigh frequencies. Thesecond term represents anoscillation of2:that is90°outofphase with theapplied force, thevelocity :i:forthisterm being inphase with theapplied force. Hence thesecond term corresponds toanabsorption ofenergy from theapplied force. The second term contains afactor ‘Yand istherefore small, if ‘Y<<wo,except near resonance. Ifweimagine adielectric medium con- sisting ofelectrons bound byelastic forces topositions ofequilibrium, then theﬁrstterm inEq.(2—180) willrepresent anelectric polarization propor- tional totheapplied oscillating electric ﬁeld, while thesecond term will represent anabsorption ofenergy from theelectric ﬁeld. Near theresonant frequency, thedielectric medium willabsorb energy, andwillbeopaque toelectromagnetic radiation. Above theresonant frequency, thedis- placement oftheelectrons isoutofphase with theapplied force, andthe 56 MOTION OFAPARTICLE INONE DIMENSION [CHAP. 2 resulting electric polarization willbeoutofphase with theapplied electric ﬁeld. The dielectric constant andindex ofrefraction willbelessthan one. Forvery high frequencies, theﬁrst term ofEq.(2—180) approaches thelastterm ofEq.(2-18), andtheelectrons behave asifthey were free. Belowthe resonant frequency, theelectric polarization willbeinphase with theapplied electric ﬁeld, andthedielectric constant andindex ofre- fraction willbegreater than one. Computing thedielectric constant from theﬁrst term inEq.(2—180), inthe same manner asforafreeelectron [seeEqs. (2—20)—(2—26)], weﬁnd, forNelec- trons perunit volume: 2 2 2 G=1+4l]\_Tf_ . (2.131) m (w0——w) +4'Yw Theindex ofrefraction forelectromagnetic waves (it=1)is n=5=(/.¢e)1/2 =6'2. (2-1s2) Forvery high orvery lowfrequencies, Eq.(2—181) becomes 2es1+i, @<<wo, (2—183) mwg 2 6é1- w>>cog. (2-184) The mean rate ofenergy absorption perunit volume isgiven byEq.(2-169): dE_Ne2E% wt” _ (H85) dt m(0)2-<»%)2+41%” The resulting dielectric constant andenergy absorption versus frequency areplotted inFig. 2-7. Thus thedielectric constant isconstant and greater than oneatlowfrequencies, increases asweapproach theresonant frequency, falls toless than one intheregion of“anomalous dispersion” where there isstrong absorption ofelectromagnetic radiation, andthen rises, approaching oneathigh frequencies. The index ofrefraction will follow asimilar curve. This isprecisely thesort ofbehavior which is exhibited bymatter inallforms. Glass, forexample, hasaconstant dielec- tricconstant atlowfrequencies; intheregion ofvisible light itsindex of refraction increases with frequency; anditbecomes opaque inacertain 2-10] THEFORCED HARMONIC OSCILLATOR 57 e 1 . dE/dt | to W0 FIG. 2-7. Dielectric constant andenergy absorption formedium containing harmonic oscillators. band intheultraviolet. X-rays aretransmitted with anindex ofrefrac- tionvery slightly lessthan one. Amore realistic model ofatransmitting medium would result from assuming several different resonant frequencies corresponding toelectrons bound with various values ofthespring con- stant lc.This picture isthen capable ofexplaining most ofthefeatures in theexperimental curves foreornvs.frequency. Notonly isthere qualita- tive agreement, buttheformulas (2—181)—(2—185) agree quantitatively with experimental results, provided theconstants N,0:0,and’Yareproperly chosen foreach material. Thesuccess ofthistheory wasoneofthereasons fortheadoption, until theyear 1913, ofthe“jelly model” oftheatom, in which electrons were imagined embedded inapositively charged jelly in which theyoscillated asharmonic oscillators. Theexperiments ofRuther- fordin1913forced physicists toadopt the“planetary” model oftheatom, butthismodel wasunable toexplain even qualitatively theoptical and electromagnetic properties ofmatter until theadvent ofquantum mechan- ics.Theresult ofthequantum-mechanical treatment isthat, fortheinter- action ofmatter andradiation, thesimple oscillator picture gives essentially correct results when theconstants areproperly chosen.* Wenow consider anapplied force F(t) which islarge only during a short time interval 6tandiszero ornegligible atallother times. Such a force iscalled animpulse, andcorresponds toasudden blow. Weassume theoscillator initially atrestata:=0,andweassume thetime 5tsoshort that themass moves only anegligibly small distance while theforce is acting. According toEq. (2—4), themomentum just after theforce is applied willequal theimpulse delivered bytheforce: ma,=pg=IFdt, (2-186) where voisthevelocity just after theimpulse, andtheintegral istaken over thetime interval 8tduring which theforce acts. After theimpulse, *SeeJohn C.‘Slater, Quantum Theory ofMatter. New York: McGraw-Hill Book Co., 1951. (Page 378.) 'i l 1 .1 J 58 MOTION orAPARTICLE INONEDIMENSION [cn.u>. 2 theapplied force iszero, andtheoscillator must move according toEq. (2—133) ifthedamping islessthan critical. Weareassuming 6tso"small that theoscillator does notmove appreciably during thistime, hence we choose 0=—(1r/2) —wlto, inorder that x=0att=to,where tois theinstant atwhich theimpulse occurs: x=Ar"sin[w1(t-¢0)]. (2-187) Thevelocity att=tois U0=w1Ae"”°. Thus A=Ee"‘°. (2-189) 011 Thesolution when animpulse pgisdelivered att==totoanoscillator at restistherefore O, t§to, 1 ___ __ .x ﬁre W'°)sin[w1(t —t0)], t>to. ( ) Here wehave neglected theshort time 6tduring which theforce acts. Weseethattheresult ofanimpulse-type force depends onlyonthetotal impulse pgdelivered, andisindependent oftheparticular form ofthefunc- tion F(t), provided only that F(t) isnegligible except during avery short time interval 6t.Several possible forms ofF(t)which have thisproperty arelisted below: . O,’ t<to, F(t)=Po/51, toSiS‘to+51, (13-191) or t>t0+at: F(t) =$2 1 —OO <t<O0, F(t) =i\%—1_rexp|:-—- ]: —oo <ti<oo. (2—193) The reader may verify that each ofthese functions isnegligible except within aninterval oftheorder of6taround to,andthat thetotal impulse delivered byeach ispo.Theexact solution ofEq.(2-86) with F(t)given byanyoftheabove expressions must reduce toEq.(2—190) when 6t—>O (seeProblem 23). 2-11] THE PRINCIPLE orSUPERPOSITION 59 2-11 Theprinciple ofsuperposition. Harmonic oscillator with arbi- trary applied force. Animportant property oftheharmonic oscillator is that itsmotion x(t), when subject toanapplied force F(t) which canbe regarded asthesum oftwoormore other forces F1(t), F2(t), ...,isthe sum ofthemotions a:1(t), x2(t), ...,which itwould have ifeach ofthe forces Fn(t)were acting separately. This principle applies tosmall mechan- icalvibrations, electrical vibrations, sound waves, electromagnetic waves, andallphysical phenomena governed bylinear differential equations. The principle isexpressed inthefollowing theorem: THEOREM IV.Letthe(ﬁnite orz'n_ﬁnite*) setoffunctions x,,(t), n=1,2,3, ...,besolutions oftheequations A ma,+ta,+la,=F,,(t), (2-194) andlet F(t)=ZF,,(t). (2-195) Then thefunction x(t)=Zac) (2-196) satisﬁes theequation mt+bi:+ha:=F(t). (2—197) Toprove thistheorem, wesubstitute Eq.(2—196) intheleftside ofEq. (2—197): mi3+b0t+Icx =mz§t,,+b2:i:,,+lcZ:a:,, 70 ‘R ‘VI =Z(ma.+be,+ta.) =2 Fn(t) =F(t). This theorem enables ustoﬁndasolution ofEq.(2—197) whenever the force F(t)canbeexpressed asasum offorces F"(t)forwhich thesolutions ofthecorresponding equations (2—194) can befound. Inparticular, whenever F(t) canbewritten asasum ofsinusoidally oscillating terms: F(t)=Z0,,cos(cant+on), (2-198) *When thesetoffunctions isinﬁnite, there arecertain mathematical restric- tions which need notconcern ushere.\ 1 l l 44 60 MOTION orAPARTICLE INoNEDIMENSION [CHAP. 2 aparticular solution ofEq.(2—197) willbe,byTheorem IVandEq.(2—164), C1, 1 . ”=Emits-...%>2+m.21"2 Sm(“"1+6"+M’(H99) 2 2_ (.00 *(.01;=tan 1-—i -B" 2'Yw,, Thegeneral solution isthen x_= Ae—'Yt cos (wlt + 2 Ch sin + an + ’ nm[(015—w§)2+4'Y2w§]”2 where Aand0are,asusual, tobechosen tomake thesolution (2—200) ﬁt theinitial conditions. Wecanwrite Eqs. (2—198) and(2—199) inadifferent form bysetting An=Cncos0", Bn=—C,, sin0". (2—201) Then F(t) =Z(Ancosw,,t—l—B”sincont), (2—202) and = Ansin(writ +I311) —Bncos (writ +B1»)_ 2_203”>;m[(<»ﬁ-was+41/2wZ]"2 ‘’ Animportant case ofthiskind isthat ofaperiodic force F(t),that is,a force such that F(t+T)=F(t), (2-20/1) where Tistheperiod oftheforce. Foranycontinuous function F(t)satis- fying Eq.(2—204) (and, infact, even foronly piecewise continuous func- tions), itcanbeshown that F(t)canalways beWritten asasum ofsinus- oidal functions: F(t) =%A0 +Z<A,, cos21%” +Bnsin , (2—205) ’!t=1 where 'ﬂl\D*1 A,,=—‘/0 F(t)cosgl.Tﬂidt, n=0,1,2,..., T (2-206) B,,= F(t)sing;-,’1?d¢, n=1,2,3,.... This result enables us,atleast inprinciple, tosolve theproblem ofthe forced oscillator foranyperiodically varying force. ThesuminEq.(2—205) 2—11] THE PRINCIPLE orSUPERPOSITION 61 iscalled aFourier series.* Theactual computation ofthesolution bythis method isinmost cases rather laborious, particularly theﬁtting ofthe constants A,0inEq.(2—200) totheinitial conditions. However, theknowl- edge that such asolution exists isoften useful initself. Ifanyofthefre- quencies 21rn/ Tcoincides with thenatural frequency <00oftheoscillator, then thecorresponding terms intheseries inEqs. (2—199) or(2—203) will berelatively much larger than therest. Thus aforce which oscillates nonsinusoidally athalfthefrequency wemay cause theoscillator toper- form anearly sinusoidal oscillation atitsnatural frequency mo. Ageneralization oftheFourier series theorem [Eqs. (2—205) and(2—206)] applicable tononperiodic forces istheFourier integral theorem, which allows ustorepresent anycontinuous (orpiecewise continuous) function F(t), subject tocertain limitations, asasuperposition ofharmonically oscillating forces. Bymeans ofFourier series andintegrals, wemay solve Eq.(2—197) foralmost anyphysically reasonable force F(t).Weshall not pursue thesubject further here. Suffice ittosaythat while themethods ofFourier series andFourier integrals areofconsiderable practical value insolving vibration problems, their greatest importance inphysics probably liesinthefactthat inprinciple such asolution exists. Many important results canbededuced without ever actually evaluating theseries or integral atall. Amethod ofsolution known asGreen's method isbased onthesolution (2—190) foranimpulse-type force. Wecanthink ofanyforce F(t)asthe sum ofaseries ofimpulses, each acting during ashort time 6tanddeliver- ing‘animpulse F(t) 6t: F(t)e2:F,,(t), (2-207) 0, if t<t,,, where tn=n6t, Frau)=Foo, iftn5t5t..'+1, (H08) o, ift>¢,.+,. As8t——>0,thesumofalltheimpulse forces F,,(t) willapproach F(t).(See Fig.2-8.) According toTheorem IVandEq.(2—190), asolution of Eq.(2—197) foraforce given byEq.(2-207) is x(t)aif e—7(t_t”) sin[w1(t-¢,,)], (2-209) *For aproof oftheabove statements and amore complete discussion of Fourier series, seeDunham Jackson, Fourier Series andOrthogonal Polynomials. Menasha, Wisconsin: George Banta Pub. Co., 1941. (Chapter 1.) 62 MOTION orAPARTICLE INONE DIMENSION [CHAP. 2 F(t) _ z Fro. 2-8. Representation ofaforce asa sum ofimpulses. Heavy curve: F(t). Light curve: Z,,F,,(t). where t,,,,§t<t,,°+1. Ifwelet6t-—>0andwrite tn=t’,Eq.(2—209) becomes t an=/ %?aW”%m@c-own mam _w 1 I Thefunction * 0, ift’>t, G(t,t') = e‘—-v(t-t’) _ _ I (2-211) -*-$1-'S11'l [(.01(t —l')], If lsl, iscalled theGreen’s function forEq.(2—197). Interms ofGreen’s function, Q .no=f oawmwan mam) Iftheforce F(t) iszero fort<to,then thesolution (2—210) willgive x(t)=0fort<to.This solution istherefore already adjusted toﬁtthe initial condition that theoscillator beatrestbefore theapplication ofthe force. Foranyother initial condition, atransient given byEq.(2—133), with appropriate values ofAand0,willhave tobeadded. The solution (2—210) isuseful instudying thetransient behavior ofamechanical sys- temorelectrical circuit when subject toforces ofvarious kinds. PROBLEMS 1.Atugofwarisheld between twoteams ofﬁvemen each. Each man weighs 160lband can initially pull ontherope with aforce of2001b-wt. Atﬁrst the teams areevenly matched, butasthemen tire, theforce with which each man pulls decreases according totheformula F=(200lb-wt)e-"', where themean tiring time ris10secforoneteam and20secfortheother. Find themotion. (g=32ft-sec-2.) What istheﬁnal velocity ofthetwo teams? Which ofourassumptions isresponsible forthis unreasonable result? PROBLEMS 63 2.Ahigh-speed proton ofelectric charge emoves with constant speed v0 inastraight linepast anelectron ofmass m,charge —e,initially atrest. The electron isatadistance afrom thepath oftheproton. (a)Assume that the proton passes soquickly that theelectron does nothave time tomove appre- ciably from itsinitial position until theproton isfaraway. Show that the component offorce inadirection perpendicular totheline along which the proton moves is 2 F=ieai »(electrostatic orgaussian units)(a2+vgt2)3/2 where aisthedistance oftheelectron from thepath oftheproton andt=0 when theproton passes closest totheelectron. (b)Assume that theelectron moves only along alineperpendicular tothepath’ oftheproton. Find theﬁnal kinetic energy oftheelectron. (c)Write thecomponent oftheforce inadirec- tion parallel totheproton velocity, andcalculate thenetimpulse inthat direc- tion delivered totheelectron. Does thisjustify theassumption inpart (b)? 3.Aparticle which had originally avelocity onissubject toaforce given byEq.(2—192). (a)Find v(t)andx(t). (b)Show that as5t—>0,themotion approaches motion atconstant velocity with anabrupt change invelocity at t=toofamount po/m. 4.Aparticle initially atrestissubject, beginning att=O,toaforce F=F0e_"cos (wt+0). (a)Find itsmotion. (b)How does theﬁnalvelocity depend on0,andonw? [Hint: Thealgebra issimpliﬁed bywriting cos(wt+ 0)interms ofcomplex expo- nential functions.] 5.Aboat with initial velocity v0isslowed byafrictional force ' F=—-bem’. (a)Find itsmotion. (b)Find thetime andthedistance required tostop.‘ 6.Ajetengine which develops aconstant maximum thrust F0isused topower aplane with africtional drag proportional tothesquare ofthevelocity. Ifthe plane starts att=0with anegligible velocity andaccelerates with maximum thrust, ﬁnditsvelocity v(t). _ 7.Find v(t)and x(t)foraparticle ofmass mwhich starts at:00=0with velocity 110,subject toaforce given byEq.(2-31) with nsé1.Find thetime tostop, and thedistance required tostop, and verify theremarks inthelast paragraph ofSection 2-4. 8.(a)Abody ofmass mslides onarough horizontal surface. The coefficient ofstatic friction is#8,and thecoefficient ofsliding friction isit.Devise an analytic function F(v) torepresent thefrictional force which hastheproper constant value atappreciable velocities andreduces tothestatic value atvery lowvelocities. (b)Find themotion under theforce you have devised ifthe body starts with aninitial velocity v0. 9.Aparticle ofmass misrepelled from theorigin byaforce inversely pro- portional tothecube ofitsdistance from theorigin. Setupandsolve theequa- tion ofmotion iftheparticle isinitially atrestatadistance nofrom theorigin. 64 MOTION orAPARTICLE INONE DIMENSION [CHAP. 2 10.(a)Amass misconnected totheorigin with aspring ofconstant la, whose length when relaxed isl.The restoring force isvery nearly proportional totheamount thespring hasbeen stretched orcompressed solong asitisnot stretched orcompressed very far. However, when thespring iscompressed too far,theforce increases very rapidly, sothat it.is impossible tocompress the spring tolessthan half itsrelaxed length. When thespring isstretched more than about twice itsrelaxed length, itbegins toweaken, and therestoring force becomes zero when itisstretched tovery great lengths. (a)Devise aforce function F(x) which represents this behavior. (Ofcourse areal spring isde- formed ifstretched toofar,sothat Fbecomes afunction ofitsprevious history, butyou aretoassume here that Fdepends only onx.) (b)Find V(x) and describe thetypes ofmotion which may occur. 11.Aparticle issubject toaforce aF-—ka: -1-Z3- (a)Find thepotential V(x), describe thenature ofthesolutions, andﬁnd the solution x(t). (b)Can you give asimple interpretation ofthemotion when E2>>ha? V(:v) +V1 I I ,1; -11 011 _VU FIGURE 2-9 12.Analpha particle inanucleus isheld byapotential having theshape shown inFig. 2-9. (a)Describe thekinds ofmotion that arepossible. (b)De- vise afunction V(:c) having this general form and having the.values '-V0 andV1atx=0and:1:=;i:x1, andﬁndthecorresponding force. 13.Derive thesolutions (2-69) and (2-70) forafalling body subject toa frictional force proportional tothesquare ofthevelocity. 14.Abody ofmass mfalls from restthrough amedium which exerts afric- tional drag be°"”'. (a)Find itsvelocity v(t). (b)What istheterminal velocity? (c)Expand your solution inapower series int,keeping terms uptot2.(d)Why does thesolution failtoagree with Eq.(1-28) even forshort times t? 15.Aprojectile isﬁred vertically upward with aninitial velocity 00.Find its motion, assuming africtional drag proportional tothesquare ofthevelocity. (Constant g.) 16.Derive equations analogous toEqs. (2-80) and (2-81) forthe mo- tion ofabody whose velocity isgreater than theescape velocity. [Hint: Set sinhﬁ =(Ex/mMG)1/2.] 17.Find themotion ofabody projected upward from theearth withavelocity equal totheescape velocity. Neglect airresistance. PROBLEMS 65 18.Find thegeneral solution forthemotion ofabody subject toalinear re- pelling force F=lax. Show that thisisthetype ofmotion tobeexpected in theneighborhood ofapoint ofunstable equilibrium. 19.Thepotential energy fortheforce between twoatoms inadiatomic mole- cule hastheapproximate form: V(x)=-§+ where :0isthedistance between theatoms anda,barepositive constants. (a) Find theforce. (b)Assuming oneoftheatoms isvery heavy andremains at restwhile theother moves along astraight line, describe thepossible motions. (c)Find theequilibrium distance andtheperiod ofsmall oscillations about the equilibrium position ifthemass ofthelighter atom ism. 20.Aparticle ofmass missubject toaforce given by 2 5 8a 28a 27aF=B(s-F+.—8)' (a)Find andsketch thepotential energy. (Bandaarepositive.) (b)Describe thetypes ofmotion which may occur. Locate allequilibrium points and de- termine thefrequency ofsmall oscillations about any which arestable. (c)A particle starts at2:=3a/2with avelocity v=—v0, where voispositive. What is thesmallest value ofU9forwhich theparticle may eventually escape toavery large distance? Describe themotion inthatcase. What isthemaximum velocity theparticle willhave? What velocity willithave when itisvery farfrom its starting point‘? 21.Aparticle ofmass mmoves inapotential well given by —Vo112(<12 +$2)_ 8a4+x4 (a)Sketch V(a:) andF(:r). (b)Discuss themotions which may occur. Locate allequilibrium points and determine thefrequency ofsmall oscillations about anythat arestable. (c)Aparticle starts atagreat distance from thepotential well with velocity votoward thewell. Asitpasses thepoint :0=a,itsuffers a collision with another particle, during which itloses afraction aofitskinetic energy. How large must ozbeinorder that theparticle thereafter remain trapped inthewell? How large must ozbeinorder that theparticle betrapped inone side ofthewell? Find theturning points ofthenew motion ifoz=1. 22.Starting with e2"=(e“)2, obtain formulas forsin20,cos20interms of sin0,cos0. 23.Find thegeneral solutions oftheequations: (a) m:ii+ bi:—kx=0, (b) mat—b:i:+ kx=O. Discuss thephysical interpretation ofthese equations andtheir solutions, assum- ingthat they aretheequations ofmotion ofaparticle.V(a:) = 66 MOTION orAPARTICLE INONEDIMENSION [CHAP. 2 24.Show that when oﬁ—’Y2isvery small, theunderdamped solution (2—133) isapproximately equal tothecritically damped solution (2—146), forashort time interval. What istherelation between theconstants C1,C2andA,0?This result suggests how onemight discover theadditional solution (2—143) inthe critical case. 25.Amass msubject toalinear restoring force -—kx anddamping —bat isdis- placed adistance sofrom equilibrium and released with zero initial velocity. Find themotion intheunderdamped, critically damped, andoverdamped cases. 26.Solve Problem 25forthecase when themass starts from itsequilibrium position with aninitial velocity v0.Sketch themotion forthethree cases. 27.Amass of1000 kgm drops from aheight of10monaplatform ofnegligible mass. Itisdesired todesign aspring anddashpot onwhich tomount theplat- form sothat theplatform willsettle toanew equilibrium position 0.2mbelow itsoriginal position asquickly aspossible after theimpact without overshooting. (a)Find thespring constant loand thedamping constant bofthedashpot. Why does theresult seem tocontradict theremarks attheendofSection 2-9? (b)Find, totwosigniﬁcant ﬁgures, thetime required fortheplatform tosettle within 1mmofitsﬁnal position. - 28.Aforce F0(1 -—e-"‘)' acts onaharmonic oscillator which isatrest at t=0.The mass ism,thespring constant lc=4ma2, andb=ma. Find the motion. Sketch x(t). *29. Solve Problem 28forthecase k=maz, b=2ma. 30.Aforce F0cos(wt-1-00)actsonadamped harmonic oscillator beginning att=0.(a)What must betheinitial values of:1:andvinorder thatthere be notransient‘? (b)If2:0=vo=0,ﬁndtheamplitude Aandphase 0ofthe transient interms ofF0,00. 31._Anundamped harmonic oscillator ofmass 1n,natural frequency wo,is initially atrestandissubject att=0toablow sothat itstarts from :00=0 with initial velocity coandoscillates freely until t=31r/24.00. From thistime on,aforce F=Bcos(wt+0)isapplied. Find themotion. 32.Anunderdamped harmonic oscillator issubject toanapplied force F=F0e_“' cos(wt—I—0). Find aparticular solution byexpressing Fastherealpart ofacomplex exponen- tialfunction andlooking forasolution for2:having thesame exponential time dependence. 33.(a)Find themotion ofadamped harmonic oscillator subject toaconstant applied force F0,byguessing a“steady-state” solution oftheinhomogeneous equation (2-86) andadding asolution ofthehomogeneous equation. 1(b)Solve thesame problem bymaking thesubstitution 2:’=ac—a,and choosing the constant asoastoreduce theequation in1;’tothehomogeneous equation (2-85). Hence show that theeffect oftheapplication ofaconstant force ismerely toshift theequilibrium position without affecting thenature oftheoscillations. >1‘Anasterisk isused, asexplained inthePreface, toindicate problems which may beparticularly diﬁicult. , ' PROBLEMS 67 34.Find themotion ofamass msubject toarestoring force —Ica:, andtoa damping force (:l:)/nng duetodrysliding friction. Show thattheoscillations are isochronous (period independent ofamplitude) with theamplitude ofoscillation decreasing by2/Jg/wg during each half-cycle until themass comes toastop. [Hint: Usetheresult ofProblem 33. When theforce hasadifferent algebraic form atdifferent times during themotion, ashere, where thesignofthedamping force must bechosen sothat theforce isalways opposed tothevelocity, itis necessary tosolve theequation ofmotion separately foreach interval oftime during which aparticular expression fortheforce istobeused, andtochoose asinitial conditions foreach time interval theﬁnal position andvelocity ofthe preceding time interval.] 35.Anundamped harmonic oscillator (7=0),initially atrest,issubject to aforce given byEq.(2~191). (a)Find x(t). (b)Foraﬁxed pg,forwhat value of6tistheﬁnal amplitude ofoscillation greatest? (c)Show that as6t—>0,your solution approaches that given byEq.(2—190). 36.Find thesolution analogous toEq.(2—190) foracritically damped har- monic oscillator subject toanimpulse pgdelivered att=to. 37.(a)Find, using theprinciple ofsuperposition, themotion ofanunder- damped oscillator ['Y=(1/3)w0] initially atrest and subject, after t==0, toaforce F=Asinwot+Bsin3w0t, where woisthenatural frequency oftheoscillator. (b)What ratio ofBtoAis required inorder fortheforced oscillation atfrequency 3wotohave thesame amplitude asthatatfrequency wo? ' 38.Find, bytheFourier-series method, thesteady-state solution forthe damped harmonic oscillator subject toaforce Fa)=0, ifnT<¢5(n+-Dr, Fo, if(1l+ %)T<:3(H-l-1)T, where nisanyinteger, andT=61r/wo, where (.00istheresonance frequency ofthe oscillator. Show thatif‘Y<<wo,themotion isnearly sinusoidal withperiod T/3. 39.Anunderdamped oscillator initially atrest isacted upon, beginning at t=0,byaforce . . F=F06_“. p Find itsmotion byusing Green’s solution (2—210). 40.Using theresult ofProblem 36,ﬁnd byGreen’s method themotion ofa critically damped oscillator initially atrestandsubject toaforce F(t). CHAPTER 3 MOTION OFAPARTICLE INTWO OR THREE DIMENSIONS 3-1Vector algebra. Thediscussion ofmotion intwoorthree dimensions isvastly simpliﬁed bytheintroduction oftheconcept ofavector. A vector isdeﬁned geometrically asaphysical quantity characterized bya magnitude andadirection inspace. Examples arevelocity, force, and position with respect toaﬁxed origin. Schematically, werepresent a vector byanarrow whose length anddirection represent themagnitude anddirection ofthevector. Weshall represent avector byaletter inbold- facetype. Thesame letter inordinary italics willrepresent themagnitude ofthevector. (See Fig. 3-1.) The magnitude ofavector may alsobe represented byvertical bars enclosing thevector symbol: A= (3-1) Two vectors areequal ifthey have thesame magnitude and direction; theconcept ofvector itself makes noreference toanyparticular location.* if FIG. 3-1. Avector Aanditsmagnitude A.(c>0) /\ FIG. 3-2. Deﬁnition ofmultiplication ofavector byascalar. (c>0) *Adistinction issometimes made between “free” vectors, which have nopar- ticular location inspace; “sliding” vectors, which may belocated anywhere along aline; and“ﬁxed” vectors, which must belocated atadeﬁnite point inspace. We prefer here toregard thevector asdistinguished byitsmagnitude anddirection alone, sothattwovectors mayberegarded asequal iftheyhave thesame magni- tudes anddirections, regardless oftheir positions inspace. 68 3-1] vncron ALGEBRA 69 Aquantity represented byanordinary (positive ornegative) number is often called ascalar, todistinguish itfrom avector. Wedeﬁne aproduct ofavector Aandapositive scalar casavector cAinthesame direction as Aofmagnitude cA.Ifcisnegative, wedeﬁne cAashaving themagnitude |c|Aandadirection opposite toA.(See Fig. 3-2.) Itfollows from this deﬁnition that ICAI=lcllAl- (3-2) Itisalsoreadily shown, onthebasis ofthisdeﬁnition, that multiplication byascalar isassociative inthefollowing sense: 3 '(cd)A=c(dA). (3-3) Itissometimes convenient tobeabletowrite thescalar totheright ofthe vector, andwedeﬁne Acasmeaning thesame vector ascA: Ac=cA. (3-4) AWedeﬁne thesum (A+B)oftwovectors AandBasthevector which extends from thetailofAtothetipofBwhen Aisdrawn with itstip atthetaﬂofB,asinFig.3-3. This deﬁnition isequivalent totheusual parallelogram rule, andismore convenient touse. Itisreadily extended tothesumofanynumber ofvectors, asinFig.3-4. Onthebasis ofthedeﬁnition given inFig.3-3,wecanreadily prove that vector addition iscommutative andassociative: A+B=B+A, (3-5) (A+B)+C=A+(B+c). (3-6) According toEq.(3-6), Wemay omit parentheses inwriting avector sum, since theorder ofadding does notmatter. From thedeﬁnitions given by Figs. 3-2and3-3, Wecanalsoprove thefollowing distributive laws: c(A+B)=cA+cB, (3-7) (c-}—d)A=cA+dA. (3-8) These statements canbeproved bydrawing diagrams representing the > B A+B \ B Ai D A A+B+C+D Fro. 3-3. Deﬁnition ofaddition of Fro. 3-4. Addition ofseveral vec- twovectors. tors. 70 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS ICHAP. 3 B. (A+B)+CorA+(B+C) FIG. 3-5. Proof ofEq.(3-6). right andleftmembers ofeach equation according tothedeﬁnitions given. Forexample, thediagram inFig. 3-5makes itevident that theresult of adding Cto(A+B)isthesame astheresult ofadding (B-1-C)toA. According toEqs. (3-3) through (3-8), thesum andproduct wehave deﬁned have most ofthealgebraic properties ofsums and products of ordinary numbers. This isthejustiﬁcation forcalling them sums and products. Thus itisunnecessary tocommit these results tomemory. Weneed only remember that wecanmanipulate these sums andproducts justaswemanipulate numbers inordinary algebra with theoneexception that theproduct deﬁned byFig.3-2canbeformed only between ascalar andavector, andtheresult isavector. Avector mayberepresented algebraically interms ofitscomponents or projections along asetofcoordinate axes. Drop perpendiculars from the tailandtipofthevector onto thecoordinate axes asinFig. 3-6. Then thecomponent ofthevector along anyaxisisdeﬁned asthelength ofthe segment cutoffontheaxis bythese perpendiculars. The component is taken aspositive ornegative according towhether theprojection ofthe tipofthevector liesinthepositive ornegative direction along theaxis Z QA M 3/ I 3; z (=1) (b) Fro. 3-6. (a)Components ofavector inaplane. (b)Components ofavector inspace. 3-1] VECTOR ALGEBRA 71 2/ AI” A Ari Axi Ix -MA. _ Z l FIG. 3-7. Diagrammatic proof oftheformula A=A,i—|—A,,j. from theprojection ofthetail. Thecomponents ofavector Aalong x-,y-, andz-axes willbewritten A,,,A,,,andAg» Thenotation (Ax, A1,,-Ag)will sometimes beused torepresent thevector A: A=(A¢, A,,,Ag). (3-9) Ifwedeﬁne vectors i,j,kofunitlength along the:c-,y-,z-axes respectively, then wecanwrite any vector asasum ofproducts ofitscomponents with i,j,k: A=A,i+A,,i+A,k. (3-10) Thecorrectness ofthisformula canbemade evident bydrawing adia- gram inwhich thethree vectors ontheright, which areparallel tothe three axes, areadded togive A.Figure 3-7shows thisconstruction for thetwo-dimensional case. . Wenow have twoequivalent ways ofdeﬁning avector: geometrically asquantity with amagnitude anddirection inspace, oralgebraically asa setofthree numbers (Ax,A1,,A,), which wecallitscomponents.* The operations ofaddition andmultiplication byascalar, which aredeﬁned geometrically inFigs. 3-2and3-3interms ofthelengths anddirections of thevectors involved, canalsobedeﬁned algebraically asoperations onthe components ofthevectors. Thus cAisthevector whose components are thecomponents ofA,each multiplied byc: cA=(cA,,, cA,,, 0A,), (3-11) *These twoways ofdeﬁning avector arenotquite equivalent asgiven here, forthealgebraic deﬁnition requires that acoordinate system besetup,whereas thegeometric deﬁnition does notrefer toanyparticular setofaxes. This ﬂaw can beremedied bymaking the algebraic deﬁnition also independent ofany particular setofaxes. This isdone bystudying how thecomponents change when theaxes arechanged, anddeﬁning avector algebraically asasetofthree quantities which transform inacertain way when theaxes arechanged. This reﬁnement willnotconcern usinthischapter. ¢72 MOTION orPARTICLE INTWO onTHREE DIMENSIONS [cn.u>. 3 y W A+B <A+B>.. B” B _ M A <iAZ-?><—B;;—> 1(A+s).—-j FIG. 3-8. Proof ofequivalence ofalgebraic and geometric deﬁnitions of vector addition. andA+Bisthevector whose components areobtained byadding the components ofAandB: A+B=(A,+B,,,AZ,+Bu,A,-I—B3). (3-12) Theequivalence ofthedeﬁnitions (3-11) and(3-12) tothecorresponding geometrical deﬁnitions canbedemonstrated bydrawing suitable diagrams. Figure 3-8constitutes aproof ofEq.(3-12) forthetwo-dimensional case. Allvectors aredrawn inFig. 3-8sothat their components arepositive; foracomplete proof, similar diagrams should bedrawn forthecases where oneorboth components ofeither vector arenegative.‘ The length of avector canbedeﬁned algebraically asfollows: IAI-(AZ+AZ+A2)“. <3-13> where thepositive square root istobetaken. Wecannow give algebraic proofs ofEqs. (3-2), (3-3), (3-5), (3-6), (3-7), and (3-8), based onthedeﬁnitions (3-11), (3-12), and (3-13). Forexample, toprove Eq.(3-7), weshow that each component oftheleft sideagrees with each component ontheright. Fortheac-component, the proof runs: [¢(A+B)]@=c(A+13).. [byEq-(3-11)] =0A,+cB,, =(¢A)= +(63).. [byEq-(3-11)] =(cA+cB),. [byEq.(3-12)] 3-1] vnoron ALGEBRA 73 -B A A—B AA—B B B FIG. 3-9. Two methods ofsubtraction ofvectors. Since allcomponents aretreated alike inthedeﬁnitions (3-11), (3-12), (3-13), thesame proof holds forthey-and2-components, andhence the vectors ontheleftandright sides ofEq.(3-7)areequal. Inview oftheequivalence ofthegeometrical andalgebraic deﬁnitions ofthevector operations, itisunnecessary, forgeometrical applications, to give both analgebraic andageometric proof ofeach formula ofvector algebra. Either ageometric oranalgebraic proof, whichever iseasiest, willsuﬁice. However, there areimportant cases inphysics where wehave toconsider sets ofquantities which behave algebraically likethecom- ponents ofvectors although they cannot beinterpreted geometrically as quantities with amagnitude anddirection inordinary space. Inorder that Wemay apply therules ofvector algebra insuch applications, itis important toknow thatallofthese rules canbeproved purely algebraically from thealgebraic deﬁnitions ofthevector operations. Thegeometric approach hastheadvantage ofenabling ustovisualize themeanings of thevarious vector notations andformulas. The algebraic approach sim- pliﬁes certain proofs, andhasthefurther advantage that itmakes possible wide applications ofthemathematical concept ofvector, including many cases where theordinary geometric meaning isnolonger retained. Wemay deﬁne subtraction ofvectors interms ofaddition andmulti- plication by—1: A—B=A+(~B) =(A,—B,,A,—-By,A,—Bz)- (3-14) The difference A-Bmay befound geometrically according toeither of thetwoschemes shown inFig.3-9. Subtraction ofvectors may beshown tohave allthealgebraic properties tobeexpected byanalogy with sub- traction ofnumbers. Itisuseful todeﬁne ascalar product (A-B) oftwovectors AandBas theproduct oftheir magnitudes times thecosine oftheangle between them (Fig. 3-10): - . A-B =ABcos0. (3-15) The scalar product isascalar ornumber. Itisalsocalled thedotproduct orinner product, andcanalsobedeﬁned astheproduct ofthemagnitude of either vector times theprojection oftheother along it.Anexample ofi 4 l 1 I 4 1 1 74 MOTION orPARTICLE INTWO onTHREE DIMENSIONS [cnx1>. 3 itsuseistheexpression forthework done when aforce Facts through a distance snotnecessarily parallel toit: W=Fscos0=.F-s. Weareentitled tocallA-B aproduct because ithasthefollowing alge- braic properties which areeasily proved from thegeometrical deﬁnition (3-15): B . A(cA)-B -1A-(cB) =c(A-B), A-(B —|-C)=A-B +A-C, Fro.3-10. Angle between twovectors.(3-16) (3-17) (3-18) (3-19)A-B=B-A, A-A=A2 These equations mean that wecantreat thedotproduct algebraically like aproduct inthealgebra ofordinary numbers, provided wekeep inmind that thetwofactors must bevectors andtheresulting product isascalar. The following statements arealso consequences ofthedeﬁnition (3-15), where i,j,andkaretheunit vectors along thethree coordinate axes: i-i=j-j=k-k=1,__ (3-20)i-1=1-1:=1;-1=0. A-B =AB, when Aisparallel toB, (3-21) A-B =0, when Aisperpendicular toB. (3-22) Notice that, according toEq.(3-22), thedotproduct oftwovectors is zeroifthey areperpendicular, even though neither vector isofzerolength. Thedotproduct canalsobedeﬁned algebraically interms ofcomponents: A-B =A,,B,, -1-A,,B,, +A,B,. (3-23) Toprove that Eq.(3-23) isequivalent tothegeometric deﬁnition (3-15), wewrite AandBintheform given byEq.(3-10), andmake useofEqs. (3-16), (3-17), (3-18), and(3-20), which follow from Eq.(3-15): A'B :7‘ + +kAz)'(i-Bx +jBg + -(i-i>A.B. +(i-i)A.B.. +(i-k)A.B. +(i-i)A..B. +i-iA..B.. +j-kA,,B, +1.-1.4.3, +1;-3.4.3,, +k-kA,B, =A,,B,,+A,,B,,+A,B,. This proves Eq.(3-23). Theproperties (3-16) to(3-20) canallbeproved 3-1] vncron ALGEBRA 75 AxB shaded area=[AxBl B.-..__.-11555555555251555555E515=5EE1E=f=5=E¢E=Z=E5E1;:¢:3!-ﬁrm-.,_ ...:. : -=-=.;=;:z5=51E=£=2:i====5=§:s=E=£=£:;==,:=¢=:;::=5rE=§=;>= ”'A t FIG. 3-11. Deﬁnition ofvector product. readily from thealgebraic deﬁnition (3-23) aswell asfrom thegeometric deﬁnition (3-15). Wecanregard Eqs. (3-21) and (3-22) asalgebraic deﬁnitions ofparallel andperpendicular. Another product convenient todeﬁne isthevector product, also called thecross product orouter product. The cross product (AXB)oftwo vectors AandBisdeﬁned asavector perpendicular totheplane ofAand Bwhose magnitude isthearea oftheparallelogram having AandBas sides. The sense ordirection of(AXB)isdeﬁned asthedirection of advance ofaright-hand screw rotated from Atoward B.(See Fig.3-11.) The length of(AXB),interms oftheangle 0between thetwovectors, isgivenby IA><Bl-ABsino (324) Note that thescalar product oftwovectors isascalar ornumber, while thevector product isanewvector. Thevector product hasthefollowing algebraic properties which canbeproved from thedeﬁnition given in Fig.3-11:* AxB=—BxA, (3-25) (cA) XB=AX(cB) =c(AXB), (3-26) AX(B+C)=(AXB)-I—(A XAC), (3-27) AxA=0, (3-28) 1AXB=0, when Aisparallel toB, (3-29) IAXB|=AB, when Aisperpendicular toB, (3-30) ixi=jxj=kxk=0, iXj=k, jXk=i, kXi=j. (3-31) *Here 0stands forthevector ofzero length, sometimes called thenullvector. Ithasnoparticular direction inspace. Ithastheproperties: A—|—0--A, A-0=O, AXO=O, A——A=0,” 0=(0,0,0). " 76 MOTION orPARTICLE INTWO onTHREE DIMENSIONS [CHAP. 3 Hence thecross product canbetreated algebraically like anordinary product with theexception thattheorder ofmultiplication must notbe changed, andprovided wekeep inmind thatthetwofactors must bevectors andtheresult isavector. Switching theorder offactors inacross product changes thesign. This istheﬁrst unexpected deviation oftherules of vector algebra from those ofordinary algebra. The reader should there- fore memorize Eq.(3-25). Equations (3-29) and (3-30), aswell asthe analogous Eqs. (3-21) and(3-22), arealsoworth remembering. (Itgoes without saying that allgeometrical and algebraic deﬁnitions should be memorized.) Inarepeated vector product like(AXB)X(CXD),the parentheses cannot beomitted orrearranged, fortheresult ofcarrying out themultiplications inadifferent order isnot, ingeneral, thesame. [See, forexample, Eqs. (3-35) and(3-36).] Notice that according toEq.(3-29) thecross product oftwovectors may benullWithout either vector being thenullvector. From Eqs. (3-25) to(3-31), using Eq.(3-10) torepresent AandB,We canprove that thegeometric deﬁnition (Fig. 3-11) isequivalent tothe following algebraic deﬁnition ofthecross product: AXB=(A,,B, —A,B,,, A,,B,, -A,,B,, A,,B,, —A,,B@). (3-32) Wecanalsowrite AXBasadeterminant: 1jk A><B= A.A.A.. (3-33) B,B,,B, Expansion oftheright sideofEq.(3-33) according totheordinary rules fordeterminants yields Eq.(3-32). Again theproperties (3-25) to(3-31) follow alsofrom thealgebraic deﬁnition (3-32). Thefollowing useful identities canbeproved: A-(B XC)=(AXB)-C, (3-34) AX(BXC)=B(A-C) —C(A-B), (3-35) (AxB)xC=B(A-C) —A(B-C), (3-36) i-(jXk)=1. (3-37) Theﬁrstthree ofthese should becommitted tomemory. Equation (3-34) allows ustointerchange dotandcross inthescalar triple product. The quantity A-(B XC)canbeshown tobethevolume oftheparallelepiped whose edges areA,B,C,with positive ornegative sign depending on whether A,B,Careinthesame relative orientation asi,j,k,that is, depending onwhether aright-hand screw rotated from Atoward Bwould 3-2] APPLICATIONS ToAsETorFORCES ACTING ONAPARTICLE 77 advance along Cinthepositive ornegative direction. The triple vector product formulas (3-35) and(3-36) areeasy toremember ifwenote that thepositive term ontheright ineach case isthemiddle vector (B)times thescalar product (A-C) oftheother two, while thenegative term isthe other vector within theparentheses times thescalar product oftheother two. Asanexample oftheuseofthevector product, therulefortheforce exerted byamagnetic ﬁeldofinduction Bonamoving electric charge q (esu) canbeexpressed as F=gvXB,c Where cisthespeed oflight andvisthevelocity ofthecharge. This equation gives correctly both themagnitude anddirection oftheforce. Thereader willremember that thesubject ofelectricity andmagnetism is fullofright- andleft-hand rules. Vector quantities whose directions are determined byright- orleft—hand rules generally turn outtobeexpressible ascross products. 3-2Applications toasetofforces acting onaparticle. According tothe principles setdown inSection 1-3,ifasetofforces F1,F2,...,F,,actona particle, thetotal force F,which determines itsacceleration, istobeob- tained bytaking thevector sum oftheforces F1,F2,...,F,,: F=F1+F2+-~+F..- (3—33) Theforces F1,F2,...,F,,areoften referred toascomponent forces, andF iscalled their resultant. The term component ishere used inamore gen- eralsense than inthepreceding section, where thecomponents ofavector were deﬁned astheprojections ofthevector onasetofcoordinate axes. When component ismeant inthissense asoneofasetofvectors whose smn isF,weshall usetheterm (vector) component. Ingeneral, unless other- wise indicated, theterm component ofavector Finacertain direction will mean theperpendicular projection ofthevector Fonalineinthat direc- tion. Insymbols, thecomponent ofFinthedirection oftheunitvector n ls F,,=n-F. (3-39) Inthissense, thecomponent ofFisnotavector, butanumber. Thecom- ponents ofFalong theat-,y-,and2-axes arethecomponents inthesense of Eq.(3-39) inthedirections i,j,andk. Iftheforces F1,F2,...,F,,aregiven, thesum may bedetermined graphically bydrawing acareful scale diagram according tothedeﬁnition ofFig. 3-3or3-4. The sum may also bedetermined analytically by 78 MOTION orPARTICLE INTwo onTHREE DIMENSIONS [CHAP. 3 F2 0 7 F1 F ) Fro. 3-12. Sum oftwoforces. drawing arough sketch ofthesum diagram andusing trigonometry to calculate themagnitude anddirection ofthevector F.If,forexample, twovectors aretobeadded, thesumcanbefound byusing thecosine and sinelaws. InFig.3-12, F1,F2,and0aregiven, and themagnitude and direction ofthesum Farecalculated from F2=F?+F3-2F1F2 cos0, (3-40) F1_F2_F_ sin,8—sina—sin0 (3-41) Note that theﬁrstofthese equations canbeobtained bysquaring, inthe sense ofthedotproduct, theequation u F=F1+F2. (3-42) Taking thedotproduct ofeach member ofthisequation with itself, we obtain F-F=F2=F1-F1 +2F1-F2 +F2~F2 =F?+F3-2F1F2 cos0. (Note that 0inFig.3-12 isthesupplement oftheangle between F1andF2 asdeﬁned byFig.3-10.) This technique canbeapplied toobtain directly themagnitude ofthesum ofany number ofvectors interms oftheir lengths and theangles between them. Simply square Eq. (3-38), and split uptheright sideaccording tothelaws ofvector algebra intoasumof squares anddotproducts ofthecomponent forces. The angle between F andanyofthecomponent forces canbefound bycrossing ordotting the component vector into Eq.(3-38). Forexample, inthecase ofasum of twoforces, wecross F1intoEq.(3-42): F1XF=F1XF1-I-F1XF2. Wetake themagnitude ofeach side, using Eqs. (3-28) and(3-24): . . F F2 F1FS11'lC!—F1F2SlIl6, OI‘ 3-2] APPLICATIONS ToASETorFORCES ACTING ONAPARTICLE 79 When asum ofmore than twovectors isinvolved, itisusually simpler to take thedotproduct ofthecomponent vector with each sideofEq.(3-38). Thevector suminEq.(3-38) canalsobeobtained byadding separately thecomponents ofF1,...,F,,along anyconvenient setofaxes: Fx=F1z+F2z+"'+Fnz; F1/=F1u+F2u+"°+Fnw (343) Fz=F1z+F2z+"'+Fnz- When asum ofalarge number ofvectors istobefound, thisislikely tobe thequickest method. The reader should usehisingenuity incombining andmodifying these methods tosuittheproblem athand. Obviously, if asetofvectors istobeadded which contains agroup ofparallel vectors, itwillbesimpler toaddthese parallel vectors first before trying toapply themethods ofthepreceding paragraph. II P (X I‘ I IO FIG. 3-13. Force Facting atpoint P. Just asthevarious forces acting onaparticle aretobeadded vectorially togive thetotal force, so,conversely, thetotal force, oranyindividual force, acting onaparticle may beresolved inanyconvenient manner into asum of(vector) component forces which may beconsidered asacting individually ontheparticle. Thus intheproblem discussed inSection 1-7 (Fig. 1-4), thereaction force Fexerted bytheplane onthebrick isresolved into anormal component Nandafrictional component f.The effect of theforce Fonthemotion ofthebrick isthesame asthat oftheforces N andfacting together. Ifitisdesired toresolve aforce Finto asum of (vector) component forces intwoorthree perpendicular directions, this canbedone bytaking theperpendicular projections ofFinthese directions, asinFig.3-6. Themagnitudes ofthevector components ofF,along aset ofperpendicular directions, arejusttheordinary components ofFinthese directions inthesense ofEq. (3-39). Ifaforce Finthexy-plane acts onaparticle atthepoint P,wedeﬁne thetorque, ormoment oftheforce Fabout theorigin O(Fig. 3-13) asthe product ofthedistance O7andthecomponent ofFperpendicular tor: N0=rFsinoz. (3-44) 80 MOTION orPARTICLE INTwo ORTHREE DIMENSIONS [CHAP. 3 Themoment N0oftheforce Fabout thepoint Oisdeﬁned aspositive when Facts inacounterclockwise direction about OasinFig. 3-13, andnega- tivewhen Factsinaclockwise direction. Wecandeﬁne inasimilar way themoment about Oofanyvector quantity located atthepoint P.The concept ofmoment willbefound useful inourstudy ofthemechanics of particles and rigid bodies. The geometrical andalgebraic properties of torques willbestudied indetail inChapter 5.Notice that torque canbe deﬁned interms ofthevector product: N0=;1=]r XFl, (3-45) where the+or-signisused according towhether thevector rXFpoints inthepositive ornegative direction along thez-axis. 1A F1F l Fl P0 r0. B FIG. 3-14. Moment ofaforce about anaxisinspace. Wecangeneralize theabove deﬁnition oftorque tothethree-dimensional case bydeﬁning thetorque ormoment ofaforce F,acting atapoint P, about anaxisAB(Fig. 3-14). Letnbeaunit vector inthedirection of AB, andletFberesolved intovector components parallel andperpendicu- lartoAB: F=F||—|—F_|_, (3-46) where F||=11(I1'F), (3-47) F_1_=F—F||. Wenowdeﬁne themoment ofFabout theaxisABasthemoment, deﬁned byEq.(3-44) or(3-45), oftheforce F1,inaplane through thepoint P 3-3] DIFFERENTIATION AND INTEGRATION orVECTORS 81 perpendicular toAB,about thepoint Oatwhich theaxisABpasses through thisplane: NAB =:l:r‘F1_Sina =i|!' XF1], (3-48) where the+or—signisused, depending onwhether rxF1isinthesame oropposite __di_rection ton.According tothis(Enition, aforce likeF|| parallel toABhasnotorque ormoment about AB. Since rXF||isper- pendicular ton, n-(r><F)-11-[r><(F||+F1)l, =n-(r XF||) +n-(r XF_|_) =n-(r XF1) =:l:j1' XFJ_j. Hence wecandeﬁne N,11;inaneater way asfollows: NAB =n-(r XF). (3-49) This deﬁnition automatically includes theproper sign, anddoes notrequire aresolution ofFintoF||andF1. Furthermore, rcannow bedrawn toP from anypoint ontheaxisAB,since acomponent ofrparallel toE,like acomponent ofFparallel toE,gives acomponent inthecross product perpendicular tonwhich disappears from thedotproduct. Equation (3-49) suggests thedeﬁnition ofavector torque orvector moment, about apoint O,ofaforce Facting atapoint P,asfollows: N0=rXF, (3-50) where risthevector from OtoP.Thevector torque N0has, according toEq.(3-49), theproperty that itscomponent inany direction isthe torque, intheprevious sense, oftheforce Fabout anaxisthrough 0inthat direction. Hereafter theterm torque willusually mean thevector torque deﬁned byEq.(3-50). Torque about anaxis Eintheprevious sense willbecalled thecomponent oftorque along AB. Wecandeﬁne the vector moment ofanyvector located atapoint P,about apoint O,byan equation analogous toEq.(3-50). 3-3Differentiation andintegration ofvectors. Avector Amay bea function ofa.scalar quantity, sayt,inthesense that with each value ofta certain vector A(t)isassociated, oralgebraically inthesense that itscom- ponents may befunctions oft: A=A(t)=lAa:(t)2A11(t)> Az(t)l- (3-51)l 1 l 82 MOTION orPARTICLE INTwo ORTHREE DIMENSIONS [CHAP. 3 The most common example isthat ofavector function ofthetime; for example, thevelocity ofamoving particle isafunction ofthetime: v(t). Other cases alsooccur, however; forexample, inEq.(3-76), thevector n isafunction oftheangle 0.Wemay deﬁne thederivative ofthevector A with respect totinanalogy with theusual deﬁnition ofthederivative of ascalar function (seeFig.3-15): dA .A(t+At)—A(t)_=1mm. 3-52 dt Alir>l0 At ( ) (Division byAthere means multiplication by1/At.) Wemay alsodeﬁne thevector derivative algebraically interms ofitscomponents: dA_ 31.4,,01.4,dA.>_.dA,, .dA,, dA,_ ,_ dt—<dt’dt’dt —1dt +171: Tkdz (553) Asanexample, ifv(t)isthevector velocity ofaparticle, itsvector accelera- tionais a=dv/dl. Examples ofthecalculation ofvector derivatives based oneither deﬁnition (3-52) or(3-53) willbegiven inSections 3-4and3-5. Thefollowing properties ofvector differentiation canbeproved by straightforward calculation from thealgebraic deﬁnition (3-53), orthey may beproved from thedeﬁnition (3-52) inthesame Way theanalogous properties areproved fordifferentiation ofascalar function: d;;,<A+B)=§+";’,§, <3-54) %(fA) =%)£A+f%» (3-55) d dA dBZﬁ(A-B) _dtB+A-W, (3-55) d dA dBa?(AxB)-WxB+AxEt-- (3-57) These results imply that diﬂerentiation ofvector sums and products obeys thesame algebraic rules asdifferentiation ofsums andproducts inordinary calculus, except, however, that theorder of(factors inthecross product must notbechanged [Eq. (3-57)]. Toprove Eq.(3-55), forexample, from thedeﬁnition (3-53), wesimply show bydirect calculation that the corresponding components onboth sides oftheequation areequal, making useofthedeﬁnitions andproperties ofthevector operations introduced 3-3] DIFFERENTIATION ANDINTEGRATION orVECTORS 83 inthepreceding section. Forthe:1:-component, theproof runs: [531<r4>l=%(14). [byEq.<3-53>: =$04.) [byE4<3-11>; _Q. Q2 [standard ruleofordi- _dtA”+fdt nary calculus: =ill’4.+r [byEq.<3-53>: ’‘($3-.I>=- [byEq-<3-11>: =g4+ -[byEq.<3-12>: Asanother example, toprove Eq.(3-56) from thedeﬁnition (3-52), we proceed asintheproof ofthecorresponding theorem forproducts ofordi- nary scalar functions. Weshall usethesymbol Atostand fortheincrement inthevalues ofanyfunction between tandt—l—At;theincrement AAofa vector Aisdeﬁned inFig. 3-15. Using thisdeﬁnition ofA,andtherules ofvector algebra given inthepreceding section, wehave A(A-B) =(A—l-AA)-(B +AB) -A-B At At =(44)-B+A-(AB)+<44)-(AB) At (AA)-B A-(AB) (AA)-(AB) At + At + At _AA AB (AA)-(AB) _ _ —EB -l"A'—AT +—"“Ft (358) When At—>0,theleftsideofEq.(3-58) approaches theleftsideofEq. ‘ (3-56), andtheﬁrsttwoterms ontheright sideofEq.(3-58) approach the A(t+M) AA A(t) Fro. 3-15. Vector increment./LA =A(t—l—At)—A(t).1 i 1 1 < 4 1 84 MOTION orPARTICLE INTWO ORTHREE DIMENSIONS [CHAP. 3 % I Z —y av 1/ .47 FIG. 3-16. The position vector rofthepoint (z,y,z). twoterms ontheright ofEq.(3-5-6), wl1ile thelastterm ontheright of Eq.(3—58) vanishes. The rigorous justiﬁcation ofthis limit process is exactly similar tothejustiﬁcation required forthecorresponding process inordinary calculus. Intreating motions inthree-dimensional space, weoften meet scalar andvector quantities wl1ich have adeﬁnite value atevery point inspace. Such quantities arefunctions ofthespace coordinates, commonly ac,y, andz.They may alsobethought ofasfunctions oftheposition vector r from theorigin tothepoint ac,y,z(Fig. 3-16). Wethus distinguish scalar point functions 7/'(r) =7/‘(xi yaz); andvector point functions A(t) =A(x; yrZ)= yrZ)!AU(x: 3/:Z):A-¢(xr yr Anexample ofascalar point function isthepotential energy V(x, y,z)ofa particle moving inthree dimensions. Anexample ofavector point func- tionistheelectric ﬁeld intensity E(x, y,z).Scalar andvector point func- tions areoften functions ofthetime taswellasofthepoint x,y,2inspace. IfWearegiven acurve Cinspace, andaVector function Adeﬁned at points along thiscurve, wemay consider thelineintegral ofAalong C’: /0A-dr. Todeﬁne theline integral, imagine thecurve Cdivided into small seg- ments, andletanysegment berepresented byavector drinthedirection ofthesegment andoflength equal tothelength ofthesegment. Then thecurve consists ofthesuccessive vectors drlaidendtoend. Now for each segment, form theproduct A-dr, where Aisthevalue ofthevector function attheposition ofthat segment. Thelineintegral above isdeﬁned asthelimit ofthesums oftheproducts A-dr asthenumber ofsegments 3-3] DIFFERENTIATION AND INTEGRATION OF VECTORS 85 increases without limit, while thelength ldrlofevery segment approaches zero. Asanexample, thework done byaforce F,which mayvary from point topoint, onaparticle which moves along acurve C’is W=Lm@ which isageneralization, tothecase ofavarying force andanarbitrary curve C,oftheformula _ W=F-s, foraconstant force acting onabody moving along astraight lineseg- ment s.Thereason forusing thesymbol drtorepresent asegment ofthe curve isthat ifristheposition vector from theorigin toapoint onthe curve, then dristheincrement inr(seeFig. 3-15) from oneendtothe other ofthecorresponding segment. Ifwewrite rintheform r=ia:+jy+kz, (3—59) then dr=idx+jdy+kdz, (3—60) where dx,dy,dzarethedifferences inthecoordinates ofthetwoendsofthe segment. Ifsisthedistance measured along thecurve from some ﬁxed point, wemay express thelineintegral asanordinary integral over the coordinate s: Lkﬁ=IAwMM, own where 0istheangle between Aandthetangent tothecurve ateach point. (See Fig. 3-17.) This formula may beused toevaluate theintegral if weknow Aandcos0asfunctions ofs.Wemay alsowrite theintegral, A odr ’\C /ii “ FIG. 3-17. Elements involved inthelineintegral. 86 MOTION orPARTICLE INTwo ORTHREE DIMENSIONS [cn».1>. 3 using Eq.(3-60), as /CA-dr =/0(A,dx +Audy+A,dz). (3-62) One ofthemost convenient ways torepresent acurve inspace istogive thethree coordinates (x,y,z)or,equivalently, theposition vector r,as functions ofaparameter swhich hasadeﬁnite value assigned toeach point ofthecurve. “The parameter sisoften, though notnecessarily, the distance measured along thecurve from some reference point, asinFig. 3-17 andinEq.(3-61). The parameter smay also bethetime atwhich amoving particle arrives atanygiven point onthecurve. Ifweknow A(r)andr(s),then thelineintegral canbeevaluated from theformula/.-/<e>d. -Qn—/(A”ds+A”.ds+A‘ds d8" The right member ofthisequation isanordinary integral over thevari- able s.1/ /dr 1 8\x w FIGURE 3-18(3-63) Asanexample ofthecalculation ofalineintegral, letuscompute the work done onaparticle moving inasemicircle ofradius aabout theorigin inthexy-plane, byaforce attracting theparticle toward thepoint (x=a, y=0)and proportional tothedistance oftheparticle from thepoint (a,0).Using thenotation indicated inFig. 3-18, wecanwrite down the following relations: i /8=%<1r-a), @=§—B=%a. D2=2a2(1 —cosa), D=2asing, F=—kD, F=kD=2kasing» s=a(1r— a). 3-4] i KINEMATICS INAPLANE 87 Using these relations, wecanevaluate thework done, using Eq.(3-61): W= F-d/C r 770=I Fcos0ds s=O __ ° 2-22 — /;=T2ka S1n2c0S2da =—4ka2 I0 sin0cos0d0 0=1r/2 =2Ica2. Inorder tocalculate thesame integral from Eq.(3-63), weexpress rand Falong thecurve asfunctions oftheparameter a: :z:=acosa, y=asina, F,=lcDcosB=2kasinzg =Ica(1 —cosa), F,,=—kD sinB=-—2ka singcos; =-ka sina. Thework isnow, according toEq.(3-63), W= /F-dr C l ' 0_ Q2211) _-/c’x=1r (Ft do:+F”da dd 0 = [—ka2(1 —cosa)sina—lca2sinacosa]da =ka Sinozdo: =2ka2. 3-4Kinematics inaplane. Kinematics isthescience which describes thepossible motions ofmechanical systems without regard tothedynami- callaws that determine which motions actually occur. Instudying the kinematics ofaparticle inaplane, weshall beconcerned with methods for describing theposition ofaparticle, andthepath followed bytheparticle, andwith methods forﬁnding thevarious components ofitsvelocity and acceleration./. . 210’ 88 MOTION orPARTICLE INTwo 0RTHREE DIMENSIONS [CHAP. 3 1/ I__________ "P <——‘§I-1 hi.;.__.| FIG. 3-19. Position vector andrectangular coordinates ofapoint Pinaplane. The simplest method oflocatiI1g aparticle inaplane istosetuptwo perpendicular axes andtospecify anyposition byitsrectangular coordi- nates as,ywith respect tothese axes (Fig. 3-19). Equivalently, wemay specify theposition vector r=(av,y)from theorigin totheposition ofthe particle. Ifwelocate aposition byspecifying thevector r,then weneed tospecify inaddition only theorigin Ofrom which thevector isdrawn. Ifwespecify thecoordinates x,y,then wemust alsospecify thecoordi- nate axes from which ac,yaremeasured. Having setupacoordinate system, wenext wish todescribe thepath ofaparticle intheplane. Acurve inthemy-plane may bespeciﬁed by giving yasafunction ofxalong thecurve, orviceversa: y=1/(Z), (3-64) or ac=x(y). (3-65) Forms (3-64) and(3-65), however, arenotconvenient inmany cases, for example when thecurve doubles back onitself. Wemay alsospecify the curve bygiving arelation between xandy, f(Iv,9)=0, (3436) such that thecurve consists ofthose points whose coordinates satisfy this relation. Anexample istheequation ofacircle: 002+;/2—a2=0. -One ofthemost convenient ways torepresent acurve isinterms ofa parameter s: xZ x(8): y= it/(8); or r=r(s). Theparameter shasaunique value ateach point ofthecurve. Ass varies, thepoint [:v(s), y(s)] traces outthecurve. The parameter smay, 3-4] KINEMATICS INAPLANE 89 forexample, bethedistance measured along thecurve from some ﬁxed point. Theequations ofacircle canbeexpressed interms ofaparameter 0 intheform as=acos0, y= asin0, where 0istheangle between thex-axis andtheradius atothepoint (x,y) onthecircle. Interms ofthedistance smeasured around thecircle, sa:= acos—,a —asinsy a Inmechanical problems, theparameter isusually thetime, inwhich case Eqs. (3-67) specify notonly thepath oftheparticle, butalso the rate atwhich theparticle traverses thepath. Ifaparticle travels with constant speed varound acircle, itsposition atanytime tmay begiven by vtx= acos—,~ a -asingiy_ a Ifaparticle moves along thepath given byEq.(3-67), wemay specify itsmotion bygiving s(t),orbyspecifying directly w=$01), 1/=1/(t), (3-68) OI‘ r=r(t). (3-69) Thevelocity andacceleration, andtheir components, aregiven by __dr__.dx v“'E_‘E+J dt’ . (3-70) 1)=@1 1)=-(£14:1 dt ” dt dv d2r .d2x .d2y‘*‘E—Eﬁ—‘Eﬁ+'W’ 371 d2x dzy (_) a":W' a”=dt7' l l I l! II90 MOTION orPARTICLE INTwo ORTHREE DIMENSIONS [cn.u>. 3 y . l. /'\\/== I,““’+""’> d“ I d9 I1(9)'IT.,,\0:I " “T “no P WW) “ d9i 6 FIG. 3-20. Plane polar coordinates. FIG. 3-21. Increments inthevec- torsnand1. Polar coordinates, shown inFig. 3-20, areconvenient inmany prob- lems. The coordinates r,0arerelated tox,ybythefollowing equations: x=rcos0, y=rsin0, (3-72) and :__ 2 21/2Te+1/>, x (H3) __ -1Z=--1ii/___ = 1ii . 0-tan x sin ($2+1/2),” cos ($2+2/2),/2 Wedeﬁne unitvectors n,1inthedirections ofincreasing 1'and0,respec- tively, asshown. The vectors n,larefunctions oftheangle 0,andare related toi,jbytheequations n= icos0—|—jsin0, _ (3-74) l= —is1n0+jcos6. Equations (3-74) follow byinspection ofFig. 3-20. Differentiating, we obtain theimportant formulas dn dlE5_1, J5_—n. (3-75) Formulas (3-75) canalsobeobtained bystudying Fig.3-21 (remembering that Inl=|lI=1).The position vector risgiven very simply interms ofpolar coordinates: r=rn(0). (3-76) Wemaydescribe themotion ofaparticle inpolar coordinates byspecifying 'r(t), 0(t),thus determining theposition vector r(t). Thevelocity vector is ._£_ﬁ @@_- v_-dt__dtn+rd0 dt_rn+r9l. (3-77) 3-5] KINEMATICS IN’THREE DIMENSIONS 91 Thus weobtain thecomponents ofvelocity inthen,ldirections: 1),=1‘, vi;=r9. (3-78) Theacceleration vector is dv .. .dnd6 . pdlQ a- '('ﬁ— dt =(r-r02)n+(rt+21-(9)1. (3-79) Thecomponents ofacceleration are a,=5‘—r62, a,=rd+2rd. (3-80) Theterm r92=vf/riscalled thecentripetal acceleration arising from motion inthe0direction. Ifr=1‘=0,thepath isacircle, anda,=-vf/r. This result isfamiliar from elementary physics. The term 230issome- times called thecoriolis acceleration. 3-5Kinematics inthree dimensions. Thedevelopment inthepreceding section forkinematics intwodimensions utilizing rectangular coordinates canbeextended immediately tothethree-dimensional case. Apoint is speciﬁed byitscoordinates 1:,y,z,withrespect tochosen ‘rectangular axes inspace, orbyitsposition vector r=(22,y,z)with respect toachosen origin. Apath inspace mayberepresented intheform oftwoequations in:0,y,andz: f(w,y,2)=0, g(1>..1/,3)=0- (8-81) Each equation represents asurface. The path istheintersection ofthe twosurfaces. Apath may alsoberepresented parametrically: w=16(8), y=1/(8), Z=Z($)- (3-82) Velocity andacceleration areagain given by v=%=iv,+iv,+kvz, (3-83) d d dv,=F:» v,,=%» v,=F:> (3-84) and d . .a=—%=Ia,+Jay+ka,, (3-85) dzx d2y d2zG,=W1 av=it-5-;_ (Z;= (3—86) Many coordinate systems other than cartesian areuseful forspecial problems. Perhaps themost widely used arespherical polar coordinates 92 MOTION orPARTICLE INTwo 0RTHREE DIMENSIONS [cnxrn 3 z k m k I 11 Z i 1 -—- :11 w p -11 Z FIG. 3-22. Cylindrical polar coordinates. andcylindrical polar coordinates. Cylindrical polar coordinates (p,(0,z) aredeﬁned asinFig.3-22, orbytheequations I :0=pcos(0, y=psin(0, 2=z, (3-87) and, conversely, P=($2+I/2)1/2, (p=tan"1 E=SlI1_1 -3 ='OOS_1 9w (Iv+1/) (w+2/) Z= Z. Asystem ofunit vectors h,m,k,inthedirections ofincreasing p,<p,z,re- spectively, isshown inFig. 3-22. kisconstant, butmandharefunc- tions of<p,justasinplane polar coordinates: h=icos<p—|—jsin <p, m=—isin<p+jcos¢, (3-89) and, likewise, p an d3;=m, -is=-11. (3-90) The position vector rcanbeexpressed incylindrical coordinates inthe form r=ph+zk. (3-91) Diﬂerentiating, weobtain forvelocity andacceleration, using Eq.(3-90): v=§,§=ph+p¢m+31:, (3-92) a=g=(ii-p¢*>h+<p¢+2/>¢>m+3:. (3-93> 3-5] KINEMATICS INTHREE DIMENSIONS 93 Since k,m,hform asetofmutually perpendicular unitvectors, anyvector Acanbeexpressed interms ofitscomponents along k,m,h: A=A’,,h+A,,m+Ask. (3-94) Itmust benoted thatsince handmarefunctions of¢,thesetofcom- ponents (A,,,A,,,,A,)refers ingeneral toaspeciﬁc point inspace atwhich thevector Aistobelocated, oratleast toaspeciﬁc value ofthecoordi- -nate <p.Thus thecomponents ofavector incylindrical coordinates, and infact inallsystems ofcurvilinear coordinates, depend notonly onthe vector itself, butalso onitslocation inspace. IfAisafunction ofa parameter, sayt,then wemay compute itsderivative bydifferentiating Eq.(3-94), butwemust becareful totake account ofthevariation ofh andmifthelocation ofthevector isalsochanging with t(e.g., ifAisthe force acting onamoving particle): Formulas (3-92) and (3-93) arespecial cases ofEq.(3-95). Aformula fordA/dt could have beenworked outalsoforthecaseofpolar coordinates intwodimensions considered inthepreceding section, andwould, infact, have beenexactly analogous toEq.(3-95) except thatthelastterm would bemissing. Spherical polar coordinates (r,0,<p)aredeﬁned asinFig.3-23 orbythe equations . I ac=rsin0cos<p, y=rsin0sin(0, 2=rcos0. (3-96) The expressions forxand yfollow ifwenote that p=rsin0,and 2 11 m h k 1- I ‘ i02 i lh ——r Z P llZ FIG. 3-23. Spherical polar coordinates. \ 1 94 MOTION OFPARTICLE INTwo ORTHREE DIMENSIONS [CHAI>. 3 useEq.(3-87) ;theformula forzisevident from thediagram. Conversely, T10:2 + y2 +z2)1/2’ - 2 21/20=a..—1<ii;/_>_. (3-97) (p=tan_1g- Unit vectors n,1,mappropriate tospherical coordinates areindicated in Fig. 3-23, where misthesame vector asincylindrical coordinates. The unit vector his_also useful inobtaining relations involving nandl.We note that k,h,n,l,alllieinonevertical plane. From theﬁgure, and Eq.(3-89), wehave n= kcos0—|—hsin0= kcos0+isin0cos<p+jsin0sin<p, l=—ksin0+hcos0= —ksin0-|-ic0s0cos(o+jcos0sin<p, (3-98) m= —isin<p+jcos<p. Bydifferentiating these formulas, ormore easily byinspection ofthedia- gram (asinFig.3-21), noting thatvariation of0,with<pandrﬁxed, corre- sponds torotation inthek,n,h,1plane, while variation of¢,with 0andr ﬁxed, corresponds torotation around thez-axis, weﬁnd an an . 1‘Kw=l, 5,=msin6, 61 61a—6-—-n, tip_mcos0, (3-99) 8 6 .§03=0, %i=—-h=—ns1n0—lcos0. Inspherical coordinates theposition vector issimply r=rn(0, (0). (3-100) Differentiating andusing Eqs. (3-99), weobtain thevelocity andaccelera- tion: v=glé=rn+r01+(Tgbsin0)m, (3—l01) a=5%=(F-r62-r¢2sinz0)n-|-(rt)+230—'7'¢2sin0cos 0)l —|—(r<,'bsin0+2r¢sin0+2r9¢ cos0)m. (3-102) 3-6] ELEMENTS orvECToR ANALYSIS 95 Again, n,m,1form asetofmutually perpendicular unitvectors, andany vector Amay berepresented interms ofitsspherical components: A=Am+A,l+A,,m. (3-103) Here again thecomponents depend notonly onAbutalsoonitslocation. IfAisafunction oft,then dA_34_T_ i€_ -E) -3?-(alt A,dt A¢S11‘10dt n E E_ E)+(d¢ +A'dt A“’°°S0dt 1 +(%‘-"+A,sina%+A,@0sa%)m. (3-104) 3-6Elements ofvector analysis. Ascalar function u(x,y,z)hasthree derivatives, which may bethought ofasthecomponents ofavector point function called thegradient ofu: gradu= = +j3;+1331- (3-193) Wemayalsodeﬁne gradugeometrically asavector whose direction isthe direction inwhich uincreases most rapidly andwhose magnitude isthe directional derivative ofu,i.e.,therate ofincrease ofuperunit distance, inthat direction. That this geometrical deﬁnition isequivalent tothe algebraic deﬁnition (3-105) can.beseen bytaking thedifferential ofu: ,8u 6u 6adu=5dx+1%dy+-3-;dz. (3—106) Equation (3—l06) hastheform ofascalar product ofgrad uwith thevector drwhose components aredrc,dy,dz: du=dr'grad u. (3-10?) Geometrically, duisthechange inuwhen wemove from thepoint r= (x,y,z)toanearby point r+dr=(a;+dz,y+dy,z+dz). ByEq. (3-15): du=|dr|[grad u|cos0, (3—108) where 0istheangle between drandgrad u.Thus ataﬁxed small dis- tance |dr|from thepoint r,thechange inuisamaximum when drisinthe same direction asgrad u,andthen: dulgfad "l—I4 I I 96 MOTION orPARTICLE INTwo 0RTHREE DIMENSIONS [CHAP. 3 This conﬁrms thegeometrical description ofgrad ugiven above. Anal- ternative geometrical deﬁnition ofgrad uisthat itisavector such that thechange inu,foranarbitrary small change ofposition clr,isgiven by Eq.(3—107). Inapurely symbolic way, theright member ofEq. (3-105) canbe thought ofasthe-“product” ofa“vector”: 6 6 8 .6 .6 6 with thescalar function u: grad u=Vu. (3—110) The symbol Vispronounced “del.” Vitself isnotavector inthegeo- metrical sense, butanoperation onafunction uwhich gives avector Vu. However, algebraically, Vhasproperties nearly identical with those ofa vector. The reason isthat thedifferentiation symbols (8/62:, 6/6y, 8/63) have algebraic properties likethose ofordinary numbers except when they actonaproduct offunctions: a aa aa aa 5.i(“+”)=aii+a;’ 6.1:6yu=6y6a:u’ 6-111) and 8 6u 5 ((1/Tl.) -—(Z(E1 provided aisconstant. However, ‘%(uv) =%v+u (3-113) Inthisonerespect differentiation operators differ algebraically from ordi- nary numbers. IfI‘)/6:1: were anumber, 6/0a:(uv) would equal either u(6/6x)12 orv(6/6:c)u. Thus wemay saythat 8/Bx behaves algebraically asanumber except that when itoperates onaproduct, theresult isasum ofterms inwhich each factor isdifferentiated separately, asinEq.(3-113). Asimilar remark applies tothesymbol V.Itbehaves algebraically asa vector, except that when itoperates onaproduct itmust betreated also asadifferentiation operation. This ruleenables ustowrite down alarge number ofidentities involving theVsymbol, based onvector identities. Weshall require very fewofthese inthistext, andshall notlistthem here. * *Foramore complete treatment ofvector analysis, seeH.B.Phillips, Vector Analysis. New York: John Wiley &Sons, 1933. 3-6] ELEIIENTs orvEcToR ANALYSIS 97 FIG. 3-24. Avolume Vbounded bya.surface S. Wecanform thescalar product ofVwith avector point function A(:c,y,z).This iscalled thedivergence ofA: -_._.§.£z % ‘Er. d1vA_VA-ax+ay+az (3-114) Thegeometrical meaning ofdivAisgiven bythefollowing theorem, called thedivergence theorem, orGauss’ theorem: fffv-Adv =[[11-Ads, (3-113) V S where Visagiven volume, Sisthesurface bounding thevolume V,andn isa.unitvector perpendicular tothesurface Spointing outfrom thevolume ateach point ofS(Fig. 3-24). Thus n-A isthecomponent ofAnormal toS,andEq.(3-115) says that the“total amount ofV-A inside V"is equal tothe“total ﬂuxofAoutward through thesurface S.”Ifvrepre- sents thevelocity ofamoving ﬂuidatanypoint inspace, then (‘In-was represents thevolume ofﬂuid ﬂowing across Spersecond. Iftheﬂuid is incompressible, thenaccording toEq.(3-115), IJIV-vdV l would represent thetotal volume ofﬂuid being produced within thevol- ume Vpersecond. Hence V-vwould bepositive atsources from which theﬂuid isﬂowing, andnegative at“sinks” intowhich itisﬂowing. We omit theproof ofGauss’ theorem [Eq.63-115)]; itmay befound inany book onvector analysis.* *See,e.g.,Phillips, op.cit.Chapter 3,Section 32. 98 IIoTIoN orPARTICLE INTwo onTHREE DIMENSIONS [cn_~u>. 3 FIG. 3-25. Asurface Sbounded byacurve O. Wecanalso form across product ofVwith avector point function A(t,3;,3).Thisiscalled thecurlofA: an-I_ _ +k(ax ay (3116) Thegeometrical meaning ofthecurlisgiven byStokes’ theorem: [[3-(v xA)dS =LA-dr, (3-117) S where Sisanysurface inspace, nistheunit vector normal toS,andCis thecurve bounding S,drbeing taken inthatdirection inwhich aman would walk around Cifhislefthand wereontheinside andhishead inthedirec- tionofn.(See Fig.3-25.) According toEq.(3—117), curlAatanypoint isameasure oftheextent towhich thevector function Acircles around that point. Agood example isthemagnetic ﬁeld around awire carrying an electric current, Where thecurlofthemagnetic ﬁeld intensity ispropor- tional tothecurrent density. Weomit theproof ofStokes’ theorem [Eq. (3-117)].* Thereader should notbebothered bythedifficulty ofﬁxing these ideas inhismind. Understanding ofnew mathematical concepts like these comes tomost people only slowly, asthey areputtouse. Thedeﬁnitions arerecorded here forfuture use. One cannot beexpected tobefamiliar with them until hehasseen how they areused inphysical problems. Thesymbolic vector Vcanalsobeexpressed incylindrical coordinates interms ofitscomponents along h,m,k.(SeeFig.3-22.) Wenote that ifu=u(,o,(,0,2), du 6a 611 *Fortheproof seePhillips, op.cit.Chapter 3,Section 29. 3—6] ELEMENTS orvncron ANALYSIS 99 and,from Eqs. (3-91) and(3-90), dr=hdp—}- mpd<p+kdz, (3—119) aresult whose geometric signiﬁcance willbeevident froin Fig.3-22. Hence, ifwe write 6 m6 6 wewillhave, since h,rn,kareasetofmutually perpendicular unitvectors, du=dr-Vu, (3—121) asrequired bythegeometrical deﬁnition ofVu=grad u.[See theremarks following Eq.(3—107).] Aformula forVcould have been worked outalso for thecase ofpolar coordinates intwo dimensions and would have been exactly analogous toEq.(3—120) except that theterm inzwould bemissing. Inapply- ingthesymbol Vtoexpressions involving vectors expressed incylindrical co- ordinates [Eq.(3—94)], itmust beremembered thattheunitvectors handmare functions of<pandsubject todiﬂerentiation when they occur after 6/6¢. . Wemayalsoﬁndthevector Vinspherical coordinates (Fig. 3-23) bynoting that 6u du Bu d1], —'5;d1‘+8? +(Q d§0, and dr=ndr+Ird0+mrsin0d<p. (3—123) Hence 8 I6 m 6 inorder that Eq.(3—12l) may hold. Again wecaution that inworking with Eq. (3-124), thedependence ofn,1,mon0,<pmust bekept inmind. Forexample, thedivergence ofavector function Aexpressed inspherical coordinates [Eq. (3—103)] is BA l6A m 6AV"-“Tr+7'5?+@155 -811,1%» )1 - )—W-—|—r(a0+A, -Frsino a¢+A,s1n0—f—|A¢cos0 16A A0 16A , 8A, 2A, ,=a,+T+-~’+ +r60 rtanﬂ rsin0 6<p __1_§ 2 _1.. Q- 11__'§A~2.“war“ A')+rsin0a0(s“‘”A‘)+rsino a¢ (Intheabove calculation, weusethefactthat l,m,nareasetofmutually per- pendicular unit vectors.) 100 MOTION orPARTICLE INTWO onTHREE DIMENSIONS [cmua 3 3-7Momentum andenergy theorems. Newton’ ssecond law, asformu- lated inChapter 1,leads, intwoorthree dimensions, tothevector equation . dzmat-25=F. (3-125) Intwodimensions, thisisequivalent totwocomponent equations, inthree dimensions, tothree, which are,incartesian coordinates, dzx dzy dzz m-c—i—£5=F,,, mW=Fy, Inthissection, weprove, using Eq.(3—125), some theorems formotion in twoorthree dimensions which arethevector analogs tothose proved in Section 2—1forone-dimensional motion. Thelinear momentum vector pofaparticle istobedeﬁned, according tocEq. (1—10), asfollows:p=mv. (3—127) Equations (3—125) and(3—126) canthen bewritten d _Q_ _ 82(mv) _dt_-F, (3128) or,incomponent form, m_ m_ Q_ _ dt_F,,, dt_F,,, dt‘_F,. (3129) IfWemultiply bydt,andintegrate from t1tot2,weobtain thechange in momentum between t1andt2: #2p2—pl=mvg —-mvl =ftFdt. (3—130) 1 The integral ontheright istheimpulse delivered bytheforce, andisa vector whose components arethecorresponding integrals ofthecom- ponents ofF.Incomponent form: E P12 —p11 Z’/;2F1dt; l 2 P112_P111:£21711 db (3—131) 1 P22 _'17:1 =/:2 Fadi- 1 Inorder toobtain anequation fortherate ofchange ofkinetic energy, weproceed asinSection 2—1,multiplying Eqs. (3-126) by11,,v,,,21,,respec- tively, toobtain %(=}mvf) =F,,v,,, %(%mvZ) =Fyvy, ;€'(%mv§) =F,v,. (3—l32) 3-8] PLANE ANDVECTOR ANGULAR MOMENTUM THEOREMS 101 Adding these equations, wehave ' d;,;[%m(vZ+vi+vb]=Fa.+Fwy+mv.. OI‘ .1 dT53(%mv2) =W=F-v. (3-133) This equation canalsobededuced from thevector equation (3—125) by taking thedotproduct with voneach side, andnoting that (%(v2)=ﬁli(v-v)= v+v = Thus, byEq.(3—132), d d2<1F-v='mV~d: =%m%—) =(-12(%mv2). Multiplying Eq.(23-133) bydt,andintegrating, weobtain theintegrated form oftheenergy theorem: T2—T1=ﬁmvg —émvf =LizF-vdt. (3—134) 1 Since vdt=dr,ifFisgiven asafunction of1',wecanwrite theright member ofEq.(3—134) asalineintegral: T2-T1=/“F-dr, (3-135) Tl where theintegral istobetaken along thepath followed bytheparticle between thepoints r1andr2.Theintegral ontheright inEqs. (3—134) and (3—135) isthework done ontheparticle bytheforce between thetimes t1andt2.Note how thevector notation brings outtheanalogy between theone- andthetwo- orthree-dimensional cases ofthemomentum and energy theorems. 3-8Plane andvector angular momentum theorems. Ifaparticle moves inaplane, wedeﬁne itsangular momentum L0about aipoint Oasthe moment ofitsmomentum vector about thepoint O,that is,astheproduct ofitsdistance from Otimes thecomponent ofmomentum perpendicular tothelinejoining theparticle toO.Thesubscript 0willusually beomitted, except when moments about more than oneorigin enter intothediscussion, butitmust beremembered that angular momentum, liketorque, refers toaparticular origin about which moments aretaken. The angular momentum Listaken aspositive when theparticle ismoving inacounter- clockwise sense with respect toO;Lisexpressed most simply interms of polar coordinates with Oasorigin. Lettheparticle have mass m.Thenl 1 1 J lll 102 MOTION OF PARTICLE IN TWO OR THREE DIMENSIONS [CHAP. 3 . UT v 1 ”B\/g m T 0 O FIG. 3-26. Components ofvelocity inaplane. itsmomentum ismv,andthecomponent ofmomentum perpendicular to theradius vector from Oismv,(Fig. 3-26), sothat, ifweuseEq.(3-78), L=rmv, =mrzd. (3—136) IfweWrite theforce interms ofitspolar components: F=nF,+lF,, (3—137) then inplane polar coordinates theequation ofmotion, Eq.(3—125), be- comes, byEq.(3-80), ma,=mi‘—mr02 =F,, (3—138) ma,=mrli+2m1‘0 =F,. (3—139) Wenow note that %=2mr1‘0 +mr2§. Thus, multiplying Eq.(3-139) byr,wehave dL d .H7=(-5(W20) =1-F,=N. (3-140) The quantity rF,isthetorque exerted bytheforce Fabout thepoint O. Integrating Eq.(3—I40), weobtain theintegrated form oftheangular momentum theorem formotion inaplane: L2-L1=my-302-mm,=f”’rF,d¢. (3-141) ii Wecangeneralize thedeﬁnition ofangular momentum toapply tothree- dimensional motion bydeﬁning theangular momentum ofaparticle about anaxisinspace asthemoment ofitsmomentum vector about thisaxis, just asinSection 3-2wedeﬁned themoment ofaforce about anaxis. 3-8] PLANE ANDvnoroa ANGULAR MOMENTUM THEOREMS 103 Thedevelopment ismost easily carried outincylindrical coordinates with thez-axis astheaxisabout which moments aretobetaken. Thegenerali- zation oftheorems (3-140) and(3-141) tothiscaseisthen easily proved in analogy with theproof given above. This development isleftasanexercise. Asaﬁnal generalization oftheconcept ofangular momentum, wedeﬁne thevector angular momentum L0about apoint Oasthevector moment of themomentum vector about O: L0=rXp=m(rxv), (3-142) where thevector ristaken from thepoint Oasorigin totheposition ofthe particle ofmass m.Again weshall omit thesubscript Owhen noconfusion canarise. The component ofthevector Linanydirection isthemoment ofthemomentum vector pabout anaxisinthat direction through O. Bytaking thecross product ofrwith both members ofthevector equa- tion ofmotion [Eq. (3-125)], weobtain. rx(mQ)=I‘><F. (3-143)dt Bytherules ofvector algebra andvector calculus, %=,%[r><(mm =rX%(mv)+§%X (mv) =rX%(mv)+vx (mv) -I><(mt) Wesubstitute thisresult inEq.(3-143): dL t=rX =N. (3-1-14) /d F Thetime rate ofchange ofthevector angular momentum ofaparticle is equal tothevector torque acting onit.Theintegral form oftheangular momentum theorem is L2-L,=ft”Ndt. (3-145) The theorems forplane angular momentum andforangular momentum about anaxisfollow from thevector angular momentum theorems bytak- ingcomponents intheappropriate direction. 104 MOTION orPARTICLE INTWO onTHREE DIMENSIONS [CHAP. 3 3-9Discussion ofthegeneral problem oftwo- andthree-dimensional motion. Iftheforce Fisgiven, ingeneral asafunction F(v,r,t).ofposition, velocity, andtime, theequations ofmotion (3-126) become asetofthree (or,intwodimensions, two) simultaneous second-order differential equa- tions: d’ ... mﬁ =F=v(xr yaZ:xv1/12;t); d2y——F"' t (e146 mﬁ _ U(x1 yrZ:xxyrZ:); _ ) dz ... mag‘ =F=(xry;z1x;y>Z:t)- Ifwearegiven theposition ro=(xo,yo,20),andthethevelocity V0= (v,,,,,v,,,,,22,0)atanyinstant to,Eqs. (3-146) giveusdzr/dtz, andfrom r,i‘,f, attime t,wecandetermine r,fashort time later orearlier att+dt,thus extending thefunctions r,i',i‘,into thepast andfuture with thehelp of Eqs. (3—146). This argument canbemade mathematically rigorous, and leads toanexistence theorem guaranteeing theexistence ofaunique solu- tion ofthese equations foranygiven position andvelocity ataninitial instant to.Wenotethatthegeneral solution ofEqs.(3—146) involves the six“arbitrary” constants xo,yo,20,0,0,0,0,0,0. Instead ofthese sixcon- stants, wemight specify anyother sixquantities from which they canbe determined. (Intwodimensions, wewillhave twosecond-order differential equations andfour initial constants.) Ingeneral, thesolution ofthethree simultaneous equations (3-146) willbemuch more diﬁicult than thesolution ofthesingle equation (2-9) forone-dimensional motion. The reason forthegreater difficulty isthat, ingeneral, allthevariables ac,y,2andtheir derivatives areinvolved inall three equations, which makes theproblem ofthesame order ofdiﬁiculty asasingle sixth-order differential equation. [Infact, thesetofEqs. (3—146) canbeshown tobeequivalent toasingle sixth-order equation] Ifeach force component involved only thecorresponding coordinate anditsderiva- tives, F,=F,(a'c, ac,t), F11 Z Fy(y; yrt): F,=F,(é, 2,t), then thethree equations (3—146) would beindependent ofoneanother. Wecould solve forx(t),y(t), z(t)separately asthree independent problems inone-dimensional motion. The most important example ofthiscase is 3-9] DISCUSSION orTHEGENERAL PROBLEM 105 probably when theforce isgiven asafunction oftime only: F=F(t) =[F,,(t), F,,(t), F,(t)]. (3—148) Theas,y,andzequations ofmotion carfthen each besolved separately by themethod given inSection 2-3. The case ofafrictional force propor- tional tothevelocity willalsobeanexample ofthetype (-3-147). Other cases willsometimes occur, forexample, thethree-dimensional harmonic oscillator (e.g., abaseball inatubful ofgelatine, oranatom inacrystal lattice), forwhich theforce is F,=-lczx, F11=—7m»/, (3—149) F,=—k,z, when theaxes aresuitably chosen. The problem now splits into three separate linear harmonic oscillator problems inx,y,andz.Inmost cases, however, wearenotsofortunate, andEq.(3—147) does nothold. Special methods areavailable forsolving certain classes oftwo- andthree-dimen- sional problems. Some ofthese willbedeveloped inthischapter. Prob- lems notsolvable bysuch methods arealways, inprinciple, solvable by various numerical methods ofintegrating setsofequations likeEqs.(3-1-16) togetapproximate solutions toanyrequired degree ofaccuracy. Such methods areeven more tedious inthethree-dimensional case than inthe one-dimensional case, and areusually impractical unless one hasthe services ofoneofthelarge automatic computing machines. When wetrytoextend theidea ofpotential energy totwo orthree dimensions, wewillﬁndthat having theforce given asF(r), afunction of ralone, isnotsufﬁcient toguarantee theexistence ofapotential-energy function V(r). Intheone-dimensional case, wefound that iftheforce is given asafunction ofposition alone, apotential-energy function canal- ways bedeﬁned byEq.(2-41). Essentially, thereason isthat inone dimension, aparticle which travels from 2:1to:02andreturns to201must return bythesame route, sothat iftheforce isafunction ofposition alone, thework done bytheforce ontheparticle during itsreturn tripmust nec- essarily bethesame asthat expended against theforce ingoing from x1 tox2.Inthree dimensions, aparticle cantravel from r1tor2andreturn byadifferent route, sothat even ifFisafunction ofr,theparticle may be acted onbyadifferent force onthereturn tripandthework done may notbethesame. InSection 3-12 Weshall formulate acriterion todetermine when apotential energy V(r) exists. When V(r) exists, aconservation ofenergy theorem stillholds, andthe total energy (T+V)isaconstant ofthemotion. However, whereas in<4 4 4 1 4 1 l l I l106 MOTION orPARTICLE INTwo onTHREE DIMENSIONS Icnxr. 3 onedimension theenergy integral isalways sufﬁcient toenable ustosolve theproblem atleast inprinciple (Section 2-5), intwoandthree dimensions thisisnolonger thecase. Ifasistheonly coordinate, then ifweknow a relation (T+V=E)between asandrt,wecansolve forit=f(x) and reduce theproblem tooneofcarrying outasingle integration. Butwith coordinates av,y,z,onerelation between zv,y,z,a':,1],2‘:isnotenough. We would need toknow ﬁvesuch relations, ingeneral, inorder toeliminate, for example, x,y,:i:,and1],andﬁndé=f(z). Inthetwo-dimensional case, we would need three relations between x,y,at,ytosolve theproblem bythis method. Toﬁndfour more relations liketheenergy integral from Eqs. (3-146) (ortwomore intwo dimensions) ishopeless inmost cases. In fact, such relations donotusually exist. Often, however, wecanﬁndother quantities (e.g., theangular momentum) which areconstants ofthemotion, andthus obtain oneortwomore relations between x,y,z,:i:,y,é,which in many cases willbeenough toallow asolution oftheproblem. Examples willbegiven later. p 3-10 Theharmonic oscillator intwoandthree dimensions. Inthis section andthenext, weconsider afewsimple problems inwhich theforce hastheform ofEqs. (3—147), sothat theequations ofmotion separate into independent equations inx,y,andz.Mathematically, wethen simply have three separate problems, each ofthetype considered inChapter 2. Theonly newfeature willbetheinterpretation ofthetln'ee solutions x(t), y(t), z(t)asrepresenting amotion inthree-dimensional space. Weﬁrstconsider brieﬂy thesolution oftheproblem ofthethree-dimen- sional harmonic oscillator without damping, whose equations ofmotion are mi?=—k,x, 1"?=—7¢t?/, (3-150) m2=—k,z. Amodel could beconstructed bysuspending amass between three per- pendicular setsofsprings (Fig. 3-27). The solutions‘ ofthese equations, Weknow from Section 2-8: at=A,cos(w,t—|—0,), wﬁ=k,/m, y=A,cos(wyt+011), wf=la,/m, (3—151) z=A,cos(w,t+0,), wf=k,/m. The sixconstants (A,,A.y,A,,0,,0,,0,)depend ontheinitial values wo,yo,20,5:0,yo,éo.Each coordinate oscillates independently withsimple 3-10] THEHAHMONIC OSCILLATOR 107 FIG. 3-27. Model ofathrcc-dimensional harmonic oscillator. harmonic motion atafrequency depending onthecorresponding restoring force cocfﬁcient, andonthemass. The resulting motion oftheparticle takes place within arectangular boxofdimensions 2A,X2A,.X2.4, about theorigin. Iftheangular frequencies 0.1,,coy,w,arecommensurable, that is,ifforsome setofintegers (n,,n,.,n,), %=‘ii’=‘B. (3-152)n, nu n, thenthepath ofthemass ininspace isclosed, andthemotion isperiodic. If(#1,,n,,,n,)arechosen sothat they have nocommon integral factor, then theperiod ofthemotion is T=27m, =21rn,, = (3_l53) oi, nu, cc, During oneperiod, thecoordinate remakes n,oscillations, thecoordinate y makes 11,,oscillations, andthecoordinate zmakes n,oscillations, sothat theparticle returns attheendoftheperiod toitsinitial position and velocity. Inthetwo-dimensional case, ifthepath oftheoscillating parti- cleisplotted forvarious combinations offrequencies cc,andmy,andvarious phases 0,and 0,,many interesting and beautiful patterns areobtained. Such patterns arecalled Lissajous figures (Fig. 3-28), andmay bepro- duced mechanically byamechanism designed tomove apencil orother writing device according toEqs. (3-151). Similar patterns may beob- tained electrically onacathode-ray oscilloscope bysweeping horizontally 108 MOTION orPARTICLE INTwo oRTHREE DIMENSIONS [CHAP. 3 y .1! 1; .1 L01=my we=Zwy y 2/ 1 ‘<,.,>\/‘v/‘v/Q0 3w, =Qwy 31»; =50:5, FIG. 3-28. Lissajous ﬁgures. and vertically with suitable oscillating voltages. Ifthefrequencies w,, my,w,areincommensurable, sothat Eq.(3-152) does nothold foranyset ofintegers, themotion isnotperiodic, andthepath ﬁllstheentire box 2A,X2A,, X2A,, inthesense that theparticle eventually comes ar- bitrarily close toevery point inthebox. The discussion canreadily be extended tothecases ofdamped andforced oscillations intwoandthree dimensions. Ifthethree constants k,,kg,Ic,areallequal, theoscillator issaidtobe isotropic, that is,thesame inalldirections. Inthiscase, thethree fre- quencies 0),,0:1,,w,areallequal andthemotion isperiodic, with each coordi- nate executing onecycle ofoscillation inaperiod. Thepath canbeshown tobeanellipse, astraight line, oracircle, depending ontheamplitudes andphases (A,,A,,,A,,0,,0,,0,). 3-11 Projectiles. Animportant problem inthehistory ofthescience of mechanics isthat ofdetermining themotion ofaprojectile. Aprojectile moving under theaction ofgravity near thesurface oftheearth moves, if 3-11] PROJECTILES 109 airresistance isneglected, according totheequation 2m%§=—mgk, (3-154) where the2-axis istaken inthevertical direction. Incomponent form: d2mi=0, (3-155) d2m#4=0, (3-156) d2mE;-"1=—mg. (3-157) Thesolutions ofthese equations are (3-158) (3—159) (3—160)x=1:0+v,,,t, yZ 1/0 + v1/Qt; z=zo+v,,,t-—ﬁgtz, or,invector form, r=ro+vot—%gt2k. (3-161) Weassume theprojectile starts from theorigin (0,0,0),with itsinitial velocity inthexz-plane, sothat v,,,,=0.This isnolimitation onthe motion oftheprojectile, butmerely corresponds toaconvenient choice of coordinate system. Equations (3-158), (3-159), (3-160) thenbecome at=v,ot, (3—162) y=0, (3-163) L z=v,ot——-%gt2. (3-164) These equations giveacomplete description ofthemotion oftheprojectile. Solving theﬁrst equation fortandsubstituting inthethird, wehave an equation forthepath inthexz-plane: z=ha: —%~%a:2. (3—165) 22,0 121° This canberewritten intheform "=~"=2__ _ (:z:—T9) - 2g_z 2g (3166)< l I I 1 110 MOTION orPARTICLE INTwo onTHREE DIMENSIONS [CHAP. 3 This isaparabola, concave downward, whose maximum altitude occurs at Z=‘£11. (3-167) andwhich crosses thehorizontal plane 2=0attheorigin andatthepoint mm=2 - (3-168) Ifthesurface oftheearth ishorizontal, xmistherange oftheprojectile. Letusnow take account ofairresistance byassuming africtional force proportional tothevelocity: .12 dmat-§=—mgk -bi? (3-169) Incomponent notation, ifweassume that themotion takes place inthe xz-plane, . - 01% dx d2 dzmd—t§=-—mg-ta? (3-171) Itshould bepointed outthat theactual resistance oftheairagainst amov- ingprojectile isacomplicated function ofvelocity, sothat thesolutions we obtain willbeonly approximate, although they indicate thegeneral nature ofthemotion. Iftheprojectile starts from theorigin att=0,thesolu- tions ofEqs. (3—170) and(3—171) are(seeSections 2-4and2-6) v,=v,,,e'_b”'”, (3-172) 1;= (1-<r””"‘), (3-173) v.=(529+v..,).e-”"'" - (3-174) mzg MU, _ mz=(-b2-+ -b—°) (1-eW)-%¢. (3-175) Solving Eq.(23-173) fortandsubstituting inEq.(3—175), weobtain an equation forthetrajectory: (.._inas_£2("W-») z— 11,0+v,,,)x b2Inmv,,, —bxi (3_176) 3-11] PROJECTILESV 111 Z \, FIG. 3-29. Trajectories formaximum range forprojectiles with various muzzle velocities. Forlowairresistance, orshort distances, when (bx)/(mv,,,) <<1,wemay expand inpowers of(bx)/(mv,,,) toobtain ‘b ze%x-%%x2-%-‘-’;x3---- (3-177) 11,0 11,0 mvxo Thus thetrajectory starts outasaparabola, butforlarger values ofx (taking 12,0aspositive), zfalls more rapidly than foraparabola. Accord- ingtoEq.(3—176), asxapproaches thevalue (mum)/b, zapproaches minus inﬁnity, i.e.,thetrajectory ends asavertical drop at:1:=(mv,,o)/b. From Eq.(3—17 4),weseethat thevertical fallattheendofthetrajectory takes place attheterminal velocity —mg/b. (The projectile may, ofcourse, return toearth before reaching thispart ofitstrajectory.) Ifwetake the ﬁrst three terms inEq.(3—177) andsolve foravwhen 2=0,wehave ap- proximately, if2:,"<<(mun)/b, _211,1), bvfv,:c,,,=%—-§%—|—---. (3—178) Thesecond term gives theﬁrst-order correction totherange duetoairre- sistance, andtheﬁrsttwoterms willgive agood approximation when the eﬂ'ect ofairresistance issmall. The extreme opposite case, when air resistance ispredominant indetermining range (Fig. 3-29), occurs when thevertical drop at2;=(mv,o) /bbegins above thehorizontal plane z=0. Therange isthen, approximately, I b 2xm >>1)- (3-179) Wecantreat (approximately) theproblem oftheeﬂ'ect ofwind onthe projectile byassuming theforce ofairresistance tobeproportional to therelative velocity oftheprojectile with respect totheair: d2r (dr ) where vwisthewind velocity. Ifv,,,isconstant, theterm bv,,,inEq.(3—180) l l > i l l112 MQTION orPARTICLE INTWO onTHREE DIMENSIONS [cmu>. 3 behaves asaconstant force added to—mgk, andtheproblem iseasily solved bythemethod above, theonlydifference being thatthere maybe constant forces inaddition tofrictional forces inallthree directions as,y,z. Theairresistance toaprojectile decreases with altitude, sothat abetter form fortheequation ofmotion ofaprojectile which rises toappreciable altitudes would be dz _dmag:;—mgk —be=”';,§, (3-181) where histheheight (sayabout ﬁvemiles) atwhich theairresistance falls to1/eofitsvalue atthesurface oftheearth. Incomponent form, mzii=—bx@"‘”‘ mi]=-bye-’”‘’ I, ’ (3-182) m2=-—mg ——bée_' . These equations aremuch harder tosolve. Since zappears inthexand yequations, wemust ﬁrst solve thezequation forz(t)andsubstitute in theother twoequations. Thezequation isnotofanyofthesimple types discussed inChapter 2.Theimportance ofthisproblem wasbrought out during theFirst World War, when itwas discovered accidentally that aiming acannon atamuch higher elevation than thatwhich hadprevi- ously been believed togivemaximum range resulted inagreat increase in therange oftheshell. The reason isthat thereduction inairresistance, ataltitudes ofseveral miles, more than makes upforthelossinhorizontal component ofmuzzle velocity resulting from aiming thegunhigher. 3-12 Potential energy. Iftheforce Facting onaparticle isafunction ofitsposition r=(:0,y,z),then thework done bytheforce when the particle moves from r1to1'2isgiven bythelineintegral /:1”F-dr.\ Itissuggested thatwetrytodeﬁne apotential energy V(r)=V(x,y,z)in analogy with Eq.(2-41) forone-dimensional motion, asthework done by theforce ontheparticle when itmoves from rtosome chosen standard point r8: V(r)=_F(r)-dr. (3-183) Such adeﬁnition implies, however, that thefunction V(r) shall beafunc- tion only ofthecoordinates (:0,y,2)ofthepoint 1'(and ofthestandard point r,,which weregard asﬁxed), whereas ingeneral theintegral onthe 3-12] POTENTIAL ENERGY - 113 right depends upon thepath ofintegration from r,tor.Only iftheintegral ontheright isindependent ofthepath ofintegration willthedeﬁnition belegitimate. Letusassume that wehave aforce function F(x,y,2)such that theline integral inEq.(3—183) isindependent ofthepath ofintegration from r,to anypoint r.Thevalue oftheintegral then depends only onr(and onr,), andEq.(3—183) deﬁnes apotential energy function V(r). The change in Vwhen theparticle moves from rtor+dristhenegative ofthework done bytheforce F: dV=—F-dr. (3-1s4) Comparing Eq.(3—184) with thegeometrical deﬁnition [Eq. (3—107)] of thegradient, weseethat '—F=grad V, (3—185)F=—VV. Equation (3—l85) may beregarded asthesolution ofEq.(3—183) forFin terms ofV.Incomponent form, OV 6V 6V F$Z*5;7 Fu=—@2 Fz:'—'$' Inseeking acondition tobesatisﬁed bythefunction F(r) inorder that theintegral inEq.(3—183) beindependent ofthepath, wenote that, since Eq.(3—28) canbeproved from thealgebraic deﬁnition ofthecross product, itmust hold alsoforthevector symbol V: i VXV=0. (3—187) Applying (VXV)tothefunction V,wehave VXVV=curl(grad V)=0. (3—l88) Equation (3—188) canreadily beveriﬁed bydirect computation. From Eqs. (3—188) and(3—185), wehave - VXF=curlF =0. (3—189) Since Eq.(3—189) hasbeen deduced ontheassumption thatapotential function exists, itrepresents anecessary condition which must besatisﬁed bytheforce function F(a:,y,z)before apotential function canbedeﬁned. Wecanshow that Eq.(3—189) isalsoasuﬂicient condition fortheexistence ofapotential bymaking useofStokes’ theorem [Eq. (3~117)1.ByStokes’ theorem, ifweconsider anyclosed path C’inspace, thework done bythe1 i4a i 1 I i 1 114 MOTION orPARTICLE INTWO ORTHREE DIMENSIONS [cn.u>. 3 I2‘ 1'1 I FIG. 3-30. Two paths between 1'1and1-2,forming aclosed path. force F(r) when theparticle travels around thispath is fF-dr= (v><F)as, (3-190) C s where Sisasurface inspace bounded bytheclosed curve C.Ifnow Eq.(3—189) isassumed tohold, theintegral ontheright iszero, andwe have, foranyclosed path C, faF-dr=0. (3-191) Butifthework done bytheforce Faround anyclosed path iszero, then thework done ingoing from r1to1'2willbeindependent ofthepath fol- lowed. Forconsider anytwopaths between 1'1andr2,andletaclosed path C’beformed going from r1to1'3byonepath andreturning to1'1 bytheother (Fig. 3-30). Since thework done around C’iszero, thework going from 1'1to:2must beequal andopposite tothat onthereturn trip, hence thework ingoing from r1tor2byeither path isthesame. Applying thisargument totheintegral ontheright inEq.(3—183), weseethat the result isindependent ofthepath ofintegration from r,tor,andtherefore theintegral isafunction V(r) oftheupper limit alone, when thelower limit r,isﬁxed. Thus Eq.(3—189) isboth necessary andsufficient forthe existence ofapotential function V(r) when theforce isgiven asafunction ofposition F(r). When curlFiszero, wecanexpress thework done bytheforce when theparticle moves from r1to1'2asthedifference between thevalues ofthe potential energy atthese points: /ti”F-dr=/r:’F-dr +fr:”F-dr =V(r1)-V(1‘2). (3-192) Combining Eq.(3—192) with theenergy theorem (3—135), wehave forany twotimes t1andt2: T1+V(r1) =T2+V(l‘2)- (3-193) I 3-12] POTENTIAL ENERGY 115 Hence thetotal energy (T+V)isagain constant, andwehave anenergy integral formotion inthree dimensions: T+V=W12+122+22)+V(w,1/,1) =E. <3-194) Aforce which isafunction ofposition alone, andwhose curlvanishes, is said tobeconservative, because itleads tothetheorem ofconservation of kinetic plus potential energy [Eq. (3—194)]. Insome -cases, aforce may beafunction ofboth position and time F(r,t).Ifatanytime tthecurlofF(r,t)vanishes, then apotential-energy function V(r,t)canbedeﬁned as - V(r,t) =_F(r,t)-dr, (3-195) andwewillhave, foranytime tsuch that VXF(r,t)=O, F(r,t)=—VV(r, t). (3—196) However, theconservation lawofenergy cannolonger beproved, for Eq.(3—192) nolonger holds. Itisnolonger true that thechange inpo- tential energy equals thenegative ofthework done ontheparticle, for theintegral which deﬁnes thepotential energy attime tiscomputed from theforce function atthat time, whereas theintegral that deﬁnes thework iscomputed using ateach point theforce function atthetime theparticle passed through that point. Consequently, theenergy T—|—Visnota constant when Fand Varefunctions oftime, andsuch aforce isnotto becalled aconservative force. When theforces acting onaparticle areconservative, Eq. (3—194) enables ustocompute itsspeed asafunction ofitsposition. Theenergy E isﬁxed bytheinitial conditions ofthemotion. Equation (3—194), like Eq.(2-44), gives noinformation astothedirection ofmotion. This lack ofknowledge ofdirection ismuch more serious intwoandthree dimensions, where there isaninﬁnity ofpossible directions, than inonedimension, where there areonly two opposite directions inwhich theparticle may move. Inonedimension, there isonly onepath along which theparticle may move. Intwoorthree dimensions, there aremany paths, andunless weknow thepath oftheparticle, Eq.(3—194) alone allows ustosayvery little about themotion except that itcanoccur only intheregion where V(x, y,z)5E.Asanexample, thepotential energy ofanelectron inthe attractive electric ﬁeldof twoprotons (ionized hydrogen molecule H2+) is V=_92-ii.(esu) (3-19.7) 1'_1 1'21 1 \ 1 J 1 116 MOTION orPARTICLE INTWO oRTHREE DIMENSIONS [CHAP. 3 --18 -23 -28 -37 _46 ' “'46 _60 . "60 _92 —-92 -184 2A 184 -1.10 —eO FIG. 3-31. Potential energy ofelectron inelectric ﬁeld oftwoprotons 2A apart. (Potential energies inunits of10-12 erg.) where r1,r2arethedistances oftheelectron from thetwoprotons. The function V(x, y)(formotion inthexy-plane only) isplotted inFig. 3-31 asacontour map, where thetwoprotons are2Aapart atthepoints y=0, x=:1;1A,andtheﬁgures onthecontours ofconstantpotential energy are thecorresponding potential energies inunits of10*” erg. Solong as E<—46 X10*” erg,theelectron isconﬁned toa‘region around one proton ortheother, andweexpect itsmotion willbeeither anoscillation through theattracting center oranorbit around it,depending oninitial conditions. (These coimnents ontheexpected motion require some physical insight orexperience inaddition towhat wecansayfrom theenergy integral alone.) For0>E>-46 X10-12 erg, theelectron isconﬁned toa region which includes both protons, andavariety ofmotions arepossible. ForE>O,theelectron isnotconﬁned toanyﬁnite region intheplane. ForE<<-46 X10_12 erg,theelectron isconﬁned toaregion where the equipotentials arenearly circles about oneproton, anditsmotion willbe practically thesame asiftheother proton were notthere. ForE<0, but[El<<46X10_12 erg,theelectron may circle inanorbit farfrom the attracting centers, anditsmotion then willbeapproximately that ofan electron bound toasingle attracting center ofcharge 2e,astheequi- potential lines farfrom theattracting centers areagain very nearly circles. 3-12] POTENTIAL ENERGY 117 Given apotential energy function V(x, y,2),Eq.(3-186) enables usto compute thecomponents ofthecorresponding force atanypoint. Con- versely, given aforce F(x,y,z),wemay compute itscurl todetermine whether apotential energy function exists forit.Ifallcomponents of curlFarezero within anyregion ofspace, then within that region, Fmay berepresented interms ofapotential-energy function as—VV. The potential energy istobecomputed from Eq.(3-183). Furthermore, since curlF=0,theresult isindependent ofthepath ofintegration, andwe may compute theintegral along anyconvenient path. Aanexample, consider thefollowing twoforce functions: (a)F,=awy, F,=—az2, F,=—a:z;2, (b)Ft=as/(yz —322), Ft=3aw(z/2 —z”),F.=-6w/Z, where aisaconstant. Wecompute thecurlineach case: .6F, BF .6F, 6F, 6F BF,fa)""F=‘(w"'a7")+1(T>?"%)+k(ﬁ—a7) =(2az)i +(2¢w)i —(wk, (b)VexF=0. Incase (a)nopotential energy exists. Incase (b)there isapotential energy function, andweproceed toﬁndit.Letustake r,=0,i.e.,take thepotential aszeroattheorigin. Since thecomponents offorce aregiven asfunctions ofx,y,z,thesimplest path ofintegration from (0,0,0)to (wo,yo,zo)along which tocompute theintegral inEq.(3-183) isonewhich follows lines parallel tothecoordinate axes, forexample asshown in Fig.3-32: (10410-=0)V(x0,y0,z0) =-/(OM) F-dr=-/CIF-dr -ICZF-dr -/can-dr. Z ($0,!/0,10) Cs (09.0)-—-z/ C1 (xo,0,0) C2 (110.1/0.0) 1; 4 FIG. 3-32. Apath ofintegration from (0,0,0)to(210,yo,20)."1 J l 1l < 4 118 MOTION or,PARTICLE INTwo oRTHREE DIMENSIONS [cn.u=. 3 Now along C1,wehave y=z=O, F,,=F,,=F,=0, dr=idx. Thus 0:07 _/'ClF-dr _/0I',dx_0. Along C2, :1:=xo, z=0, F,=aya, F,=stay”, F,=0, dr=jdy. , Thus Z U0 , Z 3[C2Fdr I0I11,dy axoyo. Along C3, it=$0: y=1'/0; Ft=az/o(z/3 —3?), Ft=3¢1wo(@/3 -Z2), F.=—-6¢woz/oz, dr=kdz. Thus fF-dr =[20F,dz=—3axoyoz§. U3 0 Thus thepotential energy, ifthesubscript zeroisdropped, is V(r,1/,F)=—<w2/3 +3¢w2/z’- Itisreadily veriﬁed that thegradient ofthisfunction istheforce given by (b)above. Infact, oneway toﬁndthepotential energy, which isoften faster than theabove procedure, issimply totrytoguess afunction whose gradient willgive therequired force. Animportant case ofaconservative force isthecentral force, aforce directed always toward oraway from aﬁxed center O,andwhose magni- tude isafunction only ofthedistance from O.Inspherical coordinates, with Oasorigin, F=nF(r). (3—19S) The cartesian components ofacentral force are(since n=r/r) .18 Fa: =;'F(T)! Ft=%F<r>. F=(F+F’+z*>”21. <3-199) Z F; =;F(T). u 3-12] POTENTIAL ENERGY 119 (To,o0,¢0)C (7'n0s,¢s) ----' F2 C1 (7'0;o3r¢8),¢ \\\\\ an’ ¢I, 0 ,1 I ,*'\ _¢' T“~~_ --~''-. FIG. 3-33. Path ofintegration foracentral force. The curlofthisforce canbeshown bydirect computation tobezero, no matter what thefunction F(r) may be.Forexample, weﬁnd aF,_d(Fm) Br_myd(Fm) 8y—xdr r8;/— rdr r’ 6F,,=yd(F(r)) 6r=xyd(F(r))_ 6:1: dr r6:1: rdr r Therefore thez-component ofcurlFvanishes, andso,likewise, dotheother twocomponents. Tocompute thepotential energy, wechoose anystand- ardpoint r,,andintegrate from r,toroalong apath (Fig. 3-33) following aradius (C1) from r,,whose coordinates are(r,,0,,<p,),tothepoint (ro,0,,<p,),then along acircle (C2) ofradius r0about theorigin tothe point‘ (T01 00: 900)‘ Along C1; dr=ndr, fF-dr=['°F(r)dr. C1 1', Along 0'2, dr=lrd0-1-mrsin0d<p, /CF-dr=0. 2 1 1 Thus s V(ro)=—/7:” F-dr=-/Cir-Fr -fC2F-dr T0 1=-F(r)dr. I 1 Thepotential energy isafunction ofralone: - 1rV(r)=V(r)=-F(r)dr. (3-200) 120 MOTION orPARTICLE INTWO oRTHREE DIMENSIONS [cILu=. 3 3-13 Motion under acentral force. Acentral force isaforce oftheform given byEq.(3—198). Physically, such aforce represents anattraction [ifF(r) <0]orrepulsion [ifF(r) >0]from aﬁxed point located atthe origin r=0.Inmost cases where twoparticles interact with each other, theforce between them is(atleast primarily) acentral force; that is,if either particle belocated attheorigin, theforce ontheother isgiven by Eq.(3—198). Examples ofattractive central forces arethegravitational force acting onaplanet duetothesun, ortheelectrical attraction acting onanelectron duetothenucleus ofanatom. Theforce between aproton oranalpha particle andanother nucleus isarepulsive central force. In themost important cases, theforce F(r) isinversely proportional tor2. This case willbetreated inthenext section. Other forms ofthefunction F(r)occur occasionally; forexample, insome problems involving thestruc- ture andinteractions ofnuclei, complex atoms, and molecules. Inthis section, wepresent thegeneral method ofattack ontheproblem ofapar- ticle moving under theaction ofacentral force. Since inallthese examples, neither ofthetwo interacting particles isactually fastened toaﬁxed position, theproblem wearesolving, like most problems inphysics, represents anidealization oftheactual problem, valid when oneoftheparticles canberegarded aspractically atrestat theorigin. This willbethecaseifoneoftheparticles ismuch heavier than theother. Since theforces acting onthetwoparticles have thesame magnitude byNewton’s third law,theacceleration oftheheavy onewill bemuch smaller than that ofthelighter one,andthemotion oftheheavy particle canbeneglected incomparison with themotion ofthelighter one. Weshall discover later, inSection 4-7,that, with aslight modiﬁcation, our solution canbemade toyield anexact solution totheproblem ofthe motion oftwointeracting particles, even when their masses areequal. Wemay note that thevector angular momentum ofaparticle under theaction ofacentral force isconstant, since thetorque is N=rXF=(rXn)F(r) =O. (3—201) Therefore, byEq.(3-144), dLE-O. (3-202) Asaconsequence, theangular momentum about any axis through the center offorce isconstant. Itisbecause many physical forces arecentral forces that theconcept ofangular momentum isofimportance. Insolving forthemotion ofaparticle acted onbyacentral force, we ﬁrstshow that thepath oftheparticle liesinasingle plane containing the center offorce. Toshow this,lettheposition 1'0andvelocity vobegiven atanyinitial time to,andchoose thex-axis through theinitial position ro 3-13] MOTION UNDER ACENTRAL FORCE 121 oftheparticle, andthez—axis perpendicular totheinitial velocity vo.Then wehave initially: x0 2 11-01: 2/0 Z Z0 = 0: 12,0=vo-i, 22,0=vo-j, 22,,=0. (3-204) The equations ofmotion inrectangular coordinates are,byEqs. (3-199), maii=§F(r), mji]=f_lF(1~), m2=fro). (s-205) Asolution ofthez-equation which satisﬁes theinitial conditions onzoand 11,0is z(t)=0. (3-206) Hence themotion takes place entirely inthemy-plane. Wecanseephys- ically that iftheforce onaparticle isalways toward theorigin, theparticle cannever acquire anycomponent ofvelocity outoftheplane inwhich it isinitially moving. Wecanalso regard thisresult asaconsequence of theconservation ofangular momentum. ByEq. (3—202), thevector L=m(rXv)isconstant; therefore both randvmust always lieina ﬁxed plane perpendicular toL. Wehave nowreduced theproblem tooneofmotion inaplane withtwo differential equations andfourinitial conditions remaining tobesatisﬁed. Ifwechoose polar coordinates r,6intheplane ofthemotion, theequations ofmotion inthe1'and0directions are,byEqs. (3-80) and(3—198), mi‘-W02=F(r), (3-207) mrd-1-2m1‘0 =0. (3-208) Multiplying Eq.(3—208) by1',asinthederivation ofthe(plane) angular momentum theorem, wehave d 2-_€l_£_ _ a(mr 0)-—dt-0. (3209) This equation expresses theconservation ofangular momentum about the origin andisaconsequence also ofEq.(13-202) above. Itmay beinte- grated togive theangular momentum integral oftheequations ofmotion: mr20 =L=aconstant. (3—210) The constant Listobeevaluated from theinitial conditions. Another integral ofEqs. (3-207) and(3—208), since theforce isconservative, is T+V=%m1‘2+gmﬁoz +v(t)=E, (3-211')1 122 MOTION orPARTICLE INTwo oRTHREE DIMENSIONS [cmua 3» where V(r) isgiven byEq.(3-200) andEistheenergy constant, tobe evaluated from theinitial conditions. Ifwesubstitute for0from Eq. (3—210), theenergy becomes 2%mi*2+ +V(r)=E. (3-212) Wecansolve for1‘: L2 1/2 _1‘=ZE-V(r)-Q-7-W) - (3-213) M Therefore . 1.1' <1 ‘2’L21,2=éz. (3-214) E—V(r) —2————mr2) The integral istobeevaluated andtheresulting equation solved forr(t). Wethen obtain 0(t)from Eq.(3—210): I 0=0+I—L—dt. (3-215)0 0mr2 Wethus obtain thesolution ofEqs. (3-207) and(3-208) interms ofthe fourconstants L,E,ro,0°,which canbeevaluated when theinitial position andvelocity intheplane aregiven. Itwillbenoted that ourtreatment based onEq.(3—212) isanalogous toourtreatment oftheone-dimensional problem based ontheenergy in- tegral [Eq. (2-44)]. The coordinate rhere plays theroleof:0,andthe9 term inthekinetic energy, when 0iseliminated byEq.(3-210), plays the roleofanaddition tothepotential energy. Wemay bring outthisanalogy further bysubstituting from Eq.(13-210) intoEq.(3-207): mi‘_Ii=F(r) (3-216)mr3 ' Ifwetranspose theterm —L2/mri’ totheright side, weobtain .. L2mr=F(r) -1--—- - (3-217)'mr3 This equation hasexactly theform ofanequation ofmotion inonedimen- sionforaparticle subject totheactual force F(r)plusa“centrifugal force” L2/mr3. Thecentrifugal force isnotreallya force atallbutapart ofthe mass times acceleration, transposed totheright side oftheequation in order toreduce theequation forrtoanequation ofthesame form asfor one-dimensional motion. Wemay callita“ﬁctitious force.” Ifwetreat 3-13] MOTION UNDER ACENTRAL FORCE 123 Eq.(3-217) asaproblem inone-dimensional motion, theeffective “poten- tialenergy” corresponding tothe“force” ontheright is ‘V’(r) =—/F(r) dr— drmr3 2=v(t)+ (3-21s) The second term in‘V’isthe“potential energy” associated with the “centrifugal force.” The resulting energy integral isjust Eq. (3—212). The reason why wehave been able toobtain acomplete solution toour problem based ononly twointegrals, orconstants ofthemotion (LandE), isthat theequations ofmotion donotcontain thecoordinate 0,sothat the constancy ofLissuﬂicient toenable ustoeliminate 0entirely from Eq. (3—207) andtoreduce theproblem toanequivalent problem inone-dimen- sional motion. Theintegral inEq.(3—214) sometimes turns outrather difficult toeval- uate inpractice, andtheresulting equation difﬁcult tosolve forr(t). Itis sometimes easier toﬁndthepath oftheparticle inspace than toﬁndits motion asafunction oftime. Wecandescribe thepath oftheparticle by giving lr(0). Theresulting equation issomewhat simpler ifwemake the substitution u=%, r=%- (3-219) Then Wehave, using Eq.(3-210), -___1_%-__2 E!T” uzdoo“ Tide ,Ldu __--7-n-E6» (3220) ,_ Ldzu L2u2 dzu"-"2;W6--7.?.102 13-221) Substituting forrandFinEq.(3—217), andmultiplying by—m/(L2u2), we have adifferential equation forthepath ororbit interms ofu(0): dzu__ m(1) _W--“-W1’ t (3222) IncaseL=0,Eq.(3—222) blows up,butweseefrom Eq.(3-210) that in thiscase 0isconstant, andthepath isastraight linethrough theorigin. 124 MOTION orPARTICLE INTWO ORTHREE DIMENSIONS ICHAP. 3 Even incases where theexplicit solutions ofEqs. (3-214) and(3—215), orEq.(3-222), aredifficult tocarry through, wecanobtain qualitative information about thermotion from theeffective potential ‘V’given by Eq.(3—218), justasintheone-dimensional case discussed inSection 2-5. Byplotting ‘V’(r), wecandecide forany total energy Ewhether the motion inrisperiodic oraperiodic, wecanlocate theturning points, and wecandescribe roughly how thevelocity 1‘varies during themotion. If ‘V’(r)hasaminimum atapoint ro,then forenergy Eslightly greater than ‘V’(T9), rmay execute small, approximately harmonic oscillations about 7'9 with angular frequency given by 1d2‘V’‘"2=E ' ‘3'223)7'0 [See thediscussion inSection 2-7concerning Eq.(2-87).] Wemust re- member, ofcourse, that atthesame time theparticle isrevolving around thecenter offorce with anangular velocity L9-—,r$- (3—224) Therateofrevolution decreases asrincreases. When thermotion is periodic, theperiod ofthermotion isnot,ingeneral, thesame asthe period ofrevolution, sothattheorbit may notbeclosed, although itis conﬁned toaﬁnite region ofspace. (See Fig. 3-34.) Incases where the rmotion isnotperiodic, then 0—>0asr——>oo,andtheparticle may or may notperform oneormore complete revolutions asitmoves toward r=oo,depending onhow rapidly rincreases. Intheevent themotion is periodic, that is,when theparticle moves inaclosed orbit, theperiod of orbital motion isrelated tothearea oftheorbit. This canbeseen as follows. Thearea swept outbytheradius from theorigin totheparticle re) FIG. 3-34. Anaperiodic bounded orbit. 3-14] ‘ INvERsE SQUARE LAW FORCE 125 FIG. 3-35. Area swept outbyradius vector. when theparticle moves through asmall angle d0isapproximately (Fig. 3-35) as=15%d0. (3-225) Hence therateatwhich area isswept outbytheradius is,byEq.(3—210), M dS . Lif=%'r0=%- (3—226) This result istrue foranyparticle moving under theaction ofacentral force. Ifthemotion isperiodic, then, integrating over acomplete period 1' ofthemotion, wehave forthearea oftheorbit I/rS—5;- (3—227) Iftheorbit isknown, theperiod ofrevolution canbecalculated from this formula. 3-14 The central force inversely proportional tothesquare ofthe distance. The most important problem inthree-dimensional motion is that ofamass moving under theaction ofacentral force inversely pro- portional tothesquare ofthedistance from thecenter: . F=gn, (3-228) forwhich thepotential energy is v(t)= (3-229) where thestandard radius r,istaken tobeinﬁnite inorder toavoid an additional constant term inV(r). Asanexample, thegravitational force (Section 1-5) between twomasses mlandm2adistance rapart isgiven by Eq. (3—228) with _ _K=—Gm1m2, G=6.67 X10-8 dyne-gm_2-cm2, (3—230) where Kisnegative, since thegravitational force isattractive. Another 126 MOTION OFPARTICLE INTWO 0RTHREE DIMENSIONS ICHAP. 3 (V10) K>O K=0 To it 7' K<QL¢0 —%(K2m/L2) K<uL=0 FIG. 3-36. Effective potential forcentral inverse square lawofforce. example istheelectrostatic force between twoelectric charges q;andQ2a distance rapart, given byEq.(3-228) with K=Q1112, (3"231) where thecharges areinelectrostatic units, andtheforce isindynes. The electrostatic force isrepulsive when qlandqzhave thesame sign, other- wise attractive. Historically, theﬁrst problems towhich Newton’s me- chanics was applied were problems involving themotion oftheplanets under thegravitational attraction ofthesun, andthemotion ofsatellites around theplanets. The success ofthetheory inaccounting forsuch motions wasresponsible foritsinitial acceptance. Weﬁrst determine thenature oftheorbits given bytheinverse square lawofforce. InFig.3-36 isplotted theeffective potential ., K L2 Forarepulsive force (K>0),there arenoperiodic motions inr;only positive total energies Earepossible, andtheparticle comes infrom r=oo toaturning point andtravels outtoinﬁnity again. Foragiven energy andangular momentum, theturning point occurs atalarger value ofrthan forK=0(noforce), forwhich theorbit would beastraight line. Foran attractive force (K<0)with L;é0,themotion isalso unbounded if E>0,butinthiscasetheturning point occurs atasmaller value ofrthan forK=0.Hence theorbits areasindicated inFig. 3-37. The light 3-14] INvERsE SQUARE LAW FORCE 127 K>O K=0 O K<0 FIG. 3-37. Sketch ofunbounded inverse square laworbits. lines inFig.3-37 represent theturning point radius orperihelion distance measured from thepoint ofclosest approach oftheparticle totheattracting orrepelling center. ForK <0,and—%K2m/L2 <E<0,thecoordinate roscillates between twotin-ning points. ForE=—1}K2m/L2, theparticle moves inacircle ofradius ro=L2/(—Km). Computation shows (see Problem 30attheendofthischapter) thattheperiod ofmalloscillations inristhesame astheperiod ofrevolution, sothat forEnear —§~K2m/L2, theorbit isaclosed curve with theorigin slightly offcenter. Weshall show later that theorbit is,infact, anellipse forallnegative values ofEif L;é0.IfL=0,theproblem reduces totheone-dimensional motion of afalling body, discussed inSection 2-6. Toevaluate theintegrals inEqs. (3-214) and (3—215) fortheinverse square lawofforce israther laborious. Weshall ﬁndthat wecanobtain all theessential information about themotion more simply bystarting from Eq.(3-222) fortheorbit. Equation (3-222) fortheorbit becomes, inthis case, ' 2%+u=- (3-233) This equation hasthesame form asthat ofaharmonic oscillator (ofunit frequency) subject toaconstant force, where 19here plays theroleoft. Thehomogeneous equation anditsgeneral solution are V2%7‘j+t=0, (3-234) u=Acos(0—00), (3-235)1 1 1 1 128 MOTION orPARTICLE INTwo oRTHREE DIMENSIONS [cH.u>. 3 where A,00arearbitrary constants. Anobvious particular soltgion ofthe inhomogeneous equation (3—233) istheconstant solution Kt=- (3-236) Hence thegeneral solution ofEq.(3—233) is ‘ 1 Ku=;=_$7+Acos(0-0,). (3-237) This istheequation ofaconic section (ellipse, parabola, orhyperbola) with focus atr=0,asweshall presently show. The constant 00determines the,orientation oftheorbit intheplane. The constant A,which may be taken aspositive (since 00isarbitrary), determines theturning points of thermotion, which aregiven by 1 mK 1 mK IfA>—mK/L2 (asitnecessarily isforK>0),then there isonly one turning point, r1,since rcannot benegative. Wecannot have A<mK/L2, since rcould then notbepositive foranyvalue of0.Foragiven E,the tuming points aresolutions oftheequation , KL2 The solutions are 1_ mK mK 2 2mE]1/ 25-""LT+11E1‘) +T2 ' (3-240) 1 mK 2mE]1/25="F"F+T ' Comparing Eq.(3—238) with Eq.(3—240), weseethat thevalue ofAin terms oftheenergy andangular momentum isgiven by 2K2 2EA2=5‘-I-4- + (3-241) The orbit isnow determined interms oftheinitial conditions. Anellipse isdeﬁned asthecurve traced byaparticle moving sothat thesum ofitsdistances from two ﬁxed points F,F’isconstant.* The *For amore detailed treatment ofconic sections, seeW.F.Osgood and W.C.Graustein, Plane andSolid Analytic Geometry. New York: Macmillan, 1938. (Chapters 6,7,8,10.) 3-14] INvERsE SQUARE LAW FORCE 129 In14$ FIG. 3-38. Geometry oftheellipse. points F,F’arecalled thefocioftheellipse. Using thenotation indicated inFig.3-38, wehave r’-1-r=2a, (3-242) where aishalfthelargest diameter (major axis) oftheellipse. Interms ofpolar coordinates with center atthefocus Fandwith thenegative :1:-axis through thefocus F’,thecosine lawgives r'2=r2-1-4a2e2 -1-4raecos0, (3—243) where asisthedistance from thecenter oftheellipse tothefocus. eis called theeccentricity oftheellipse. Ife==0,thefocicoincide andthe ellipse isacircle. Asel—>1,theellipse degenerates intoaparabola or straight linesegment, depending onwhether thefocus F’recedes toin- ﬁnity orremains aﬁnite distance from F.Substituting r’from Eq.(3-242) inEq.(3-243), weﬁnd _a(1-£2) _"1C ’—1J.%a' (3244) This istheequation ofanellipse inpolar coordinates with theorigin atone focus.‘ Ifbishalf thesmallest diameter (minor axis), wehave, from Fig.3-38, b=a(1-8)”? (3-245) The area oftheellipse canbeobtained inastraightforward way byin- tegration: S=1rab. (3-246) Ahyperbola isdeﬁned asthecurve traced byaparticle moving sothat thedifference ofitsdistances from twoﬁxed fociF,F’isconstant (Fig. 3-39). Ahyperbola hastwo branches deﬁned by r’—r=2a (-1-branch), (3—247)r’—r=—2a (—branch).1 -\ 1 1 130 MOTION orPARTICLE INTWO oRTHREE DIMENSIONS [cnzua 3 D \ I’\\ I 1\ I \ , r\ ,,rI r T I F.0 F I\ ‘ F 3; 3‘ ,' \ as-1 |'_a II ‘1 \\ I /I \\/ \ ' \ +branch —branch _ FIG. 3-39. Geometry ofthehyper- FIG-. 3-40. Geometry ofthepara- bola. bola. Weshall callthebranch which encircles Fthe-1-branch (leftbranch inthe ﬁgure), andthebranch which avoids F,the—branch (right branch inthe ﬁgure). Equation (3—243) holds alsoforthehyperbola, buttheeccen- tricity eisnowgreater than one. Theequation ofthehyperbola becomes inpolar coordinates: a(e2 —1)"=111%" (H48) (The -1-sign refers tothe+branch, the—sign tothe—branch.) The asymptotes ofthehyperbola (dotted lines inFig.3-39) make anangle oz with theaxisthrough thefoci,where aisthevalue of0forwhich risinﬁnite: cosoz==1: (3-249) Aparabola isthecurve traced byaparticle moving sothatitsdistance from aﬁxed lineD(thedirectrix) equals itsdistance from aﬁxed focus F. From Fig.3-40, wehave G ’—1+cos0’ (3-250) where aisthedistance from thefocus Ftothedirectrix D. Wecanwrite theequations forallthree conic sections inthestandard form 3 é=B-1-Acos0, (3—251) 3-14] INvERsE SQUARE LAW FORCE 131 where Aispositive, andBandAaregiven asfollows: B>A,ellipse, S 1 eB~ A- <3'252> B=A,parabola, B=l. A=1; Gan G G 0<B<A,hyperbola, -1-branch, 1 e B—- -9 AZ , —-A <B<0,hyperbola, —branch, ThecaseB<—Acannot occur, since rwould then notbepositive forany value of0.Ifweallow anarbitrary orientation ofthecurve with respect totheas-axis, then Eq.(3-251) becomes %=B+Am@—%% Qma where 00istheangle between theas-axis andthelinefrom theorigin tothe perihelion (point ofclosest approach ofthecurve totheorigin). Itwill benoted that inallcases Ae-W - (3—257) Foranellipse orhyperbola, Ba= ' (3—258) Equation (3-23?) fortheorbit ofaparticle ‘under aninverse square law force hastheform ofEq. (3-256) foraconic section, with [ifweuse Eq.(3—241)] __m§B_ Ly (3—259) 1/2(B:+2s@) - 132 MOTION orPARTICLE INTwo onTHREE DIMENSIONS [cHA1>. 3 Theeccentricity oftheorbit, byEq.(3-257), is 2E|L2 1/2 q 5=(1 —|— ' (-3-260) Foranattractive force (K<0),theorbit isanellipse, parabola, orhyper- bola, depending onwhether E<0,E=0,orE>0;ifahyperbola, itis the+branch. Forarepulsive force (K>0),wemust have E>0,and theorbit canonly bethe—branch ofahyperbola. These results agree with ourpreliminary qualitative discussion. Forelliptic andhyperbolic orbits, thesemimajor axisaisgiven by K(Z— ' (3—261) Itiscurious that thisrelation does notinvolve theeccentricity ortheangu- larmomentum; theenergy Edepends only onthesemimajor axisa,and viceversa. Equations (3—260) and(3—261) may beobtained directly from Eq.(3—239) fortheturning points ofthermotion. Ifwesolve thisequa- tionforr,weobtain theturning points _K K.2 L2:|1/2 M—5ET +m ' (H62) Themaximum andminimum radii foranellipse are 7'1,2 =(1(1 :1:6), andtheminimum radius forahyperbola is r1=a(eIF1), (3-264) where theupper sign isforthe-1-branch andthelower sign forthe—- branch. Comparing Eqs. (3—263) and (3—264) with Eq.(3-262), wecan read offthevalues ofaande.Thus ifweknow that thepath isanellipse orhyperbola, wecanﬁndthesizeandshape from Eq.(3-239), which fol- lows from thesimple energy method oftreatment, without going through theexact solution oftheequation fortheorbit. This isauseful point to remember. 3-15 Elliptic orbits. The Kepler problem. Early intheseventeenth century, before Newton’s discovery ofthelaws ofmotion, Kepler an- nounced thefollowing three laws describing themotion oftheplanets, de- duced from theextensive andaccurate observations ofplanetary motions byTycho Brahe: 3-15] ELLIPTIC oRBITs. THE KEPLER PROBLEM 133 (1)Theplanets move inellipses with thesunatonefocus. (2)Areas swept outbytheradius vector from thesuntoaplanet in equal times areequal. (3)Thesquare oftheperiod ofrevolution isproportional tothecube ofthesemimajor axis. The second lawisexpressed byourEq.(3—226), andisaconsequence of theconservation ofangular momentum; itshows that theforce acting on theplanet isacentral force. The ﬁrst lawfollows, aswehave shown, from thefactthat theforce isinversely proportional tothesquare ofthe distance. Thethird lawfollows from thefactthat thegravitational force isproportional tothemass oftheplanet, aswenow show. Inthecase ofanelliptical orbit, wecanﬁndtheperiod ofthemotion from Eqs. (3-227) and(3-246): 2 2 1/2T=%1m1> =2%”-1ra2(1 -8)‘/2= .(3-265) or,using Eq.(3-261), 12=41r2a3 - (3-266) Inthecaseofasmall body ofmass mmoving under thegravitational at- traction [Eq. (3—230)] ofalarge body ofmass M,thisbecomes 2T2=51%a3. (3-267) The coeﬁicient ofIa3isnow aconstant forallplanets, inagreement with Kepler's third law. Equation (3-267) allows usto“weigh” thesun, ifwe know thevalue ofG,bymeasuring theperiod andmajor axisofanyplane- tary orbit. This hasalready been worked outinChapter 1,Problem 9, foracircular orbit. Equation (3—267) now shows that theresult applies alsotoelliptical orbits ifthesemimajor axisissubstituted fortheradius. Wehave shown that Kepler's laws follow from Newton’s laws ofmotion andthelawofgravitation. The converse problem, todeduce thelawof force from Kepler’s laws andthelawofmotion, isaneasier problem, anda very important onehistorically, foritwasinthisway that Newton de- duced thelawofgravitation. Weexpect that themotions oftheplanets should show slight deviations from Kepler's laws, inview ofthefactthat thecentral force problem which wassolved inthelastsection represents an idealization oftheactual physical problem. Intheﬁrst place, aspointed outinSection 3-13, wehave assumed that thesunisstationary, whereas actually itmust wobble slightly duetotheattraction oftheplanets going1 1 1 1 1 1 1 1 1 1 134 MOTION orPARTICLE INTwo oRTHREE DIMENSIONS [CHAP. 3 around it.This effect isvery small, even inthecaseofthelargest planets, andcanbecorrected forbythemethods explained later inSection 4-7. Inthesecond place, agiven planet, saytheearth, isacted onbythegravi- tational pulloftheother planets, aswellasbythesun. Since themasses ofeven theheaviest planets areonly afewpercent ofthemass ofthesun, thiswillproduce small butmeasurable deviations from Kepler’s laws. The expected deviations canbecalculated, andthey agree with thevery pre- ciseastronomical observations. Infact, theplanets Neptune andPluto were discovered asaresult oftheir effects ontheorbits oftheother planets. Observations oftheplanet Uranus forabout sixty years after itsdiscovery in1781 showed unexplained deviations from thepredicted orbit, even after corrections were made forthegravitational effects oftheother known planets. Byacareful andelaborate mathematical analysis ofthedata, Adams andLeverrier were able toshow that thedeviations could beac- counted forbyassuming anunknown planet beyond Uranus, andthey cal- culated theposition oftheunknown planet. The planet Neptune was promptly discovered inthepredicted place. The orbits ofthecomets, which areoccasionally observed tomove in around thesunandoutagain, are,atleast insome cases, very elongated ellipses. Itisnotatpresent known whether anyofthecomets come from beyond thesolar system, inwhich casetheywould, atleast initially, have parabolic orhyperbolic orbits. Even those comets whose orbits areknown tobeelliptical have rather irregular periods duetotheperturbing gravita- tional pull ofthelarger planets near which they occasionally pass. Be- tween close encounters with thelarger planets, acomet willfollow fairly closely apath given byEq.(3—256), butduring each such encounter, its motion willbedisturbed, sothat afterwards theconstants A,B,and 00 willhave values different from those before theencounter. Asnoted inSection 3-13, weexpect ingeneral "that thebounded orbits arising from anattractive central force F(r) willnotbeclosed (Fig. 3-34). Closed orbits (except forcircular orbits) arise only where theperiod of radial oscillations isequal to,orisanexact rational multiple of,theperiod ofrevolution. Only forcertain special forms ofthefunction F(r), ofwhich theinverse square lawisone,willtheorbits beclosed. Any change inthe inverse square law, either achange intheexponent ofroranaddition to F(r) ofaterm notinversely proportional tor2,willbeexpected tolead to orbits that arenotclosed. However, ifthechange isvery small, then the orbits ought tobeapproximately elliptical. Theperiod ofrevolution will then beonly slightly greater orslightly lessthan theperiod ofradial oscilla- tions, and theorbit will beapproximately anellipse whose major axis rotates slowly about thecenter offorce. Asamatter offact, aslow pre- cession ofthemajor axisoftheorbit oftheplanet Mercury hasibeen ob- served, with anangular velocity of41seconds ofarcpercentury, over and 3-16] HYPERBOLIC oRBITs. THE RUTHERFORD PROBLEM 135 above theperturbations accounted forbythegravitational effects ofthe other planets. Itwasonce thought that thiscould beaccounted forby thegravitational effect ofdust inthesolar system, butitcanbeshown that theamount ofdust isfartoosmall toaccount fortheeffect. Itis now fairly certain that theeffect isduetoslight corrections toNewton’s theory ofplanetary motion required bythetheory ofrelativity.* Theproblem ofthemotion ofelectrons around thenucleus ofanatom would bethesame asthat ofthemotion ofplanets around thesun, if Newtonian mechanics were applicable. Actually, themotion ofelectrons must becalculated from thelaws ofquantum mechanics. Before thedis- covery ofquantum mechanics, Bohr’ wasable togive afairaccount ofthe behavior ofatoms byassuming that theelectrons revolve inorbits given byNewtonian mechanics. Bohr’s theory isstilluseful asarough picture ofatomic structuretj 3-16 Hyperbolic orbits. TheRutherford problem. Scattering cross section. The hyperbolic orbits areofinterest inconnectionwith themo- tion ofparticles around thesunwhich may come from orescape toouter space, andalsoinconnection with thecollisions oftwocharged particles. Ifalight particle ofcharge qlencounters aheavy particle ofcharge Q2at rest, thelight particle willfollow ahyperbolic trajectory pasttheheavy particle, according totheresults obtained inSection 3-14. Inthecase ofcollisions ofatomic particles, theregion inwhich thetrajectory bends from oneasymptote totheother isvery small (afewangstrom units or less), andwhat isobserved isthedeﬂection angle O=1r—2a(Fig. 3-41) between thepaths oftheincident particle before andafter thecollision. Figure 3-41 isdrawn forthecaseofarepelling center offorce atF,butthe ﬁgure may also betaken torepresent thecase ofanattracting center at F’.ByEqs. (3-249) and(3-260), 21/2tang =cota =(e2-1)_1/2 = ~ (3—268) Lettheparticle have aninitial speed vo,andletitbetraveling insuch a direction that, ifundeﬁected, itwould passadistance sfrom thecenter of force (F). Thedistance siscalled theimpact parameter forthecollision. Wecanreadily compute theenergy andangular momentum interms of *A.Einstein andL.Infeld, TheEvolution ofPhysics. New York: Simon and Schuster, 1938. (Page 253.) Foramathematical discussion, seeR.C.Tolman, Relativity, Thermodynamics, andCosmology. Oxford: Oxford University Press, 1934. (Section 83.) '1'M.Born, Atomic Physics, tr.byJohn Dougall. New York: Stechert, 1936. (Chapter 5.) F1 1 1 1 1 136 MOTION orPARTICLE INTwo oRTHREE DIMENSIONS [cnA1>. 3 \ TVLVU FIG. 3-41. Ahyperbolic orbit. thespeed andimpact parameter: E=emit, (3-269) L=mvos. (3-270) Substituting inEq.(3-268), wehave forthescattering angle O: tan9==Q - (3-271) 2 msvo Ifalight particle ofcharge qlcollides with aheavy particle ofcharge Q2, thisis,byEq.(23-231), tan9-=11$} (3-272) 2 msvo Inatypical scattefing experiment, astream ofcharged particles may be shot inadeﬁnite direction through athinfoil. Many oftheparticles emerge from thefoilinadifferent direction, after being deﬂected orscat- tered through anangle Cbyacollision with aparticle within thefoil. ToputEq. (3-272) inaform inwhich itcanbecompared with experiment, wemust eliminate theimpact parameter s,which cannot bedetermined experimentally. Intheexperiment, thefraction ofincident particles scat- tered through various angles ®isobserved. Itiscustomary toexpress theresults interms ofacross section deﬁned asfollows. IfNincident particles strike athinfoilcontaining nscattering centers perunit area, the average. number dNofparticles scattered through anangle between 9and 3-16] HYPERBOLIC ORBITS. THE RUTHERFORD PROBLEM 137 A - 3’ do lg”lls FIG. 3-42. Cross section forscattering. O-1-d@isgiven interms ofthecross section do"bytheformula dN7V—_ndc. (3—273) do’iscalled thecross section forscattering through anangle between E)and 6+dE),andcanbethought ofastheeffective areasurrounding thescat- tering center which theincident particle must hitinorder tobescattered through anangle between (E)andO+d®.Forifthere isa“target area.” do"around each scattering center, then thetotal target area inaunit area isndo’. IfNparticles strike oneunit area, theaverage number striking thetarget area isNndc,andthis, according toEq.(3—273), isjustdN,the number ofparticles scattered through anangle between E)andO-1-dE). Now consider anincident particle approaching ascattering center F asinFigs. 3-41 and 3-42. Iftheimpact parameter isbetween sand s-1-ds,theparticle willbescattered through anangle between Oand C-1-d®,where Oisgiven byEq.(3—272), andd®isgiven bythedifferen- tialofEq.(3—272): , 1 __1911121 _— "T/82113 d8- The area ofthering around Fofinner radius s,outer radius s-1-ds,at which theincident particle must beaimed inorder tobescattered through anangle between OandC-1-d8,is .. do"=21rsds. (3—275) Substituting forsfrom Eq.(3—272), andfordsfrom Eq.(3-274) (omitting 138 MOTION orPARTICLE INTwo oRTHREE DIMENSIONS [cHAP. 3 do=472%2 do. (3-276)thenegative sign), weobtain This formula canbecompared with do’determined experimentally asgiven byEq.(3-273). Formula (3-276) wasdeduced byRutherford andused in interpreting hisexperiments onthescattering ofalpha particles bythin metal foils. Hewasable toshow that theformula agrees with hisexperi- ments with ql=2e(charge onalpha particle),* andQ2=Ze(charge on atomic nucleus), solong astheperihelion distance (a asinFig.3-41) is larger than about 10-12 cm,which shows that thepositive charge onthe atom must beconcentrated within aregion ofradius lessthan 10"” cm. This wastheorigin ofthenuclear theory oftheatom. The perihelion distance canbecomputed from formula (3-262) orbyusing theconserva- tionlaws forenergy andangular momentum, andisgiven by 21/21,=l21%[1 +<1+ (3-277) mqiqz Thesmallest perihelion distance forincident particles ofagiven energy occurs when L=0(s=0),andhasthevalue 1-1......=$’%- 13-218) Hence ifthere isadeviation from Coulomb’s lawofforce when thealpha particle grazes orpenetrates thenucleus, itshould show upﬁrstasadevia- tionfrom Rutherford’s law[Eq. (3-276)] atlarge angles ofdeﬂection C,and should show upwhen theenergy Eislarge enough sothat E>1%. (3-279)1‘0 where roistheradius ofthenucleus. Theearliest measurements ofnuclear radiiwere made inthiswaybyRutherford, andturnouttobeoftheorder ofl0‘12 cm. The above calculation ofthecross section isstrictly correct only when thealpha particle impinges onanucleus much heavier than itself, since the scattering center isassumed toremain ﬁxed. This restriction canbere- moved bymethods tobediscussed inSection 4-8. Alpha particles also collide with electrons, buttheelectron issolight that itcannot appreciably deﬂect thealpha particle. Thecollision ofanalpha particle with anucleus *Here estands forthemagnitude oftheelectronic charge. 3-17] MoTIoN orAPARTICLE INANELECTRoMAGNETIc FIELD 139 should really betreated bythemethods ofquantum mechanics. Thecon- ceptofadeﬁnite trajectory withadeﬁnite impact parameter sisnolonger valid inquantum mechanics. Theconcept ofcross section isstillvalid in quantum mechanics, however, asitshould be,since itisdeﬁned interms of experimentally determined quantities. The ﬁnal result forthescattering cross section turns outthesame asourformula (3-276). *Itisafortunate coincidence inthehistory ofphysics that classical mechanics gives theright answer tothisproblem. 3-17 Motion ofaparticle inanelectromagnetic field. Thelawsdeter- mining theelectric and magnetic ﬁelds duetovarious arrangements of electric charges and currents arethesubject matter ofelectromagnetic theory. Thedetermination ofthemotions ofcharged particles under given electric andmagnetic forces isaproblem inmechanics. Theelectric force onaparticle ofcharge qlocated atapoint ris F=qE(r), (3-280) where E(r) istheelectric ﬁeld intensity atthepoint r.The electric ﬁeld intensity may beafunction oftime aswell asofposition inspace. The force exerted byamagnetic ﬁeld onacharged particle atapoint rdepends onthevelocity voftheparticle, andisgiven interms ofthemagnetic induction B(r) bytheequationzj F=gv><B(r), (3-281) where c=3X101° cm/sec isthevelocity oflight, andallquantities are ingaussian units, i.e.,qisinelectrostatic units, Binelectromagnetic units (gauss), andvandFareincgsunits. Inmks units, theequation reads F=qvXB(r). (3-282) Equation (3-280) holds foreither gaussian ormks units. Weshall base ourdiscussion onEq.(3-281) (gaussian units), buttheresults arereadily transcribed intomksunits byomitting cwherever itoccurs. Thetotal electromagnetic force acting onaparticle duetoanelectric ﬁeld intensity E andamagnetic induction Bis F=qE+gvxB. "(3-283) *D.Bohm, Quantum Theory. New York: Prentice-Hall, 1951. (Page 537). 1'G.P.Harnwell, Principles ofElectricity andElectromagnetism, 2nded.New York: McGraw-Hill, 1949. (Page 302.) 140 MOTION orPARTICLE INTwo onTHREE DIMENSIONS [cHAP. 3 Ifanelectric charge moves near thenorth pole ofamagnet, themagnet willexert aforce onthecharge given byEq.(3—281); andbyNewton’s third lawthecharge should exert anequal and opposite force onthe magnet. This isindeed found tobethecase, atleast when thevelocity oftheparticle issmall compared with thespeed oflight, ifthemagnetic ﬁeld duetothemoving charge iscalculated andtheforce onthemagnet computed. However, since themagnetic induction Bisdirected radially away from thepole, andtheforce Fisperpendicular toB,theforces on thecharge andonthepolearenotdirected along thelinejoining them, asinthecase ofacentral force. Newton’s third lawissometimes stated inthe“strong” form inwhich action andreaction arenotonly equal and opposite, butaredirected along thelinejoining theinteracting particles. Formagnetic forces, thelawholds only inthe“weak” form inwhich nothing issaidabout thedirections ofthetwoforces except that they areopposite. This istrue notonly oftheforces between magnets andmoving charges, butalsoofthemagnetic forces exerted bymoving charges ononeanother. Ifthemagnetic ﬁeld isconstant intime, then theelectric ﬁeld intensity canbeshown tosatisfy theequation vxE=o. (3-284) The proof ofthisstatement belongs toelectromagnetic theory andneed notconcern ushere.*. Wenote, however, that thisimplies that forstatic electric andmagnetic ﬁelds, theelectric force onacharged particle iscon- servative. Wecantherefore deﬁne anelectric potential ¢(r)=~{E-dr, (3-285) such that E=—V¢. (3—286) Since Eistheforce perunit charge, ¢willbethepotential energy perunit charge associated with theelectric force: V(r)=q¢(I)- (3-237) Furthermore, since themagnetic force isperpendicular tothevelocity, it candonowork onacharged particle. Consequently, thelawofcon- servation ofenergy holds foraparticle inastatic electromagnetic ﬁeld: . T+q¢=E, (3—288) where Eisaconstant. *Harnwell, op.cit.(Page 340.) 3-17] MOTION orAPARTICLE INANELECTRCMAGNETIC FIELD 141 Agreat variety ofproblems ofpractical andtheoretical interest arise involving themotion ofcharged particles inelectric andmagnetic ﬁelds. Ingeneral, special methods ofattack must bedevised foreach type of problem. Weshall discuss twospecial problems which areofinterest both fortheresults obtained andforthemethods ofobtaining those results. Weﬁrstconsider themotion ofaparticle ofmass m,charge q,inauni- form constant magnetic ﬁeld. Letthez-axis bechosen inthedirection of theﬁeld, sothat B(r,t)=Bk, (3—289) where Bisaconstant. Theequations ofmotion arethen, byEq.(3-281), mi?=%y, my=-—%:t, mé=0. (3-290) According tothelastequation, the2-component ofvelocity isconstant, andweshall consider thecase when v,=0,andthemotion isentirely in thexy-plane. The ﬁrst twoequations arenothard tosolve, butwecan avoid solving them directly bymaking useoftheenergy integral, which in thiscasereads emf=E. (3-291) Theforce isgiven by: F=%vxk, (3-292) F=%2- (3-293) The force, and consequently theacceleration, istherefore ofconstant magnitude and perpendicular tothevelocity. Aparticle moving with constant speed vandconstant acceleration aperpendicular toitsdirection of1motion moves inacircle ofradius rgiven byEq.(3-80): 2a=re”=5=5- (3-294) T Tn Wesubstitute forFfrom Eq.(3-293) andsolve forr: cmv=-—- 3-295 rqB ( ) Theproduct Bristherefore proportional tothemomentum andinversely proportional tothecharge. 142 MOTION orPARTICLE INTwo 0RTHREE DIMENSIONS [cnAP. 3 This result hasmany practical applications. Ifacloud chamber isplaced ina uniform magnetic ﬁeld, onecanmeasure themomentum ofacharged particle by measuring theradius ofcurvature ofitstrack. The same principle isused ina beta-ray spectrometer tomeasure themomentum ofafastelectron bythecurva- ture ofitspath inamagnetic ﬁeld. Inamass spectrometer, aparticle isac- celerated through aknown difference ofelectric potential, sothat, byEq.(3-288), itskinetic energy is %"W2 =q(¢o -¢1)- (3-296) Itisthen passed through auniform magnetic ﬁeldB.Ifqisknown, andr,B, (qﬁg-—¢1)aremeasured, wecaneliminate vbetween Eqs. (3—295) and(3—296), andsolve forthemass: m=ii. (3-297)2¢2(4>o -—¢1) There aremany variations ofthis basic idea. The historic experiments of J.J.Thomson which demonstrated theexistence oftheelectron were essentially ofthistype, andbythem Thomson succeeded inshowing that thepath traveled byacathode rayisthat which would befollowed byastream ofcharged particles, allwith thesame ratio q/m. Inacyclotron, charged particles travel incircles in auniform magnetic ﬁeld, andreceive increments inenergy twice perrevolution bypassing through analternating electric ﬁeld. Theradius rofthecircles there- foreincreases, according toEq.(3—295), until amaximum radius isreached, at which radius theparticles emerge inabeam ofdeﬁnite energy determined by Eq.(3-295). Thefrequency vofthealternating electric ﬁeldmust bethesame asthefrequency vofrevolution oftheparticles, which isgiven by v=2JI'7‘1I. (3—298) Combining thisequation with Eq.(3—295), Wehave _2. _ 11_2mm (3299) Thus ifBisconstant, visindependent ofr,andthisisthefundamental principle onwhich theoperation ofthecyclotron isbased.* Inthebetatron, electrons travel incircles, and themagnetic ﬁeld within thecircle ismade toincrease. Since Bischanging with time, VXEisnolonger zero; thechanging magnetic ﬂuxinduces avoltage around thecircle such thatanetamount ofwork isdone ontheelectrons bytheelectric ﬁeld asthey travel around thecircle. Thebetatron issodesigned that theincrease ofBattheelectron orbit isproportional tothe increase ofmv,sothat rremains constant. *According tothetheory ofrelativity, themass ofaparticle increases with velocity atvelocities near thespeed oflight, and consequently thecyclotron cannot accelerate particles tosuch speeds unless visreduced orBisincreased as theparticle velocity increases. [Itturns outthat Eq.(3-295) stillholds in relativity theory.] 3-17] MOTION orAPARTICLE INANELECTROMAGNETIC FIELD 143 Finally, weconsider aparticle ofmass m,charge q,moving inauniform constant electric ﬁeld intensity Eandauniform constant magnetic induc- tion B.Again letthez-axis bechosen inthedirection ofB,andletthe y-axis bechosenso that Eisparallel totheyz-plane: B=Bk, E=E,,j+E,k, (3~300) where B,E”,E,areconstants. Theequations ofmotion, byEq.(3—283), are mi=5?y, (3-301) mi]=-9651+qE,,, (3-302) m2=qE,. (3—303) Thez-component ofthemotion isuniformly accelerated: Z=Z0+20:+%1%K. (3-304) Tosolve the:1:andyequations, wedifferentiate Eq.(3—301) andsubstitute inEq.(3—302) inorder toeliminate ii. 2 B _.-2"?‘=- +qE,,. (3-305) Bymaking thesubstitutions (.0=E, (s-306)mc _¥1_1"71/ _ a-m, (3307) wecanwrite Eq.(3—305) intheform d2- .W’?+0:22;=aw. (3-308) This equation hasthesame form astheequation foraharmonic oscillator with angular frequency wsubject toaconstant applied “force” aw,except that atappears inplace ofthecoordinate. The corresponding oscillator problem wasconsidered inChapter 2,Problem 33.The solution inthis case willbe a sit=5—|—A,,cos(wt+0,), (3—309) where A,,and0,,arearbitrary constants tobedetermined. Byeliminating 513 144 MOTION orPARTICLE INTwo 0RTHREE DIMENSIONS [CHAP. 3 from Eqs. (3—30l) and(3—302), inasimilar way, weobtain asolution for1]: ~ 1]=A,cos(wt+0,). (3—3l0) ’ » Wegetasandybyintegrating Eqs. (3-309) and(3—310): x=0,+if+$8111(wt+0,), (3-311) y=_0,,+%sin(wt+0,). (3-312) Now adiﬁiculty arises, forwehave sixconstants A,,A,,,0,,0,,C',,and(7,, tobedetermined, and only four initial values wo,yo,so,gotodetermine them. Thetrouble isthatWeobtained thesolutions (3—311) and(3—3l2) by differentiating theoriginal equations, anddifferentiating anequation may introduce new solutions that donotsatisfy theoriginal equation. Con- sider, forexample, thevery simple equation x=3. Differentiating, weget 1:;=0, whose solution is :2:=C. Now only foroneparticular value oftheconstant Cwillthissatisfy the original equation. Letussubstitute Eqs. (-3-311) and(3—3l2) or,equiva- lently, Eqs. (Z-S-309) and(3—310) intotheoriginal Eqs. (3—301) and(3—302), using Eqs. (3—306) and(3—307): B. B~q?A,S111(wt+0,)=%A,,cos (wt+0,), (3-313) -$11,,sin(wt+0,)=-161311,cos(wt+0,). (3-314) These twoequations willhold only ifA,,A,,,0,,and0,,arechosen sothat A,=A,,, (3—315) sin(wt+0,)=—-cos (wt—|—0,), (3—316) cos(wt+0,)=sin(wt+0,). (3—317) Thelatter twoequations aresatisﬁed if 0,,=0,+ (3-318) 3-17] MOTION orAPARTICLE INANELECTROMAGNETIC FIELD 145 y ’ WWZ V FIG. 3-43. Orbits inthemy-plane ofacharged particle subject toamagnetic ﬁeld inthe2-direction andanelectric ﬁeld inthey-direction.$ Letusset A,=A,,=wA, (3—319) 0,=o, (3-320) 0,,=0+ (3-321) Then Eqs. (3-31 1)and(3—3l2) become 2:=0,+Asin(wt+06)+ (3-322) y=C,+Acos(wt+0). (3—323) There arenow only four constants, A,0,C',,Cy’tobedetermined bythe initial values xo,yo,5:0,go.Thez-motion is,ofcourse, given byEq.(3—304). IfE,=0,theany-motion isinacircle ofradius Awith angular velocity co about thepoint (0,,Cy); this isthemotion considered intheprevious example. The eﬁect ofE,istoaddtothisuniform circular motion a uniform translation inthex-direction! Theresulting path inthemy-plane willbeacycloid having loops, cusps, orripples, depending ontheinitial conditions andonthemagnitude ofE,(Fig. 3-43). This problem isof interest inconnection with thedesign ofmagnetrons. 146 MOTION orPARTICLE INTWO oRTHREE DIMENSIONS [CHAP. 3 PROBLEMS 1.Prove, onthebasis ofthegeometric deﬁnitions oftheoperations ofvector algebra, thefollowing equations. Inmany cases adiagram willsuffice. (a)Eq. (3-7), (b)Eq.(3-17), (c)Eq.(3-26), (d)Eq.(3-27), *(e) Eq.(3-35). 2.Prove, onthebasis ofthealgebraic deﬁnitions oftheoperations ofvector algebra interms ofcomponents, thefollowing equations: (a)Eq.(3-8), (b)Eq. (3-17), (c)Eq.(3-27), (d)Eq.(3-34), (e)Eq.(3-35). '3.Derive Eq.(3-32) bydirect calculation, using Eq.(3-10) torepresent A andB,andmaking useofEqs. (3-25) to(3-31). 4.(a)Prove thatA-(BXC)isthevolume oftheparallelepiped whose edges areA,B,Cwith positive ornegative signaccording towhether aright-hand screw rotated from Atoward Bwould advance along Cinthepositive ornegative direc- tion. A,B,Careany three vectors notlying inasingle plane. (b)Usethis result toprove Eq.(3-34) geometrically. Verify thattheright andleftmembers ofEq.(3-34) areequal insign aswell asinmagnitude. 5.Prove thefollowing inequalities. Give ageometric andanalgebraic proof (interms ofcomponents) foreach: (=1) |A+Bl3IA!+IBI- (b) IA-Bl SIAIIBI- (c) IAXBl3IAIIBI- 6.(a)Obtain aformula analogous toEq.(3-40) forthemagnitude ofthe sumofthree forces F1,F2,F3,interms ofF1,F2,F3,andtheangles 012,023,031 between pairs offorces. [Usethesuggestions following Eq.(3—40).] (b)Obtain aformula inthesame terms fortheangle a1,between thetotal force andthecomponent force F1. 7.Prove Eqs. (3-54) and(3-55) from thedeﬁnition (3-52) ofvector diﬂ’eren- tiation. 8.Prove Eqs. (3-56) and(3-57 )from thealgebraic deﬁnition (3-53) ofvector differentiation. ;9.Give suitable deﬁnitions, analogous toEqs. (3-52) and(3-53), forthe integral ofavector function A(t) with respect toascalar t:. ta /1A(t)dt- Write asetofequations likeEqs. (3—54)—(3—57) expressing thealgebraic proper- tiesyouwould expect such anintegral tohave. Prove that onthebasis ofeither deﬁnition 0 d&/0 A(t) dt=A(t). 10.A45°isosceles right triangle ABC hasahypotenuse ABoflength 4a.A particle isacted onbyaforce attracting ittoward apoint Oonthehypotenuse a distance afrom thepoint A.Theforce isequal inmagnitude tok/r2, where ris PROBLEMS 147 thedistance oftheparticle from thepoint O.Calculate thework done bythis force when theparticle moves from AtoC’toBalong thetwolegsofthetriangle. Make thecalculation byboth methods, that based onEq.(3-61) andthat based onEq.(3-63). 11.(a)Aparticle inthemy-plane isattracted toward theorigin byaforce F=k/y, inversely proportional toitsdistance from the:2:-axis. Calculate the work done bytheforce when theparticle moves from thepoint :1:=0,y=ato thepoint :v=2a,y=0along apath which follows thesides ofarectangle consisting ofasegment parallel tothea:-axis from z=0,y=ato:1:=2a, y=a,andavertical segment from thelatter point totheav-axis. (b)Calculate thework done bythesame force when theparticle moves along anellipse of semiaxes a,2a.[Hint: Set:1:=2asin0,y=acos0.] 12.(a)Find thecomponents ofd3r/dt3 inspherical coordinates. (b)Find thecomponents ofd2A/dt2 incylindrical polar coordinates, where thevector A isafunction oftandislocated atamoving point. *l3. (a)Plane parabolic coordinates f,haredeﬁned interms ofcartesian coordinates a:,ybytheequations 1=1-h.y=2<rh>"2. where fandharenever negative. Find fandhinterms of2:andy.Letunit vectors f,hbedeﬁned inthedirections ofincreasing fandhrespectively. That is,fisaunitvector inthedirection inwhich apoint would move ifitsf-coordinate increases slightly while itsh-coordinate remains constant. Show thatfandhare perpendicular atevery point. [Hint: f=(idz:+jdy)[(dx)2 +(dy)2]‘1/2, when df>0,dh=0.Why?] (b)Show that fandharefunctions off,h,and ﬁnd their derivatives with respect tofand h.Show that r=fl/2(f—|— h)1/2f—|— hl/2(f —|—h)1/2h. Find thecomponents ofvelocity andacceleration inparabolic coordinates. 14.Aparticle moves along theparabola 1/2=4f3—4fow. where f0isaconstant. Itsspeed visconstant. Find itsvelocity andacceleration components inrectangular andinpolar coordinates. Show thattheequation of theparabola inpolar coordinates is 1'cos2 g=fo. What istheequation ofthisparabola inparabolic coordinates (Problem 13)? 15.Aparticle moves with varying speed along anarbitrary curve lying inthe my-plane. Theposition oftheparticle istobespeciﬁed bythedistance sthepar- ticle hastraveled along thecurve from some ﬁxed point onthecurve. Let-r(s) beaunitvector tangent tothecurve atthe point sinthedirection ofincreasing s. Show that Q_2 ds_r’ 148 MOTION orPARTICLE INTwo 0RTHREE DIMENSIONS [CHAP. 3 where v(s)isaunitvector normal tothecurve atthepoint s,andr(s)istheradius ofcurvature atthepoint s,deﬁned asthedistance from thecurve tothepoint of intersection oftwonearby normals.* Hence derive thefollowing formulas forthe velocity andacceleration oftheparticle: .2 v=é-r, a=I§-r+§7-v. 16.Using theproperties ofthevector symbol V,derive thevector identities: curl(curl A)=grad (divA)—V2A, ‘ ugrad v=grad (uv) —vgrad u. Then write outthezr-components ofeach sideofthese equations andprove by direct calculation that they areequal ineach case. (One must bevery careful, inusing theﬁrstidentity incurvilinear coordinates, totake proper account ofthe dependence oftheunit vectors onthecoordinates.) 17.Calculate curlAincylindrical coordinates. 18.Give asuitable deﬁnition oftheangular momentum ofaparticle about an axisinspace. Taking thespeciﬁed axisasthe2-axis, express theangular momen- tum interms ofcylindrical coordinates. Iftheforce acting ontheparticle has cylindrical components F,,F,,F,,,,prove that thetime rate ofchange ofangular momentum about thez-axis isequal tothetorque about that axis. . 19.Amoving particle ofmass mislocated byspherical coordinates r(t),0(t), <p(t). The force acting onithasspherical components F,,F9,F,. Calculate the spherical components oftheangular momentum vector andofthetorque vector about theorigin, andverify bydirect calculation that theequation dL _ dtTN follows from Newton’s equation ofmotion. 20.Solve forthenext term beyond those given inEqs. (3-177) and (3-178). 21.Aprojectile istobeﬁred from theorigin inthezcz-plane (z-axis vertical) with muzzle velocity 00tohitatarget atthepoint 2:=mo,z=0.(a)Neglect- ingairresistance, ﬁnd thecorrect angle ofelevation ofthegun. Show that, in general, there aretwosuch angles unless thetarget isatorbeyond themaximum range. (b)Find theﬁrst-order correction totheangle ofelevation duetoair resistance. 22.Aprojectile isﬁredfrom theorigin withinitial velocity vo=(v,,,,22,0,v.0). The wind velocity isv,,,=wj.Solve theequations ofmotion (3—180) forx,y,z asfunctions oft.Find thepoint x1,y1atwhich theprojectile willreturn tothe horizontal plane, keeping only‘ ﬁrst-order terms inb.Show that ifairresistance andwind velocity areneglected inaiming thegun, airresistance alone willcause theprojectile tofallshort ofitstarget afraction 4bv,,,/ 3mgofthetarget distance, *W.F.Osgood, Introduction totheCalculus. New York: Macmillan, 1937, p.259. g PROBLEMS 149 and that thewind causes anadditional miss inthey-coordinate ofamount 2bwv§.,/(ma2)- 23.Determine which ofthefollowing forces areconservative, and ﬁnd the potential energy forthose which are: (a) F,=6abz3y -—20ba:3g/2, F,=6abzz3 —10ba:4y, F,=18ab:cz2y. (b) F,=18abyz3 —20b2:3y2, F,=18ab:rz3 —10b:v4y, F,=6a,ba:yz2. (0) F=iF,(:I:) -|-jF,(y) —|—kF,(Z). 24.Determine thepotential energy foranyofthefollowing forces which are conservative : (a) =2aw(z3 +1/3). Ft=2wy(z3 +113)+3ay2(<v2 +yz), =3az2(x2 +(1/2). =apzcos<p, F,=G.p2sin<p, F,=21122. €I€ $131315?‘=—2ar sin0cos<p, F9=—arcos0cos(0, =arsin0sin<p. 25.Aparticle isattracted toward thez-axis byaforce proportional tothe square ofitsdistance from themy-plane andinversely proportional toitsdistance from thez-axis. Add anadditional perpendicular force insuch away asto make thetotal force conservative, andﬁndthepotential energy. Besureto write expressions fortheforces andpotential energy which aredimensionally consistent. 26.Find thecomponents offorce forthefollowing potential-energy functions: (a) V=ax;/2z3. (b) V'=-H012. (c) V=ékwz+lvktz/2+%k-z2- - 27.Find theforce ontheelectron inthehydrogen molecule ionforwhich the potential is e2 82 V=———-—11'1 1'2 where r1isthedistance from theelectron tothepoint y=z=0,x=—a, and1'2isthedistance from theelectron tothepoint y=z=0,:2:=a. 28.Show thatF=nF(r) (where nisaunitvector directed away from the origin) isaconservative force byshowing bydirect calculation that theintegral I"F-dr '1 along anypath between r1and1'2depends only onr1andT2.[Hint:Express F anddrinspherical coordinates.] 150 MOTION orPARTICLE INTWO 0RTHREE DIMENSIONS [CI-IAP. 3 29.The potential energy foranisotropic harmonic oscillator is V=aw. Plot theeffective potential energy forther-motion when aparticle ofmass m moves with this potential energy and with angular momentum Labout the origin. Discuss thetypes ofmotion that arepossible, giving ascomplete adescrip- tion asispossible without carrying outthesolution. Find thefrequency of revolution forcircular motion and thefrequency ofsmall radial oscillations about thiscircular motion. Hence describe thenature oftheorbits which diﬂer slightly from circular orbits. 30.Find thefrequency ofsmall radial oscillations about steady circular mo- tion fortheeffective potential given byEq. (3—232) foranattractive inverse square lawforce, andshow that itisequal tothefrequency ofrevolution. 31.Find r(t),0(t)fortheorbit oftheparticle inProblem 29. Compare with theorbits found inSection 3-10 forthethree-dimensional harmonic oscillator. 32.Aparticle ofmass mmoves under theaction ofacentral force whose poten- tialis Vo)=Kr4, K>0. Forwhat energy andangular momentum willtheorbit beacircle ofradius a about theorigin? What istheperiod ofthiscircular motion? Iftheparticle is slightly disturbed from thiscircular motion, what willbetheperiod ofsmall radial oscillations about r=a? 33.According toYukawa’s theory ofnuclear forces, theattractive force be- tween aneutron andaproton hasthepotential V(r)=%. K<0. (a)Find theforce, andcompare itwith aninverse square lawofforce. (b)Dis- cussthetypes ofmotion which canoccur ifaparticle ofmass mmoves under such aforce. (C)Discuss how themotions willbeexpected todiffer from the corresponding types ofmotion foraninverse square lawofforce. (d)Find L andEformotion inacircle ofradius a.(e)Find theperiod ofcircular motion and theperiod ofsmall radial oscillations. (f)Show that thenearly circular orbits arealmost closed when aisvery small. 34.(a)Discuss bythemethod oftheeffective potential thetypes ofmotion to beexpected foranattractive central force inversely proportional tothecube of theradius: KF(r)=-T-3. K>O. (b)Find theranges ofenergy andangular momentum foreach type ofmotion. (c)Solve theorbital equation (3-222), andshow that thesolution isoneofthe forms: PROBLEMS 151 -=A@0s[ﬁ(6 -00)]. P<1) "ll-lﬁhlil-lib-‘ﬁl—l—=Acosh[B(0—00)]. (2) -=Asinh[;8(0-00)]. (3) -=A(0—00), (4) 1ino (5)-=—e. 1 To (d)Forwhat values ofLandEdoes each oftheabove types ofmotion occur? Express theconstants AandBinterms ofEandLforeach case. (e)Sketch a typical orbit ofeach type. 35.(a)Discuss thetypes ofmotion that canoccur foracentral force K K’F(r)=—,—2+-,§- Assume that K>0,andconsider both signs forK’. (b)Solve theorbital equation, andshow that thebounded orbits have theform (ifL2>—mK') 2 r=a(1—e)_ 1+ecosa0 (c)Show thatthisisaprecessing ellipse, determine theangular velocity ofpre- cession, andstate whether theprecession isinthesame orintheopposite direc- tiontotheorbital angular velocity. 36.Acomet isobserved adistance of1.00X108kmfrom thesun,travel- ingtoward thesunwith avelocity of51.6 kmpersecond atanangle of45° with theradius from thesun. Work outanequation fortheorbit ofthecomet inpolar coordinates with origin atthesunand a:-axis through theobserved position ofthecomet. (The mass ofthesunis2.00 X103° kgm.) 37.Itwillbeshown inChapter 6(Problem 5)that theeffect ofauniform dis- tribution ofdust ofdensity pabout thesunistoaddtothegravitational attrac- tionofthesunonaplanet. ofmass manadditional attractive central force F’=-—mkr, where 41:- 16—-3- PG. (a)Ifthemass ofthesunisM,ﬁndtheangular velocity ofrevolution ofthe planet inacircular orbit ofradius ro,andﬁnd theangular frequency ofsmall radial oscillations. Hence show thatifF’ismuch lessthan theattraction due 152 MOTION orPARTICLE INTWO 0RTHREE DIMENSIONS [CHAP. 3 tothesun,anearly circular orbit willbeapproximately anellipse whose major axisprecesses slowly with angular velocity _2 @112. 0Jp— 1l'[\ M (b)Does theaxisprecess inthesame orintheopposite direction totheorbital angular velocity? Look upMandtheradius oftheorbit ofMercury, andcal- culate thedensity ofdust required tocause aprecession of41seconds ofarc percentury. _38. Itcanbeshown (Chapter 6,Problems 15and 19)that thecorrection to thepotential energy ofamass mintheearth’s gravitational ﬁeld, duetothe oblate shape oftheearth, isapproximately, inspherical coordinates, relative tothepolar axisoftheearth, 2 V’=——-—-—-nm1,_l:_gR (1—3cos2 0), where Misthemass oftheearth and2R,2R(1 —11)aretheequatorial and polar diameters oftheearth. Calculate therate ofprecession oftheperigee (point ofclosest approach) ofanearth satellite moving inanearly circular orbit intheequatorial plane. Look upthemass oftheearth andtheequatorial andpolar diameters, andestimate therateofprecession indegrees perrevolution forasatellite 400miles above theearth. *39. Calculate thetorque onanearth satellite duetotheoblateness potential energy correction given inProblem 38.Asatellite moves inacircular orbit of radius rwhose plane isinclined sothat itsnormal makes anangle awith the polar axis. Assume that theorbit isvery little affected inonerevolution, and calculate theaverage torque during arevolution. Show thattheeffect ofsuch a torque istomake thenormal totheorbit precess inacone ofhalfangle ozabout thepolar axis, andﬁndaformula fortherateofprecession indegrees perrevolu- tion. Calculate therate forasatellite 400miles above theearth, using suitable values forM,17,andR. 40.(a)Asatellite istobelaunched from thesurface oftheearth. Assume theearth isasphere ofradius R,andneglect friction with theatmosphere. The satellite istobelaunched atanangle orwith thevertical, with avelocity v0,so astocoast without power until itsvelocity ishorizontal atanaltitude h1above theearth’s surface. Ahorizontal thrust isthen applied bythelaststage rocket soastoaddanadditional velocity A111tothevelocity ofthesatellite. Theﬁnal orbit istobeanellipse with perigee h1(point ofclosest approach) andapogee hg(point farthest away) measured from theearth’s surface. Find therequired initial velocity v0andadditional velocity A01, interms ofR,a,h1,hg,and g, theacceleration ofgravity attheearth’s surface. (b)Write aformula forthechange 6h1inperigee height duetoasmall error 66intheﬁnal thrust direction, toorder (6/3)2. ' 41.Two planets move inthesame plane incircles ofradii r1,T2about the sun. Aspace probe istobelaunched from planet 1with velocity v1relative totheplanet, soastoreach theorbit ofplanet 2.(The velocity v1istherelative PROBLEMS 153 velocity after theprobe hasescaped from thegravitational ﬁeld oftheplanet.) Show thatv1isaminimum foranelliptical orbit whose perihelion andaphelion arer1andrz.Inthat case, ﬁnd v1,and therelative velocity v2between the space probe andplanet 2iftheprobe arrives atradius T2attheproper time to intercept planet 2.Express your results interms ofr1,r2,andthelength of theyear Y1ofplanet 1.Look uptheappropriate values ofr1and1'2,andesti- mate v1fortrips toVenus andMars from theearth. 42.Arocket isinanelliptical orbit around theearth, perigee r1,apogee 1'2, measured from thecenter oftheearth. Atacertain point initsorbit, itsengine isﬁred forashort time soastogive avelocity increment Avinorder toput therocket onanorbit which escapes from theearth with aﬁnal velocity U0 relative totheearth. (Neglect anyeffects duetothesunandmoon.) Show that Avisaminimum ifthethrust isapplied atperigee, parallel totheorbital velocity. Find Avinthat case interms oftheelliptical orbit parameters e,a, theacceleration gatadistance Rfrom theearth’s center, andtheﬁnal velocity vq.Can youexplain physically why Avissmaller forlarger e? 43.Asatellite moves around theearth inanorbit which passes across the poles. The time atwhich itcrosses each parallel oflatitude ismeasured sothat thefunction 0(t)isknown. Show how toﬁnd theperigee, thesemimajor axis, andtheeccentricity ofitsorbit intermsof 0(t), andthevalue ofgatthesurface oftheearth. Assume theearth isasphere ofradius R. . 44.Itcanbeshown that theorbit given bythespecial theory ofrelativity for aparticle ofmass mmoving under apotential energy V(r) isthesame astheorbit which theparticle would follow according toNewtonian mechanics ifthepoten- tialenergy were [E-v<>12V“)'WT’ where Eistheenergy (kinetic plus potential), andcisthespeed oflight. Discuss thenature oftheorbits foraninverse square lawofforce according tothetheory ofrelativity. Show bycomparing theorbital angular velocity with thefrequency ofradial oscillations fornearly circular motion that thenearly circular orbits, when therelativistic correction issmall, areprecessing ellipses, and calculate theangular velocity ofprecession. ' 45.Aparticle ofmass mmoves inanelliptical orbit ofmajor axis2a,eccentric- itye,insuch away that theradius totheparticle from thecenter oftheellipse sweeps outarea ataconstant rate dSE-0' andwith period -rindependent ofaande.(a)Write outtheequation oftheellipse inpolar coordinates with origin atthecenter oftheellipse. (b)Show that the force ontheparticle isacentral force, andﬁndF(r)interms ofm,1'. 46.Arocket moves with initial velocity votoward themoon ofmass M, radius 1'0.Find thecross section 11forstriking themoon. Take themoon tobe atrest, andneglect allother bodies. 154 MOTION orPARTICLE INTWO oRTHREE DIMENSIONS [CHAP. 3 47.Show thatforarepulsive central force inversely proportional tothecube oftheradius, F(r)=,53, K>O, theorbits areoftheform (1)given inProblem 34,andexpress /3interms ofK,E, L,andthemass moftheincident particle. Show that thecross section for scattering through anangle between G)and G)+d®foraparticle subject to thisforce is ,1,=2'15 "'i—(’9__ d@_ mug ®2(2n' —®)2 48.Avelocity selector forabeam ofcharged particles ofmass m,charge e, istobedesigned toselect particles ofaparticular velocity vo.Thevelocity selector utilizes auniform electric ﬁeld Einthe2:-direction and auniform magnetic ﬁeld Binthey-direction. The beam emerges from anarrow shtalong they-axis and travels inthez-direction. After passing through thecrossed ﬁelds foradistance Z,thebeam passes through asecond slitparallel totheﬁrst andalsointheyz-plane. (a)Ifaparticle leaves theorigin with avelocity 110atasmall angle with the z-axis, ﬁnd thepoint atwhich itarrives attheplane z=Z.Assume that the initial angle issmall enough sothat second-order terms intheangle may be neglected. (b)What isthebestchoice ofE,Binorder thataslarge afraction aspossible oftheparticles with velocity voarrive atthesecond slit,while particles ofother velocities miss theslitasfaraspossible? - (C)Iftheslitwidth ish,what isthemaximum velocity deviation 5vfrom v0forwhich aparticle moving initially along thez-axis canpass through the second sht? Assume that E,Bhave thevalues chosen inpart (b). 49.Aparticle ofcharge qinacylindrical magnetron moves inauniform mag- netic ﬁeld B=Bk, and anelectric ﬁeld, directed radially outward orinward from acentral wire along thez-axis, a E=5h, where pisthedistance from thez-axis, andhisaunitvector directed radially outward from thez-axis. The constants aand Bmay beeither positive or negative. (a)Setuptheequations ofmotion incylindrical coordinates. (b) Show that thequantity B mp2¢+gin2 =K isaconstant ofthemotion. (c)Using thisresult, giveaqualitative discussion, based ontheenergy integral, ofthetypes ofmotion thatcanoccur. Consider all cases, including allvalues ofa,B,K,andE.(c)Under what conditions can circular motion about theaxisoccur? (d)What isthefrequency ofsmall radial oscillations about thiscircular motion? CHAPTER 4 THE MOTION OFASYSTEM OFPARTICLES 4-1Conservation oflinear momentum. Center ofmass. Weconsider inthischapter thebehavior ofmechanical systems containing twoormore particles acted upon byinternal forces exerted bytheparticles upon one another, andbyexternal forces exerted upon particles ofthesystem by agents notbelonging tothesystem. Weassume theparticles tobepoint masses each speciﬁed byitsposition (x,y,z)inspace, likethesingle par- ticle whose motion wasstudied inthepreceding chapter. Letthesystem wearestudying contain Nparticles, andletthem be numbered 1,2,...,N.Themasses oftheparticles wedesignate by m1,mg,...,mN.Thetotal force acting onthekthparticle willbethesum oftheinternal forces exerted onparticle lcbyalltheother (N—1)parti- clesinthesystem, plusanyexternal force which may beapplied toparticle lc.Letthesum oftheintemal forces onparticle lcbeF}-,,andletthetotal extemal force onparticle lobeFi.Then theequation ofmotion ofthe kthparticle willbe ' m;,1‘;,=Fi+F;§, lc=1,2,...,N. (4-1) The Nequations obtained byletting lcinEqs. (4-1) mmover thenum- bers 1,...,Naretheequations ofmotion ofoursystem. Since each of these Nequations isitself avector equation, wehave ingeneral asetof3N simultaneous second-order differential equations tobesolved. The solu- tionwillbeasetoffunctions 1';,(t) specifying themotion ofeach particle in thesystem. The solution willdepend on6N“arbitrary” constants speci- fying theinitial position andvelocity ofeach particle. The problem of solving thesetofequations (4-1) isvery diﬂicult, except incertain special cases, and nogeneral methods areavailable forattacking theN-body problem, even inthecase where theforces between thebodies arecentral forces. The two-body problem canoften besolved, asweshall see,and some general theorems areavailable when theinternal forces satisfy certain conditions. Ifpk=mkvk isthelinear momentum ofthekthparticle, wecanwrite Eqs. (4-1) intheform %‘=F7;+1=,';, k=1,...,N. (4-2) 1551 1 1 156 THEMOTION orASYSTEM orPARTICLES [CHAR 4 Summing theright andleftsides ofthese equations over alltheparticles, wehave Ndpk dN N8 Ni ZW=;,;Zv1.=)'_jF,.+ZFt. (4-3) k=1 lc=1 k=l k=1 Wedesignate byPthetotal linear momentum oftheparticles, andbyF thetotal external force: N N P=ZPk=2mkvk, (4"4)k=1 k=1 NF=ZF5. (4-5) k==1 Wenow make theassumption, tobejustiﬁed below, that thesum ofthe internal forces acting onalltheparticles iszero: N -ZF;=o. (4-c) k=1 When Eqs. (4-4), (4-5), and(4-6) aresubstituted inEq.(4-3), itbecomes dPE_F. (4-7) This isthemomentum theorem forasystem ofparticles. Itstates that the time rate ofchange ofthetotal linear momentum isequal tothetotal external force. Animmediate corollary istheconservation theorem for linear momentum, which states that thetotal momentum Pisconstant when noexternal forces act. i Wenow trytojustify theassumption (4-6). Ourﬁrst proof isbased onNewton’s third law. Weassume that theforce acting onparticle lo duetoalltheother particles canberepresented asasumofseparate forces duetoeach oftheother particles: T‘ Z Ff:—>k; \ (4“8) lsek , where F§_,,, istheforce onparticle kduetoparticle l.According toNew- ton’s third law, theforce exerted byparticle lonparticle Icisequal and opposite tothat exerted bylconZ: F;;-1= —F€_.,.. (4-9) 4-1] CONSERVATION orLINEAR MOMENTUM. CENTER orMASS 157 Equation (4-9) expresses Newton’s third lawinwhat wemay calltheweak form; that is,itsays that theforces areequal andopposite, butdoes not imply that theforces actalong thelinejoining thetwoparticles. Ifwe now consider thesum inEq.(4-6), wehave N . N .ZFZ,=ZZr; ,,. (4-10) -> k=l k=1 laﬁk Thesum ontheright isover allforces acting between allpairs ofparticles inthesystem. Since foreach pair ofparticles lo,l,twoforces F}',_,, and F§_,,, appear inthetotal sum, andbyEq.(4-9) thesum ofeach such pair iszero, thetotal sum ontheright inEq.(4-10) vanishes, andEq.(4-6) isproved. Thus Newton’s third law,intheform (4-9), issufﬁcient toguarantee the conservation oflinear momentum forasystem ofparticles, anditwasfor this purpose that thelawwasintroduced. The lawofconservation of momentum has, however, amore general validity than Newton’s third law, asweshall seelater. Wecanderive assumption (4-6) onthebasis of asomewhat weaker assumption than Newton’s third law. Wedonotneed toassume that theparticles interact inpairs. Weassume only that the internal forces aresuch that they would dononetwork ifevery particle in thesystem should bedisplaced thesame small distance 8rfrom itsposition atanyparticular instant. Animagined motion ofalltheparticles inthe system iscalled avirtual displacement. The motion described, inwhich every particle moves thesame small distance 6r,iscalled asmall virtual translation ofthesystem. Weassume, then, that inany small virtual translation 6roftheentire system, theinternal forces would dononet work. From thepoint ofview ofthegeneral ideaofconservation ofenergy, thisassumption amounts tolittle more than assuming that space ishomo- geneous. Ifwemove thesystem toaslightly different position inspace without otherwise disturbing it,theinternal state ofthesystem should be unaﬁected, hence inparticular thedistribution ofvarious kinds ofenergy within itshould remain thesame andnonetwork canhave been done by theinternal forces. Letususethisidea toprove Eq.(4-6). The Work done bytheforce inasmall virtual translation 8ris 6W1» =F1‘;-6r. (4-11) Thetotal work done byalltheinternal forces is N N _aw=ZaW,,=ar-(Z mg), (4-12) k=1 k=1 ‘ 158 THE MOTION orASYSTEM orPARTICLES [cn.u>. 4 where wehave factored out8rfrom thesum, since itisthesame forall particles. Assuming that 6W=0,wehave 6r-<ﬁ =0. (4-13) lc=1 Since Eq.(4-13) must hold forany5r,Eq.(4-6) follows. WecanputEq.(4-7) inanilluminating form byintroducing thecon- cept ofcenter ofmass ofthesystem ofparticles. The vector Rwhich locates thecenter ofmass isdeﬁned bytheequation NMR=Zm,,r,,, (4-14) k=1 where Misthetotal mass: NM=Zmk. (4-15) k-=1 The coordinates ofthecenter ofmass aregiven bythecomponents of Eq.(4-14): X1N Y1N z1N 4-16 mkxk; — mkyk; _Ml§l7nltzli" ( ) Thetotal momentum deﬁned byEq.(4-4) is,interms ofthecenter ofmass, NP=Zmks,=MR, (4-17) It1 sothatEq.(4-7) canbewritten MR=F. (4-1s) This equation hasthesame form astheequation ofmotion ofaparticle of mass Macted onbyaforce F.Wethus have theimportant theorem that [when Eq.(4-6) holds] thecenter ofmass ofasystem ofparticles moves like asingle particle, whose mass isthetotalmass ofthesystem, acted onbyaforce equal tothetotal external force acting onthesystem. ' 4-2Conservation ofangular momentum. Letuscalculate thetimerate ofchange ofthetotal angular momentum ofasystem ofNparticles rela- tivetoapoint Qnotnecessarily ﬁxed inspace. The vector angular mo- mentum ofparticle kabout apoint Q,notnecessarily theorigin, istobe deﬁned according toEq.(3-142): Lkq =m),(1'], —-IQ) X(fk-—fq), (4-19) 4-2] CONSERVATION OFANGULAR MOMENTUM 159 where IQistheposition vector ofthepoint Q,and(rk—rQ)isthevector from Qtoparticle lo.Note that inplace ofthevelocity £7,wehave written thevelocity (iv,—iq)relative tothepoint Qasorigin, sothat LkQ istheangular momentum ofmkcalculated asifQwere aﬁxed origin. Thisisthemost useful waytodeﬁne theangular momentum about amoving point Q.Taking thecross product of(rk—rQ)with theequa- tion ofmotion (4-2) forparticle lo,asinthederivation ofEq.(3-144), we obtain (1,,-IQ)><%=(1'),-IQ)><F);+(r,,-IQ)>< (4-20) Wenow differentiate Eq.(4-19): ' =(Ik -—IQ) Xgal-:16 -|—m;,(i';, —IQ) X(I1, -—-IQ) -—m;,(r;, —IQ) XIQ. (4-21) Thesecond term ontheright vanishes. Therefore, byEq.(4-20), d—TIé%-Q =(I1,—IQ) XF7;+(I),—IQ) X —m1,(I], —IQ) XIQ. 1(4-22) Thetotal angular momentum andtotal external torque about thepoint Q aredeﬁned asfollows:N. LQ=ZLkQ2 (4-23) k=1 N.NQ=Z(r,,-rQ)><FZ. (4-24) k1 Summed over allparticles, Eq.(4-22) becomes, ifweuseEq.(4-14), ‘ZLQ N N "MR " 1 -dT= Q+Z(fk—fQ) ><F1=— (-1'0) ><1'o- (4-25)k=1 The lastterm willvanish iftheacceleration ofthepoint Qiszero oris along thelinejoining Qwith thecenter ofmass. Weshall restrict the discussion tomoments about apoint Qsatisfying thiscondition: ' (R-rQ)><rq=o. (4-26) The most important applications willbetocases where Qisatrest, or where Qisthecenter ofmass. Ifwealso assume that thetotal internal lI l 1 )160 THEMOTION orAsvsrnm orPARTICLES [c£u.r. 4 torque vanishes: N Z(rt.—rt)><Ft=0, (4-21) kil then Eq.(4-25) becomes d1-Q_W _NQ. (4-28) This istheangular momentum theorem forasystem ofparticles. An immediate corollary istheconservation theorem forangular momentum, which states that thetotal angular momentum ofasystem ofparticles is constant ifthere isnoexternal torque onthesystem. Inorder toprove Eq.(4-27) from Newton’s third law,weneed toassume astronger version ofthelawthan that needed inthepreceding section, namely, that theforce F§,_,, isnotonly equal andopposite toF§_,,,, but that these forces actalong thelinejoining thetwoparticles; that is,the twoparticles canonly attract orrepel each other. Weshall assume, as intheprevious section, that isthesum offorces duetoeach ofthe other particles: N 1 N 1 2(r;,—rQ) xF;’Z= 226,,-—rQ) xFZ_,;, It-1 It-1 lqhk = (rt-re)><FL).+(rt-ro)><F1541]- '°== (4-29) Inthesecond step, thesum oftorques hasbeen rearranged asasum of pairs oftorques duetopairs offorces which, according toNewton’s third law, areequal andopposite [Eq. (4—9)], sothat N _ N 2:(1’k-1'0) ><Fi=E lO=l l6= =-M=-as I-‘Mp-I I-I"W F‘r-I4_(rt—rt)—(rt-rQ)l><FL». R‘Ms I-INWI-*>-I/-\"\P?‘ =Q-rt)><FL). (4-30) The vector (r1,—-rl)hasthedirection ofthelinejoining particle lwith particle k.IfF§_,,, acts along thisline, thecross product inEq.(4-30) vanishes. Hence ifweassume Newton’s third lawinthestrong form, then assumption (4-27) canbeproved. - Alternatively, byassuming that nonetwork isdone bytheinternal forces inasmall virtual rotation about anyaxisthrough thepoint Q,we canshow that thecomponent oftotal internal torque inanydirection is zero, andhence justify Eq.(4-27). 4-2] CONSERVATION orANGULAR MOMENTUM 161 L x ‘><F """"""""" ~- »’ “\/' \\I \' I . ‘~ ‘E' ~ - . ____ ___¢ L L+dL Q F FIG. 4-1. Motion ofasimple gyroscope. Asanapplication ofEq.(4-28), weconsider theaction ofagyroscope ortop. Agyroscope isarigid system ofparticles symmetrical about an axisandrotating about that axis. The reader canconvince himself that when thegyroscope isrotating about aﬁxed axis, theangular momentum vector ofthegyroscope about apoint Qontheaxisofrotation isdirected along theaxisofrotation, asinFig.4-1. Thesymmetry about theaxis guarantees thatanycomponent oftheangular momentum L),ofparticle Itthatisperpendicular totheaxiswillbecompensated byanequal and opposite component duetothediametrically opposite particle. Letus choose thepoint Qwhere thegyroscope axisrests onitssupport. Ifnow aforce Fisapplied downward onthegyroscope axis (e.g., theforce of gravity), thetorque (rXF)duetoFwillbedirected perpendicular tor andtoL,asshown inFig.4-1. ByEq.(4-28) thevector dL/dt isinthe same direction, asshown intheﬁgure, andthevector Ltends toprecess around theﬁgure inacone under theaction oftheforce F.Now the statement that Lisdirected along thegyroscope axisisstrictly true only ifthegyroscope issimply rotating about itsaxis. Ifthegyroscope axis itself ischanging itsdirection, then thislatter motion willcontribute an additional component ofangular momentum. If,however, thegyroscope isspinning very rapidly, then thecomponent ofangular momentum along itsaxiswillbemuch greater than thecomponent duetothemotion ofthe axis, andLwillbevery nearly parallel tothegyroscope axis. Therefore thegyroscope axismust alsoprecess around thevertical, remaining essen- tially parallel toL.Acareful analysis oftheoff-axis components ofL shows that, ifthegyroscope axisisinitially stationary inacertain direction andisreleased, itwillwobble slightly down andupasitprecesses around thevertical. This willbeshown inChapter 11.The gyroscope does not “resist anychange initsdirection,” asissometimes asserted, fortherate A 162 THE MOTION orASYSTEM orPARTICLES [CHAP. 4 ofchange initsangular momentum isalways equal totheapplied torque, justastherate ofchange oflinear momentum isalways equal totheap- plied force. Wecanmake thegyroscope turn inanydirection weplease byapplying theappropriate torque. Theimportance ofthegyroscope as adirectional stabilizer arises from thefact that theangular momentum vector Lremains constant when notorque isapplied. The changes in direction ofawell-made gyroscope aresmall because theapplied torques aresmall andLisvery large, sothat asmall dLgives noappreciable change indirection. Furthermore, agyroscope only changes direction while a torque isapplied; ifitshifts slightly duetooccasional small frictional torques initsmountings, itstops shifting when thetorque stops. Alarge nonrotating mass, ifmounted likeagyroscope, would acquire only small angular velocities duetofrictional torques, butonce setinmotion bya small torque, itwould continue torotate, andthechange inposition might eventually become large. 4-3Conservation ofenergy. Inmany cases, thetotal force acting on anyparticle inasystem ofparticles depends only onthepositions ofthe particles inthesystem: F],=' F],(I1,1'2,...,IN), fO= 1,2,...,N. (4-31) Theexternal force F1,,forexample, might depend ontheposition r;,of particle k,andtheinternal force might depend onthepositions ofthe other particles relative toparticle k.Itmay bethat apotential fimction V(r1, r2,...,rN)exists such that 6V 8V 6V Fin:-"E1 F]W=—%1 F1”?--—5;;6-> k=1,...,N. (4-32) Conditions tobesatisﬁed bytheforce functions F;,(r1, ...,rN)inorder forapotential Vtoexist canbeworked out, analogous tothecondition (3-189) forasingle particle. The result israther unwieldy andoflittle practical importance, andweomit this development here. Ifapotential energy exists, wecanderive aconservation ofenergy theorem asfollows. ByEq.(4-32), theequations ofmotion ofthekthparticle are mk =— r mk =* ! mk%5=—%- Multiplying Eqs. (4-33) bymm,vim,vi“,respectively, andadding, wehave foreach ks d 2 6Vdz), 6Vdyk 6Vdz], E(%'”""’°)+ax,, at+6y;,, at+02). dt=0’ 7°:1""’N' “'34) 4-3] CONSERVATION orENERGY V163 This istobesummed over allvalues ofk: aN Nava ava ava3;; (tmtvi) +kl(MZ“+ayk3,”+azkdz;=0.(4-35) =1 =1 Thesecond term inEq.(4-35) isdV/dt: av__ N(avanavay.av012,.) E?_,26:0),at+8y),at+a2,,at’ (H56) andtheﬁrstterm isthetime derivative ofthetotal kinetic energy :M=NI“§ T= ,,a,%. (4-37) Consequently, Eq.(4-35) canbewritten gin’+v)=0. (4-as) Hence weagain have aconservation ofenergy theorem, T+V=E, (4-39) where Eisconstant. Iftheinternal forces arederivable from apotential- energy function V,asinEq.(4-32), buttheexternal forces arenot, the energy theorem willbe N%(T+v)=Zrt-v,,. (4-40) k=1 Suppose theinternal force acting onanyparticle kcanberegarded as thesum offorces duetoeach oftheother particles, where theforce F§_,,, onhduetoldepends only ontherelative position (rk—rl)ofparticle k with respect toparticle l: =ZFi_>k(1'1= —1'z)- (441) leek 1 Itmaybethatthevector function F§_,,,(r;, —-r;)issuchthatwecandeﬁne apotential-energy function I Vkl(rkl) =—[IHFi_.k(1'laz)'d1'kz, (4-42) where Ikj =Ik—1'1. This willbetrue ifFf_,,, isaconservative force inthesense ofChapter 3, 164 THEMOTION orASYSTEM orPARTICLES [CHAI-'. 4 that is,if _ curlFL”, =0, (4-44) where thederivatives arewith respect to20),),y;,;,2),).Thegravitational and electrostatic forces between pairs ofparticles areexamples ofconservative forces. IfF§_,k isconservative, sothat V1,;canbedeﬁned, then* ,- 3V1,; .6V1,; 3V1,;F_,=-—- —--k-—Zk 139%: J6?/kl 321,: __.6V;,j _.dV],j ___ dV1,1_ — 16.10;, J8y], k82;, (4-45) IfNewton’s third law(weak form) holds, then i ___ 1‘ _.6V;,, .6Vj,j 6V;,;Fla->1 — Fl->k —1-—~axkl +1?/H -l"kfzkl _'___.dV1,1_ .6V,,, _ (9V],1_ — 1Ox; J6y, k62, (4-46) Thus V1,;willalsoserve asthepotential-energy function fortheforce F},_,,. Wecannow deﬁne thetotal internal potential energy V‘forthesystem of particles asthesumofV1,;overallthepairs ofparticles: -M=~Pi‘_M-<2 v‘(r..---.rN)= _,.,(r,-rt). (4-41) - k= = Itfollows from Eqs. (4-41), (4-45), and(4-46), that theinternal forces are given by ,- .aV" .aV" aV" F],=—l.Tvk—]M—k5Zc-: k=1,...,N. Inparticular, iftheforces between pairs ofparticles arecentral forces, the potential energy V1,z(1‘;,,) foreach pair ofparticles depends only onthe distance r;,)between them, andisgiven byEq.(3—200); theinternal forces ofthesystem arethen conservative, andEq.(4-48) holds. The energy theorem (4-40) willbevalid forsuch asystem ofparticles. Iftheexternal forces arealsoconservative, their potential energy canbeadded toV‘,and thetotal energy isconstant. Ifthere isinternal friction, asisoften thecase, theinternal frictional forces depend ontherelative velocities oftheparticles, andtheconserva- tionlawofpotential plus kinetic energy nolonger holds. *N0t8 that V(I),1) =V($),1, 1/1,1, 21,1) =V($], —"IE1,y),—yj,Z],—Z1), SOthat 6V/6x), =6V/8.721,; =—6V/ox), Bt0. 4-4] CRITIQUE orTHECONSERVATION LAws 165 4-4Critique oftheconservation laws. Wemaydivide thephenomena towhich thelaws ofmechanics have been applied intothree major classes. The motions ofcelestial bodies—stars, satellites, planets—are described with extremely great precision bythelaws ofclassical mechanics. Itwas inthisﬁeld that thetheory hadmany ofitsimportant early successes. Themotions ofthebodies inthesolar system canbepredicted with great accuracy forperiods ofthousands ofyears. Thetheory ofrelativity pre- dicts afewslight deviations from theclassically predicted motion, butthese aretoosmall tobeobserved except inthecase oftheorbit ofMercury, where relativity andobservation agree inshowing aslow precession ofthe axisoftheelliptical orbit around thesunatanangular velocity ofabout 0.01 degree percentury. The motion ofterrestrial bodies ofmacroscopic and microscopic size constitutes thesecond major division ofphenomena. Motions inthisclass areproperly described byNewtonian mechanics, without anysigniﬁcant corrections, butthelaws offorce areusually very complicated, andoften notprecisely known, sothat thebeautifully precise calculations ofcelestial mechanics cannot beduplicated here. The third class ofphenomena isthemotion of“atomic” particles: molecules, atoms, electrons, nuclei, protons, neutrons, etc. Early attempts todescribe themotions ofsuch particles were based onclassical mechanics, andmany phenomena inthisclass canbeunderstood andpredicted onthis basis. However, theﬁner details ofthebehavior ofatomic particles can only be.properly described interms ofquantum mechanics and, forhigh velocities, relativistic quantum mechanics must beintroduced. Wemight addafourth class ofphenomena, having todowith theintrinsic structure oftheelementary particles themselves (protons, neutrons, electrons, etc.). Even quantum mechanics fails todescribe such phenomena correctly, and physics isnow struggling toproduce anewtheory which willdescribe this class ofphenomena. - Theconservation lawforlinear momentum holds forsystems ofcelestial bodies aswell asforbodies ofmacroscopic andmicroscopic size. The gravitational and mechanical forces acting between such bodies satisfy Newton’s third law, atleast toahigh degree ofprecision. Linear momen- tumisalsoconserved inmost interactions ofparticles ofatomic size,except when high velocities orrapid accelerations areinvolved. Theelectrostatic forces between electric charges atrestsatisfy Newton’s third law,butwhen thecharges areinmotion, their electric ﬁelds propagate with thevelocity oflight, sothat iftwocharges areinrapid relative motion, theforces be- tween them may notatanyinstant beexactly equal andopposite. Ifa fastelectron moves past astationary proton, theproton “sees” theelectron always alittle behind itsactual position atanyinstant, andtheforce on theproton isdetermined, notbywhere theelectron is,butbywhere itwas 166 THE MOTION OFASYSTEM OFPARTICLES [CHAP. 4 amoment earlier. When electric charges accelerate, they may emit electro- magnetic radiation andlosemomentum insodoing. Itturns outthat the lawofconservation ofmomentum canbepreserved alsoinsuch cases, but only byassociating momentum with theelectromagnetic ﬁeld aswell as with moving particles. Such aredeﬁnition ofmomentum goes beyond the original limits ofNewtonian mechanics. Celestial bodies andbodies ofmacroscopic ormicroscopic sizeareob- viously notreally particles, since they have astructure which formany purposes isnotadequately represented bymerely giving tothebody three position coordinates x,y,z.Nevertheless, themotion ofsuch bodies, in problems where their structure canbeneglected, iscorrectly represented bythelawofmotion ofasingle particle, mi‘=F. (4-49) This isoften justiﬁed byregarding themacroscopic body asasystem ofsmaller particles satisfying Newton’s third law. Forsuch asystem, thelinear momentum theorem holds, andcanbewritten intheform of Eq.(4-18), which hasthesame form asEq.(4-49). This isavery con- venient way ofjustifying theapplication ofEq.(4-49) tobodies ofmacro- scopic orastronomical size, provided ourconscience isnottroubled bythe factthataccording tomodem ideas itdoesnotmake sense. Iftheparticles ofwhich thelarger body iscomposed aretaken asatoms andmolecules, then intheﬁrstplace Newton’s third lawdoes notinvariably hold forsuch particles, andinthesecond place weshould apply quantum mechanics, not classical mechanics, totheir motion. Themomentum theorem (4-18) can bederived forbodies made upofatoms byusing thelaws ofelectrodynam- icsandquantum mechanics, butthisliesoutside thescope ofNewtonian mechanics. Hence, forthepresent, wemust take thelawofmotion (4-49), asapplied tomacroscopic and astronomical bodies, asafundamental postulate initself, whose justiﬁcation isbased onexperimental grounds orontheresults ofdeeper theories. Thetheorems proved inSection 4-1 show that this postulate gives aconsistent theory ofmechanics inthe sense that if,from bodies satisfying thispostulate, weconstruct acom- posite body, thelatter body willalsosatisfy thepostulate. The lawofconservation ofangular momentum, asformulated inSec- tion 4-2forasystem ofparticles, holds forsystems ofcelestial bodies (regarded asparticles) andforsystems ofbodies ofmacroscopic sizewhen- ever effects duetorotation oftheindividual bodies canbeneglected. When rotations oftheindividual bodies enter into themotion, then a conservation lawforangular momentum stillholds, provided weinclude theangular momentum associated with suchrotations; thebodies arethen nolonger regarded asparticles ofthesimple type considered inthepre- ceding sections whose motions arecompletely described simply byspecify- 4-4] CRITIQUE OFTHE CONSERVATION LAWS 167 ingthefunction r(t)foreach particle. The total angular momentum of thesolar system isvery nearly constant, even ifthesun, planets, and satellites areregarded assimple particles whose rotations canbeneglected. Tidal forces, however, convert some rotational angular momentum into orbital angular momentum oftheplanets andsatellites, andsorotational angular momentum must beincluded ifthelawofconservation ofangular momentum istohold precisely. Some change inangular momentum occurs duetofriction with interplanetary dust androcks, buttheeffect istoo small tobeobserved, andcould inany case beincluded byadding the angular momentum oftheinterplanetary matter tothetotal. Thelawofconservation oftotal angular momentum, including rotation, ofastronomical andterrestrial bodies canbejustiﬁed byregarding each body asasystem ofsmaller particles whose mutual forces satisfy Newton’s third law(strong form). Theargument ofSection 4-2then gives thelaw ofconservation oftotal angular momentum, therotational angular momen- tum ofabody appearing asordinary orbital angular momentum (rxp)of theparticles ofwhich itiscomposed. This argument issubject tothesame criticism asapplied above tothecase oflinear momentum. Ifthe“par- ticles” ofwhich abody iscomposed areatoms andmolecules, then Newton’s third lawdoes notalways hold, particularly initsstrong form; moreover, thelaws ofquantum mechanics apply tosuch particles; andinaddition atoms andmolecules alsopossess rotational angular momentum which must betaken _intoaccount. Even theelementary particles—electrons, protons, neutrons, etc.—possess anintrinsic angular momentum which isnotassociated with their orbital motion. This angular momentum is called spin angular momentum from itsanalogy with theintrinsic angular momentum ofrotation ofamacroscopic body, andmust beincluded ifthe total istosatisfy aconservation law. Thus wenever arrive attheideal simple particle ofNewtonian mechanics, described byitsposition r(t)alone. Weareleftwith thechoice ofaccepting theconservation lawofangular momentum asabasic postulate, orappealing foritsjustiﬁcation totheories which gobeyond classical mechanics. The gravitational forces acting between astronomical bodies arecon- servative, sothat theprinciple ofconservation ofmechanical energy holds very accurately inastronomy. Inprinciple, there isasmall lossofmechan- icalenergy inthesolar system duetofriction with interplanetary dust and rocks, buttheeffect istoosmall toproduce anyobservable effects onplane- tary motion, even with thehigh precision with which astronomical events arepredicted andobserved. There isalsoavery gradual butmeasurable lossofrotational energy ofplanets andsatellites duetotidal friction. For terrestrial bodies ofmacroscopic ormicroscopic size, friction usually plays animportant part, andonly incertain special cases where friction may be neglected cantheprinciple ofconservation ofenergy intheform (4-39) 168 ATHE MOTION OFASYSTEM OFPARTICLES [cHAP. 4 oreven (4-40) beapplied. However, itwasdiscovered byJoule that we canassociate energy with heat insuch away that thelawofconservation ofenergy ofasystem ofbodies stillapplies tothetotal kinetic plus poten- tialplusheat energy. Ifweregard abody ascomposed ofatoms andmole- cules, itsheat energy turns outtobekinetic andpotential energy ofran- dom motion ofitsatoms andmolecules. The electromagnetic forces on moving charged particles arenotconservative, and anelectromagnetic energy must beassociated with theelectromagnetic ﬁeld inorder topre- serve theconservation lawofenergy. Such extensions oftheconcept of energy toinclude heat andelectromagnetic energy are,ofcourse, outside thedomain ofmechanics. When thedeﬁnition ofenergy issuitably ex- tended toinclude notonly kinetic energy, butenergy associated with the electromagnetic ﬁelds andanyother force ﬁelds which may act,then a lawofconservation ofenergy holds quite generally, inclassical, relativistic, andquantum physics. The conservation laws ofenergy, momentum, andangular momentum arethecornerstones ofpresent-day physics, being generally valid inall physical theories. Itseems atpresent anidleexercise toattempt toprove them formaterial bodies within theframework ofclassical mechanics by appealing toanoutmoded picture ofmatter asmade upofsimple New- tonian particles exerting central forces upon oneanother. The conserva- tionlawsareinasense notlawsatall,butpostulates which weinsist must hold inanyphysical theory. If,forexample, formoving charged particles, weﬁndthat thetotal energy, deﬁned as(T—l—V),isnotconstant, wedo notabandon thelaw, butchange itsmeaning byredeﬁning energy toin- clude electromagnetic energy insuch away astopreserve thelaw. We prefer always tolook forquantities which areconserved, and agree to apply thenames “total energy,” “total momentum,” “total angular mo- mentum” only tosuch quantities. The conservation ofthese quantities isthen notaphysical fact, butaconsequence ofourdetermination tode- ﬁnethem inthisway. Itis,ofcourse, astatement ofphysical fact, which may ormay notbetrue, toassert that such deﬁnitions ofenergy, momen- tum, andangular momentum canalways befound. This assertion, has sofarbeen true; adeeper justiﬁcation willbesuggested attheendof Section 9-6. 4-5Rockets, conveyor belts, andplanets. There aremany problems that canbesolved byappropriate applications oftheconservation laws oflinear momentum, angular momentum, and energy. Insolving such problems, itisnecessary todecide which conservation laws areappropriate. The conservation laws oflinear and angular momentum or,rather, the theorems (4-7)and(4-28) ofwhich they arecorollaries, arealways appli- cable toanyphysical system provided allexternal forces andtorques are 4-5] ROCKETS, CONVEYOR BELTS, AND PLANETS 169 taken into account, and application ofoneortheother isappropriate whenever theexternal forces ortorques areknown. Thelawofconserva- tion ofkinetic plus potential energy isapplicable only when there isno conversion ofmechanical energy into other forms ofenergy. Wecannot usethelawofconservation ofenergy when there isfriction, forexample, unless there isaway todetermine theamount ofheat energy produced. The conservation laws ofenergy, momentum, andangular momentum refer always toadeﬁnite ﬁxed system ofparticles. Inapplying thecon- servation laws, caremust betaken todecide justhow much isincluded in thesystem towhich they aretobeapplied, andtoinclude alltheenergy andmomentum ofthissystem inwriting down theequations. One may choose thesystem arbitrarily, including andexcluding whatever particles may beconvenient, butifanyforces actfrom outside thesystem onparti- clesinthesystem, these must betaken intoaccount. Atypical problem inwhich thelawofconservation oflinear momentum isapplicable istheconveyor beltproblem. Material isdropped continu- ously from ahopper onto amoving belt, anditisrequired toﬁndtheforce Frequired tokeep thebeltmoving atconstant velocity v(Fig. 4-2). Let therateatwhich mass isdropped onthebeltbedm/dt. Ifmisthemass ofmaterial onthebelt, andMisthemass ofthebelt (which really does notﬁgure intheproblem), thetotal momentum ofthesystem, beltplus material onthebeltandinthehopper, is P=(m-1-M)v. (4-50) Weassume that thehopper isatrest; otherwise themomentum ofthe hopper anditscontents must beincluded inEq.(4-50). The linear mo- mentum theorem requires that dP dmF—-'E'—l)W' This gives theforce applied tothebelt. Thepower supplied bytheforce is F1) =92% =;%(my2) = +M)v2]. (4-52) This istwice therateatwhich thekinetic energy isincreasing, sothat the ".'.*5-3' .<:’~i:§:§{' /as.2 V ,4, m ?_.> F “ Y-v 1,K ._ d Qe§§<;.,-,5-,‘ﬁ ';=,--=,%. ''cg,,\»_ﬁi;-,-»- Fro. 4-2. Aconveyor belt. 170 THEMOTION OFASYSTEM OFPARTICLES [crrA1>. 4 conservation theorem ofmechanical energy (4-40) does notapply here. VVhere istheexcess halfofthepower going? The equation ofmotion ofarocket canbeobtained from thelawof conservation ofmomentum. Letthemass oftherocket atanygiven in- stant beM,andletitsspeed bevrelative tosome ﬁxed coordinate system. Ifmaterial isshot outoftherocket motor with anexhaust velocity urela- tivetotherocket, thevelocity oftheexhaust relative totheﬁxed coordi- nate system isv+u.Ifanexternal force Falsoactsontherocket, then thelinear momentum theorem reads inthiscase: %(Mv) -(v+u)5%=F. (4-53) Theﬁrstterm isthetime rateofchange ofmomentum oftherocket. The second term represents therate atwhich momentum isappearing inthe rocket exhaust, where— (dM/dt) istherate atwhich matter isbeing ex- hausted. The conservation lawapplies toadeﬁnite ﬁxed system ofpar- ticles. Ifweﬁxourattention ontherocket atanymoment, wemust remember that atatime dtlater thissystem willcomprise therocket plus thematerial exhausted from therocket during that time, andboth must beconsidered incomputing thechange inmomentum. The equation can berewritten:dv_ dM(4-54) Theﬁrstterm ontheright iscalled thethrust oftherocket motor. Since dM/dt isnegative, thethrust isopposite indirection totheexhaust veloc- ity. The force Fmay represent airresistance, oragravitational force. Letussolve thisequation forthespecial case where there isnoexternal force: M%=Au9%-ti- (4-55) Wemultiply bydt/M andintegrate, assuming that uisconstant: Mv—vo=—uln-Z-‘Tm (4-56) The change ofspeed inanyinterval oftime depends only ontheexhaust velocity andonthefraction ofmass exhausted during that time interval. This result isindependent ofanyassumption astotherateatwhich mass isexhausted. Problems inwhich thelawofconservation ofangular momentum isuse- fulturn upfrequently inastronomy. The angular momentum ofthe galaxy ofstars, orofthesolar system, remains constant during thecourse 4-6] COLLISION PROBLEMS 171 ofitsdevelopment provided nomaterial isejected from thesystem. The effect oflunar tides isgradually toslow down therotation oftheearth. Astheangular momentum oftherotating earth decreases, theangular momentum ofthemoon must increase. The magnitude ofthe(orbital) angular momentum ofthemoon is L=mr2w, (4-57) where misthemass, wistheangular velocity, andristheradius ofthe orbit ofthemoon. Wecanequate themass times thecentripetal accelera- tion tothegravitational force, toobtain therelation mrw2 =% » (4-58) where Misthemass oftheearth. Solving thisequation forwandsubsti- tuting inEq.(4-57), weobtain L=(GMm2r)1/2. (4-59) Therefore, asthemoon’s angular momentum increases, itmoves farther away from theearth. (Inattempting todetermine therate ofrecession ofthemoon byequating thechange ofLtothechange oftheearth’s rota- tional angular momentum, itwould benecessary todetermine how much oftheslowing down oftheearth’s rotation bytidal friction isduetothe moon andhow much tothesun. The angular momentum ofthemoon plustherotational angular momentum oftheearth isnotconstant because ofthetidal friction duetothesun. The total ‘angular momentum ofthe earth-moon system about thesunisvery nearly constant except forthe very small effect oftides raised onthesunbytheearth.) 4-6Collision problems. Many questions concerning collisions of particles canbeanswered byapplying theconservation laws. Since the conservation laws arevalid alsoinquantum mechanics,* results obtained with their usearevalid forparticles ofatomic andsubatomic size, aswell asformacroscopic particles. Inmost collision problems, thecolliding particles aremoving atconstant velocity, freeofanyforce, forsome time before andafter thecollision, while during thecollision they areunder the action oftheforces which they exert ononeanother. Ifthemutual forces during.the collision satisfy Newton’s third law, then thetotal linear mo- mentum oftheparticles isthesame before and after thecollision. If Newton’s third lawholds inthestrong form, thetotal angular momentum *P.A.M.Dirac, ThePrinciples ofQuantum Mechanics, 3rded. Oxford: Oxford University Press, 1947. (Page 115.) 172 THE MOTION orAsYsTEM orPARTICLES [crIA1>. 4 isconserved also. Iftheforces areconservative, kinetic energy iscon- served (since thepotential energy before andafter thecollision isthesame). Inanycase, theconservation laws arealways valid ifwetake intoaccount alltheenergy, momentum, andangular momentum, including that asso- ciated with anyradiation which may beemitted andincluding anyenergy which isconverted from kinetic energy intoother forms, orviceversa. Weconsider ﬁrst acollision between twoparticles, 1and2,inwhich thetotal kinetic energy andlinear momentum areknown tobeconserved. Such acollision issaid tobeelastic. Ifwedesignate bysubscripts 1and 2thetwoparticles, andbysubscripts IandFthevalues ofkinetic energy andmomentum before andafter thecollision respectively, theconservation laws require P11+P21=PIF+P2F, (4430) T11-l"T21=T1F'+T2F- (4-61) Equation (4-61) canberewritten interms ofthemomenta andmasses of theparticles: Pi! P31_pi» pip_ 2m1 +2mg ___2'm.1 +2771.2 Tospecify anymomentum vector p,wemust specify three quantities, which may beeither itsthree components along anysetofaxes, oritsmag- nitude anddirection (thelatter speciﬁed perhaps byspherical angles 0,(,0). Thus Eqs. (4-60) and(4-62) represent four equations involving theratio ofthetwomasses andtwelve quantities required tospecify themomenta involved. Ifnine ofthese quantities aregiven, theequations canbesolved fortheremaining four. Inatypical case, wemight begiven themasses and initial momenta ofthetwoparticles, andtheﬁnal direction ofmotion of oneoftheparticles, sayparticle 1.Wecould then ﬁndtheﬁnal momen- tum P21‘ofparticle 2,andthemagnitude oftheﬁnal momentum ply(or equivalently,‘ theenergy) ofparticle 1.Inmany important cases, themass ofoneoftheparticles isunknown, andcanbecomputed from Eqs. (4-60) and(4-62) ifenough isknown about themomenta andenergies before and after thecollision. Note that theinitial conditions alone arenotenough todetermine theoutcome ofthecollision from Eqs. (4-60) and(4-62); we must know something about themotion after thecollision. The initial conditions alone would determine theoutcome ifwecould solve theequa- tions ofmotion ofthesystem. Consider acollision ofaparticle ofmass ml,momentum p11, with a particle ofmass m2atrest. This isacommon case. (There isactually no lossofgenerality inthisproblem, since, aswepointed outinSection 1-4 andwillshow inSection 7-1,ifmgisinitially moving with auniform veloc- 4-6] COLLISION PROBLEMS 173 PIF ml -3"______ ml P11 "12 02 mg P2F Fro. 4-3. Collision ofparticle mlwith particle mgatrest. ityV21,Newton’s laws areequally applicable inacoordinate system mov- ingwith uniform velocity v21,inwhich m2isinitially atrest.) Letmlbe “scattered” through anangle dl;that is,let191betheangle between its ﬁnal anditsinitial direction ofmotion (Fig. 4-3). The momentum P21? must lieinthesame plane aspl;andplysince there isnocomponent of momentum perpendicular tothisplane before thecollision, andthere must benone after. Letp2;.~make anangle 02with thedirection ofpl1.We write outEq.(4-60) incomponents along andperpendicular topl1: pl;=P111‘cos:91-]-pg)!cos192, (4-63) 0= P114‘ Sin 1,1 — pzp sin 1,2. Equation (4-62) becomes, inthepresent case, 2___2 2 =__ P214‘ 1 m2 Iftwoofthequantities (PIF/P11; P2F/P11, 171)172,ml/m2) areknown, theremaining three canbefound. Ifthemasses, theinitial momentum pl1,andtheangle dlareknown, forexample, wecansolve for plF,P217, 02asfollows. Transposing theﬁrst term ontheright tothe leftsideinEqs. (4-63) and(4-64), squaring, andadding, weeliminate 132: 4 pi;+Fir—2111111114 cos191=pin (4-66) After substituting thisinEq.(4-65), wecansolve forplF: 2 1/2PIF __ mi ml 2 m2—ml] P11_mi+m2cos013:i('m1 +m2) cos‘,1+ml-l"m2 , (4-67) and P21" cannow befound from Eq. (4-66), and 62from Eq. (4-63). 174 THE MOTION OFASYSTEM 0FPARTICLES [cnAP. 4 Ifml>m2,thequantity under theradical iszero forall=19",,where 17",isgiven by 2 <><>s2a,,,=1-1%. ()g1’1]',5;l" (4-cs) ml Ifdl>0",(and dl51r),then plF/pl1iseither imaginary ornegative, neither ofwhich isallowable physically, sothat 0”,represents themaxi- mum angle through which mlcanbescattered. Ifml>>mg,thisangle isvery small, asweknow from experience. Fordl<dm,there aretwo values ofplF/pl1,thelarger corresponding toaglancing collision, the smaller toamore nearly head-on collision; 192willbedifferent forthese two cases. The case dl=0may represent either nocollision atall (plF=pl1)orahead-on collision. Inthelatter case, 155:“--‘_"‘2 a=0 M=i_2’”2 -4-69 P11 m1+m2, 2 , P11 m1'l‘m2 ( ) Ifml=m2,(Eqs.(4-67), (4-cc), and(4-64)reduce to %§=coscl, gf=sincl, 02= -cl) (4-70) clnowvaries from dl=0fornocollision tocl=1r/2forahead-on collision inwhich theentire momentum istransferred toparticle 2.(Actu- ally,clisundeﬁned ifpm=0,butcl—>1r/2 andplF—>0asthecolli- sionapproaches ahead-on collision.) Ifml<m2,allvalues ofdlfrom 0 to1rarepossible, andgive apositive value forplF/pll iftheplus signis chosen inEq.(4-67). Theminus signcannot bechosen, since itleads toa negative value forpl,-/pll. Ifdl=0,then ply=pll; thisisthecase when there isnocollision. The case dl=1rcorresponds toahead-on collision, forwhich PIE M2—mi4.." "1""4 2m (4-71) = -‘*2=°- Ifmlisunknown, buteither P11orTllcanbemeasured orcalculated, ob- servation oftheﬁnal momentum ofparticle 2(whose mass isassumed known) issufficient todetermine ml. Asanexample, ifTlI=pf;/2ml isknown, and T2l.~ ismeasured forahead-on collision, mlisgiven by Eq.(4-69) or(4-71): E —‘ 27111 —_ [<21-11I __ )2 _ :|1/2 . _‘m2_-—T2F 14-—T2F 1 1 (472) Wethus determine mltowithin oneoftwopossible values. Ifresults for 4-6] COLLISION PROBLEMS 175 acollision with another particle ofdifferent mass m2,orforadiﬁerent scattering angle, areknown, mlisdetermined uniquely. Essentially this method wasused byChadwick toestablish theexistence oftheneutron.* Unknown neutral particles created inanuclear reaction were allowed to impinge onmatter containing various nuclei ofknown masses. The ener- gies oftwokinds ofnuclei ofdifferent masses m2,mgprojected forward byhead-on collisions were measured. Bywriting Eq. (4-72) forboth cases, theunknown energy TlIcould beeliminated, andthemass mlwas found tobepractically equal tothat oftheproton. Wehave seen that ifweknow theinitial momenta oftwo colliding particles ofknown masses, andtheangle ofscattering dl(or02),allother quantities involved inthecollision canbecalculated from theconserva- tion laws. Topredict theangles ofscattering, wemust know notonly theinitial momenta andtheinitial trajectories, butalsothelawofforce between theparticles. Anexample isthecollision oftwoparticles acted onbyacentral inverse square lawofforce, tobetreated inSection 4-8. Such predictions canbemade forcollisions ofmacroscopic orastronomical bodies under suitable assumptions astothelawofforce. Foratomic par- ticles, which obey quantum mechanics, thiscannot bedone, although we canpredict theprobabilities ofobserving various angles dl(or62)for given initial conditions; thatis,wecanpredict cross sections. Inallcases where energy isconserved, therelationships between energies, momenta, andangles ofscattering developed above arevalid except atparticle veloc- ities comparable with thevelocity oflight. Inthelatter case, Eqs. (4-60), (4-61), (4-63), and(4-64) arestillvalid, buttherelativistic relationships between mass, momentum, andenergy must beused, instead ofEq.(4-62). Wequote without proof therelation between mass, momentum, and energy asgiven bythetheory ofrelativityzj L2_ T2_ _ 2m_T+2mc2 ’ (473) where cisthespeed oflight, andmistherestmass oftheparticle, that is, themass when theparticle isatrest. The relativistic relations between kinetic energy, momentum, andvelocity are T=mc2 -1). (4-74) 1—v2c2) = v , _ p \/1—(v2/02) (475) *J.Chadwick, Nature, 129, 312(1932). TP.G.Bergmann, Introduction totheTheory ofRelativity. New York: Prentice- Hall, 1946. (Chapter 6.) 176 THE MOTION oFASYSTEM orPARTICLES [cHA1>. 4 which reduce totheclassical relations (2-5) and (3-127), when v<<c. Unless visnearly equal toc,thesecond term ontheright inEq.(4-73) ismuch smaller than theﬁrst, andthisequation reduces totheclassical one. With thehelp ofEq.(4-73), theconservation laws canbeapplied to collisions involving velocities near thespeed oflight. Atoms, molecules, and nuclei possess internal potential and kinetic energy associated with themotion oftheir parts, andmay absorb orre- lease energy oncollision. Such inelastic collisions between atomic particles aresaid tobeoftheﬁrst hind, orendoergic, ifkinetic energy oftransla- tional motion isabsorbed, and ofthesecond kind, orexoergic, ifkinetic energy isreleased intheprocess. Itmayalsohappen thatinanatomic ornuclear collision, theﬁnal particles after thecollision arenotthesame astheinitial particles before collision. Forexample, aproton may collide with anucleus andbeabsorbed while aneutron isreleased andﬂiesaway. There areagreat many possible types ofsuch processes. Two particles may collide andstick together toform asingle particle or,conversely, a single particle may suddenly break upinto twoparticles which ﬂyapart. Two particles may collide andform twoother particles which ﬂyapart. Orthree ormore particles may beformed intheprocess and ﬂyapart after thecollision. Inallthese cases, thelawofconservation ofmomen- tumholds, andthelawoflconservation ofenergy alsoifwetake into account theinternal energy oftheatoms andmolecules. Weconsider here acaseinwhich aparticle ofmass mlcollides with aparticle ofmass ma atrest(Fig. 4-4). Particles ofmasses mgandm4leave thescene ofthe collision atangles 193and 194with respect totheoriginal direction of motion ofml. Letkinetic energy Qbeabsorbed intheprocess (Q>0 foranendoergic collision; Q=0foranelastic collision; Q<0foran P3 m3 \3ml P1 m2‘nu _____ =94 '"‘ P4 Fre. 4-4. Collision ofmlwith mgatrest, resulting intheproduction ofm3 andm4. 4-6] COLLISION PROBLEMS 177 exoergic collision). Then, applying theconservation lawsofenergy and momentum, wewrite pl=pl;cos63+plcos04, (4-76) 0=p3sin03-—p4sin04, (4-77) T1 =T3 +T4 +Q. Since kinetic energy canbeexpressed interms ofmomentum, ifthemasses areknown, wemay ﬁndanythree ofthequantities pl,p3,pl,173,04,Qin terms oftheother three. Inmany cases plisknown, p3and03aremeas- ured, anditisdesired tocalculate Q.Byeliminating :94from Eqs. (4-76) and(4-77), asintheprevious example, weobtain pi=Pi+P5—2mm cos193- (4-79) This maynowbesubstituted inEq.(4-78) togiveQinterms ofknown quantities: 2 2 2 2_Q=T1_T3_T4=21:;_21;;__pl+113 213P1I1acos03, 1 3 4 ( OI‘ 1/2Q=T.(1__T,(l+1:)+2(M%) ....».. 4m m m (4-s1) Every step uptothesubstitution forTl,T3,andT4isvalid alsoforpar- ticles moving atvelocities oftheorder ofthevelocity oflight. Athigh velocities, therelativistic relation (4-73) between Tandpshould beused inthelaststep. Equation (4-81) isuseful inobtaining Qforanuclear reaction inwhich anincident particle mlofknown energy collides with a nucleus m2,with theresult that aparticle ml,isemitted whose energy and direction ofmotion canbeobserved. Equation (4-81) allows ustodeter- mine Qfrom these known quantities, taking intoaccoimt theeffect ofthe slight recoil oftheresidual nucleus ml,which isusually difﬁcult toobserve directly. Collisions ofinert macroscopic bodies arealways inelastic andendoergic, kinetic energy being converted toheatbyfrictional forces during theim- pact. Kinetic energy oftranslation may also beconverted into kinetic energy ofrotation, andconversely. (Exchanges ofrotational energy are included inQintheprevious analysis.) Such collisions range from the nearly elastic collisions ofhard steel balls, towhich theabove analysis of elastic collisions applies when rotation isnotinvolved, tocompletely in- elastic collisions inwhich thetwobodies stick together after thecollision. 178 THE MOTION orASYSTEM orPARTICLES [crma 4 Letusconsider acompletely inelastic collision inwhich abullet ofmass ml,velocity v1strikes andsticks inanobject ofmass mgatrest. Letthe velocity ofthetwoafter thecollision beV2.Evidently theconservation ofmomentum implies that V2beinthesame direction asv1,andwehave: ’m1V1 =(ml+m2)V2- (4-82) Thevelocity after thecollision is _ ml _ v2-ml+m2v1- (+83) Energy isnotconserved insuch acollision. The amount ofenergy con- verted intoheat is Q=%m1"i —%(m1 +m2)v§ =%m1vi '(4"84) Inahead-on collision oftwobodies inwhich rotation isnotinvolved, it wasfound experimentally byIsaac Newton that theratio ofrelative veloc- ityafter impact torelative velocity before impact isroughly constant for anytwogiven bodies. Letbodies m1,m2, traveling with initial velocities v11,v21along theac-axis, collide and rebound along thesame axis with velocities v11.-,1221?. Then theexperimental result isexpressed bythe equation* vw—vw=e(v1r —1121), (4-85) where theconstant eiscalled thecoeﬁicient ofrestitution, andhasavalue between 0and1.Ife=1,thecollision isperfectly elastic; ife=0,itis completely inelastic. Conservation ofmomentum yields, inanycase, m1U1[ +‘H’!/2112] '= 77’!/1171p +m2U2F. _ Equations (4-85) and(4-86) enable ustoﬁndtheﬁnal velocities 1111»and U21?forahead-on collision when theinitial velocities areknown. 4-7Thetwo-body problem. Weconsider inthissection themotion ofa system oftwo particles acted onbyinternal forces satisfying Newton’s third law(weak form), andbynoexternal forces, orbyexternal forces satisfying arather specialized condition tobeintroduced later. Weshall ﬁndthat thisproblem canbeseparated intotwosingle-particle problems. *More recent experiments show that eisnotreally constant, butdepends on theinitial velocities, onthemedium inwhich thecollision takes place, andon thepasthistory ofthebodies. Foramore complete discussion with references, seeG.Barnes, “Study ofCollisions, ”Am; J.Phys. 26,5(January, 1958). 4-7] THE TWO-BODY PROBLEM 179 Themotion ofthecenter ofmass isgoverned byanequation (4-18) ofthe same form asthat forasingle particle. Inaddition, weshall ﬁndthat the motion ofeither particle, with respect totheother asorigin, isthesame asthemotion with respect toaﬁxed origin, ofasingle particle ofsuitably chosen mass acted onbythesame internal force. This result willallow application oftheresults ofSection 3-14 tocases where themotion ofthe attracting center cannot beneglected. Letthetwoparticles have masses mlandm2,andletthem beacted on byexternal forces Fj,F3,andinternal forces F‘,, exerted byeach parti- cleontheother, andsatisfying Newton’s third law: F1=—F§. (4-87) Theequations ofmotion forthesystem arethen 'm1i"1 = +Fi, (4-88) mgfg=F;+F5. (+89) Wenow introduce achange ofcoordinates: R= (4-90) r=rl—r2. (4-91) Theinverse transformation is _ i_ _rl-R+m1+m2r, (492) _ _ ml _1'2--R ml+m2r, (493) where Risthecoordinate ofthecenter ofmass, and1'istherelative coor- dinate ofmlwith respect tom2. (See Fig.4-5.) Adding Eqs. (4-88) and (4-89) andusing Eq.(4-87), weobtain theequation ofmotion forR: (ml+m2)ﬁ =Fi+FE» (4_94) Multiplying Eq.(4-89) byml,andsubtracting from Eq.(4-88) multiplied 7"-2 1’ I2 0.111. mi0 '1 FIG. 4-5. Coordinates forthetwo-body problem. 180 THEMOTION orASYSTEM orPARTICLES [oHA1>. 4 bymg,using Eq.(4-87), weobtain theequation ofmotion forr: u i Fe Fe mim21' =(mi+m2)F1 +m1m2 — (4‘95) Wenow assume that Fi Fie _=__, mlmg A(4-ac) andintroduce theabbreviations * M=mi+m2, (4*97) , 4_g8 Hm1_,_m2 () 1 F=Fi+ (4-99) Equations (4-94) and(4-95) then take theform ofsingle-particle equa- tions ofmotion: _ MR=F, (4-100) tr=Fl. (4-101) Equation (4—100) isthefamiliar equation forthemotion ofthecenter of mass. Equation (4—101) istheequation ofmotion foraparticle ofmass u acted onbytheinternal force that particle 2exerts onparticle 1. Thus themotion ofparticle 1asviewed from particle 2isthesame asif particle 2were ﬁxed andparticle 1hadamass /.¢(piscalled thereduced mass). Ifoneparticle ismuch heavier than theother, nisslightly less than themass ofthelighter particle. Iftheparticles areofequal mass, ;.¢ ishalfthemass ofeither. Wemay now apply theresults ofSection 3-14 toanytwo-body problem inwhich thetwoparticles exert aninverse square lawattraction orrepulsion oneach other, provided theexternal forces are either zero orareproportional tothemasses, asrequired byEq.(4-96). Equation (4-96) issatisﬁed iftheexternal forces aregravitational forces exerted bymasses whose distances from thetwo bodies mlandm2are much greater than thedistance rfrom mltomg.Asanexample, themotion oftheearth-moon system canbetreated, toagood approximation, bythe method ofthissection, since themoon ismuch closer totheearth than either istothesun(ortotheother planets). Atomic particles areacted onbyelectrical forces proportional totheir charges, andhence Eq.(4-96) holds ordinarily only iftheexternal forces arezero. There isalsotheless important case where thetwoparticles have thesame ratio ofcharge to mass, andareacted onbyexternal forces duetodistant charges. Wemay remark here that although Eqs. (4-88) and(4-89) arenotthecorrect equa- tions fordescribing themotions ofatomic particles, theintroduction ofthe _-/4-8] CENTER-OF-MASS COORDINATES 181 coordinates R,r,andthereduction ofthetwo-body problem totwoone- body problems canbecarried outinthequantum-mechanical treatment in awayexactly analogous totheabove classical treatment, under thesame assumptions about theforces. Itisworth remarking that thekinetic energy ofthetwo-body system canbeseparated intotwoparts, oneassociated with eachofthetwoone- body problems intowhich wehave separated thetwo-body problem. The center-of-mass velocity andtherelative velocity are, according toEqs. (4-90)—(4-93), related totheparticle velocities by V=R= , m1-l-m2 v=f'=vl—v2,(4—102) (4-103) or vl=V+—1:—1v, (4—104)' V2=V'—%2V. (4—105) Thetotal kinetic energy is » T=%m111i -|-imzvg =QMV”+gm)”. The angular momentum cansimilarly beseparated intotwoparts:(4—106) L=mi(1'1 XV1)+m2(!'2 XV2) =M(R XV)—|—p.(1' XV). (4—107) The total linear momentum is,however, just P=m1V1 +m2V2 = There isnoterm p.Vinthetotal linear momentum. 4-8Center-of-mass coordinates. Rutherford scattering byacharged particle offinite mass. Bymaking useoftheresults ofthepreceding sec- tion, wecansolve atwo-body scattering problem completely, ifweknow theinteraction force between thetwoparticles, bysolving theone-body equation ofmotion forthecoordinate r.The result, however, isnotin avery convenient form forapplication. The solution r(t)describes the motion ofparticle 1with respect toparticle 2asorigin. Since particle 2 itself willbemoving along some orbit, thisisnotusually avery convenient way ofinterpreting themotion. Itwould bebetter todescribe themotion ofboth particles bymeans ofcoordinates rl(t), r2(t)referred tosome ﬁxed l l 7l l182 THEMOTION orASYSTEM orPARTICLES [CHAP. 4 origin. Usually oneoftheparticles isinitially atrest; weshall take itto beparticle 2,andcallitthetarget particle. Particle 1,approaching the target with aninitial velocity vl1,weshall calltheirwident particle. The twoparticles aretobelocated byvectors rland1'2relative toanorigin with respect towhich thetarget particle isinitially atrest. Weshall call thecoordinates rl,r2thelaboratory coordinate system. The translation from thecoordinates R,rtolaboratory coordinates is most conveniently carried outintwosteps. Weﬁrstintroduce acenter-of- mass coordinate system inwhich theparticlesare located byvectors rl,rg with respect tothecenter ofmass asorigin: ri=r-Ri1’ (4-109) r;=r2—R, and, conversely, rl=rl+R,, (4-110) 1'2=1'2+ , The relation between thecenter-of-mass coordinates andtherelative co- ordinate risobtained from Eqs. (4-92) and(4-93): i m2 H'1=mT.’=m"I,Z_m,I=*it (41-111) 2 mi-l-M2 m2’ Theposition vectors oftheparticles relative tothecenter ofmass arecon- stant multiples oftherelative coordinate 1'.The center ofmass hasthe advantage over particle 2,asanorigin ofcoordinates, inthat itmoves with uniform velocity incollision problems where noexternal forces are assumed toact. Inthecenter-of-mass coordinate system thetotal linear momentum is zero, andthemomenta piandpgofthetwoparticles arealways equal and opposite. Thescattering angles 1?‘,and0Qbetween thetwoﬁnal_directions ofmotion andtheinitial direction ofmotion ofparticle 1arethesupple- ments ofeach other, asshown inFig.4-6. Wenow determine therelation between thescattering angle 9inthe equivalent one-body problem andthe scattering angle dlinthelaboratory coordinate system (Fig. 4-7). Thevelocity oftheincident particle inthe center-of-mass systemlis related totherelative velocity intheone-body problem, according toEq.(4-111), by v’l=%1tr. (4-112) 4-8] CENTER-OF—MASS COORDINATES 183 ml . T Dir mi /A01 ii I92 i M1 ‘,1 m2ciE 1i.0 P11 P21 dmzm2 Pir Fro. 4-6. Two-particle collision in FIG. 4-7. Orbits fortwo-body colli- center-of-mass coordinates. sioninthelaboratory system. l Since these twovelocities arealways parallel, theangle ofscattering 0‘, oftheincident particle inthecenter-of-mass system isequal totheangle ofscattering 9intheone-body problem. Theincident particle velocities inthecenter-of-mass andlaboratory systems arerelated by[Eq. (4-110)] n=ﬁ+W mm) where theconstant velocity ofthecenter ofmass canbeexpressed interms oftheinitial velocity inthelaboratory system byEq.(4—102): =_.__""1_. =L V ml+m2vll m2vll. (4—114) The relation expressed byEq.(4—113) isshown inFig. 4-8, from which therelation between 1?‘,=6anddlcanbedetermined: 1' . tanol=.—"‘?-sl’l§)—, (4-115) vlpcos6+V V ‘iv 1»- Flo. 4-8. Relation between velocities inlaboratory and center-of-mass co- ordinate systems.71!‘ Vll > \\ l184 THE MOTION orASYSTEM orPARTICLES [CHAP- 4 or,withthehelpofEqs. (4-112) and(4—114), _ .9=____S-ii-£5’-—-, 4.-116tan 1 cos9+(M101/MQUF) ( ) where vland vFaretheinitial and ﬁnal relative speeds, and wehave substituted vlforvl1,since initially therelative velocity isjustthevelocity oftheincident particle. Ifthecollision iselastic, theinitial andﬁnal speeds arethesame andEq.(4-116) reduces to:, '9to=--ﬂ-——- 4-117an1cos9+(mi/ma) ( ) Asimilar relation for62canbeworked out. Iftheincident particle ismuch heavier than thetarget particle, then olwillbevery small, nomatter what value 9may have. This corre- sponds totheresult obtained inSection 4-6, that dlcannever belarger than 0,,given byEq.(4-68), ifml>mg. Ifml=m2,then Eq.(4-117) iseasily solved fordl: sin9 2sin(9/2) cos(9/2) 9tan0l= =~- =tan-»cos9+1 2cos? (9/2) 2 ol=lo. ‘ (4-11s) Since 9may always have anyvalue between Oand1rwithout violating the conservation laws inthecenter-of-mass system, themaximum value ofdl inthiscase is1r/2, inagreement with thecorresponding result ofSection 4-6. Ifthetarget mass m2ismuch larger than theincident mass ml,then tandlitan9;this justiﬁes rigorously ourapplication tothis case of Eq. (3-276) fortheRutherford cross section, deduced inChapter 3for theone-body scattering problem with aninverse square lawforce. According totheabove developments, Eq.(3-276) applies alsotothe two-body problem foranyratio ml/m2 ofincident mass totarget mass, but9must beinterpreted astheangle ofscattering interms ofrelative coordinates, orelseinterms ofcenter-of-mass coordinates. That is,do’ inEq.(3-276) isthecross section forascattering process inwhich the relative velocity vafter thecollision makes anangle between 9and9-|—d9 with theinitial velocity. Since itisthelaboratory scattering angle dlthat isordinarily measured, wemust substitute for9andd9inEq.(3-276) their values interms ofdlanddﬁlasdetermined from Eq.(4-117). This ismost easily done incaseml=mg,when, byEq.(4-118), theRutherford scattering cross section {Eq. (3-276)] becomes 2 do‘=11% 2-(lag-in 21rsindlddl. (4-119) 2p.vo sindl l \ > 4-9] THE N-BODY PROBLEM 185 4-9TheN-body problem. Itwould bevery satisfactory ifwecould arrive atageneral method ofsolving theproblem ofanynumber ofparti- clesmoving under theforces which they exert ononeanother, analogous to themethod given inSection 4-7bywhich thetwo-body problem wasre- duced totwoseparate one-body problems. Unfortunately nosuch general method isavailable forsystems ofmore than twoparticles. This does not mean that such problems cannot besolved. The extremely accurate cal- culations ofthemotions oftheplanets represent asolution ofaproblem involving thegravitational interactions ofaconsiderable number ofbodies. However, these solutions arenotgeneral solutions oftheequations of motion, likethesystem oforbits wehave obtained forthetwo-body case, butarenumerical solutions obtained byelaborate calculations forspeciﬁed initial conditions andholding over certain periods oftime. Even thethree- body problem admits ofnogeneral reduction, say,tothree one-body prob- lems, ortoanyother manageable setofequations. Till i '1 CID. 0 0 1'1 Tic 0 R ”"= '1» 0 Fro. 4-9. Center-of-mass andinternal coordinates ofasystem ofparticles. However, wecanpartially separate theproblem ofthemotion ofa system ofparticles intotwoproblems: ﬁrst, toﬁndthemotion ofthecenter ofmass, andsecond, toﬁndtheinternal motion ofthesystem, that is,the motion ofitsparticles relative tothecenter ofmass. Letusdeﬁne thein- ternal coordinate vector r};ofthekthparticle asthevector from thecenter ofmass tothelathparticle (Fig. 4-9): r,';=r,,-R, k=1,...,N, (4-120) r,,=R+r,';, k=1,...,N. (4-121) Inview ofthedeﬁnition (4-14) ofthecenter ofmass, theinternal co- ordinates r};satisfy theequation N -Zmp‘):=0. (4-122) k=l 186 THE MOTION orASYSTEM OFPARTICLES [CHAP. 4 Wedeﬁne thecenter-of-mass velocity andtheinternal velocities: l l v=R, (4-123) vi;=r,';=v,,-v. (4-124) Thetotal internal momentum ofasystem ofparticles (i.e., themomentum ‘relative tothecenter ofmass) vanishes byEq.(4—122): 5 ~-.‘l':,N IEm,,v,‘;=o. (4-125) k=1 Weﬁrst show that thetotal kinetic energy, momentum, andangular momentum caneach besplit upintoapart depending onthetotal mass M andthemotion ofthecenter ofmass, andaninternal part depending only ontheinternal coordinates andvelocities. Thetotal kinetic energy ofthe system ofparticles is ’Il‘4=rel!-‘5Pi‘ T= vi. (4-126) Bysubstituting forvl,from Eq.(4—124), andmaking useofEq.(4—125), wecansplit Tintotwoparts: . N i - I -T=Z11'mk(V2 +2V-vt+vi?) k-1 .TM=Nil-'3.2 2= 1.1/2+Zsmnl?+Zmkv-vlt = k=1 k=1 N N 2 .2 . =%MV +2smtvt +V-Z mivi k=1 k=1 N .=szm/2 +Z%m;,v;',2. ' (4-127) Ic=1 The total linear momentum is,ifwemake useofEqs. (4-124) and (4-125), L N P=Z mkVk k=1- zv N _ =2 m;,V +Z ’I1lkV]1§ k=1 k=1 =MV. (4—128) Theinternal linear momentum iszero. 4-9] THE N-BODY PROBLEM 187 Thetotal angular momentum about theorigin is,ifweuseEqs. (4—121), (4-122), (4—124), and(4-125), NL=Em,,(r,,xvl) k=1 N I=Z:ml,(RxV+r,',xV+Rxv,’,+rl‘,xv,‘,) l0=1 . N u=Zmk(RxV)+< r,Z)xV+Rx<Z:ml,v7,) k=1 = k=12 §M=§ N +Zmk(fi= XVi) k=1 =M(Rxv)+iv:m,.(r;;><v,';). (4-129) k=1 Notice that theinternal angular momentum depends only ontheinternal coordinates andvelocities andisindependent oftheorigin about which L isbeing computed (and from which thevector Risdrawn). The position ofparticle lcwith respect toparticle lisspeciﬁed bythe vector _ _ rl,—rl=ri—II. (4-130) The relative positions oftheparticles with respect toeach other depend only ontheinternal coordinates rl,,andlikewise therelative velocities, so that theinternal forces willbeexpected todepend only ontheinternal coordinates rl,,andpossibly ontheinternal velocities. Ifthere isapoten- tialenergy associated with theinternal forces, itlikewise willdepend only ontheinternal coordinates. Although theforces, energy, momentum, andangular momentum can each besplit intotwoparts, apart associated with themotion ofthecenter ofmass andaninternal partdepending only ontheinternal coordinates and velocities, itmust notbesupposed that theinternal motion andthecenter- of-mass motion aretwocompletely separate problems. Themotion ofthe center ofmass, asgoverned byEq.(4-18), isaseparate one-body problem when theexternal force Fisgiven. However, inmost cases Fwilldepend tosome extent ontheinternal motion ofthesystem. Theinternal equa- tions ofmotion contain theexternal forces except inspecial cases and, furthermore, they alsodepend onthemotion ofthecenter ofmass. Ifwe substitute Eqs. (4—121) inEqs. (4-1), andrearrange, wehave mm;=Fl‘;+F7,-mlii. (4-131) 188 THEMOTION orASYSTEM orPARTICLES ICHAP. 4 There aremany cases, however, inwhich agroup ofparticles forms a system which seems tohave some identity ofitsownindependent ofother particles andsystems ofparticles. Anatomic nucleus, made upofneutrons andprotons, isanexample, asisanatom, made upofnucleus andelectrons, oramolecule, composed ofnuclei andelectrons, orthecollection ofparticles which make upabaseball. Inallsuch cases, itturns outthat theinternal forces aremuch stronger than theexternal ones, andtheacceleration Ris small, sothat theinternal equations ofmotion (4—131) depend essentially only ontheinternal forces, andtheir solutions represent internal motions which arenearly independent oftheexternal forces andofthemotion of thesystem asawhole. The system viewed externally then behaves like asingle particle with coordinate vector R,mass M,acted onbythe (external) force F,butaparticle which has, inaddition toits“orbital” energy, momentum, andangular momentum associated with themotion ofitscenter ofmass, anintrinsic orinternal energy andangular momentum associated with itsinternal motion. Theorbital andintrinsic parts ofthe energy, momentum, and angular momentum canbeidentiﬁed inEqs. (4-127), (4-128), and(4-129). Theinternal angular momentum isusually called spin andisindependent oftheposition orvelocity ofthecenter of mass relative totheorigin about which thetotal angular momentum isto becomputed. Solong astheexternal forces aresmall, thisapproximate representation ofthesystem asasingle particle isvalid. Whenever the external forces arestrong enough toaffect appreciably theinternal motion, theseparation into problems ofinternal and oforbital motions breaks down andthesystem begins toloseitsindividuality. Some ofthecentral problems atthefrontiers ofpresent-day physical theories areconcerned with bridging thegapbetween aloose collection ofparticles andasystem with suﬁicient individuality tobetreated asasingle particle. 4-10 Two coupled harmonic oscillators. Avery commonly occurring type ofmechanical system isoneinwhich several harmonic oscillators interact with oneanother. Asatypical example ofsuch asystem, con- sider themechanical system shown inFig. 4-10, consisting oftwomasses ml,m2fastened toﬁxed supports bysprings whose elastic constants are kl,kg,andconnected byathird spring ofelastic constant k3.Wesuppose themasses arefreetomove only along theno-axis; they may, forexample, slide along larail. Ifspring kgwere notpresent, thetwomasses would I r 5”‘ ,,2’! \\‘ kl : ml kg m2 :kg FIG. 4-10. Asimple model oftwocoupled harmonic oscillators. 4-10] Two COUPLED HARMONIC OSCILLATORS 189 vibrate independently insimple harmonic motion with angular frequencies (neglecting damping) oi’,=,/%1, ago= (4-132) Wewish toinvestigate theeffect ofcoupling these twooscillators to- gether bymeans ofthespring loll.Wedescribe thepositions ofthetwo masses byspecifying thedistances zlanda:2that thesprings lclandI02 have been stretched from their equilibrium positions. Weassume for simplicity that when springs klandk2arerelaxed (xl=x2=0),spring k3isalsorelaxed. The amoimt bywhich spring ksiscompressed isthen (xl—l—x2). The equations ofmotion forthemasses ml,mg(neglecting friction) are ‘"7/1121 = —lO]_Il31 — lC3(Il?]_ +IE2), mgig = —k2Z2 '- k3(IE1 +132). Werewrite these intheform ‘m1551 +kiilli +793932 =0, (4—135) m2§52 +7¢'2$2 +793901 =0» (4-135) where kl=kl-!—I03, (4—137) kg=k2+k3. (4-138) Wehave two second-order linear diﬁerential equations tosolve simul- taneously. Ifthethird terms were notpresent, theequations would be independent ofoneanother, andwewould have independent harmonic vibrations ofxlandac;atfrequencies (.010 = kt0029 =' Z; ' These arethefrequencies with which each mass would vibrate iftheother were held ﬁxed. Thus theﬁrst effect ofthecoupling spring issimply to change thefrequency ofindependent vibration ofeach mass, duetothe factthat each mass isnow held inposition bytwosprings instead ofone. Thethird terms inEqs. (4-135) and(4—136) giverisetoacoupling between themotions ofthetwomasses, sothat they nolonger move independently. 190 THE MOTION oFASYSTEM or‘PARTICLES [CHAP. 4 Wemay solve Eqs. (4—135), (4-136) byanextension ofthemethod of Section 2-8applicable toanysetofsimultaneous linear differential equa- tions with constant coefﬁcients. Weassume that xl=ole“, , (4-141) $2=0261"‘, (4-142) where Cl,C2areconstants. Note that thesame time dependence isas- sumed forboth xland1:2,inorder that thefactor ep‘willcancel outwhen wesubstitute inEqs. (4-135) and(4—136): (mipz +k'1)C1 +70302 =0» (4443) (ma?+rec.+no.=0. <4-144) Wenow have two algebraic equations inthethree unknown quantities Cl,C2,p.Wenote that either Eq.(4—143) or(4—144) canbesolved for theratio C2/Cl: 4 2 / QZ=_ =__%_. (4_145) C1 ks m2? -l"kh Thetwovalues ofC2/C’lmust beequal, andwehave anequation forp: ' 2 i =+ , (4.446) I93 mgp + which may berearranged asaquadratic equation inp2,called thesecular equation: mim2P4 +(mzki +mik’2)P2 +(kiké -kg)=0, (4-147) whose solutions are .__1_k_'l Hm k'.)2_ an isll” p_ 2(ml +mg :|: 4:m1 +m2 mlmg +m1m2 =_1(w2 +wz):|: (Q2 _wz)2+_kg'__j|1/2. (4_148)2 10 20 4 10 20 mlmz Itisnothard toshow that thequantity inbrackets islessthan thesquare oftheﬁrst term, sothat wehave twonegative solutions forp2. Ifwe assume that wlllZ0:20, thesolutions forp2are P2=-4-vi =“(win +%A¢°2)l 2 2 I2 ,12 (41-149)P=—¢°2 =—('-"20 —2A‘-° )» 4-10] TWO COUPLED HARMONIC OSCILLATORS 191 where 2 2 2 4K4 1/2Aw -_=(0010 —(.029) 1+ é —11 (4:-150) with theabbreviation K2= <4-151) mlmg where Kisthecoupling constant. Ifwlo=0:20, Eq.(4—150) reduces to Awz =21:2. (4-152) Thefour solutions forpare P=:E’I:O)1, :.|:’l:¢02. Ifp2=-10%, Eq.(4—145) canbewritten Q_Lu2_2_2&2/L";C1 ks ((0 (.010) 2 21— 1 —K2 m andifp2=—w§, itcanbewritten 0' m Aw2 m3=-,;f<<»%-wit)=—§;(-2-,/If (4-155) Bysubstituting from Eq.(4—153) inEqs. (4—141), (4—142), wegetfour solutions ofEqs. (4—135) and (4—136) provided theratio C2/C1 ischosen according toEq.(4—-154) or(4-155). Each ofthese solutions involves one arbitrary constant (C1orC2). Since theequations (4—135), (4—l36) are linear, thesum ofthese four solutions willalsobeasolution, andisinfact thegeneral solution, foritwill contain four arbitrary constants (say C11 Cir C22 .0’ _.w _A2 . A2 _.w x1=Ole’ ‘t+Cﬂe ’"—5%‘/%jC2e'”" —-2%"Znm—iC§e '2‘, (41-156) Z2=éég Cleiaut +$2? C/1e—iul1t +Czeiaazyt +C58-1'w2l_ (4—157) Inorder tomake 1:1and$2real, wechoose 01=%A1e“’*, Ca=%A1e-‘"1, (+158) 02=%A2@"”=*, cg=g-A26-"°=, (4-159) 192 THEMOTION orASYSTEM orPARTICLES [CHAP¢ 4 sothat ACO2 m2 131==A1 0OS((.01t +01) —'Ta.‘ A2 COS (wgt +02), 2 x2=£‘ if-niA1cos(wlt+01)+A2cos(wzt+02). (4-161)2 2:42 m This isthe general solution, involving the four arbitrary constants A1,A2,01,02. Weseethat themotion ofeach coordinate isasuper- position oftwoharmonic vibrations atfrequencies 0:1and(.02. The os- cillation frequencies arethesame forboth coordinates, buttherelative amplitudes arediﬁerent, andaregiven byEqs. (4—154) and(4—155). IfA1orA2iszero, only onefrequency ofoscillation appears. The re- sulting motion iscalled anormal mode ofvibration. The normal mode of highest frequency isgiven by Z131=A1 COS ((.O1t -|-01), AL02 ml222=5-K? -‘ A10OS(w1t +91), (4—163) 2 ‘J1’=mi,+%Aw2. (4-164) Thefrequency ofoscillation ishigher thanwm. Byreferring toFig.4-10, weseethat inthismode ofoscillation thetwomasses m1andm2areoscil- lating outofphase; that is,their displacements areinopposite directions. Themode ofoscillation oflower frequency isgiven by A602 mgx1=—Ta, A2cos(wgt+02), (4—165) $2 = A2 COS (wgt + 02), cog=Q20—%Aw2. (11-167) Inthis mode, thetwo masses oscillate inphase atafrequency lower than (029. Themost general motion ofthesystem isgiven byEqs. (11-160), (4—161), andisasuperposition ofthetwonormal modes ofvibration. Theeffect ofcoupling isthus tocause both masses toparticipate inthe oscillation ateach frequency, andtoraise thehighest frequency andlower thelowest frequency ofoscillation. Even when both frequencies are initially equal, thecoupling results intwofrequencies ofvibration, one higher and onelower than thefrequency without coupling. When the coupling isvery weak, i.e.,when 'K2<<%<<»%o—wit), <4-168) 4-10] TWO COUPLED HARMONIC OSCILLATORS 193 then Eq.(4-1-50) becomes 24Aw2iTL; (+169) 6°10—Q20 Forthehighest frequency mode ofvibration, theratio oftheamplitude of vibration ofmass m2tothat ofmass mlisthen $2:Awz [mli_ K2 ’ml_ (4470) xl 21:2 m2 wfo_0,30 mg Thus, unless mg<<ml,themass m2oscillates atmuch smaller amplitude than ml. Similarly, itcanbeshown that forthelow-frequency mode of vibration, mloscillates atmuch smaller amplitude than m2. Iftwooscil- lators ofdifferent frequency areweakly coupled together, there aretwo normal modes ofvibration ofthesystem. Inonemode, theoscillator of higher frequency oscillates atafrequency slightly higher than without coupling, andtheother oscillates weakly outofphase atthesame fre- quency. Intheother mode, theoscillator oflowest frequency oscillates at afrequency slightly lower than without coupling, andtheother oscillates weakly andinphase atthesame frequency. Atornear resonance, when thetwonatural frequencies wloand0:20areequal, thecondition forweak coupling [Eq.(4-168)] isnotsatisﬁed even when thecoupling constant is verysmall. Aw2isthengiven byEq.(4—152), andweﬁndforthetwonor-. malmodes ofvibration: , Q=1,/E, (4-171) $1 mg 0,2=will=sK2. (4-172) The twooscillators oscillate inoroutofphase with anamplitude ratio depending only ontheir mass ratio, andwith afrequency higher orlower than theuncoupled frequency byanamount depending onthecoupling constant. Aninteresting special case isthecase oftwo identical oscillators (ml=mg,kl=I02)coupled together. The general solution (4—160), (4—-161) is,inthiscase, asl=Alcos(colt+0l)—A2cos(wgt+02), (4—173) $2 = A1 COS (wlt + 01) + A2 COS (w2t + 02), where wland0:2aregiven byEq.(4—172). IfA2=0,wehave thehigh- frequency normal mode ofvibration, andifAl=0,wehave thelow-fre- quency normal mode. Letussuppose that initially m2isatrestinits equilibrium position, while mlisdisplaced adistance Afrom equilibrium 194 THEMOTION orASYSTEM OFPARTICLES [cn.u=. 4 andreleased att=0.The choice ofconstants which ﬁtsthese initial conditions is 01:02:01 4-175A1=—A2=%A, () sothat Eqs. (4—173), (-1-174) become xl=%A(coswlt+coswzt), (4—l76) :02=—1§A(coswlt—cos(1)20, (11-177) which canberewritten intheform xl=Acos t)cos t)» (4—178) $2=-Asin z)Sin ¢)- (4-179) Ifthecoupling issmall, wland0:2arenearly equal, and:z:landx2oscillate rapidly attheangular frequency (wl+0:2)/2 -*-wlé0:2,with anam- plitude which varies sinusoidally atangular frequency (wl—<02)/2. The motion ofeach oscillator isasuperposition ofitstwonormal-mode motions, which leads tobeats, thebeat frequency being thedifference between the twonormal-mode frequencies. This isillustrated inFig. 4—11,where os- cillograms ofthemotion ofx2areshown: (a)when thehigh-frequency normal mode alone isexcited, (b)when thelow-frequency normal mode isexcited, and(c)when oscillator mlalone isinitially displaced. InFig. 4—12, oscillograms ofxland :02asgiven byEqs.- (4-178), (4—179) are shown. Itcanbeseen that theoscillators periodically exchange their energy, duetothecoupling between them. Figure 4—13 shows thesame motion when thesprings klandkgarenotexactly equal. Inthiscase, oscillator mldoes notgiveupallitsenergy tomgduring thebeats. Figure 4-14 shows that theeffect ofincreasing thecoupling istoincrease thebeat frequency wl—(.02[Eq. (4—172)]. Ifafrictional force acts oneach oscillator, theequations ofmotion (-1-135) and(4—136) become m1Ii§1 +b1ZiI1 +10,1131 +kgilig =0, mgﬁg +bgillg +kéﬂg +[(331131 =0, 4-10] TWO COUPLED HARMONIC OSCILLATORS 195 (a) (b) (*1) (C) (b) FIG. 4-11. Motion ofcoupled har- FIG. _4—12. Motion oftwoidentical monic oscillators. (a)High-frequency coupled oscillators. normal mode. (b)Low-frequency nor- mal mode. (c)mlinitially displaced. (#1) (=1) (b) (b) FIG. 4-13. Motion oftwononidenti- FIG. 4-14. Motion oftwo coupled calcoupled oscillators. oscillators. (a)Weak coupling. (b) Strong coupling. 196 THE MOTION orASYSTEM orPARTICLES [CHAP. 4 where blandblaretherespective friction coefﬁcients. The substitution (4—141), (4—142) leads toafourth-degree secular equation forp: m1m2I14 +(mzbi -1-m1b2)P3 +(mzki +Ynikb -'1'b1b2)P2 +(51795 -1-b2k'1)P +(kikﬁ —kg)=0-(4482) This equation cannot besolved soeasily asEq.(4-147). The four roots forpare,ingeneral, complex, andhave theform (ifblandb2arenottoo large) 1 A p=—-'Yl :1:iwl, . 4-183 p=' '—'Y2 :|: 7402. ( ) That theroots have thisform with 'Yland‘Y2positive canbeshown (though noteasily) algebraically from astudy ofthecoeﬂicients inEq.(4-182). Physically, itisevident that theroots have theform (4—183), since thiswill leadtodamped vibrations, theexpected result offriction. Ifblandb2are large enough, oneorboth ofthepairs ofcomplex roots may become apair ofreal negative roots, thecorresponding normal mode ormodes being overdamped. Apractical solution ofEq.(4—182) can, ingeneral, beob- tained) onlybynumerical methods when numerical values fortheconstants aregiven, although anapproximate algebraic solution canbefound when thedamping isvery small. The problem ofthemotion ofasystem oftwocoupled harmonic oscil- lators subject toaharmonically oscillating force applied toeither mass can besolved bymethods similar tothose which apply toasingle harmonic oscillator. Asteady-state solution canbefound inwhich both oscillators oscillate atthefrequency oftheapplied force with deﬁnite amplitudes and phases, depending ontheir masses, thespring constants, thedamping, and theamplitude andphase oftheapplied force. Thesystem isinresonance with theapplied force when itsfrequency corresponds toeither ofthetwo normal modes ofvibration, andthemasses then vibrate atlarge amplitudes limited only bythedamping. Thegeneral solution consists ofthesteady- state solution plus thegeneral solution oftheunforced problem. Asuper- position principle canbeproved according towhich, ifanumber offorces actoneither orboth masses, thesolution isthesum ofthesolutions with each force acting separately. This theorem canbeused totreat theprob- lemofarbitrary forces acting onthetwomasses. Other types ofcoupling between theoscillators arepossible inaddition tocoupling bymeans ofaspring asintheexample above. Theoscillators may becoupled byfrictional forces. Asimple example would bethecase where onemass slides overtheother, asinFig.4-15. Weassume thatthe force offriction isproportional totherelative velocity ofthetwomasses. W !<-—$1—>I ml ‘gxig.4-10] TWO COUPLED rmnmomo oscrnmvrons 197 |__x2___: k2 kl k2 FIG. 4-15. Frictional coupling. FIG. 4-16. Coupling through amass.aé § Q>6” Theequations ofmotion ofmlandmgarethen m1§51 =-7611111 —b(¢1 +932), (4-184) m2:'é2 =—lC2il32 '-' +I531), 01‘ 7711121 +bIi31 +1611131 +big =0, ‘"1252 + big +102132 + bill = 0- Thecoupling isexpressed inEqs. (4-186), (4-187) byaterm intheequation ofmotion ofeach oscillator depending onthevelocity oftheother. The oscillators may alsobecoupled byamass, asinFig.4-16. Itislefttothe reader tosetuptheequations ofmotion. (See Problem 26attheendof thischapter.) Two oscillators may becoupled insuch away that theforce acting on onedepends ontheposition, velocity, oracceleration oftheother, oron anycombination ofthese. Ingeneral, allthree types ofcoupling occur to some extent; aspring, forexample, hasalways some mass, andissubject tosome internal friction. Thus themost general pairofequations fortwo coupled harmonic oscillators isoftheform M1531 -1'51551 +791111 +"M32 +bciiz+k¢$2 =0, (4-1-88) ‘M2552 +112512 +762112 +"@551 +bail +196131 =0- (4-189) These equations canbesolved bythemethod described above, with similar results. Two normal modes ofvibration appear, ifthefrictional forces arenottoogreat. Equations oftheform (4—188), (4—189), orthesimpler special cases con- sidered inthepreceding discussions, arisenotonlyinthetheory ofcoupled mechanical oscillators, butalsointhetheory ofcoupled electrical circuits. Applying Kirchhoff’s second lawtothetwomeshes ofthecircuit shown in Fig. 4-17, with mesh currents il,1'2around thetwomeshes asshown, We obtain (L+L011+(R+R011++({)q1 +La+Ra.+§q2=0. (4-1!-)0) 198 THE MOTION orASYSTEM or‘PARTICLES [CHAP. 4 R1 R2 R C L1 L2 il 1'2 llLII C1 C2 FIG. 4-17. Coupled oscillating circuits. and (L+L2)<i2 +(R+R2)(l2 + +%)q2 -1"Lfli'1‘R91+%q1 =0» (41-191) where qlandQ2arethecharges built uponClandC2bythemesh currents iland1'2.These equations have thesame form asEqs. (4—188), (4-189), andcanbesolved bysimilar methods. Inelectrical circuits, thedamping isoften fairly large, andﬁnding thesolution becomes aformidable task. Thediscussion ofthissection canbeextended tothecaseofanynumber ofcoupled mechanical orelectrical harmonic oscillators, with analogous results. The algebraic details become almost prohibitive, however, unless wemake useofmore advanced mathematical techniques. Wetherefore postpone further discussion ofthisproblem toChapter 12. Allmechanical andelectrical vibration problems reduce inthelimiting case ofsmall amplitudes ofvibration toproblems involving) oneorseveral coupled harmonic oscillators. Problems involving vibrations ofstrings, membranes, elastic solids, andelectrical andacoustical vibrations intrans- mission lines, pipes, orcavities, canbereduced toproblems ofcoupled oscillators, andexhibit similar normal modes ofvibration. Thetreatment ofthebehavior ofanatom ormolecule according toquantum mechanics results inamathematical problem identical with theproblem ofcoupled harmonic oscillators, inwhich theenergy levels play theroleofoscillators, andexternal perturbing inﬂuences play theroleofthecoupling mechanism. 199 PROBLEMS 1.Formulate andprove aconservation lawfortheangular momentum about theorigin ofasystem ofparticles conﬁned toaplane. 2.Water ispoured intoabarrel attherateof120lbperminute from aheight of16ft. The barrel weighs 25lb,andrests onascale. Find thescale reading after thewater hasbeen pouring into thebarrel foroneminute. 3.Ascoop ofmass m1isattached toanarm oflength landnegligible weight. Thearmispivoted sothatthescoop isfreetoswing inavertical arcofradius l. Atadistance ldirectly below thepivot isapileofsand. Thescoop islifted until thearm isata45°angle with thevertical, andreleased. Itswings down and scoops upamass mgofsand. Towhat angle with thevertical does thearmofthe scoop riseafter picking upthesand? This problem istobesolved byconsidering carefully which conservation laws areapplicable toeach part oftheswing ofthe scoop. Friction istobeneglected, except that required tokeep thesand inthe scoop. 4.(a)Aspherical satellite ofmass m,radius a,moves with speed vthrough atenuous atmosphere ofdensity p.Find thefrictional force onit,assuming that thespeed oftheairmolecules canbeneglected incomparison with v, and that each molecule which isstruck becomes embedded intheskin ofthe satellite. (b)Iftheorbit isacircle 400kmabove theearth (radius 6360 km), where p=10'“ kgm/m“3, and ifa=1m,m=100kgm, ﬁnd thechange inaltitude andthechange inperiod ofrevolution inoneweek. 5.Atwo-stage rocket istobebuilt capable ofaccelerating a100-kgm payload toavelocity of6000 m/sec infreeﬂight. (Inatwo-stage rocket, theﬁrst stage isdetached after exhausting itsfuel, before thesecond stage isﬁred.) Assume thatthefuelused canreach anexhaust velocity of1500m/sec, andthatstruc- tural requirements imply that anempty rocket (without fuelorpayload) will weigh 10°70 asmuch asthefuelitcancarry. Find theoptimum choice ofmasses forthetwostages sothat thetotal take-off weight isaminimum. Show that it isimpossible tobuild asingle-stage rocket which willdothejob. 6.Arocketis tobeﬁred vertically upward. Theinitial mass isMQ,theexhaust velocity —uisconstant, and therate ofexhaust —(dM/dt) =Aisconstant. After atotal mass AM isexhausted, therocket engine runs outoffuel. Neglect- ingairresistance andassuming that theacceleration gofgravity isconstant, setupandsolve theequation ofmotion, andshow thatifM0,u,andAMare ﬁxed, then thelarger therate ofexhaust A,that is,thefaster ituses upitsfuel, thegreater themaximum altitude reached bytherocket. 7.Auniform spherical planet ofradius arevolves about thesuninacircular orbit ofradius 1'0,androtates about itsaxis with angular velocity wo,normal totheplane oftheorbit. Due totides raised ontheplanet, itsangular velocity ofrotation isdecreasing. Find aformula expressing theorbit radius rasafunc- tion ofangular velocity wofrotation atany later orearlier time. [You will need formulas (5-9) and (5~9l) from Chapter 5.]Apply your formula tothe earth, neglecting theeffect ofthemoon, and estimate how much farther the earth willbefrom thesunwhen thedayhasbecome equal tothepresent year. Iftheeffect ofthemoon were taken into account, would thedistance begreater orless? 200 THE MOTION orASYSTEM OFPARTICLES [on.»u>. 4 *8.Amass mofgasanddebris surrounds astarofmass M.The radius ofthe star isnegligible incomparison with thedistances totheparticles ofgasand debris. Thematerial surrounding thestarhasinitially atotal angular momentum L,andatotal kinetic and potential energy E.Assume that m<<M,sothat thegravitational ﬁelds duetothemass marenegligible incomparison with that ofthestar. Due tointernal friction, thesurrounding material continually loses mechanical energy. Show that there isamaximum energy AEwhich canbelost inthisway, andthatwhen thisenergy hasbeen lost,thematerial must alllie onacircular ring around thestar (but notnecessarily uniformly distributed). Find AEandtheradius ofthering. (You willneed tousethemethod ofLa- grange multipliers.) 9.Aparticle ofmass m1,energy T11collides elastically with aparticle ofmass mg,atrest. Ifthemass mgleaves thecollision atanangle rigwith theoriginal direction ofmotion ofm1,ﬁnd theenergy TZFdelivered toparticle mg. Show that Tgpisamaximum forahead-on collision, andthat inthiscase theenergy lostbytheincident particle inthecollision is l 4m1'm2T"-T"= T"- 10.Acloud-chamber picture shows thetrack ofanincident particle which makes acollision andisscattered through anangle 171.The track ofthetarget particle makes anangle 0gwiththedirection oftheincident particle. Assuming thatthecollision waselastic andthatthetarget particle wasinitially atrest,ﬁnd theratio m1/m2 ofthetwomasses. (Assume small velocities sothattheclassical expressions forenergy andmomentum may beused.) 11.Show that anelastic collision corresponds toacoeﬂicient ofrestitution e=1,that is,show that forahead-on elastic collision between twoparticles, Eq.(4-85) holds with e=1. 12.Calculate theenergy lossQforahead-on collision between aparticle of mass m1,velocity v1with aparticle ofmass mgatrest, ifthecoeﬂicient ofrestitu- tion ise. 13.Aparticle ofmass ml,momentum pl]collides elastically with aparticle ofmass mg,momentum I121going intheopposite direction. Ifm1leaves thecolli- sionatanangle :91withitsoriginal course, ﬁnditsﬁnalmomentum. 14.Find therelativistic corrections toEq.(4-81) when theincident particle m1andtheemitted particle m3move with speeds near thespeed oflight. Assume that therecoil particle m4ismoving slowly enough sothat theclassical relation between energy andmomentum canbeused forit. 15.Aparticle ofmass m1,momentum p1collides with aparticle ofmass mgat rest. Areaction occurs from which two particles ofmasses m3andm4result, which leave thecollision atangles 03and04with theoriginal path ofm1. Find theenergy Qabsorbed inthereaction interms ofthemasses, theangles, andp1. 16.Anuclear reaction whose Qisknown occurs inaphotographic plate in which thetracks oftheincident particle m1andthetwoproduct particles mg andm4canbeseen. Find theenergy oftheincident particle interms ofm1, mg,m4,Q,andthemeasured angles 03and04between theincident track and thetwoﬁnal tracks. What happens ifQ=0? PROBLEMS 201 17.TheCompton scattering ofx-rays canbeinterpreted astheresult ofelastic collisions between x-ray photons andfreeelectrons. According toquantum theory, aphoton ofwavelength Ahasakinetic energy hc/X, andalinear momen- tum ofmagnitude h/)\, where hisPlanck’s constant andcisthespeed oflight. IntheCompton effect, anincident beam ofx-rays ofknown wavelength A1ina known direction isscattered inpassing through matter, andthescattered radia- tionatanangle :91totheincident beam isfound tohave alonger wavelength AF, which isafunction oftheangle 01.Assuming anelastic collision between aninci- dent photon andanelectron ofmass matrest, setuptheequations expressing conservation ofenergy andmomentum. Usetherelativistic expressions forthe energy andmomentum oftheelectron. Show that thechange inx-ray wave- length is hp—M=%(1—cos171), andthat theejected electron appears atanangle given by tang2=_iiLi_ . [1+(h/>~nrw)](1 —cos191) 18.Work outacorrection toEq.(3—267) which takes intoaccount themotion ofthecentral mass Munder theinﬂuence oftherevolving mass m.Apairofstars revolve about eachother, soclose together thattheyappear inthetelescope asa single star. Itisdetermined from spectroscopic observations thatthetwostars areofequal mass andthateach revolves inacircle with speed vandperiod 1- under thegravitational attraction oftheother. Find themass mofeach tarby using your formula. 19.Show that iftheincident particle ismuch heavier than thetarget particle (mi>>mg),theRutherford scattering cross section da[Eq.(3-27 6)]inlaboratory coordinates isapproximately 2 2g q1q2 47 s. d“_ 2 221122 221/22"'sm"1 “M121n2vQ [1—-(1——'7131) ](1—'7131) if7191 <1,where ’Y=m1/mg. Otherwise, dc=0. 20.Find anexpression analogous toEq.(-1-116) fortheangle ofrecoil ofthe target particle (02inFig.4-7) interms ofthescattering angle 6intheequivalent one-body problem. Show that, foranelastic collision, 192=%(1I'—59)- 21.Assume that mg>>m1,andthat 6=01+6,inEq.(4-117). Find a formula for5interms of191.Show that theﬁrst-order correction totheRuther- ford scattering cross section [Eq. (3-276)], duetotheﬁnite mass ofmg,vanishes. 22.Setuptheequations ofmotion forFig. 4-10, assuming that therelaxed length ofeach spring isl,andthat thedistance between thewalls is3(l—|—a),so that thesprings arestretched, even intheequilibrium position. Show that the equations canbeputinthesame form asEqs. (4—135) and(-1-136). 202 THEMOTION orASYSTEM orPARTICLES [cn.u>. 4 23.Forthenormal mode ofvibration given byEqs. (4-162) and(4—163), ﬁnd theforce exerted onm1through thecoupling spring, andshow thatthemotion of:01satisﬁes theequation forasimple harmonic oscillator subject tothisdriving force. 1 24.The system ofcoupled oscillators shown inFig. 4-10 issubject toanap- plied force F=F0coswt, applied tomass m1. Setuptheequations ofmotion andﬁndthesteady-state solution. Sketch theamplitude andphase oftheoscillations ofeach oscillator as functions ofw. 25.Find thetwonormal modes ofvibration forapairofidentical damped coupled harmonic oscillators [Eqs. (4—180), (4—181)]. That is,m1=mg,b1=bg, k1=kg.[Hint:IfI63=0,youcancertainly ﬁndthesolution. Youwillﬁndthis point helpful infactoring thesecular equation.] 26.Setuptheequations ofmotion forthesystem shown inFig.4—16. The relaxed lengths ofthetwosprings areZ1,lg.Separate theproblem intotwo problems, oneinvolving themotion ofthecenter ofmass, andtheother involving the“internal motion” described bythetwocoordinates x1,xg.Find thenormal modes ofvibration. CHAPTER 5 RIGID BODIES. ROTATION ABOUT AN AXIS. STATICS 5-1Thedynamical problem ofthemotion ofa.rigid body. Inorder to apply thetheorems ofthepreceding chapter tothemotion ofarigid body, weregard arigid body asasystem ofmany particles whose positions rela- tivetooneanother remain ﬁxed. Wemay deﬁne arigid body asasystem ofparticles whose mutual distances areallconstant. The forces which hold theparticles atﬁxed distances from oneanother areinternal forces, andmay beimagined asexerted byrigid weightless rods connected be- tween allpairs ofparticles. Forces likethiswhich maintain certain ﬁxed relations between theparticles ofasystem arecalled forces ofconstraint. Such forces ofconstraint canalways beregarded assatisfying Newton’s third law(strong form), since theconstraints could bemaintained byrigid rods fastened totheparticles byfrictionless universal joints. Wemay therefore apply thetheorems ofconservation oflinear andangular momen- tlmi tothemotion ofarigid body. Foraperfectly rigid body, thetheorem ofconservation ofmechanical energy holds also, since wecanshow by Newton’s third lawthat theforces ofconstraint donowork inarigid mo- tion ofthesystem ofparticles. Thework done bytheforce exerted bya moving rodonaparticle atoneendisequal andopposite tothework done bytheforce exerted bytherodonaparticle attheother end, since both particles have thesame component ofvelocity inthedirection oftherod (Fig. 5—1): F2->1'V1 +F1->2'V2 =F2-»1'V1 —F2->1'V2 (5—1) =F2_>1°(V1~- V2) =0. Weshall base ourderivation oftheequations ofmotion ofarigid body onthese conservation laws. Noactual solid body iseverperfectly rigid, sothat ourtheory ofthemotion ofrigid bodies willbeanidealized ap- proximation tothemotion ofactual bodies. However, inmost applica- tions thedeviation ofactual solid bodies from true rigidity isnotsig- niﬁcant. Inalikespirit isourassumption that theideal rigid body can beimagined asmade upofideal point particles held atﬁxed distances from oneanother. Asolid body ofordinary sizeiscomposed ofsuch alarge number of atoms andmolecules that formost purposes itismore convenient torepre- 203 204 men) BODIES. ROTATION ABOUT ANAXIS. STATICS [CHAP. 5 .v1 V1-—-V2 V2 ’I‘\‘ F1->2 I ‘ "12I ‘\III ‘ml F2-> 1 FIG. 5-1. Forces exerted bytwoparticles connected byarigid rod. sent itsstructure byspecifying theaverage density pofmass perunit volume ateach point inthebody. Thedensity isdeﬁned by. dMP—3'17' (5-2) where dMisthetotal mass inavolume dVwhich istobechosen large enough tocontain alarge number ofatoms, yetsmall enough sothatthe properties ofthematerial arepractically uniform within thevolume dV. Only when adVsatisfying these tworequirements canbechosen inthe neighborhood ofapoint inthebody canthedensity pbeproperly deﬁned atthat point. Sums over alltheparticles, such asoccur intheexpressions fortotal mass, total momentum, etc., canbereplaced byintegrals over the volume ofthebody. Forexample, thetotal mass is M=Zm,-=[f/pdv. (5-3) '1 (body) Further examples willappear inthefollowing sections. Inorder todescribe theposition ofarigid body inspace, sixcoordinates areneeded. Wemay, forexample, specify thecoordinates (001,1/1,21) of some point P1inthebody. Any other point Pgofthebody adistance r from P1will then liesomewhere onasphere ofradius rwith center at (x1,y1, zl). Wecanlocate P2onthis sphere with two coordinates, for example, thespherical coordinate angles 0g,<pgwith respect toasetof axes through thepoint (x1,yl,zl). Any third point P3adistance a950 from thelinethrough P1andP2must nowlieonacircle ofradius aabout thisline. Wecanlocate P3onthiscircle with onecoordinate. Wethus require atotal ofsixcoordinates tolocate thethree points P1,Pg,P3of 5-1] THEDYNAMICAL PROBLEM or‘THEMOTION orARIGID BODY 205 thebody, andwhen three noncollinear points areﬁxed, thelocations of allpoints ofarigid body areﬁxed. There aremany possible ways ofchoos- ingsixcoordinates bywhich theposition ofabody inspace canbespeciﬁed. Usually three ofthesixcoordinates areused asabove tolocate some point inthebody. The remaining three coordinates determine theorientation ofthebody about thispoint. Ifabody isnotconnected toanysupports, sothat itisfreetomove in anymanner, itisconvenient tochoose thecenter ofmass asthepoint to belocated bythree coordinates (X,Y,Z),orbythevector _R.Themotion ofthecenter ofmass Risthen determined bythelinear momentum theo- rem, which canbeexpressed intheform (4-18): Mii=F, (5-4) where Misthetotal mass andFisthetotal external force. Theequation fortherotational motion about thecenter ofmass isgiven bytheangular momentum theorem (4-28): dLE—N, (5~5) where Listheangular momentum andNisthetorque about thepoint R. Iftheforce Fisindependent oftheorientation ofthebody inspace, asin thecase ofabody moving inauniform gravitational ﬁeld, themotion of thecenter ofmass isindependent oftherotational motion, andEq.(5-4) isaseparate equation which canbesolved bythemethods ofChapter 3. Ifthetorque Nisindependent oftheposition Rofthecenter ofmass, or ifR(t) isalready known, sothat Ncanbecalculated asafunction oftime andoftheorientation ofthebody, then therotational motion about the center ofmass may bedetermined from Eq.(5-5). Inthemore general case, when FandNeach depend onboth position andorientation, Eqs. (5-4) and (5-5) must besolved simultaneously assixcoupled equations insome suitable setofcoordinates; thiscaseweshall notattempt totreat, although after thereader hasstudied Chapter 11,hewillbeabletosetup forhimself thesixequations which must besolved. Ifthebody isconstrained byexternal supports torotate about aﬁxed point O,then moments andtorques aretobecomputed about that point. Wehave tosolve Eq.(5-5) fortherotation about thepoint O.Inthiscase Eq. (5-4) serves only to"determine theconstraining force required to maintain thepoint Oatrest. Thediﬂiculty inapplying Eq.(5-5) liesinthechoice ofthree coordinates todescribe theorientationof thebody inspace. The ﬁrst thought that comes tomind istochoose azero position forthebody, andtospecify anyother orientation byspecifying theangles ofrotation <02,‘Pu:$02,about 206 RIGID BODIES. ROTATION ABOUT ANAXIS. STATICS [cnAr. 5 three perpendicular axes, required tobring thebody tothis orientation. However, alittle experimenting with asolid body willconvince anyone that nosuitable coordinates ofthis sort exist. Consider, forexample, theposition speciﬁed byox=90°, <p,,=90°, <p,=0.Ifabody isﬁrst rotated 90°about thex-axis, andthen 90°about they-axis, theﬁnal posi- tion willbefound tobedifferent from that resulting from a90°rotation about they-axis followed bya90°rotation about theac-axis. Itturns out that nosimple symmetric setofcoordinates canbefound todescribe the orientation ofabody, analogous tothecoordinates 01:,y,zwhich locate the position ofapoint inspace. Wetherefore postpone toChapter 11the treatment oftherather difficult problem oftherotation ofabody around apoint. Weshall discuss here only thesimple problem ofrotation about aﬁxed axis. * 5-2Rotation about anaxis. Itrequires onl'y onecoordinate tospecify theorientation ofabody which isfreetorotate only about aﬁxed axis. Lettheﬁxed axisbetaken asthez-axis, andletalineOT1inthebody, through theaxisandlying in(orparallel to)thexy-plane, bechosen. We ﬁxtheposition ofthebody byspecifying theangle 0between thelineOI ﬁxed inthebody and thex-axis. Choosing cylindrical coordinates to locate each particle inthebody, wenowcompute thetotal angular momen- tum aboutthe z-axis. (See Fig. 5-2.) Weshall write r,-instead ofp,-to represent thedistance ofparticle m,-from thez-axis, inorder toavoid con- fusion with thedensity p: L=Em.»»?¢.-. (5-6) ‘L Letﬁrbetheangle between thedirection ofthelineOAinthebody and thedirection oftheradius from thez-axis totheparticle m,-.Then, fora Z EA . Fm. 5-2. Coordinates ofaparticle inarigid body. 5-2] ROTATION ABOUT ANAXIS 207 rigid body, B,isconstant, and <01‘=9+Bi; (5-7) ‘ ¢i=9- (5*8) Substituting inEq.(5-6), wehave L=Zm;r?9 i = mgr?) 9 =1.6, (5-9) where I,=Zmyrg. (5-10) 1~ ~ . Thequantity I,isaconstant foragiven body rotating about agiven axis, andiscalled themoment ofinertia about that axis. AWemay alsoexpress I,asanintegral over thebody: 1,=ff]pr2dV. (5-11) _ (body) Itissometimes convenient tointroduce theradius ofgyration kgdeﬁned by theequation Mk?=1,; (5-12) that is,kgisaradius such that ifallthemass ofthebody were situated a distance lo,from theaxis, itsmoment ofinertia would be1,. Using Eq.(5-9), wemay write thecomponent ofEq.(5-5) along the axisofrotation intheform dLW=Igd-N3, (5-13) where N3isthetotal external torque about theaxis. Equation (5-13) is theequation ofmotion forrotation ofarigid body about aﬁxed axis. It hasthesame form asEq.(2-1) forthemotion ofaparticle along astraight line. The problem ofrotation ofabody about aﬁxed axisistherefore equivalent totheproblem treated inChapter 2.Allmethods andresults ofChapter 2canbeextended directly tothepresent problem according to thefollowing scheme ofanalogy: u 208 RIGID BODIES. ROTATION ABOUT ANAXIS. STATICS [onAr. 5 Rectilinear motion Rotation about aﬁxed axis position: as angular position: 6 velocity: v=a‘: angular velocity: w=0 acceleration: a=55 angular acceleration: oz=9 force : F torque : N, mass: m moment ofinertia: I, potential energy: potential energy: V(x)=-F(x)da: V(a)=-/;’N.(e)do dV dV kinetic energy: T=§~m.i:2 kinetic energy: T=111,02 linear momentum: p=ma‘: angular momentum: L=I,6 The only mathematical difference between thetwoproblems isthat the moment ofinertia I,depends upon thelocation oftheaxisinthebody, while themass ofabody doesnotdepend onitsposition oronitsmotion. This doesnotaffect thetreatment ofrotation about asingle ﬁxed axis. Therotational potential andkinetic energies deﬁned bytheequations, 1/(0)=-/0”N,(0)d6, (5-14) dVN,=—W» (5-15) T=-3-1,02, (5-16) arenotmerely analogous tothecorresponding quantities deﬁned byEqs. (2-41), (2-47), and(2-5) forlinear motion. They are,infact, equal tothe potential andkinetic energies, deﬁned inChapters 2and4,ofthesystem ofparticles making uptherigid body. The potential energy deﬁned by Eq.(5-14), forexample, isthework done against theforces whose torque isNZ,when thebody isrotated through theangle 0—0,.The kinetic energy deﬁned byEq.(5-16) isjustthesumoftheordinary kinetic energies ofmotion oftheparticles making upthebody. Theproof ofthese state- ments isleftasanexercise. 5-3The simple pendulum. Asanexample ofthetreatment ofrota- tional motion, weconsider themotion ofasimple pendulum, consisting of amass msuspended from aﬁxed point 0byastring orweightless rigid rod 5-3] THE SIMPLE PENDULUM 209 l m "I9 FIG. 5-3. The simple pendulum. oflength Z.Ifastring supports themass m,wemust suppose that it remains taut, sothat thedistance lfrom mtoOremains constant; other- wise wecannot treat thesystem asarigid one. Weconsider only motions ofthependulum inonevertical plane, inorder tobeable toapply the simple theory ofmotion about asingle ﬁxed axis through O.Wethen have (Fig. 5-3) I,=ml2, (5-17) N,=—mgl sin0, (5-18) where thez-axis isanaxisthrough Operpendicular totheplane inwhich thependulum isswinging. Thetorque istaken asnegative, since itacts insuch adirection astodecrease theangle 0.Substituting intheequation ofmotion (5-13), weﬁnd 5=--gsin0. (5-19) This equation isnoteasy tosolve. If,however, weconsider only small oscillations ofthependulum (say 0<<1r/2), then sin0i0,andwecan write 9+-‘Z10-0. (5-20) This isofthesame form asEq.(2-84) fortheharmonic oscillator. Its solution is 0=Kcos(wt+B), (5-21) where _21/2 _ . 5_(l), (522) andKandBarearbitrary constants which determine theamplitude and phase oftheoscillation. Notice that thefrequency ofoscillation is 210 ruoro BODIES.) ROTATION ABOUT ANAXIS. STATICS [cruun 5 independent oftheamplitude, provided theamplitude issmall enough so that Eq.(5-20) isagood approximation. This isthebasis fortheuse ofapendulum toregulate thespeed ofaclock. Wecantreat theproblem ofmotion atlarge amplitudes bymeans of theenergy integral. The potential energy associated with thetorque given byEq.(5-18) is 9V(0)=-A—mglsin0d0 =—mgl cos0, (5-23) where wehave taken 0,=1r/2forconvenience. Wecould have written down V(0) right away asthegravitational potential energy ofamass m, referred tothehorizontal plane through Oasthelevel ofzero potential energy. Theenergy integral is %ml20'2 —mglcos0=E. (5-24) Wecould prove that Eisconstant from theequation ofmotion (5-13), butweneed not, since theanalogy described inthepreceding section guarantees that alltheorems forone-dimensional linear motion willhold in their analogous forms forrotational motion about anaxis. Thepotential energy V(0)isplotted inFig.5-4. Weseethatfor—mgl <E<mgl,the motion isanoscillating one, becoming simple harmonic motion forE slightly greater than —mgl. ForE>mgl, themotion isnonoscillatory; 0steadily increases orsteadily decreases, with 9oscillating between amaxi- mum andminimum value. Physically, when E>mgl, thependulum has enough energy toswing around inacomplete circle. (Inthis case, of course, themass must beheld byarigid rodinstead ofastring, unless 9is very large.) This motion isstill aperiodic one, thependulum making onecomplete revolution each time 0increases ordecreases by21r. In V +mgl 01 1 1 ° 1eA1 1 —31r —21r —1l' 11' 2-1r 311' —mgl_ FIG. 5-4. Potential energy forsimple pendulum. 5-3] THESIMPLE PENDULUM 211 either case, theattempt tosolve Eq.(5-24) for19leads totheequation 0do _@)1/2 £5(E/mgl +cos0)‘/2 —(l t' (5_25) The integral ontheleftmust beevaluated interms ofelliptic functions. The period ofthemotion canbeobtained byintegrating between appro- priate limits. When themotion isoscillatory (E<mgl), themaximum value Kof0isgiven, according toEq.(5-24), by - E=—mgl cosK. (5-26) Equation (5-25) becomes, inthiscase, 0 d0 2g1/2=(T)" <5-27> which canalsobewritten ' 0 d0 g1/2=2(1)‘- (H8) Theangle 0oscillates between thelimits =|=x. Wenowintroduce anew variable (5which runs from 0to21rforonecycle ofoscillation of0: . '02 1.0sin(p= =asmé-1 (5-29) where 5=sin (5-so) With these substitutions, Eq.(5-28) canbewritten P dw _(2)1/2 _ /I,(1-52sin2¢)1/2_z" (531) where wehave taken 60=0,forconvenience. Theintegral isnowina standard form forelliptic integrals. When aissmall, theintegrand canbe expanded inapower series ina2: ‘/0?[1+-1.52sinz5+-_--111.,»=<91/2 1. (5-32) This canbeintegrated term byterm: 12 (p-|—Q-a2(2<p —sin21,0) +'°'= It- (5-33) 212 RIGID BODIES. ROTATION ABOUT ANAxis. STATICS [CHAP. 5 Theperiod ofthemotion isobtained bysetting <p=2-Ir: 1=21(3)1/2(1 +g+-- (5-34) Thus astheamplitude ofoscillation becomes large, theperiod becomes slightly longer than forsmall oscillations, aprediction which isreadily veriﬁed experimentally bysetting uptwopendulums ofequal length and setting them toswinging atunequal amplitudes. Equation (5-33) canbe solved approximately for<pbysuccessive approximations, andtheresult substituted inEq.(5-29), which canbesolved for0bysuccessive approxi- mations. Theresult, toasecond approximation, is Ks Ks 6iK+E sinw’t+T55sin3w’t, (5-35) 1/2 2~»'-5'-(1) <-36> Ifweneglect terms inK2andK3,thissolution agrees with Eq.(5-21). At larger amplitudes insecond approximation, thefrequency isslightly lower than atsmall amplitudes, andthemotion of0contains asmall third har- monic term.where 5-4Thecompound pendulum. Arigid body suspended andfreeto swing about anaxisiscalled acompound pendulum. Weassume that the axisdoes notpass through thecenter ofmass, andwespecify theposition ofthebody bytheangle 0between avertical lineandaperpendicular line drawn from apoint 0ontheaxis, through thecenter ofmass G’(Fig. 5-5). Inorder tocompute thetotal torque exerted bygravity, weanticipate a 0\ 0 h l G hi OI mo Fro. 5-5. Thecompound pendulum. 5-4] THECOMPOUND PENDULUM 213 theorem, tobeproved later, that thetotal torque isthesame asifthe total gravitational force were applied atthecenter ofmass G.Wethen have, using Eqs. (5-12) and(5-13), Mk§;('i=—Mgh sino, (5-37) where histhedistance OG. This equation isthesame asEq.(5-19) fora simple pendulum oflength Z,ifwetake l=g- (5-38)h The point O’adistance lfrom Oalong thelinethrough thecenter of mass Giscalled thecenter ofoscillation. Ifallthemass Mwere atO’, themotion ofthependulum would bethesame asitsactual motion, for anygiven initial conditions. Ifthedistance WCish’,wehave z=h+5', " (5-39) hh’=5?,-52. (5-40) Itwillbeshown inthenext section that themoment ofinertia about anyaxisequals themoment ofinertia about aparallel axisthrough the center ofmass GplusMha,where histhedistance from theaxistoG’. Letkgbetheradius ofgyration about G’.Wethenhave 765=763+hz, (5-41) sothat Eq.(5-40) becomes hh’=11%. (5-42) Since this equation issymmetrical inhandh’,weconclude that ifthe body were suspended about aparallel axisthrough O’,thecenter ofoscilla- tion would beatO.The acceleration gofgravity canbemeasured very accurately bymeasuring theperiod ofsmall oscillations ofapendulum and using Eq.(5-22). Ifacompound pendulum isused, theradius ofgyration must beknown, ortheperiod measured about twoaxes, preferably O,O’, sothat theradius ofgyration canbeeliminated from theequations. Consider arigid body suspended from anaxisabout which itisfreeto move. Letitbestruck ablow atapoint 0'adistance lfrom theaxis, thedirection oftheblow being perpendicular tothelineTfrom the axistoO’.Place O’sothat theline%_'passes through thecenter ofmass G’,andleth,h’bethedistances E5’?(Fig. 5-6). Theimpulse delivered atthepoint O’bytheforce F’during theblow is J’=/11"at. (5-43) 214 RIGID Booms. ROTATION ABOUT ANAxis. STATICS [CHAP. 5 0F h .6“, hi 0/ l<” FIG. 5-6. Rigid body pivoted atOandstruck ablow atO’. Attheinstant theblow isstruck, aforce Fwill, ingeneral, have tobe exerted onthebody atthepoint Oontheaxisinorder tokeep Oﬁxed. Theimpulse delivered tothebody atOis Q J=/Fdt. (5-14) Anequal andopposite impulse —Jisdelivered bythebody tothesupport atO.Themomentum theorem forthecomponent Poflinear momentum ofthebody inthedirection ofFis: . %=5,‘-’-i(Mhe) =F+F’, (5-45) where 9istheangular velocity ofthebody about O.From thiswehave, forthemomentum justafter theblow, Mh0 =J-|—J’, (5-46) assuming that thebody isinitially atrest. The conservation theorem of angular momentum about Ois: dL_d g-_dt—E7(Mkoﬂ) -F’l. (5-47) Integrating, wehave, fortheangular momentum justafter theblow, M1130 =J’l. (5-4s) Weeliminate 0'between Eqs. (5-46) and(5-48): hl=5?,(1+ (5-49) 5-5] COMPUTATION orCENTERS orMASS ANDMOMENTS orINERTIA 215 Wenow askforthecondition that noimpulsive force beexerted onthe axisatOattheinstant oftheblow, i.e.,J=0: hl=5%,. (5-50) This equation isidentical with Eq.(5-38) andmay alsobeexpressed in thesymmetrical form [Eq. (5—42)] hh’=kg. _(5-51) The point O’atwhich ablow must bestruck inorder that noimpulse be feltatthepoint Oiscalled thecenter ofpercussion relative toO.Wesee that thecenter ofpercussion isthesame asthecenter ofoscillation relative toO,andthat Oisthecenter ofpercussion relative toO’.Ifthebody is unsupported, andisstruck atO’,itsinitial motion willbearotation about O.Forexample, abatter tries tohitabaseball atthecenter ofpercussion relative tohishands. Iftheballhitsvery farfrom thecenter ofpercus- sion, theblow istransmitted tohishands bythebat. 5-5Computation ofcenters ofmass andmoments ofinertia. Wehave given inSection 4-1thefollowing deﬁnition ofcenter ofmass forasystem ofparticles: R=%2m,-1',-. (5-52) Forasolid body, thesum may beexpressed asanintegral: 1R=H//[pr dV, (5-53) or,incomponent form, X,=%/[fps av, (5-54) Y=T1[f[[pydV, (5-55) Z=%f[[pZ dV. (5-55) Theintegrals canbeextended either over thevolume ofthebody, orover allspace, since p=0outside thebody. These equations deﬁne apoint G ofthebody whose coordinates are(X,Y,Z).Weshould ﬁrst prove that thepoint Gthus deﬁned isindependent ofthechoice ofcoordinate system. Since Eq.(5-52) or(5-53) isinvector form, andmakes noreference to anyparticular setofaxes, thedeﬁnition ofGcertainly does notdepend on 216 RIGID Booms. ROTATION ABOUT ANAxis. STATICS [cnAi>. 5 FIG. 5-7. Location ofcenter ofmass relative totwodifferent origins. anyparticular choice ofdirections fortheaxes. Weshould prove, how- ever, that Gisindependent alsoofthechoice oforigin. Consider asystem ofparticles, andletanyparticle m,-belocated byvectors r,~andI}with respect toanytwoorigins 0andO’.Ifaisthevector from OtoO’,the relation between r,-and1'}is(Fig. 5-7) L r,~=rt+a. (5-57) Thecenters ofmass G,G’withrespect toO,O’arelocated bythevectors RandR’,where R’isdeﬁned by 112'=HZ (5-53) Using Eq.(5-57), wecanrewrite Eq.(5-58): 1R’=Mzm,~(r,- —a) =R—a. (5-59) Thus RandR’arevectors locating thesame point with respect toOandO’ sothat GandG’arethesame point. General theorems liketheoneabove canbeproved either forasystem of particles orforabody described byadensity p.Whichever point ofview isadopted inanyproof, aparallel proof canalways begiven from the other point ofview. Much ofthelabor involved inthecalculation oftheposition ofthe center ofmass from Eqs. (5-54), (5-55), (5-56) canoften beavoided by theuseofcertain laborsaving theorems, including thetheorem proved 5-5] COMPUTATION OF CENTERS OFMASS AND MOMENTS OFINERTIA 217 above which allows usafree choice ofcoordinate axes andorigin. We have ﬁrstthefollowing theorem regarding symmetrical bodies: THEOREM. Ifabody issymmetrical with respect toaplane, its center ofmass liesinthatplane. (5-60) When wesayabody issymmetrical with respect toaplane, Wemean that forevery particle ononesideoftheplane there isaparticle ofequal mass located atitsmirror image intheplane. Foracontinuously distributed mass, wemean that thedensity atany point equals thedensity atits mirror image intheplane. Choose theorigin intheplane ofsymmetry, andlettheplane ofsymmetry bethemy-plane. Then incomputing Zfrom Eq. (5-56) [or(5—52)], foreach volume element (orparticle) atapoint (x,y,2)above themy-plane, there is,bysymmetry, avolume element of equal mass atthepoint (x,y,—z)below themy-plane, andthecontribu- tions ofthese twoelements totheintegral inEq.(5-56) willcancel. Hence Z=0,andthecenter ofmass liesinthemy-plane. This proves Theorem (5-60). Thetheorem hasanumber ofobvious corollaries: Ifabody issymmetrical intwoplanes, itscenter ofmass lieson their lineofintersection. (5-61) Ifabody issymmetrical about anaxis, itscenter ofmass lieson thataxis. (5-62) Ifabody issymmetrical inthree planes with onecommon point, thatpoint isitscenter ofmass. (5-63) Ifabody hasspherical symmetry about apoint (i.e., ifthedensity depends only onthedistance from thatpoint), thatpoint isits center ofmass. (5-64) These theorems enable u-stolocate thecenter ofmass inunediately insome cases, andtoreduce theproblem toacomputation ofonly oneortwo coordinates ofthecenter ofmass inother cases. One should beonthe lookout forsymmetries, and usethem tosimplify theproblem. Other cases notincluded inthesetheorems willoccur (e.g., theparallelepiped), where itwillbeevident that certain integrals willbeequal orwillcancel, andthecenter ofmass canbelocated without computing them. Another theorem which often simpliﬁes thelocation ofthecenter of mass isthat ifabody iscomposed oftwoormore parts whose centers of mass areknown, then thecenter ofmass ofthecomposite body canbe computed byregarding itscomponent parts assingle particles located at their respective centers ofmass. Letabody, orsystem ofparticles, be composed ofnparts ofmasses M1,...,M,,.Letanypart M1,becomposed ofN1,particles ofmasses mki, ...,m;,N,,, located atthepoints rm,...,r;,Nk. 218 RIGID BODIES. ROTATION ABOUT ANAxis. STATICS [cIiAi>. 5 Then thecenter ofmass ofthepart M1,islocated atthepoint 1"1Rk=—- mkzfkz and Na Mk —Z mgl. z=1 Thecenter ofmass oftheentire body islocated atthepoint 1 1|. N], R=17E2mkzfkz , k=1 l=1 where 1|. N1, M—2 Z mm. k=1 l=1 ByEq.(5-65), Eq.(5-67) becomes . 1"R=— MR M IE It I0) andbyEq.(5-66), Eq.(5-68) becomes 1r==§j11, kil Equations (5-69) and(5-70) arethemathematical statement ofthetheo- remtobeproved. _\__..__..\--__\\\ ,__\__Q;\..__-_-II111I6cm , \~’/I i ‘. I m \_’// 10om I |<—4 cm->| FIGURE 5-8(5-55) (5-66) (5-67) (5-cs) (5-69) (5-70) 5-5] COMPUTATION orCENTERS orMASS ANDMOMENTS orINERTIA 219 Asanexample, letusconsider auniform rectangular block with acylin- drical holedrilled out,asshown inFig.5-8. Bythesymmetry about the twovertical planes bisecting theblock parallel toitssides, weconclude that thecenter ofmass liesalong thevertical lineEthrough thecenters ofthetopandbottom faces. Letthecenter ofmass oftheblock liea distance Zbelow A,andletthedensity oftheblock bep.Iftheholewere notcutout, themass oftheblock would be6cmX4cmX10cmXp, anditscenter ofmass would beatthemidpoint ofXE, 5cmfrom A.The mass ofthematerial drilled outis1rcm2 X6cmXp,anditscenter of mass, before itwas removed, was onTB, 2cmbelow A. Hence the theorem (5-69) above allows ustowrite (6cmX4cm>< 10cmXp) X'5cm= (1rcm2X6cmXp) X2cm +6cmX (4cmX 10cm—1rcm2) XpXZ. Thesolution forZ is Z_6X4X10X5—1rX6X2cm _ 6X(4X10—1r) ' Asasecond example, welocate thecenter ofmass ofahemisphere of radius a.Bysymmetry, ifthedensity isuniform, thecenter ofmass lies ontheaxisofsymmetry, which wetake asthez-axis. Wehave then to in __ __ ___Tdp do ,” /’ (Z12.. _____=..____“l_____-.-FINII1*1@I|'11II'1'1’.1III II § 11\I ii.|:V1'i/‘'55I:s1I§" ll £ (5) (5) rdd ____ '1a2_,2;i_‘_~_ fl‘.1? 7‘\SlIl 0drp r’a’ (c) (d) FIG. 5-9. Methods ofintegrating overahemisphere. 220 RIGID BODIES. ROTATION ABOUT ANAxis. STATICS [CHAP. 5 compute only theintegral inEq.‘(5-56). The integral canbesetupin rectangular, cylindrical, orspherical coordinates (Fig. 5-9): 5ti.Z-.1 a (a2_z2)l/2 (a2_z2_u2)1/2 Rectangular: Z=— = =_(a2_z2)l/2 /x=_(a2_z2_u2)l/2 pzdxdydz. a 21 (oz-:2)”: Cylindrical: Z=if / f pzrdrdodz. Z=0 ¢=0 1‘=0 a 1r/2 211' Spherical: Z=iI / f (prcos0)r2sin0drd0d<p.Mr=0 o=0 ¢=o Any oneofthese expressions canbeused toevaluate Zforanydensity distribution. Ifpisuniform, wecanalsobuild upthe-hemisphere outof rings ordisks andsave oneortwointegrations. Forexample, building up thehemisphere outofdisks perpendicular tothez-axis (this isequivalent tocarrying outtheintegration over rand<pincylindrical coordinates), we canwrite ,, ‘ Z=i/z=0 zp1r(a2 —zz)dz-(A)<-A- where theintegrand iszptimes thevolume ofadiskofthickness dz,radius (a2 _ z2)l/2,‘ When thedensity pisuniform, thecenter ofmass ofabody depends only onitsgeometrical shape, andisgiven by R=T1,]/frdv. (5-72) V The point Gwhose coordinate Risgiven byEq.(5-72) iscalled thecen- troid ofthevolume V.Ifwereplace thevolume Vbyanarea Aorcurve G inspace, weobtain formulas forthecentroid ofanarea orofacurve: R=-31-[1]r5,4, (5-73) R=g[Crds, (5-74) where sisthelength ofthecurve C.Thefollowing twotheorems, dueto Pappus, relate thecentroid ofanarea orcurve tothevolume orarea swept outbyitwhen itisrotated about anaxis: 5-51 co:-IPU'I‘.i'rIoN orcEx'rI-zns 01+‘M.-\ss AND MOMENTS onIXERTIA 221 I ___-— 3 dsto FIG. 5-10. Pappus’ ﬁrst theorem. FIG. 5-ll. Sphere formed byrota- tingasemicircle. THEOREM 1.Ifaplane curve rotates about anaxis initsown plane which doesnotintersect it,thearea ofthesurface ofrevolu- tionwhich itgenerates isequal tothelength ofthecurve multiplied bythelength ofthepath ofitscentroid. (5-75) THEOREM 2.Ifaplane area.rotates about anarcisinitsownplane which doesnotintersect it,thevolume generated isequal tothearea times thelength ofthepath ofitscentroid. (5-76) The proof ofTheorem 1isvery simple, with thenotation indicated in Fig.5-10: A=fa255ds=25[Cyds=21rYs, (5-7?) where Yisthey-coordinate ofthecentroid ofthecurve C,andsisits length. Theproof ofTheorem 2issimilar andislefttothereader. These theorems may beused todetermine areas andvolumes ofﬁgures sym- metrical about anaxis when thecentroids ofthegenerating curves or areas areknown, andconversely. Welocate, forexample, theposition of thecenter ofmass ofauniform semicircular diskofradius a,using Pappus’ second theorem. Ifthedisk isrotated about itsdiameter, thevolume of thesphcrc generated, byPappus’ theorem (Fig. 5-11), is 2gem“= (211rY), from which weobtain 4Y= (5-73) Themoment ofinertia Iofabody about anaxisisdeﬁned byEq.(5-10) : I=Emgrf, (5-79) 222 mom BODIES. ROTATION ABOUT ANAXIS. STATICS [cmm 5 P r r’ G R 0 FIG. 5-12. Location ofpoint Pwith respect topoints OandG. or 1=fffp1‘2av, (5-so) where risthedistance from each point orparticle ofthebody tothegiven axis. Weﬁrst prove several laborsaving theorems regarding moments ofinertia: PARALLEL AXIS THEOREM. Themoment ofinertia ofabody about anygiven axisisthemoment ofinertia about aparallel axis through thecenter ofmass, plusthemoment ofinertia about the given axis ifallthemass ofthebody were located atthecenter of mass. (5-81) Toprove this theorem, letI0bethemoment ofinertia about a.2-axis through thepoint O,andletIGbethemoment ofinertia about aparallel axis through thecenter ofmass G.Let1'andr’bethevectors toany point Pinthebody, from 0andG,respectively, andletRbethevector from OtoG.The components ofthese vectors willbedesignated by (x,y,z),(x’,y’,z’),and(X,Y,Z).Then, since (Fig. 5—12) r=1"+R, weseethat 2- x2+v2=(x'+X>2+<y'+Y>2 =w'2+1/'2+X2+Y2+2Xx’+2Yy', sothatthemoment ofinertia 10is It=fffeﬁ+mpdv =ff/(#2 +2/’2)pdV+(X2+Y2)/f/pav+2Xf/[tut av +2Y//fy'p dV. _ (5-82) 5-5] COMPUTATION orCENTERS orMASS AND MOMENTS OFINERTIA 223 The ﬁrst integral isIG,andtheintegral inthesecond term isthetotal mass Mofthebody. Theintegrals inthelasttwoterms arethesame as theintegrals occurring inEqs. (5-54) and (5—55), anddeﬁne thex-and y-coordinates ofthecenter ofmass relative toG.Since Gisthecenter of mass, these integrals arezero, andwehave m=m+MmH4%. t ww This isthemathematical statement oftheParallel Axis Theorem. Ifwe know themoment ofinertia about anyaxis, andcanlocate thecenter of mass, wecanusethistheorem todetermine themoment ofinertia about anyother parallel axis. The moment ofinertia ofacomposite body about any axis may be found byadding themoments ofinertia ofitsparts about thesame axis, a statement which isobvious from thedeﬁnition ofmoment ofinertia. This fact canbeputtouseinthesame way astheanalogous result forthe center ofmass ofacomposite body. Abody whose mass isconcentrated inasingle plane iscalled aplane lamina. Wehave thefollowing theorem foraplane lamina: PERPENDICULAR AXIS THEOREM. Thesumofthemoments of inertia ofaplane lamina about anytwoperpendicular axesinthe plane ofthelamina isequal tothemoment ofinertia about anaxis through theirpoint ofintersection perpendicular tothelamina. (5-84) Theproof ofthistheorem isvery simple. Consider anyparticle ofmass m inthexy-plane. Itsmoments ofinertia about thex-andy-axes are I,=myz, I,,=mxz. (5-85) Adding these, wehave themoment ofinertia ofmabout thez-axis: n+n=mﬁ+w=n. ww Since themoment ofinertia ofanylamina inthexy-plane isthesum of themoments ofinertia oftheparticles ofwhich itiscomposed, wehave theorem (5-84). Weillustrate these theorems byﬁnding themoments ofinertia ofa uniform circular ringofradius a,mass M,lying inthexy-plane (Fig. 5-13). Themoment ofinertia about az-axis perpendicular totheplane ofthering through itscenter iseasily computed: I,=Maz. i (5-87) The moments I,andI,,areevidently equal, andwehave, therefore, byth 5-s4,e°ren1( ) 1,==s1;==%B[a2. (5-ss) 224 RIGID BODIES. ROTATION ABOUT ANAXIS. STATICS [onA1>. 5 Z Z r 2/ a dr x' A V F10. 5-13. Aringofradius a. FIG. 5-14. Finding themoment of inertia ofadisk. asin0 __ I/' acos0 V FIG. 5-15. Finding themoment ofinertia ofasolid sphere. Themoment ofinertia about anaxisAtangent totheringis,bytheParallel Axis Theorem, 2 2 IA=I,+Ma =§Ma. (5-89) Themoment ofinertia ofasolid body canbesetupinwhatever coordi- nate system may beconvenient fortheproblem athand. Ifthebody is uniform andofsimple shape, itsmoment ofinertia canbecomputed by considering itasbuilt upoutofrods, rings, disks, etc. Forexample, the moment ofinertia ofacircular disk about anaxis perpendicular toit through itscenter canbefound byregarding thedisk asmade upofrings (Fig. 5-14) andusing Eq.(5-87): “ 4 I,=/Ir2p21rr dr=7%‘: =§Ma2. (5-90)o Themoment ofinertia ofasolid sphere canbecalculated from Eq.(5-90) byregarding thesphere asmade upofdisks (Fig. 5-15): 02-2 5I=/93% (p1ra2SlIl20)tutcoso)=8%?=slut’. (5~91) 5-6] STATICS ormom BODIES 225 Abody with apiece cutoutcanbetreated bysetting itsmoment of inertia equal tothemoment ofinertia oftheoriginal body minus the moment ofinertia ofthepiece cutout,allmoments being taken, ofcourse, about thesame axis. 5-6Statics ofrigid bodies. The equations ofmotion ofarigid body areEqs. (5-4) and(5-5): MR=E (5-92) dL0_ ._-E_ mo. (5-93) Equation (5-92) determines themotion ofthecenter ofmass, located by thevector R,interms ofthesum ofallexternal forces acting onthebody. Equation (5-93) determines therotational motion about apoint O,which may bethecenter ofmass orapoint ﬁxed inspace, interms ofthetotal external torque about thepoint O.Thus ifthetotal external force acting onarigid body andthetotal external torque about asuitable point are given, itsmotion isdetermined. This would notbetrue ifthebody were notrigid, since thenitwould bedeformed bytheexternal forces inaman- nerdepending ontheparticular points atwhich they areapplied. Since weareconcerned only with extemal forces throughout this section, we may omit thesuperscript e.Itisonly necessary togive thetotal torque about anyonepoint O,since thetorque about anyother point O’canthen befound from thefollowing formula: ZN10’ =2Nro+(1'0—1'0’) XZFt‘, (5'94) ‘I Z ‘L where 1'0,rotarevectors drawn tothepoints O,O’from anyconvenient origin. That is,thetotal torque about O’isthetotal torque about Oplus thetorque about O’ifthetotal force were acting atO.The proof of Eq.(5-94) isvery simple. Letr,~bethevector from theorigin tothepoint atwhich Ftacts. Then" ZNt"o' =2_(1't'—1'0') XFt" =Z:(1‘t'-1'0-I-Io—I0') XFt' =Z(1't'—1'0) XFt'-l-z(1’0—1'0') XFt' =ZNro+(1'o —1'0’) XZR‘- 226 RIGID BODIES. ROTATION ABOUT ANAXIS. STATICS [cn.u>. 5 P F ,P F 1-,, IF O FIG. 5-16. Thetorque isindependent ofwhere along itslineofaction aforce acts. If,inparticular, arigid body isatrest, theleftmembers ofEqs. (5-92) and(5-93) arezero, andwehave ZF.=0, (5-95) 2N.=o. (5-95) These aretheconditions tobesatisﬁed bytheexternal forces andtorques inorder forarigid body tobeinequilibrium. They arenotsuﬁicient to guarantee that thebody isatrest, foritmight stillbeinuniform transla- tional and rotational motion, butifthebody isinitially atrest, itwill remain atrestwhen these conditions aresatisﬁed. Itissuﬁicient forthe total torque inEq.(5-96) tobezeroabout anypoint, since then, by Eq.(5-94), itwillbezeroalsoabout every other point ifEq.(5-95) holds. Incomputing thetorque duetoaforce F,itisnecessary toknow not only thevector F(magnitude and direction), butalso thepoint Pof thebody atwhich theforce acts. But ifwedraw alinethrough Pin thedirection ofF,then ifFacts atanyother point P’ofthisline, its torque willbethesame, since, from thedeﬁnition ofthecross product, itcanbeseen (Fig. 5-16) that Ip XF=I'p' XF. (5-97) (The areas oftheparallelograms involved areequal.) Thelinethrough P inthedirection ofFiscalled thelineofaction oftheforce. Itisoften convenient incomputing torques toremember that theforce may becon- sidered toactanywhere along itslineofaction. Adistinction issome- times made inthisconnection between “free” and “sliding” vectors, the force‘ being a“sliding” vector. Theterminology islikely toprove confus- ing, however, since asfarasthemotion ofthecenter ofmass isconcerned [Eq. (5—92)], theforce isa“free” vector, i.e.,may actanywhere, whereas incomputing torques, theforce isa“sliding” vector, andforanonrigid body, each force must belocalized atthepoint where itacts. Itisbetter todeﬁne vector, aswedeﬁned itinSection 3-1, asaquantity having magnitude anddirection, Without reference toanyparticular location in space. Then, inthecaseofforce, weneed forsome purposes tospecify not 5-6] smrrcs orRIGID BODIES 227 C \ it FIG. 5-17. Asingle force Cwhose torque isthesumofthetorques ofAandB. only theforce vector Fitself, butinaddition thepoint orlineonwhich theforce acts. Atheorem duetoVarignon states that ifC=A+B,then themoment ofCabout anypoint equals thesum ofthemoments ofAandB,provided A,B,andCactatthesame point. The theorem isanimmediate conse- quence ofthevector identity given byEq.(3-27): rXC=rXA+rXB, if C=A+B. (5-98) This theorem allows ustocompute thetorque duetoaforce byadding thetorques duetoitscomponents. Combining Varignon’s theorem with theresult ofthepreceding paragraph, wemay reduce thetorque dueto twoforces A,Bacting inaplane, asshown inFig. 5-17, tothetorque duetothesingle force C,since both AandBmay beconsidered toactat theintersection oftheir lines ofaction, andEq.(5-98) then allows usto addthem. Wecould now addCsimilarly toanythird force acting inthe plane. This process canbecontinued solong asthelines ofaction ofthe forces being added arenotparallel, andisrelated toamore general theorem regarding forces inaplane tobeproved below. Since, forarigid body, themotion isdetermined bythetotal force and total torque, weshall calltwosystems offorces acting onarigid body equivalent ifthey give thesame total force, and thesame total torque about every point. Inview ofEq.(5-94), twosystems offorces arethen equivalent ifthey give thesame total force, andthesame total torque about anysingle point. Itisofinterest toknow, foranysystem offorces, what isthesimplest system offorces equivalent toit. Ifasystem offorces F,-acting atpoints r,-isequivalent toasingle force F acting atapoint r,then theforce Facting atrissaidtobetheresultant of thesystem offorces F,~.IfFistheresultant ofthesystem offorces F,-, then wemust have F=ZF,-, (5-99) 1 (r-to)><F=Z(I,-to)>< (5-100) ‘I 228 RIGID BODIES. ROTATION ABOUT ANAXIS. STATICS [CHAP. 5 where r0isanypoint about which moments aretaken. ByEq.(5-94), if Eq.(5-99) holds, andEq.(5—100) holds foranypoint 1'0,itholds forevery point r0.Theforce —Facting at1'iscalled theequilibrant ofthesystem; iftheequilibrant isadded tothesystem offorces, theconditions forequi- librium aresatisﬁed. Anexample ofasystem offorces having aresultant isthesystem of gravitational forces acting onabody near thesurface oftheearth. We shall show that theresultant inthiscase acts atthecenter ofmass. Let theacceleration ofgravity beg.Then theforce acting onaparticle m,-is F;=m,-g. (51-101) Thetotal force is F=Zm.-g=Mg, (5-102) where Misthetotal mass. Thetotal torque about anypoint Ois,with O asorigin, ZN10=Z(ItXmtg) ' i = (mart XE) -(>2XK =MRxg =RxMg, (5—103) where Risthevector from Otothecenter ofmass. Thus thetotal torque isgiven bytheforce Mgacting atthecenter ofmass. Because ofthis result, thecenter ofmass isalsocalled thecenter ofgravity. Weshall see inthenext chapter that, ingeneral, thisresult holds only inauniform gravitational ﬁeld, i.e.,when gisthesame atallpoints ofthebody. If thesystem offorces acting onarigid body hasaresultant, theforces may bereplaced bythisresultant indetermining themotion ofthebody. Asystem offorces whose sum iszero iscalled acouple: ZF,=o. (5-104) Acouple evidently hasnoresultant, except inthetrivial case where the total torque iszero also, inwhich case theresultant force iszero. By Eqs. (5-94) and(5-104), acouple exerts thesame total torque about every point: 2 N50’ = N10. (5—105) 5-6] STATICS orRIGID BODIES 229 P Y %F FIG. 5-18. Asimple couple. Thus acouple ischaracterized byasingle vector, thetotal torque, and allcouples with thesame total torque areequivalent. The simplest sys- tem equivalent toanygiven couple, ifweexclude thetrivial case where thetotal torque iszero, isapairofequal andopposite forces F,—F, acting atpoints P,P’separated byavector r(Fig. 5-18) such that ZNw=r><F. (5-105) Equation (5-106) states that themoment ofthegiven couple about O equals themoment ofthecouple (F,—F) about P’;thetwo systems aretherefore equivalent, since thepoint about which themoment ofa couple iscomputed isimmaterial. The force Fandthepoints PandP’ arebynomeans uniquely determined. Since only thecross product rXF isdetermined byEq.(5-106), wecanchoose Parbitrarily; wecanchoose thevector Farbitrarily except that itmust lieintheplane perpendicular to thetotal torque; andwecanthen choose rasanyvector lying inthesame plane anddetermining with Faparallelogram whose area isthemagnitude ofthetotal torque. Theproblem ofﬁnding thesimplest system equivalent toanygiven sys- temofforces issolved bythefollowing theorems: THEOREM I.Every system offorces isequivalent toasingle force through anarbitrary point, plus acouple (either orbothofwhich may bezero). (5-107) Toprove this, weshow how toﬁndtheequivalent single force andcouple. Letthearbitrary point Pbechosen, letthesumofalltheforces inthesys- tembe andlettheir total torque about thepoint PbeN.Then, ifwe letthesingle force FactatP,andaddacouple whose torque isN,wehave asystem equivalent totheoriginal system. Since thecouple canbecom- posed oftwoforces, oneofwhich may beallowed toactatanarbitrary point, wemay letoneforce ofthecouple actatthepoint P,and add itto Ftogetasingle force acting atPplus theother force ofthecouple. This proves THEOREM II.Any system offorces canbereduced toanequiva- lentsystem which contains atmost twoforces. (5-108) 230 RIGID Booms. ROTATION ABOUT ANAxis. STATICS [CHAP. 5 Thefollowing theorem canbeproved intwoways: THEOREM III. Asingle nonzero force andacouple inthesame plane (i.e., such thatthetorque vector ofthecouple isperpendicular tothesingle force) have aresultant and, conversely, asingle force isequivalent toanequal force through anyarbitrary point, plus a couple. (5—109) Since acouple with torque Nisequivalent toapairofequal andopposite forces, F,—F,where Fmay bechosen arbitrarily intheplane perpendicular toN,Wemay always choose Fequal tothesingle force mentioned inthe theorem. Furthermore, wemay choose thepoint ofaction ofFarbitrarily. Given asingle nonzero force Facting atP,andacouple, weform acouple (F,-—F) equivalent tothegiven couple, andlet—FactatP;Fand—F then cancel atP,andtheremaining force Fofthecouple isthesingle re- sultant. Theconverse canbeproved byasimilar argument. Theother method ofproof isasfollows. Letthegiven force Factata point P,andletthetotal torque ofthecouple beN.Then thetorque of thesystem about thepoint PisN.Wetake anyvector 1',intheplane perpendicular toN,which forms with Faparallelogram ofarea N,and letP’bethepoint displaced from Pbythevector r.Ifthesingle force F actsatP’,thetorque about Pwillthen beN,andhence thissingle force is equivalent totheoriginal force Facting atPplusthecouple. Wecan combine Theorems IandIIItoobtain THEOREM IV. Every system offorces isequivalent toasingle force plus acouple whose torque isparallel tothesingle force. (Or,alternatively, every system offorces isequivalent toacouple plus asingle force perpendicular totheplane ofthecouple.) (5-110) Toprove this, weuseTheorem Itoreduce anysystem toasingle force plus acouple, anduseTheorem IIItoeliminate anycomponent ofthe couple torque perpendicular tothesingle force. The point ofapplication ofthesingle force mentioned inTheorem IVisnolonger arbitrary, asits lineofaction willbeﬁxed when weapply Theorem III. Either thesingle force orthecouple may vanish inspecial cases. Forasystem offorces in aplane, alltorques about anypoint intheplane areperpendicular tothe plane. Hence Theorem IVreduces to THEOREM V.Any system offorces inaplane hasaresultant, unless itisacouple. (5—111) Inpractice, thereduction ofacomplicated system offorces toasimpler system isaproblem whose simplest solution isusually obtained byan ingenious application ofthevarious theorems andtechniques mentioned inthissection. Onemethod which always works, andwhich isoften the 5-7] STATICS orSTRUCTURES 231 simplest ifthesystem offorces isvery complicated, istofollow thepro- cedure suggested bytheproofs oftheabove theorems. Find thetotal force Fbyvector addition, andthetotal torque Nabout some conven- iently chosen point P.Then Facting atP,plus acouple oftorque N, together form asystem equivalent totheoriginal system. IfFiszero, the original system reduces toacouple. IfNisperpendicular toF,thesystem hasaresultant, which canbefound byeither ofthemethods indicated in theproof ofTheorem III. IfNisnotperpendicular toF,andneither is zero, then thesystem hasnoresultant, andcanbereduced toasystem of twoforces, asinthederivation ofTheorem II,ortoasingle force anda couple whose torque isparallel toit,asinTheorem IV. Itisamatter of taste, orofconvenience forthepurpose athand, which ofthese latter re- ductions isregarded asthesimplest. Infact, fordetermining themotion ofabody, themost convenient reduction iscertainly just thereduction given byTheorem I,with thearbitrary point taken asthecenter ofmass. 5-7Statics ofstructures. The determination oftheforces acting at various points inasolid structure isaproblem ofutmost importance in allphases ofmechanical engineering. There aretwoprincipal reasons for wanting toknow these forces. First, theengineer must besure that the materials andconstruction aresuch aswillwithstand theforces which will beacting, without breaking orcrushing, and usually without suffering permanent deformation. Second, since noconstruction materials are really rigid, butdeform elastically andsometimes plastically when subject toforces, itisnecessary tocalculate theamount ofthisdeformation, and totake itintoaccount, ifitissigniﬁcant, indesigning thestructure. When deformation orbreaking ofastructure isunder consideration, thestruc- ture obviously cannot beregarded asarigid body, andweareinterested intheactual system offorces acting onandinthestructure. Theorems regarding equivalent systems offorces arenotofdirect interest insuch problems, butareoften useful astools inanalyzing parts ofthestructure which may, toasufficient approximation, beregarded asrigid, orinsug- gesting possible equivalent redistributions offorces which would subject thestructure tolessobjectionable stresses while maintaining itinequi- librium. Ifastructure isatrest, Eqs. (5-95) and(5-96) areapplicable either to thestructure asawhole, ortoanypart ofit.Itmust bekept inmind that theforces and torques which aretobeincluded inthesums arethose which areexternal toandacting onwhichever part ofthestructure isunder consideration. Ifthestructure ismoving, themore general equations (5-92) and (5-93) areapplicable. Either pair ofvector equations repre- sents, ingeneral, sixcomponent equations, orthree ifallforces lieina single plane. (Why three?) Itmay bethat thestructure issoconstructed 232 RIGID Booms. ROTATION ABOUT ANAxis. STATICS [CHAP. 5 l 0 F1 A B W FIG. 5-19. The ﬂagpole problem. that when certain oftheexternal forces and-their points ofapplication are given, alltheinternal forces andtorques acting oneach part ofthestruc- ture canbedetermined byappropriate applications ofEqs. (5-95) and (5-96) (inthecase ofastructure atrest). Such astructure issaidtobe statically determinate. Anelementary example isshown inFig.5-19, which shows ahorizontal ﬂagpole ABhinged atpoint Atoawall andsupported byacable BC. Aforce Wactsonthepole asshown. When theforce W and thedimensions ofthestructure aregiven, itisasimple matter to apply Eqs. (5-95) and(5-96) tothepole andtocalculate theforce F1ex- erted bythecable andtheforce F2acting through thehinge. Many ex- amples ofstatically determinate structures aregiven inanyelementary physics textbook. Suppose now that thehinge atAinFig.5-19 were replaced byawelded joint, sothat theﬂagpole would support theload even without thecable BC,provided thejoint atAdoes notbreak. Then, given only theweight W,itisevidently impossible todetermine theforce F1exerted bythe cable; F1may have anyvalue from zerotoarather large value, depending onhow tightly thecable isdrawn upandonhow much stress isapplied tothejoint atA.Such astructure issaid tobestatically indeterminate. Astatically indeterminate structure isoneinwhich theforces acting on itsparts arenotcompletely determined bytheexternal forces, butdepend also onthedistribution ofstresses within thestructure. Toﬁnd the internal forces inanindeterminate structure, wewould need toknow the elastic characteristics ofitsparts andtheprecise way inwhich these parts aredistorted. Such problems areusually farmore difficult than problems involving determinate structures. Many methods ofcalculating internal forces inmechanical structures have been developed forapplication toen- gineering problems, and some ofthese areuseful inawide variety of physical problems. 5-8Stress andstrain. Ifanimaginary surface cutsthrough anypart ofasolid structure (arod, string, cable, orbeam), then, ingeneral, the material ononesideofthissurface willbeexerting aforce onthematerial Fr->l l—>r 35-8] STRESS ANDSTRAIN 233 A sI itli (> X Fl—>r; Fr—>l | (b) FFL->r l ’lFr->l l (<1) FIG. 5-20. Stresses inabeam. (a)Compression. (b)Tension. (c)Shear. ontheother side, andconversely, according toNewton’s third law. These internal forces which actacross any surface within thesolid arecalled stresses. The stress isdeﬁned astheforce perunit area acting across any given surface inthematerial. Ifthematerial oneach sideofanysurface pushes onthematerial ontheother sidewithaforce perpendicular tothe surface, thestress iscalled acompression. Ifthestress isapullperpen- dicular tothesurface, itiscalled atension. Iftheforce exerted across the surface isparallel tothesurface, itiscalled ashearing stress. Figure 5-20 illustrates these stresses inthecase ofabeam. The vector labeled F)_,, represents theforce exerted bythelefthalfofthebeam ontheright half, andtheequal andopposite force F,_,) isexerted onthematerial onthe leftbythematerial ontheright. Astress atanangle toasurface canbe resolved intoashear component andatension orcompression component. Inthemost general case, thestress may actinanydirection relative to thesurface, andmay depend ontheorientation ofthesurface. The de- scription ofthestate ofstress ofasolid material inthemost general case israther complicated, andisbest accomplished byusing themathe- matical techniques oftensor algebra tobedeveloped inChapter 10.We shall consider here only cases inwhich either thestress isapure com- pression, independent oftheorientation ofthesurface, orinwhich only onesurface isofinterest atanypoint, sothat only asingle stress vec- torisneeded tospecify theforce perunitarea across that surface. Ifweconsider asmall volume AVofanyshape inastressed material, thematerial within thisvolume willbeacted onbystress forces exerted across thesurface bythematerial surrounding it.Ifthematerial isnot perfectly rigid, itwillbedeformed sothat thematerial inthevolume AV may have adifferent shape andsizefrom that which itwould have ifthere were nostress. This deformation ofastressed material iscalled strain.( 1 l 4 I l I 234 RIGID Booms. ROTATION ABOUT ANAxis. STATICS [cBA1>. 5 Thenature andamount ofstrain depend onthenature andmagnitude of thestresses andonthenature ofthematerial. Asuitable deﬁnition of strain, stating howitistobemeasured, willhave tobemade foreach kind ofstrain. Atension, forexample, produces anextension ofthematerial, andthestrain would bedeﬁned asthefractional increase inlength. Ifawire oflength landcross-sectional area Aisstretched tolength l+Albyaforce F,thedeﬁnitions ofstress andstrain are stress =F/A, (5—112) strain =Al/l. (5—113) Itisfound experimentally that when thestrain isnottoolarge, thestress isproportional tothestrain forsolid materials. This isHooke’s law, and itistrue forallkinds ofstress andthecorresponding strains. Itisalso plausible ontheoretical grounds forthereasons suggested inthepreliminary discussion inSection 2-7. Theratio ofstress tostrain istherefore constant foranygiven material ifthestrain isnottoolarge. Inthecase ofexten- sionofamaterial inonedirection duetotension, thisratio iscalled Young's modulus, andis stress FlY=$351=Ti" <5‘11‘*> Ifasubstance issubjected toapressure increment Ap,theresulting deformation willbeachange involume, andthestrain willbedeﬁned by strain = (5-115) Theratio ofstress tostrain inthiscase iscalled thebulkmodulus B: B= = -%, (5-115) Where thenegative signisintroduced inorder tomake Bpositive. Inthecaseofashearing stress, thestress isagain deﬁned byEq.(5-112), where Fistheforce acting across andparallel tothearea A.Theresult- ingshearing strain consists inamotion ofAparallel toitself through a distance Al,relative toaplane parallel toAatadistance Axfrom A(Fig. 5-21). Theshearing strain isthen deﬁned by strain =gala=tan0, (5—117) where 0istheangle through which alineperpendicular toAisturned as aresult oftheshearing strain. Theratio ofstress tostrain inthiscase is 5-9] EoUiLiBRiUM orFLEXIBLE STRINGS ANocABLEs 235 F ~A —F I-—Ax—~j FIG. 5-21. Shearing strain. called theshear modulus n: stress _F"—aim—ﬁt" (5418) Anextensive study ofmethods ofsolving problems instatics isoutside thescope ofthistext. Weshall restrict ourselves inthenext three sec- tions tothestudy ofthree special types ofproblems which illustrate the analysis ofaphysical system, todetermine theforces which actupon its parts andtodetermine theeffect ofthese forces indeforming thesystem. 5-9Equilibrium ofﬂexible strings andcables. Anideal ﬂexible string isonewhich willsupport nocompression orshearing stress, noranybending moment, sothat theforce exerted across anypoint inthestring canonly beatension directed along thetangent tothestring atthat point. Chains andcables used inmany structures canberegarded formost purposes as ideal ﬂexible strings. Letusﬁrst take avery simple problem inwhich astring ofnegligible weight issuspended between twopoints P0andP2,andaforce F1actsata point P1onthestring (Fig. 5-22). Let1'0bethetension inthesegment _1%T{, and1'1thetension inthesegment Letloandl1bethelengths ofthese segments ofthestring, andletZ02bethedistance between P0and P2. The angles a,)8between thetwo segments ofstring andthelineﬁz- aredetermined bythecosine law: 12+?-12 l%e+z%-12 _A2___9___i, = , _ cosoz- 210,02 cosB 211,02 (5119) 236 RIGID Booms. ROTATION ABOUT ANAxis. STATICS [cnAi>. 5 P2 102 W Po W liQ 1,, T1 T0 P1 (B+'Y) T1F. (N+ ‘(V _0‘) F1 To Fro. 5-22. Aﬂexible string heldatthree points. sothat theposition ofthepoint P1isindependent oftheforce F1,provided thestring does notstretch. Since thebitofstring atthepoint P1isin equilibrium, thevector sum ofthethree forces F1,1'0,and1'1acting on thestring atP1must vanish, sothat these forces form aclosed triangle, asindicated inFig. 5-22. The tensions arethen determined interms of theangle between theforce F1andthedirection ofthelineE, bythe sinelaw: sin(I3-|-'Y) _ sin(7—a)_ _ T9—F1 : T1—F1i——i-'Sin(a+ﬁ) (5120) Now suppose that thestring stretches according toHooke’s law, sothat lo=lt(1+km), l1=l’1(1+kn), (5—121) where l,',,llaretheimstretched lengths, andlcisaconstant [1/Itwould be Young’s modulus, Eq.(5—114), multiplied bythecross-sectional area of thestring]. The unknown quantities 1'11,1'1,lo,andl1canbeeliminated from Eqs. (5-119) bysubstitution from Eqs. (5-120) and (5-121). We then have tworather complicated equations tobesolved fortheangles ozand13.The solution must becarried outbynumerical methods when numerical values ofZ6,li,lo,ZO2,F1,and'Yaregiven. When ozandBare found, 1'0,1'1,lo,andl1canbefound from Eqs. (5—120) and(5—121). One wayofsolving these equations bysuccessive approximations istoassume ﬁrstthat thestring does notstretch, sothat lo=Z6,ll=lj,andtocalcu- late aand Bfrom Eqs. (5—119), and 1'0,1'1from Eqs. (5—120). Using these values of1'0,11,wethen calculate lo,l1from Eqs. (5—121). The new values oflo,Z1canbeused inEqs. (5—119) togetbetter values foroz,Bfrom which better values of1'11,1'1canbecalculated. These canbeused toget stillbetter values forZ11,l1from Eqs. (5—121), andsoon.Asthisprocess isrepeated, thesuccessive calculated values ofoz,)6,1'1,,1'1,lo,l1willcon- verge toward thetrue values. Ifthestring stretches only very little, the 5-9] EouiLiBRiUM orFLEXIBLE STRINGS AND cABLEs 237 llFl Tsin05' ('4\/A OO{I}5°'-1 Q3 FIG. 5-23. Aﬂexible string hanging under itsown weight. ﬁrstfewrepetitions willbesufficient togivevery close values. Themethod suggested here isanexample ofavery general class ofmethods ofsolution ofphysical problems bysuccessive approximations. Itisanexample of what arecalled relaxation methods ofsolving statics problems. Wenext consider astring acted onbyforces distributed continuously along thelength ofthestring. Apoint onthestring willbespeciﬁed by itsdistance sfrom oneend, measured along thestring. Letf(s)bethe force perunitlength atthepoint s,that is,theforce onasmall segment of length dsisfds. Then thetotal force acting onthelength ofstring" be- tween theends=0andthepoint siszeroifthestring isinequilibrium: F1,+fat+1(8)=o, (5-122) where F0isthesupporting force attheends=0,andf(s) isavector whose magnitude isthetension atthepoint s,oriented inthedirection of increasing s.Bydifferentiating Eq.(5-122) with respect tos,weobtain adifferential equation for1'(s): d'r%_-r. (5-123) The simplest andmost important application ofEq.(5-123) istothe case ofastring having aweight wperunit length. Ifthestring isacted onbynoother forces except attheends, itwillhang inavertical plane, which wetake tobethexy-plane, with thex-axis horizontal andthey-axis vertical. Let0betheangle between thestring andthex-axis (Fig. 5-23). Then thehorizontal andvertical components ofEq.(5-123) become: 5;(1'sin0)=w, (5—124) dis(1'cos0)=O. (5-125) 238 RIGID Booms. ROTATION ABOUT ANAxis. STATICS [ciiAi=. 5 Equation (5—125) implies that 1'cos0=C. (5—126) Thehorizontal component oftension isconstant, asitshould besince the external forces onthestring areallvertical, except attheends. Bydivid- ingEq.(5—124) byC,andusing Eq.(5-126), weeliminate thetension: dtan6 w75- —5- (5—127) Ifwerepresent thestring byspecifying thefunction y(x), wehave the relationsdAtan0=3%=1/, (5-128) ds=[(5)2+(di/)2l"2 =die+5'2)“, <5-129) sothat Eq.(5-127) becomes -55,-'=:55+v’2)”2- <5-130) This canbeintegrated, ifwisconstant: _..d._v' _2 _ /la+?/2),” -[Cdx, (5131) sinh_1 1/=196%’+oz, (5-132) where ozisaconstant. Wesolve fory’: d .y’=%=sinh +Cl)- (5—133) This canbeintegrated again, andweobtain Cy=B—|—Ecosh +Ct)- (5-134) Thecurve represented byEq.(5-134) iscalled acatenary, andistheform in which auniform string willhang ifacted onbynoforce other than itsown weight, except attheends. The constants C’,)6,and aaretobechosen so that yhastheproper value attheendpoints, andsothat thetotal length ofthestring hastheproper value. Thetotal length is z=fat=fr“(1+1/2)1'2dt =f“cosh +5.)at =g[eini1 +51)—sinh + (5-135) .\ . 5-10] EQUILIBRIUM or‘soLIo BEAMS 239 5-10 Equilibrium ofsolid beams. Ahorizontal beam subject tovertical forces isoneofthesimplest examples ofastructure subject toshearing forces andbending moments. Tosimplify theproblem, weshall consider only thecase when thebeam isunder nocompression ortension, andwe shall assume that thebeam issoconstructed andtheforces soapplied that thebeam bends inonly onevertical plane, without anytorsion (twisting) about theaxis ofthebeam. Weﬁnd ﬁrst thestresses within thebeam from aknowledge oftheexternal forces, andthen determine thedistortion ofthebeam duetothese stresses. Points along thebeam willbelocated byacoordinate xmeasured hori- zontally from theleftendofthebeam (Fig. 5-24). Letvertical forces F1,...,F,,actatthedistances x1,...,x,,from theleftend. Aforce willbetaken aspositive ifitisdirected upward. LetAA’ beaplane perpendicular tothebeam atany distance xfrom theend. According toTheorem I(5—107), ofSection 5-6, thesystem offorces exerted across theplane AA’ bythematerial ontheright against that ontheleftis equivalent toasingle force Sthrough anypoint intheplane, andacouple oftorque N. (Note that inapplying Theorem I,wearetreating the plane AA’ asarigid body, that is,weareassuming that thecross-sectional plane AA’ isnotdistorted bytheforces acting onit.)Inthecase weare considering there isnocompression ortension andallforces arevertical, so thatSisdirected vertically. Weshall deﬁne theshearing force Sasthe vertical force acting across AA’fromright toleft;Swillbetaken aspositive when thisforce isdirected upward, negative when itisdownward.* By Newton’s third law, theforce acting across AA’ from lefttoright is—-S. Since weareassuming notorsion about theaxisofthebeam (x-axis), and since alltheforces arevertical, thetorque Nwillbedirected horizontally andperpendicular tothebeam. Weshall deﬁne thebending moment N F2 FIG. 5-24. Forces acting onabeam..8’ ..,q___>-“Q7: 17>? *This sign convention forSisinagreement with signconventions throughout thisbook, where theupward direction istaken aspositive. Sign conventions for shearing force andbending moment arenotuniform inphysics andengineering texts, andonemust becareful inreading theliterature tonote what sign con- vention isadopted byeach author. 240 RIGID Booms. ROTATION ABOUT ANAxis. STATICS [CHAP. 5 asthetorque exerted from right toleftacross AA’ about ahorizontal axis intheplane AA’; Nwillbetaken aspositive when ittends torotate the plane AA’ inacounterclockwise direction. Since Sisvertical, thetorque willbethesame about anyhorizontal axisintheplane AA’. The shearing force Sand bending moment Ncanbedetermined by applying theconditions ofequilibrium [Eqs. (5-95) and (5—96)] tothe part ofthebeam totheleftoftheplane AA’. The total force andtotal torque about ahorizontal axis intheplane AA’ are,ifweneglect the weight ofthebeam, ZF.+S=0, (5-135) x,-<2 —N1,-Z(x-x,-)F,~+N=0, (5-197) a:,'<:c where thesums aretaken over allforces acting totheleftofAA’ andN0 isthebending moment, ifany, exerted bytheleftendofthebeam against itssupport. The torque N0willappear only ifthebeam isclamped or otherwise fastened atitsleftend. The force exerted byany clamp or other support attheendistobeincluded among theforces F,~. Ifthe beam hasaweight wperunit length, this should beincluded inthe equilibrium equations: zF,~—[:wdx+S=0, (5—138) x1'<:c —N11-Z(tt-.5,-)F,~+/0”(x-x’)wdx’+N=0.(5-139) I,"<x The shearing force andbending moment atadistance xfrom theendare therefore s=_ZF,-+/“ode, (5-140) :c1'<a: 0 N=N0+Z(x-x,-)F,--/0“(x-x’)wdx'. (5-141) x,'<x Ifthere isanyadditional force distributed continuously along thebeam, this canbeincluded inwasanadditional Weight perunit length. Ifthe beam isfreeatitsends, theshearing force andbending moment must be zero attheends. IfwesetS=N=0attheright endofthebeam, equations (5—140) and(5—141) may besolved fortwooftheforces acting onthebeam when theothers areknown. Ifthebeam isfastened orclamped ateither end, SandNmay have anyvalues there. Equations (5-140) and (5—141) determine Sand Neverywhere along thebeam when all 5-10] EQUILIBRIUM orsoLio BEAMS 241 A A OI 0 0' 5 0 A A’ (=1) (b) A 0’ §‘ A0 1,A, (c)\\is x FIG. 5-25. Distortion ofabeam byshearing andbending. (a)Undistorted beam. (b)Beam inshear. (c)Beam bent andinshear. theforces areknown, including theforce and torque exerted through theclamp, ifany, ontheleftend. The shearing force andbending mo- ment may beplotted asfunctions ofxwhose slopes atany point are obtained bydifferentiating Eqs. (5-140) and(5—141): 3;:=w, (5-142) dN 1776-=ZF,--[0wdx’=-s. (5-143) x,'<:c The shearing force increases by—F, from lefttoright across apoint x,- where aforce F,-acts. Letusnow consider thedistortion produced bytheshearing forces and bending moments inabeam ofuniform cross section throughout its length. InFig.5—25(a) isshown anundistorted horizontal beam through which aredrawn ahorizontal lineO0’andavertical plane AA’. InFig. 5—25(b) thebeam isunder ashearing strain, theeffect ofwhich istoslide thevarious vertical planes relative tooneanother sothat thelineO0’ makes anangle 0with thenormal totheplane AA’. According toEq. (5-118), theangle 0isgiven interms oftheshearing force Sandtheshear modulus nby: o=%, (5-144) 242 RIGID Booms. ROTATION ABOUT ANAxis. STATICS lCHAP. 5 Al 0' B (2-Al) A BlY g neural l 0 (r+Ar) layer <0A, ' FIG. 5-26. Strains inabent beam. where Aisthecross-sectional area, andwehave made theapproximation tan0é0,since 0willbevery small. InFig.5—25(c), weshow thefurther effect of"bending thebeam. The plane AA’ now makes anangle <pwith thevertical. Itisassumed that thecross-sectional surface AA’ remains plane andretains itsshape when thebeam isunder stress, although this may notbestrictly truenear thepoints where forces areapplied. Inorder todetermine cp,weconsider twoplanes AA’ andBB’initially vertical and asmall distance lapart. When thebeam isbent, AA’ andBB’willmake angles <pandtp+A<pwith thevertical (Fig. 5-26). Duetothebending, theﬁbers ontheoutside ofthecurved beam willbestretched andthose on theinside willbecompressed. Somewhere within thebeam willbeaneu- trallayer ofunstretched ﬁbers, andweshall agree todraw thelineO0’so that itliesinthisneutral layer. Alinebetween AA’ andBB’ parallel to O0’andadistance zabove O0’willbecompressed toalength l-Al, where (seeFig.5-26) ~ Al=zAcp. (5—145) Thecompressive force dFexerted across anelement ofarea dAadistance z above theneutral layer O0’willbegiven byEq.(5-114)interms ofYoung’s modulus: dF_ Al_ E or,ifweletl=ds,aninﬁnitesimal element oflength along thelineO0’, dF___ dga_H—Y2E (5-147) This equation isimportant inthedesign ofbeams, asitdetermines the stress ofcompression ortension atany distance zfrom theneutral layer. Thetotal compressive force through thecross-sectional area Aofthebeam 5-10] EQUILIBRIUM orsoLIo BEAMs 243 willbe . ' F=£[dF= YZ—:£[zdA. (5-148) Since weareassuming nonettension orcompression ofthebeam, F=0, and /IZJA =0. (5-149) A This implies thattheneutral layer contains thecentroid oftheareaAof thebeam, andwemay require that O0’bedrawn through thecentroid of thecross-sectional area ofthebeam. The bending moment exerted by theforces dFis N=[/ear: Y%//221111 A A _ 2d_¢, __YkAds (5—150) where k2=i[1fZ2dA, (5-151) andhistheradius ofgyration ofthecross-sectional area ofthebeam about ahorizontal axisthrough itscentroid. The differential equation for<pis therefore d N<5-52> Lettheupward deﬂection ofthebeam from ahorizontal x-axis bey(x), measured tothelineOO’(Fig. 5-25). Then y(x) istobedetermined by solving theequation %=an<0+ti. <5-153) when 0andcphave been determined from Eqs. (5—144) and(5—152). If weassume that both 0andtoarevery small angles, Eqs. (5-152) and (5-153) become d Nif=W-ii, (5-154) dy_ _ 244 RIGID BODIES. ROTATION ABoU'r ANAxis. STATICS [ciiAi>. 5 When there arenoconcentrated forces F1along thebeam, wemay differ- entiate Eq.(5—155) andmake useofEqs. (5-154), (5-144), (5-142), and (5-143) toobtain dzy w Nas-H+I/tTA' <5"15"‘> d4y 1d2w was=mRF"vex" <5-157) Ifbending canbeneglected, asinashort, thick beam, Eq.(5—156) with N=0becomes asecond-order differential equation tobesolved fory(x). Foralonger beam, Eq.(5—157) must beused. These equations canalso beused when concentrated loads F;arepresent, bysolving them foreach segment ofthebeam between thepoints where theforces F,areapplied, andﬁtting thesolutions together properly atthese points. The solutions oneither sideofapoint x,-where aforce F,isapplied must bechosen so that y,<p,Narecontinuous across x,-,while S,dN/dx, dy/dx, d3y/dx3 increase across thepoint x,~byanamount determined byEqs. (5—140), (5-143), (5—155), and (5-156). The solution ofEq.(5—156) willcontain twoarbitrary constants, andthatofEq.(5-157), four, which aretobe determined bytheconditions attheendsofthebeam orsegment ofbeam. Asanexample, weconsider auniform beam ofweight W,length L, clamped inahorizontal position (i.e., sothat (5=0)*atitsleftend (x=0),andwith aforce F1=-W’ exerted onitsright end(x=L). Inthiscase, Eq.(5—157) becomes try_ Wan-—rm? <5*1~”8> Thesolution is W914 3 2?/=— +B03111 +iiczx "l"C199+00- (5-159) Todetermine theconstants C0,C1,C2,C3,wehave attheleftendofthe beam: y=C0=0, (5-160) dz/_ ___S___W+W' where wehave used Eqs. (5—155) and (5-144). Weneed twomore con- ditions, which may bedetermined inavariety ofways. The easiest way *Thecondition <p=0means that theplane AA’ isvertical; that is,thebeam would behorizontal ifthere were noshearing strain. 5-11] EQUILIBRIUM orFLUIDS 245 inthiscaseistoapply Eq.(5-156) anditsderivative attheleftendofthe beam: 2 d_y_ _W_W’L +lWL dx2_C2”HAL Yk2A ’ (H62) as;/_ _1dN_ s W’+W E_C3_*Y1t2A dx—_Yh2A=Yk2A ’(H63) where Wehave used Eq.(5—143). Thedeﬂection ofthebeam atanypoint xisthen .__ L3 |:Wx2(1 ____+ 2 W’x2<1 1y‘ Yh2A 4L2 "“ 317 “éi 1 W’—--<1~2%)+%l- <5-s>cowhis §b'@»l—l FilblaP‘§Q/+ Thedeﬂection atx=Lis L3 1 1 I L 1 I The ﬁrst term ineach equation isthedeﬂection duetobending, andthe second isthatduetoshear. Theﬁrstterm isproportional toL3,andin- versely proportional tok2.The second term isproportional toLandin- dependent oflc.Hence bending ismore important forlong, thin beams, andshear ismore important forshort, thick beams. Ouranalysis here is probably notvery accurate forshort, thick beams, since, aspointed out above, some ofourassumptions arenotvalid near points ofsupport or points where loads areapplied (where “near” means relative tothecross- sectional dimensions ofthebeam). 5-11 Equilibrium ofﬂuids. Aﬂuid isdeﬁned asasubstance which will support noshearing stress when inequilibrium. Liquids andgases ﬁtthis deﬁnition, andeven very viscous substances likepitch, ortar,orthemate- rialintheinterior oftheearth, willeventually come toanequilibrium in which shearing stresses areabsent, ifthey areleftundisturbed forasuﬂi- ciently long time. The stress F/Aacross anysmall area Ainaﬂuid in equilibrium must benormal toA,andinpractically allcases itwillbea compression rather than atension. Weﬁrst prove that thestress F/Anear anypoint intheﬂuid isinde- pendent oftheorientation ofthesurface A.Letanytwodirections be given, andconstruct asmall triangular prism with twoequal faces A1= A2perpendicular tothetwogiven directions. Thethird faceA3istoform with A1andA2across section having theshape ofanisosceles triangle (Fig. 5-27). LetF1,F2,F2bethestress forces perpendicular tothefaces 246 RIGID BODIES. ROTATION ABoU'r ANAxis. STATICS [oiiAP. 5 A1A2 F1 F2 K A3 F F2 FIG. 5-27. Forces onatriangular prism inaﬂuid. A1,A2,A2. Iftheﬂuid intheprism isinequilibrium, F1 +F2 +F3 = Theforces ontheendfaces oftheprism need notbeincluded here, since they areperpendicular toF1,F2,andF3,andmust therefore separately addtozero. Itfollows from Eq.(5—166), andfrom theway theprism has been constructed, that F1,F2,and F3must form anisosceles triangle (Fig. 5-27), andtherefore that F1=F2. (5-167) Since thedirections ofF1andF2areanytwodirections intheﬂuid, and since A1=A2,thestress F/Aisthesame inalldirections. Thestress in aﬂuid iscalled thepressure p: _ﬁ_&. _ 11-A1_A2 (5168) Now suppose that inaddition tothepressure theﬂuid issubject toan external force fperunit volume ofﬂuid, that is,anysmall volume dVin theﬂuid isacted onbyaforce fdV. Such aforce iscalled abody force,' fisthebody force density. The most common example isthegravita- tional force, forwhich f=Pg, (5—169) where gistheacceleration ofgravity, andpisthedensity. Ingeneral, thebody force density may differ inmagnitude and direction atdifferent points intheﬂuid. Intheusual case, when thebody force isgiven by Eq.(5—l69), gwillbeconstant andfwillbeconstant indirection; ifpis constant, fwillalsobeconstant inmagnitude. Letusconsider twonearby points P1,P2intheﬂuid, separated byavector dr.Weconstruct acylinder oflength drandcross-sectional area dA,whose endfaces contain thepoints P1andP2.Then thetotal component offorce inthedirection ofdracting 5-11] EQUILIBRIUM orFLUIDS 247 ontheﬂuid inthecylinder, since theﬂuid isinequilibrium, willbe f-drdA +p1dA—p2dA=0, where p1andp2arethepressures atP1andP2.Thedifference inpressure between twopoints adistance drapart istherefore dp=p2—p1=f-dr. (5-170) Thetotal difference inpressure between twopoints intheﬂuidlocated by vectors r1and1'2willbe P2—p.=fr"r-dr, <5-111) 1 where thelineintegral ontheright istobetaken along some path lying entirely within theﬂuid from r1tor2.Given thepressure p1atr1,Eq. (5—171) allows ustocompute thepressure atanyother point 1'2which can bejoined to1'1byapath lying within theﬂuid. Thedifference inpressure between any two points depends only onthebody force. Hence any change inpressure atanypoint inaﬂuid inequilibrium must beaccom- panied byanequal change atallother points ifthebody force does not change. This isPascal’s law. According tothegeometrical deﬁnition (3-107) ofthegradient, Eq. (5—170) implies that f=Vp. (5-172) The pressure gradient inaﬂuid inequilibrium must beequal tothebody force density. This result shows that thenetforce perunit volume dueto pressure is—Vp. The pressure pisasort ofpotential energy perunit volume inthesense that itsnegative gradient represents aforce perunit volume duetopressure. However, theintegral ofpdVover avolume does notrepresent apotential energy except invery special cases. Equa- tion (5-172) implies that thesurfaces ofconstant pressure intheﬂuid are everywhere perpendicular tothebody force. According toEqs. (3-187) and(5-172), theforce density fmust satisfy theequation I VXf=0. (5-173) This istherefore anecessary condition onthebody force inorder forequi- librium tobepossible. Itisalsoasuﬂicient condition forthepossibility ofequilibrium. This follows from thediscussion inSection 3-12, forif Eq.(5-173) holds, then itispermissible todeﬁne afunction p(r) bythe equation p(r)=1».+/:1-dr, <5-114) 248 RIGID BODIES. ROTATION ABOUT ANAxis. STATICS [CHAP. 5 where p1isthepressure atsome ﬁxed point 1'1,andtheintegral may be evaluated along anypath from r1torwithin theﬂuid. Ifthepressure in theﬂuid atevery point rhasthevalue p(r) given by(5-174), then Eq. (5-172) willhold, andthebody force fperunit volume willeverywhere bebalanced bythepressure force —Vp perunitvolume. Equation (5-174) therefore deﬁnes anequilibrium pressure distribution foranybody force satisfying Eq.(5-173). Theproblem ofﬁnding thepressure within aﬂuid inequilibrium, ifthe body force density f(r)isgiven, isevidently mathematically identical with theproblem discussed inSection 3-12 ofﬁnding thepotential energy fora given force function F(r). Weﬁrst check that Vxfiszeroeverywhere within theﬂuid, inorder tobesure that anequilibrium ispossible. We then take apoint r1atwhich thepressure isknown, anduseEq.(5—174) to ﬁndthepressure atanyother point, taking theintegral along anyconven- ientpath. Thetotal body force acting onavolume Voftheﬂuid is rt=[![fdV. (5-115) Thetotal force duetothepressure onthesurface AofVis F1,=-HepdA, (5-115) A where nistheoutward normal unit vector atanypoint onthesurface. These twomust beequal andopposite, since theﬂuid isinequilibrium: F1,=—-F1,. (5-177) Equation (5-176) gives thetotal force duetopressure onthesurface ofthe volume V,whether ornotVisoccupied byﬂuid. Hence weconclude from Eq.(5—177) that abody immersed inaﬂuid inequilibrium isacted onby aforce F1,duetopressure, equal andopposite tothebody force F1,which would beexerted onthevolume Vifitwere occupied byﬂuid inequi- librium. This isArchimedes’ principle. Combining Eqs. (5—172), (5-175), (5-176), and(5-177), wehave /[Hp AA=/[fvp dV. (5-17s) A V This equation resembles Gauss’ divergence theorem [Eq. (3—115)], except that theintegrands arenpandVpinstead ofn-A andV-A. Gauss’ the- orem can,infact, beproved inavery useful general form which allows usto replace thefactor ninasurface integral byVinthecorresponding volume 5-11] EQUILIBRIUM orFLUIDS 249 integral without anyrestrictions ontheform oftheintegrand except that itmust besowritten that thedifferentiation symbol Voperates onthe entire integrand.* Given thisresult, wecould start with Eqs. (5-175), (5-176), and(5-177), anddeduce Eq.(5—172): F1,+F,, =[!ffdV —[{[npdA =/ff(i- Vp)dV=0. (5-119) V Since thismust hold foranyvolume V,Eq.(5—172) follows. Sofarwehave been considering only thepressure, i.e.,thestress, ina ﬂuid. The strain produced bythepressure within aﬂuid isachange in volume perunit mass oftheﬂuid or,equivalently, achange indensity. IfHooke’s lawissatisﬁed, thechange dVinavolume Vproduced bya small change dpinpressure canbecalculated from Eq. (5—116), ifthe bulk modulus Bisknown: <1V__Q. _ -I7— B (5180) Ifthemass ofﬂuid inthevolume VisM,then thedensity is M andthechange dpindensity corresponding toaninﬁnitesimal change dV involume is‘given by dp dV. —=———1 5-182 P V ( ) sothat thechange indensity produced byasmall pressure change dpis dp dp -——=—-- 5-183 g P B ( ) After aﬁnite change inpressure from potop,thedensity willbe PdP=Poexp(I-5') (5—184) ' P0 Inanycase, thedensity ofaﬂuid isdetermined byitsequation ofstate in *Fortheproof ofthis theorem, seePhillips, Vector Analysis. New York: John Wiley andSons, 1933. (Chapter III,Section 34.) 250 RIGID Booms. ROTATION ABOUT ANAxis. STATICS [CHAP. 5 terms ofthepressure andtemperature. Theequation ofstate foraper- fectgasis pV=RT, I (5—185) where Tistheabsolute temperature, Visthevolume permole, andRis theuniversal gasconstant: R=8.314><107erg-deg'1 0-mole-1. (5—186) Bysubstitution from Eq.(5—l8l), weobtain thedensity interms ofpres- sure andtemperature: _1l'£ _ p--RT, (5187) where Misthemolecular weight. Letusapply these results tothemost common case, inwhich thebody force isthegravitational force onaﬂuid inauniform vertical gravitational ﬁeld [Eq. (5-169)]. Ifweapply Eq.(5—173) tothiscase, wehave Vxf=Vx(pg)=0. (5-188) Since gisconstant, thedifferentiation implied bytheVsymbol operates onlyonp,andwecanmove thescalar pfrom onefactor ofthecross product totheother toobtain: (VP)X2=0, (5—l89) that is,thedensity gradient must beparallel tothegravitational ﬁeld. The density must beconstant onanyhorizontal plane within theﬂuid. Equation (5—l89) may also bederived from Eq. (5—188) bywriting outexplicitly thecomponents ofthevectors Vx(pg)and(Vp) Xg,and verifying that they arethesame.* According toEq.(5-172), thepres- sure isalso constant inanyhorizontal plane within the-ﬂuid. Pressure anddensity aretherefore functions only ofthevertical height zwithin the ﬂuid. From Eqs. (5—172) and (5-169) weobtain adifferential equation forpressure asafunction of2: gig=-pg. (5—190) Iftheﬂuid isincompressible, andpisuniform, thesolution is v=Po—Pea (5—191) *Equation (5—l89) holds also inanonuniform gravitational ﬁeld, since VXg=O,byEq.(6-21). PROBLEMS 251 where paisthepressure atz=0.Iftheﬂuid isaperfect gas,either porp may beeliminated from Eq.(5—190) bymeans ofEq.(5—187). Ifwe eliminate thedensity, wehave dz»_ My _dz_ RTp. (5192) Asanexample, ifweassume thattheatmosphere isuniform intemperature andcomposition, wecansolve Eq.(5—l92) fortheatmospheric pressure asafunction ofaltitude: Mp=poexp(——ﬁg,z)- (5—193) PROBLEMS 1.(a)Prove that thetotal kinetic energy ofthesystem ofparticles making up arigid body, asdeﬁned byEq.(4-37), iscorrectly given byEq.(5-16) when the body rotates about aﬁxed axis. (b)Prove thatthepotential energy given by Eq.(5-14) isthetotal work done against theexternal forces when thebody is rotated from 0,to0,ifN,isthesumofthetorques about theaxisofrotation due totheexternal forces. 2.Prove, starting with theequation ofmotion (5-13) forrotation, thatifN, isafunction of0alone, then T-|—Visconstant. 3.Awheel ofmass M,radius ofgyration lo,spins smoothly onaﬁxed horizontal axleofradius awhich passes through aholeofslightly larger radius atthehubof thewheel. The coefﬁcient offriction between thebearing surfaces is;1..Ifthe wheel isinitially spinning with angular velocity wo,ﬁndthetime andthenumber ofturns thatittakes tostop. 4.The balance wheel ofawatch consists ofaring ofmass M,radius a,with spokes ofnegligible mass. The h8,i1‘SpI‘il1g exerts arestoring torque N,=—k0. Find themotion ifthebalance wheel isrotated through anangle 00andreleased. 5.Anairplane propeller ofmoment ofinertia Iissubject toadriving torque N=N0(1 —|—ozcoswot), andtoafrictional torque due-toairresistance N;=—b6. Find itssteady-state motion. ‘ 6.Amotor armature weighing 2kgm hasaradius ofgyration of5cm. Its no-load speed is1500 rpm. Itiswound sothat itstorque isindependent ofits speed. Atfullload, itdraws acurrent of2amperes at110volts. Assume that theelectrical eﬁiciency is80%, andthat thefriction isproportional tothesquare oftheangular velocity. Find thetime required forittocome uptoaspeed of 1200 rpm after being switched onwithout load. 7'.Derive Eqs. (5-35) and(5-36). 252 RIGID BODIES. ROTATION ABOUT ANAXIS. STATICS [CHAP. 5 8.Assume thatasimple pendulum suﬁers africtional torque —mb19 dueto friction atthepoint ofsupport, andafrictional force —b2v onthebobdueto airresistance, where visthevelocity ofthebob. The bobhasamass m,andis suspended byastring oflength l.Find thetime required fortheamplitude to damp to1/eofitsinitial (small) value. How should m,lbechosen ifitisdesired that thependulum swing aslong aspossible? How should m,lbechosen ifit isdesired that thependulum swing through asmany cycles aspossible? 9.Acompound pendulum isarranged toswing about either oftwoparallel axes through twopoints O,0'located onalinethrough thecenter ofmass. The distances h,h’from O,O’tothecenter ofmass, andtheperiods 1-,1-’ofsmall amplitude vibrations about theaxes through OandO’aremeasured. OandO’ arearranged sothat each isapproximately thecenter ofoscillation relative to theother. If-r=1',ﬁndaformula forginterms ofmeasured quantities. If 1"=-r(1+5),where 6<<1,ﬁndacorrection tobeadded toyour previous formula sothat itwillbecorrect toterms oforder 6. 10.Abaseball batheld horizontally atrestisstruck atapoint O’byaball which delivers ahorizontal impulse J’perpendicular tothebat. Letthebatbe initially parallel tothea;-axis, andletthebasbeall betraveling inthenegative direction parallel tothey-axis. Thecenter ofmass Gofthebatisinitially atthe origin, andthepoint O’isatadistance h’from G.Assuming that thebatislet gojust astheballstrikes it,andneglecting theeffect ofgravity, calculate and sketch themotion x(t), y(t)ofthecenter ofmass, andalsoofthecenter ofper- cussion, during theﬁrst fewmoments after theblow, sayuntil thebathas rotated aquarter turn. Comment onthedifference between theinitial motion ofthecenter ofmass andthatofthecenter ofpercussion. 11.Acircular disk ofradius aliesinthemy-plane with itscenter attheorigin. Thehalfofthediskabove thex-axis hasadensity aperunitarea, andthehalf below thea:-axis hasadensity 20.Find thecenter ofmass G,andthemoments of inertia about the:2:-,y-,and2-axes, andabout parallel axes through G.Make as much useoflaborsaving theorems aspossible. 12.(a)Work outaformula forthemoments ofinertia ofacone ofmass m, height h,andgenerating angle oz,about itsaxisofsymmetry, andabout anaxis (1 _..... ____ T1| I Fro. 5—28. Frustum ofacone. PROBLEMS 253 @V.~_\ FIG. 5-29. How much thread canbewound onthisspool‘? through theapex perpendicular totheaxisofsymmetry. Find thecenter ofmass ofthecone. (b)Usethese results todetermine thecenter ofmass ofthefrustum ofacone, shown inFig.5-28, andtocalculate themoments ofinertia about hori- zontal axes through each base andthrough thecenter ofmass. The mass ofthe frustum isM. 13.How many yards ofthread 0.03inch indiameter canbewound onthe spool shown inFig.5-29? 14.Given thatthevolume ofaconeisone-third theareaofthebasetimes the height, locate byPappus’ theorem thecentroid ofaright triangle whose legsare oflengths aandb. 15.Prove that Pappus’ second theorem holds even iftheaxisofrevolution intersects thesurface, provided that wetake asvolume thedifference inthe volumes generated bythetwoparts intowhich thesurface isdivided bytheaxis. What isthecorresponding generalization oftheﬁrst theorem? 16.Find thecenter ofmass ofawire bent into asemicircle ofradius a.Find thethree radii ofgyration about x-,y-,andz-axes through thecenter ofmass, where zisperpendicular totheplane ofthesemicircle andasbisects thesemicircle. Useyour ingenuity toreduce thenumber ofcalculations required toaminimum. 17.(a)Find aformula fortheradius ofgyration ofauniform rodoflength l about anaxisthrough oneendmaking anangle awith therod. (b)Using this result, ﬁnd themoment ofinertia ofanequilateral triangular pyramid, con- structed outofsixuniform rods, about anaxis through itscentroid andoneof itsvertices. 18.Find theradii ofgyration ofaplane lamina intheshape ofanellipse of semimajor axis a,eccentricity e,about itsmajor andminor axes, andabout a4third axisthrough onefocus perpendicular totheplane. 19.Forces 1kgm-wt, 2kgm-wt, 3kgm-wt, and 4kgm-wt actinsequence clockwise along thefour sides ofasquare 0.5X0.5m2. The forces aredirected inaclockwise sense around thesquare. Find theequilibrant. 20.Aniceboat hasaﬁatsailintheshape ofaright triangle with avertical legoflength aalong themast andahorizontal legoflength balong theboom. The force onthesailacts atitscentroid andisgiven byF=Ic[n-(w—v)]n, 254 RIGID BODIES. ROTATION ABOUT ANAXIS. STATICS [CHAP. 5 411» 31b 81b lft 81b 71b FIG. 5-30. Asystem offorces acting onacube. where nisaunitvector normal tothesail,wisthewind velocity, visthevelocity oftheboat, andIcisaconstant. The sailmakes anangle awith thecenter line oftheboat. The angle ozmay have anyvalue uptothat forwhich Fbecomes zero. The center lineoftheboat makes anangle 5with thedirection (—w) from which thewind isblowing. The runners areparallel tothecenter line. The coefﬁcient offriction along therunners isp.,and there isaforce Nper- pendicular totherunners suﬂicient toinsure thatvisparallel totherunners and constant indirection. Find vasafunction ofoz,B,wandthemass mofthe boat. Find Nandthepoint atwhich itacts. What value ofamakes vamaxi- mum ifp.isvery small? i 21.(a)Reduce thesystem offorces acting onthecube shown inFig.5-30 to anequivalent single force acting atthecenter ofthecube, plus acouple com- posed oftwoforces acting attwoadjacent corners. (b)Reduce thissystem to asystem oftwoforces, andstate where these forces act. (c)Reduce thissystem toasingle force plus atorque parallel toit. 22.(a)Acable isconnected inastraight linebetween twoﬁxed points. By exerting asidewise force Watthecenter ofthecable, aconsiderably greater force 1-canbeapplied tothesupport points ateach endofthecable. Find a formula for1'interms ofW,and thearea Aand Young’s modulus Yofthe cable, assuming thattheangle through which thecable ispulled issmall. (b) Show that this assumption iswell satisﬁed ifW=100lb,A=3in2, and Y=60,000lb-in“2. Find 1'. 23.Acable istobeespecially designed tohang vertically andtosupport a _load Watadistance Zbelow thepoint ofsupport. The cable istobemade ofa material having aYoung’s modulus Yandaweight wperunit volume. Inas- much asthelength lofthecable istobefairly great, itisdesired tokeep the weight ofthecable toaminimum bymaking thecross-sectional area A(z) ofthe cable, ataheight zabove thelower end, justgreat enough tosupport theload beneath it.The cable material cansafely support aload just great enough to stretch it1%. Determine thefunction A(z) when thecable issupporting the given load. PROBLEMS 255 l-—Z1> FIG. 5-31. Asuspension bridge. 24.Acable 20ftlong issuspended between twopoints AandB,15ftapart. The lineABmakes anangle of30°with thehorizontal (Bhigher). Aweight of 2000 lbishung from apoint C’8ftfrom theendofthecable atA.(a)Find the position ofpoint C,andthetensions inthecable, ifthecable does notstretch. (b)Ifthecable is§inch indiameter andhasaYoung’s modulus of5X105lb- in"2, ﬁnd theposition ofpoint C’and thetensions, taking cable stretch into account. Carry outtwosuccessive approximations, andestimate theaccuracy ofyour result. 25.(a)Acable oflength l,weight wperunit length, issuspended from the points sv=;l=aonthe:0-axis. They-axis isvertical. Byrequiring that y=0at as==l;a,andthat thetotal length ofcable bel,show that oz=0inEq.(5—134), andsetupequations tobesolved for/3andC’.(b)Show thatthesame results canbeobtained foraandCbyrequiring thatthecable besymmetrical about the y-axis, andthattheforces atitsends balance theweight ofthecable. 26.Abridge ofweight wperunitlength istobehung from cables ofnegligible weight, asshown inFig.5-31. Itisdesired todetermine theshape ofthesuspen- sioncables sothat thevertical cables, which areequally spaced, willsupport equal weights. Assume thatthevertical cables aresoclosely spaced thatwecan regard theweight wperunit length ascontinuously distributed along the suspension cable. Theproblem then differs from that treated inthetext, where thestring hadaweight wperunit length salong thestring, inthat here there isaweight wperunit horizontal distance x.Setupadifferential equation for theshape y(a:) ofthesuspension cable, and solve fory(:r) iftheends areat thepoints y=0,2:==|=§D, andifthemaximum tension inthecable isto be010- _ 27.Acable oflength l,weight wperunit length, issuspended from points :1:==I:aontheac-axis. They-axis isvertical. Aweight Wishung from themid- point ofthecable. Setuptheequations from which B,oz,andC’aretobede- termined. 28.Aseesaw ismade ofaplank ofwood ofrectangular cross section 2X12in2 and 10ftlong, weighing 60lb. Young’s modulus is1.5X1061b-in“2. The plank isbalanced across anarrow support atitscenter. Two children weighing 100lbeach sitonefoot from theends. Find theshape oftheplank when itis balanced inastationary horizontal position. Neglect shear. 29.Anempty pipe ofinner radius a,outer radius b,ismade ofmaterial with Young’s modulus Y,shear modulus n,density p.Ahorizontal section oflength Lisclamped atboth ends. Find thedeﬂection atthecenter. Find theincrease indeﬂection when thepipe isﬁlled with aﬂuid ofdensity p0. 256 RIGID Booms. ROTATION ABOUT ANAXIS. STATICS [cnA1>. 5 30.AnI-beam hasupper andlower ﬂanges ofwidth a,connected byacenter webofheight b.Thewebandﬂanges areofthesame thickness c,assumed negligi- blewith respect toaandb,andaremade ofamaterial with Young’s modulus Y, shear modulus n.Thebeam hasaweight W,length L,andrests onsupports at each end. Aload W’rests onthemidpoint ofthebeam. Find thedeﬂection of thebeam atitsmidpoint. Separate thedeﬂection into terms duetoshear and tobending, andinto terms duetothebeam weight Wandtheload W’. 31.Ifthebulk modulus ofwater isB,andtheatmospheric pressure atthesur- faceoftheocean ispg,ﬁndthepressure asafunction ofdepth intheocean, taking into account thecompressibility ofthewater. Assume that Bisconstant. Look upBforwater, andestimate theerror thatwould bemade atadepth of5miles ifthecompressibility were neglected. 32.Find theatmospheric pressure asafunction ofaltitude ontheassumption that thetemperature decreases with altitude, thedecrease being proportional to thealtitude. , CHAPTER 6 GRAVITATION 6-1Centers ofgravity forextended bodies. You willrecall that we formulated thelawofgravitation inSection 1-5. Any twoparticles of masses m1andm2,adistance rapart, attract each other with aforce whose magnitude isgiven byEq.(1—11): GmmF=-ii, (6-1) where G=6.67 X10_8 dyne-cm2-gm*2, (6—2) asdetermined bymeasurements oftheforces between large lead spheres, carried outbymeans ofadelicate torsion balance. Equation (6-1) can bewritten inavector form which gives both thedirection andmagnitude oftheattractive forces. Let1'1and1'2betheposition vectors ofthetwo particles. Then thegravitational force onmgduetomlis F1->2 =I (1'1""1'2)- A (6-3) The vector (r1—r2)gives theforce thecorrect direction, anditsmagni- tude isdivided outbytheextra factor Ir,—r2|inthedenominator. The lawofgravitation asformulated inEq.(6-3) isapplicable only to particles ortobodies whose dimensions arenegligible compared with the distance between them; otherwise thedistance [r1—r2|isnotprecisely deﬁned, norisitimmediately clear atwhat points andinwhat directions theforces act. Forextended bodies, wemust imagine each body divided into pieces orelements, small compared with thedistances between the bodies, andcompute theforces oneach oftheelements ofonebody due toeach oftheelements oftheother bodies. Consider now anextended body ofmass Mandaparticle ofmass m atapoint P(Fig. 6-1). Ifthebody ofmass Misdivided intosmall pieces ofmasses m,-,each piece isattracted toward mbyaforce which weshall callF,~.Now thesystem offorces F,canberesolved according toTheorem IofSection 5-6(5—107) into asingle force through anarbitrary point, plus acouple. Letthissingle force beF: F=ZF.~. <6-1) 257 258 en.\v1'rA'r1o>; lcnsr. 6 ‘Bl/7"" -r FIG. 6-1. Gravitational attraction between aparticle andanextended body. andletthearbitrary point betaken asthepoint P.Since none ofthe forces F;exerts anytorque about P,thetotal torque about Piszero, and thecouple vanishes. The system offorces therefore hasaresultant F acting along alinethrough themass m.Theforce acting onmis—F, since Newton’s third lawapplies toeach oftheforces F,inEq. (6-4). Welocate onthislineofaction ofFapoint Gadistance rfrom Psuch that GmM|F|=ﬁr‘ (5-5) Then thesystem ofgravitational forces between thebody Mandthepar- ticle misequivalent tothesingle resultant forces FonMand-—Fonm which would actifallthemass ofthebody Mwere concentrated atG. The point Giscalled thecenter ofgravity ofthebody Mrelative tothe point P;G’isnot, ingeneral, atthecenter ofmass ofbody Bf,noreven onthelinejoining Pwiththecenter ofmass. Theparts ofthebody close toPareattracted more strongly than those farther away, whereas in ﬁnding thecenter ofmass, allparts ofthebody aretreated alike. Further- more, theposition ofthepoint Gwilldepend ontheposition ofP.When Pisfaraway compared with thedimensions ofthebody, theacceleration ofgravity duetomwillbenearly constant over thebody and, inthis case, weshowed inSection 5-6that G’willcoincide with thecenter of mass. Also, inthecase ofauniform sphere oraspherically symmetrical distribution ofmass, weshall show inthenext section that thecenter of gravity always liesatthecenter ofthesphere. Therelative character of theconcept ofcenter ofgravity makes itoflittle useexcept inthecase of asphere orofabody inauniform gravitational ﬁeld. Fortwoextended bodies, nounique centers ofgravity caningeneral bedeﬁned, even relative toeach other, except inspecial cases, aswhen the bodies arefarapart, orwhen oneofthem isasphere. The system of 6-2] GRAVITATIONAL FIELD AND GRAVITATIONAL POTENTIAL 259 gravitational forces oneither body duetotheother may ormay nothave aresultant; ifitdoes, thetworesultants areequal andopposite andact along thesame line. However, even inthiscase, wecannot deﬁne deﬁ- nite centers ofgravity G1,G2forthetwobodies relative toeach other, since Eq.(6-5) speciﬁes only thedistance The general problem ofdetermining thegravitational forces between bodies isusually best treated bymeans oftheconcepts oftheﬁeld theory ofgravitation discussed inthenext section. 6-2Gravitational field and gravitational potential. The gravitational force Fmacting onaparticle ofmass matapoint r,duetoother particles miatpoints r,-,isthevector sum oftheforces duetoeach oftheother particles acting separately: If,instead ofpoint masses mi,wehave mass continuously distributed in space with adensity p(r), theforce onapoint mass matris Theintegral may betaken over theregion containing themass whose attraction wearecomputing, orover allspace ifweletp=0outside this region. Now theforce Fmisproportional tothemass m,and we deﬁne thegravitational ﬁeld intensity (orsimply gravitational ﬁeld) g(r), atanypoint rinspace, duetoanydistribution ofmass, astheforce per unit mass which would beexerted onanysmall mass matthat point: gm= (es) where Fmistheforce that would beexerted onapoint mass matthe point r.Wecanwrite formulas forg(r)forpoint masses orcontinuously distributed mass: g(,-)=E , (5_9) , lrw'_Ila g(r)=[// dV’. (6-10) The ﬁeld g(1') hasthedimensions ofacceleration, and isinfact the acceleration experienced byaparticle atthepoint r,onwhich noforces actother than thegravitational force. 260 GRAVITATION [CHAP- 6 The calculation ofthegravitational ﬁeld g(r)from Eq.(6-9) or(6-10) isdifﬁcult except inafewsimple cases, partly because thesum andin- tegral callfortheaddition ofanumber ofvectors. Since thegravitational forces between pairs ofparticles arecentral forces, they areconservative, asweshowed inSection 3-12, andapotential energy canbedeﬁned for aparticle ofmass msubject togravitational forces. Fortwoparticles m andm,-,thepotential energy isgiven byEqs. (3-229) and(3-230): -—Gmm-Vm-=-4 - 6-11 M,It_rt, <> Thepotential energy ofaparticle ofmass matpoint rduetoasystem of particles miisthen \ —G'mm,-Vm(1') = (6-12) Wedeﬁne thegravitational potential 9(r) atpoint rasthenegative ofthe potential energy perunit mass ofaparticle atpoint 1';[This choice of signin9(r)isconventional ingravitational theory.] so)=— <6-13> Forasystem ofparticles, ;G'em=Z <6-14> Ifp(r)represents acontinuous distribution ofmass, itsgravitational poten- tialis 9(r)=///‘§'3§r)T\ av’. (6-15) Because itisascalar point function, thepotential g(r) iseasier towork with formany purposes than istheﬁeld g(r). Inview oftherelation (3—185) between force andpotential energy, gmay easily becalculated, when 9isknown, from therelation g=V9. (6-16) Theinverse relation is T 9(r)=/ g-dr. (6-17) Thedeﬁnition of9(r), likethat ofpotential energy V(r), involves anarbi- trary additive constant or,equivalently, anarbitrary point r,atwhich 9=0.Usually r,istaken ataninﬁnite distance from allmasses, asin Eqs. (6-14) and(6-15). V 6-2] GRAVITATIONAL FIELD AND GRAVITATIONAL POTENTIAL 261 ‘ P0.a . To""" -» Fro. 6-2. Method ofcomputing potential ofaspherical shell. The concepts ofgravitational ﬁeld and gravitational potential are mathematically identical tothose ofelectric ﬁeld intensity and electro- static potential inelectrostatics, except that thenegative signinEq.(6-13) isconventional ingravitational theory, and except that allmasses are positive andallgravitational forces areattractive, sothat theforce law hastheopposite signfrom that inelectrostatics. Thesubject ofpotential theory isanextensive one,andwecangivehere only avery brief introduc- tory treatment. Asanexample oftheuseoftheconcept ofpotential, wecalculate the potential duetoathin homogeneous spherical shell ofmatter ofmass M, density <1perunit area, andradius a: M=41ra2o'. (6-18) The potential atapoint Piscomputed byintegrating over asetofring elements asinFig. 6-2. The potential ofaring ofradius asin0,width ad0,allofwhose mass isatthesame distance rfrom P,willbe dg=G0'(21ra :in0)ad0, andthetotal potential atPofthespherical shell is _ T‘Go(21ra sin0)ad0 at_/0V, _MG’I” sined0 _2 0(rﬁ+a2—2am cos0)1/2 MG=5,;[(m +<1)—[To—all (6-19) Wehave twocases, according towhether Pisoutside orinside theshell: ~ 9(P)= 1'02a,9(P)= 1,,5a.(6-20) 262 GRAVITATION [CHAP. 6 Thus outside theshell thepotential isthesame asforapoint mass Mat thecenter oftheshell. The gravitational ﬁeld outside aspherical shell is then thesame asifallthemass oftheshell were atitscenter. Thesame statement then holds forthegravitational ﬁeld outside any spherically symmetrical distribution ofmass, since thetotal ﬁeld isthesum ofthe ﬁelds duetotheshells ofwhich itiscomposed. This proves thestatement made intheprevious section; aspherically symmetrical distribution of mass attracts (and therefore isattracted by)anyother mass outside itas ifallitsmass were atitscenter. Inside aspherical shell, thepotential is constant, and itfollows from Eq. (6-16) that thegravitational ﬁeld is there zero. Hence apoint inside aspherically symmetric distribution of mass atadistance 'rfrom thecenter isattracted asifthemass inside the sphere ofradius rwere atthecenter ;themass outside thissphere exerts nonetforce. These results would besomewhat more difﬁcult toprove by computing thegravitational forces directly, asthereader canreadily verify. Indeed, ittook Newton twenty years! The calculation ofthe force ofattraction onthemoon bytheearth described inthelastsection ofChapter 1wasmade byNewton twenty years before hepublished his lawofgravitation. Itislikely that hewaited until hecould prove an assumption implicit inthat calculation, namely, that theearth attracts any body outside itasifallthemass oftheearth were concentrated atitscenter. 6-3Gravitational field equations. Itisofinterest toﬁnddifferential equations satisﬁed bythefunctions g(r) and9(r). From Eq. (6-16) it follows that Vxg=0. (6-21) When written outinanycoordinate system, thisvector equation becomes asetofthree partial differential equations connecting thecomponents of thegravitational ﬁeld. Inrectangular coordinates, §le_%_ %__Q91_ %_‘1%_ _ 6y 62—O’ 62 6x_O’ 61: 6y_0' (622) These equations alone donotdetermine thegravitational ﬁeld, forthey aresatisﬁed byevery gravitational ﬁeld. Todetermine thegravitational ﬁeld, weneed anequation connecting gwith thedistribution ofmatter. Letusstudy thegravitational ﬁeld gduetoapoint mass m.Consider anyvolume Vcontaining themass m,andletnbetheunit vector normal ateach point tothesurface Sthat bounds V(Fig. 6-3). Letuscompute thesurface integral I=f/n-gas. (6-23) s 6-3] GRAVITATIONAL FIELD nouirrroxs 263 ~___ ‘\__ Flo. 6-3. Amass inenclosed inavolume V. The physical orgeometric meaning ofthisintegral canbeseen ifwein- troduce theconcept oflines offorce, drawn everywhere inthedirection ofg,andinsuch amanner that thenumber oflines persquare centimeter atanypoint isequal tothegravitational ﬁeld intensity. ‘Then Iisthe number oflines passing outthrough thesurface S,andiscalled theﬂux ofgthrough S.The element ofsolid angle dS2subtended attheposition ofmbyanclement ofsurface dSisdeﬁned asthearea swept outona sphere ofunit radius byaradius from mwhich sweeps over thesurface element db“. This area is £19= . (644) From Fig.6-3, wehave therelation ,_.,.gZ_ (6_2_.;,, When useismade ofthese tworelations, theintegral I[Eq. (6—23)] be- comes I=If-me do=—41rmG‘. (c-26) s The integral Iisindependent oftheposition ofmwithin thesurface S. This result isanalogous tothecorresponding result inelectrostatics that there are41'rlines offorce coming from every unit charge. Since the gravitational ﬁeld ofanumber ofmasses isthesum oftheir individual ﬁelds, wehave, forasurface Ssurrounding asetofmasses m,-: I=ffn-g as=-Zam,-0. (es-27) S i. Foracontinuous distribution ofmass within S,thisequation becomes Us-gas =-fffieopev. (ezs) S ‘V 264 GRAVITATION [crma 6 Wenow apply Gauss’ divergence theorem [Eq. (3—115)] totheleftsideof thisequation:ffn-gds=[[[v-gdv. (e29) s ,v Subtracting Eq.(6-28) from Eq.(6-29), wearrive attheresult H/(v-g +410;»)av=0. (6-so) V Now Eq.(6-30) must hold foranyvolume V,andthiscanonly betrue if theintegrand vanishes: V-g=-41rGp. (6-31) This equation incartesian coordinates hastheform 6 8 85%;++ai;=-41rGp<x, 1/,z>. <6-32> When p(x,y,z)isgiven, thesetofequations (6-22) and (6-32) canbe shown todetermine thegravitational ﬁeld (g,,,gy,g,)uniquely, ifweadd theboundary condition thatg—>0asIr]——>oo.Substituting from Eq. (6-16), wegetanequation satiﬁed bythepotential: V29 =-41rGp, (6-33) or a2 a2 a”5%+$5+5;;=—41rGp. (6-34) This single equation determines 9(x,y,z)uniquely ifweaddthecondition that <3—>0as|r{—>oo.This result wequote from potential theory with- outproof. The solution ofEq.(6-33) is,infact, Eq.(6-15). Itisoften easier tosolve thepartial differential equation (6-34) directly than tocom- pute theintegral inEq.(6-15). Equations (6-33), (6-16), and(6-8) to- gether constitute acomplete summary ofNewton’s theory ofgravitation, aslikewise doEqs. (6-31), (6-21), and(6-8); that is,alltheresults ofthe theory canbederived from either ofthese setsofequations. Equation (6-33) iscalled P0iss0n’s equation. Equations ofthisform turn upfrequently inphysical theories. Forexample, theelectrostatic potential satisﬁes anequation ofthesame form, where pistheelectric charge density. Ifp=0,Eq.(6-33) takes theform v29=0. (6-35) This iscalled Laplace’s equation. Anextensive mathematical theory of PROBLEMS 265 Eqs. (6-33) and (6-35) hasbeen developed.* Adiscussion ofpotential theory is,however, outside thescope ofthistext. PROBLEMS 1.(a)Given Newton’s laws ofmotion, andKepler’s ﬁrsttwolaws ofplanetary motion (Section 3-15), show thattheforce acting onaplanet isdirected toward thesunandisinversely proportional tothesquare ofthedistance from thesun. (b)UseKepler's third lawtoshow that theforces ontheplanets arepropor- tional totheir masses. (c)Ifthissuggests toyouauniversal lawofattraction between anytwomasses, useNewton’s third lawtoshow thattheforce must be proportional toboth masses. 2.(a)Find thegravitational ﬁeld andgravitational potential atanypoint z onthesymmetry axis ofauniform solid hemisphere ofradius a,mass M.The center ofthehemisphere isatz=0.(b)Locate thecenter ofgravity ofthe hemisphere relative toapoint outside itonthez-axis, and show that as z—->:|=w, thecenter ofgravity approaches thecenter ofmass. 3.Assuming that theearth isasphere ofuniform density, with radius a, mass M,calculate thegravitational ﬁeldintensity andthegravitational potential atallpoints inside andoutside theearth, taking 9=0ataninﬁnite distance. 4.Assuming that theinterior oftheearth canbetreated asanincompressible ﬂuid inequilibrium, (a)calculate thepressure within theearth asafunction of distance from thecenter. (b)Using appropriate values fortheearth’s mass and radius, calculate thepressure intons persquare inch atthecenter. 5.Show that ifthesunwere surrounded byaspherical cloud ofdust ofuniform density p,thegravitational ﬁeldwithin thedust cloud would be MG 41r r E= -'(7T+?PG"');I where Misthemass ofthesun, andrisavector from thesuntoanypoint inthe dust cloud. 6.Assume thatthedensity ofastarisafunction onlyoftheradius rmeasured from thecenter ofthestar, andisgiven by‘I’ _ Ma2 P_2m'(r2—|— a2)2 ’ where Misthemass ofthestar, andaisaconstant which determines thesizeof thestar. Find thegravitational ﬁeld intensity andthegravitational potential as functions ofr. *O.D.Kellogg, Foundations ofPotential Theory. Berlin: J.Springer, 1929. 1'Theexpression forpischosen tomake theproblem easy tosolve, notbecause ithasmore than aremote resemblance tothedensity variations within anyactual star. 266 GRAVITATION [CHAP- 6 7.Setuptheequations tobesolved forthepressure asafunction ofradius inaspherically symmetric mass Mofgas,assuming that thegasobeys theperfect gaslaws andthat thetemperature isknown asafunction ofradius. 8.(a)Assume that ordinary cold matter collapses, under apressure greater than acertain critical pressure P0,toastate ofvery high density p1.Aplanet of mass Misconstructed ofmatter ofmean density p0initsnormal state. Assuming uniform density andconditions ofﬂuid equilibrium, atwhat mass M0andradius Towillthepressure atthecenter reach thecritical value pg? (b)IfM>M0,the planet willhave avery dense coreofdensity p1surrounded byacrust ofdensity pg.Calculate theresulting pressure distribution within theplanet interms ofthe radius 1'1ofthecore andtheradius 7'2oftheplanet. Show that ifMissomewhat larger than M0,then theradius r2oftheplanet islessthan T0.(The planet Jupiter issaidtohave amass very nearly equal tothecritical mass M0,sothat ifitwere heavier itmight besmaller.) 9.Find thepressure and temperature asfunctions ofradius forthestar of Problem 6ifthestariscomposed ofaperfect gasofatomic weight A. 10.Find thedensity andgravitational ﬁeld intensity asafunction ofradius inside asmall spherically symmetric planet, toorder (1/B2), assuming that the bulk modulus Bisconstant. Themass isMandtheradius isa.[Hint: Calculate g(r) assuming uniform density; then ﬁnd theresulting pressure p(r), and the density p(r)toorder (1/B). Recalculate g(r)using thenewp(r), andproceed bysuccessive approximations toterms oforder (1/B2).] 11.Consider aspherical mountain ofradius a,mass M,ﬂoating inequilibrium intheearth, andwhose density ishalfthatoftheearth. Assume thataismuch lessthan theearth’s radius, sothattheearth’s surface canberegarded asﬂatin theneighborhood ofthemountain. Ifthemountain were notpresent, thegravi- tational ﬁeldintensity neartheearth’s surface would bego.(a)Find thediﬁer- ence between goandtheactual value ofgatthetopofthemountain. (b)Ifthe topofthemountain iseroded ﬁat,level with thesurrounding surface oftheearth, andifthisoccurs inashort time compared with thetime required forthemoun- tain toﬂoat inequilibrium again, ﬁnd thedifference between goandtheactual value ofgattheearth’s surface atthecenter oftheeroded mountain. 12.(a)Find thegravitational potential andtheﬁeld intensity duetoathin rodoflength landmass Matapoint adistance rfrom thecenter oftherodina direction making anangle 0with therod. Assume that 1'>>l,and carry the calculations only tosecond order inl/r. (b)Locate thecenter ofgravity ofthe rodrelative tothespeciﬁed point. 13.(a)Calculate thegravitational potential ofauniform circular ring of matter ofradius a,mass M,atadistance rfrom thecenter oftheringinadirec- tion making anangle 0with theaxisofthering. Assume that r>>a,andcal- culate thepotential only tosecond order ina/r. (b)Calculate tothesame ap- proximation thecomponents ofthegravitational ﬁeld oftheringatthespeciﬁed point. 14.Asmall body with cylindrical symmetry hasadensity p(r,6)inspherical coordinates, which vanishes forr>a.The origin r=0liesatthecenter of mass. Approximate thegravitational potential atapoint r,0farfrom thebody PROBLEMS 267 (r>>a),byexpanding inapower series in(a/1), andshow that ithastheform cw)=§+%P2<@<>sv>+§§1>3<cosv>+---, where P2(cos 0),P3(cos0)arequadratic and cubic polynomials incos0that donotdepend onthebody, andQ,Eareconstants which depend onthemass distribution. Find expressions forP2,P3,Q,andE,andshow that Qisofthe order ofmagnitude Ma2,andEoftheorder Ma3.Itisconventional tonormalize P2sothat theconstant term is-—%, andP3sothat thelinear term is—%cos0. The parameters Q,Earethen called thequadrupole moment and theoctopole moment ofthebody. Thepolynomials P2,P3,...aretheLegendre polynomials. 15.Theearth hasapproximately theshape ofanoblate ellipsoid ofrevolution whose polar diameter 2a(1 ——11)isslightly shorter than itsequatorial diameter 2a. (11=0.0034.) Todetermine toﬁrst order in17,theeffect oftheearth’s oblateness onitsgravitational ﬁeld, wemay replace theellipsoidal earth bya sphere ofradius Rsochosen astohave thesame volume. The gravitational ﬁeld oftheearth isthen theﬁeld ofauniform sphere ofradius Rwith themass oftheearth, plus theﬁeld ofasurface distribution ofmass (positive ornegative), representing themass perunit area which would beadded orsubtracted toform theactual ellipsoid. (a)Show that therequired surface density is,toﬁrst order in17, <1=-Q-|7ap(1 ——3cosz 0), where J0isthecolatitude, and pisthevolume density oftheearth (asumed uniform). Since thetotal mass thusadded tothesurface iszero, itsgravitational ﬁeld willrepresent theeffect oftheoblate shape oftheearth. (b)Show that theresulting correction tothegravitational potential atavery great distance r>>afrom theearth is,toorder (a3/r3), 1MG259-51;-7§‘i(1 -300320), (r>>a). 16.(a)UseGauss’ theorem (6-26) todetermine thegravitational ﬁeld inside and outside aspherical shell ofradius a,mass M,uniform density. (b)Cal- culate theresulting gravitational potential. 17.(a)Find thegravitational ﬁeld atadistance acfrom aninﬁnite plane sheet ofdensity :1perunit area. (b)Compare this result with theﬁeld just outside aspherical shell ofthesame surface density. What part oftheﬁeld comes from theimmediately adjacent matter andwhat part from more distant matter? 18.Show that thegravitational ﬁeld equations (6-21), (6-31), and (6-33) aresatisﬁed bytheﬁeld intensity and potential which you calculated inProb- lem3. *19. (a)Show that 59found inProblem 15(b) satisﬁes Laplace’s equation (6-35). This, together with thefactthat 69hasthesame angular dependence 268 GRAVITATION [CHAP. 6 asthemass density which produces it,suggests that theformula given for89 mayactually bevalid everywhere outside theearth. (b)Toshow this, consider Poisson’s equation (6-33) with p=f(r)(1 —3cos20).Show that asolution 9=h(r)(1 —3cosz0)willsatisfy Eq.(6-33) with thisform ofp,provided (121. 2dh 6ha§+;t"w-"“"'Gf- i (c)Show that h=r“3satisﬁes thisequation intheregion where f=0.Can you complete theproof that theformula for5Qfound inProblem 15(b) isin factvalid everywhere outside theearth? CHAPTER 7 MOVING COORDINATE SYSTEMS 7-1Moving origin ofcoordinates. Letapoint inspace belocated by vectors r,r*with respect totwoorigins ofcoordinates 0,0*,andlet0*be located byavector hwith respect to 0(Fig. 7-1). Then therelation be- tween thecoordinates rand r*is given by 1,, r=r*+h, (7-1) 1' r*=r—h. (7-2) .0! Interms ofrectangular coordinates, with axes as*,y*,z*parallel toaxes :0,y,z,respectively, these equations canbewritten: w=w*+h», 2/=y*+h... z=z*+h=; (7-3) a:*=a:-h,, y*=y—h,,, z*=z—h,. (7-4)h 0 Fro. 7-1. Change oforigin ofcoordi- nates. ,Now iftheorigin 0*ismoving with respect totheorigin 0,which we regard asﬁxed, therelation between thevelocities relative tothetwosys- tems isobtained bydifferentiating Eq.(7-1): dr dr* dhA"=a—7zr+a =v*+v;,, (7-5) where vandv*arethevelocities ofthemoving point relative toOand0*, andv;,isthevelocity of0*relative to0.Wearesupposing that the axes :c*,y*,z*remain parallel tox,y,z.This iscalled atranslation ofthe starred coordinate system with respect totheunstarred system. Written outincartesian components, Eq. (7-5) becomes thetime derivative of Eq.(7-3). Therelation between relative accelerations is a_d_2r_d2r*_|_d2h _dtz—dt2 dt2 =8*+811- (7-5) Again these equations caneasily bewritten outinterms oftheir rectangu- larcomponents. 269 270 MOVING coonnrnxrn SYSTEMS [CHAIM 7 Newton’s equations ofmotion hold intheﬁxed coordinate system, so that wehave, foraparticle ofmass msubject toaforce F: d2mJ;=r. (7-7) Using Eq.(7-6), wecanwrite thisequation inthestarred coordinate sys- tem: 2 m% +ma), =F. (7-8) If0*ismoving atconstant velocity relative toO,then ah=0,andwe have d2r* Thus Newton's equations ofmotion, ifthey hold inanycoordinate system, hold also inanyother coordinate system moving with uniform velocity relative totheﬁrst. This istheNewtonian principle ofrelativity. It implies that, sofarasmechanics isconcerned, Wecannot specify any unique ﬁxed coordinate system orframe ofreference towhich Newton’s laws aresupposed torefer; ifwespecify onesuch system, anyother system moving with constant velocity relative toitwilldoaswell. This property ofEq.(7-7) issometimes expressed bysaying that Newton’s equations of motion remain invariant inform, orthat they arecovariant, with respect to uniform translations ofthecoordinates. Theconcept offrame ofreference isnotquite thesame asthat ofacoordinate system, inthatif wemake a change ofcoordinates that does notinvolve thetime, wedonotregard this asachange offrame ofreference. Aframe ofreference includes allcoordi- nate systems atrest with respect toanyparticular one. The principle ofrelativity proposed byEinstein asserts that therelativity principle isnot restricted tomechanics, butholds forallphysical phenomena. Thespecial theory ofrelativity istheresult oftheapplication ofthisprinciple toall types ofphenomena, particularly electromagnetic phenomena. Itturns outthat thiscanonly bedone bymodifying Newton’s equations ofmotion slightly and, infact, even Eqs. (7-5) and(7-6) require modiﬁcation)‘ Foranymotion of0*,wecanwrite Eq.(7-8) intheform 2 m =F—~mah. (7-10) This equation hasthesame form astheequation ofmotion (7-7) inaﬁxed coordinate system, except that inplace oftheforce F,wehave F-—mah. TP.G.Bergmann, Introduction totheTheory ofRelativity. New York: Prentice- Hall, 1946. (Part 1.) 7-2] ROTATING COORDINATE SYSTEMS 271 The term —ma;, wemay callaﬁctitious force. Wecantreat themotion ofamass mrelative toamoving coordinate system using Newton’s equa- tions ofmotion ifweaddthisﬁctitious force totheactual force which acts. From thepoint ofview ofclassical mechanics, itisnotaforce atall,but part ofthemass times acceleration transposed totheother side ofthe equation. The essential distinction isthat therealforces Facting onm depend onthepositions andmotions ofother bodies, whereas theﬁctitious force depends ontheacceleration ofthestarred coordinate system with respect totheﬁxed coordinate system. Inthegeneral theory ofrelativity, terms like—ma;, areregarded aslegitimate forces inthestarred coordi— nate system, onthesame footing with theforce F,sothat inallcoordinate systems thesame lawofmotion holds. This, ofcourse, canonly bedone ifitcanbeshown how todeduce theforce —ma;, from thepositions and motions ofother bodies. The program isnotsosimple asitmay seem from thisbrief outline, andmodiﬁcations inthelaws ofmotion arerequired tocarry itthrough.T 7-2Rotating coordinate systems. Wenow consider coordinate systems ac,y,2and20*,y*,2*whose axes arerotated relative tooneanother asin Fig. 7—2, Where, forthepresent, theorigins ofthetwosetsofaxes coin- cide. Introducing unit vectors i,j,kassociated with axes x,y,z,and unit vectors i*,j*,k*associated with axes as*,11*,2*,wecanexpress the position vector 1'interms ofitscomponents along either setofaxes: r=xi+yj—l—zk, (7-11) r=:z:*i* -|—y*j* +z*k*. (7-12) Note that since theorigins now coincide, apoint isrepresented bythe same vector rinboth systems; only thecomponents ofraredifferent along thediﬁerent axes. The relations between thecoordinate systems zz* k k* . 5* J 1 |1- ya- U l’ Q7 $18 FIG. 7-2. Rotation ofcoordinate axes. ’rBergmann, op.cit.(Part 2.) 272 MOVING ooonnmxrn svsrnms icnxr. 7 canbeobtained bytaking thedotproduct ofeither thestarred orthe unstarred unit vectors with Eqs. (7-11) and (7-12). Forexample, ifwe compute i-r,j-r,k-r,from Eqs. (7-11) and(7-12) andequate theresults, weobtain w=w*(i*-i) +y*(i*'i) +Z*(k*'i), ' y==v*(i*'i) +1/*(i*'i) +z*(k*'i), (7-13) z=x*(i*-k) —|—y*(j*-k) +z*(k*-k). The dotproducts (i*-i), etc., arethecosines oftheangles between the corresponding axes. Similar formulas forx*,y*,2*interms ofx,y,zcan easily beobtained bythesame process. These formulas arerather compli- cated andunwieldy, andweshall fortunately beable toavoid using them inmost cases. Equations (7-11), (7-12), and(7-13) donotdepend onthe factthat thevector 1'isdrawn from theorigin. Analogous formulas apply interms ofthecomponents ofanyvector Aalong thetwosetsofaxes. Thetime derivative ofanyvector Awasdeﬁned byEq.(3-52): @_. A(t—|-At)-—A(t)_ _ dtTAllI—l}0 At (714) Inattempting toapply thisdeﬁnition inthepresent case, weencomiter adiﬂiculty ifthecoordinate systems arerotating with respect toeach other. Avector which isconstant inonecoordinate system isnotcon- stant intheother, butrotates. Thedeﬁnition requires ustosubtract A(t)from A(t+At). During thetime At,coordinate system :c*,y*,2*has rotated relative toac,y,z,sothat attime t+At,thetwosystems willnot agree astowhich vector is(orwas) A(t), i.e.,which vector isinthesame position that Awasinattime t.Theresult isthat thetime derivative of agiven vector willbedifferent inthetwocoordinate systems. Letususe d/dt todenote thetime derivative with respect totheunstarred coordinate system, which weregard asﬁxed, andd*/dt todenote thetime derivative with respect totherotating starred coordinate system. Wemake this distinction with regard tovectors only; there isnoambiguity with regard tonumerical quantities, andwedenote their time derivatives byd/dt, or byadot, which willhave thesame meaning inallcoordinate systems. Letthevector Abegiven by A=A,i+A,,j+A,k, (7-15) A==A§i* +A§,"j* +AZ‘k*. (7-16) The unstarred time derivative ofAmay beobtained bydifferentiating Eq.(7-15), regarding i,j,kasconstant vectors intheﬁxed system: %=A;+A.,i+A.k. (7-17) 7-2] ROTATING COORDINATE SYSTEMS 273 Similarly, thestarred derivative ofAisgiven interms ofitsstarred com- ponents by d*A '*-* '*.* ' * 'dT =A11 +Av] -I‘ . Wemay regard Eqs. (7-17) and(7-18) asthedeﬁnitions ofunstarred and starred time derivatives ofavector. Wecanalso obtain aformula for d/dt instarred components bytaking theunstarred derivative ofEq. (7-16), remembering that theunit vectors i*,j*,k*aremoving relative totheunstarred system, andhave time derivatives: '* '* *‘j,—',‘=A:i*+A’;i*+A:1<*+A";%-+113%+A:%~ <1-19> Asimilar formula could beobtained ford*A/dt interms ofitsunstarred components. Letusnowsuppose thatthestarred coordinate system isrotating about some axis OQthrough theorigin, with anangular velocity w(Fig. 7-3). Wedeﬁne thevector angular velocity wasavector ofmagnitude wdirected along theaxisOQinthedirection ofadvance ofaright-hand screw rotat- ingwith thestarred system. Consider avector Batrestinthestarred system. Itsstarred derivative iszero, andwenowshow thatitsunstarred derivative is “ET?=or><B. <1-20) Inorder tosubtract B(t) from B(t+At),wedraw these vectors with their tails together, and itwillbeconvenient toplace them with their tails ontheaxisofrotation. (The time derivative depends only onthecom- ponents ofBalong theaxes, andnotontheposition ofBinspace.) We Q co Bsin0,3 3(1) B(t+Al) U O Fro. 7-3. Time derivative ofarotating vector. 274 MOVING COORDINATE SYSTEMS lcnar. 7 ﬁrst verify from Fig. 7-3that thedirection ofdB/dt isgiven correctly by Eq.(7-20), recalling thedeﬁnition [Eq. (3-24) andFig.3-11] ofthecross product. Themagnitude ofdB/dt asgiven byEq.(7-20) is \%\=Ia;><Bl=wBsin0. (7-21) This isthecorrect formula, since itcanbeseen from Fig. 7-3that, when Atissmall, IABI =(Bsin0)(coAt). When Eq.(7-20) isapplied totheunit vectors i*,j*,k*,Eq.(7-19) be- comes, ifwemake useofEqs. (7-18) and(7-16): %=% +A1‘(w Xi*)+A’§(w Xj*)—|—A’§(w Xk*) d*A=W -l—wXA. (7-22) Thisisthefundamental relationship between timederivatives forrotating coordinate systems. Itmay beremembered bynoting that thetime de- rivative ofanyvector intheunstarred coordinate system isitsderivative inthestarred system plus theunstarred derivative itwould have ifit were atrestinthestarred system. Equation (7-22) applies even when theangular velocity vector wischanging inmagnitude anddirection with time. Taking thederivative ofright and leftsides ofEq. (7-22), and applying Eq.(7-22) again toAandd*A/dt, wehave forthesecond time derivative ofanyvector A: d2A_d<d*A> an11...W_E 7?+“"sr+n"A d*2A d*A d*A d =—d-Z5—+w><W+wX<-J-t-—+wXA)+E‘%’XA d*2A d*A d =-W-+2wX—t-i-',—+wX(wXA)+7‘;’XA. Inview ofEq.(3-29), thestarred andunstarred derivatives ofanyvector parallel totheaxisofrotation arethesame, according toEq.(7-22). In particular, dc»_d*w H_'5' 7-2] ROTATING COORDINATE SYSTEMS 275 Itistobenoted that thevector coonboth sides ofthisequation isthe angular velocity ofthestarred system relative totheunstarred system, although itstime derivative iscalculated with respect totheunstarred system ontheleftside, and with respect tothestarred system onthe right. Theangular velocity oftheunstarred system relative tothestarred system willbe—w. Wenow show that therelations derived above forarotating coordinate system areperfectly general, inthat they apply toany motion ofthe starred axes relative totheunstarred axes. Lettheunstarred rates of change ofthestarred unit vectors begiven interms ofcomponents along thestarred axes by di* W E dt * %=aa1i* -1-0321* +<Isak*-=a11i* +a12i* +¢l13k*» =a21i* "l"0221* +<l2ak*, (7-24) Bydifferentiating theequation i*-i* =1, (7-25) weobtain -=0:‘gt-i*=0. (7-26) From thisandthecorresponding equations forj*andk*,wehave an = (J/22 = 0,33 = Bydifferentiating theequation i*-k* =0, (7-28) weobtain di* .dk* dt 2—dt 0'29) From thisandtheother twoanalogous equations, wehave a31=—a1s, 1112=-1121, 1123="-as2- (7-30) Letavector wbedeﬁned interms ofitsstarred components by: wt=1123, NZ=llai, 09’:=1112- (7‘31)1 i l i 276 MOVING COORDINATE SYSTEMS [CHAP- 7 Equations (7-24) cannow berewritten, with thehelp ofEqs. (7-27), (7-30), and(7-31), intheform di*_ .., 'E—(|)Xl, %=coXj*, (7-32) * % =avXk*. According toEq.(7-20), these time derivatives ofi*,j*,k*arejustthose tobeexpected ifthestarred lmit vectors arerotating with anangular velocity w.Thus nomatter how thestarred coordinate axes may be moving, wecandeﬁne atanyinstant anangular velocity vector w,given byEq.(7-31), such that thetime derivatives ofanyvector relative tothe starred andunstarred coordinate systems arerelated byEqs. (7-22) and (7-23). Letusnowsuppose thatthestarred coordinate system ismoving sothat itsorigin 0*remains ﬁxed attheorigin Ooftheﬁxed coordinate system. Then anypoint inspace islocated bythesame position vector rinboth coordinate systems [Eqs. (7-11) and(7-12)]. Byapplying Eqs. (7-22) and(7-23) totheposition vector 1',weobtain formulas fortherelation between velocities andaccelerations inthetwocoordinate systems: d d*1' i=3-l-NXI, d2r d*2r d*r dwW:-1-2-itT+a>X(wXr)—}-2uXW+ZZ-XI. (7-34) Formula (7-34) iscalled Coriolis’ theorem. The ﬁrst term ontheright is theacceleration relative tothestarred system. The second term iscalled thecentripetal acceleration ofapoint inrotation about anaxis (centripetal means “toward thecenter”). Using thenotation inFig. 7-4, wereadily verify that wX(wX1')points directly toward andperpendicular tothe axisofrotation, andthat itsmagnitude is |wX(wXr)|=wzrsinﬂ U2 =Y0 , (7-35) where v=wrsin0isthespeed ofcircular motion and(rsin0)isthedis- tance from theaxis. The third term ispresent only when thepoint ris moving inthestarred system, andiscalled thecoriolis acceleration. The 7-2] ROTATING oooanrrurrn SYSTEMS 277 WXT l’ la) T FIG. 7-4. Centripetal acceleration. last term vanishes foraconstant angular velocity ofrotation about a ﬁxed axis. Ifwesuppose that Newton’s lawofmotion (7-7) holds intheunstarred coordinate system, weshall have inthestarred system: d*2r d*r dwmT1l,7+'rnwX(wXr)+2mmX—Jt—+mEXr-F. (7-36) Transposing thesecond, third, andfourth terms totheright side, weob- tainanequation ofmotion similar inform toNewton’s equation ofmotion: 2 m%;=F—mwx(wxr)—2nwx%—m%?xr. (7-37) The second term ontheright iscalled thecentrifugal force (centrifugal means “away from thecenter ”);thethird term iscalled thecoriolis force. Thelastterm hasnospecial name, andappears only forthecase ofnon- uniform rotation. Ifweintroduce theﬁctitious centrifugal and coriolis forces, thelaws ofmotion relative toarotating coordinate system arethe same asforﬁxed coordinates. Agreat deal ofconfusion hasarisen regard- ingtheterm “centrifugal force.” This force isnotarealforce, atleast inclassical mechanics, andisnotpresent ifwerefer toaﬁxed coordinate system inspace. Wecan,however, treat arotating coordinate system as ifitwere ﬁxed byintroducing thecentrifugal andcoriolis forces. Thus a particle moving inacircle hasnocentrifugal force acting onit,butonly a force toward thecenter which produces itscentripetal acceleration. How- ever, ifweconsider acoordinate system rotating with theparticle, inthis system theparticle isatrest, andtheforce toward thecenter isbalanced bythecentrifugal force. Itisvery often useful toadopt arotating coordi- nate system. Instudying theaction ofa.cream separator, forexample, it isfarmore convenient tochoose acoordinate system inwhich theliquid 278 MOVING COORDINATE SYSTEMS ICHAP. 7 isatrest, andusethelaws ofdiffusion tostudy thediffusion ofcream toward theaxisunder theaction ofthecentrifugal force ﬁeld, than totry tostudy themotion from thepoint ofview ofaﬁxed observer watching thewhirling liquid. ‘ Wecantreat coordinate systems insimultaneous translation androta- tionrelative toeach other byusing Eq.(7-1) torepresent therelation be- tween thecoordinate vectors randr*relative toorigins O,0*notneces- sarily coincident. Inthederivation ofEqs. (7-32), noassumption was made about theorigin ofthestarred coordinates, andtherefore Eqs. (7-22) and(7-23) may stillbeused toexpress thetime derivatives ofanyvector with respect totheunstarred coordinate system interms ofitstime de- rivatives with respect tothestarred system. Replacing dr*/dt, d2r*/dt in Eqs. (7-5) and(7-6) bytheir expressions interms ofthestarred deriva- tives relative tothestarred system asgiven byEqs. (7-33) and (7-34), weobtain fortheposition, velocity, andacceleration ofapoint with re- spect tocoordinate systems inrelative translation androtation: 1'=1'*—l—h, (7-38) dr d*r* dh a=T+QXI*+W1 dz: d*2r* , d*r* d @1211 %=-—d—t-2-—|—wX(wX1’*)+2wX-K-1-—(§Xr*+—&F* 7-3Laws ofmotion ontherotating earth. Wewrite theequation of motion, relative toacoordinate system ﬁxed inspace, foraparticle of mass msubject toagravitational force mgandanyother nongravitational forces F: d2 =F+mg. (7-41) Now ifwerefer themotion oftheparticle toacoordinate system atrest relative totheearth, which rotates with constant angular velocity w,and ifwemeasure theposition vector rfrom thecenter oftheearth, wehave, byEq.(7-34): dz: 2=m%+mwX (..,><r)+2m...><%‘, (7-42) which canberearranged intheform d*2r d*r m—gt7=F—|—m[g—wX(wXr)]—2mwX—ﬁ-- 7-3] LAWS orMOTION ONTHE ROTATING EARTH 279 This equation hasthesame form asNewton’s equation ofmotion. We have combined thegravitational andcentrifugal force terms because both areproportional tothemass oftheparticle andboth depend only onthe position oftheparticle ;intheir mechanical effects these twoforces are indistinguishable. Wemay deﬁne theeffective gravitational acceleration geatanypoint ontheearth’s surface by: g,,(r) =g(r) —wX(wXr). (7-44) The gravitational force which wemeasure experimentally onabody of mass matrest’[ ontheearth’s surface ismg.,. Since —wX(wXr)points radially outward from theearth’s axis, g,atevery point north ofthe equator willpoint slightly tothesouth oftheearth’s center, ascanbe seen from Fig. 7-5. Abody released near theearth’s surface willbegin tofallinthedirection ofgs,thedirection determined byaplumb lineis that ofge,andaliquid willcome toequilibrium with itssurface perpen- dicular toge.This iswhy theearth hassettled into equilibrium inthe form ofanoblate ellipsoid, ﬂattened atthepoles. Thedegree ofﬂattening isjustsuch astomake theearth’s surface atevery point perpendicular to g,(ignoring local irregularities). Equation (7-43) cannow bewritten V d*”r d*r mW—F+mg¢—2mwXW- Thevelocity andacceleration which appear inthisequation areunaffected ifwerelocate ourorigin ofcoordinates atanyconvenient point atthesur- face oftheearth; hence thisequation applies tothemotion ofaparticle of mass matthesurface oftheearth relative toalocal coordinate system at restontheearth’s surface. The only unfamiliar term isthecoriolis force —mx(wx r) Sf; ge FIG. 7-5. Effective acceleration ofgravity ontherotating earth. TAbody inmotion issubject alsotothecoriolis force. 280 MOVING ooonnrrwrs SYSTEMS [cmun 7 which acts onamoving particle. The reader canconvince himself bya fewcalculations that thisforce iscomparatively small atordinary veloci- tiesd*r/dt. Itwillbeinstructive totryworking outthedirection ofthe coriolis force forvarious directions ofmotion atvarious places onthe earth’s surface. The coriolis force isofmajor importance inthemotion oflarge airmasses, andisresponsible forthefact that inthenorthern hemisphere tornados andcyclones circle inthedirection south toeast to north towest. Inthenorthern hemisphere, thecoriolis force acts tode- ﬂect amoving object toward theright. Asthewinds blow toward alow pressure area, they aredeﬂected totheright, sothat they circle thelow pressure area inacounterclockwise direction. Anairmass circling inthis way willhave alowpressure onitsleft,andahigher pressure onitsright. This isjust what isneeded tobalance thecoriolis force urging ittothe right. Anairmass canmove steadily inonedirection only ifthere isa high pressure totheright ofittobalance thecoriolis force. Conversely, apressure gradient over thesurface oftheearth tends todevelop winds moving atright angles toit.Theprevailing westerly winds inthenorthern temperate zone indicate that theatmospheric pressure toward theequator isgreater than toward thepoles, atleast near theearth’s surface. The easterly trade winds intheequatorial zone areduetothefactthat anyair mass moving toward theequator willacquire avelocity toward thewest duetothecoriolis force acting onit.Thetrade winds aremaintained by highpressure areas oneither sideoftheequatorial zone. 7-4TheFoucault pendulum. Aninteresting application ofthetheory ofrotating coordinate systems istheproblem oftheFoucault pendulum. TheFoucault pendulum hasabobhanging from astring arranged toswing freely inanyvertical plane. Thependulum isstarted swinging inadeﬁ- nitevertical plane anditisobserved that theplane ofswinging gradually precesses about thevertical axis during aperiod ofseveral hours. The bobmust bemade heavy, thestring very long, andthesupport nearly frictionless, inorder that thependulum cancontinue toswing freely for long periods oftime. Ifwechoose theorigin ofcoordinates directly below thepoint ofsupport, atthepoint ofequilibrium ofthependulum bobof mass m,then thevector rwillbenearly horizontal, forsmall amplitudes of oscillation ofthependulum. Inthenorthern hemisphere, wpoints inthe general direction indicated inFig.7-6, relative tothevertical. Writing -r forthetension inthestring, wehave astheequation ofmotion ofthebob, according toEq.(7-45): - d*2 d* m#=r+mg¢—2mwX?,r- (7-46) Ifthecoriolis force were notpresent, thiswould betheequation fora 7-4] THE FOUCAULT PENDULUM 281 simple pendulum onanonrotating earth. The coriolis force isvery small, lessthan 0.1% ofthegravita- tional force ifthevelocity is5mi/hr orless,anditsvertical component is therefore negligible incomparison with thegravitational force. (Itis thevertical force which determines themagnitude ofthetension inthe 0, k' string.) However, thehorizontal 0 m component ofthecoriolis force is r perpendicular tothevelocity d*r/dt, mgl andasthere arenoother forces in this direction when thependulum FIu- 7'6- TheFoucault Pendulum- swings toandfro,itcanchange the nature ofthemotion. Any force with ahorizontal component perpendic- ular tod*r/dt willmake itimpossible forthependulum tocontinue to swing inaﬁxed vertical plane. Inorder tosolve theproblem including thecoriolis term, weusetheexperimental result asaclue, andtrytoﬁnd anew coordinate system rotating about thevertical axis through the point ofsupport atsuch anangular velocity that inthis system the coriolis terms, oratleast their horizontal components, aremissing. Let usintroduce anewcoordinate system rotating about thevertical axis with constant angular velocity k9,where kisavertical unit vector. We shall callthisprecessing coordinate system theprimed coordinate system, and denote thetime derivative with respect tothis system byd’/dt. Then weshall have, byEqs. (7-33) and(7-34):' ‘I’ an4'W=Et£+(lkXr, (7-47) d*2r .1”: d'rE2-=7t;+S22kx(kxr)+2£lkXa- (7-48) Equation (7-46) becomes 2 m-(%1-,t?1’= r+mg,-—-2mwX(%+QkXI') i 2 dlr—mQkX(kXr)—2mQkX-1? =-r+mg,—2m£2uX(kXr)—mS22kX(kXr) d'r—-2m(w + XE ' 282 MOVING COORDINATE SYSTEMS [cmu>. 7 Weexpand thetriple products bymeans ofEq.(3-35): d'2rmg}; =-r+mge —'m(2Slw-I“—|— 822k-r)k .dl+'m.(2f2k-w +S22)r-2m(w+kc)><d;-(7-50) Every vector ontheright sideofEq.(7-50) liesinthevertical plane con- taining thependulum, except thelastterm. Since, forsmall oscillations, d'r/dt ispractically horizontal, wecanmake thelastterm lieinthisvertical plane alsobymaking (w—|—kfl)horizontal._ Wetherefore require that k-(w +kQ)=0. (7-51) This determines S2: S2=—wcos0, (7-52) where wistheangular velocity oftherotating earth, Qistheangular velocity oftheprecessing coordinate system relative totheearth, and6is theangle between thevertical andtheearth’s axis, asindicated inFig.7-6. Thevertical isalong thedirection of——g,, andsince thisisvery nearly the same asthedirection of—g(seeFig.7-5), 0willbepractically equal tothe colatitude, thatis,theangle between randwinFig.7-5. Forsmall oscilla- tions, if£2isdetermined byEq.(7-52), thecross product inthelastterm ofEq.(7-50) isvertical. Since allterms ontheright ofEq.(7-50) now lieinavertical plane containing thependulum, theacceleration d’2r/dtz ofthebobintheprecessing system isalways toward thevertical axis, and ifthependulum isinitially swinging toandfro,itwillcontinue toswing toandfrointhesame vertical plane intheprecessing coordinate system. Relative totheearth, theplane ofthemotion precesses with angular velocity $2ofmagnitude andsense given byEq.(7-52). Inthenorthern hemisphere, theprecession isclockwise looking down. Since thelastthree terms ontheright inEq.(7-50) aremuch smaller than theﬁrst two, theactual motion intheprecessing coordinate system ispractically thesame asforapendulum onanonrotating earth. Even at large amplitudes, where thevelocity d’r/dt hasavertical component, care- fulstudy willshow that thelastterm inEq. (7-50), when Qischosen according toEq. (7-52), does notcause any additional precession relative totheprecessing coordinate system, butmerely causes thebobtoswing in anarcwhich passes slightly east ofthevertical through thepoint ofsup- port. Attheequator, S1iszero, andtheFoucault pendulum does notpre- cess; bythinking about itamoment, perhaps youcanseephysically why thisisso.Atthenorth orsouth pole, S2==l=w,andthependulum merely swings inaﬁxed vertical plane inspace while theearth turns beneath it. 7-5] LA1vroR’s THEOREM 283 Note that wehave been able togive afairly complete discussion ofthe Foucault pendulum, byusing Coriolis’ theorem twice, without actually solving theequations ofmotion atall. 7-5Larmor’s theorem. Thecoriolis force inEq.(7-37) isofthesame form asthemagnetic force acting onacharged particle (Eq. 3-281), in that both aregiven bythecross product ofthevelocity oftheparticle with avector representing aforce ﬁeld. Indeed, inthegeneral theory of relativity, thecoriolis forces onaparticle inarotating system canbere- garded asduetotherelative motion ofother masses intheuniverse ina way somewhat analogous tothemagnetic force acting onacharged par- ticle which isduetotherelative motion ofother charges. The similarity inform ofthetwoforces suggests that theeffect ofamagnetic ﬁeld ona system ofcharged particles may becanceled byintroducing asuitable rotating" coordinate system. This idea leads toLarmor’s theorem, which westate ﬁrst, andthen prove: LARMoR’s THEOREM. Ifasystem ofcharged particles, allhaving thesame ratio q/mofcharge tomass, aeted anbytheir mutual (central) forces, and byacentral force toward acommon center, issubject inaddition toaweak uniform magnetic ﬁeld B,itspossible motions willbethesame asthemotions itcould perform without themagnetic ﬁeld, superposed upon aslow pre- cession oftheentire system about thecenter offorce with angular velocity 0,=-5!";B. (7-53) The deﬁnition ofaweak magnetic ﬁeld willappear astheproof isde- veloped. Weshall assume that alltheparticles have thesame charge q andthesame mass m,although itwillbeapparent that theonly thing that needs tobeassumed isthat theratio q/m isconstant. Practically the only important applications ofLarmor’s theorem aretothebehavior of anatom inamagnetic ﬁeld. The particles here areelectrons ofmass m, charge q=—e,acted upon bytheir mutual electrostatic repulsions and bytheelectrostatic attraction ofthenucleus. Letthecentral force acting ontheIcthparticle beFZ,andletthesumof theforces duetotheother particles be Then theequations ofmotion ofthesystem ofparticles, intheabsence ofamagnetic ﬁeld, are dzfk E imE=F;,+F;,, hi: 1,...,N, (7-54) where Nisthetotal number ofparticles. The force FZdepends only on thedistance ofparticlek from thecenter offorce, which weshall take as origin, andtheforces F},depend only onthedistances oftheparticles from 284 MOVING COORDINATE SYSTEMS [cn.u>. 7 oneanother. When themagnetic ﬁeldisapplied, theequations ofmotion become, byEq.(3—281): d2n._ 0¢qdrk _m—(F—Fk+Fk+EE'XB, k—1,...,N. Inorder toeliminate thelastterm, weintroduce astarred coordinate sys- temWith thesame origin, rotating about thisorigin with angular velocity w.Making useofEqs. (7-33) and(7-34), wecanwrite theequations of motion inthestarred coordinate system: *2 _ m%=F,§+F,1—-mwx(wxr,,)+%(wxr,,)xB d* B+%><(Q7+2'mw)- (7-56) Wecanmake thelastterm vanish bysetting ____L _w- 2mcB. (757) Equation (7-56) then becomes d*2"‘—F‘F‘ 92BB k—1 N 75s m-;itT— k+ It-I-11;; X( xrk); —,---» -(_) Theforces F};and depend only onthedistances oftheparticles from the origin andontheir distances from oneanother, andthese distances willbe thesame inthestarred andunstarred coordinate systems. Therefore, if weneglect thelast term, Eqs. (7—58) have exactly thesame form in terms ofstarred coordinates asEqs. (7-54) have inunstarred coordinates. Consequently, their solutions willthen bethesame, andthemotions of thesystem expressed instarred coordinates willbethesame asthemotions ofthesystem expressed inunstarred coordinates intheabsence ofamag- netic ﬁeld. This isLarmor’s theorem. Thecondition that themagnetic ﬁeld beweak means that thelastterm inEq.(7~58) must benegligible incomparison with theﬁrst twoterms. Notice thattheterm weareneglecting isproportional toB2,whereas the term inEq.(7-55) which wehave eliminated isproportional toB.Hence, forsuﬂiciently weak ﬁelds, theformer may benegligible even though the latter isnot. Thelastterm inEq.(7—58) may bewritten intheform 2$5Bx (Bxr;,)= 1moX(wX1';,). (1-59) Another way offormulating thecondition foraweak magnetic ﬁeld isto 7—6] THE nnsrrucrmn THREE-BODY PROBLEM 285 saythat theLarmor frequency w,given byEq.(7—57), must besmall com- pared with thefrequencies ofthemotion intheabsence ofamagnetic ﬁeld. The reader who hasunderstood clearly theabove derivation should be able toanswer thefollowing two questions. The cyclotron frequency, given byEq.(3——299), forthemotion ofacharged particle inamagnetic ﬁeld istwice theLarmor frequency, given byEq.(7-57). Why does not Larmor’s theorem apply tothecharged particles inacyclotron? Equa- tion (7—58) canbederived without anyassumption astotheorigin ofco- ordinates inthestarred system. Why isitnecessary that theaxisofrota- tion ofthestarred coordinate system pass through thecenter offorce of thesystem ofparticles‘? 7-6Therestricted three-body problem. Wepointed outinSection 4-9 that thethree-body problem, inwhich three masses move under their mutual gravitational forces, cannot besolved inanygeneral way. Inthis section wewillconsider asimpliﬁed problem, therestricted problem of three bodies, which retains many features ofthemore general problem, among them thefactthat there isnogeneral method ofsolving it.Inthe restricted problem, wearegiven twobodies ofmasses M1andM2that revolve incircles under their mutual gravitational attraction andaround their cormnon center ofmass. Thethird body ofverysmall mass mmoves inthegravitational ﬁeldofM1andM2. Wearetoassume thatmisso small thattheresulting disturbance ofthemotions ofM1andM2canbe neglected. Wewill further simplify theproblem byassuming that m remains intheplane inwhich M1andM2revolve. The problem thus reduces toaone-body problem inwhich wemust ﬁndthemotion ofmin thegiven (moving) gravitational ﬁeld oftheother two. Anobvious ex- ample would bearocket moving inthegravitational ﬁelds oftheearth and themoon, which revolve very nearly incircles about their common center ofmass. IfM1and M2areseparated byadistance a,then according tothe results ofSection 4—7,their angular velocity isdetermined byequating the gravitational force tomass times acceleration inthereduced problem, in which M1isatrestandM2hasmass ,4asgiven by"Eq. (4-98): we= (7-co) sothat 11,2=Q.li.tl _ (7_61) The center ofmass divides thedistance ainto segments that arepropor- tional tothemasses. 286 MOVING COORDINATE SYSTEMS [CI-IAP. 7 Wenow introduce acoordinate system rotating with angular velocity wabout thecenter ofmass ofM1andM2. Inthissystem, M1andM2are atrest, andwewilltake them tobeontheac-axis atthepoints _ M2 _ M1{I11 — M1 +M2 (Z, IE2 -— M1 +M2 G. The angular velocity coistaken tobealong thez-axis. Then mmoves in thexy-plane, anditsequation ofmotion is *2 * m%:=F1—|—F2—mw><(w><r)—-2mw><%, (7-63) where F1andF2arethegravitational attraction ofM1and M2onm. Written interms ofcomponents, thetwoequations become ____ MG(x—x) __ MG’(x-20) (M+M)G'x ,x— Km__1x1)2+1/L13/2 [(9,_2x2)2+1/Z13/2+ 1as2JFZQ?/, .. MG MG (M +M )G . y:_[($—$012Ell‘1/213/2 —[(90—~1v2)22-ii/213/2 + 1a32y—2°)” (7-64) Note thatthemass mcancels inthese equations. Since thecoriolis force isperpendicular tothevelocity, itdoes no ‘work’ inthismoving coordinate system. Moreover, thecentrifugal force haszero curlandcanbederived from the‘potential energy’ V.=—%mw2(:1c2 +1/2). (7-65) Therefore thetotal ‘energy’ inthemoving coordinate system isaconstant ofthemotion: ‘E’=%m(1'=2 +272)+‘V’, (7-66) where ‘V, __ mM1G __ MMZG _ ' [(-"v-—$1)-2+2/211/2 Kw—w2)2+2/211/2 m(M +M)G(w2 +2/2) _ . (7_67) The energy equation (7—66) enables ustomake certain statements about thekinds oforbits that may bepossible. Inorder tosimplify thealgebra, letusset E=I/<1, 11=y/<1, (7-68) _ M2 ___ M1 _ __ £1‘ M-1 +M27 £2 * M1 +M2 —’ E1 7—6] THE RESTRICTED THREE-BODY PROBLEM 287 Then Eq.(7-67) canbewritten as ¢V,_m(M1+M2)Gi E2 ‘T a [(5—‘§1)2+172]‘/2 " r5‘E2‘F"ill"-79 Inorder toseethenature ofthisfunction, letusﬁrstlookforitssingular points, where 6‘V’/65and6‘V’/61;both vanish: as-s> so-s>"Ks—2z1>2+1213/2 +l(.<.=-1292 +2121=’»/2 _5=°' z 2_[(2—$057+ 11213/2+us—22)}-277+11213/2_”=0'(7-71) Apoint (ac,y)forwhich these equations aresatisﬁed isanequilibrium point forthemass m(intherotating coordinate system), since Eqs. (7-64) areevidently satisﬁed ifmisatrestatthispoint. Weﬁrstconsider points onthe11='0 axis. The second equation isthen satisﬁed, andtheﬁrst becomes InFig. 7-7, weplot thefunction ‘V’,asgiven byEq.(7-70), along the 11=0axis. Theroots ofEq.(7-72) arethemaxima of‘V’(£, 0)inFig.7-7, where itcanbeseen that there arethree such roots. Letuscallthem 5,1,£3,£0asintheﬁgure. Each istheroot ofaquintic equation which may bederived from Eq. (7-72). Itisnotdifﬁcult toshow that 62‘V’/65 an=0,62‘V’/652 <0,and62‘V’/6112 >0atthesepoints A,B, T11]: £14 52 €lB £1 £l(j' Et-> I . _ M M1 FIG. 7-7. Aplot of‘V’(E, 0). 288 MOVING COORDINATE srsrnms [CHAP. 7 1/ . 4VV 5V6V7 VB VVQW V8 10 V7 Vs W V4V3V7“ V’A M V4 V5 Vs V1 V It E V10 Vs VBV7V5V4 FIG. 7-8. Equipotential contours for‘V’(:c, y).51Q3—-x andC.Ifweexpand ‘V’inaTaylor series about anyoneofthese points, andconsider only thequadratic terms, weseethat thecurves ofconstant ‘V’arehyperbolas inthe£1;-plane intheneighborhood ofpoints A,B,C, asshown inFig. 7-8, where weplot thecontours ofconstant ‘V’. These points aresaddlepoints of‘V’;that is,‘V’hasalocal maximum along the £-axis andaminimum along alineperpendicular tothe5-axis ateach of these points A,B,C’.If1;750,itcanbefactored from thesecond of Eqs. (7-71). Wethen multiply thesecond ofEqs. (7-71) by(15—-£1) andsubtract from theﬁrst ofthese equations. After some manipulation andusing Eq.(7-69), weobtain (5-E2)’+122=1, (7-73) and, similarly, (E—£1)’+1:2=1- (7-74) These equations show that there aretwo singular points D,E,offthe 11=Oaxis, which lieatunit distance from (£1,0)and (£2,0)which are 7—6] THE RESTRICTED THREE-BODY PROBLEM 289 themselves separated byaunit distance. Byexpanding ‘V’inaTaylor series about point DorE,wecanshow that curves ofconstant ‘V’are ellipses intheneighborhood ofDorE,andthat ‘V’hasamaximum atD and E.Knowing thebehavior near thesingular points, wecaneasily sketch thegeneral appearance ofthecontours ofconstant ‘V’,asshown inFig. 7-8. Thecurves arenumbered inorder ofincreasing ‘V’. A Ifthiswere aﬁxed coordinate system, wecould immediately conclude that equilibrium points A,B,C,D,Eareallunstable, since theforce —V‘V’ isdirected away from each equilibrium point when misatsome nearby points. However, this argument does nothold here because it neglects thecoriolis force inEqs. (7-64). Ifweexpand theright members ofEqs. (7-64) inpowers ofthedisplacements (say x—rap,y—yp) from oneoftheequilibrium points (say D),andretain only linear terms, wemay determine approximately themotion near theequilibrium point. Ifthisisdone near point D(orE),forexample, weﬁnd that inlinear approximation, themotion near Disstable ifoneofthemasses M1or M2contains more than about 96% ofthetotal mass (M1 +M2). (See Problem 16.) Formotions very near topoint D,wemay expect thelinear approximation toyield asolution which isvalid forvery long times. Whether those motions which arestable inlinear approximation aretruly stable, inthesense’ thatthey remain nearpoint Dforalltime, isoneof theunsolved problems ofclassical mechanics. This matter isdiscussed further attheendofSection 12—6.* Itisnotdiﬁicult toshow that, even inlinear approximation, theequi- librium points A,B,Careunstable. Ifthemotion inlinear approximation isunstable, then theexact solution iscertainly unstable. That is,regard- lesofhow close misinitially totheequilibrium point (but notatit),it willnot,ingeneral, remain asclose butwillmove exponentially away, at least atﬁrst. The neglected nonlinear terms may, ofcourse, eventually prevent thesolution from going more than some ﬁnite distance from the equilibrium point. The only rigorous statements wecanmake about themotion ofm,for very long periods oftime, arethose which canbederived from theenergy equation (7-66). Given aninitial position and velocity ofm,wecan calculate ‘E’. Theorbit then must remain intheregion where ‘V’3‘E’. For example, motions which start near either mass M1orM2, with ‘E’<V3,must remain conﬁned toaregion near that mass. Motions with ‘E’>V5maygotoarbitrarily large distances; whether they actually do,wecannot sayfrom energy arguments. However, thestudies which *Amore complete discussion oftheproblem ofthree bodies, onamore ad- vanced level than thepresent text, willbefound inAurel Wintner, TheAnalytical Foundations ofCelestial Mechanics. Princeton: Princeton University Press, 1947. 290 MOVING COORDINATE SYSTEMS [crnu=. 7 have sofarbeen made ofthethree-body problem make itvery plausible, though ithasnotbeen proved, that except forspecial cases (e.g., M2=0) orforspecial initial conditions, most orbits eventually wander throughout theregion that is‘energetically’ allowed. Ifwecould ﬁnd another constant ofthemotion, sayF(x, y,a':,y),we could solve theproblem bymethods likethose used inChapter 3forthe central force problem, where theangular momentum isalso constant. Unfortunately, noother such constant isknown, andinview ofthelast sentence ofthepreceding paragraph, itseems likely that none exists. This problem hasbeen studied very extensively.* Faced with thissituation, wemay turn tothepossibility ofcomputing particular orbits from given initial conditions. This canbedone either analytically, byapproximation methods, ornumerically, andinprinciple canbedone toanydesired accuracy andforanydesired ﬁnite period of time. . InChapter 12weshall discuss aclosely related special caseofthethree- body problem. PROBLEMS 1.(a)Solve theproblem ofthefreely falling body byintroducing atranslating coordinate system with anacceleration g.Setupandsolve theequations of motion inthisaccelerated coordinate system andtransform theresult back to acoordinate system ﬁxed relative totheearth. (Neglect theearth’s rotation.) (b)Inthesame accelerated coordinate system, setuptheequations ofmotion forafalling body subject toanairresistance proportional toitsvelocity (rela- tivetotheﬁxed air). 2.Amass misfastened byaspring (spring constant lo)toapoint ofsupport which moves back and forth along thex-axis insimple harmonic motion at frequency w,amplitude a.Assuming themass moves only along the:1:-axis, set upandsolve theequation ofmotion inacoordinate system whose origin isat thepoint ofsupport. 3.Generalize Eq.(5-5) tothecase when theorigin ofthecoordinate system ismoving, byadding ﬁctitious torques duetotheﬁctitious force oneach particle. Express theﬁctitious torques interms ofthetotal mass M,thecoordinate R* ofthecenter ofmass, andah.Compare your result with Eq.(4-25). 4.Derive aformula ford3A/dt3 interms ofstarred derivatives relative toa rotating coordinate system. 5.Westerly winds blow from west toeastinthenorthern hemisphere with an average speed v.Ifthedensity oftheairisp,what pressure gradient isrequired tomaintain asteady ﬂow ofairfrom west toeastwith thisspeed? Make reason- able estimates ofvandp,andestimate thepressure gradient inlb-in_2-mile '1. *SeeA.Wintner, op.cit. PROBLEMS 291 6.(a)Ithasbeen suggested that birds may determine their latitude by sensing thecoriolis force. Calculate theforce abird must exert inlevel ﬂight at30mi/hr against thesidewise component ofcoriolis force inorder toﬂyina straight line. Express your result ing’s,that is,asaratio ofcoriolis force to gravitational force, asafunction oflatitude anddirection ofﬂight. (b)Ifthebird’s ﬂight path isslightly circular, acentrifugal force will be present, which willaddtothecoriolis force andproduce anerror inestimated latitude. At45°Nlatitude, how much may theﬂight path bend, indegrees per mile ﬂown, ifthelatitude istobedetermined within ;l;100 miles? (Assume the sidewise force ismeasured asprecisely asnecessary!) 7.Abody isdropped from restataheight habove thesurface oftheearth. (a)Calculate thecoriolis force asafunction oftime, assuming ithasanegligible effect onthemotion. Neglect airresistance, andassume hissmall sothat gt canbetaken asconstant. (b)Calculate thenetdisplacement ofthepoint of impact duetothecalculated coriolis force. *8.Find theanswer toProblem 7(b) bysolving forthemotion inanon- rotating coordinate system. What approximations areneeded toarrive atthe same result? 9.Agyroscope consists ofawheel ofradius r,allofwhose mass islocated on therim. Thegyroscope isrotating with angular velocity 9about itsaxis, which isﬁxed relative totheearth’s surface. Wechoose acoordinate system atrest relative totheearth whose z-axis coincides with thegyroscope axisandwhose origin liesatthecenter ofthewheel. Theangular velocity woftheearth lies inthexz-plane, making anangle awiththegyroscope axis. Find theas-,y-,andz-components ofthetorque Nabout theorigin, duetothe coriolis force inthexyz-coordinate system, acting onamass montherimofthe gyroscope wheel whose polar coordinates inthemy-plane arer,0.Use this result toshow that thetotal coriolis torque onthegyroscope, ifthewheel has amass M,is N=jMr2w6 sinoz. This equation isthebasis fortheoperation ofthegyrocompass. *10. Amass mofaperfect gasofmolecular weight M,attemperature T,is placed inacylinder ofradius a,height h,andwhirled rapidly with anangular velocity coabout theaxis ofthecylinder. Byintroducing acoordinate system rotating with thegas, and applying thelaws ofstatic equilibrium, assuming that allother body forces arenegligible compared with thecentrifugal force, show that_2157’—M”°exp2RT ’ where pisthepressure, 1'isthedistance from theaxis, and mMw2 ”°=21rhRT[eXp (M12222/2Rr) -11' ,_____ 292 MOVING COORDINATE SYSTEMS lcnxr. 7 *11. Aparticle moves inthexy-plane under theaction ofaforce F=—lcr, directed toward theorigin. Find itspossible motions byintroducing acoordinate system rotating about thez-axis with angular velocity wchosen sothat the centrifugal force just cancels theforce F,andsolving theequations ofmotion inthiscoordinate system. Describe theresulting motions, andshow that your result agrees with that ofProblem 31,Chapter 3. 12.A-ball ofmass mslides without friction onahorizontal plane atthesur- face oftheearth. Show that itmoves likethebob ofaFoucault pendulum oflength equal totheearth’s radius, provided itremains near thepoint of tangency. 13.Thebobofapendulum isstarted soastoswing inacircle. Bysubstitut- inginEq.(7-46), ﬁndtheangular velocity andshow that thecontribution due tothecoriolis force isgiven very nearly byEq. (7-52). Neglect thevertical component ofthecoriolis force, after showing that itiszero ontheaverage for theassumed motion. 14.Anelectron revolves about aﬁxed proton inanellipse ofsemimajor axis 10"’; cm. ‘Ifthecorresponding motion occurs inamagnetic ﬁeld of10,000 gausses, show that Larmor’s theorem isapplicable, andcalculate theangular velocity ofprecession oftheellipse. 15.Write down apotential energy forthelastterm inEq.(7—58). Ifthe plane oftheorbit inProblem 14isperpendicular toB,andiftheorbit isvery nearly circular, calculate (bythemethods ofChapter 3)therate ofprecession oftheellipse duetothelastterm inEq.(7—58) intherotating coordinate system. Isthisprecession tobeadded toorsubtracted from that calculated inProblem 14? 16.Find thethree second derivatives of‘V’with respect toE,1;forthepoint DinFig. 7-7. Expand theequations ofmotion (7-64), keeping terms linear in E’=E—£0and 1;’=17—no. Using themethod ofSection 4-10, ﬁnd the condition onM1,M2inorder that thenormal modes ofoscillation bestable. IfM1>M2, what istheminimum value ofM1/(M1 -1-M2)? 17.Prove thestatements made inSection 7-6regarding thesecond deriva- tives of‘V’atpoints A,B,andC’inFig. 7-7. Expand theequations ofmotion about points AandB,keeping terms linear in1)and5'=E-E15. Show by themethod ofSection 4-10 that some ofthesolutions areunstable forany values ofthemasses. (You cannot ﬁndthesecond derivatives explicitly, butthe proof depends only ontheir signs.) *18. (a)Write outthequintic equation which must besolved for£4inFig.7-7. Show that ifM2=0,thesolution is£4=——1. (b)Solve numerically for$4 totwo decimal places fortheearth-moon system. (c).Find theminimum launching velocity from thesurface oftheearth forwhich itis‘energetically’ possible forarocket toleave theearth-moon system. Compare with theescape velocity from theearth. 19.Two planets, each ofmass Mandradius R,revolve incircles about each other atadistance aapart. Find theminimum velocity with which arocket might leave oneplanet toarrive attheother. Show that therocket must have a PROBLEMS 293 larger velocity than would becalculated ifthemotion oftheplanets were neg- lected. .20.(a)Locate allﬁxed points inthelimiting caseM2->0,andsketch Fig.7-7 forthiscase. Show that theresults inSection 7-6applied tothiscase arecon- sistent with thecomplete solution given inSection 3-14. (b)Show from thisexample forwhich thecomplete solution isknown, that theminimum ‘energetically’ possible launching velocity forescape calculated asinProblem 18(c) isnotnecessarily thetrueminimum escape velocity. CHAPTER 8 INTRODUCTION TO THE MECHANICS OF CONTINUOUS MEDIA Inthis chapter webegin thestudy ofthemechanics ofcontinuous media, solids, ﬂuids, strings, etc. Insuch problems, thenumber ofparti- clesissolarge that itisnotpractical tostudy themotion ofindividual particles, and weinstead regard matter ascontinuously distributed in space and characterized byitsdensity. Weareinterested primarily in gaining anunderstanding oftheconcepts andmethods oftreatment which areuseful, rather than indeveloping indetail methods ofsolving practical problems. Intheﬁrst four sections, weshall treat thevibrating string, using concepts which areadirect generalization ofparticle mechanics. In theremainder ofthechapter, themechanics ofﬂuids willbedeveloped in awaylessdirectly related toparticle mechanics. 8-1Theequation ofmotion forthevibrating string. Inthissection we shall study themotion ofastring oflength l,stretched horizontally and fastened ateach end, andsetintovibration. Inorder tosimplify the problem, weassume thestring vibrates only inavertical plane, andthat theamplitude ofvibration issmall enough sothat each point onthestring moves only vertically, andsothat thetension inthestring does notchange appreciably during thevibration. Weshall designate apoint onthestring bygiving itshorizontal distance zvfrom theleft-hand end(Fig. 8-1). Thedistance thepoint 2;hasmoved from thehorizontal straight linerepresenting theequilibrium position of thestring willbedesignated byu(x). Thus anyposition oftheentire string istobespeciﬁed byspecifying thefunction u(x) for0§x§l. This isprecisely analogous, inthecaseofasystem ofNparticles, tospeci- fying thecoordinates 11:,-,y,~,z,~,fori=1,...,N. Inthecase ofthe string, xisnotacoordinate, butplays thesame roleasthesubscript i;it designates apoint onthestring. Our idealized continuous string has 1' 9 u (u+du) 0T1<x+<a> 1 Fro. 8-1. Thevibrating string. I 294 8-1] THE EQUATION OFMOTION FOR THE VIBRATING STRING 295 inﬁnitely many points, corresponding totheinﬁnitely many values ofx between Oandl.Foragiven point x,itisu(x) that plays theroleofa coordinate locating that point, inanalogy with thecoordinates :0,-,y,-,z,-of particle 1'.Just asamotion ofthesystem ofparticles istobedescribed byfunctions ac,-(t), y,-(t), z,~(t), locating each particle atevery instant of time, soamotion ofthestring istobedescribed byafunction u(:c,t), locating each point aconthestring atevery instant oftime. Inorder toobtain anequation ofmotion forthestring, weconsider a segment ofstring oflength dzbetween :1:andac+dx.Ifthedensity ofthe string perunit length is0',then themass ofthissegment iscrdx. The velocity ofthestring atanypoint is6u/6t, anditsslope isdu/600. The vertical component oftension exerted from right toleftacross anypoint in thestring is Tu=1'sin0, (8-1) where 0istheangle between thestring andthehorizontal (Fig. 8-1). We areassuming that 0isvery small and, inthiscase, L Tsin0-'=1'tan0='r%- (8-2) Thenetupward force dFduetothetension, onthesegment dxofstring, isthedifference inthevertical component Tubetween thetwoends ofthe segment: dF Z[Tul:c+d:c _[Tulz ,6 6u—ax(1'E)dx. IfWedonotlimit ourselves tovery small slopes 6u/Ox, then asegment of string may alsohave anethorizontal component offorce duetotension, andthesegment willmove horizontally aswellasvertically, apossibility wewish toexclude. Ifthere is,inaddition, avertical force fperunit length, acting along thestring, theequation ofmotion ofthesegment dx willbe(8-3) 2 a'dx%=‘%(T%)d:c+fdx. (8-4) Forahorizontal string acted onbynohorizontal forces except atitsends, and forsmall amplitudes ofvibration, thetension isconstant, and Eq. (8-4) canberewritten: a2u_6214, 296 THE MECHANICS orCONTINUOUS MEDIA [cn.u>. 8 The force fmay bethegravitational force acting onthestring, which is usually negligible unless thetension isvery small. The force fmay also represent anexternal force applied tothestring tosetitinto vibration. Weshall consider only thecasef=0,andwerewrite Eq.(8-5) intheform ’ 62u 162u5;"25as=°’ (H) 6-(§)‘”~ <H> The constant chasthedimensions ofavelocity, andweshall seeinSec- tion8-3that itisthevelocity with which awave travels along thestring. Equation (8-6) isapartial diﬂerential equation forthefimction u(a:,t); itisthemathematical expression ofNewton’s lawofmotion applied to thevibrating string. Weshall want toﬁndsolutions u(:z:,t)toEq.(8-6), forany given initial position uo(x) ofthestring, and any given initial velocity v@(x) ofeach point along thestring. Ifwetake theinitial instant att=0,thismeans that wewant asolution u(:c,t)which satisﬁes the initial conditions:where u(xr =710(37): 6[£14,=110(93)- Thesolution must alsosatisfy theboundary czmditizms:(8-8) 7/'(0r t)=1/“(la t)=or which express thefactthat thestring istiedatitsends. From thenature ofthephysical problem, weexpect that there should bejustonesolution u(:v,t)ofEq.(8-6) which satisﬁes Eqs. (8-8) and(8-9), andthissolution willrepresent themotion ofthestring with thegiven initial condition. Itistherefore reasonable toexpect thatthemathematical theory ofpartial diﬂerential equations willlead tothesame conclusion regarding thenum- berofsolutions ofEq.(8-6), and indeed itdoes. 8-2Normal modes ofvibration forthevibrating string. Weshall ﬁrst trytoﬁndsome solutions ofEq.(8-6) which satisfy theboundary condi- tions (8-9), without regard totheinitial conditions (8-8). This isanalo- gous toourtreatment oftheharmonic oscillator, inwhich weﬁrstlooked forsolutions ofacertain type andlater adjusted these solutions toﬁtthe initial conditions oftheproblem. Themethod ofﬁnding solutions which 8-2] NORMAL MODES orVIBRATION FORTHEVIBRATING srnmo 297 weshall useiscalled themethod ofseparation ofvariables. Itisoneof thefewgeneral methods sofardevised forsolving partial diﬁerential equa- tions, andmany important equations canbesolved bythismethod. Un- fortunately, itdoes notalways work. Inprinciple, anypartial differential equation canbesolved bynumerical methods, butthelabor involved in doing soisoften prohibitive, even forthemodern large-scale automatic computing machines. The method ofseparation ofvariables consists inlooking forolutions oftheform “(ﬁei)=X(=v)@(i), (8-10) that is,uistobeaproduct ofafunction Xofasandafunction 9oft.The derivatives ofuwillthen be . 621» d2X 62.. d2E)<n—2‘®W’ as-Xv‘ <8-11> Ifthese expressions aresubstituted inEq.(8-6), andifwedivide through by®X, then Eq.(8-6) canberewritten: 8 62d2X 11126)xd—x2='ew' (H2) The leftmember ofthisequation isafunction only ofa:,andtheright member isafunction only oft.Ifwehold tﬁxed andvary :0,theright member remains constant, andtheleftmember must therefore beinde- pendent of:0.Similarly, theright member must actually beindependent oft.Wemay setboth members equal toaconstant. Itisclear onphysical grounds that this constant must benegative, fortheright member of Eq.(8-12) istheacceleration ofthestring divided bythedisplacement, and theacceleration must beopposite tothedisplacement orthestring willnot return toitsequilibrium position. Weshall calltheconstant —¢-:2: 1die 2d2XsW=-‘"2’ isW="“’2- <8-13> Theﬁrstofthese equations canberewritten as ‘ d2®W+we=0, (8-14) which werecognize astheequation fortheharmonic oscillator, whose gen- eralsolution, intheform most suitable forourpresent purpose, is 9=Acoswt+Bsinwt, (8-15) 298 THE‘MECHANICS orCONTINUOUS MEDIA [CHAP. 8 where AandBarearbitrary constants. Thesecond ofEqs. (8-13) hasa similar form: d2X (02%~ +c—2X-0, (8-16) andhasasimilar solution: X=Ccos‘? —]—Dsin (8-17) The boundary condition (8-9) canhold foralltimes tonly ifXsatisﬁes theconditions X(O)==C=0, x(z)=0cos‘%l+Dsin‘%l=0. (8-18) The ﬁrst ofthese equations determines C,andthesecond then requires that sin‘%l=0. (s-19) This willhold only ifwhasoneofthevalues w,,=$, n=1,2,3,.... (s-20) Had wetaken theseparation constant inEqs. (8-13) aspositive, Wewould have obtained exponential solutions inplace ofEq.(8-17), anditwould have been impossible tosatisfy theboundary conditions (8-18). Thefrequencies 11,,=w,,/211' given byEq.(8-20) arecalled thenormal frequencies ofvibration ofthestring. Foragiven n,weobtain asolution by substituting Eqs. (8-15) and(8-17) inEq.(8-10), andmaking useofEqs. (8-18), (8-20): u(:v,t)=Asin7%cos% +Bsingsin-1? 1 (8-21) where wehave setD=1.This iscalled anormal mode ofvibration of thestring, and isentirely analogous tothenormal modes ofvibration which wefound inSection 4-10 forcoupled harmonic oscillators. Each point onthestring vibrates atthesame frequency w,,with anamplitude which varies sinusoidally along thestring. Instead oftwocoupled oscil- lators, wehave aninﬁnite number ofoscillating points, andinstead oftwo normal modes ofvibration, wehave aninﬁnite number. The initial position andvelocity att=0ofthenthnormal mode of vibration asgiven byEq.(8-21) are 8-2] NORMAL MODES orVIBRATION FOR THE VIBRATING STRING 299 u0(x) =Asin$, (s-22) n1rcB .n1ra:v0(x) =f sin-l—- Only forthese very special types ofinitial conditions willthestring vibrate inoneofitsnormal modes. However, wecanbuild upmore general solu- tions byadding solutions; forthevibrating string, liketheharmonic oscil- lator, satisﬁes aprinciple ofsuperposition. Letu1(x, t)andu2(as,t)beany two solutions ofEq.(8-6) which satisfy theboundary conditions (8-9). Then thefunction u(x: t)=ul(-1:2 t)+u2(x> t) also satisﬁes theequation ofmotion andtheboundary conditions. This isreadily veriﬁed simply bysubstituting u(:c,t)inEqs. (8-6) and (8-9), andmaking useofthefactthat u1(x, t)andu2(ac,t)satisfy these equations. Amore general solution ofEqs. (8-6) and(8-9) istherefore tobeobtained byadding solutions ofthetype (8-21), using different constants AandB foreach normal frequency: u(x,t)=2(A,,sin2cosQ+BusinLi?sin -(8-23) 10-1 Theinitial position andvelocity forthissolution are u0(x) =2A,,sin[lag, n=1 vo(x) =2g sin n=1(8-24) Whether ornotEq.(8-23) gives ageneral solution toourproblem depends onwhether, with suitable choices oftheinﬁnite setofconstants A,,,B,,, wecanmake thefunctions u0(x) and v0(x) correspond toany possible initial position andvelocity forthestring. Ourintuition isnotvery clear onthispoint, although itisclear that wenowhave agreat variety ofpossi- blefunctions u0(x) andv0(x). Theanswer isprovided bytheFourier series theorem, which states that anycontinuous function u0(x) for(0<at<Z), which satisﬁes theboundary conditions (8-9), canberepresented bythe sum ontheright inEq.(8-24), iftheconstants A,,areproperly chosen.* *R.V.Churchill, Fourier Series andBoundary Value Problems. New York: McGraw-Hill, 1941. (Pages 57-70.) Even functions with aﬁnite number ofdis- continuities canberepresented byFourier series, butthispoint isnotofgreat interest inthepresent application. 300 THEMECHANICS orCONTINUOUS MEDIA lcnxr. 8 Similarly, with theproper choice oftheconstants B”,anycontinuous func- tion v0(x) for/(0 <:1:<l)canberepresented.* The expressions forA,, andB,,are,inthiscase, z 2 .A,,=if u0(x)s1n 1%dx, ° (8-25) z 2 . B1, = 1)g(It) SID. $ dill. The most general motion ofthevibrating string istherefore asuperposi- tionofnormal modes ofvibration atthefundamental frequency v1=c/21 anditsharmonics 11,,=nc/2l. 8-3Wave propagation along astring. Equations (8-14) and(8-16) have alsothecomplex solutions " o=Ae=H"", (s-26) X=¢*"<“"”". (s-27) Hence Eq.(8-6) hascomplex solutions oftheform ta,t)=A@*‘<"/"><”*°‘>. (s-2s) Bytaking therealpart, orbyadding complex conjugates anddividing by2,weobtain therealsolutions ta,t)=A603%(x-ct), (s-29) ta,t)=Acos‘i(x+ct). (s-30) Bytaking imaginary parts, orbysubtracting complex conjugates and dividing by2i,wecould obtain similar solutions with cosines replaced by sines. These solutions donotsatisfy theboundary conditions (8-9), but they areofconsiderable interest inthat they represent waves traveling down thestring, aswenow show. Aﬁxed point :0onthestring willoscillate harmonically intime, accord- ingtothesolution (8-29) or(8-30), with amplitude Aandangular fre- quency w.Atany given instant t,thestring will beintheform ofasinus- *TheFourier series theorem wasquoted inSection 2-11 inaslightly different form. Theconnection between Eqs. (8-24) and(2-205) istobemade byreplacing tby:0andTby2linEq.(2—205). Both sineandcosine terms arethen needed to represent anarbitrary function u0(x) intheinterval (0<2:<2l),butonly sine terms areneeded ifwewant torepresent ’lL()(:0)only intheinterval (0<:1:<Z). [Cosine terms alone would alsodoforthisinterval, butsineterms areappropriate ifu0(a:) vanishes atx=0andx=l.] . 8-3] WAVE PROPAGATION ALONG ASTRING 301 oidal curve with amplitude Aandwavelength A(distance between successive maxima) : 21rcA-?- (8-31) Wenowshow thatthispattern moves along thestring withvelocity c,to theright insolution (8-29), andtotheleftinsolution (8-30). Let E=x—ct, (8-32) sothat Eq.(8-29) becomes u=Acosw?£, (8-33) where 5iscalled thephase ofthewaverepresented bythefunction u.For aﬁxed value of£5,uhasaﬁxed value. Letusconsider ashort time interval dtandﬁndtheincrement dz:required tomaintain aconstant value ofE: d£=dx—cdt=0. (8-34) Now ifda:anddthave theratio given byEq.(8-34), %=c, (8-35) then thevalue ofuatthepoint :0+da:attime t+dtwillbethesame as itsvalue atthepoint :1:attime t.Consequently, thepattern moves along thestring with velocity cgiven byEq.(8-7). Theconstant cisthephase velocity ofthewave. Similarly, thevelocity dz/dt forsolution (8-30) is-c. Itisoften convenient tointroduce theangular wave number lcdeﬁned by theequation co 21r|k|___2=T, (8-36) where lcistaken aspositive forawave traveling totheright, andnegative forawave traveling totheleft. Then both solutions (8-29) and (8-30) canbewritten inthesymmetrical form u=Acos(lea:—wt). (8-37) The angular wave number lcismeasured inradians percentimeter, justas theangular frequency wismeasured inradians persecond. The expres- sionforuinEq.(8-37) istherealpart ofthecomplex function ,u=Ae“'°'_“”. (s-as) This form isoften used inthestudy ofwave motion. 302 THE MECHANICS orCONTINUOUS MEDIA [CHAP~ 8 The possibility ofsuperposing solutions oftheform (8-29) and (8-30) with various amplitudes andfrequencies, together with theFourier series theorem, suggests amore general solution oftheform Wet)=f(w—ct)+q(w+rt), (8-39) where f(5,1)andg(r)) arearbitrary functions ofthevariables 5=ac—ct, and 11=or+ct.Equation (8-39) represents awave ofarbitrary shape traveling totheright with velocity c,andanother traveling totheleft. Wecanreadily verify that Eq. (8-39) gives asolution ofEq. (8-6) by calculating thederivatives ofu: 6u_df6E+dg6n__df+dg 6x—d£6x dqax“dg an’ n_n+n6:202“452 due’ 1 %_ﬂi€ @?l__if_ @6t—d£6t+dn6tT cd£+cd11' 2 2 2 "7 When these expressions aresubstituted inEq.(8-6), itissatisﬁed iden- tically, nomatter what thefunctions f(E)andg(r)) may be,provided, of course, that they have second derivatives. Equation (8-39) is,infact, the most general solution oftheequation (8-6); thisfollows from thetheory ofpartial differential equations, according towhich thegeneral solution ofasecond-order partial differential equation contains twoarbitrary func- tions. Wecanprove thiswithout resorting tothetheory ofpartial differ- ential equations byassuming thestring tobeofinﬁnite length, sothat there arenoboundary conditions toconcern us,andbysupposing that the initial position andvelocity ofallpoints onthestring aregiven bythe functions u0(x), v0(x). Ifthesolution (8-39) istomeet these initial con- ditions, wemust have, att=0: “(em0)=f(%)+9(%)=uo(vv), (8-40) [%‘]t=0 =[—c%J; +c%]t=0 =v0(x). (8-41) Att=0,5=1;=x,sothat Eq.(8-41) canberewritten: %[—f<x>+go>1= <8-12> 8-3] WAVE PROPAGATION ALONG ASTRING 303 which canbeintegrated togive {I -f(w) +g(r)=%(0vo(w)dw+0- (8-43) Byadding andsubtracting Eqs. (8-40) and(8-43), weobtain thefunctions fandg: an=s(u0c>-§/01»0<w>dx—0)» (H4) !I(=v)=%(uo(1v) +g/E)vo($)div+C)- The constant Ccanbeomitted, since itwillcancel outinu=f+g,and wecanreplace xby5and1;respectively inthese equations: Ef(s)=e(u0<s> -§/on»0<s>dz), (8-45) g(r)=%(uo(ﬂ) + vo(11)dn) I This gives asolution toEq.(8-6) foranyinitial position andvelocity of thestring. Associated with awave u=f(x—ct), (8-46) there isaﬂow ofenergy down thestring, asweshow bycomputing the power delivered from lefttoright across anypoint xonthestring. The power Pistheproduct oftheupward velocity ofthepoint xandtheup- ward force [Eq. (8-2)] exerted bythelefthalf ofthestring ontheright halfacross thepoint :0: 66P=-15%3;" (s-47) Ifuisgiven byEq.(8-46), thisis df2P=orE » (8-48) which isalways positive, indicating that thepower flow isalways from lefttoright forthewave (8-46). Forawave traveling totheleft, Pwill benegative, indicating aﬂow ofpower from right toleft. Forasinusoidal wave given byEq.(8-37), thepower is P=kw'rA2 sin2(kx—wt), (8-49) or,averaged over acycle, (P),,, =%kw1-A2. (8-50) 304 THEMECHANICS orCONTINUOUS MEDIA [cn.u>. 8 Wenow consider astring tiedatx=0andextending totheleftfrom av=0toa:=——oo. Thesolution (8-39) must now satisfy theboundary condition 4 M0,!) =f(—vi) +y(¢¢)=9, (8-51) OX‘/<-a=—g<a . <8-52> forallvalues ofE.The initial values u0(x) andv0(a:) willnow begiven only fornegative values ofx,andEqs. (8-45) willdeﬁne f(E)andg(r))only fornegative values of5and n.The values of_f(£) andg(17) forpositive values of£and11canthenbefound from Eq.(8-52): f(E)=-9(-E), g(r)=—f(—11)- (8-53) Letusconsider awave represented byf(x—ct)traveling toward theend as=0.Aparticular phase £0,forwhich thewave amplitude isf(£0),will attime tobeatthepoint -730=E0+6150- (8"54) Letussuppose that £0andtoaresochosen that moisnegative. Atalater time t1,thephase £0willbeatthepoint I $1= £0"l'Ctl =$9+C(t1 —to). Att1=to——(xo/c), 11:1=0and thephase £0reaches theendofthe string. Atlater times x1willbepositive, andf(x1-—ctl)willhave no physical meaning, since thestring does notextend topositive values ofx. Now consider thephase '00,oftheleftward traveling wave g(x+ct), deﬁned by 110=1?-|-615=-50- (3-56) Theamplitude oftheleftward wave g(110) forthephase noisrelated tothe amplitude oftherightward wave f($0)forthecorresponding phase £0by Eq.(8-53): 9(m>) =-f(€o)- (8-57) Attime t1,thephase nowillbeatthepoint IE2=‘110—Oil =—1?Q "'C(t1i— tn). If(t1—to)>-:00/c, x2isnegative andg(n0) represents awave ofequal and opposite amplitude tof(E0),traveling totheleft. Thus thewave f(x——ct)isreﬂected outofphase atas=0andbecomes anequal and opposite wave traveling totheleft. (SeeFig.8-2.) Thetotal distance 8-4] STRING ASLIMITING CASE orSYSTEM orPARTICLES 305 i» ' /flgh ,2 - O I M 0 9('In) O l-to_ t=ll FIG. 8-2. Awave reﬂected at:0=0. traveled bythewave during thetime (t1—-to),from at=motoso=0 andback to:1:=x2is,byEq.(8-58), —$o "$2=6(t1—to), (8"59) asitshould be. Thesolution (8-39) canalsobeﬁtted toastring ofﬁnite length fastened atav=0and2:=l.Inthiscase, theinitial position andvelocity u0(:c) andv0(x) aregiven only for(03:1:3l).Thefunctions f(£)andg(r))are then deﬁned byEq.(8-45) only for(0353l,031;3l).Ifwedeﬁne f(£)andg(11)fornegative values of5and11byEq.(8-53), interms oftheir values forpositive 5and1;,then theboundary condition (8-51) willbesatis- ﬁedat:1:=0.Byanargument similar tothat which ledtoEq.(8-53), wecanshow thattheboundary condition (8-9) forx=lwillbesatisﬁed if, forallvalues of£and11, f(E+ l)=—9(l-5), 901+1)=—f(l—11)- Bymeans ofEqs. (8-53) and(8-60), wecanﬁndf(£)andg(r))forallvalues ofEand 1;,once their values aregiven [byEqs. (8-45)] for0353l, 03113l.Thus weﬁndasolution forthevibrating string oflength lin terms ofwaves traveling inopposite directions andcontinuously being reﬂected ata:=0,andat=l.The solution isequivalent tothesolution given byEqs. (8-23) and(8-25) interms ofstanding sinusoidal waves.(8-60) 8-4Thestring asa caseofasystem ofparticles. Intheﬁrst three sections ofthischapter, wehave considered anidealized string char- acterized byacontinuously distributed mass with density 0'andtension 'r. Anactual string ismade upofparticles (atoms andmolecules); ourtreat- ment ofitascontinuous isvalid because oftheenormously large number of particles inthestring. Atreatment ofanactual string which takes into account theindividual atoms would behopelessly difficult, butweshall consider inthissectionian idealized model ofastring made upofaﬁnite number ofparticles, each ofmass m. Figure 8-3shows this idealized string, inwhich anattractive force 1'acts between adjacent particles 306 THE MECHANICS orCONTINUOUS MEDIA [CHAP. 8 T , T ul ug ug |-fr-1 FIG. 8-3. Astring made upofparticles. along theline joining them. The interparticle forces aresuch that in equilibrium thestring ishorizontal, with theparticles equally spaced a distance hapart. Thestring isoflength (N+1)h,with N-I—2particles, thetwoendparticles being fastened attheac-axis. TheNparticles which arefreetomove arenumbered 1,2,...,N,andtheupward displacement ofparticle jfrom thehorizontal axiswillbecalled u,-.Itwillbeassumed that theparticles move only vertically andthat only small vibrations are considered, sothat theslope ofthestring isalways small. Then the equations ofmotion ofthissystem ofparticles are 2. . _ . ._ .mdd,’§’=¢“’*‘h ”’-Tu’ hu’-1, j=1,...,N, (s-61) where theexpression ontheright represents thevertical components ofthe forces 1'between particle jandthetwoadjacent particles, andwearesup- posing thattheforces 1'areequal between allpairs ofparticles. Now let usassume that thenumber Nofparticles isvery large, andthat thedis- placement ofthestring issuch that atanytime t,asmooth curve u(x,t) canbedrawn through theparticles, sothat u(jh,I5)=“r(t)-A (8-62) Wecanthen represent thesystem ofparticles approximately asacontin- uous string oftension 1',andoflinear density <1= (s-63) The equations ofmotion (8-61) canbewritten intheform d2u,- _1'1(ui+1 -u; u,-—u,-_1)W—<1hT“h h 6'64) Now iftheparticles aresufﬁciently close together, weshall have, approxi- mately, "i+1 —Wi h ax $=(J'+1/2),‘, W—H1-1 é[Q] h 313:=(j-1/2)h’ 8-4] STRING AsLIMITING CASE orsYsTEM orPARTICLES 307 andhence 1 u azu ‘ ‘ _ uj-l-1 _— uj _ uj '— J‘-1) é[ ] _ h( h h 01621=-in ( ) The function u(x,t)therefore, when hisvery small, satisﬁes theequation 62u "r62uas-55;’ <8“) which isthesame asEq.(8-6) forthecontinuous string. Thesolutions ofEqs. (8-61) when Nislarge willbeexpected toapproxi- mate thesolutions ofEq.:(8-6). Ifwewere unable tosolve Eq.(8-6) otherwise, onemethod ofsolving itnumerically would betocarry outthe above process inreverse, soastoreduce thepartial differential equation (8-67) tothesetofordinary differential equations (8-61), which could then besolved bynumerical methods. The solutions ofEqs. (8-61) areof some interest intheir own right. Letusrewrite these equations inthe form I d2- 2 A .rnFt%'—+7"lCu_,--;7’(u,-_|_1—|-u,~_1)=0, _1=1,...,N. (sass) These aretheequations forasetofharmonic oscillators, each coupled to thetwoadjacent oscillators. Weareled,either byourmethod oftreat- ment ofthecoupled oscillator problem orbyconsidering ourresults forthe continuous string, totryasolution oftheform u,-=a,-ei""". (8-69) Ifwesubstitute thistrial solution inEqs. (8-68), thefactor e*"“" cancels out, andwegetasetofalgebraic equations: 2 .(%'-—'rruo2)a,-—%a,~_,_1—%a,-_1=0, _7= l,...,N. (8-70) This isasetoflinear difference equations which could besolved fora,-+1 interms ofa,-anda,-_1. Since ao=0,ifa1isgiven wecanﬁndthevalues oftheremaining constants a,-bysuccessive applications ofthese equations. Aneater method ofsolution istonotice theanalogy between thelinear difference equations (8-70) and thelinear differential equation (8-16), andtotrythesolution a,-=Ad", j=1,...,N. (s-71) When thisissubstituted inEqs. (8-70), weget,after canceling thefactor Aefpj : . -rP _<2-Z-W2)-E(J+e*1’)=0, (8-12) 308 THEMECHANICS orCONTINUOUS MEDIA [CHAP. 8 or M2 cosp =1- (8-73) 41'1/200¢- » (8-74)Ifwislessthan there willberealsolutions forp.Letasolution begiven by p=lch, 03lch31r. (8-75) Then another solution is p=—kh. (8-76) Allother solutions forpdiffer from these bymultiples of21r,andinview oftheform ofEq.(8-71), they lead tothesame values ofa,-,sowecanre- strict ourattention tovalues ofpgiven byEqs. (8-75) and(8-76). Ifwesubstitute Eq.(8-71) inEq.(8-69), making useofEq.(8-75), we have asolution ofEqs. (8-68) intheform u,-=A@='=“'""'-"'>. (8-77) Since thehorizontal distance ofparticle jfrom theleftendofthestring is $1=ih,. (8-78) weseethat thesolution (8-77) corresponds toourprevious solution (8-38) forthecontinuous string, andrepresents traveling sinusoidal waves. By combining thetwo complex conjugate solutions (8-77 )and using Eq. (8-78), weobtain therealsolution u,-=Acos(kxj —wt), (8-79) which corresponds toEq.(8-37). Wethus have sinusoidal waves which may travel ineither direction with thevelocity [Eq. (8-36)] hm3 c=%=W, (s-so) where pisgiven byEq.(8-73). Ifw<<we[Eq. (8-74)], then pwillbe nearly zero, andwecanexpand cospinEq.(8-73) inapower series: __f_A __mhw21 2-1 Tr» h1/2Irl-w(%) » <8-81> 8-4] srnmo AsLIMITING cAsE orSYSTEM orPARTICLES 309 and , 0s('1)2, (8-s2) Tn which agrees with Eq. (8-7) forthecontinuous string, inview ofEq. (8-63). However, forlarger values ofw,thevelocity cissmaller than for thecontinuous string, andapproaches ha. 2h"2°=7=r(t) (H3) asw -+w,.(wc=ooforthecontinuous string forwhich mh=ahz=0.) Since thephase velocity given byEq.(8-80) depends upon thefrequency, wecarmot superpose sinusoidal solutions toobtain ageneral solution ofthe form (8-39). Ifawave ofother than sinusoidal shape travels along the string, thesinuosidal components intowhich itmay beresolved travel with different velocities, andconsequently theshape ofthewave changes asit moves along. This phenomenon iscalled dispersion. When w>40¢,Eq.(8-73) hasonly complex solutions forp,oftheform ' p=1r=1:iv. (8-84) These leadtosolutions u,-oftheform u,-=(—1)jAe*'” coswt. (8-85) There isthen nowave propagation, butonly anexponential decline in amplitude ofoscillation totheright ortotheleftfrom anypoint which may besetinoscillation. The minimum wavelength [Eq. (8—36)] which isallowed byEq.(8-75) is >.,,=%=2h. (8-sc) O Itisevident that awave ofshorter wavelength than thiswould have no meaning, since there would notbeenough particles inadistance lessthan X,todeﬁne thewavelength. The wavelength A,,corresponds tothefre- quency wc,forwhich u,-=A¢""eﬂ=‘"=' =(-1)"A6-W. (s-87) Adjacent particles simply oscillate outofphase with amplitude A. Wecanbuild upsolutions which satisfy theboundary conditions ‘Mo=u1v+1 =0 (8"33) byadding andsubtracting solutions oftheform (8-77). Wecan, bysuit- ably combining solutions oftheform (8-77), obtain thesolutions . 310 THE MECHANICS orCONTINUOUS MEDIA [CHAP. 8 u,-=Asinpjcoswt-|-Bsinpjsinwt+Ccospjcoswt +Dcospjsinwt. (8-89) Inorder tosatisfy theconditions (8-88), wemust set C’=D=0, 7171' p_]VTi' n:1r2:"';N! where thelimitation n3Narises from thelimitation onpinEq.(8-75). Thenormal frequencies ofvibration arenow given byEq.(8-73): 22 1/w,.=[ﬁ(1_c0s%>] ,n=1,2,...,N. (8-91) Ifn<<N,wecanexpand thecosine inapower series, toobtain .[n2ﬂ_2,,_ T/2 “’"=mh(N+1)2 =$(§)1/2. [z=(N+1)h], (8-92) which agrees with Eq.(8-20) forthecontinuous string. Aphysical model which approximates fairly closely thestring ofparticles treated inthissection canbeconstructed byhanging weights matintervals halong astretched string. The mass mofeach weight must belarge in comparison with that ofalength hofthestring. 8-5General remarks onthepropagation ofwaves. Ifwedesignate byFtheupward component offorce duetotension, exerted from leftto right across anypoint inastretched string, andbyvtheupward velocity ofanypoint onthestring, then, byEq.(8-2), wehave F=-1%, (8-93) du ByEq.(8-4), ifthere isnoother force onthestring, wehave 6v 16F 'E-—E-)5, (8-95) andbydifferentiating Eq.(8-93) with respect tot,assuming 1-isconstant 8-5] GENERAL REMARKS ONTHE PROPAGATION orWAvEs 311 intime, weobtain ‘ill-—'T99- (8-96)at“ Bx Equations (8-95) and (8-96) areeasily understood physically. The ac- celeration ofthestring willbeproportional tothedifference intheupward force Fattheends ofasmall segment ofstring. Likewise, since Fispro- portional totheslope, thetime rate ofchange ofFwillbeproportional to thedifference inupward velocities oftheends ofasmall segment ofthe string. The power delivered from lefttoright across anypoint inthe string is P=Fv. (8-97) Equations (8-95) and(8-96) aretypical ofmany types ofsmall ampli- tude wave propagation which occur inphysics. There aretwoquantities, inthiscase Fandv,such that thetime rate ofchange ofeither ispropor- tional tothespace derivative oftheother. Forlarge amplitudes, theequa- tions forwave propagation may become nonlinear, andneweffects likethe development ofshock fronts may occur which arenotdescribed bythe equations wehave studied here. When there isdispersion, linear terms in vandF,orterms involving higher derivatives than theﬁrst, may appear. Whenever equations oftheform (8-95) and(8-96) hold, awave equation oftheform (8-6) canbederived foreither ofthetwoquantities. Forex- ample, ifwedifferentiate Eq.(8-95) with respect tot,andEq.(8-96) with respect toas,assuming 0,1tobeconstant, weobtain 62F_ 62v_ a2» 6x6t_“TE _-59?’ OI‘ 62v 182v552“ETa=°' <8-98> where 0=(91/S (8-99) 62F 162Fas"ziW=°- <8-1°°>Similarly, wecanshow that Usually oneofthese twoquantities canbechosen soastobeanalogous to aforce (F),andtheother tothecorresponding velocity (v),andthen the power transmitted willbegiven byanequation likeEq.(8-97). Likewise, allother quantities associated with thewave motion satisfy awave equa- tion, as,forexample, u,which satisﬁes Eq.(8-6). 312 THEMECHANICS orCONTINUOUS MEDIA [CHAP. 8 Asafurther example, theequations foraplane sound wave traveling in theac-direction, which willbederived inSection 8-10, canbewritten in theform 6v 18p’, 6p’ 6v, where p’istheexcess pressure (above atmospheric), visthevelocity, in thex-direction, oftheairatanypoint, andwhere pisthedensity andB thebulk modulus. Thephysical meaning ofthese equations isclear almost without further discussion ofthemotion ofgases. Both p’andvsatisfy wave equations, easily derived from Eqs. (8-101): 621)’ 162p’ 6222 16222Er"aaY=9 5a"aaY=@ @*m c=(gym: (8-103) andthepower transmitted intheat-direction perunit area iswhere P=p’v. (8—104) Inthecaseofaplane electromagnetic wave traveling inthex-direction andlinearly polarized inthey-direction, theanalogous equations canbe shown tobe(gaussian units) <’B=__iE_v .<?E11__‘?1iTi‘ °ax’ at_‘ax’ (8405) where E,andB,arethey-andz-components ofelectric andmagnetic ﬁeld intensities, andcisthespeed oflight. Thecomponents E,andB,satisfy wave equations with wave velocity c,andthepower transmitted inthe x-direction perunit area is 1 EyBg _- P_Tm (8-106) Asaﬁnal example, onatwo-wire electrical transmission line, thevolt- ageEacross thelineandthecurrent ithrough thelinesatisfy theequa- tions BE 1Bi Oi 1GEE”_6%’ a--ta’ 9”” where Cistheshunt capacitance perunitlength, andListheseries in- ductance perunitlength. Again wecanderive wave equations foriandE 8-6] KINEMATICS orMovING FLUIDS 313 withthewave velocity 11,2 c= , (8—108) andagain thepower transmitted intheac-direction is P=Ei. (8—l09) Thus thestudy ofwave propagation inastring isapplicable to-a wide variety ofphysical problems, many ofthem ofgreater practical andtheo- retical importance than thestring itself. Inmany cases, ourdiscussion ofthestring asmade upofanumber ofdiscrete particles isalsoofinterest. Theelectrical transmission line, forexample, canbeconsidered alimiting case ofaseries oflow-pass ﬁlters. Anelectrical network made upofseries inductances andshunt capacitances canbedescribed byasetofequations ofthesame form asourEqs. (8-61), with analogous results. Inthecase ofsound waves, weareledbyanalogy toJexpect that atvery high fre- quencies, when thewavelength becomes comparable tothedistance be- tween molecules, thewave velocity willbegin todepend onthefrequency, andthat there“ willbealimiting frequency above which nowave propaga- tionispossible. 8-6Kinematics ofmoving ﬂuids. Inthissection weshall develop the kinematic concepts useful instudying themotion ofcontinuously distrib- uted matter, with particular reference tomoving ﬂuids. Oneway of describing themotion ofaﬂuid would betoattempt tofollow themotion ofeach individual point intheﬂuid, byassigning coordinates 2:,y,ztoeach ﬂuid particle andspecifying these asfunctions ofthetime. Wemay, forex- ample, specify agiven ﬂuid particle byitscoordinates, wo,yo,20,atanini- tialinstant t=to.We_canthen describe themotion oftheﬂuid bymeans offunctions a:(a:0, yo,zo,t),1/(mo, yo,zo,t),z(x0, yo,zo,t)which determine thecoordinates x,y,zattime toftheﬂuid particle which wasatx0,yo,20 attime to.This would beanimmediate generalization oftheconcepts of particle mechanics, andofthepreceding treatment ofthevibrating string. This program originally. duetoEuler leads totheso-called “Lagrangian equations” ofﬂuid mechanics. Amore convenient treatment formany purposes, duealsotoEuler, istoabandon theattempt tospecify thehis- tory ofeach ﬂuid particle, andtospecify instead thedensity andvelocity oftheﬂuid ateach point inspace ateach instant oftime. This isthe method which weshall follow here. Itleads tothe“Eulerian equations” ofﬂuid mechanics. Wedescribe themotion oftheﬂuid byspecifying the density p(.r,y,z,t)andthevector velocity v(a:,y,2,t),atthepoint ax,y,z atthetime t.Wethus focus ourattention onwhat ishappening atapar- ticular point inspace ataparticular time, rather than onwhat ishappening toaparticular ﬂuid particle. 314 THEMECHANICS orCONTINUOUS MEDIA [CI-IAP. 8 Any quantity which isused indescribing thestate oftheﬂuid, forex- ample thepressure p,willbeafunction [p(;z:, y,2,t)]ofthespace coordi- nates x,y,zandofthetime t;that is,itwillhave adeﬁnite value ateach point inspace andateach instant oftime. Although themode ofdescrip- tionwehave adopted focuses attention onapoint inspace rather than on aﬂuid particle, weshall notbeable toavoid following theﬂuid particles themselves, atleast forshort time intervals dt.Foritistotheparticles, andnottothespace points, that thelaws ofmechanics apply. Weshall be interested, therefore, intwotime rates ofchange foranyquantity, sayp. The rate atwhich thepressure ischanging with time ataﬁxed point in space willbethepartial derivative with respect totime (op/6t); itisitself afunction ofx,y,z,andt.Therateatwhich thepressure ischanging with respect toapoint moving along with theﬂuid willbethetotal derivative dp 6p 6pda: 6pdy 8pdz dt=at+6xdtJ’6ydt+dzdl’ (8410) where dx/dt, dy/dt, dz/dt arethecomponents oftheﬂuid velocity v.The change inpressure, dp,occurring during atime dt,attheposition ofa moving ﬂuid particle which moves from ac,y,ztorc+dx,y+dy,z+dz during thistime, willbe dp ="p($ +dz: fl"l"dya Z+dzrt + —P($: yrZ:0 Iﬁr ﬁr 62> 62>—axdx +aydy +azdz-I—atdt, andifdt—>0,thisleads toEq.(8-110). Wecanalsowrite Eq.(S-110) in theforms: I dp_8p 8p 8p Hp _ dt_6t+v”8x+v"8y+v‘6z (8111) and Q_62 . - _ dt_at+v Vp, (8112) where thesecond expression isashorthand fortheﬁrst, inaccordance with theconventions forusing thesymbol V.The total derivative dp/dt is alsoafunction ofac,y,z,andt.Asimilar relation holds between partial andtotal derivatives ofanyquantity, andwemay write, symbolically, d 6 32— &+v-V, where total andpartial derivatives have themeaning deﬁned above. Letusconsider now asmall volume 6Vofﬂuid, andweshall agree that 5Valways designates avolume element which moves with theﬂuid, so 8-6] KINEMATICS orMovING FLUIDS 315 ‘U2 -- oi-1} 1)::8' ‘1:.\.g\\I\\\Vr___________I ->1_:\.-\IQII Q4IQ1 5x FIG. 8-4. Amoving, expanding element ofﬂuid. that italways contains thesame ﬂuid particles. Ingeneral, thevolume 6Vwillthen change with time, andwewish tocalculate thisrateofchange. Letusassume that 5Visintheform ofarectangular boxofdimensions ﬁx,5y,52(Fig. 8-4): 6V=5x5y62. (8-114) Theas-component ofﬂuid velocity v,may bedifferent attheleftandright faces ofthebox. Ifso,6xwillchange with time atarate equal tothe difference between these twovelocities: d 0-(E,8.1:=5%Bx, and, similarly, i_% 8-115 dt5y_ay5y, ( ) d 6v,-8=— .dtZézaz Thetime rate ofchange of6Visthen d d d d;ﬁ6V= 6y¢SzE6ac+6a:<$za6y+6x6@/(E62 _(6:12 +6y+Oz 6%6y6g’ andﬁnally, %6V =V-v 6V. (8—116) This derivation isnotvery rigorous, butitgives aninsight into the meaning ofthedivergence V-v. Thederivation canbemade rigorous by keeping careful track ofquantities that were neglected here, likethede- 316 THEMECHANICS 01-‘co.\"r1xU01:s MEDIA [cn.u=. B pendence of11,upon yandz,andshowing thatwearrive atEq.(S-116) inthelimit as6V—>0.However, there isaneasier waytogiveamore rigorous proof ofEq.(8-116). Letusconsider avolume Vofﬂuidwhich iscomposed ofanumber ofelements 5V: V=Z6V. (S-117) Ifwesum theleftsideofEq.(8—l16), wehave d d _dVZ5aV=aEZav_ - (s-11s) The summation signhere really represents anintegration, since wemean topass tothelimit 6V——>0,butthealgebraic steps inEq.(S-118) would look rather unfamiliar iftheintegral signwere used. Nowletussum the right sideofEq.(8-116), thistime passing tothelimit andusing thein- tegral sign, inorder thatwemay apply Gauss’ divergence theorem [Eq. (3-115)}; ZV-v6V =ff./‘V-vdV V =Us-v dS, (8-119) S where Sisthesurface bounding thevolume V,andnistheoutward normal unit vector. Since n-vistheoutward component ofvelocity ofthesurface element dS,thevolume added toVbythemotion ofd-Sina. time altwillben-vdtdS(Fig. 8-5), andhence thelastlineinEq.(S-119) istheproper expression fortherateofincrease involume: dV-‘E=Us-v as. (8-120) s Therefore Eq. (S-116) must bethecorrect expression fortherate of vdi Fm. 8-5. Increase ofvolume duetomotion ofsurface. 8-6] KINEMATICS orMOVING FLUIDS 317 increase ofavolume element, since itgives thecorrect expression forthe rate ofincrease ofanyvolume Vwhen summed over V.Note that the proof isindependent oftheshape of6V.Wehave incidentally derived an expression forthetime rate ofchange ofavolume Vofmoving ﬂuid: %=flfv-v dV. (s-121) Iftheﬂuid isincompressible, then thevolume ofevery element ofﬂuid must remain constant: %av=0, (8-122) andconsequently, byEq.(8—116), v-v=0. (s-123) Noﬂuid isabsolutely incompressible, butformany purposes liquids may beregarded aspractically soand, asweshall see,even thecompressibility ofgases may often beneglected. Now themass ofanelement ofﬂuid is 8m=p6V, (8—124) andthiswillremain constant even though thevolume anddensity may not: d _d _E6m-E2(p8V)—O. (8—l25) Letuscarry outthedifferentiation, making useofEq.(8—116): Q d8V_ Q _ _5Vdt+p—T -5Vdt—|—pVv6V-0, or,when 6Visdivided out, %+pV-V =0. (8—126) Byutilizing Eq.(8-113), wecanrewrite thisinterms ofthepartial deriva- tives referred toaﬁxed point inspace: %+ v-Vp +pV-v =0. 318 THE MECHANICS OFCONTINUOUS MEDIA [CHAP. 8 Thelasttwoterms canbecombined, using theproperties ofVasasymbol ofdifferentiation: %-l-V-(pv) =O. (8—127) This istheequation ofcontinuity forthemotion ofcontinuous matter. It states essentially that matter isnowhere created ordestroyed; themass 5minanyvolume 6Vmoving with theﬂuid remains constant. Weshall make frequent useintheremainder ofthischapter oftheproperties ofthesymbol V,which were described brieﬂy inSection 3-6. The operator V hasthealgebraic properties ofavector and, inaddition, when aproduct isin- volved, itbehaves likeadifferentiation symbol. The simplest way toperform thissort ofmanipulation, when Voperates onaproduct, isﬁrst towrite asum ofproducts ineach ofwhich only onefactor istobedifferentiated. The factor tobedifferentiated may beindicated byunderlining it.Then each term may be manipulated according totherules ofvector algebra, except that theunderlined factor must bekept behind theVsymbol. When theunderlined factor istheonly onebehind theVsymbol, orwhen allother factors areseparated outbyparen- theses, theunderline may beomitted, asthere isnoambiguity astowhat factor istobedifferentiated bythecomponents ofV.Asanexample, therelation between Eqs. (8—126) and(S-127) ismade clear bythefollowing computation: V-(gv) +V'(P!) (Vg)-v+ PV-y (VP)-v+ PV‘V V-Vp —l—pV-V. (S-128)V-(nv) Any formulas arrived atinthisway canalways beveriﬁed bywriting outboth sides interms ofcomponents, andthereader should dothisafewtimes tocon- vince himself. However, itisusually farlesswork tomake useoftheproperties oftheVsymbol. Wenow wish tocalculate therate ofﬂow ofmass through asurface S ﬁxed inspace. LetdSbeanelement ofsurface, andletnbeaunit vector normal todS.Ifweconstruct acylinder bymoving dSthrough adistance vdtinthedirection of—v,then inatime dtallthematter inthiscylinder willpass through thesurface dS(Fig. 8—6). The amount ofmass inthis cylinder is pn-v dtdS, where n-vdtisthealtitude perpendicular totheface dS. The rate of ﬂow ofmass through asurface Sistherefore %=éfpn-vdS =L/n-(pv) dS. (8-129) 8-6] KINEMATICS orMOVING FLUIDS 319 dS i—>n FIG. 8-6. Flow ofﬂuid through asurface element. Ifn-vispositive, themass ﬂow across Sisinthedirection ofn;ifn-v isnegative, themass ﬂow isinthereverse direction. Weseethat pv,the momentum density, isalso themass current, inthesense that itscom- ponent inanydirection gives therate ofmass ﬂow perunit area inthat direction. Wecannow give afurther interpretation ofEq. (8—127) by integrating itover aﬁxed volume Vbounded byasurface Swith outward normaln: /,[f%§dV+fI[fv-(pv)dv =0.g (s-130) Since thevolume Vhere isaﬁxed volume, wecantake thetime differenti- ation outside theintegral intheﬁrstterm. IfWeapply Gauss’ divergence theorem tothesecond integral, wecanrewrite thisequation: %fffpdV =-[fa-(pods. (8-131) V S This equation states that therateofincrease ofmass inside theﬁxed vol- ume Visequal tothenegative oftherateofﬂowofmass outward across thesurface. This result emphasizes thephysical interpretation ofeach term inEq.(8—127). Inparticular, thesecond term evidently represents therate ofﬂow ofmass away from anypoint. Conversely, bystarting with theself-evident equation (8—131) andworking backwards, wehave anindependent derivation ofEq.(8—127). Equations analogous toEqs. (8—126), (8—127), (8—129), and (8—131) apply tothedensity, velocity, andrate ofﬂow ofanyphysical quantity. Anequation oftheform (8-12?) applies, forexample, totheﬂowofelectric charge, ifpisthecharge density andpvtheelectric current density. 320 THEMECHANICS orCONTINUOUS MEDIA [CHAP. 8 L. v/K\@ it FIG. 8-7. Meaning ofnonzero curlv.(a)Avortex. (b)Atransverse velocity gradient. Thecurlofthevelocity Vxvisaconcept which isuseful indescribing ﬂuid ﬂow. Tounderstand itsmeaning, wecompute theintegral ofthe normal component ofcurlvacross asurface Sbounded byacurve C’.By Stokes’ theorem (3—117), thisis f/n»-(v ><v)as=[Cv-dr, (s-132) S where thelineintegral istaken around C’inthepositive sense relative to thenormal n,aspreviously deﬁned. Ifthecurve C’S111‘I‘011I).dS avortex in theﬂuid, sothatvisparallel todraround C’(Fig. 8-7), then thelinein- tegral ontheright ispositive andmeasures, inasense, therateatwhich theﬂuid iswhirling around thevortex. Thus VXvisasortofmeasure oftherate ofrotation oftheﬂuid perunit area; hence thename curl v. Curl vhasanonzero value intheneighborhood ofavortex intheﬂuid. Curl vmay alsobenonzero, however, inregions where there isnovortex, that is,where theﬂuid does notactually circle apoint, provided there is atransverse velocity gradient. Figure 8—7illustrates thetwocases. In each case, thelineintegral ofvcounterclockwise aroimd thecircle Cwill have apositive value. Ifthecurlofviszeroeverywhere inamoving ﬂuid, theﬂow issaid tobeirrotational. Irrotational ﬂow isimportant chieﬂy because itpresents fairly simple mathematical problems. Ifatanypoint Vxv=0,then anelement ofﬂuid atthat point will have nonet angular velocity about that point, although itsshape and sizemay be changing. Wearrive atamore precise meaning ofcurl vbyintroducing aco- ordinate system rotating with angular velocity av.Ifv’designates the velocity oftheﬂuid relative totherotating system, then byEq.(7-33), v=v'-I-wxr, where risavector from theaxisofrotation (whose location does notmat- terinthisdiscussion) toapoint intheﬂuid. Curl visnow 8-71 EQUATIONS orMOTION sonANIDEAL FLUID 321 VXV=VXv'—|—VX(wXr) ==VXv'+wV-1'—w-V1‘ ==VXv'—|—3w—w =VXV'-I-2w, where thesecond linefollows from Eq.(3-35) forthetriple cross product, andthethird linebydirect calculation ofthecomponents inthesecond andthird terms. Ifweset ' w=avxv, (s-133) then VXv’=0. (8—134) Thus ifVXvas0atapoint P,then inacoordinate system rotating with angular velocity w=%VXv,theﬂuid ﬂowisirrotational atthepoint P. Wemay therefore interpret %VXvastheangular velocity oftheﬂuid near anypoint. IfVXvisconstant, then itispossible tointroduce a rotating coordinate system inwhich theﬂow isirrotational everywhere. 8-7Equation ofmotion foranideal ﬂuid. Fortheremainder ofthis chapter, except inthelastsection, weshall consider themotion ofanideal ﬂuid, thatis,one'inwhich there arenoshearing stresses, even when the ﬂuid isinmotion. Thestress within anideal ﬂuid consists inapressure p alone. This isamuch greater restriction inthecase ofmoving ﬂuids than inthecase ofﬂuids inequilibrium (Section 5-11). Aﬂuid, bydeﬁnition, supports noshearing stress when inequilibrium, butallﬂuids have some viscosity andtherefore there arealways some shearing stresses between layers ofﬂuid inrelative motion. Anideal ﬂuid would have noviscosity, andourresults forideal ﬂuids willtherefore apply only when theviscosity isnegligible. Letussuppose that, inaddition tothepressure, theﬂuid isacted on byabody force ofdensity fperunit volume, sothat thebody force acting onavolume element 6Vof-ﬂuid isf6V. Weneed, then, tocalculate the force density duetopressure. Letusconsider avolume element 6V= 6:08y62intheform ofarectangular box(Fig. 8-8). The force dueto pressure ontheleftface oftheboxisp5y6z,andactsinthea:-direction. The force duetopressure ontheright face oftheboxisalsop6y62,and acts intheopposite direction. Hence thenet2:-component offorce 6F, ontheboxdepends upon thedifference inpressure between theleftand right faces ofthebox: 517', =(— g5.2:)6y8.2. (8-135) 322 THEMECHANICS OFCONTINUOUS MEDIA [CHAP- 8 1. .433’ _-.->0 M P Q0N \\"3\\.3\\0}_____| >\III|II Q’|‘§|I 5.18 II; FIG. 8-8. Force onavolume element duetopressure. Asimilar expression may bederived forthecomponents offorce inthe y-and2-directions. Thetotal force ontheﬂuid intheboxduetopressure isthen__-n_-an.<12)51?“) 1890 Jay kaz ‘W =——Vpav. (8-136) Theforce density perunit volume duetopressure istherefore —Vp. Thisresult wasalsoobtained inSection 5-11 [Eq.(5—172)]. Wecannow write theequation ofmotion foravolume element 5V ofﬂuid: paV%=rav-Vpav. (8-137) This equation isusually written intheform p% —l-Vp=f. (8—138) Bymaking useoftherelation (8—113), wemay rewrite thisinterms of derivatives ataﬁxed point: 1 6v 1 f Ft‘-i-V‘VV-i-3V1): I3-v where f/pisthebody force perunit mass. This isEuler’s equation of motion foramoving ﬂuid. Ifthedensity pdepends only onthepressure p,weshall calltheﬂuid homogeneous. This deﬁnition does notimply that thedensity isuniform. Anincompressible ﬂuid ishomogeneous ifitsdensity isuniform. Acom- 8-8] CONSERVATION LAWS FOR FLUID MOTION 323 pressible ﬂuid ofuniform chemical composition anduniform temperature throughout ishomogeneous. When aﬂuid expands orcontracts under the inﬂuence ofpressure changes, work isdone byorontheﬂuid, andpart of thiswork may appear intheform ofheat. Ifthechanges indensity occur suﬁiciently slowly sothat there isadequate time forheat ﬂow tomaintain thetemperature uniform throughout theﬂuid, theﬂuid may beconsidered homogeneous within themeaning ofourdeﬁnition. Therelation between density andpressure isthen determined bytheequation ofstate ofthe ﬂuid orbyitsisothermal bulk modulus (Section 5-11). Insome cases, changes indensity occur sorapidly that there isnotime foranyappreciable ﬂow ofheat. Insuch cases theﬂuid may alsobeconsidered homogeneous, andtheadiabatic relation between density andpressure ortheadiabatic bulk modulus should beused. Incases between these twoextremes, the density willdepend notonly onpressure, butalsoontemperature, which, inturn, depends upon therate ofheat ﬂow between parts oftheﬂuid at different temperatures. Inahomogeneous ﬂuid, there arefour unknown functions tobede- termined ateach point inspace andtime, thethree components ofvelocity v,andthepressure p.Wehave, correspondingly, fourdifferential equations tosolve, thethree components ofthevector equation ofmotion (8—139), andtheequation ofcontinuity (8—127). The only other quantities ap- pearing inthese equations arethebody force, which isassumed tobe given, and thedensity p,which canbeexpressed asafunction ofthe pressure. Ofcourse, Eqs. (8—139) and(8-12?) have atremendous variety ofsolutions. Inaspeciﬁc problem wewould need toknow theconditions attheboundary oftheregion inwhich theﬂuid ismoving andthevalues ofthefunctions vandpatsome initial instant. Inthefollowing sections, weshall conﬁne ourattention tohomogeneous ﬂuids. Intheintermediate case mentioned attheendofthelastparagraph, where theﬂuid isin- homogeneous andthedensity depends onboth pressure andtemperature, wehave anadditional unknown function, thetemperature, andwewill need anadditional equation determined bythelawofheat ﬂow. Weshall notconsider this case, although itisavery important oneinmany problems. 8-8Conservation lawsforﬂuid motion. Inasmuch asthelawsofﬂuid motion arederived from Newton’s laws ofmotion, wemay expect that appropriate generalizations oftheconservation laws ofmomentum, energy, andangular momentum alsohold forﬂuid motion. Wehave already had anexample ofaconservation lawforﬂuid motion, namely, theequation ofcontinuity [Eq. (8-127) or(8—131)], which expresses thelawofcon- servation ofmass. Mass isconserved alsoinparticle mechanics, butwe didnotﬁnditnecessary towrite anequation expressing thisfact.l 324 THEMECHANICS orCONTINUOUS MEDIA [CHAP. 8 Aconservation lawinﬂuid mechanics may bewritten inmany equiva- lentforms. Itwillbeinstructive tostudy some ofthese inorder togeta clearer idea ofthephysical meaning ofthevarious mathematical expres- sions involved. Letpbethedensity ofany physical quantity: mass, momentum, energy, orangular momentum. Then thesimplest form of theconservation lawforthis quantity willbeequation (8—125), which states that theamount ofthisquantity inanelement 6Vofﬂuid remains constant. Ifthequantity inquestion isbeing produced atarate Qper unit volume, then Eq.(8-125) should begeneralized: %(pav)=Qav. (s-140) This isoften called aconservation lawforthequantity p.Itstates that thisquantity isappearing intheﬂuid atarateQperunitvolume, ordisap- pearing ifQisnegative. Inthesense inwhich wehave used theterm in Chapter 4,this should notbecalled aconservation law except when Q=O.Byaderivation exactly likethat which ledtoEq.(8—127), we canrewrite Eq.(8—140) asapartial differential equation: t‘§,§+v-on=Q- <8-141) This isprobably themost useful form ofconservation law. Themeaning oftheterms inEq.(8-141) isbrought outbyintegrating each term over a ﬁxed volume Vandusing Gauss’ theorem,* asinthederivation ofEq. (8—131): %fV[[pdv+fSfn-vpds=fI[/Qdv. (s-142) According tothediscussion preceding Eq. (8—129), this equation states that therate ofincrease ofthequantity within V,plus therate ofﬂow outward across theboundary S,equals therate ofappearance dueto sources Within V.Another form oftheconservation lawwhich issome- times useful isobtained bysumming equation (8—140) over avolume V moving with theﬂuid: Zgip av)=$2,» av=Zoav. (8-143) *Ifpisavector, asinthecase oflinear orangular momentum density, then a generalized form ofGauss’ theorem [mentioned inSection 5-11 inconnection with Eq.(5—178)] must beused. 8-8] CONSERVATION LAWS FOR FLUID MOTION 325 Ifwepass tothelimit 5V—>0,thesummations become integrations: 5%/Z/pdV=[![QdV. (s-144) The surface integral which appears intheleftmember ofEq. (8—142) does notappear inEq.(8-144); since thevolume Vmoves with theﬂuid, there isnoﬂow across itsboundary. Since Eqs. (8-1/10), (8—141), (8—142), and(8-144) areallequivalent, itissuﬂicient toderive aconservation law inanyoneofthese forms. Theothers then follow. Usually itiseasiest to derive anequation oftheform (8-140), starting with theequation of motion intheform (8—138). Wecanalsostart with Eq.(8—139) andde- riveaconservation equation intheform (8-141), butabitmore manipula- tionisusually required. Inorder toderive aconservation lawforlinear momentum, weﬁrst note that themomentum inavolume element 8Vispv8V. The mo- mentum density perunit volume istherefore pv,andthisquantity will play theroleplayed bypinthediscussion ofthepreceding paragraph. In order toobtain anequation analogous toEq.(8—140), westart with the equation ofmotion intheform (8—138), which refers toapoint moving with theﬂuid, andmultiply through bythevolume 8Vofasmall ﬂuid element: p5V%‘Zr +Vp6V=f5V. (8-145) Since p6V=6misconstant, wemay include itinthetime derivative: gig»av)=(r-Vp)av. (s-146) Themomentum ofaﬂuid element, lmlike itsmass, isnot,ingeneral, con- stant. This equation states that thetime rate ofchange ofmomentum ofamoving ﬂuid element isequal tothebody force plus theforce dueto pressure acting upon it.The quantity f—Vphere plays theroleofQ inthepreceding general discussion. Equation (8—146) canberewritten inanyoftheforms (8—141), (S-142), and(8—144). Forexample, wemay write itintheform (8—144): %f;fpvdv=f!fidv_f!fvpdv. (8-147) Wecannow apply thegeneralized form ofGauss’ theorem [Eq. (5-178)] tothesecond term ontheright, toobtain4 l < 1 l J < 1 l i 1 326 THEMECHANICS orCONTINUOUS MEDIA [CHAP. 8 %/1!}-pvdV=II![fdV+[gf—npdS, <8-148) where Sisthesurface bounding V. This equation states that thetime rate ofchange ofthetotal linear momentum inavolume Vofmoving ﬂuid isequal tothetotal external force acting onit.This result isanimmediate generalization ofthelinear momentum theorem (4-7) forasystem ofparticles. The internal forces, inthecase ofaﬂuid, arerepresented bythepressure within theﬂuid. By theapplication ofGauss’ theorem, wehave eliminated thepressure within thevolume V,leaving only theexternal pressure across thesurface ofV. Itmay beasked how wehave managed toeliminate theinternal forces without making explicit useofNewton’s third law, since Eq. (S-138), from which westarted, isanexpression only ofNewton’s ﬁrst twolaws. The answer isthat theconcept ofpressure itself contains Newton’s third lawimplicitly, since theforce duetopressure exerted from lefttoright across anysurface element isequal andopposite totheforce exerted from right toleftacross thesame surface element. Furthermore, thepoints of application ofthese twoforces arethesame, namely, atthesurface ele- ment. Both forces necessarily have thesame lineofaction, andthere isno distinction between theweak andstrong forms ofNewton’s third law. Theinternal pressures willtherefore alsobeexpected tocancel outinthe equation forthetime rateofchange ofangular momentum. Asimilar remark applies totheforces duetoanykindofstresses inaﬂuidorasolid; Newton’s third lawinstrong form isimplicitly contained intheconcept of stress. Equations representing theconservation ofangular momentum, analo- gous term byterm with Eqs. (8—140) through (8—144), canbederived by taking thecross product ofthevector rwith either Eq.(8-138) or(8—139), andsuitably manipulating theterms. Thevector rishere thevector from theorigin about which moments aretobecomputed toanypoint inthe moving ﬂuid orinspace. This development isleftasanexercise. The lawofconservation ofangular momentum isresponsible forthevortices formed when aliquid ﬂows outthrough asmall hole inthebottom ofa tank. The only body force here isgravity, which exerts notorque about thehole, anditcanbeshown that ifthepressure isconstant, ordepends only onvertical depth, there isnonetvertical component oftorque across anyclosed surface duetopressure. Therefore theangular momentum of anypart oftheﬂuid remains constant. Ifaﬂuid element hasanyangular momentum atallinitially, when itissome distance from thehole, itsangu- larvelocity willhave toincrease ininverse proportion tothesquare ofits distance from thehole inorder foritsangular momentum toremain con- stant asitapproaches thehole. 8-8] CONSERVATION LAWS FORFLUID MOTION 327 Inorder toderive aconservation equation fortheenergy, wetake the dotproduct ofvwith Eq.(8—146), toobtain %(ipt2 av)=v~(f_Vp)av. (s-149) This istheenergy theorem intheform (8—140). Inplace ofthedensity p, wehave here thekinetic energy density %pv2. The rate ofproduction of kinetic energy perunit volume is Q=v-(f—Vp). (8—l50) Inanalogy with ourprocedure inparticle mechanics, weshall now try todeﬁne additional forms ofenergy soastoinclude asmuch aspossible oftheright member ofEq.(8—149) under thetime derivative ontheleft. Wecanseehow torewrite thesecond term ontheright bymaking useof Eqs. (8—113) and(8—116): d _@ d6V ZlZ(p'W)“ dz‘W+p dr =%av+v~Vpav+pV-vav, (s-151) sothat —v-Vp av=-%(pav)+';_fav+pV-Vav. (s-152) Letusnow assume that thebody force fisagravitational force: f=Pg=PV9, (8—153) where 9isthegravitational potential [Eq. (6—16)], i.e.,thenegative poten- tialenergy perunit mass duetogravitation. The ﬁrst term ontheright inEq.(8—149) isthen d 8v-fav=(v-vg)p av=(7?-£)p 6V _i _§ _-dt(pg6V) pat 6V, (8154) since p6V=6misconstant. With thehelp ofEqs. (8—152) and(8—154), Eq.(8-149) canberewritten: d 8 8E[(%Pv2 +1»—/>9)W]=(alt)—/1;?) 6V+rv-v 6V-(8—155) The pressure phere plays theroleofapotential energy density whosel 328 THE MECHANICS orCONTINUOUS MEDIA [cnA1>. 8 negative gradient gives theforce density duetopressure [Eq. (8—136)]. The time rate ofchange ofkinetic energy plus gravitational potential energy plus potential energy duetopressure isequal totheexpression on theright. Ordinarily, thegravitational ﬁeld ataﬁxed point inspace will not change with time (except perhaps inapplications tomotions ofgasclouds inastronomical problems). Ifthepressure atagiven point inspace is constant also, then theﬁrstterm ontheright vanishes. What isthesig- niﬁcance ofthesecond term‘? For anincompressible ﬂuid, V-v=O, and thesecond term would vanish also. Wetherefore suspect that it represents energy associated with compression andexpansion oftheﬂuid element 6V. Letuscheck thishypothesis bycalculating thework done in changing thevolume oftheelement 6V. Thework dWdone bytheﬂuid element 6V,through thepressure which itexerts onthesurrounding ﬂuid when itexpands byanamount d6V,is dW=pd5V. (8—156) Therateatwhich energy issupplied bytheexpansion oftheﬂuid element is,byEq.(8—116), dW d5V W =PT =pv'V 5V, which isjust thelastterm inEq.(8—155). Sofar,allourconservation equations arevalid foranyproblem involving ideal ﬂuids. Ifwerestrict ourselves tohomogeneous ﬂuids, that is,ﬂuids whose density depends only onthepressure, wecandeﬁne apotential energy associated with theex- pansion and contraction oftheﬂuid element 6V. Weshall deﬁne the potential energy u6montheﬂuid element 6Vasthenegative work done through itspressure onthesurroimding ﬂuid when thepressure changes from astandard pressure pgtoanypressure p.Thepotential energy per unit mass uwillthen beafunction ofp: P u8m=—/ pd6V. (8—158) P0 Thevolume 6V=6m/pisafunction ofpressure, andwemay rewrite this invarious forms: Pd ,,=/rm Po P2 =[P3‘fldp (8-159)1»./>2div PP =~01[MP3 p’ 8-9] STEADY 1-mow 329 Where thelaststep makes useofthedeﬁnition ofthebulk modulus [Eq. (5—116)]. The time rate ofchange ofuis,byEqs. (8—158) or(8—159) and(8—116), d6 d6VLtltlnl =—p7t——- =——-pV-v av. (s-160) Wecannow include thelastterm ontheright inEq.(S-155) under the time derivative ontheleft: §,'»’;[<%/M +p—ps+Pu)W1=-P?)W<8-161) Theinterpretation ofthisequation isclear from thepreceding discussion. Itcanberewritten inanyoftheforms (8—141), (8—l42), and (S-144). IfpandQareconstant atanyﬁxed point inspace, then thetotal kinetic plus potential energy ofaﬂuid element remains constant asitmoves along. Itisconvenient todivide by6m=p6Vinorder toeliminate refer- ence tothevolume element: dv2 p >_16p 89 a(§—l-3-9-l—u —;E"—5Z' (8-162) This isBernoulli’s theorem. Theterm 69/6t ispractically always zero; wehave kept itmerely tomake clear themeaning oftheterm (1/p)(6p/6t), which plays asimilar roleandisnotalways zero. When both terms onthe right arezero, asinthecase ofsteady ﬂow, wehave, forapoint moving along with theﬂuid, 222 p5+B~—9+u=aconstant. (8—163) Other things being equal, that isifu,9,andpareconstant, thepressure of amoving ﬂuid decreases asthevelocity increases. Foranincompressible ﬂuid, panduarenecessarily constant. Theconservation laws oflinear andangular momentum apply notonly toideal ﬂuids, butalso, when suitably formulated, toviscous ﬂuids and even tosolids, inview oftheremarks made above regarding Newton’s third lawandtheconcept ofstress. The lawofconservation ofenergy (8—162) willnotapply, however, toviscous ﬂuids, since theviscosity is duetoaninternal friction which results inalossofkinetic andpotential energies, unless conversion ofmechanical toheat energy byviscous friction isincluded inthelaw. [Equation (8—155) applies inanycase.] 8-9Steady ﬂow. Bysteady ﬂow ofaﬂuid wemean amotion ofthe ﬂuid inwhich allquantities associated with theﬂuid, velocity, density, pressure, force density, etc., areconstant intime atanygiven point in 330 THE MECHANICS OFCONTINUOUS MEDIA [crnua 8 space. Forsteady ﬂow, allpartial derivatives with respect totime canbe setequal tozero. The total time derivative, which designates thetime rateofchange ofaquantity relative toapoint moving with theﬂuid, will notingeneral bezero, but, byEq.(8—113)l, willbe %=v-V. (s-164) The path traced outbyanyﬂuid element asitmoves along iscalled a streamline. Astreamline isalinewhich isparallel ateach point (as,y,z) tothevelocity v(:z:,y,z)atthat point. Theentire space within which the ﬂuid isﬂowing canbeﬁlled with streamlines such that through each point there passes oneand only onestreamline. Ifweintroduce along any streamline acoordinate swhich represents thedistance measured along thestreamline from anyﬁxed point, wecanregard anyquantity associated with theﬂuid asafunction ofsalong thestreamline. The component of thesymbol Valong thestreamline atanypoint isd/ds, asweseeifwe choose acoordinate system whose :v—axis isdirected along thestreamline atthat point. Equation (8—164) cantherefore berewritten: d d This equation isalsoevident from thefactthatv=ds/dt. Forexample, Eq.(8—162), inthecaseofsteady ﬂow, canbewritten: , T 5%? +3-9+U.)=0. (s-166) The quantity inparentheses istherefore constant along astreamline. The equation ofcontinuity (8—127) inthecase ofsteady ﬂow becomes V-(pv) =0. (8—167) Ifweintegrate this equation over aﬁxed volume V,and apply Gauss’ theorem, wehave [[11-(pv) as=0, (8-168) S where Sistheclosed surface bounding V.This equation simply states that thetotal mass ﬂowing outofanyclosed surface iszero. Ifweconsider allthestreamlines which pass through any (open) sur- faceS,these streamlines form atube, called atubeofﬂow (Fig. 8—9). The walls ofatube ofﬂow areeverywhere parallel tothestreamlines, sothat noﬂuid enters orleaves it.Asurface Swhich isdrawn everywhere per— pendicular tothestreamlines andthrough which passes each streamline in 8-9] srmnr FLOW 331 F10. 8-9. Atube offlow. atube ofﬂow, willbecalled a.cross sectio/n ofthetube. Ifweapply Eq. (8—168) totheclosed surface bounded bythewalls ofatube ofﬂow and twocross sections S1andS2,then since nisperpendicular tovover the walls ofthetube, andnisparallel orantiparallel tovover thecross sec~ tions, wehave /fpvds -ffpvds =0, (s-169) O1‘ S1 S’ ffpvas=I=aconstant, (s-170) S where Sisanycross section along agiven tube ofﬂow. Theconstant I iscalled theﬂuid current through thetube. The energy conservation equation (8—l61), when rewritten intheform (8—141), becomes, inthecaseofsteady ﬂow, V-Ktpvz +P—P9+pu)v]=0- (8—171) This equation hasthesame form asEq.(8—167), andwecanconclude in thesame way that theenergy current isthesame through anycross sec- tionSofatube ofﬂow: fI(%pv2 -1-p—pg+pu)v dS=a.constant. (S-172) s Thisresult isclosely related toEq.(8—166). Iftheﬂowisnotonlysteady, butalsoirrotational, then VXv=0 (S-173) everywhere. This equation isanalogous inform toEq.(3—189) foraconservative force, andwecanproceed asinSection 3-12 tohowthatifEq.(8-173)holds, it 332 THE MECHANICS orCONTINUOUS MEDIA [cnAi>. 8 ispossible todeﬁne avelocity potential function ¢(z,y,z)bytheequation ¢(r)=/'v-dr, (s-174) 1's where r,isanyﬁxed point. Thevelocity atanypoint willthen be v=V¢. (8-17 5) Substituting thisinEq.(8—167), wehave anequation tobesolved for¢: V-(pV¢) =0. (S-176) Inthecases usually studied, theﬂuid canbeconsidered incompressible, andthis becomes V2¢ =0. (8—177) This equation isidentical inform with Laplace’s equation (6—35) forthegravita- tional potential inempty space. Hence thetechniques ofpotential theory may beused tosolve problems involving irrotational ﬂow ofanincompressible ﬂuid. 8-10 Sound waves. Letusassume aﬂuid atrest with pressure p0, density po,inequilibrium under theaction ofabody force fo,constant in time. Equation (S-139) then becomes 1 fo— =-—- 8-178P0vpo Po ( ) Wemay note that thisequation agrees with Eq.(5—172) deduced inSec- tion 5-11 foraﬂuid inequilibrium. Letusnow suppose that theﬂuid is subject toasmall disturbance, sothat thepressure anddensity atany point become P=Po+P’, (8—179) P=P0+P’, (8—180) where p’<<pandp’<<p.Weassume that theresulting velocity vand itsspace andtime derivatives areeverywhere very small. Ifwesubsti- tute Eqs. (8—179) and (8—180) intheequation ofmotion (8—139), and neglect higher powers than theﬁrstofp’,p’,vandtheir derivatives, mak- inguseofEq.(8-178), weobtain lg=—F10Vp’. (8—181) Making asimilar substitution inEq.(8—127), weobtain I 65%=——p0V-v -—v-Vpo. '(S-182) 8-10] SOUND wavns 333 Letusassume that theequilibrium density poisuniform, ornearly so,so thatVpoiszeroorverysmall, andthesecond term canbeneglected. The pressure increment p’anddensity increment p’arerelated bythe bulk modulus according toEq.(5-183): 5=71- s-183 P0B () This equation maybeused toeliminate either p’orp’from Eqs. (8—181) and(8-182). Letuseliminate p’from Eq.(8—182): I ap -— I O _ Equations (8-181) and(8—184) arethefundamental differential equations forsound waves. Theanalogy with theform (S-101) forone-dimensional waves isapparent. Here again wehave twoquantities, p’andv,such that thetime derivative ofeither isproportional tothespace derivatives ofthe other. Infact, ifv=iv,andif11,,andp’arefunctions ofxalone, then Eqs. (8—181) and(8-184) reduce toEqs. (8-101). Wemay proceed, inanalogy with thediscussion inSection 8-5, to eliminate either vorp’from these equations. Inorder toeliminate v, wetake thedivergence ofEq.(8—181) andinterchange theorder ofdiffer- entiation, again assuming ponearly uniform: a 1(T,(V-v)=-;)—(;V2p'. (s-185) Wenow differentiate Eq.(8-184) with respect tot,andsubstitute from Eq.(8—185): 1a”'V21)’_E5T1;=0, (8-186) c=(3)1/2 - (8-187) This isthethree-dimensional wave equation, asweshall show presently. Formula (8-187) forthespeed ofsound waves wasﬁrst derived byIsaac Newton, and applies either toliquids orgases. Forgases, Newton as- sumed that theisothermal bulk modulus B=pshould beused, butEq. (8-187) does notthen agree with theexperimental values forthespeed of ound. The sound vibrations aresorapid that they should betreated as adiabatic, andtheadiabatic bulk modulus B='Ypshould beused, where 'Yistheratio ofspeciﬁc heat atconstant pressure tothat atconstantwhere 334 THEMECHANICS orCONTINUOUS MEDIA , [CHAP. 8 volume.* Formula (8—187) then agrees with theexperimental values ofc. Ifweeliminate p’byasimilar process, weobtain awave equation forv: 182v Inderiving Eq.(8—188), itisnecessary tousethefactthat VXv=0.It follows from Eq.(8—181) that VXvisinanycaseindependent oftime, so that thetime-dependent part ofvwhich ispresent inasound wave isir- rotational. [We could addtothesound wave asmall steady ﬂow with VXV;é0,without violating Eqs. (8—181) and(8—182).] Inorder toshow that Eq.(8—186) leads tosound waves traveling with speed c,wenote ﬁrstthat ifp’isafunction ofxandtalone, Eq.(8—186) becomes a’'1a’'-(,-ml;-g%=0. (s-189) This isofthesame form astheone-dimensional wave equation (8-6), and therefore hassolutions oftheform p’=f(a:—ct). (8—190) Thisiscalled aplane wave, foratanytimetthephase x—ctandthepres- sure p’areconstant along anyplane (:1:=aconstant) parallel totheya- plane. Aplane wave traveling inthedirection oftheunit vector nwill begiven by Vp’=f(n-r —ct), (8—191) where ristheposition vector from theorigin toanypoint inspace. To seethat thisisawave inthedirection n,werotate thecoordinate system until the:7:-axis liesinthisdirection, inwhich case Eq.(8-191) reduces toEq.(8—190). The planes f=aconstant, atanytime t,arenow per- pendicular ton,andtravel inthedirection ofnwith velocity c.Wecan seefrom theargument just given that thesolution (8-191) must satisfy Eq.(8-186), orwemay verify thisbydirect computation, foranycoor- dinate system: Vp' =géV5=gn, (8—192) where £5=n-r—ct, (8—193) *Millikan, Roller, and Watson, Mechanics, Molecular Physics, Heat, and Sound. Boston: Ginn andCo., 1937. (Pages 157, 276.) 8-10] sormn WAVES 335 and, similarly, d2 d2 d2 V21)’ =HE‘-in-V5 =éfhll. =(Fir 2/ 2 2 2852-=%(5':-9%) =02g. (8-195) sothat Eq.(8-186) issatisﬁed, nomatter what thefunction f(.§)may be. Equation (8-188) willalsohave plane wave solutions: v=h(n’-r —ct), (8-196) corresponding towaves traveling inthedirection n’with velocity c,where hisavector function of5’=n’-r —ct.Toanygiven pressure wave of theform (8-191) willcorrespond avelocity wave oftheform (8—196), re- lated toitbyEqs. (8-181) and(8—182). Ifwecalculate 6v/6t from Eq. (8—196), andVp’from Eq.(S-191), andsubstitute inEq.(8-181), wewill have db__n 511'. _Tr'<Bp..>1/2 dz <8197) Equation (8-197) must hold atallpoints ratalltimes t.Theright mem- berofthisequation isafunction of5andisconstant foraconstant 5. Consequently, theleftmember must beconstant when Eisconstant, and must beafunction only of5,which implies that 5’=5(oratleast that 5'isafunction ofE),andhence n’=n.This isobvious physically, that thevelocity wave must travel inthesame direction asthepressure wave. Wecannow set5'=E,andsolve Eq.(8-197) forh: nh_W f, (8-198) where theadditive constant iszero, since both p’andvarezeroinaregion where there isnodisturbance. Equations (8—198), (8—196), and (8-190) imply that foraplane sound wave traveling inthedirection n,thepressure increment andvelocity arerelated bytheequation Iv=MW n, (s-199) where v,ofcourse, ishere thevelocity ofaﬂuid particle, notthat ofthe wave, which isan.Thevelocity oftheﬂuid particles isalong thedirection ofpropagation ofthesound wave, sothat sound waves inaﬂuid arelongi- tudinal. This isaconsequence ofthefactthat theﬂuid willnotsupport ashearing stress, andisnottrue ofsound waves inasolid, which may be either longitudinal ortransverse. 336 THE MECHANICS OFCONTINUOUS MEDIA [cmu>. 8 Aplane wave oscillating harmonically intime with angular frequency w may bewritten intheform p’=Acos(k-r-wt)=ReAe’(k""‘"’), (8-200) where k,thewave vector, isgiven by k=$11. (s-201) Ifweconsider asurface perpendicular tonwhich moves back and forth with theﬂuid asthewave goes by,thework done bythepressure across thissurface inthedirection ofthepressure is,perunit area perunit time, P=pv. (8—202) Ifvoscillates with average value zero, then since p=po+p’,where pois constant, theaverage power is Pay=<1/v>..= <8-203)<p..B>1/2 where wehave made useofEq.(8—199). This gives theamount ofenergy perunit area persecond traveling inthedirection n. Thethree-dimensional wave equation (8-186) hasmany other solutions corresponding towaves ofvarious forms whose wave fronts (surfaces of constant phase) areofvarious shapes, andtraveling invarious directions. Asanexample, weconsider aspherical wave traveling outfrom theorigin. Therateofenergy ﬂowisproportional top’2(asmall portion ofaspherical wave may beconsidered plane), andweexpect that theenergy ﬂow per unit area must falloffinversely asthesquare ofthedistance, bythe energy conservation law. Therefore p’should beinversely proportional to thedistance rfrom theorigin. Wearehence ledtotryawave oftheform p’=%-yo»-ct). (s-204) This will represent awave ofarbitrary time-dependence, whose wave fronts, £=r—ct=aconstant, arespheres expanding with thevelocity c.Itcanreadily beveriﬁed bydirect computation, using either rectangular coordinates, orusing spherical coordinates with thehelp ofEq. (3—124), that thesolution (8—204) satisﬁes thewave equation (8-186). Aslight diﬂiculty isencountered with theabove development ifweattempt to apply toasound wave theexpressions forenergy ﬂowandmass ﬂowdeveloped inthetwopreceding sections. Therate ofﬂow ofmass perunit area persecond, 8-11] NORMAL VIBRATIONS orFLUID INARECTANGULAR BOX 337 byEqs. (8—199), (8—180), and(8—183), is I I_ 2P'°"'”°(1+B)<p0B>1/2”‘ Weshould expect that pvwould beanoscillating quantity whose average value iszero forasound wave, since there should benonetﬂow ofﬂuid. Ifweaverage theabove expression, wehave 1/2 (Pv)av = (<p’2>av +B<p'>av)n! sothat there isasmall netﬂow ofﬂuid inthedirection ofthewave, unless ,2<z>’>...=- <8-205) IfEq.(8—205) holds, sothat there isnonetﬂow ofﬂuid, then itcanbeshown that, tosecond-order terms inp’andv,theenergy current density given byEq. (8—161) is,ontheaverage, forasound wave, /2 ((%Pv2 +11—P9+pu)v)=v = n, (8—206) inagreement with Eq.(8-203). When approximations aremade intheequa- tions ofmotion, wemay expect thatthesolutions will atisfy theconervation laws only tothesame degree ofapproximation. Byadding second-order (or higher) terms like(8—205) toaﬁrst-order solution, wecanofcourse satisfy the conservation laws tosecond-order terms (orhigher). 8-11 Nonnal vibrations ofﬂuid inarectangular box. The problem of thevibrations ofaﬂuid conﬁned within arigid boxisofinterest notonly because ofitsapplications toacoustical problems, butalso because the methods used canbeapplied toproblems inelectromagnetic vibrations, vibrations ofelastic solids, wave mechanics, andallphenomena inphysics which aredescribed bywave equations. Inthissection, weconsider a ﬂuid conﬁned toarectangular boxofdimensions L,L,,L,. Weproceed asinthesolution oftheone-dimensional wave equation in Section 8-2. Weﬁrstassume asolution ofEq.(8-186) oftheform P’=U(Iv,1/,z)@(i)- (8—207) Substitution inEq.(8-186) leads totheequation 1 1@126) ﬁV2U =‘xi *8-Z-5-' Again weargue that since theleftsidedepends only onac,y,andz,andthe4 1 J l 1 I 338 THE MECHANICS orCONTINUOUS MEDIA [cHA1>. 8 right sideonly ont,both must beequal toaconstant, which weshall call —w2/c2: d2® Ft? +(.02® =0, 2wU+%U=o(8—209) (8-210) Thesolution ofEq.(8—209) canbewritten: ®= OI‘ (9 where AandBareconstant. oftheform (8—200). WeareAcoswt+Bsinwt, (8-21 1) A6-""", (8-212) Theform (S-212) leads totraveling waves concerned here with standing waves, andwe therefore choose theform (8—211). Inorder tosolve Eq.(8-210), weagain usethemethod ofseparation ofvariables, andassume that U($,y,Z)=X(1>)Y(y)Z(Z)- (8-213) Substitution inEq.(8-210) leads totheequation 1d2X 1d2Y 1d2Z 0:2 - Ydue?T7 dyz+2 dz”=_c2. 6-214) This canhold forallx,y,zonly ifeach term ontheleftisconstant. We shall callthese constants —k§,, —k§, —kf, sothat d2X d2Y d2ZT+rix=0,W+sir=0,3?-+ kiz=0,(s-215)$2 where k2+2 2_ (iik,,+I0;_C,- (8-216) Thesolutions ofEqs. (8—215) inwhich weareinterested are X=C,cosk,x+D,sink,,a:, Y=0,,coskyy-l-D1,sinkyy, (8—217) Z=C’,coskzz—l—D,sinkzz. Ifwechoose complex exponential solutions forX,Y,Z,and(9,wearrive at thetraveling wave solution (8—200), where lc,,,kg,k,arethecomponents ofthewave vector k. 8-11] NORMAL VIBRATIONS orFLUID INARECTANGULAR BOX 339 Wemust now determine theappropriate boundary conditions tobe applied atthewalls ofthebox, which weshall take tobethesixplanes :0=O,av=L,,,y=0,y=Ly,z=0,z=L,.Thecondition isevidently that thecomponent ofvelocity perpendicular tothewall must vanish at thewall. Atthewall :1:=0,forexample, 11,,must vanish. According to Eq.(8-181)’ n__1n.at"po6x (8_218) Wesubstitute forp’from Eqs. (8—207), (8-211), (8—213), and(8—217): %-Q3=—1% (Acoswt+Bsinwt)(—C, sink,x+D,cos70,1). (3-219)Integrating, wehave v,==—kilz (Asinwt-Bcoswt)(—C, sinkzx+D,coskzx) “Po (8-220) plusafunction ofav,y,z,which vanishes, since wearelooking foroscillating solutions. Inorder toensure that v,vanishes atas=0,wemust set D,=0,i.e.,choose thecosine solution forXinEq.(8—217). Thismeans that thepressure p’must oscillate atmaximum amplitude atthewall. This isperhaps obvious physically, andcould have been used instead ofthe condition 1),,=0,which, however, seems more self-evident. Thevelocity component perpendicular toawall must have anode atthewall, andthe pressure must have anantinode. Similarly, thepressure must have an antinode (maximum amplitude ofoscillation) atthewall x=L,,: Icosk,L,=:l=1, (s-221) sothat k,=%'5, z=o,1,2,... (s-222) Byapplying similar considerations tothefour remaining Walls, wecon- clude that D,,=D,=0,and r,=%, m=O,1,2,..., 1! (8—223) r,=’il:, n=0,1,2,.... Foreach choice ofthree integers Z,m,n,there isanormal mode ofvibra- . I 340 THE MECHANICS onCONTINUOUS MEDIA [cnA1>. 8 tionoftheﬂuid inthebox. Thefrequencies ofthenormal modes ofvibra- tionaregiven byEqs. (8—216), (S-222), and(8—223): Z2 "L2 n2 1/2 wzmn =TF6 +F+ ' (8~224) I y Z Thethree integers l,m,ncannot allbezero, forthisgives w=0anddoes notcorrespond toavibration oftheﬂuid. Ifwecombine these results with Eqs. (8—217), (8—213), 8-211), and(8—207), wehave forthenormal mode ofvibration characterized bythenumbers Z,m,n: p’=(Acoswlmnt —|-Bsinwlmnt) coslgcosEl cos%»(8—225)L, L1, L, where wehave suppressed thesuperﬂuous constant C¢C,,C,. The corre- sponding velocities are l1r . .l1rx m1ry mrzv=a Asinw t—Bcosw ts1n——cos————cos— :0 Lxpowlmn ( lmn lmn ) LE Ly L2 ' m1r . l1rx.mvry mrzv=i——— Asmw; t—Bcosw; t)cos——s1n——cos—» u Lllpfiwlmn ( mu mn LI! LU LI mr . hm: m1ry .n-rrzv=i—— Asinw t—Bcosw tcos——cosis1n—- 2 Lzpowlmn ( lmn lmn ) Lt Ly Lz (8—226) These four equations give acomplete description ofthemotion oftheﬂuid foranormal mode ofvibration. Thewalls x=0,x=L1,andthe(l—1) equally spaced parallel planes between them arenodes for22,,and anti- nodes forp’,U”,andvz.Asimilar remark applies tonodal planes parallel totheother Walls. Itwillbeobserved that thenormal frequencies arenot,ingeneral, har- monically related tooneanother, asthey were inthecase ofthevibrating string. If,however, oneofthedimensions, sayL,,,ismuch larger than the other two, sothat theboxbecomes along square pipe, then thelowest frequencies willcorrespond tothecasewhere m=n=0andZisasmall integer, andthese frequencies areharmonically related. Thus, inapipe, theﬁrst fewnormal frequencies above thelowest willbemultiples ofthe lowest frequency. This explains why itispossible togetmusical tones from anorgan pipe, aswellasfrom avibrating string. Ourtreatment here applies only toaclosed organ pipe, andasquare oneatthat. Thetreat- ment ofaclosed circular pipe isnotmuch more difﬁcult than theabove treatment andthegeneral nature oftheresults issimilar. Theopen ended 8-12] SOUND WAVES INPIPES 341 pipe is,however, much more difficult totreat exactly. The diﬂiculty lies inthedetermination oftheboundary condition attheopen end;indeed, nottheleast ofthedifficulties isindeciding just where theboundary is. Asarough approximation, onemay assume that theboundary isaplane surface across theendofthepipe, andthat thissurface isapressure node. Theresults arethen similar tothose fortheclosed pipe, except that ifone endofalong pipe isclosed andoneopen, theﬁrst fewfrequencies above thelowest arealloddmultiples ofthelowest. Thegeneral solution oftheequations forsound vibrations inarectangu- larcavity canbebuilt up,asinthecaseofthevibrating string, byadding normal mode solutions oftheform (8—225) forallnormal modes ofvibra- tion. The constants AandBforeach mode ofvibration canagain be chosen toﬁttheinitial conditions, which inthiscasewillbeaspeciﬁcation ofp’and6p’/dt (orp’andv)atallpoints inthecavity atsome initial instant. Weshall notcarry outthis development here. [Intheabove discussion, wehave omitted thecase Z=m=n=0,which corresponds toaconstant pressure increment p’.Likewise, weomitted steady velocity solutions v(x,y,2)which donotoscillate intime. These solutions would have tobeincluded inorder tobeable toﬁtallinitial conditions.] Forcavities ofother simple shapes, forexample spheres andcylinders, themethod ofseparation ofvariables used intheabove example works, butinthese cases instead ofthevariables ac,y,2,coordinates appropriate totheshape oftheboundary surface must beused, forexample spherical orcylindrical coordinates. Inmost cases, except forafewsimple shapes, themethod ofseparation ofvariables cannot bemade towork. Approxi- mate methods canbeused when theshape isvery close tooneofthesimple shapes whose solution isknown. Otherwise theonly general methods of solution arenumerical methods which usually involve aprohibitive amount oflabor. Itcanbeshown, however, that thegeneral features ofourre- sults forrectangular cavities hold forallshapes; that is,there arenormal modes ofvibration with characteristic frequencies, andthemost general motion isasuperposition ofthese. 8-12 Sound waves inpipes. Aproblem ofconsiderable interest isthe problem ofthepropagation ofsound waves inpipes. Weshall consider apipe whose axisisinthez-direction, andwhose cross section isrectangu- lar,ofdimensions L,L,,. This problem "isthesame asthat ofthepreceding section except that there arenowalls perpendicular tothez-axis. Weshall apply thesame method ofsolution, theonly difference being that theboundary conditions now apply only atthefour walls x=O, 2:=LI,y=0,y=Ly.Consequently, wearerestricted inourchoice of thefunctions X(ac)and Y(y), just asinthepreceding section, byEqs. (8—217), (S-222), and(8—223). There arenorestrictions onourchoice of 342 THEMECHANICS orCONTINUOUS MEDIA [CHAP- 8 solution oftheZ-equation (8—215). Since weareinterested insolutions representing thepropagation ofwaves down thepipe, wechoose theex- ponential form ofsolution forZ: z=e"’°=‘, (s-227) andwechoose thecomplex exponential solution (8—212) for®.Oursolu- tionforp’,then, foragiven choice oftheintegers l,m,is - l1rx mvrp’=ReAe“'°"_“'” cos—cosifL, L1, =Acoshicos-m—ﬂ-2 cos(kzz—wt). (8—228)L, L1, This represents aharmonic wave, traveling inthez-direction down the pipe, whose amplitude varies over thecross section ofthepipe according totheﬁrst twocosine factors. Each choice ofintegers Z,mcorresponds towhat iscalled amade ofpropagation forthepipe. (The choice l=O, m=Oisanallowed choice here.) Foragiven l,mandagiven frequency co, thewave number k,isdetermined byEqs. (8—216), (8—222), and(8—223): The plus sign corresponds toawave traveling inthe—|—z-direction, and conversely. ForZ=m=O,thisisthesame astherelation (8—201) fora wave traveling with velocity cinthez-direction inaﬂuid ﬁlling three- dimensional space. Otherwise, thewave travels with thevelocity w l1rc 2 mvrc 2_1/2 .-<_>_(._)] .C’ |k,| “l QL, (0111, () which isgreater than canddepends onw.There isevidently aminimum fruencyeq [(m)2wzm= E +T; (8"231) below which nopropagation ispossible intheZ,mmode; fork,would be imaginary, andtheexponent inEq.(8—227) would bereal, sothat instead ofawave propagation wewould have anexponential decline inamplitude ofthewave inthez-direction. Note thesimilarity ofthese results tothose obtained inSection 8-4forthediscrete string, where, however, there was anupper rather than alower limit tothefrequency. Since cmdepends on w,weagain have thephenomenon ofdispersion. Awave ofarbitrary 8—13] THE MACH NUMBER 343 shape, which canberesolved into sinusoidally oscillating components of various frequencies w,willbedistorted asittravels along thepipe because each component willhave adifferent velocity. Weleave asanexercise the problem ofcalculating theﬂuid velocity v,andthepower ﬂow, associated with thewave (8—228). Similar results areobtained forpipes ofother than rectangular cross section. Analogous methods and results apply totheproblem ofthe propagation ofelectromagnetic waves down awave guide. This isone reason forourinterest inthepresent problem. 8-13 The Mach number. Suppose wewish toconsider twoproblems inﬂuid ﬂow having geometrically similar boimdaries, butinwhich the dimensions oftheboundaries, ortheﬂuid velocity, density, orcompressi- bility aredifferent. Forexample, wemay wish toinvestigate theﬂow ofa ﬂuid intwopipes having thesame shape butdifferent sizes, orwemay be concerned with theﬂow ofaﬂuid atdifferent velocities through pipes of thesame shape, orwith theﬂow ofﬂuids ofdifferent densities. Wemight beconcerned with therelation between thebehavior ofanairplane andthe behavior ofascale model, orwith thebehavior ofanairplane atdifferent altitudes, where thedensity oftheairisdifferent. Two such problems in- volving boundaries ofthesame shape weshall callsimilar problems. Under what conditions willtwosimilar problems have similar solutions? Inorder tomake thisquestion more precise, letusassume that foreach problem acharacteristic distance soisdeﬁned which determines thegeo- metrical scale oftheproblem. Inthecase ofsimilar pipes, somight bea diameter ofthepipe. Inthecaseofanairplane, somight bethewing span. Wethen deﬁne dimensionless coordinates x’,y’,2'bytheequations xi=33/30; y,=Z//80> Z,=z/s0- Theboundaries fortwosimilar problems willhave identical descriptions in terms ofthedimensionless coordinates x’,y’,z’;only thecharacteristic distance sowillbedifferent. Inasimilar way, letuschoose acharacteristic speed voassociated with theproblem. The speed vomight betheaverage speed ofﬂow ofﬂuid inapipe, orthespeed oftheairplane relative tothe stationary airatadistance from it,orvomight bethemaximum speed of anypart oftheﬂuid relative tothepipe ortheairplane. Inanycase, we suppose that voissochosen that themaximum speed ofanypart ofthe ﬂuid isnotvery much larger than vo.Wenowdeﬁne adimensionless veloc- ityv',andadimensionless time coordinate t’: V’=V/vo, (S-233) t,Z 1)0t/80. 344 THE MECHANICS OFCONTINUOUS MEDIA [CHAP. 8 Wenow saythat twosimilar problems have similar solutions ifthesolu- tions areidentical when expressed interms ofthedimensionless velocity v’ asafunction ofx’,y’,z’,andt’.The ﬂuid ﬂow pattern willthen bethe same inboth problems, differing only inthedistance andtime scales de- termined bysoandvo.Weneed alsotoassume acharacteristic density poandpressure po.Inthecaseoftheairplane, these would bethedensity andpressure oftheundisturbed atmosphere; inthecase ofthepipe, they might betheaverage density andpressure, orthedensity andpressure at oneendofthepipe. Weshall deﬁne adimensionless pressure increment p”asfollows: p”=ti? (s-235) P090 Weshall now assume that thechanges indensity oftheﬂuid aresmall enough sothat wecanwrite dP=Po+ﬁg(P—P0), (23-236) where higher order terms intheTaylor series forphave been neglected. Bymaking useofthedeﬁnition (8-235) forp”,andofthebulk modulus B asgiven byEq.(5-183), thiscanbewritten h P=P0(1 +M229”), (8-237) were 1/2M=to(P52) = (s-238) Here Mistheratio ofthecharacteristic velocity votothevelocity ofsound candiscalled theMach number fortheproblem. Inasimilar way, wecan expand 1/p,assuming that |p—po|<<po: 11 2—=—1—M”. 8-239 P,,0( P) () With thehelp ofEqs. (8-237) and(8-239), wecanrewrite theequation ofcontinuity andtheequation ofmotion interms ofthedimensionless variables introduced byEqs. (8-232) to(8-235). The equation ofcon- tinuity (8—127), when wedivide through bytheconstant povo/so and collect separately theterms involving M,becomes V’-v’+M2gt’+v'-(p"v')] =0, (s-240) where .6 .6 8 8-14] VISCOSITY 345 The equation ofmotion (8—139), when wedivide through byvg/so, be- comes, inthesame way, av’ I II 21/ /1! sofTﬂ+V'VV+(1—Mp)Vp (8~242) O Equations (8-240) and(8—242) represent four differential equations tobe solved forthefour quantities p’,v’,subject togiven initial andboundary conditions. Ifthebody forces arezero, orifthebody forces perunit mass f/paremade proportional to22%/so, then theequations fortwosimilar prob- lems become identical iftheMach number Misthesame forboth. Hence, similar problems willhave similar solutions ifthey have thesame Mach number. Results ofexperiments onscale models inwind tunnels canbe extrapolated tofull-sized airplanes ﬂying atspeeds with corresponding Mach numbers. IftheMach number ismuch lessthan one, theterms in M2inEqs. (8—240) and(S-242) canbeneglected, andthese equations then reduce totheequations foranincompressible ﬂuid, asisobvious either from Eq.(8—240) or(8-237). Therefore atﬂuid velocities much lessthan thespeed ofsound, even airmay betreated asanincompressible ﬂuid. Ontheother hand, atMach numbers near orgreater than one, thecom- pressibility becomes important, even inproblems ofliquid ﬂow. Note that theMach number involves only thecharacteristic velocity vo,andthe velocity ofsound, which inturndepends onthecharacteristic density po andthecompressibility B.Changes inthedistance scale factor sohave no effect onthenature ofthesolution, nordochanges inthecharacteristic pressure poexcept insofar asthey affect poandB. Itmust beemphasized that these results areapplicable only toideal ﬂuids, i.e.,when viscosity isunimportant, andtoproblems where theden- sityoftheﬂuid does notdiffer greatly atanypoint from thecharacteristic density po.Thelatter condition holds fairly wellforliquids, except when there iscavitation (formation ofvapor bubbles), andforgases except at very large Mach numbers. 8-14 Viscosity. Inmany practical applications ofthetheory ofﬂuid ﬂow, itisnotpermissible toneglect viscous friction, ashasbeen done in thepreceding sections. When adjacent layers ofﬂuid aremoving past one another, thismotion isresisted byashearing force which tends toreduce their relative velocity. Letusassume that inagiven region thevelocity oftheﬂuid isinthex—direction, and that theﬂuid isﬂowing inlayers parallel tothexz-plane, sothat 21,,isafunction ofyonly (Fig. 8-10). Letthepositive y-axis bedirected toward theright. Then if80,/6y is positive, theviscous friction willresult inapositive shearing force F, acting from right toleftacross anarea Aparallel tothexz-plane. The 346 THEMECHANICS orCONTINUOUS MEDIA [cH.u>. 8 Z All’ J / >y (E FIG. 8-10. Velocity distribution inthedeﬁnition ofviscosity. coeﬁicient ofviscosity 1;isdeﬁned astheratio oftheshearing stress tothe velocity gradient:_F»/A. _"—am <8243) When thevelocity distribution isnotofthissimple type, thestresses due toviscosity aremore complicated. (See Section 10-6.) Weshall apply this deﬁnition totheimportant special case ofsteady ﬁow ofaﬂuid through apipe ofcircular cross section, with radius a.We shall assume laminar ﬂow; that is,weshall assume that theﬂuid ﬂows in layers, ascontemplated inthedeﬁnition above. Inthiscase, thelayers arecylinders. Thevelocity iseverywhere parallel totheaxisofthepipe, which wetake tobethez-axis, andthevelocity v,isafunction only ofr, thedistance from theaxisofthepipe. (See Fig. 8-11.) Ifweconsider a cylinder ofradius randoflength Z,itsarea willbeA=2-zrrl, andaccord- ingtothedeﬁnition (8—243), theforce exerted across thiscylinder bythe ﬂuid outside ontheﬂuid inside thecylinder is F,=i7(21rrl) ‘ff (s-244) Since theﬂuid within thiscylinder isnotaccelerated, ifthere isnobody force theviscous force must bebalanced byadifference inpressure 8-14] vrscosrrr 347 2 11 r ______<i_______-____-__-31< .___._____..___.-.___-i1|>-----< ____________'________-___-P<!~| Q<2* FIG. 8-11. Laminar ﬂow inapipe. between thetwoends ofthecylinder: Ap(7r7‘2) +F,=0, (s-245) where Apisthedifference inpressure between thetwoends ofthecylinder adistance lapart, andweassume that thepressure isuniform over the cross section ofthepipe. Equations (8-244) and(8-245) canbecombined togive adifferential equation for11,: dc,___rAp_ _W— —~—2nl (8246) Weintegrate outward from thecylinder axis: 7): A T ’/vodv,=—2T€)l/érdr, 2Av,=to-YT,’-5, _ (s-247) where voisthevelocity attheaxisofthepipe. Weshall assume that the ﬂuid velocity iszero atthewalls ofthepipe: __ _a2Ap__ _ [vzlr=a W U0 “W 07 although thisassumption isopen toquestion. 348 THE MECHANICS OFCONTINUOUS MEDIA [CHAP. 8 Then2_aA7’ _ to_4",, (s249) and U;=2-2(a2—7'2). (s-250) Thetotal ﬂuid current through thepipe is 1=//pv,as=21¢»f“Mdt. (s-251, O Wesubstitute from Eq.(8—250) andcarry outtheintegration: I 1ra4Ap—=—-- 8-252 P 8,, ( ) This formula iscalled Poiseuille’s law. Itaffords aconvenient andsimple way ofmeasuring 11. Although wewillnotdevelop now thegeneral equations ofmotion for viscous ﬂow, wecanarrive ataresult analogous tothat inSection 8-13, taking viscosity into account, without actually setting uptheequations forviscous ﬂow. Suppose that weareconcerned, asinSection 8-13, with twosimilar problems inﬂuid ﬂow, andletso,vo,po,pobeacharacteristic distance, velocity, pressure, anddensity, which again deﬁne thescale in anyproblem. However, letussuppose that inthiscase viscosity istobe taken into account, sothat theequation ofmotion (8—139) isaugmented byaterm corresponding totheforce ofviscous friction. Wedonotat present know theprecise form ofthisterm, butatanyrateitwillconsist of11multiplied byvarious derivatives ofvarious velocity components, and divided byp[since Eq. (8-139) hasalready been divided through byp].When weintroduce thevelocity v’,andthedimensionless coordi- nates x’,y’,z’,t’,asinSection 8-13, anddivide theequation ofmotion by03/so,wewillobtain justEq.(8-242), augmented byaterm involving thecoefficient ofviscosity. Since alltheterms inEq.(8—242) aredimen- sionless, theviscosity term willbealso, andwillconsist ofderivatives of components ofv’with respect toac’,y’,z’,multiplied bynumerical factors andbyadimensionless coeﬂicient consisting of1;times some combination ofvoandso,anddivided byp=po(1 +M2p”) [Eq. (8—237)]. Now the dimensions of17,asdetermined byEq.(8-243), are mass["1= M53) and the only combination ofpo,vo,and sohaving these dimensions 8-14] vIscosITY 349 ispovoso. Therefore theviscosity term willbemultiplied bythecoeffi- cient I where RistheReynolds number, deﬁned by R=3%? (s-255) 77 Wecannow conclude that when viscosity isimportant, twosimilar prob- lems willhave thesame equation ofmotion indimensionless variables, and hence similar solutions, only iftheReynolds number R,aswell asthe Mach number M,isthesame forboth. IftheMach number isvery small, then compressibility isunimportant. IftheReynolds number isvery large, then viscosity may beneglected. Itturns outthat there isacritical value ofReynolds number foranygiven problem, such that thenature ofthe ﬂow isvery different forRlarger than thiscritical value than forsmaller values ofR.Forsmall Reynolds numbers, theﬂow islaminar, asthe viscosity tends todamp outanyvortices which might form. Forlarge Reynolds numbers, theﬂow tends tobeturbulent. This willbethecase when theviscosity issmall, orthedensity, velocity, orlinear dimensions arelarge. Note that theReynolds number depends onso,whereas the Mach number does not, sothat thedistance scale ofaproblem isim- portant when theeffects ofviscosity areconsidered. Viscous effects are more important onasmall scale than onalarge scale. Itmay benoted that theexpression (8-255) fortheReynolds number, together with thefactthat Eq.(8—139) isdivided by113/so toobtain the dimensionless equation ofmotion, implies that theviscosity term tobe added toEq.(8-139) hasthedimensions of(1;vo)/ (posﬁ). This, inturn, implies that theviscous force density must beequal to1;times asum of second derivatives ofvelocity components with respect toas,y,and z. This isperhaps also evident from Eq. (8—243), since incalculating the total force onaﬂuid element, thedifferences instresses onopposite faces oftheelement will beinvolved, and hence asecond differentiation of velocities relative toas,y,andzwillappear intheexpression fortheforce. Anexpression fortheviscous force density willbedeveloped inChapter 10. 350 ATHE MECHANICS orCONTINUOUS MEDIA [CHAP. 8 PROBLEMS 1.Astretched string oflength listerminated attheendat=lbyaring of negligible mass which slides without friction onavertical rod. (a)Show that the boundary condition atthisendofthestring is 8u _0. (b)Iftheendat=0istied, ﬁndthenormal modes ofvibration. 2.Find theboundary condition andthenormal modes ofvibration inProb- lem1ifthering atoneendhasaﬁnite mass m.What isthesigniﬁcance ofthe limiting cases m=0andm=w? 3.Themidpoint ofastretched string oflength lispulled adistance u=Z/10 from itsequilibrium position, sothat thestring forms twolegs ofanisosceles triangle. The string isthen released. Find anexpression foritsmotion bythe Fourier series method. 4.Apiano string oflength l,tension 1-,anddensity 0,tiedatboth ends, and initially atrest, isstruck ablow atadistance afrom oneendbyahammer of mass mand velocity vo. Assume that thehammer rebounds elastically with velocity -110, and that itsmomentum lossistransferred toashort length Al ofstring centered around ac=a.Find themotion ofthestring bytheFourier series method, assuming that Alisnegligibly small. Iftheﬁnite length ofAl were taken intoaccount, what sortofeffect would thihave onyour result? If itisdesired thatnoseventh harmonic ofthefundamental frequency bepresent (itissaidtobeparticularly unpleasant), atwhat points amay thestring be struck? 5.Astring oflength listied atav=l.The endatx=0isforced tomove sinusoidally sothat u(0,t)=Asinwt. (a)Find thesteady-state motion ofthestring; that is,ﬁnd asolution inwhich allpoints onthestring vibrate with thesame angular frequency w.(b)How would youﬁndtheactual motion ifthestring were initially atrest? 6.Aforce oflinear density f(:v,t)=f0sin$coswt, where nisaninteger, isapplied along astretched string oflength l.(a)Find thesteady-state motion ofthestring. [Hint: Assume asimilar time andspace dependence foru(x,t),andsubstitute intheequation ofmotion.] (b)Indicate how onemight solve themore general problem ofaharmonic applied force _f(x,t)=f0(:v) coswt, where fo(a:) isanyfunction. 7.Assume thatthefriction oftheairaround avibrating string canberepre- sented asaforce perunit length proportional tothevelocity ofthestring. Set PROBLEMS 351 uptheequation ofmotion forthestring, andﬁndthenormal modes ofvibration ifthestring istiedatboth ends. 8.Find themotion ofahorizontal stretched string oftension 1-,density cr, andlength l,tiedatboth ends, taking intoaccount theweight ofthestring. The string isinitially held straight and horizontal, and dropped. [Hint: Find the steady-state “motion” andaddasuitable transient.] 9.Along string isterminated atitsright endbyamassless ring which slides onavertical rodandisimpeded byafrictional force proportional toitsvelocity. Setupasuitable boundary condition anddiscuss thereﬂection ofawave atthe end. How does thereﬂected wave behave inthelimiting cases ofvery large and very small friction? Forwhat value ofthefriction constant isthere noreﬂected wave‘? 10.Discuss thereﬂection ofawave traveling down along string terminated byamassless ring, asinProblem 1. 11.Find asolution toProblem 3bysuperposing waves f(x—ct)and g(a:—|—ct)insuch awayastosatisfy theinitial andboundary conditions. Sketch theappearance ofthestring attimes t=0,it/c, it/c, andl/c. 12.(a)Along stretched string oftension 1-anddensity <11istied at2:=0 toastring ofdensity 02. Ifthemass oftheknot isnegligible, show that u anddu/6a: must bethesame onboth sides oftheknot. (b)Awave Acos(kn: —wt)traveling toward theright ontheﬁrst string isincident onthejunction. Show that inorder tosatisfy theboundary conditions attheknot, there must beareﬂected wave traveling totheleftintheﬁrststring andatransmitted wave traveling totheright inthesecond string, both ofthe same frequency astheincident wave. Find theamplitudes andphases ofthe incident andreﬂected waves. ‘ (c)Check your result inpart (b)bycalculating thepower inthetransmitted and reﬂected waves, and showing that thetotal isequal tothepower inthe incident wave. s 13.Derive directly from Eq.(8—139) anequation expressing theconservation ofangular momentum inaform analogous toEq.(8—141). 14.Derive anequation expressing thelawofconservation ofangular mo- mentum foraﬂuid inaform analogous toEq.(8—140). From this, derive equa- tions analogous toEqs. (8—l41), (8—142), and (8-144). Explain thephysical meaning ofeach term ineach equation. Show that theinternal torques dueto pressure canbeeliminated from theintegrated forms, and derive anequation analogous toEq.(8—148). 15.Derive andinterpret thefollowing equation: ;%_U7(%pv2 —-P9—Pu)dV+[fn-v(%nv2 —P9—pu)dS V S=—//We —/// S V where Visaﬁxed volume bounded byasurface Swith normal n. 352 THE MECHANICS orCONTINUOUS MEDIA [CHAP. 8 16.(a)Amass ofinitially stationary airat45°Nlatitude ﬂows inward toward alow-pressure spot atitscenter. Show that thecoriolis torque about thelow- pressure center depends only ontheradial component ofvelocity. Hence, show that iffrictional torques areneglected, theangular momentum perunit mass at radius rfrom thecenter depends only onrandontheinitial radius roatwhich theairisstationary, butdoes notdepend onthedetails ofthemotion. . (b)Calculate theazimuthal component ofvelocity around thelowasafunc- tion ofinitial andﬁnal radius. Ifthis were areasonable model ofatornado, what would betheinitial radius roiftheairat264ftfrom thecenter hasa velocity of300mi/hr? 17.(Evaluate thepotential energy uperunit mass asafunction ofpfora perfect gasofmolecular weight ll!attemperature T.Forthesteady isothermal ﬂow ofthisgasthrough apipe ofvarying cross section andvarying height above theearth, ﬁnd expressions forthepressure, density, andvelocity ofthegasas functions ofthecross section Softhepipe, theheight h,andthepressure po andvelocity voatapoint inthepipe atheight h=0where thecross section isSo.Assume p,v,andpuniform over thecross section. 18.Work Problem 17foranincompressible ﬂuid ofdensity po. 19.The function ¢=a/r,where aisaconstant andristhedistance from a ﬁxed point, satisﬁes Laplace’s equation (8-17 7),except atr=0,because ithas thesame form asthegravitational potential ofapoint mass. Ifthisisavelocity potential, what isthenature oftheﬂuid ﬂow towhich itleads? 20.(a)Verify bydirect computation thatthespherical wave (8-204) satisﬁes thewave equation (8-186). (b)Write ananalogous expression foracylindrical wave ofarbitrary time dependence, traveling, outfrom thez-axis, independent ofzand with cylindrical symmetry. Make theamplitude depend onthedis- tance from theaxisinsuch away astosatisfy therequirement ofconservation ofenergy. Show that such awave cannot satisfy thewave equation. (Itisa general property ofcylindrical waves that they donotpreserve their shape.) *21. Show that thenormal mode ofvibration given byEqs. (8—225) and(8—226) canberepresented asasuperposition ofharmonically oscillating plane waves traveling inappropriately chosen directions with appropriate phase relation- ships. Show that inthenormal vibrations ofaﬂuid inabox, thevelocity oscil- lates 90°outofphase with thepressure atanypoint. How canthisbereconciled with thefactthat inaplane wave thevelocity andpressure areinphase? 22.Find thenormal modes ofvibration ofasquare organ pipe with oneend open andtheother closed, ontheassumption that theopen endisapressure node. 23.(a)Calculate theﬂuid velocity vforthewave given byEq.(8—228). (b) Calculate themean rate ofpower ﬂow through thepipe. *24. Show that theexpression (8—228) forasound wave inapipe canberepre- sented asasuperposition ofplane waves traveling with speed cinappropriate directions, andbeing reﬂected atthewalls. Explain, interms ofthisrepresenta- tion, why there isaminimum frequency foranygiven mode below which awave cannot propagate through thepipe inthismode. 25.Ifthesound wave given byEq.(8—228) isincident onaclosed endofthe pipe atz=0,ﬁndthereﬂected wave. PROBLEMS 353 26.Develop thetheory ofthepropagation ofsound waves inacircular pipe, using cylindrical coordinates andapplying themethod ofseparation ofvariables. Carry thesolution asfarasyoucan. You arenotrequired tosolve theequation fortheradial partofthewave, butyoushould indicate thesortofsolutions you would expect toﬁnd. 27.Aﬂuid ofviscosity 11ﬂows steadily between two inﬁnite parallel plane walls adistance lapart. Thevelocity oftheﬂuid iseverywhere inthesame direc- tion, anddepends only onthedistance from thewalls. The total ﬂuid current between thewalls inanyunit length measured along thewalls perpendicular to thedirection ofﬂowisI.Find thevelocity distribution andthepressure gradient parallel tothewalls, assuming that thepressure varies only inthedirection of ﬂow. 28.Prove that theonly combination ofpo,vo,sohaving thedimensions of viscosity ispovoso. ! ! 4 1l < CHAPTER 9 LAGRANGE’S EQUATIONS 9-1Generalized coordinates. Direct application ofNewton’s laws to amechanical system results inasetofequations ofmotion interms ofthe cartesian coordinates ofeach oftheparticles ofwhich thesystem iscom- posed. Inmany cases, these arenotthemost convenient coordinates in terms ofwhich tosolve theproblem ortodescribe themotion ofthesystem. Forexample, intheproblem ofthemotion ofasingle particle acted onby acentral force, which wetreated inSection 3-13, wefound itconvenient to introduce polar coordinates intheplane ofmotion oftheparticle. The reason wasthat theforce inthiscasecanbeexpressed more simply interms ofpolar coordinates. Again inthetwo-body problem, treated inSection 4-7, wefound -itconvenient toreplace thecoordinates rl,r2ofthetwo particles bythecoordinate vector Rofthecenter ofmass, andtherelative coordinate vector rwhich locates particle 1with respect toparticle 2.We hadtworeasons forthischoice ofcoordinates. First, themutual forces which theparticles exert oneach other ordinarily depend ontherelative coordinate. Second, inmany cases weareinterested inadescription of themotion ofoneparticle relative totheother, asinthecase ofplanetary motion. Inproblems involving many particles, itisusually convenient tochoose asetofcoordinates which includes thecoordinates ofthecenter ofmass, since themotion ofthecenter ofmass isdetermined byarelatively simple equation (4-18). InChapter 7,wefound theequations ofmotion ofaparticle interms ofmoving coordinate systems, which aresometimes more convenient tousethan theﬁxed coordinate systems contemplated inNewton’s original equations ofmotion. Weshall include coordinate systems ofthesort described above, to- gether with cartesian coordinate systems, under thename generalized co- ordinates. Asetofgeneralized coordinates isanysetofcoordinates by means ofwhich thepositions oftheparticles inasystem may bespeciﬁed. Inaproblem requiring generalized coordinates, wemaysetupNewton’s equations ofmotion interms ofcartesian coordinates, andthen change to thegeneralized coordinates, asintheproblems studied inprevious chap- ters. Itwould bevery desirable andconvenient, however, tohave agen- eralmethod forsetting upequations ofmotion directly interms ofany convenient setofgeneralized coordinates. Furthermore itisdesirable to have uniform methods ofwriting down, and perhaps ofsolving, the equations ofmotion interms ofanycoordinate system. Such amethod wasinvented byLagrange andisthesubject ofthischapter. 354 9-1] GENERALIZED COORDINATES 355 Ineach ofthecases mentioned intheﬁrst paragraph, thenumber of coordinates inthenew system ofcoordinates introduced tosimplify the problem wasthesame asthenumber ofcartesian coordinates ofallthe particles involved. Wemay, forexample, replace thetwocartesian co- ordinates x,yofaparticle moving inaplane bythetwopolar coordinates r,0,orthethree space coordinates ac,y,zbythree spherical orcylindrical coordinates. Orwemay replace thesixcoordinates x1,yl,zl,1:2,yo,.22 ofapair ofparticles bythethree coordinates X,Y,Zofthecenter of mas plusthethree coordinates ac,y,zofoneparticle relative totheother. Orwemay replace thethree coordinates ofaparticle relative toaﬁxed system ofaxes bythree coordinates relative tomoving axes. (Avector counts asthree coordinates.) Inourtreatment oftherotation ofarigid body about anaxis (Section 5-2), wedescribed theposition ofthebody interms ofthesingle angular coordinate 0.Here wehave acase where wecanreplace agreat many cartesian coordinates, three foreach particle inthebody, byasingle coordinate 0.This ispossible because thebody isrigid andisallowed to rotate only about aﬁxed axis. Asaresult ofthese twofacts, theposition ofthebody iscompletely determined when wespecify theangular position ofsome reference lineinthebody. Theposition ofafreerigid body can bespeciﬁed bysixcoordinates, three tolocate itscenter ofmass, andthree todetermine itsorientation inspace. This isavast simpliﬁcation com- pared with the3Ncartesian coordinates required tolocate itsNparticles. Arigid body isanexample ofasystem ofparticles subject toconstraints, that is,conditions which restrict thepossible setsofvalues ofthecoordi- nates. Inthecase ofarigid body, theconstraint isthat thedistance between any two particles must remain ﬁxed. Ifthebody canrotate only about aﬁxed axis, then inaddition thedistance ofeach particle from theaxisisﬁxed. This isthereason why specifying thevalue ofthesingle coordinate 0issufﬁcient todetermine theposition ofeach particle inthe body. Weshall postpone thediscussion ofsystems likethiswhich involve constraints until Section 9-4. Inthissection, andthenext, weshall set upthetheory ofgeneralized coordinates, assuming that there areasmany generalized coordinates ascartesian coordinates. Weshall then ﬁnd, in Section 9-4, that this theory applies also tothemotion ofconstrained systems. VVhen wewant tospeak about aphysical system described byasetof generalized coordinates, without specifying forthemoment justwhat the coordinates are,itiscustomary todesignate each coordinate bytheletter q with anumerical subscript. Asetofngeneralized coordinates would be written asq1,q2, ...,q,,. Thus aparticle moving inaplane may be described bytwo coordinates ql,qo,which may inspecial cases bethe cartesian coordinates as,y,orthepolar coordinates r,0,oranyother suit- 356 LAoRANoE’s EQUATIONS [CHAP. 9 able pair ofcoordinates. Aparticle moving inspace islocated bythree coordinates, which may becartesian coordinates x,y,z,orspherical coordinates r,0,(p,orcylindrical coordinates p,z,<p,or,ingeneral '11,<12,Q3- The conﬁguration ofasystem ofNparticles may bespeciﬁed bythe 3Ncartesian coordinates x1,yl,zl,mo,yo,.22,...,xN,yN,zNofitspar- ticles, orbyanysetof3Ngeneralized coordinates ql,qo,...,q31v. Since foreach conﬁguration ofthesystem, thegeneralized coordinates must have some deﬁnite setofvalues, thecoordinates ql,...,q3Nwillbefunc- tions ofthecartesian coordinates, and possibly also ofthetime inthe case ofmoving coordinate systems: ql=q1(11, 1/1: Z11$2: f/22 ---2yN: zNit)1 q2=q2(x1, 3/1,...........,zN;t), (9_1) qazv=q3N(7311 1'11,----------,ZN}t)- Since thecoordinates ql,...,q3Nspecify theconﬁguration ofthesystem, itmust bepossible alsotoexpress thecartesian coordinates interms of thegeneralized coordinates: $1=$1(<11i¢12,---1qaN;t), 2/1=y1(q1, -----,q3N§t), (9_2) ZN=ZN(q1a-----1 qsN; t)- IfEqs. (9-1) aregiven, they may besolved forml,yl,...,2Ntoobtain Eqs. (9-2), andviceversa. The mathematical condition that this solution be(theoretically) possible is that theJacobian determinant ofEqs. (9-1) bedifferent from zero atallpoints, ornearly allpoints: an 3231 6:61aqszv 6901 3l13N 32/1an 3211‘E 391 ‘E 621;;a(q1r '--2q3N) = a(w1: yly --'2ZN)9'50. (9-3) in 621,1u-0 621v Ifthisinequality does nothold, then Eqs. (9-1) donotdeﬁne alegitimate set ofgeneralized coordinates. Inpractically allcases ofphysical interest, itwill 9-1] GENERALIZED COORDINATES 357 beevident from thegeometrical deﬁnitions ofthegeneralized coordinates whether ornotthey arealegitimate setofcoordinates. Thus weshall not have anyoccasion toapply theabove testtoourcoordinate systems. [For a derivation ofthecondition (9-3), seeW.F.Osgood, Advanced Calculus, New York: Macmillan, 1937, p.129.] Asanexample, wehave theequations (3-72) and (3-73) connecting thepolar coordinates r,0ofasingle particle inaplane with itscartesian coordinates ac,y.Asanexample ofamoving coordinate system, we consider polar coordinates inwhich thereference axis from which 0is measured rotates counterclockwise with constant angular velocity w (Fig. 9-1): T=($2+1/°')”2, 0=tan_1 g—wt, (9-4) and conversely, at=rcos(0—l—wt), y=rsin(0—l—wt). (9-5) Asanexample ofgeneralized coordinates forasystem ofparticles, we have thecenter ofmass coordinates X,Y,Zandrelative coordinates ac,y,zoftwoparticles ofmasses mlandmo,asdeﬁned byEqs. (4-90) and(4-91), where X,Y,Zarethecomponents ofR,andzv,y,zarethe components ofr.Because thetransformation equations (4-90) and (4-91) donotcontain thetime explicitly, weregard thisasaﬁxed co- ordinate system, even though x,y,zarethecoordinates ofmlreferred toamoving origin located onmo. Therulewhich deﬁnes thecoordinates X,Y,Z,x,y,zisthesame atalltimes. Ifasystem ofparticles isdescribed byasetofgeneralized coordinates ql,...,qoN, weshall callthetime derivative q,,,ofanycoordinate qk, thegeneralized velocity associated with this coordinate. The generalized velocity associated with acartesian coordinate x,-isjustthecorrespond- 1/ r 0wt I Fro. 9-1. Arotating polar coordinate system. 358 I.AeRANoE’s EQUATIONS [CHAP. 9 ingcomponent rt,ofthevelocity oftheparticle located bythat coordinate. The generalized velocity associated with anangular coordinate 0isthe corresponding angular velocity (9.The velocity associated with theco- ordinate Xinthepreceding example isX,theas-component ofvelocity ofthecenter ofmass. Thegeneralized velocities canbecomputed interms ofcartesian coordinates andvelocities, andconversely, bydifferentiating Eqs. (9-1) or(9-2) with respect totaccording totherules fordifferentiat- ingimplicit functions. Forexample, thecartesian velocity components canbeexpressed interms ofthegeneralized coordinates and velocities bydifferentiating Eqs. (9-2): 3N . 6171 . 0201it=2-q.+—,k=l Bqk 6t 5 (9-6) 3N, 621v‘ . 621V 1- —— + —i IZ”,2,6q;,q'° at Asanexample, wehave, from Eqs. (9-5): at=rcos (0+wt)—rdsin(0—I—wt)-—rwsin(0-]—wt),9-7 y=rsin(0+wt)+r0cos(0+wt)+rwcos(0—]-wt). () The kinetic energy ofasystem ofNparticles, interms ofcartesian coordinates, is Il"l=NIH3 T= At?+if+é?)- (as) Bysubstituting from Eqs. (9-6), weobtain thekinetic energy interms ofgeneralized coordinates. Ifwerearrange theorder ofsummation, the result is azvazv azv T==22%/lkzékélz +ZBic. +To, (9-9) k=1z=1 k=1 where N r__ _61;,6x; 6y,-6y,- dz,62¢) _ AH— m'(6q1¢ dqz+anéqi+6qtaqz , (910) N___ _19$," ail}; 31/; 61!./1; 82,- 32¢‘) _B,._;m(aq a+q +aq 8,, (911)= 1 loi 31,,Gt lo To=l(%)2+(3)2+(‘T%)2l- <9-12> Il"]=NIH§ 9-1] GENERALIZED COORDINATES 359 The coefficients Ara, Bk, and Toarefunctions ofthe coordinates ql,..., q3N, and also oftforamoving coordinate system. IfA,,;is zeroexcept when k=l,thecoordinates aresaidtobeorthogonal. Thecoef- ﬁcients B1,andToarezero when tdoes notoccur explicitly inEqs. (9-1), i.e.,when thegeneralized coordinate system does notchange with time. Weseethat thekinetic energy, ingeneral, contains three setsofterms: T=T2+T1"l"T0, (9-13) where T2contains terms quadratic inthegeneralized velocities, T1con- tains linear terms, and Toisindependent ofthevelocities. The terms T1andToappear only inmoving coordinate systems; forﬁxed coordinate systems, thekinetic energy isquadratic inthegeneralized velocities. Asanexample, inplane polar coordinates [Eqs. (3-72)], thekinetic energy is T=%”"l(i2 +272) =—§—(mr2 +mr2(§2), (9-14) asmay beobtained bydirect substitution from Eqs. (3-72), orasa special case ofEq. (9-9), where 6 6 .6-:=cost), 5;=—rsin0, 9-15) al=sin0 %=rcos0. (6r ’ 80 Ifwetake themoving coordinate system deﬁned byEqs. (9-5), weﬁnd, bysubstituting from Eqs. (9-7), orbyusing Eq. (9-9), T=%'"(5?2 +172) =%(mr2 +mrzdz) +mrzwd -]—§~mr2w2. (9-16) Inthiscase, aterm linear in9andaterm independent of1‘and9appear. The kinetic energy forthetwo-particle system canalsoeasily bewritten down interms ofX,Y,Z,x,y,z,deﬁned byEqsl (4-90) and (4-91). Instead ofﬁnding thekinetic energy ﬁrst incartesian coordinates and then translating into generalized coordinates, asintheexamples above, itisoften quicker towork outthekinetic energy directly interms of generalized coordinates from aknowledge oftheir geometrical meaning. Itmay then bepossible tostart aproblem from thebeginning with a suitable setofgeneralized coordinates without writing outexplicitly the transformation equations (9-1) and (9-2) atall. Forexample, wemay obtain Eq. (9-14) immediately from thegeometrical meaning ofthe4 i l l 360 LAoRANoE’s EQUATIONS [cHAr>. 9 coordinates r,6(seeFig.3-20) bynoticing that thelinear velocity associ- ated with achange inrisrandthat associated with achange in0isrd. Since thedirections ofthevelocities associated with rand6areperpen- dicular, thesquare ofthetotal velocity is 92=i=2+T292, (9-17) from which Eq.(9-14) follows immediately. Care must betaken inapplying thismethod ifthevelocities associated with changes ofthevarious coordinates arenotperpendicular. Forexam- ple,letusconsider apair ofcoordinate axes u,'wmaking anangle orless than 90°with each other, asinFig.9-2. Letuandwbethesides ofapar- allelogram formed bythese axes andbylines parallel totheaxes through themass masshown. Letaand bbeunit vectors inthedirections of increasing uandw.Using uandwascoordinates, thevelocity ofthe mass mis v=ua-]—wb. (9-18) Thekinetic energy is ‘ L T=smv-v =smut’+%m1b2 +mowcos.9. (9-19) This isanexample ofasetofnonorthogonal coordinates inwhich across product term inthevelocities appears inthekinetic energy. The reason forusing theterm orthogonal, which means perpendicular, isclear from this example. When systems ofmore than oneparticle aredescribed interms of generalized coordinates, itisusually safest towrite outthekinetic energy ﬁrst incartesian coordinates and transform togeneralized coordinates. However, insome cases, itispossible towrite thekinetic energy directly ingeneral coordinates. Forexam- ple,ifarigid body rotates about an axis, weknow that the kinetic energy is%Iw2, where wistheangu- /1 larvelocity about that axisandIis l\ themoment ofinertia. Also, wecan usethetheorem proved inSection 4-9that thetotal kinetic energy of asystem ofparticles isthekinetic energy associated with thecenter of mass plus that associated with the internal coordinates. [See Eq. FIG. 9-2. Anonorthogonal coordi- (4"127)»l A5anexample; thekinetic natesystem. energy ofthetwo-particle system inM 9-1] GENERALIZED COORDINATES 361 terms ofthecoordinates X,Y,Z,ax,y,z,deﬁned byEqs. (4-90) and(4-91) is T=%M(X2 +Y2+Z2)+4/»(-i2+92+:22). <9-29> where Manditaregiven byEqs. (4-97) and (4-98). The result shows that this isanorthogonal coordinate system. Ifthelinear velocity of each particle inasystem canbewritten down directly interms ofthe generalized coordinates and velocities, then thekinetic energy canim- mediately bewritten down. Wenow note that thecomponents ofthelinear momentum ofparticle i, according toEq.(9-8), are, . 6T . 6T . 6Tpix: m$¢=,g1_' Pu/= ml/i"=5,Z' piz= mZi=5§;' (9-21) Inthecase ofaparticle moving inaplane, thederivatives ofTwith respect torand 0,asgiven byEq. (9-14), are _-_<’T _2-_<E _ p,—mr-5?: po_mr0_a6., (922) where p,isthecomponent oflinear momentum inthedirection ofin- creasing r,andp,istheangular momentum about theorigin. Similar results willbefound forspherical andcylindrical coordinates inthree dimensions. Infact, itisnothard toshow that foranycoordinate q), which measures thelinear displacement ofanyparticle orgroup ofpar- ticles inagiven direction, thelinear momentum ofthat particle orgroup inthegiven direction is8T/dot; andthat forany coordinate q),which measures theangular displacement ofaparticle orgroup ofparticles about anaxis, their angular momentum about that axisis8T/6(1),. This suggests that wedeﬁne thegeneralized momentum pkassociated with the coordinate q),by* 6T=__. .2Pk aqk (93) Ifq),isadistance, pkisthecorresponding linear momentum. Ifqkisan *The kinetic energy Tisdeﬁned byEq.(9-9) asafunction ofQ1,...,Qaiv; q1,...,q31v, andperhaps oft.The derivatives ofthisfunction Twith respect tothese variables willbedenoted bythesymbols forpartial diﬂerentiation. Since <11,...,q3N; 111,...,Qszvareallfunctions ofthetime tforany given motion ofthesystem, Tisalso afunction oftalone forany given motion. The derivative ofTwith respect totime inthissense willbedenoted byd/dt. The same remarks apply toanyother quantity which may bewritten asafunction ofthecoordinates andvelocities andperhaps oft,andwhich isalso afunction oftalone foranygiven motion.l4 l i J 1l l l 362 LAcRANcE’s EQUATIONS [CIIAP. 9 angle, pkisthecorresponding angular momentum. Inother cases, pk willhave some other corresponding physical signiﬁcance. According to Eq. (9-9), thegeneralized momentum pkis 3Npk=ZAklql+B... (9-24) l=-1 Inthecase ofthecoordinates X,Y,Z,x,y,zforthetwo-particle sys- tem, thisdeﬁnition gives PX=MX, py=MY, pz=MZ, Pr=I455: Pu=P-ll; P2=I-"5: (9-25) where pX,py,pzarethecomponents ofthetotal linear momentum of thetwo particles, andp1,,p,,,p,arethelinear momentum components intheequivalent one-dimensional problem in:0,y,2towhich thetwo- body problem wasreduced inSection 4-7. Weshall seeinthenext sec- tion that theanalogy between thegeneralized momenta p1,and the cartesian components oflinear momentum canbeextended totheequa- tions ofmotion ingeneralized coordinates. Ifforces F1,, F1,, F1,, ...,FN, actontheparticles, thework done bythese forces iftheparticles move from thepositions x1,y1,21,...,2N tonearby points x1—]—6001,y1+8y1,z1—]—621,..., 2N—]—621vis 1v .6W=Z(F...ax.+F...an+F.~.at.-). (9-26) i=1 The small displacements 620,,6y,-,62,-may beexpressed interms ofgen- eralized coordinates: azvan 5‘= —-"5 xi gaqk qk; .1-—3N6”‘a (9-27) yi * qk; N3 621'6-= —-5 Z1 - qh) where 5q1, ...,5q3N arethedifferences inthegeneralized coordinates associated with thetwo sets ofpositions oftheparticles. Wecallthis avirtual displacement ofthesystem because itisnotnecessary that it represent anyactual motion ofthesystem. Itmay beanypossible mo- tionofthesystem. Inthecase ofamoving coordinate system, weregard thetime asﬁxed; that is,wespecify thechanges inposition interms of 9-1] GENERALIZED COORDINATES 363 thecoordinate system ataparticular time t.Ifwesubstitute Eqs. (9-27) inEq.(9-26), wehave, after rearranging terms: 3N .6W=ZQtat.‘ <9-28) kZ 1 where 1v6231' dy,~ 621'Qk=E(Fiz(Tﬁ+FiubE+Fiz5E)‘ (9'-29) nI-1 The coefﬁcients Q1,depend ontheforces acting ontheparticles, onthe coordinates q1,...,q;»,N, andpossibly alsoonthetime t.Inview ofthe similarity inform between Eqs. (9-26) and (9-28), itisnatural tocall thequantity Q),thegeneralized force associated with thecoordinate q;,. Wecandeﬁne thegeneralized force Q1,directly, without reference tothe cartesian coordinate system, asthe coefﬁcient which determines the work done inavirtual displacement inwhich qkalone changes: 5W=Qk54k, (9—30) where 6Wisthework done when thesystem moves insuch awaythat q),increases by6q;,,allother coordinates remaining constant. Notice that thework inEq.(9-26), andtherefore also inEq.(9-30), istobe computed from thevalues oftheforces forthepositions x1,...,zN,or q1,...,q3N;_that is,wedonottake account ofanychange intheforces during thevirtual displacement. Iftheforces F1,, ...,FN, arederivable from apotential energy V(a:1, ...,2N)[Eqs. (4-32)], then 5W=—5V N8V 8V 8V. =1—Z 5131‘ + 61/1; -]" 521;) ' i=1 I yl ‘L IfVisexpressed interms ofgeneralized coordinates, then 6W=—aV 3”aV==- —-of 9-32 gaqk qr ( ) Bycomparing thiswith Eq.(9-28), weseethat ' 6VV Qk——E1 (9-33) which shows that inthissense alsothedeﬁnition ofQ1,asageneralizedI11 I l 364 LAoRANoE’s EQUATIONS [CHAP. 9 force isanatural one. Equation (9-33) may also beveriﬁed bydirect calculation of6V/6q1,: gl/'_ av6x,~_,_6V 6y,-_,_6V 62,-) 599 3961'aqk 31/1aqk 32¢aqk _IM= _'M2/'\__— =<Fizaqk+Fiyaqk+Fizaqk =—-Qt Asanexample, letuscalculate thegeneralized forces associated with thepolar coordinates r,0,foraparticle acted onbyaforce F=iF,—]-jF,,=nF,+1F,. (9-34) Ifweusethedeﬁnition (9-29), wehave, using Eqs. (9-15): _Q <22 a-aM+aM =F,cos0 —]—F,,sin0 =Fr; 6 6a=a£+m% =—rF,, sin0—]-rF,,cos0 Z 7'Fg. Weseethat Q,isthecomponent offorce inther-direction, andQ1is thetorque acting toincrease 0.Itisusually quicker tousethedeﬁni- tion(9-30), which enables ustobypass thecartesian coordinates altogether. Ifweconsider asmall displacement inwhich rchanges tor+6r,with 0 remaining constant, thework is 6W=F,6r, (9-36) from which theﬁrst ofEqs. (9-35) follows. Ifweconsider adisplace- ment inwhich risﬁxed and 0increases by66,thework is 6W=For66, (9-37) from which thesecond ofEqs. (9-35) follows. Ingeneral, ifq),isaco- ordinate which measures thedistance moved bysome part ofthemechani- calsystem inacertain direction, andifF1,isthecomponent inthisdirec- tionofthetotal force acting onthispart ofthesystem, then thework done 9-2] LAoEANoE’s EQUATIONS 365 when qkincreases by6qk,allother coordinates remaining constant, is » 6W=Fk6qk. (9-38) Comparing thiswith Eq.(9-30), wehave Qt=Fa (9—39) Inthiscase, thegeneralized force Qkisjusttheordinary force Fk. Ifqk measures theangular rotation ofacertain part ofthesystem about a certain axis, andifNkisthetotal torque about that axisexerted onthis part ofthesystem, then thework done when qkincreases by6qkis 6W=Nk6qk. (9-40) Comparing thiswith Eq.(9-30), wehave Qk=N11- (9-41) The generalized force Qkassociated with anangular coordinate qkisthe corresponding torque. 9-2Lagrange’s equations. Theanalogy which ledtothedeﬁnitions of generalized momenta and generalized forces tempts ustosuspect that thegeneralized equations ofmotion willequate thetime rateofchange ofeach momentum pktothecorresponding force Qk. Tocheck this suspicion, letuscalculate thetime rate ofchange ofpk: is_1dt_dtaqk (9'42> Wewillneed tostart with Newton’ sequations ofmotion incartesian form: 'm¢55t =Fit; "W91" =Fill; =11'''1N] mﬁi =F1'2- Therefore weexpress Tincartesian coordinates [Eq. (9—8)]. Wethen have aT N as. .017,- .as.) aqk -— m, Q31 +it/1bg +Z1, ! where :21,y1,...,211;aregiven asfunctions ofq1,...,qoN; Q1,...,q31v;t byEqs. (9-6). Since 620,-/oqk and6x,-/at arefunctions only ofq1,...,q3N; t,wehave, bydifferentiating Eqs. (9-6): 366 LAcRANcE’s EQUATIONS [CHAP. 9 8131' 6x,--—-ii, aélk 3% ‘191_alt.-_ ._ aqk-—aqk7 [i_1,...,N,lc-1,...,3N] (9-45) %=‘E. 3% aqk Bysubstituting from Eqs. (9-45) inEq. (9-44), and differentiating again with respect tot,weobtain dpk _ N 6.731" ..61/i ..621;) dt _21mi xlaq +1/itaq +zZaq,-= k k k N,Claw; .dal/i .(I325)1"i——* 1"—"— 1"-—"—' 9—46(ac dt6qk+y dl6qk+z (lloqk ( ) According toNewton’s equations ofmotion (9-43), and thedeﬁnition (9-29), theﬁrst term inEq.(9-46) is N N gml<£laqk+ylaqk+ztaqk F1Zaqk+F1U6qk+Fl8aqk =91, H <9-11) The derivatives appearing inthelastterm inEq.(9-46) arecalculated asfollows: d611;»; 3N 321),; , 62$; 6<3” 6.77,; 31111; 67;;__ i :_ 2 + I. 2 .1+ —(— 1' *3 1 . dtoqk ,=aqkaq,” aqka»: aqk,=16q1q ‘t aqk - (9-48) where wehave made useofEq. (9-6). Similar expressions hold fory andz.Thus thelastsum inEq.(9-46) is N dt6qk+y‘dt6qk+z’dt6qk N -35% .391' .65¢ 3N1 .2 -2 .2=gmi $1"aqk+?/1" aqk+3iaqk =6qk;=:12mi'($r +1/1" -I-Z1") GT —3911 (9-4) Wehave ﬁnally: M_ E _ _dt—Qk+aqkr la-1,...,3N. (950) 9-2] LAGRANGE’S EQUATIONS ' 367 Ouroriginal expectation wasnotquite correct, inthat wemust addto thegeneralized force Qkanother term 6T/élqk inorder togettherate of change ofmomentum pk.Toseeitsmeaning, consider thekinetic energy ofaparticle interms ofplane polar coordinates, asgiven byEq.(9-14). Inthiscase, ‘g=mr92, (9-51) andifwemake useofEqs. (9-22) and(9-35), theequation ofmotion (9-50) forqk=ris mi‘=F,+mm? (9-52) Ifwecompare thiswith Eq.(3—207), which results from adirect applica- tion ofNewton’s lawofmotion, weseethat theterm 6T/6r ispart of themass times acceleration which appears here transposed totheright side oftheequation. Infact, 6T/61' isthe “centrifugal force” which must beadded inorder towrite theequation ofmotion forrintheform ofNewton’s equation formotion inastraight line. Had webeen abitmore clever originally, weshould have expected that some such term might have tobeincluded. Wemay call6T/6q;, a“ﬁctitious force” which ap- pears ifthekinetic energy depends onthecoordinate q1,.This willbe thecase when thecoordinate system involves “curved” coordinates, that is,ifconstant generalized velocities Q1,...,q3Nresult incurved motions ofsome parts ofthemechanical system. Equations (9-50) areusually written intheform d8T 6TE(ﬁ)—%_Q,,, k_1,...,3N., (9-5.3) Ifapotential energy exists, sothat theforces Q1,arederivable from apotential energy function [Eq. (9—33)], wemay introduce theLagrangian function L(q1;--':q3N; Q1,---iﬁszv; t)=T'_V: where Tdepends onboth q1,...,q3NandQ1,...,qw, butVdepends only onq1,...,q3N(and possibly t),sothat d6L d6Ta2@__%'3fq,c, (955) aL 6T aV 6T67%-'E—@-E+Qt. (9-56) Hence Eqs. (9-53) canbewritten inthis case intheform i(22)_%_ k—-1 3N 9-57 dtaqk aqk_o, . () 368 LAGRANGE'S EQUATIONS [CHAP. 9 Innearly allcases ofinterest inphysics (although notinengineering), theequations ofmotion canbeWritten intheform (9-57). The most important exception isthecase where frictional forces areinvolved, but such forces donotusually appear inatomic orastronomical problems. Since Lagrange’s equations have been derived from Newton’s equations ofmotion, they donotrepresent anewphysical theory, butmerely adiffer- entbutequivalent way ofexpressing thesame laws ofmotion. Asthe example ofEqs. (9-52) and (3—207) illustrates, theequations wegetby Lagrange’s method canalsobeobtained byadirect application ofNewton’s lawofmotion. However, incomplicated cases itisusually easier towork outthekinetic energy andtheforces orpotential energy ingeneralized coordinates, andwrite theequations inLagrangian form. Particularly in problems involving constraints, asweshall seeinSection 9-4, theLa- grangian method ismuch easier toapply. The chief value ofLagrange’s equations is,however, probably atheoretical one. From themanner in which they were derived, itisevident that Lagrange’s equations (9-57) or (9-53) hold inthesame form inanysystem ofgeneralized coordinates. It canalsobeveriﬁed bydirect computation (seeProblem 24)that ifEqs. (9-57) hold inanycoordinate system foranyfunction L(q1, ...,q;>,N;q1, ...,q3N; t),then equations ofthesame form hold inanyother coordinate system. TheLagrangian function Lhasthesame value, foranygiven set ofpositions andvelocities oftheparticles, nomatter inwhat coordinate system itmay beexpressed, buttheform ofthefunction Lmay bedifferent indifferent coordinate systems. Thefactthat Lagrange’s equations have thesame form inallcoordinate systems islargely responsible fortheir theoretical importance. Lagrange’s equations represent auniform way of writing theequations ofmotion ofasystem, which isindependent ofthe kind ofcoordinate system used. They form astarting point formore ad- vanced formulations ofmechanics. Indeveloping thegeneral theory of relativity, inwhich cartesian coordinates may noteven exist, Lagrange’s equations areparticularly important. 9-3Examples. Weﬁrst consider asystem ofparticles ml,...,my, located bycartesian coordinates, andshow that inthiscase Lagrange’s equations become theNewtonian equations ofmotion. The kinetic en- ergy is ~ T=—1-(vb?+12%+@?>, (9-58)Z'M*lg? and 6T 8T 6T553—i—(%-—bZ=O, (9-59) 6T . QT . 3T _55=mﬂo ,%=mi?/1', (E,=mizt (9-69) 9-4] SYSTEMS SUBJECT T0CONSTRAINTS 369 The generalized force associated with each cartesian coordinate isjust theordinary force, asweseeeither from Eq. (9-29), orbycomparing Eq.(9-28) with Eq.(9-26). Hence theequations ofmotion (9-53) are i(<’_T)_ﬁ_ .--._F.dtax, ax," m‘”‘_ "" d 6T .. . m,-y,~=F,-,,, [Z=1,...,N] (9-61) /'\Q;Q.=Q,i§‘°'ﬁvsLu)1 d 6T .._ E2 ""—' 77142,-F”. Foraparticle moving inaplane, thekinetic energy inpolar coordi- nates isgiven byEq.(9-14), andtheforces Q,andQ,byEqs. (9-35). The Lagrange equations are mi‘-W02=F,, (94a2) gi(mr29) =rF,,. (9-63) These equations were obtained inSection 3-13 byelementary methods. Wenowconsider therotating coordinate system deﬁned byEqs. (9-4) or(9-5). Thekinetic energy isgiven byEq.(9-16), andthegeneralized forces Q,andQ,willbethesame asintheprevious example. Lagrange’s equations inthiscase are ml‘—mr92 —-2mwr9 —nuozr =F,, (9-64) %(mrztl) +2mwri* =rF,. (9-65) The reader should verify that thethird term ontheleftinEq. (9-64) isthenegative ofthecoriolis force inther-direction duetotherotation ofthecoordinate system, and that thefourth term isthenegative of thecentrifugal force. The second term inEq.(9-65) isthenegative of thecoriolis torque inthe0—direction. Thus thenecessary ﬁctitious forces areautomatically included when wewrite Lagrange’s equations ina moving coordinate system. Itmust benoticed, however, that weusethe actual kinetic energy [Eq. (9—16)] with respect toacoordinate system atrest, expressed interms oftherotating coordinates, andnotthekinetic energy asitwould appear intherotating system ifweignored themotion ofthecoordinate system. A 9-4Systems subject toconstraints. Oneimportant class ofmechanical problems inwhich Lagrange’s equations areparticularly useful comprises systems which aresubject toconstraints. 370 LAGRANGE’S EQUATIONS [cnxrn 9 Arigid body isagood example ofasystem ofparticles subject to constraints. Aconstraint isarestriction onthefreedom ofmotion ofa system ofparticles intheform ofacondition which must besatisﬁed by their coordinates, orbytheallowed changes intheir coordinates. For example, avery simple hypothetical rigid body would beapair ofpar- ticles connected byarigid weightless rodoflength l.These particles are subject toaconstraint which requires that they remain adistance Zapart. Interms oftheir cartesian coordinates, theconstraint is [(502—$1)2 +(92*1/1)2 +(Z2—YZ1)2l1/2 =,l- (9456) Ifweusethecoordinates X,Y,Zofthecenter ofmass and spherical coordinates r,0,¢tolocate particle 2with respect toparticle 1asorigin, theconstraint takes thesimple form: 'r=l. (9-67) There arethus only ﬁvecoordinates X,Y,Z,0,<plefttodetermine. Each constraint which canbeexpressed intheform ofanequation like(9-66) enables ustoeliminate oneofthecoordinates bychoosing coordinates in such amanner that oneofthem isheld constant bytheconstraint. Fora rigid body, theconstraints require thatthemutual distances ofallpairs ofparticles remain constant. Forabody containing Nparticles, there areQ-N(N—1)pairs ofparticles. However, itisnothard toshow that itissufficient tospecify themutual distances of3N-6pairs, ifN23. Hence wecanreplace the3Ncartesian coordinates oftheNparticles by 3N—6mutual distances, 3coordinates ofthecenter ofmass, and3coor- dinates describing theorientation ofthebody. Since the3N—6mutual distances areallconstant, theproblem isreduced tooneofﬁnding the motion interms ofsixcoordinates. Another example ofasystem subject toaconstraint isthat ofabead sliding onawire. The wire issituated along acertain curve inspace, andtheconstraints require that theposi- tion ofthebead lieonthis curve. Since thecoordinates ofthepoints along aspace curve satisfy two equations (e.g., theequations oftwo surfaces which intersect along thecurve), there aretwoconstraints, and wecanlocate theposition ofthebead byasingle coordinate. (Can you suggest asuitable coordinate?) Ifthewire ismoving, wehave amoving constraint, and our single coordinate isrelative toamoving system of reference. Constraints which canbeexpressed intheform ofanequa- tionrelating thecoordinates arecalled holonomic. Alltheabove examples involve holonomic constraints. Constraints may also bespeciﬁed byarestriction onthevelocities, rather than onthecoordinates. For example, acylinder ofradius a, rolling andsliding down aninclined plane, with itsaxisalways horizontal, 9-4] SYSTEMS SUBJECT TOCONSTRAINTS 371 / /Q ” Fro. 9-3. Acylinder rolling down FIG. 9-4. Adiskrolling onahori- anincline. .» zontal plane. 'i canbelocated bytwocoordinates sand 0,asinFig. 9-3. The coordi- nate smeasures thedistance thecylinder hasmoved down theplane, andthecoordinate 0istheangle that aﬁxed radius inthecylinder has rotated from theradius tothepoint ofcontact with theplane. Now suppose that thecylinder isrolling without slipping. Then thevelocities .§and9must berelated bytheequation ' 5=ad, (9-68) which may alsobeWritten ds=ad0. (9-69) This equation canbeintegrated: 8-a0‘=0, (9-70) where Cisaconstant. This equation isofthesame type asEq.(9-66), andshows that theconstraint isholonomic, although itwasinitially ex- pressed interms ofvelocities. Ifaconstraint onthevelocities, likeEq. (9-68), canbeintegrated togive arelation between thecoordinates, like Eq.(9-70), then theconstraint isholonomic. There aresystems, however, inwhich such equations ofconstraint cannot beintegrated. Anexample isadisk ofradius arolling onahorizontal table, asinFig.9-4. Forsim- plicity, weassume that thedisk cannot tipover, andthat thediameter which touches thetable isalways vertical. Four coordinates arerequired tospecify theposition ofthedisk. The coordinates xandylocate the point ofcontact ontheplane; theangle <pdetermines theorientation of theplane ofthedisk relative tothezv-axis; andtheangle 0istheangle between aradius ﬁxed inthedisk andthevertical. Ifwenow require that thedisk rollwithout slipping (itcan also rotate about thevertical axis), thisimplies twoequations ofconstraint. Thevelocity ofthepoint ofcontact perpendicular totheplane ofthedisk must bezero: atsin <p+ycos <p=0, (9-71) 372 LAGRANGE’S EQUATIONS [CHAPQ 9 andthevelocity parallel totheplane ofthedisk must be atcos(p-3]singo=a0. (9-72) Itisnotpossible tointegrate these equations togettworelations between thecoordinates ac,y,0,go.Toseethis, wenote that byrolling thedisk without slipping, andbyrotating itabout avertical axis, wecanbring thedisk toanypoint ac,y,with anyangle <pbetween theplane ofthedisk andtheac-axis, andwith anypoint onthecircumference ofthedisk in contact with thetable, i.e.,anyangle 0.Forifthedisk isatanypoint ac,y,andthedesired point onthecircumference isnotincontact with the table, wemay rollthedisk around acircle whose circumference isof proper length, sothat when itreturns tox,y,thedesired point willbein contact with thetable. Itmay then berotated tothedesired angle ¢. This shows that thefour coordinates ac,y,0,<pareindependent ofone another, andthere cannot beanyrelation between them. Itmust there- forebeimpossible tointegrate Eqs. (9-71) and(9-72), andconsequently thisisanexample ofanonholonomic constraint. The number ofindependent ways inwhich amechanical system can move without violating anyconstraints which may beimposed iscalled thenumber ofdegrees offreedom ofthesystem. Tobemore precise, the number ofdegrees offreedom isthenumber ofquantities which must be speciﬁed inorder todetermine thevelocities ofallparticles inthesystem foranymotion which does notviolate theconstraints. Forexample, a single particle moving inspace hasthree degrees offreedom, butifitis constrained tomove along acertain curve, ithasonly one. Asystem ofNfreeparticles has3Ndegrees offreedom, arigid body has6degrees offreedom (three translational and three rotational), and arigid body constrained torotate about anaxis hasonedegree offreedom. The disk shown inFig. 9-4hasfour degrees offreedom ifitisallowed to sliponthetable, because weneed then tospecify ab,g,9,¢>.Butifthe disk isrequired torollwithout slipping, there areonly two degrees of freedom, because ifqbandanyoneofthevelocities ct,g,0aregiven, the remaining two canbefound from Eqs. (9-71) and (9-72). The disk is only freetoroll, andtorotate about avertical axis. Forholonomic sys- tems, thenumber ofdegrees offreedom isequal totheminimum number ofcoordinates required tospecify theconﬁguration ofthesystem when coordinates held constant bytheconstraints areeliminated. Nonholo- nomic constraints occur insome problems inwhich bodies rollwithout slipping, butthey arenotofvery great importance inphysics. Weshall therefore restrict ourattention toholonomic systems. Foraholonomic system ofNparticles subject tocindependent con- straints, wecanexpress theconstraints ascrelations which must hold 9-4] SYSTEMS SUBJECT TOCONSTRAINTS 373 between the3Ncartesian coordinates (including possibly thetime ifthe constraints arechanging with time): h1(x1; Z/17' ''2ZN; Z alr h2(x1; ylr '-'2zNi t)=a2; 1 I h6(x1: ylr -''1ZN; t)=av: where h1,...,hearecspeciﬁed flmctions. The number ofdegrees of freedom willbe f=3N-—c. (9-74) AsEqs. (9-73) areindependent, wemay solve them forcofthe3Ncar- tesian coordinates interms oftheother 3N—ccoordinates and the constants a1,...,ac.Thus only 3N—ccoordinates need bespeciﬁed, and theremainder can befound from Eqs. (9-73) iftheconstants a1,...,a,areknown. Wemay take asgeneralized coordinates these 3N—ccartesian coordinates andthecquantities a1,...,acdeﬁned by Eqs. (9-73), and held constant bytheconstraints. Orwemay deﬁne 3N—cgeneralized coordinates q1,..., qf,inany convenient way: ql='q1(%1, y1!' ''1zNit): q?=‘q2(@1, ylr '''2zN; t); qf=qf(x1; ll/1: ''-2ZN; t)- Equations (9-73) and (9-75) deﬁne asetof3Ncoordinates q1,...,qf; a1,...,a,,andareanalogous toEqs. (9-1). They may besolved forthe cartesian coordinates: :01=x1(q1,...,q,»; a1,...,ac; t), :1/1=1/1(q1, '''rqfi a1) ---;ac; t); ' Z1v=Z1v(q1,---,q/;<11,---Ah;t)- Now letQ1,...,Q),Q;+1, ...,Q,-+0 bethegeneralized forces corre- sponding tothecoordinates q1,...,qf;a1,...,ac.Wehave then aset ofLagrange equations fortheconstrained coordinates and another for theunconstrained coordinates: d6T 6T-_--= k=1,...,, 9-77dt aqk Q/61 .f ( ) d6T 6T .———-—= - = ...' = .9-dt Qf-l-.7! J 17 Icl6+f ( 374 LAGRANGE’S EQUATIONS [CHAP- 9 The importance ofthis separation oftheproblem into two groups of equations isthat theforces ofconstraint canbesochosen that they do nowork unless theconstraints areviolated, asweshall show inthenext paragraph. Ifthisistrue, then according tothedeﬁnition (9-30) ofthe generalized force, theforces ofconstraint donotcontribute tothegen- eralized force Q1,associated with anunconstrained coordinate qk.Since thevalues oftheconstrained coordinates a1,...,acareheld constant, wecansolve Eqs. (9-77) forthemotion ofthesystem interms ofthe coordinates q1,...,qf,treating a1,...,a,asgiven constants, without knowing theforces ofconstraint. This isagreat advantage, forthe forces ofconstraint depend upon how thesystem ismoving, andcannot, ingeneral, bedetermined until after themotion hasbeen found. All weusually know about theconstraining forces isthat they have whatever values arerequired tomaintain theconstraints. Having solved Eqs. (9-77) forq1(t), ...,q;(t), wemay then, ifwewish, substitute these functions inEqs. (9-78) and calculate theforces ofconstraint. This may bea matter ofconsiderable interest totheengineer who needs toverify that theconstraining members arestrong enough towithstand theconstrain- ingforces. Lagrange’s equations thus reduce theproblem ofﬁnding the motion ofanyholonomic system with fdegrees offreedom totheproblem ofsolving fsecond-order differential equations (9-77). When wespeak ofthegeneralized coordinates, theconstrained coordinates a1,..., a, may ormay notbeincluded, asconvenient. Ifabead slides onafrictionless wire, thewire canonly exert constrain- ingforces perpendicular toitself, sothat nowork isdone onthebead solong asitstays onthewire.* Ifthere isfriction, wecanseparate -the force onthebead into acomponent perpendicular tothewire which holds thebead onthewire without doing any work, and africtional component along thewire which does work andwilltherefore have tobe included inthegeneralized force associated with motion along thewire. Ifthe.frictional component depends ontheperpendicular component, as itdoes fordrysliding friction, then wecannot solve Eqs. (9-77) ﬁrst, independently ofEqs. (9-78), andonegreat advantage oftheLagrangian method islost. Iftwoparticles areheld aﬁxed distance apart byarigid rod, then byNewton’s third law, theforce exerted bytherodonone particle isequal and opposite tothat ontheother. Itwas shown in *Ifthewire ismoving, theforce exerted bythewire may dowork onthe bead, butthevirtual displacements interms ofwhich thegeneralized forces have been deﬁned aretobeimagined astaking place ataﬁxed instant oftime, and forsuch adisplacement which does notviolate theconstraints, nowork isdone. Hence even inthecase ofmoving constraints, theconstraining forces donot appear inthegeneralized forces associated with theunconstrained coordinates. 9-5] EXAMPLES orSYSTEMS SUBJECT TOCONSTRAINTS 375 Section 5-1that nonetwork isdone onthesystem bytherodsolong as theconstraint isnotviolated, that is,solong astherodisnotstretched orcompressed. Asimilar situation willbefound inallother cases; the constraints could always bemaintained byforces which donowork. Iftheforces Q1,...,Q;arederivable from apotential energy func- tion, then Wecandeﬁne aLagrangian function L(q1, ...,q,-;q'1,...,Q/) which may insome cases depend ont,andwhich may alsodepend onthe constants a1,...,ac.The ﬁrst fLagrange equations (9-77) canthen bewritten intheform d6L 6L—-—,———=(), k=1,...,. 9-79dwqt ac. f <) 9-5Examples ofsystems subject toconstraints. Asimple mechanical system involving constraints istheAtwood’s machine shown inFig. 9-5. Weights m1,m2areconnected byarope oflength lover aﬁxed pulley. Weassume theweights move only vertically, sothat wehave only one degree offreedom. Wetake ascoordinates thedistance 2:ofm1below thepulley axle, andl,thelength oftherope. The coordinate liscon- strained tohave aconstant value, andcould beleftoutofconsideration from thestart ifwewish only toﬁnd themotion. Ifwealso want to ﬁnd thetension inthestring, wemust include lasacoordinate. The kinetic energy is T=aw+%m2(l-92- <9-80> Theonly forces acting onm1andm2arethetension 1'intherope andthe force ofgravity. The work done when :7:increases by6.1:,lremaining iﬁ1-! mi!) M2 my Fro. 9-5. Atwood’s machine.T I 376 LAGRANGE,S EQUATIONS [CHAP. 9 constant, is 8W=(m1g —7)620—-(mzg —T)6x =(mi—m2)!I59¢=Q»59¢, (9-81) sothat Q9=(mi—m2)g- (9-82) Note that Q,isindependent of7'.The work done when Zincreases by 51,acremaining constant, is 6W=(m2g —1')51=Q161, (9-83) sothat Q1=7'I'L2§ -7'. (9-84) Notice that inorder toobtain anequation involving theforce ofcon- straint 7',wemust consider amotion which violates theconstraint. This isalso true ifwewish tomeasure aforce physically; wemust allow at least asmall motion inthedirection oftheforce. TheLagrange equations ofmotion are(since i=I=0) % —2.,-Z,"=(ml+mar=(m.-mag. <9-85> d6T 6T . E? —-67- =-771.23? =mgg —T. The ﬁrst equation istobesolved toﬁndthemotion: 1:=mo+vot+%%:—_Tf:%: gt2. (9-87) The second equation canthen beused toﬁndthetension 7'necessary to maintain theconstraint: 21'='m2(9+ri)=fir”-,;—2 9- (9-88) Inthis case thetension isindependent oftime and canbefound from Eqs. (9-85) and(9-86) immediately, although inmost cases thecon- straining forces depend onthemotion andcanbedetermined only after themotion isfound. Equations (9-85) and(9-86) have anobvious phys- icalinterpretation andcould bewritten down immediately from elemen- tary considerations, aswas done inSection 1-7. Aproblem oflittle practical importance, butwhich isquite instructive, isthat inwhich onecylinder rolls upon another, asshown inFig. 9-6. Thecylinder ofradius aisﬁxed, andthecylinder ofradius aarolls around itunder theaction ofgravity. Suppose wearegiven that thecoeﬁicient 9-5] EXAMPLES orSYSTEMS summer TOCONSTRAINTS 377 \ FIG. 9-6. One cylinder rolling onanother. ofstatic friction between thecylinders is/.4,thecoefﬁcient ofsliding friction iszero,* andthat themoving cylinder starts from restwith its center vertically above thecenter oftheﬁxed cylinder. Weshall assume that theaxisofthemoving cylinder remains horizontal during themotion. Itisadvisable inallproblems, andessential inthisone, tothink carefully about themotion before attempting toﬁndthemathematical solution. Itisclear that themoving cylinder cannot rollalltheway around the ﬁxed cylinder, forthenormal force Fwhich isexerted bytheﬁxed cyl- inder onthemoving onecanonly bedirected outward, never inward. Therefore atsome point, themoving cylinder willﬂyofftheﬁxed one. The point atwhich itﬂies offisthepoint atwhich F=0. (9-s9) Furthermore, thecylinder cannot continue torollwithout slipping right uptothepoint atwhich itﬂies off,forthefrictional force fwhich pre- vents slipping islimited bythecondition fS/JLF, (9-90) andwillcertainly become toosmall toprevent, slipping before thepoint atwhich Eq. (9-89) holds. The motion therefore isdivided into three parts. Atﬁrst thecylinder rolls without slipping through anangle 01 determined bythecondition f=pF. (9-91) *This implies that themoving cylinder either rolls without slipping, ifthe static friction isgreat enough, orslips without anyfriction atall. The lattcr assumption ismade tosimplify theproblem. 378 LAGRANGE’S EQUATIONS [crm1>. 9 Beyond theangle 01,thecylinder slides without friction until itreaches theangle 02determinedlay Eq. (9-89), after which itleaves theﬁxed cylinder andfallsfreely. Wemay anticipate some mathematical difficulties with theinitial part ofthemotion duetothefact that theinitial posi- tion ofthemoving cylinder isoneofunstable equilibrium. Physically there isnodifficulty, since theslightest disturbance willcause thecylinder torolldown, butmathematically there may beadiﬂiculty which wemust watch outfor,inasmuch astheneeded slight disturbance willnotappear intheequations. Letusﬁndthat part ofthemotion when themoving cylinder rolls with- outslipping. There isthen only onedegree offreedom, and weshall specify theposition ofthecylinder bytheangle 0between thevertical and thelineconnecting thecenters ofthetwocylinders. Inorder tocompute thekinetic energy, weintroduce theauxiliary angle <pthrough which the moving cylinder hasrotated about itsaxis. The condition that thecylin- derrollwithout slipping leads totheequation ofconstraint: 1 ad=aa(¢ -9), (9-92) which canbeintegrated intheform _ (1—|—a)0=mp. (9-93) Ifwewere concerned onlywiththerolling motion, wecould nowproceed tosetuptheLagrange equation for0,butinasmuch asweneed toknow theforces ofconstraint Fandf,itisnecessary tointroduce additional co- ordinates which aremaintained constant bythese constraining forces. Thefrictional force fmaintains theconstraint (9-93), andanappropriate coordinate is ___°‘i'i_. _ "Y-0 1+0‘ (994) Solong asthecylinder rolls without slipping, ‘Y=0;'Ymeasures theangle ofsliparound theﬁxed cylinder. The normal force Fmaintains thedis- tance rbetween thecenters ofthecylinders: r=a-|-aa=(1—|—a)a. (9-95) Thekinetic energy oftherolling cylinder istheenergy associated with the motion ofitscenter ofmass plus therotational energy about thecenter of mass: T=%'m(1‘2 +r292)+%I¢2- (9-96) After substituting (0from Eq.(9-94), andsince I=%rna’a’, forasolid cylinder ofradius aa,wehave T=gm? -1-%mr2(i2 -1-ﬁm(1 +a)2a2(92 -—2'79—|—'72). (9-97) 9-5] EXAMPLES orSYSTEMS SUBJECT TOCONSTRAINTS 379 Theequations ofconstraint [Eq. (9-95) and7=0]must notbeused until after theequations ofmotion arewritten dovsm. The generalized forces aremost easily determined with thehelp ofEq.(9-30); they are* Q,=mgrsin0, (9-98) Q1=—f<1(1 +<1). (9-99) Q,=F—mgcos0. (9-100) The Lagrange equations for0,‘Y,andrarenow m[r2 -|-%a2(1 -|—a)2]5 +2mrr0 —~%ma2(1 -l—oz)2'l; =mgrsin0,(9-101) —-%ma2(1 +(1)29 -|—%ma2(1 +a)2'f’ =—fa(1 —|-oz), (9—102) mi‘—mrﬂz =F—mgcos0. (9—103) Wecannow insert theconstraints 'Y=0andr=(1+a)a,sothat these equations become 3-(1+a)2ma29 =(1+a)mga sin0, (9—104) f=-§(1+a)mad, (9—105) F=mgcos0—(1+a)ma92. (9—106) Had weignored theterms involving ‘iinthekinetic energy, the0equation, which determines themotion, would have come outcorrectly, butthe equation fortheconstraining force fwould have been missing aterm. This happens when theconstrained coordinates arenotorthogonal tothe unconstrained coordinates, since across term ('90) then appears inthe kinetic energy. The equation ofmotion (9—104) canbesolved bytheenergy method. The total energy, solong asthecylinder rolls without slipping, is +}(1+a)2ma2d2 +(1+a)mga cos0=E, (9-107) and isconstant, ascaneasily beshown from Eq. (9—104), and aswe know anyway since thegravitational force isconservative andtheforces ofconstraint donowork. Since themoving cylinder starts from rest at0=0, E=(1-1-a)mga. (9-108) *The reader willﬁnditaninstructive exercise toverify these formulas. 380 LAGRANGE’S EQUATIONS [CHAP. 9 Wesubstitute this inEq. (9-107) and solve for9: 1/20=2 sin (9-109) where .2 Wecannow integrate toﬁnd 0(t):/”<9”/‘0 1‘ F 0dt, 1/2[Intan = 7. (9-112) When wesubstitute thelower limit 0=0,weruninto adiﬂiculty, for ln0=—-ool This istheexpected diﬂiculty duetothefact that 0=0 isapoint ofequilibrium, albeit unstable. Ifthere isnodisturbance what- ever, itwilltake aninﬁnite time forthecylinder torollofftheequilibrium point. Letussuppose, however, thatitdoes rolloffduetosome slight disturbance, andletustakethetime t=0asthetimewhen theangle 0 hassome small value 00.There isnownodifficulty, andwehave 1/2 tan2=(tan %9>exp t]- (9-113) Ast—>oo,0-—>211',andthemoving cylinder rolls alltheway around theﬁxed one, iftheconstraints continue tohold. The rolling constraint holds, however, only solong asEq. (9-90) holds. When wesubstitute from Eqs. (9-105), (9-106), and (9—109), Eq. (9-90) becomes §mgsin0§%,umg(7 cos0—4). (9-114) At0=0,this certainly holds, sothat thecylinder does initially roll, aswehave supposed. At0=1r/2, however, itcertainly does nothold, since theleftmember isthen positive andtheright, negative. Theangle 01 atwhich slipping begins isdetermined bytheequation sin01=;L(7cos01—4), (9—115) whose solution is 2s21332"2cos0,= . (9-116) 9-6] CONSTANTS OF THE MOTION AND IGNORABLE COORDINATES 381 Thesecond part ofthemotion, during which themoving cylinder slides without friction around theﬁxed one,canbefound bysolving Eqs. (9-101) and (9-102) for9(t), ’Y(t), with f=0andwith only thesingle constraint r=(1+a)a, andwith initial values 9=91,9=91,determined from Eqs. (9-116) and (9-109). The solution canbefound without essential difﬁculty, andtheangle 92atwhich themoving cylinder leaves theﬁxed onecanthen bedetermined from Eqs. (9-106) and(9-89). These calcu- lations arelefttothereader. 9-6Constants ofthemotion andignorable coordinates. Weremarked inChapter 3that onegeneral method forsolving dynamical problems is tolook forconstants ofthemotion, that is,functions ofthecoordinates and velocities which areconstant intime. One common case inwhich such constants canbefound arises when thedynamical system ischarac- terized byaLagrangian function inwhich some coordinate qtdoes not occur explicitly. The corresponding Lagrange equation (9-57) then re- duces to d9La -0. (9-117) This equation canbeintegrated immediately: (€—gc=pk=aconstant. (9—118) Thus, whenever acoordinate q1,does notoccur explicitly intheLagrangian function, thecorresponding momentum pkisaconstant ofthemotion. Such acoordinate qkissaid tobeignorable. Ifqkisignorable, wecan solve Eq.(9—118) for<11,interms oftheother coordinates andvelocities, and oftheconstant momentum pk,and substitute intheremaining Lagrange equations toeliminate q,,andreduce byonethenumber ofvari- ables intheproblem; (qkwas already missing from theequations, since itwas assumed ignorable.) When theremaining variables have been found, they canbesubstituted inEq. (9-118), togive 4,,asafunction oft;q1,isthen obtained byintegration. Ifallbutoneofthecoordinates areignorable, theproblem canthus bereduced toaone-dimensional problem andsolved bytheenergy integral method, ifLdoes notdepend onthetime texplicitly. Forexample, inthecase ofcentral forces, thepotential energy depends only onthedistance rfrom theorigin, sothat ifweusepolar coordinates r,9inaplane, Visindependent of9.Since Tisalso independent of9 according toEq.(9-14) (Tdepends ofcourse on9),wewillhave aL a6-,,_(Q(T-V)_0, (9-119) 382 LAGRANGE’S EQUATIONS [CHAP. 9 andhence 6L 259-=mr9=pa=aconstant, (9-120) aresult which weobtained inSection 3-13 byadifferent argument. We seethat theconstancy ofpgisaresult ofthefactthat thesystem issym- metrical about theorigin, sothat Lcannot depend on9.Ifasystem of particles isacted onbynoexternal forces, then ifwedisplace thewhole system inany direction, without changing thevelocities and relative positions oftheparticles, there willbenochange inTorV,orinL.If X,Y,andZarerectangular coordinates ofthecenter ofmass, and if theremaining coordinates arerelative tothecenter ofmass, sothat changing Xcorresponds todisplacing thewhole system, then 6Lif—0, (9-12 1) andtherefore PX,thetotal linear momentum inthew-direction, willbe constant, aresult weproved inSection 4-1byadifferent method. Itisofinterest toseehow toshow from Lagrange’s equations that the total energy isaconstant ofthemotion. Inorder toﬁndanenergy inte- gral oftheequations ofmotion inLagrangian form, itisnecessary to know how toexpress thetotal energy interms oftheLagrangian func- tionL.Tothisend,letusconsider asystem described interms ofaﬁxed system ofcoordinates, sothat thekinetic energy Tisahomogeneous quadratic function ofthegeneralized velocities (11,...,Q;[i.e., T1= To=0inEq. (9—13)]. ByEuler’s theorem,* wehave ’aT'—.-=2T. 9-122 lgqkaqk ( ) Thus if L=T2—V, (9-123) where Visafunction ofthecoordinates q1,...,q;alone, then, byEq. (9—122), ’aLZq,,a—,-L=T+V=E. (9-124) k=1 qk Wenow consider thetime derivative oftheleftmember ofEq.(9-124). Forgreater generality, weshall atﬁrst allow Ltodepend explicitly ont. *W.F.Osgood, Advanced Calculus. New York: Macmillan, 1937.(Page 121.) Thereader unfamiliar with Euler’s theorem canreadily verify Eq.(9-122) for himself bysubstituting forT=T2from Eq.(9-9). 9-6] CONSTANTS orTHEMOTION ANDIGNORABLE COORDINATES 383 Inthecase Wehave considered, Ldoes notdepend explicitly ont.There arecases, however, when asystem issubject toexternal forces that change with time andthat canbederived from apotential Vthat varies with time. Anexample would beanatom subject toavarying external electric ﬁeld. Insuch cases, theequations ofmotion canbewritten intheLa- grangian form (9-57) with theLagrangian depending explicitly onthe time t.Inthecase ofmoving coordinate systems also, theLagrangian may depend onthetime even though theforces areconservative. The time derivative oftheleftmember ofEq. (9—124) is E qk5q—'“L)—2[ﬁk'E'l'qlc;i‘i(%c‘>'_'(E(.lk_‘ﬁqk:|—g ’.aaL 6L] aL aL =,2q'°l9(6q1.)* aqk _35=_atl($125)d<i: aL _’ aL .d aL aL aL aL k=1 '2 k= U-¢ IfLdoes notdepend explicitly ont,theright sideofEq.(9—125) iszero, and 2qk5_——L=aconstant. (9-126) k-1 qk When Lhastheform (T2—V),asinastationary coordinate system, thisistheconservation ofenergy theorem. Regardless oftheform ofL, Eq.(9-126) represents anintegral ofLagrange’s equations (9-57), when- ever Ldoes notcontain texplicitly, buttheconstant quantity onthe leftisnotalways thetotal energy. Note theanalogy between thecon- servation ofgeneralized momentum pkwhen Lisindependent ofqk,and theconservation ofenergy when Lisindependent oft.There aremany ways inwhich therelation between time andenergy isanalogous tothe relation between acoordinate andthecorresponding momentum. Wehave seen that thefamiliar conservation laws ofenergy, momentum, andangular momentum canberegarded asconsequences ofsymmetries exhibited bythemechanical systems towhich they apply; that is,they are consequences ofthefactthat theLagrangian function L,which determines theequations ofmotion, isindependent oftime andoftheposition and orientation oftheentire system inspace. This result, derived here for classical mechanics, holds generally throughout physics. Inquantum mechanics andinrelativity theory, even when weinclude electromagnetic andother kinds offorce ﬁelds, conservation laws areassociated with sym- metries inthefundamental equations. Wemight, forexample, deﬁne energy asthat quantity which isconstant because thelaws ofphysics are always thesame (ifindeed they arel). I 384 LAGRANGE’S EQUATIONS [crnu>. 9 9-7Further examples. The spherical pendulum isasimple pendulum freetoswing through theentire solid angle about apoint. Thependulum bobisconstrained tomove onaspherical surface ofradius R.Welocate thebob bythespherical coordinates 9,<p(Fig. 9-7). Wemay include thelength Rofthependulum asacoordinate ifwewish toﬁndtheten- sion inthestring, butweomit ithere, asweareconcerned only with ﬁnding themotion. Ifthebobswings above thehorizontal, wewillsup- pose that itstillremains onthesphere, which would betrue ifthestring were replaced byarigid rod. Otherwise theconstraint disappears when- ever acompressional stress isrequired tomaintain it,since astring will support only atension andnotacompression. Thevelocity ofthebobis v=R91+Rsin9(bm. (9—127) Hence thekinetic energy is T=211102=2-mR202 +%mR2$1112092. (9-128) Thepotential energy duetogravity, relative tothehorizontal plane, is V=mgR cos9. (9—129) Hence theLagrangian function is L=T—V=2mR292 -|—%mR2 sin29 (52—-mgR cos9.(9—130) TheLagrange equations are %(mR29) —mR2¢2 sin9cos9—mgR sin9=0, (9—131) gt(111122$11120.,>)=0. (9-132) The coordinate (0isignorable, andthesecond equation canbeintegrated immediately : mR2 sin29 ¢=pk=aconstant. (9—133) Also, since %=0, (9-134) thequantity L .L .0%,,-+11-L=211112202 +2111122$11120¢2+mgR110$0(9-135) isconstant, byEq.(9-126). Werecognize thequantity ontheright as thetotal energy, asitshould be,since weareusing aﬁxed coordinate 9-7] FURTHER EXAMPLES 385 Z (V7 w y $9 x R 9 M0R 01 0»-0 m I1 10m T/2 7,, 1 n —MgR pk=0 FIG. 9-7. Aspherical pendulum. Fro. 9-8. Eﬂ’ective potential ‘V’(9) forspherical pendulum. system. Calling thisconstant E,andsubstituting for¢from Eq.(9—133), wehave 2 %mR292 + +mgR cos9=E. (9—136) Wemay introduce aneffective potential ‘V’(9) forthemotion: 2‘V’(9)=mgR00$0+ . (9-137) sothat %mR292 =E-—‘V’(9). (9—138) Since theleftmember cannot benegative, themotion isconﬁned tothose values of9forwhich ‘V’(9) gE.Theeffective potential ‘V’(9) isplotted inFig. 9-8. Weseethat forpk=0,‘V’(9) isthepotential curve fora simple pendulum, with aminimum at9=7randamaximum at9=0. ForE=—mgR, thependulum isatrestat9=1r.FormgR >E> —mgR, thependulum oscillates about 9=1r.ForE>mgR, thependulum swings inacircular motion through thetopand bottom points 9=0 and7r.When pk-50,themotion isnolonger that ofasimple pendulum, and‘V’(9) nowhasaminimum atapoint 90between 1r/2and1r,and rises toinﬁnity at9=0and9=rr.The larger pk,thelarger themini- mum value of‘V’(9), andthecloser 911isto7r/2. IfE=‘V’(91,), then 9 isconstant andequal to90,andthependulum swings inacircle about thevertical axis. Aspk—>oo,thependulum swings more and more nearly inahorizontal plane. ForE>‘V’(90), 9oscillates between a maximum and minimum value while thependulum swings about the vertical axis. The reader should compare these results with hismechan- ical intuitions orhisexperience regarding themotion ofaspherical 386 LAGRANGPYS EQUATIONS [cn.u>. 9 pendulum. The solution ofEq. (9—138) for9(t)cannot becarried out interms ofelementary functions, butwecantreat circular andnearly circular motions very easily. The relation between pkand90foruniform circular motion ofthependulum about thez-axis is d‘V’ . p2cos9[ﬁlo =—-mgR sin90—E =0. (9—139) Itisevident from this equation that 90>7r/2, andthat 911—>7r/2 as pk—>oo. Bysubstituting from Eq. (9-133), weobtain arelation be- tween 1band 90foruniform circular motion: -2__9__l_. _ ‘R_R(—-cos 90) (9140) The energy foruniform circular motion atanangle 90,ifweuseEqs. (9—136) and (9-139), and thefact that 9=0,is -i . 2<-11> For anenergy slightly larger than E0,and anangular momentum pk given byEq.(9—139), theangle 9willperform simple harmonic oscilla- tions about thevalue 911. Forifweset d2‘V’ mgR »11=[Flea =:58} (1+3111111201,), (9-142) then, forsmall values of9—911,wecanexpand ‘V’(9) inaTaylor series: ‘V’(0) -E0+211(0—0k)2. (9-143) Theenergy equation (9—138) now becomes 211112202 +%k(9-0.,)2=E-E0. (9-144) This istheenergy foraharmonic oscillator with energy E’—E0,coordi- nate 9—90,mass mR2, spring constant lc.The frequency ofoscillation in9istherefore given by k 1320‘*2=W=ii (2245) This oscillation in9issuperposed upon acircular motion &I'0\1Ild the z-axis with anangular velocity ¢given byEq. (9-133); 1bwill vary slightly as9oscillates, butwillremain very nearly equal totheconstant value given byEq.(9-140). Itisofinterest tocompare 11':andw: 9-7] FURTHER EXAMPLES 387 -211%=m,W1.,' (2429 Since 90>7r/2, thisratio islessthan 1,sothat w>¢,andthependulum wobbles upanddown asitgoes around thecircle. At90=1r/2, 1b=w, andthependulum moves inacircle whose plane istilted slightly from the horizontal; thiscase occurs only inthelimit ofvery large values ofpk.It isclear physically that when pkissolarge that gravity may beneglected, themotion canbeacircle inanyplane through theorigin. Canyoushow thismathematically? Near 90=0,11>=2113,sothat 9oscillates twice per revolution andthependulum bobmoves inanellipse whose center ison thez-axis. This corresponds tothemotion ofthetwo-dimensional har- monic oscillator discussed inSection 3-10, with equal frequencies inthe twoperpendicular directions. _ Asalast example, weconsider asystem inwhich there aremoving constraints. Abead ofmass mslides without friction onacircular hoop ofradius a.The hoop liesinavertical plane which isconstrained to rotate about avertical diameter with constant angular velocity w.There isjust onedegree offreedom, andinasmuch aswearenotinterested in theforces ofconstraint, wechoose asingle coordinate 9which measures theangle around thecircle from thebottom ofthevertical diameter to thebead (Fig. 9-9). Thekinetic energy isthen T=1}-ma292 +1§~ma2w2 sinz9, (9—147) andthepotential energy is 9 V=-mga cos9. (9-148) TheLagrangian function is L=2ma292 +%ma2w2 sinz9-1-mgacos9. ,(9—149) Q“ ‘V, +0101» 7r/2 O la .,, co5wk 9E -mga w>we "10 FIG. 9-9. Abead sliding onarota- FIG. 9-10. Effective potential en- ting hoop. ergy forsystem shown inFig. 9-9.1»-9 388 LAGRANGE’S EQUATIONS [cn,u>. 9 The Lagrange equation ofmotion caneasily bewritten out, butthisis unnecessary, forwenotice that 9L91'-°» andtherefore, byEq.(9-126), thequantity 9%€-—_L=§*ma292 —irmagwg sing9—mgacos9=‘E’ (9—150) isconstant. The constant ‘E’isnotthetotal energy T-1-V,forthe middle term hasthewrong sign. The total energy isevidently notcon- stant inthis case. (What force does thework which produces changes inT+V?) Wemay note, however, that wecaninterpret Eq.(9-149) asaLagrangian function interms ofaﬁxed coordinate system with the middle term regarded aspart ofaneffective potential energy: ‘V’(9) =—§ma2w2 sinz9—mgacos9. (9—151) Theenergy according tothisinterpretation is‘E’. Theﬁrstterm in‘V’(9) isthepotential energy associated with thecentrifugal force which must beadded ifweregard therotating system asﬁxed. The effective poten- tialisplotted inFig. 9-10. The shape ofthepotential curve depends on whether coisgreater orlessthan acritical angular velocity w.=(0/<1)1’2- (9-15?) Itislefttothereader toshow this, andtodiscuss thenature ofthemotion ofthebead inthetwocases. 9-8Electromagnetic forces andvelocity-dependent potentials. Ifthe forces acting onadynamical system depend upon thevelocities, itmay bepossible toﬁnd afunction U(q1, ...,qf;01,...,0,;t)such that d9U BU=-——-———1 k= ... . Q1d,6,,8,, 1..1 9-153) Ifsuch afunction Ucanbefound, then wecandeﬁne aLagrangian function 2 L=T—U, - (9-154) sothat theequations ofmotion (9-53) canbewritten intheform (9-57): d8L 6L——_-——-=0, l<;=1,...,. 1 dtaqk 3911 f (9_55) 9-8] ELECTROMAGNETIC FORCES, VELOCITY-DEPENDENT POTENTIALS 389 Thefunction Umaybecalled avelocity-dependent potential. Ifthere are alsoforces derivable from anordinary potential energy V(q1, ...,qf),V may beincluded inU,since Eq.(9-153) reduces toEq.(9-33) forthose terms which donotcontain thevelocities. The function Umay depend explicitly onthetime t.Ifitdoes not, andifthecoordinate system isa ﬁxed one, then Lwillbeindependent oft,andthequantity ’aLE= '—_—L, 9-156 lgqtaqk ( ) willbeaconstant ofthemotion, according toEq.(9-126). Inthiscase, wemay saythat theforces areconservative even though they depend on thevelocities. Itisclear from this result that itcannot bepossible to express frictional forces intheform (9—153), forthetotal energy isnot constant when there isfriction unless weinclude heat energy, andheat energy cannot bedeﬁned interms ofthecoordinates and velocities q1,...,qf;q'1,...,11,,and hence cannot beincluded inEq. (9—156). Itisnothard toshow that ifthevelocity-dependent parts ofUarelinear inthevelocities, asthey areinallimportant examples, theenergy E deﬁned byEq.(9—156) isjustT+V,where Vistheordinary potential energy andcontains theterms inUthatareindependent ofthevelocities. Asanexample, aparticle ofcharge qsubject toaconstant magnetic ﬁeldBisacted onbyaforce (guassian units) F=gv>1B, (9-157) OI‘ Fa: ='%(yBz '_éB1l)s F,=g(es,-103,), (9-158) F,=5(113,,-113.). Equations (9—158) have theform (9-153) if U=9(1.012.+11012,,+1',). (9-159) a:Bc Itis,infact, possible toexpress theelectromagnetic force intheform (9—153) forany electric and magnetic ﬁeld. The electromagnetic force onaparticle ofcharge qisgiven byEq.(3-283): F=qE+gv>1B. (9-100) 390 LAoRANoE’s EQUATIONS [CHAP. 9 Itisshown inelectromagnetic theory* that foranyelectromagnetic ﬁeld, itispossible todeﬁne ascalar function ¢(x,y,z,t)andavector function A(:r, y,z,t)such that 16A B=VXA. (9—162) The function 11>iscalled thescalar potential, andAiscalled thevector potential. Ifthese expressions aresubstituted inEq.(9—160), weobtain F=-qv¢-§%+§v><(v><A). (9-163) The lastterm canberewritten using formula (3-35) forthetriple cross product: F=—qV¢ -g%%-gv-VA +5v(v-A). (9-164) [The components ofvare(dc,1],2)andareindependent of:0,y,z,sothat v isnotdifferentiated bytheoperator V.] The twomiddle terms canbe combined according toEq.(8-113): F=—qV¢-3%+Z1v(v-A), (9-105) where dA/dt isthetime derivative ofAevaluated attheposition ofthe moving particle. Itmay now beveriﬁed bydirect computation that the potential function U=q¢—gv-A, (9-166) when substituted inEqs. (9—153), with q1,q2,q3=:0,y,z,yields the components oftheforce Fgiven byEq.(9-165). Itisalsoeasy toshow that theenergy Edeﬁned byEq. (9—156) with L=T—Uis E=T+04>. (9-167) IfAand ¢areindependent oft,then Lisindependent oftinaﬁxed coordinate system and theenergy Eisconstant, aresult derived bymore elementary methods inSection 3-17 [Eq. (3—288)]. When there isavelocity dependent potential, itiscustomary todeﬁne themomentum interms oftheLagrangian function, rather than interms *See, e.g., Slater and Frank, Electromagnetism. New York: McGraw-Hill Book Co., 1947. (Page 87.) 9-9] 1.AeRANeE’s EQUATIONS FORTHEVIBRATING STRING 391 ofthekinetic energy: 9LP11— - (9—168) Ifthepotential isnotvelocity dependent, then thisdeﬁnition isequivalent toEq.(9-23). Inanycase, itis0L/04,. whose time derivative occurs in theLagrange equation forqk,and which isconstant ifqkisignorable. Inthecase ofaparticle subject toelectromagnetic forces, themomentum components pk,pk,pkwillbe,byEqs. (9-168) and(9-166), pk=mi+gA11. 0..-m0+§A... <9-169) pk=mé—|—gA2. Thesecond terms play theroleofapotential momentum. Itappears that gravitational forces, electromagnetic forces, andindeed allthefundamental forces inphysics canbeexpressed intheform (9—153), forasuitably chosen potential function U.(Frictional forces wedonot regard asfundamental inthissense, because they areultimately reducible toelectromagnetic forces between atoms, andhence areinprinciple also expressible intheform (9—153) ifweinclude allthecoordinates ofthe atoms andmolecules ofwhich aphysical system iscomposed.) Therefore theequations ofmotion ofanysystem ofparticles canalways beexpressed intheLagrangian form (9-155), even when velocity dependent forces are present. Itappears that there issomething fundamental about theform ofEqs. (9-155). Oneimportant property ofthese equations, aswehave already noted, isthat they retain thesame form ifwesubstitute anynew setofcoordinates forq1,...,qf.This canbeveriﬁed byastraightforward, ifsomewhat tedious, calculation. Further insight into thefundamental character oftheLagrange equations must await thestudy ofamore ad- vanced formulation ofmechanics utilizing thecalculus ofvariations, which isbeyond thescope ofthisbook.* 9-9Lagrange’s equations forthevibrating string. The Lagrange method canbeextended alsotothemotion ofcontinuous media. Weshall consider only thesimplest example, thevibrating string. Using thenota- tion ofSection 8-1, wecouldtake u(x) asasetofgeneralized coordinates analogous toqk.Inplace ofthesubscript lcdenoting thevarious degrees offreedom, wehave theposition coordinate :1:denoting thevarious points *See, e.g., H.Goldstein, Classical Mechanics. Reading, Mass.: Addison- Wesley, 1950. (Chapter 2.) 392 LAGRANGE’S EQUATIONS [CHAP. 9 onthestring. The number ofdegrees offreedom isinﬁnite foranideal continuous string. The generalization, oftheLagrange method todeal with acontinuous index :1:denoting thevarious degrees offreedom intro- duces mathematical complications which wewish toavoid here.* There- forewemake useofthepossibility ofrepresenting thefunction u(x) asa Fourier series. According totheFourier series theorem quoted inSection 8-2, ifthe string istied attheends x=0,l,wecanrepresent itsposition u(x) by theseries (8-24): °° .k1ru(:e) =2qksinTx- (9-170) Thecoeﬁicients qkaregiven byEq.(8-25): z qk=an u(x) sin@d:z:, lc=1,2,3,.... (9-171) Since thecoeﬂicients qkgive acomplete description oftheposition ofthe string, they represent asuitable setofgeneralized coordinates. When the string vibrates, thecoordinates qkbecome functions oft: 1191,1)=2111(1)$111 (9-172) k=1 Wehave stillaninﬁnite number ofcoordinates qk,butthey depend onthe discrete subscript I0andcanbetreated exactly likethegeneralized coordi- nates considered earlier inthischapter. Since thestring could inprinciple betreated asasystem with avery large number ofparticles, andsince we areallowed todescribe thesystem byanysuitable setofgeneralized co- ordinates, weneed only express theLagrangian function interms ofthe coordinates qkinorder towrite down theequations ofmotion. Weﬁrstneed tocalculate thekinetic energy, which isevidently I 2T=I211 1111. (9-173)o Ifwedifferentiate Eq.(9—172) with respect totandsquare, weobtain R‘ 1-IM2 Q.M2 0-IIQ.02 _.I0.'= kq,-s1n-1lEs1n‘mTx- (9-174) *For atreatment ofthis problem, seeH.Goldstein, op.cit.(Chapter 11.) 9-9] LAGRANGE’S EQUATIONS FORTHEVIBRATING STRING 393 Wenowmultiply by20da:andintegrate from 0tolterm byterm.* Since 1 .lc1rx .j7r:c 1l,j= I0,/(9)s1n—Z—s1n—l-dz ={Qj#k’ (9_175) theresult obtained is T=Z110149 (9-170) 11-1 Wenext calculate thegeneralized force Qk.Ifcoordinate qkincreases by 6qk,while therestareheld ﬁxed, apoint :0onthestring moves upadis- tance given byEq.(9-170): 011=6qk$111 (9-177) The upward force onanelement da:ofstring isgiven byEq.(8-3). The work done istherefore z W=QkBqk=/L%(1%)1s11111. (9-17s) Wesubstitute for6u/67: from Eq.(9-170), andfor6ufrom Eq.(9-177), andintegrate term byterm, toobtain (assuming 7'constant): Q11=-2-l'r qt. (9-179) Theforces Qkareobviously derivable from thepotential energy function, °° 1rk22V=Z111-1-qk. (9-180) I1-1 Itwillbeinstructive tocalculate Vdirectly bycalculating thework done against thetension -rinmoving thestring from itsequilibrium position to theposition u(x). Atthesame time weshall verify that thiswork isin- dependent ofhowwemove thestring totheposition u(a0). Letu(:z:,t)be theposition ofthestring atanytime twhile thestring isbeing moved to *Inorder todifferentiate andintegrate inﬁnite series term byterm, andto rearrange orders ofsummation, asweshall dofreely inthis section, wemust require that theseries allconverge uniformly. This willbethecase ifu(:c, t) and itsderivatives arecontinuous functions. (For aprecise statement and derivation oftheconditions formanipulation ofinﬁnite series, seeatext on advanced calculus, e.g., W.Kaplan, Advanced Calculus. Reading, Mass.: Addison-Wesley, 1952. Chapter 6.) 394 LAGRANGE’S EQUATIONS [cmua 9 u(aa). [The function u(x,t)isnotnecessarily asolution oftheequation of motion, since wewish toconsider anarbitrary manner ofmoving thestring from u=0tou=u(00).] Att=0,thestring isinitsequilibrium position: u(00,0)=0. (9-181) Lett=t1bethetime thestring arrives atitsﬁnal position: u(00,t1)=u(a:). (9—182) The work done against thevertical components oftension [Eq. (8-3)] during theinterval dtis l 9 9u 0u 6lV= Weintegrate byparts, remembering that uanddu/9t are0atac=0,l: I 2dudu dl/—~‘/;:=0T%&%diI3dt z 9 9u2-dtav/;=o -if dx. (9-183) Thetotal work done isthen ii V=/9 dV tr-0 l lla 2=1/..21(12.2)cl.-. l8 2=[0 11111, (9-184) where inthelastexpression, u=u(:c) corresponds totheﬁnal position of thestring. Theresult depends only ontheﬁnal position ofthestring—an independent proof that thetension forces areconservative. The work done against thetension isstored aspotential energy inthe stretched string. Bysubstituting inEq. (9-184) from Eq. (9-170), we again canobtain Eq.(9—180). InEq.(8-61) forastring ofparticles,<the right member contains twoterms that represent thevertical components offorce between adjacent pairs ofparticles. Athird way ofderiving the potential energy istoﬁndthepotential energy function between apair of 9-9] LAGRANGE'S EQUATIONS FOR THE VIBRATING STRING 395 particles which yields thisforce. Itmust then beshown that, when this issummed over allpairs ofadjacent particles, theresult approaches Eq. (9-184) inthelimit h—>0. The Lagrangian function forthevibrating string cannow bewritten as co 2L=T—V=211111,: -111 qfl (9-185) k=1 Theresulting Lagrange equation forqkis .. lﬁ291111+111(9)11=0. <9-186) whose general solution is 2A qk=Akcoswkt-1-Bksinwkt, (9-187) where 1/2rrk'r rrkc» ‘"2=7(5) -T' (27188) This result canbesubstituted inEq.(9-172) toobtain thesolution u(:z:,t)=2(Aksin1%coswkt—[—Bksin@sinwkt) 1(9-189) 11 I-I which isinagreement with Eq.(8-23). Ifu=u0(x) anddu/9t =vO(0c) aregiven att=0,wecanuseEqs. (9—171) and (9-187) toﬁnd the constants Ak,Bk: 1 l A1.=11(0)=§/0011(1)si11@d1. l (9-190) Bk=2"-Q) = v0(:z:) sin@dx,kl0 l wk 0) inagreement with Eqs. (8-25). The coordinates qkdeﬁned byEqs. (9—l70) and (9-171) arecalled the normal coordinates forthevibrating string. Each coordinate evidently represents onenormal mode ofvibration. Thenormal coordinates arealso very useful intreating thecase where aforce f(x,t)isapplied along the string (seeProblem 26attheendofthischapter). Mathematically, the normal coordinates have theproperty that theLagrangian Lbecomes a sum ofterms, each term involving only onedegree offreedom. Thus innormal coordinates theproblem issubdivided into separate problems, oneforeach degree offreedom. 396 LAGRANGE-,S EQUATIONS [on1u>. 9 Itwas, ofcourse, rather fortunate that thecoordinates qrwhich were chosen atthebeginning oftheproblem turned outtobethenormal co- ordinates. Ingeneral, this does nothappen. Forexample, consider a string Whose density varies along itslength according to ' 0'=0'0+asin?- (9—191) This string isheaviest near itscenter. Wewillusethesame coordinates qkasdeﬁned byEqs. (9—l70) and (9—171). Wesubstitute Eqs. (9—191) and(9—172) inEq.(9—173) and, instead ofEq.(9—176), weobtain (after some calculation), s ZN”Nb-'‘i T=Z kjdkéi, (9-192) lc=1 = where 41 102 . . Tkj=%lG0+?a > lfk=], __41¢ kj . ._T““1ruh+02—11[<k—1>2—11’‘”°¢1’(9193) andk,jareboth even orboth odd; otherwise T1,;=0. Forthis string, theq;/sareevidently notnormal coordinates. Inthe Lagrange equations theq;/s, with lceven, areallcoupled together, asare those with kodd. Theproblem isthen much more diﬂicult, andasolution willnotbeattempted here. 9-10 Hami1ton’s equations. The discussion inthis section will be restricted tomechanical systems obeying Lagrange’s equations intheform (9-57). TheLagrangian Lisafunction ofthecoordinates qk,oftheveloci- tiesq,,,andperhaps oft.Thestate ofthemechanical system atanytime, that is,thepositions andvelocities ofallitsparts, isspeciﬁed bygiving the generalized coordinates and velocities qk,qk.Lagrange’s equations are second-order equations which relate theaccelerations ﬁktothecoordinates andvelocities. Thestate ofthesystem could equally well bespeciﬁed by giving thecoordinates qkand themomenta pkdeﬁned byEq. (9——168): p,,=Q, k=1,2,...,f. (9-194)aqk * These equations specify pkinterms ofq1,...,qf;Q1,...,Q1.They can, H1principle, besolved forq,,interms ofq1,...,qf;pl,...,pf. 9-10] HAM11.'roN’s EQUATIONS 397 Itisaninteresting exercise totrytowrite equations ofmotion interms ofthecoordinates qkand momenta pk. Note ﬁrst that, byuseofthe deﬁnition (9—194) and theequations ofmotion (9—57), wehave ’aL aL )aLdL= <—,d‘-—d —atQéqkq"+éqkq’°+as ’ .. aL=Z(Pkdqk+Pkdqo+5dt. (H95)1==1 Wenext deﬁne afunction H(q1, ...,q;;p1, ...,pf;t)by fH=Zpm—L, (H96) k=1 where forthevelocities q,,wesubstitute their expressions interms ofco- ordinates andmomenta. Then wehave arr=fl(qkdpk -pkdqk)—%d¢. (9—197) k==l Thedeﬁnition (9—196) ischosen sothatdHdepends explicitly upon dpk, dqk,anddt.Byinspection ofEq.(9—197),- weseethat -_<’_Ii -__?lY_ _qk-— apkr p],—- aqkr k— 1,...,f, and 6H 6L-5;-—3;- (9—199) Equations (9—198) arethedesired equations ofmotion that express Q1,and pkinterms ofthecoordinates andmomenta. Equations (9—198) areHami1ton’s equations ofmotion foramechanical system. Thefunction H,deﬁned byEq.(9—196), iscalled theHamiltonian function. Weseefrom Eq.(9—124) that when Visafimction only ofthe coordinates, forastationary coordinate system, Hisjustthetotal energy expressed interms ofcoordinates andmomenta. Foramoving coordinate system, where Tisgiven byEq.(9-13), theHamiltonian is H=T2+V——T0, (9—200) with T2expressed interms ofcoordinates andmomenta. According to Section 9-8, Hwillalso bethetotal energy inastationary coordinate system when electromagnetic forces arepresent. 398 LAGRANGE’S EQUATIONS [CHAP. 9 When Ldoes notcontain thetime explicitly, neither does Haccording toEq.(9—199), asisalsoobvious from theway inwhich Hwasdeﬁned. According toEq.(9—125), Hisaconstant ofthemotion inthiscase. This canalsobeproved directly from Eqs. (9—198), since itiseasy toshow that %=%, (9-201) asthereader may verify. Ifanycoordinate qkdoes notappear explicitly inH,then Eqs. (9—198) give pk=aconstant, (9—202) inagreement with Eq.(9—118). Since Hdoes notcontain qk,wemay take pkasagiven constant, andthe2(f—1)equations (9—198) fortheother coordinates andmomenta arethen theHamiltonian equations forasys- temoff—1degrees offreedom. Thus degrees offreedom corresponding tocoordinates that donotappear inHsimply drop outoftheproblem. This istheorigin oftheterm “ignorable coordinate.” After theremaining equations ofmotion have been solved forthenonignorable coordinates and momenta, anyignorable coordinate isgiven byEqs. (9—198) asanintegral over t: pz ac)=mo)+fgidt. <9-208)0Pk Hamilton’s equations aresimply anew formulation ofNewton’s laws ofmotion. Insimple cases, they reduce toequations which could have been Written immediately from Newton’s laws. Intheharmonic oscillator, for example, with coordinate ac,themomentum is p=m:i;. (9—204) The Hamiltonian function istherefore 2H=T+V=Q-n+gm”. (9-205) Equations (9—198) become ._Q, .=__ _x-m p kx. (9206) Theﬁrstofthese isthedeﬁnition ofp,andthesecond isNewton’s equation ofmotion. Although they areofcomparatively little value asameans ofWriting theequations ofmotion ofasystem, Hamilton’s equations areimportant fortwo general reasons. First, they provide auseful starting point in setting upthelaws ofstatistical mechanics andof“quantum mechanics. 9-11] LIOUVILLE’S THEOREM 399 Hamilton orginally developed hisequations byanalogy with asimilar mathematical formulation which hehadfound useful inoptics. Itisnot surprising that Hamilton’s equations should form thestarting point for wave mechanics! Second, there areanumber ofmethods ofsolution of mechanical problems based onHamilton’s formulation oftheequations of motion. Itisclear from theway inwhich they were derived that Hamil- ton’s equations (9—198), likeLagrange’s equations, arevalid forany set ofgeneralized coordinates q1,..., q;together with thecorresponding momenta pl,...,pf,deﬁned byEq.(9—194). InfactHamilton’s equations arevalid foramuch wider class ofcoordinate systems obtained byde- ﬁning new coordinates andmomenta ascertain functions oftheoriginal coordinates andmomenta. This isthebasis fortheutility ofHamilton’s equations inthesolution ofmechanical problems. Afurther discussion of these topics isbeyond thescope ofthisbook.* Wewill, however, prove onegeneral theorem inthenext section which gives some insight intothe importance ofthevariables pkandqk. 9-11 Liouville’s theorem. Wemay regard thecoordinates q1,...,qf asthecoordinates ofapoint inanf-dimensional space, theconﬁguration space ofthemechanical system. Toeach point intheconﬁguration space there corresponds aconﬁguration oftheparts ofthemechanical system. Asthesystem moves, thepoint q1,...,q;traces apath intheconﬁgura- tion space. This path represents thehistory ofthesystem. Ifwewish tospecify both theconﬁguration andthemotion ofasystem atanygiven instant, wemust specify thecoordinates andvelocities, or equivalently, thecoordinates andmomenta. The 2f-dimensional space whose points arespeciﬁed bythecoordinates andmomenta q1,...,qf; pl,...,pfiscalled thephase space ofthemechanical system. Asthesys- temmoves, thephase point q1,...,qf;pl,...,pftraces outapath inthe phase space. The velocity ofthephase point isgiven byHamilton’s equations (9—198). Each phase point represents apossible state ofthemechanical system. Letusimagine that each phase point isoccupied bya“particle” which moves according topthe equations ofmotion (9—198). These particles trace outpaths that represent allpossible histories ofthemechanical system. Thetheorem ofLiouville states that thephase “particles” move asanincompressible ﬂuid. More precisely, thephase volume occupied by asetof“particles” isconstant. . Toprove Liouville’s theorem, wemake useoftheorem (8—121) gener- alized toaspace of2fdimensions. Wemay ‘either generalize theargument which ledtoEq. (8—116), orwemay usethegeneralization ofGauss’ *SeeH.Goldstein, op.cit. (Chapters 7,8,9.) 400 LAGRANGE'S EQUATIONS [cnlun 9 divergence theorem which isvalid inanynumber ofdimensions. Ineither case, Wehave foravolume Vinphase space, moving with the“particles”: H;/.../-’ in),...,,,,,...,, 9,207d,— V;aqk+apk ql qrm P/,( ) which isEq.(8—121) written forthe2f-dimensional phase space. Wenow substitute thevelocities from Hamilton’s equations (9—198): dv_/../’ (_@fa__ﬁ_)d ...,,d...,,-0 di_ V Q aqk31% 31%aqk ql qfpl pf '. (9—208) This isLiouville’s theorem,"and itshould benoted that thistheorem holds even when Hdepends explicitly ont. Inthecase ofaharmonic oscillator, thephase space isaplane with co- ordinate axes xandp.The phase points move around ellipses H=con- stant, given byEq.(9—205), andwith velocities asgiven byEq.(9—206). According toLiouville’s theorem, themotion isthat ofatwo-dimensional incompressible ﬂuid. Inparticular, asetofpoints that lieinaregion of area Awillatanylater time lieinanother region ofarea A. Liouville’s theorem makes thecoordinates andmomenta more useful formany purposes than coordinates andvelocities. Because ofthis theorem, theconcept ofphase space isanimportant toolinstatistical mechanics. Imagine alarge number ofmechanical systems identical toa given one, butwith different initial conditions. Leteach system berepre- sented byapoint intheir common phase space, andletthesepoints move according toHamilton’s equations. The statistical properties ofthiscol- lection ofsystems may bespeciﬁed atanytime tbygiving thedensity p(q1, ...,qf;pl,...,pf;t)inthephase space ofsystem points perunit volume. Liouville’s theorem implies that thedensity pintheimmediate neighborhood ofany system point must remain constant asthat point moves through thephase space. (Why?) Ifwedeﬁne statistical equi- librium asadistribution inwhich pisconstant intime ateach ﬁxed point inthephase space, then clearly thenecessary andsufﬁcient condition for equilibrium isthat pbeimiform along theﬂow lines ofthesystem points. (Why?) Wehave been able inthis section togive only anarrow glimpse ofthe power oftheHamiltonian methods. 401 PROBLEMS 1.Coordinates u,waredeﬁned interms ofplane polar coordinates r,0bythe equations V u=ln(r/a) —0cot §',' w=In(r/a) 0tan f, 1 where aandIareconstants. Sketch thecurves ofconstant uandofconstant 'w. Find thekinetic energy foraparticle ofmass minterms ofu,w,11,11:.Find expressions forQ“,Q,,,interms ofthepolar force components F,,F0.Find pk,pw. Find theforces Q",Q”required tomake theparticle move with constant speed é along aspiral ofconstant u=uo. 2.Two masses‘ m1and mgmove under their mutual gravitational attraction inauniform external gravitational ﬁeld whose acceleration isg.Choose as coordinates thecartesian coordinates X,Y,Zofthecenter ofmass (taking Zin thedirection ofg),thedistance rbetween m1and mg,and thepolar angles 0and<pwhich specify thedirection oftheline from mltomg. Write expres- sions forthekinetic energy, thesixforces QX,...,Q,,and thesixmomenta. Write outthesixLagrange equations ofmotion. 3.(a)Setuptheexpression forthekinetic energy ofaparticle ofmass min terms ofplane parabolic coordinates f,h,asdeﬁned inProblem 13ofChapter 3. Find themomenta p;andpk. (b)Write outtheLagrange equations inthese coordinates iftheparticle isnotacted_on byanyforce. . 4.(a)Find theforces Q;andQkrequired tomake theparticle in‘Problem 3 move along aparabola f=fo=aconstant, with constant generalized velocity h=ho,starting from h=0att=0.(b)Find thecorresponding forces F, andF,relative toacartesian coordinate system. 5.(a)SetuptheLagrange equations ofmotion inspherical coordinates r,0,(,0, foraparticle ofmass msubject toaforce whose spherical components are F,,Fa,Fk. (b)SetupLagrange equations ofmotion forthesame particle inasystem of spherical coordinates rotating with angular velocity wabout thez-axis. (c)Identify thegeneralized centrifugal and coriolis forces ‘Q/, ‘Q0’, and ‘Q,,’ bymeans ofwhich theequations intherotating system canbemade totake the same form asintheﬁxed system. Calculate thespherical components ‘F,’, ‘F9’, ‘F,,’ofthese centrifugal and coriolis forces, and show that your results agree with theexpressions derived inChapter 7. 6.SetuptheLagrangian function forthemechanical system shown in Fig. 4-16, using thecoordinates x,x1,2:2asshown. Derive theequations of motion, and show that they areequivalent totheequations that would be written down directly from Newton’s lawofmotion. 7.Choose suitable coordinates and write down theLagrangian function for therestricted three-body problem. Show that itleads totheequations ofmo- tion obtained inSection 7—6. 8.Masses mand2maresuspended from astring oflength l1which passes over apulley. Masses 3mand4maresimilarly suspended byastring oflength 402 LAGRANGE’S EQUATIONS [CHAP~ 9 Z2over another pulley. These twopulleys hang from theends ofastring oflength Z3over athird ﬁxed pulley. SetupLagrange’s equations, andﬁndtheaccelera- tions andthetensions inthestrings. 9.Amassless tube ishinged atoneend. Auniform rodofmass m,length l, slides freely init.Theaxisabout which thetube rotates ishorizontal, sothat the motion isconﬁned toaplane. Choose asuitable setofgeneralized coordinates, oneforeach degree offreedom, andsetupLagrange’s equations. 10.SetupLagrange’s equations forauniform door whose axisisslightly out ofplumb. What istheperiod ofsmall vibrations? 11.Adouble pendulum isformed bysuspending amass mgbyastring of length lgfrom amass m1which inturn issuspended from aﬁxed support bya string oflength Z1. (a)Choose asuitable setofcoordinates, and write the Lagrangian function, assuming thedouble pendulum swings inasingle vertical plane. (b)Write outLagrange’s equations, andshow that they reduce totheequa- tions forapair ofcoupled oscillators ifthestrings remain nearly vertical. (c)Find thenormal frequencies forsmall vibrations ofthedouble pendulum. Describe thenature ofthecorresponding vibrations. Find thelimiting values ofthese frequencies when mi>>mg,and when m2>>m1. Show that these limitingvalues aretobeexpected onphysical grounds byconsidering thenature ofthenormal modes ofvibration when either mass becomes vanishingly small. 12.Aladder rests against asmooth wall andslides without friction onwall andﬂoor. Setuptheequation ofmotion, assuming thattheladder maintains contact w_ith"thewall. Ifinitially theladder isatrestatanangle ozwith the ﬂoor, atwhat angle, ifany,willitleave thewall? 13.Oneendofauniform rodofmass Mmakes contact with asmooth vertical wall, theother with asmooth horizontal ﬂoor. Abead ofmass mandnegligible dimensions slides ontherod. Choose asuitable setofcoordinates, setupthe Lagrangian function, andwrite outtheLagrange equations. Therodmoves ina single vertical plane perpendicular tothewall. 14.Aring ofmass Mrests onasmooth horizontal surface andispinned ata point onitscircumference sothat itisfreetoswing about avertical axis. A bugofmass mcrawls around thering with constant speed. (a)Setuptheequa- tions ofmotion, taking thisasasystem with twodegrees offreedom, with the force exerted bythebugagainst thering tobedetermined from thecondition that hemoves with constant speed. (b)Now setuptheequation ofmotion, taking thisasasystem with onede- gree offreedom, thebugbeing constrained tobeatacertain point onthering ateach instant oftime. Show that thetwo formulations oftheproblem are equivalent. 15.Apendulum bobofmass missuspended byastring oflength Zfrom a point ofsupport. Thepoint ofsupport moves toandfroalong ahorizontal :1:-axis according totheequation 1;=acoswt. Assume that thependulum swings only inavertical plane containing thex-axis. Lettheposition ofthependulum bedescribed bytheangle 0which thestring ...-ls m m ‘l lll., PROBLEMS 403 makes with alinevertically downward. (a)SetuptheLagrangian function and write outtheLagrange equation. (b)Show that forsmall values of0,theequation reduces tothat ofaforced harmonic oscillator, andﬁndthecorresponding steady-state motion. How does theamplitude ofthesteady-state oscillation depend onm,l,a,andco? 16.Apendulum bobofmass missuspended byastring oflength lfrom acar ofmass Mwhich moves without friction along ahorizontal overhead rail. The pendulum swings inavertical plane containing therail. (a)SetuptheLagrange equations. (b)Show that there isanignorable coordinate, eliminate it,anddis- cuss thenature ofthemotion bytheenergy method. 17.Find thetension inthestring forthespherical pendulum discussed in Section 9-7, asafunction ofE,p,,,and 0.Determine, foragiven Eandpk, theangle 01atwhich thestring willcollapse. 18.Aparticle ofmass mslides over theinner surface ofaninverted cone of half-angle oz.The apex ofthecone isattheorigin, andtheaxisofthecone ex- tends vertically upward. The only force acting ontheparticle, other than the force ofconstraint, istheforce ofgravity. (a)Setuptheequations ofmotion, using ascoordinates thehorizontal distance poftheparticle from theaxis, and theangle <pmeasured inahorizontal circle around thecone. Show that <pis ignorable, anddiscuss themotion bythemethod oftheeﬁective potential. (b)Foragiven radius po,ﬁndtheangular velocity ‘P0ofrevolution inahori- zontal circle, and theangular frequency toofsmall oscillations about this cir- cular motion. Show that thesmall oscillations areawobbling oranup-and- down spiraling motion, depending onwhether theangle ozisgreater than orless than theangle oz,=sin_1\/§ 19.Aﬂyball governor forasteam engine isshown inFig. 9-11. Two balls, each ofmass m,areattached bymeans offour hinged arms, each oflength l, tosleeves which slide onavertical rod. Theupper sleeve isfastened totherod; thelower sleeve hasmass Mandisfreetoslide upanddown therodastheballs FIG. 9~11. Aﬂyball governor. 404 LAGRANGE’S EQUATIONS [CHAP. 9 move outfrom ortoward therod. The rod-and-ball system rotates with con- stant angular velocity w.(a)Setuptheequation ofmotion, neglecting theWeight ofthearms androd. Discuss themotion bytheenergy method. (b)Determine thevalue oftheheight zofthelower sleeve above itslowest point asafunction ofwforsteady rotation oftheballs, andﬁnd thefrequency ofsmall oscillations of2about thissteady value. 20.Discuss themotion ofthegovernor described inProblem 19iftheshaft isnotconstrained torotate atangular velocity w,butisfreetorotate, without anyexternally applied torque. (a)Find theangular velocity ofsteady rotation foragiven height zofthesleeve. (b)Find thefrequency ofsmall vibrations about thissteady motion. (c)How does thismotion differ from that ofProb- lem19? 21.Arectangular coordinate system with axes 2:,y,2isrotating with uniform angular velocity wabout thez-axis. Aparticle ofmass mmoves under theaction ofapotential energy V(x, y,z).(a)SetuptheLagrange equations ofmotion. (b)Show that these equations canberegarded astheequations ofmotion ofa particle inaﬁxed coordinate system acted onbytheforce —VV, andbyaforce derivable from avelocity dependent potential U.Hence ﬁnd avelocity de- pendent potential forthecentrifugal andcoriolis forces. Express Uinspherical coordinates r,0,go,1‘,9,¢,andverify that itgives risetotheforces ‘Q/, ‘Q9’,‘Q,,’ found inProblem 5. 22.Show that auniform magnetic ﬁeld Binthez-direction canberepresented incylindrical coordinates (Fig. 3-22) bythevector potential A=§Bp m. Write outtheLagrangian function foraparticle insuch aﬁeld. Write down theequations ofmotion, andshow that there arethree constants ofthemotion. Compare with Problem 49ofChapter 3. 23.The kinetic part oftheLagrangian function foraparticle ofmass min relativistic mechanics is. L1.=—mc2[1 —~(v/c)2]1/2. Show that thisgives theproper formula (4-75) forthecomponents ofmomentum. Show that ifthepotential function forelectromagnetic forces Eq.(9-166) is subtracted, andifAandctdonotdepend explicitly ont,then T+q¢iscon- stant, with Tgiven byformula (4-74). *24. Show bydirect calculation that ifEqs. (9—155) hold forsome function L(q1, ...,qf;Q1,...,q,;t),and weintroduce new coordinates qf,...,q}", where ql#=fl=(qT2-":q.t§t>r 19:12"-rfr then *i<i__ =0,1- ‘itMi Bqi where L*(q’f, ...,q}";qf,...,qf;t)=L(q1, ...,qf;111,...,q,;t)isobtained bysubstitution offk(q’f, ...,qjf;t)forqk. PROBLEMS 405 25.Derive formula (9-184) bywriting down apotential energy which gives theinterparticle forces forthestring ofparticles studied inSection 8-4,and passing tothelimit h—->0. 26.Astretched string issubject toanexternally applied force oflinear density f(a:,t).Introduce normal coordinates qk,and ﬁnd anexpression forthegen- eralized applied force Qk(t). UsetheLagrangian method tosolve Problem 6(a), Chapter 8. 27.Solve Problem 7,Chapter 8,byusing thecoordinates qkdeﬁned by Eqs.~ (9-170) and(9-171). 28.Write down theHamiltonian function forthespherical pendulum. Write theHamiltonian equations ofmotion, andderive from them Eq.(9—136). *29. Work outtherelativistic Hamiltonian function foraparticle subject to electromagnetic forces, using theLagrangian function given inProblem 23. Write outtheHamiltonian equations ofmotion andshow that they areequiva- lenttotheLagrange equations. 30.Work outtheHamiltonian function H(qk,pk)forthevibrating string, starting from Eq.(9-185). Write down theequations which relate themomenta pktothefunction u(x,t)which describes themotion ofthestring. Hence show that H=T+V,with TandVgiven byEqs. (9—173) and(9—184). 31.Write down theHamiltonian function forProblem 2.Write outHamil- ton’s equations. Identify theignorable coordinates andshow thatthere remain twoseparate onedegree offreedom problems, each ofwhich canbesolved (in principle) bytheenergy method. What arethecorresponding twopotential- energy functions? 32."A beam ofelectrons isdirected along thez-axis. Theelectrons areuni- formly distributed over thebeam cross section, which isacircle ofradius ao, andtheir transverse momentum components (pk,p,,)aredistributed uniformly inacircle (inmomentum space) ofradius po. Iftheelectrons arefocused by some lenssystem soastoform aspot ofradius a1,ﬁndthemomentum distribu- tion ofelectrons arriving atthespot. 33.Agroup ofparticles allofthesame mass m,having initial heights and vertical momenta lying inthesquare —a§z§a,—b§p3b,fallfreely intheearth’s gravitational ﬁeld foratime t.Find theregion inthephase space within which they lieattime t,andshow bydirect calculation that itsarea isstill 4ab. 34.Inanelectron microscope, electrons scattered from anobject ofheight 20arefocused byalens atdistance D0from theobject andform animage of height 21atadistance D1behind thelens. The aperture ofthelens isA.Show bydirect calculation that thephase area inthe(2,p,)phase plane occupied by electrons leaving theobject (and destined topass through thelens) isthesame asthephase area occupied byelectrons arriving attheimage. Assume that Z0<<D0and Z1<<D1. . CHAPTER 10 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS Inthischapter weshall develop thealgebra oflinear vector functions, ortensors, asamathematical toolwhich isuseful intreating many prob- lems. Inparticular, weshall need tensors inthestudy ofthegeneral motion ofarigid body andintheformulation oftheconcept ofstress in asolid, orinaviscous ﬂuid. . 10-1 Angular momentum ofarigid body. The equation ofmotion for therotation ofarigid body isgiven byEq.(5-5) andrestated here: %=N, (urn where Listheangular momentum andNisthetorque about apoint P which may beeither ﬁxed orthecenter ofmass ofthebody. InSection 5-2westudied therotation ofarigid body about aﬁxed axis. Inorder totreat thegeneral problem oftherotation ofabody about apoint P, wemust ﬁndtherelation between theangular momentum vector Land theangular velocity vector w. Consider abody made upofpoint masses mksituated atpoints rk relative toanorigin ofcoordinates atP.Wehave shown inSection 7-2 that themost general motion ofthebody about thepoint Pisarotation with angular velocity w,andthat thevelocity vkofeach particle inthe body isgiven by vk=w><rk. (10-2) Wesum theangular momentum given byEq.(3—142) over allparticles: N L=Z mkfk XV1,; k=1 N=Zmm.><(...><rk). (10-3) k-1 Equation (10-3) expresses Lasafunction ofw,L(w). Bysubstitution in Eq.(10-3) itisreadily veriﬁed that thefunction L(w), foranytwovectors w,00',andanyscalar c,satisﬁes thefollowing relations: L(cw) =cL(w), ) (10-4) L(w —|—co’) =L(w) —|—L(w'). (10-5) 406 10-2] TENSOR ALGEBRA 407 Avector function L(w) with theproperties (10-4), (10-5) iscalled a linear vector function. Linear vector functions areimportant because they occur frequently inphysics, andbecause they have simple mathematical properties. Inorder todevelop ananalogy between Eq.(10-3) andEq.(5-9) for thecase ofrotation about anaxis, wemake useofEq.(3-35): L=ﬁ[mkriw —mkrk(rk -w)]. (10-6) k=1 The factor wisindependent oflcandcanbefactored from thesum over theﬁrstterm. Inapurely formal way, wemay alsofactor wfrom thesum over thesecond term: (1.N N L= 2 mkr;€> (.0— 'WL]kI'kI']k> '(0. =1 k=1 The second term hasnomeaning, ofcourse, since thejuxtaposition rkrk oftwovectors hasnotyetbeen deﬁned. Weshall trytosupply amean- inginthenext section. 10-2 Tensor algebra. The dyad product ABoftwovectors isdeﬁned bythefollowing equation, where Cisanyvector: _ (AB) -C=A(B -C). (10-8) The right member ofthisequation isexpressed interms ofproducts de- ﬁned inSection 3-1. Theleftmember is,bydeﬁnition, thevector given by theright member. Note that thedyad ABisdeﬁned only interms ofitsdot product with anarbitrary vector C.Wecanreadily show, from deﬁnition (10-8), that multiplication ofavector byadyad isalinear operation in thesense that (AB)-(¢C)=¢l(AB) ~C], (10-9) (AB)'(C+D)=(AB)-C+(AB)'13- (10-10) Forﬁxed vectors A,B,thedyad ABtherefore deﬁnes alinear vector func- tion F(C): F(C) =(AB) -C. (10-11) Thedyad ABisanexample ofalinear vector operator, that is,itrepresents anoperation which may beperformed onanyvector Ctoyield anew vector (AB) -C,which isalinear function ofC. 408 TENSOR ALGEBRA. INERTIA ANDsrnnss TENSORS [cnxr. 10 Alinear vector operator isalsocalled atensor.* Tensors willberepre- sented bysans-serif boldface capitals, A,B,C,etc. Wemay forexample letTbethetensor represented bythedyad AB: T=AB. (10-12) Themeaning ofthetensor Tisspeciﬁed bythedeﬁnition,'|' T-C=A(B -C), (10-13) which gives theresult ofapplying Ttoanyvector C.Wecanform more general linear vector operations bytaking sums ofdyads. The sum of twodyads, ortensors S,T,isdeﬁned asfollows: (S—|—T)-C=Si-C—l—T-C. (10-14) Note that alldeﬁnitions ofalgebraic operations ontensors, liketheabove deﬁnition of(S+T),areformulated interms oftheapplication ofthe tensors toanarbitrary vector C.Thesum ofoneormore dyads iscalled adyadic. According tothedeﬁnition (10-14), thedyadic (AB+DE) operating onCyields thevector i (AB+DE) -C=A(B -C)+D(E -C), (10-15) Wecanreadily show that thesum oftwo linear operators isalinear operator; therefore dyadics arealsolinear vector operators andwehave for anydyadic ortensor T, T-(cC) =c(T-C), (10-16) T-(C+D)=T-C-|—T-D. (10-17) The linearity relations (10-16), (10-17), together with thedeﬁnition (10-14), guarantee that dyad products, sums oftensors, anddotproducts oftensors with vectors satisfy alltheusual algebraic rules forsums and products. Wecanalsodeﬁne adotproduct ofadyad with avector onthe leftintheobvious way, C-(AB) =(C-A)B, I (10-18) *More precisely, alinear vector operator may becalled asecond-rank tensor, todistinguish itfrom third- andhigher-rank tensors obtained aslinear combina- tions oftriads ABC, etc. Weshall beconcerned inthisbook only with second- rank tensors, which weshall refer tosimply astensors. TThe result ofapplying atensor Ttoavector Cisoften denoted byTC, without thedot. Weshall usethedotthroughout thisbook. 10-2] TENSOR ALGEBRA 409 andcorrespondingly forsums ofdyads. Note that thedotproduct ofa dyadic with avector isnotcommutative; T-C=C-T (10-19) does nothold ingeneral. Wecandeﬁne, inanobvious way, aproduct cT ofatensor byascalar, with theexpected algebraic properties (seeProblem 1). Avery simple tensor isgiven bythedyadic l=ii+ii -I-kk, A (10-20) where i,j,kare-asetofperpendicular miit vectors along :1:-,y-,andz- axes. Wecalculate, using thedeﬁnitions (10-14) and(10-8), 1-A=iA,,+5.4,,+kA,=A. (10-21) The tensor ‘Iiscalled theunit tensor; itmay bedeﬁned astheoperator which, acting onanyvector, yields that vector itself. Evidently Iisone ofthespecial cases forwhich I-A=A-1. (10-22) Ifcisanyscalar, theproduct cliscalled aconstant tensor, andhasthe property (c'l)-A=A-(cl)=cA. (10-23) Using thedeﬁnitions above, wecannow write Eq.(10-7) intheform ' L=I-co, (10-24) where Iistheinertia tensor oftherigid body, deﬁned by N |=2 (7I'LkT]%'| —mkrkrk). k=1 The inertia tensor Iistheanalog, forgeneral rotations, ofthemoment of inertia forrotations about anaxis. Note that Landwarenotingeneral parallel. Wewillstudy theinertia tensor inmore detail after wehave developed thenecessary properties oftensors. Ifwewrite allvectors interms oftheir components, C=C',,i+Cuj+C',k, (10-26) then itisclear that bymultiplying outdyad products andcollecting terms, anydyadic canbewritten intheform: 410 TENSOR ALGEBRA. INERTIA AND srnnss TENSORS [CHAP. 10 1=T..ii+r..i1+ T..ik +Tyxji+T1/all+Tuzjk +Tnki +T,,,kj +Tzzkka (10-27) Just asanyvector Acanberepresented byitsthree components (Ak, A,,, A,,), soanydyadic canbespeciﬁed bygiving itsnine components Tm, ...,Tu.These may conveniently bewritten intheform ofasquare array ormatrix: ﬁe“?Tm: Txy T= Tow T2/11 Tzx Tzy Asanexample, thereader may verify that thecomponents oftheinertia tensor (10-25) are N N Ian: =2 7'nk(?/lg +31%): I111 =_'2 777/kxlcl/kw etc- lc 1 k 1> (10-28) Inorder tosimplify writing thetensor components, itoften willbecon- venient tonumber thecoordinate axes x1,x2,x3instead ofusing x,y,2: 2:=1:1, y=wk, z=$3. (10-30) Weshall write thecorresponding unit vectors ase,~: I=81, j=62, kZ63. Equations (10-26) and(10-27) cannow bewritten as 3c=0.3,, (10-32) and 31=ZT,-,-e,-e,-. (10-33) 11.1=1 Another advantage ofthisnotation isthat itallows thediscussion tobe generalized tovectors andtensors inaspace ofanynumber ofdimensions simply bychanging thesummation limit. Byusing thedeﬁnitions ofdyad products andsums, wecanexpress the components ofthevector T-Cinterms ofthecomponents ofTandC: 3. A(T'C);=ZT1705, 9 (10-34) j=1 10-2] TENSOR ALGEBRA 411 asthereader should verify. Similarly, 3(c-1). =ZC,-T,-,-. (1<»35) i=1 Wenote that, byEq.(10-33), Ti," =81''(T'6]‘) =(65'T)'e,-. (10-36) Wemay omit theparentheses, since theorder inwhich themultiplications arecarried outdoes notmatter. Wecannowshow that anylinear vector function canberepresented by adyadic. LetF(C) beanylinear function ofC.Consider ﬁrst thecase when Cisaunitvector e,~,andletT,-,~bethecomponents ofFinthat case: 3F(e,-)=ZT,-,-e,-. (10-37) i=1 Now anyvector Ccanbewritten as 3c=ZC’,-e,-. (10-33) i=1 Byuseofthelinear property ofF(C), wehave therefore 3F(C)=ZF<0.e.~) ]=1 3=Z0.-Fe.-)i=1 3=ZC’,-T,~,-e,-. (10-39) i.j=1 Thus thecomponents ofF(C) canbeexpressed interms ofthenumbers Tiji . .3 lF(C)lt =2_T'¢1'Ca'- (1040)i=1 Ifwedeﬁne thedyadic3 T= 2 T,-,-e,-e,-, (10-4:1) _§.H. \-I weseefrom Eqs. (10-40) and(10-34) that r(c)=1-c. (10-42) 412 TENSOR ALGEBRA. INERTIA AND srnnss TENSORS [cn.\1>. 10 Thus theconcepts ofdyadic and linear vector operator ortensor are identical, and areequivalent totheconcept oflinear vector, function inthesense that every linear vector function deﬁnes acertain tensor or dyadic, andconversely. Wecandeﬁne adotproduct oftwotensors asfollows: (T-S)-C=T-(S-C). (10-43) Application oftheoperator T-Stoanyvector means ﬁrst applying S, andthen T.Wenow calculate interms ofcomponents, using thedeﬁnition (10-43), IN"___§~'[\4,,S01(T'$)'C=T' jkCk6j = ijsjkckei‘ == = T,~,~S,-k) Ck]e,-. (10-44) Q‘.-M» I-1TilPrl"J=~= D-I Comparing thisresult with Eq.(10-34), weseethat a (T'5)at=2Ti.1‘Sjk- (19-45)1-=1 Equation (10-45) alsoresults ifwesimply evaluate T-Sinaformal way bywriting thedotanddyad products andcollecting terms: $@oQ@-._.M0:EM“, W|-lT-5= T,-,-Sk;e,-e,- '61,61 = Tiisileﬁz, (10-46) andthisshows that ourdeﬁnition (10-43) isconsistent with theordinary rules ofalgebra. IfT,Sarewritten asmatrices according toEq.(10-28), then Eq.(10-45) istheusual mathematical ruleformultiplying matrices. Wecansimilarly show that thedeﬁnition (10-14) implies that tensors areadded byadding their component matrices according totherule: (T+5)i'j=Tij"l"Sij- (10-47) Sums andproducts oftensors obey alltheusual rules ofalgebra except that dotmultiplication, ingeneral, isnotcoimnutativez 10-2] TENSOR ALGEBRA r+s=s+L T-($+P)=T-S+T-P, T-(S-P)=(T-S)-P, rr=r1=r andsoon,but T-S9'5S-T, ingeneral. Itisuseful todeﬁne thetranspose T‘ofatensor Tasfollows: T‘-C=C-T. Interms ofcomponents, T3=Tkn3 new new uoem (10-51) (10-52) (10-53) (10-54) The transpose isoften written T,butthenotation T‘ispreferable for typographical reasons. Thefollowing properties areeasily proved: 0+9@=V+9, (T-S)‘=S‘-T‘, (T‘)'=T. Atensor issaidtobesymmetric if T‘=T.(10-55) (10-56) (10-57) (10-58) Forexample, theinertia tensor, given byEq.(10-25) issymmetric. For asymmetric tensor, T,-1=T,-,-. (10-59) Asymmetric tensor may bespeciﬁed bysixcomponents ;theremaining three arethen determined byEq.(10-59). Atensor issaidtobeantisymmetric if T‘=—-T. The components ofanantisymmetric tensor satisfy theequation Ti,‘ =—T5,".(10-60) (10-61) Evidently thethree diagonal components T,-,-areallzero, andifthree off- diagonal components aregiven, thethree remaining components aregiven 414 TENSOR ALGEBRA. INERTIA AND srnnss TENSORS [CHAP. 10 byEq. (10-61). Anantisymmetric tensor hasonly three independent components (inthree-dimensional space). Anexample isthelinear operator deﬁned by T-C= wXC, (10-62) where wisaﬁxed vector. Comparing Eq.(10-62) with Eq.(7-20), we seethat theoperator Tcanbeinterpreted asgiving thevelocity ofany vector Crotating with anangular velocity w.Comparing Eq. (10-62) with Eq.(10-34), weseethat thecomponents ofTare: T11 =T22 =T33 =0, T =—T =co 21 12 3, (10_63) T32="T23 =011, T13 =‘T31 =w2~ Q, Since anantisymmetric tensor, likeavector, hasthree independent com- ponents, wemay associate with every antisymmetric tensor Tavector w(inthree-dimensional space only!) whose components arerelated to those ofTbyEq.(10-63). The operation T-willthen beequivalent to wX,according toEq.(10-62). Given anytensor T,wecandeﬁne asymmetric andanantisymmetric tensor by T.=%(T+T‘), (10-64) Ta=%(T—T‘), (10-65) such that T=T,—l—T,,. (10-66) Wesaw, inthepreceding paragraph, that anantisymmetric tensor could berepresented geometrically byacertain vector w.WewillseeinSec- tion 10-4 how torepresent asymmetric tensor. Since antisymmetric and symmetric tensors have rather different geometric properties, tensors which occur inphysics areusually either symmetric orantisynnnetric rather than acombination ofthetwo. Inthree-dimensional space, the introduction ofanantisymmetric tensor canalways beavoided bythe useoftheassociated vector. Itistherefore notacoincidence that thetwo principal examples oftensors inthis chapter, theinertia tensor andthe stress tensor, areboth symmetric. 10-3 Coordinate transformations. Wesawintheprevious section that atensor Tmay bedeﬁned geometrically asalinear vector operator by specifying theresult ofapplying Ttoanyvector C.Alternatively, the 10-3] COORDINATE TRANSFORMATIONS 415 tensor may bespeciﬁed algebraically bygiving itscomponents T,-,-. A discrepancy exists between thetwo deﬁnitions ofatensor, inthat the algebraic deﬁnition appears todepend upon thechoice ofaparticular co- ordinate system. Asimilar discrepancy inthecase ofavector wasnoted inSection 3-1. Wewillnow remove thediscrepancy bylearning how to transform thecomponents ofvectors and tensors when thecoordinate system ischanged. Wewill restrict thediscussion torectangular co- ordinates. Letusconsider twocoordinate systems, rl,2:2,3:3,andx{,xé,22$,hav- ingthesame origin. The coordinates ofapoint inthetwosystems are related byEqs. (7-13): 3 x',-=Za,~,~ar:,-, (10-67) j=1 where 11,-;=6'5'6," (10-68) isthecosine oftheangle between thex,l-andxi-axes. Likewise, 3 1;]:=Z(155313;-. (10-69) i-1 Therelations between theprimed andunprimed components ofanyvector 3 3c=Zcw,=Z0-3, (10-70) Ji=1 j=1 7 may beobtained inasimilar manner bydotting e§-ore,-intoEq.(10-70): 3 Ct=Zm-1'01", (10-71) j=1 . 30,=7».-,-0'.-. (10-72) Wecannow deﬁne avector algebraically asasetofthree components (C1,C2,C3)which transform likethecoordinates (x1,702,703)when theco- ordinate system ischanged. Byreferring toallcoordinate systems, this deﬁnition avoids giving preferential treatment toanyparticular coordinate system. Inthesame way, theprimed andunprimed components ofa tensor T= meter = ilejel (10-73)QM“'5-3M"‘Q 416 TENSOR ALGEBRA. INERTIA AND srnnss TENSORS [cn.u=. 10 arerelated by[seeEq.(10-36)] - 5»..l"1°’Q Tile=91''T'eh= ijaklT1'l: (10‘74) 5:.H".M‘°Q Til =6,:-T-61= ij(lklT(ik. (10-75) Atensor may bedeﬁned algebraically asasetofnine components (T51) that transform according totherule given inEqs. (10-74) and (10-75). Note thedistinction between atensor andamatrix. The concept ofa matrix ispurely mathematical; matrices arearrays ofnumbers which may beadded andmultiplied according totherules (10-45) and(10-47). The concept ofatensor isgeometrical; atensor may berepresented inany particular coordinate system byamatrix, butthematrix must betrans- formed according toadeﬁnite ruleifthecoordinate system ischanged. The coeﬂlcients a,-,-deﬁned byEq.(10-68) arethecomponents ofthe unit vectors efintheunprimed system,‘ andconversely: 3 6';= Z ¢l¢j€j, J'—1 and . 8 OJ‘=2 0,7655. i=1 Since ei,eé,el,areasetofperpendicular unit vectors, Weseethat the numbers a,-,-must satisfy theequations: 3 6';'6;,=Zaijakj =511;, (10-78) _1'=1 where 6,-kisashorthand notation for am={O ifi96la, (1049) 1 f k ii= . There aresixrelations (10-78) among thenine coeﬂicients a,-k. Hence, ifthree oftheconstants a.-,~arespeciﬁed, therestmaybedetermined from Eqs. (10-78). Itisclear that three independent constants must bespeciﬁed tolocate theprimed axes relative totheunprimed (orvice versa). For thexi-axis may point inany direction and two coordinates arethere- forerequired tolocate it.Once theposition ofthexi-axis isdetermined, 10-3] COORDINATE TRANSFORMATIONS 417 theposition ofthexé-axis, which may beanywhere inaplane perpendicu- lartoxi,may bespeciﬁed byonecoordinate. Theposition ofthexé-axis isthen determined (except forsign). Wecanwrite additional relations between thea,-,-’s bytheuseofsuch relations as ‘ Gj'81=511, CaXeﬁ=:l:€§, C1'(C2 XG3) =:l:1, 8170. (10-80) Since atleast three ofthea,-,-’s must beindependent, itisclear that the relations obtained from Eqs. (10-80) arenotindependent butcould be obtained algebraically from Eqs. (10-78). Aninteresting relation isob- tained from G11 (121 031 e1'(92X93)=(112 (122 1132 ==1=11 (10‘81) (113 1123 (133 where theresult is+1iftheprimed and unprimed systems areboth right- orboth left-handed andis—1ifoneisright-handed andtheother left-handed. Hence thedeterminant |a,~,-|is+1or—1according towhether thehandedness ofthecoordinate system isorisnotchanged. In'a left-handed system, thecross product istobedeﬁned using theleftin place oftheright hand. [InEq.(10-81), thetriple product ontheleftistobe evaluated intheprimed system.] The algebraic deﬁnition isthen thesame ineither case: (AXB)=(A233 —A332, A331 -A133, A132 —A2B1)- (10-32) This deﬁnition implies that thecross product AXBoftwoordinary vectors is notitself anordinary vector, since itsdirection reverses when wechange the handedness ofthecoordinate system. Anordinary vector thathasadirection independent ofthecoordinate system iscalled apolar vector. Avector whose sense depends upon thehandedness ofthecoordinate system iscalled anaxial vector orpseudovector. The angular velocity vector 0:isanaxial vector, andso isanyother vector whose sense isdeﬁned bya“right-hand rule.” The vector associated with an(ordinary) antisymmetric tensor isanaxial vector. Thecross product wXCofanaxial with apolar vector isitself apolar vector. The dis- tinction between axial andpolar vectors arises only ifwewish toconsider both right- andleft-handed coordinate systems. Intheapplications inthisbook, we need only consider rotations ofthecoordinate system. Since rotations donot change thehandedness ofthesystem, weshall notbeconcerned with thisdis- tinction. 418 TENSOR ALGEBRA. INERTIA AND srnnss TENSORS [CHAP. 10 The transformation deﬁned byEqs. (10-67), (10-71), and (10-74), where thecoeﬂicients satisfy Eq.(10-78), iscalled orthogonal. Asthename implies, anorthogonal transformation enables ustochange from oneset ofperpendicular unit vectors toanother. Theright member ofEq.(10-71) isformally similar totheright mem- berofEq. (10-34). This suggests analternative interpretation ofEqs. (10-71). Letusdeﬁne atensor Awith components A5," =(Iii, (10-83) andconsider thevector C’=A-C. (10-84) The components CofC’aregiven byEq. (10-71). Similarly, byEq. (10-72), * C=A‘-C’. (10-85) Thus Eqs. (10-71) and(10-72) may beinterpreted alternatively asrepre- senting theresult ofoperating with thetensors A,A‘upon thevectors C, C’,respectively. Intheoriginal interpretation, C’J-,Cfarecomponents of thesame vector Cintwodifferent coordinate systems. Inthealternative interpretation, C’,-,Cfarethecomponents oftwodifferent vectors C,C’ inthesame coordinate system. Weareprimarily interested i11theﬁrst interpretation, inwhich these equations represent acoordinate trans- formation. However, thelatter interpretation will often beuseful in deriving certain algebraic properties ofEqs. (10-71), (10-72), which, of course, areindependent ofhow wechoose tointerpret them. Incase the primed axes areﬁxed inarotating rigid body, either interpretation isuse- ful. Iftheprimed axes initially coincide with theunprimed axes, then we may interpret Eqs. (10-71), (10-72) asexpressing thetransformation from onecoordinate system totheother. Alternatively, wemay interpret A asthetensor which represents theoperation ofrotating thebody from itsinitial toitspresent position, i.e.,avector ﬁxed inthebody andini- tially coinciding with Cwillberotated soastocoincide with C’=A-C. Making useofEqs. (10-84), (10-85), and(10-43), wededuce that, A‘-(A-C) =(A‘-A)-c=c. (10-s0) Hence, byEq.(10-22), At-A=‘I, (10-87) andsimilarly A-A‘=I. (10-88) Atensor having thisproperty issaid tobeorthogonal. Equation (10-87) isevidently equivalent toEq.(10-78). Inthesecond interpretation, Eqs. (10-74) and(10-75) canbewritten as 10-3] COORDINATE TRANSFORMATIONS 419 T’=A-T-A‘, (10-89) T=A‘-T’-A. (10-90) The orthogonal tensor istheonly example weshall have ofatensor with adeﬁnite geometrical signiﬁcance, which isneither symmetric noranti- symmetric; ithas,instead, theorthogonality property given byEq.(10-87). Inview ofthefactthatthevarious vector operations were deﬁned with- outreference toacoordinate system, itisclear thatallalgebraic rules for computing sums, products, transposes, etc., ofvectors andtensors willbe unaffected byanorthogonal transformation ofcoordinates. Thus, for example, . (B+C)?"=B?+03', (10-91) 30-<1);=IET:-102. (1392)=1 (T')'.-1 =T93 (10-93) Wecanalso verify directly theabove equations, and others likethem, byusing thetransformation equations andtherules ofvector andtensor algebra. This ismost easily done bytaking advantage ofthesecond inter- pretation ofthetransformation equations. Forexample, wecanprove Eq.(10-93) bynoting that 1(r‘)'=A-1'-A‘ [byEq.(10-39)] =A-(A-T)‘ [byEqs.(10-50) and(10-57)] =[(A-T)-A‘]‘ [byEqs.(10-50) and(10-57)] =(1')‘,Q.E.D. [byEq.(10-39)]. Any property orrelation between vectors andtensors which isexpressed inthesame algebraic form inallcoordinate systems hasageometrical meaning independent ofthecoordinate system andiscalled aninvariant property orrelation. Given atensor T,wemay deﬁne ascalar quantity called thetrace ofT asfollows:331(1)=ZT,-.-. (10-94) =11». Since this deﬁnition isinterms ofcomponents, wemust show that the trace ofTisthesame inallcoordinate systems. Intheprimed system, wehave 420 TENSOR ALGEBRA. INERTIA ANDSTRESS TENSORS [cn.u>. 10 3v(t)=ZT2,- i=1 EM»as»aw‘3M"QQ= ijailTjl [byEq. (10—74)] = ijajl] T,1 [rearranging sums] = jg551 [BISinEq. = ,-,-,Q.E.D. [byEq.(10—79)]. Another invariant scalar quantity associated with atensor isthedeter- minant --T11 T12 T13 Z T21 T22 T23 ! Tai T32 Tas asmayalsobeveriﬁed bydirect computation. Letusnowstudy theresult ofcarrying outtwocoordinate transforma- tions insuccession. The primed coordinates aredeﬁned byEq.(10—67), interms oftheunprimed coordinates. Letdouble-primed coordinates be deﬁned by @-M~= D-lQa4\s. xi,’= 202. (10-96) Wesubstitute for from Eq. (10~67) toobtain thedouble-primed co- ordinates interms oftheunprimed coordinates: :M~»:M~Iii-:M~»$5!= iciaijwj = ‘ ' aim-a,-,-] 117]‘ 3=Za§¢’,~:c,-, (10-97) i=1 where thecoefﬁcients ofthetransformation av—>:0"aregiven by $-M~= PiQ§1\@-Q@-Q.. at?= (lass) 10-4] DIAGONALIZATION or‘ASYMMETRIC TENSOR 421 Thus thematrix ofcoefficients afjisobtained bymultiplying thematrices aléi,ai.1-according totheruleformatrix multiplication. Ifweinterpret the transformation coefﬁcients asthecomponents oftensors A,A’,A”,we then seefrom Eqs. (10-45) and(10-98) that A”=A’-A. (10-99) This result alsofollows immediately from, Eq.(10-84), applied twice, and wetherefore have analternative way toderive Eq.(10-98). 10-4 Diagonalization ofasymmetric tensor. The constant tensor, deﬁned byEq.(10-23), hasinevery coordinate system* thematrix: c00 cl=0c0- (10-100) 00c Anonconstant tensor may, inaparticular coordinate system, have the matrix: ’ T1 0 0 1=<0 T20) (10-101) 00T3‘ Thetensor Tisthen.said tobeindiagonal form. WedonotcallTadiagonal tensor, because theproperty (10-101) applies only toaparticular co- ordinate system ;after achange ofcoordinates [Eq. (10—74)], Twillusually nolonger beindiagonal form. IfTisindiagonal form, then itseﬁect on avector isgiven simply by (T-C);=T,C,-, i=1,2,3. (10-102) The importance ofthediagonal form liesinthefollowing fundamental theorem: Any symmetric tensor canbebrought into diagonal form by anorthogonal transformation. Thediagonal elements arethen unique except fortheir order, andthecorresponding axes are ' unique except fordegeneracy. (10-103) Before proving this important theorem, letustrytounderstand its signiﬁcance. The theorem states that, given any symmetric tensor T, wecanalways choose thecoordinate axes sothat Tisrepresented bya diagonal matrix. Furthermore, this canbedone inessentially only one way; there isonly onediagonal form (10-101) foragiven tensor T,except *SeeProblem 10attheendofthischapter. 422 rnnsoa ALGEBRA. INERTIA ANDsrnnss TENSORS ICHAP. 10 fortheorder inwhich thediagonal elements T1,T2,T3appear, andeach element isassociated with aunique axisinspace, except fordegeneracy, that is,except when twoorthree ofthediagonal elements areequal. The axes e1,e2,e3inthecoordinate system inwhich thetensor hasadiagonal form arecalled itsprincipal axes. Thediagonal elements T1,T2,T3are called theeigenvalues orcharacteristic values ofT.Whenever wewrite a tensor element with asingle subscript, weshall mean ittobeaneigenvalue. Any vector Cparallel toaprincipal axis iscalled aneigenvector ofT. Aneigenvector, according toEq.(10-102), hastheproperty that opera- tion byTreduces tomultiplication bythecorresponding eigenvalue: T-C=T,C, 1 (10-104) where T,-istheeigenvalue associated with theprincipal axis e,-parallel toC. The theorem (10-103) allows ustopicture asymmetric tensor Tasa setofthree numbers attached tothree deﬁnite directions inspace. If wethink ofTasapplied toeach vector Cinthevector space, then formula (10-102) shows that theeffect isastretching oracompression along each principal axis, together With areﬂection ifT,-isnegative. InSection 10-2 wesawthat asymmetric tensor may bespeciﬁed bygiving sixcomponents T,~,-inanyarbitrarily chosen coordinate system. Wenowseethatwecan alternatively specify Tbyspecifying theprincipal axes (this requires three numbers, aswehave seen), andthethree associated eigenvalues. Iftwoorthree oftheeigenvalues areequal, wesaytheeigenvalue is doubly ortriply degenerate. Iftheeigenvalue istriply degenerate, the tensor clearly isaconstant tensor [Eq. (10—100)] anddiagonal inevery coordinate system. The principal axes arenolonger unique; any axis isaprincipal axis. Every vector isaneigenvector ofaconstant tensor. Iftwoeigenvalues areequal, sayT1=T2,then ifweconsider rotating thecoordinate axes inthee1e2-plane, wecanseethat thetensor willre- main indiagonal form; thefour elements referring tothisplane behave likeaconstant tensor inthat plane. Again theprincipal axes arenot unique, since twoofthem may lieanywhere inthee1e2-plane. Thethird axis e3associated with thenondegenerate eigenvalue T3,however, is unique. Wecanprove that every axis inthee1e2-plane isaprincipal axisbyconsidering theeffect ofTonanyvector \ C=C1G1 -I"C262 inthisplane. Inview ofEq.(10—102),_ ifT1=T2,wehave T'C=T1C1G1 +1120282 =T1C, (10-106) 10-4] DIAGONALIZATION orASYMMETRIC TENSOR 423 sothat Cisaneigenvector ofT.Every vector inthee1e2-plane isan eigenvector ofTwith eigenvalue T1. Ifwelike, wemay saythat there isaprincipal plane associated with adoubly degenerate eigenvalue. Wewillnow, prove thetheorem (10-103) byshowing how theprincipal axes canbefound. Letasymmetric tensor Tbegiven interms ofitscom- ponents T,-,~insome coordinate system, which wewillcalltheinitial co- ordinate system. Toﬁndaprincipal axis, wemust look foraneigenvector ofT.LetCbesuch aneigenvector, andT’thecorresponding eigenvalue. Wecanrewrite Eq.(10-104) intheform " (T—T"l) -C=0. (10-107) Ifwewrite thisequation interms ofcomponents, weobtain (T11 _T’)01 +T1202 +T1303 =0, T2101 +(T22 —T’)02 +T2303 =0, (10408) T3101 +T3202 —|—(T33 ——T’)C'3 =0. These equations fortheunknown vector Chave, ofcourse, thetrivial solution C=0.Ifwewrite thesolution forC1interms ofdeterminants, weseethat C=0istheonly solution unless thedeterminant T11—"T’ T12 T13 T21 T22 —T’ T23 =0, (10-109) T31 T32 T33'*T’ inwhich case thesolution forC’;isindeterminate. Inthiscase itisshown inthetheory oflinear equations* that Eqs. (10-108) have alsonontrivial solutions C’,-.Itisclear that Eqs. (10-108) cannot determine thenumbers C’1,C2,C3uniquely, butonly their ratios tooneanother, C1:C2: C3.This isalso clear from Eq. (10-107), from which webegan. Geometrically, only thedirection ofCisdetermined, notitsmagnitude (nor itssense). Equation (10-109), called thesecular equation, represents acubic equation tobesolved fortheeigenvalue T’.Ingeneral there willbethree roots Ti,Té,Té.Given anyroot T’,wecanthen substitute itinEqs. (10-108) andsolve fortheratios C'1:C2:C'3. Any vector whose components arein theratio C1:C’2:C3isaneigenvector ofTcorresponding totheeigenvalue T’.Foreach eigenvalue T,’-,wecanthen take aunit vector e;-along the direction ofthecorresponding eigenvectors. The axes ei,eé,eéarethen *See, forexample, Knebelman and Thomas, Principles ofCollege Algebra. New York: Prentice-Hall, Inc., 1942. (Chapter IX,Theorem 10.) 424 TENSOR ALGEBRA. INERTIA ANDsrnnss TENSORS [c11A1>. 10 theprincipal axes ofT.When wesolve Eqs. (10-108) forthecomponents ofe;-foragiven T;-,Wegetthree numbers a,-1(=C',~ forT’=T;.)which arethecomponents ofe}along theaxes e,-oftheinitial coordinate system: 3eg»=2a,-1-e,-. (10-110) i=1 This isjustEq.(10-76), hence thenumbers a,-,1arethecoefficients ofthe orthogonal transformation from theinitial coordinate system totheprin- cipal axes. Wesaythat thetransformation with coefﬁcients a,~,-diago- nalizes T. Inorder tobesure wecancarry outtheabove program, wemust prove three lemmas, asthereader may have noted.’ First, wemust prove that theroots T’ofthesecular equation (10-109) arereal; otherwise wecannot ﬁndrealsolutions ofEqs. (10-108) forC1,C2,C3.Second, wemust prove that thevectors e},obtained from Eqs. (10-108) forthedifferent eigen- values T},areperpendicular ;otherwise Wedonotobtain asetofperpen- dicular unit vectors. Third, wemust show that inthedegenerate case, two(orthree) perpendicular unit vectors e}canbefound that correspond toadoubly (ortriply) degenerate eigenvalue. ~ LEMMA 1.Theroots ofthesecular equation (10-109) fora symmetric tensor arereal. (10-111) Equation (10-109) isobtained from theeigenvalue equation (10-104): . T-C=T’C. (10-112) Toprove thelemma, letusﬁrst allow T’tobecomplex. Wewillneed alsotoallow thecomponents C1ofthevector Ctobecomplex. Avector Cwith complex components, hasnogeometric meaning intheusual sense, ofcourse, butwecanregard allthealgebraic deﬁnitions ofthevarious vector operations asapplying alsotovectors with complex components. The various theorems ofvector algebra willhold also forvectors with complex components. [There isoneexception tothese statements. The length ofacomplex vector cannot bedeﬁned byEq.(3-13), butinstead must bedeﬁned by |A|=(A*-A)%. (10-113) This deﬁnition willnotberequired here.] Wewilldenote byC*thevector whose components arethecomplex conjugates ofthose ofC.Letusmulti- plyC*into Eq.(10-112): c*-1-c =r'(c*-c). (10-114) 10-4] DIAGONALIZATION orASYMMETRIC TENSOR 425 Ifwetake thecomplex conjugate ofthisequation, wehave C-T-C*=T’*(C* -C), (10-115) since Tisreal, andinview ofEq.(3-18). Now bydeﬁnition (10-53), C*-T=T‘-C*. (10-116) Hence - C*-T-C =(C*-T)-C =(T’-C*)-C [byEq.(10—53)] =C-T’-C* [byEq.(3-18)]. (10-117) Forasymmetric tensor, T=T‘,sothat theleftmembers ofEqs. (10-114) and(10-115) areequal, and T’=T’-*, (10-118) sothat T’isreal. LEMMA 2.Theeigenvectors ofasymmetric tensor correspond- ingtodiﬁerent eigenvalues areperpendicular. (10-119) Toprove thislemma, letusassume that T1,T§aretwoeigenvalues ofT corresponding totheeigenvectors C1,C2: T-C1=T’1C1, (10-120) T-C2=T’2C2. (10-121) Wemultiply C2intoEq.(10-120), andC1intoEq.(10-121): C2-T-C1=T’1(C2 -C1), (10-122) C1-T-C2=T2(C2 -C1). (10-123) Since Tissymmetric, theleftmembers areequal, andwehave (T’1—T2)(C2 -C1)=0. (10-124) Iftheeigenvalues T1,Téareunequal, theeigenvectors C2,C1areper- pendicular. LEMMA 3.Inthecase ofdouble ortriple degeneracy, Eqs. (10-108) have twoorthree mutually perpendicular solutions forthevector C. (10-125) Fortheproof of(10-125), suppose that Ti=Tg. Ifwesubstitute T’=TiinEqs. (10-108), then bythetheorem referred tointhe 426 TENSOR ALGEBRA. INERTIA ANDsrnnss TENSORS [CHAP. 10 footnote onpage 423, there isatleast onenontrivial solution C1,C2, C3. Leteibeaunit vector parallel tothevector (C1,C2,C3). Then 1-e’1=T’1e’1. (10-120) Now, choose anypair ofperpendicular unit vectors eg’,e§’perpendicular toef,anduseEqs. (10-68) and (10-74) totransform thecomponents of Tintothedouble-primed coordinate system e1,eg’,e§,’.Byacomparison ofEqs. (10-126) and(10-36), weseethat wemust get T’1’1=T’1, T2’1=0, T§'1=0. (10-127) Since Tissymmetric, itsdouble-primed components must therefore be given by T’100 1=0T2’2T2},- (10-123) 0 T23 T23 Furthermore, thesecular equation T’1—T’ 0 0 0 T22—T’ T23 =0 (10-129) 0 T2’3 Ti:’a-T’ must have thesame roots asEq.(10-109). This istruesince theleftmem- bers ofboth equations arethedeterminants ofthesame tensor (T—T’I), expressed intheunprimed anddouble-primed coordinate systems, andwe noted attheendofSection 10-3 that thedeterminant ofatensor hasthe same value inallcoordinate systems. Ifweexpand thedeterminant (10-129) byminors oftheﬁrstrow, weobtain (T’1-T’)T52_T’ “'3 =0. (10-130) T’2’3 T§’3—T’ Since Tiisadouble ortriple root ofthisequation, itmust bearoot ofthe equation 1/__ I II‘T22 T T2“ =0. (10-131) T23 T3’3-T’ Therefore theequations (T32—T'1)(-"2’ +T2’30is’ =0,- (10-132) T5902’ +(Ti3’3—T'1)C'%’ =0, 10-4] DIAGONALIZATION orASYMMETRIC TENSOR 427 have anontrivial solution which deﬁnes aneigenvector (0,Ci’,Ci’)inthe egeii’-plane with theeigenvalue Ti. Wehave therefore asecond lmit eigenvector eéparallel to(0,Ci’,C3’)andperpendicular toei.Ifwetake athird unitvector eiperpendicular toei,ei,then inthisprimed coor- dinate system, wemust have T11 Z T1; T21 Z 07 T31 Z 0) (10-133) T12 =02 T22 =TII; T’32 =0- Thus Tmust have thecomponents T'1 0 0 T=0 T'1 0 » (10-134) 0 0 T3 andei,ei,ei,,areprincipal axes. IfTiwere atriple root ofEq.(10-109), itwould alsobeatriple root ofthesecular equation T'1—T’ 0 0 0 T'1—T’ 0 =(T’1—T’)(T’1 —T’)(T§ —T’)=0. 0 0 T3—T’ . (10-135) Therefore Ti=Ti,andwehave three perpendicular eigenvectors cor- responding tothetriple root Ti=Ti=Ti. Theabove three lemmas complete theproof ofthefundamental theorem (10-103). The algebra inthissection may begeneralized tovector spaces ofany number ofdimensions, with analogous results regarding theexistence of principal axes ofasymmetric tensor. AttheendofSection 10-3 wenoted that thetrace andthedeterminant of atensor Thave thesame value inallcoordinate systems. Weseefrom Eq. (10-101) that thetrace isthesum oftheeigenvalues ofT: WT) =T1+T2+T3, (10-136) andthedeterminant istheproduct oftheeigenvalues: det(T) =T1T2T3. (10-137) Wecanform athird invariant scalar quantity associated with asymmetric tensor bysumming theproducts ofpairs ofeigenvalues: M(T) =T1T2 -[-TZT3 —[-T3T1. (10-138) 428 TENSOR ALGEBRA. INERTIA ANDsrnnss TENSORS A[CHAP. 10 Wecanevaluate M(T)inanycoordinate system bysolving thesecular equation (10-109) forthethree roots T1,T2,T3andusing Eq.(10-138). Thesolution of Eq.(10-109) canbeavoided bynoting that thesum (10-138) must bethe coefficient ofT’inEq.(10-109), which isthesum ofthediagonal minors ofthe determinant ofT: T11T12 T22T23 T33T31 M(T) = + -|" ' (19-139)T21T22 Ta2T33 T13T11 Wecould also show bydirect calculation that M(T)asgiven byEq.(10-139) hasthesame value after acoordinate transformation given byEq.(10-74). Foranytensor T,thedeterminant of(T-—T’‘|)must have thesame value inallcoordinate systems. Therefore, inparticular, theroots T’ofEq.(10-109) willbethesame inallcoordinate systems, even foratensor Tthat isnotsym- metric. Westillcalltheroots T’theeigenvalues ofT.IfTisnotsymmetric, oneeigenvalue willberealandtheother two willbeaconjugate complex pair. Forthereal eigenvalue wecanﬁnd aneigenvector. Forthecomplex eigen- values wecannot, ingeneral, ﬁnd eigenvectors. (That is,notunless weadmit vectors with complex components, which have only algebraic signiﬁcance. Even then, wecannot prove ingeneral that theeigenvectors areorthogonal.) Inany case, theexpressions given byEqs. (10-136), (10-137), (10-138), and (10-139) arestillrealandindependent ofthecoordinate system. Asanexample ofthediagonalization procedure, letusdiagonalize the tensorT=AA+BD+DB, which obviously issymmetric. Wewilltake A=4ae1, B=7ae2 —l—ae3, D=ae2—ae3. Thetensor Tisthen represented inthiscoordinate system bythematrix: 1022 0 0 T=014112 —6a2 - 0 —6a2 —2a2 Inthiscase, thesecular equation (10-109) is 1022-T’ 0 0 0 1422-T’ -ca’ =(10.12-:1") 0 —6a2 -2a2 —T’ ><(T’2+12112:!" -042‘)=0. 10-4] DIAGONALIZATION orASYMMETRIC TENSOR 429 Theroots (necessarily real) are T1=1002, T2=1002, T5=-402. Equations (10-108), forthedoubly degenerate root T’=16a2, are 0=0, —2a2C2 —6a2C3 =0, —6a2C2 —18a2C3 =0. Clearly, C1isarbitrary, andthelasttwoequations areboth satisﬁed if C2=——3C3. Therefore anyvector oftheform C=C1e1 —3C3e2 +C3e3 isaneigenvector forarbitrary C1,C3. Thus wehave atwo-parameter family ofpossible eigenvectors, from which wemay select forei,ei anytwoperpendicular unit vectors. Wewilltake ell.=e1; 3 1e’=-e ——-e.2\/10 2\/10 3 Wecould have guessed eifrom theform ofT.The reader should verify that eiandeiand, infact, anyvector inthee{e2-plane satisfy Eq. (10-104) with T;=16a2. ForT’=—4a2, Eqs. (10-108) become 200201 =0, 1811202 —0020 3=0, —6a2C2 +20203 =0. Now there isjustaone-parameter family ofsolutions ofwhich there isoneunit eigenvector (except forsign): e3=i°2-bi-93»V10 V10 where positive signs were chosen sothat ei,ei,eiwould form aright- handed system. The vector eiiisperpendicular totheeiei-plane asit 430 TENSOR ALGEBRA. INERTIA ANDsrnsss rnnsons [CHAP. 10 must beaccording tolemma 1.Itmay beveriﬁed that ei,isaneigen- vector ofTwith theeigenvalue —4a2. Byreference toEq.(10-76), wemay write thecoefﬁcients ofthetrans- (121 (122 (Z23 = 0 . (131 G/32 G33 0 1/\/10 Thereader should verify that these coeflicients satisfy Eqs. (10-78); that thevectors eiareproperly transformed according toEqs. (10-71) and (10-72), which inthiscase are 3 3 5512=2aktetu", 6%=Zam"5170 i=1 k=1formation totheprincipal axes ofT: (G11 G/12 (113) 0 0 ) where ejiistheithcomponent ofe,’-intheunprimed coordinate system and 6,-1,isthekthcomponent ofeiintheprimed coordinate system; and also that Tisproperly transformed according toEq. (10-74) from its original form toitsdiagonal form. -10-5 Theinertia tensor. Theinertia tensor ofarigid body isgiven by Eq.(10-25). Forabody ofdensity p(x,y,z),wemayrewrite theinertia tensoras 1.,=[p(r2'l -11)dV, (10-140) where wehave used thesubscript “o”toremind usthat theinertia tensor iscalculated with respect toasetofaxeswith origin atO.Wewillomit the subscript except when thediscussion concerns more than oneorigin. The diagonal components ofIarejust themoments ofinertia [Eq. (5-80)] about thethree axes:1..=[ffP(1/2+Z2)dv. 1,,=ff/p(z2 +02)dV, (10-141) In=/[[002 +1/2)dV- Theoff-diagonal components, oftencalledproducts ofinertia, are 1.,=1,.=-fffpxydv, "I2,=1.2=—/'//pyz dV, (10-142) 1..=I...=—[/[pew dV. 10-5] THE INERTIA TENSOR 431 Since wemay useEq. (10-74) tocalculate thecomponents ofthe inertia. tensor relative toanyother setofaxes through thesame origin O, weseefrom Eqs. (10-74) and(10-141) thatthemoment ofinertia about any axis through O,inadirection designated bytheunit vector n,is In=n-I-n. (10-143) Itoften iseasier tocalculate thecomponents oftheinertia tensor with respect toaconveniently chosen setofaxesandthen useEq.(10-143), than tocalculate Indirectly, iftheaxisnisnotanaxisofsymmetry ofthe body. Wecanobtain auseful analog totheParallel Axis Theorem (5-81) for themoment ofinertia bycalculating theinertia tensor lorelative toan arbitrary origin ofcoordinates Ointerms oftheinertia tensor |grela- tivetothecenter ofmass G.Letrandr’beposition vectors ofanypoint Pinthebody relative toOandG’respectively, andletRbethecoordinate ofGrelative toO(Fig. 5-12), r=r’+R. (10-144) Then wehave, from Eq.(10-140), 1.,=[f[p1<r'+11)-0' +R>1—0'+R><v+R>1dv =/f/p[(r'-r')1- r’r']dV+ [(R-R)! -RR]f[/pdv +21[R-f[[p1"dV] —[fffpr'dv]R -R/I/pr’dV. (10-145) Inview ofthedeﬁnition (5-53) ofthecenter ofmass, wehave [Ups av=0. (10-140) Equation (10-145) therefore reduces to 1,,=10+M(R21 -RR). (10-147) Note that both thestatement andproof ofthis theorem areinprecise analogy with theParallel Axis Theorem (5-83) forthemoment ofinertia. Itisevident from thedeﬁnition (10-140) thattheinertia tensor ofa composite body may beobtained bysumming theinertia tensors ofits parts, allrelative tothesame origin. Ifabody rotates, thecomponents ofitsinertia tensor, relative tosta- tionary axes, will change with time. The components relative toaxes ﬁxed inthebody, ofcourse, willnotchange ifthebody isrigid. Wemay think oftheinertia tensor Iasrotating with thebody. Ifthe(constant) 432 TENSOR ALGEBRA. INERTIA ANDsrnnss TENSORS [cn111>. 10 components along axes ﬁxed inthebody aregiven, the(changing) com- ponents along stationary axes arethen given byEq. (10-89), where A represents thetransformation from body axes tospace axes. The most convenient setofaxes inthebody formost purposes aretheprincipal axes oftheinertia tensor, alsocalled theprincipal axes ofthebody. The eigenvalues oftheinertia tensor arecalled theprincipal moments ofin- ertia. Wewilllearn more inthenext chapter ofthedynamical signiﬁcance oftheprincipal axes, butwemay note here that, according toEq.(10-24), ifthebody rotates about aprincipal axis, theangular momentum isparal- leltotheangular velocity. Wemay always choose arbitrary axes, compute I,andthen usethemethod ofSection 1-4toﬁndtheprincipal axes. Itis often possible, however, tosimplify theproblem bychoosing tobegin with acoordinate system inwhich oneoralloftheaxes areprincipal axes. Inmany cases, abody willhave some symmetry, sothat wecansee that certain oftheproducts ofinertia (10-142) willvanish iftheaxes are chosen inacertain way. Forexample, wecanprove thefollowing theorem: Any plane ofsymmetry ofabody isperpendicular toaprin- cipal axis. (10-148) Ifwechoose theyz-plane astheplane ofsymmetry, then p(-_x) yaZ)=p(x: yaz)' Itiseasy toshow that because ofEq.(10-149) theintegrals (10-142) for I1,andI,,willvanish. Therefore thex-axis isaprincipal axisinthiscase. Inasimilar way, wecanprove thetheorem: Any axisofsymmetry ofabody isaprincipal axis. Theplane perpendicular tothisaxis isaprincipal plane corresponding toadegenerate principal moment ofinertia. (10-150) Asphere, orabody with spherical symmetry has, evidently, aconstant inertia tensor. . Asanexample, consider theright triangular pyramid shown inFig. 10-1. The components oftheinertia tensor relative totheaxes (x,y,z) aretobecalculated from theformula: ii“ '1-§' ¢—v—§# y2+z2 -xy —zx I=/‘ /l f p —xy 22+x2 —yz dxdydz,==0 11-0 (i=0 _zx _yz $2+ya where each component ofIistobeobtained byevaluating theindicated integral over thecorresponding component ofthematrix. The density pisgiven interms ofthemass Mby M=iasp. 10-5] THE INERTIA TENSOR 433 2 A 3 4 E“ a ___-wy av--------- \\\\\ \\\ a //1/ //Z 2: FIG. 10-1. Aright triangular pyramid. Because ofthesymmetry between xandy,itisnecessary toevalute only thefour integrals J1=/ffpxzdxdydz =f/fpyzdwdydx =-B5-Mag, J2=/ffpﬁdxdydz =;f;;Ma2, J3=ff/lpxydxdydz =$6M112, J4=[ﬂpxzdxdydz =ﬂfpyzdxdydz =ﬁ;Ma2. Theinertia tensor isthen given by J1+J2 —J3 —-J4 13-2 —3 Mag l= —J3 J1+J2 —J4 =-2 13-3 E-, J4 J4 2J 3 3 8__ _ 1 _ _ where thenotation means that each element ofthematrix istobemul- tiplied byMa2/40.Letusﬁndtheprincipal axes. Bysymmetry [theorem (10-148)] theaxisx"shown inFig. 10-1 isaprincipal axis. Letusthere- fore ﬁrst transform totheaxes x",y”,z.The coeﬁicients ofthetrans- formation are,byEq.(10-68), V a,,~, a,,~,, a,,~, 1/\/2 -1/\/2 0 a,/I, a,/1,, a,/I, =1/\/2 1/\/2 0- a,,, aw an 0 0 1 \ 434 TENSOR ALGEBRA. INERTIA ANDSTRESS TENSORS [CHAP. 10 Using Eq.(10-74), wenow calculate theinertia tensor components along thex”-,y"-,andz-axes: 15 0 0 2 1=<011 -3\/5) M“0-s\/5 s Weseethat thex"-axis isindeed aprincipal axis. Thesecular equation is 15->. 0 0 20 11->.—3\/i=0, T’=%—)\,' 0 -s\/5 s->. andtheroots are p >.,'=15, A,,’=5,>.,,=14, OI‘ T,’=sM112, T1,’=4Ma2, T,’=51,,Ma2. Equations (10-108) canbesolved forthecomponents oftheunit vectors i’,j’,k’interms ofi”,j",k: il=ill, 1'=15/P31"+s~/Ek. kr=_%\/gin + Asasecond example, letusﬁndtheinertia tensor about thepoint Oof theobject shown inFig. 10-2. Theobject iscomposed ofthree ﬂatdisks 2 O, Ill ‘F 9, . 1/’ FIG. 10-2. Three disks. Fro. 10-3. Acircular disk with its principal axes. 10-5] THE INERTIA TENSOR 435 ofmass Mandradius a.Bysymmetry, theprincipal axes aretheindi- cated axesx,y,z.Weﬁrstcalculate theinertia tensor ofasingle disk about itscenter, relative toitsprincipal axes x’,y’,z’,asshown inFig. 10-3. Themoment ofinertia IZ’isgiven byEq.(5-90), andthemoments ofinertia I,,',I,1arehalf I,»,bythePerpendicular Axis Theorem (5-84). Wecantherefore write theinertia tensor ofadisk, relative toitsprincipal axes x’,y’,z’,as 1OO 2 1,,=<010)% (10-151) OO2 Forthebottom disk, theprincipal axes areparallel tox,y,z,andweneed only apply theorem (10-147) toobtain itsinertia tensor relative to"the x-,y-,z-axes with origin atO: 1,,=1,,+M(3a21 -3a2kk) 13 O0 2=(..1.». 0 02 Fortheright-hand disk, withtheaxesx’,y’,z’oriented asshown, weﬁrst apply theorem (10-147) toobtain theinertia tensor about O,relative to axe parallel tox’,y’,z’: 500 2M |a(x’1l'z’) =(0 10)Ta ' 006 _ Thetransformation from x’-,y’-,z’-axes tox-,y-,z-axes isgiven by axz’ any’ asa’ A 1 aux’ avg’ avg’ 1 0 an’ azy‘ an’ 0 "- Wenow useEq.(10-74). Itisperhaps easier tocarry outtheprocess in twosteps, according toEq.(10-89) 2*1-‘ ‘°"‘NIH05M1-I NP$0ea V 5 ‘O O 2M A'|o(:¢'1/z’) =<0 '% Ta' '0-%\/5 3 *Matrices may bemultiplied conveniently according totherule (10-45) bynoting that theelement (T-5).-1.isObtained bysumming theproducts of pairs ofelements across rowiinTanddown column hinS. 436 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [cmua 10 Now 0 (A'|o(a:'y'z’)) 'At=<0 3\/3’) <00- 302x/3 2 =|0(==uz) = _ 2ca "*“w|\-O /‘\eo43 MU!<»-12¢<=ooOI—‘NFC Wemay interpret thisalgebra asacomputation ofI,relative toanew setofaxes. Alternatively, wemay interpret Aasatensor which rotates thedisk through anangle of60°about thex-axis; l,,(,,',,',,', isthen the moment ofinertia ofadiskwhose principal axes areparallel tox,y,z,and thealgebra isacomputation oftheeffect ofrotating thedisk toitsﬁnal position. Theleft-hand disk, correspondingly, hastheinertia tensor M 2 |o(zyz) = 0 J19" — . 0-\/3 Z7U1 "*"o vPl=o'F{OCAD V Theinertia tensors ofthethree disksmay now beadded andweobtain 23 0 0 2.,,=(0220)MTa. 0 0 6% Letuscalculate themoment ofinertia oftheobject shown inFig. 10-2 about they’-axis through O.ByEq.(10-143), wehave 1,,=5'-|,,-5'=10%Ma2. Wecould usetheorem (10-147) toobtain Iabout thecenter ofgravity G, which isattheintersection ofthe2-and2’-axes. Itisclear from symmetry that anyaxisperpendicular tooneofthethree disks through itscenter is aprincipal axisrelative toG.This canonly betrueiftheinertia tensor rel- ative toGhasadouble degeneracy intheyzy’z’-plane. Thereader should check thisbycarrying outthetranslation ofl,,tothecenter ofmass G. Itshould benoted that theprincipal axes oftheinertia tensors ofabody relative totwodifferent points OandO’,ingeneral, willnotbeparallel, asexperimentation with Eq.(10-147) willshow. Thekinetic energy Tofarotating rigid body canalsobeexpressed con- veniently interms oftheinertia tensor. From Eqs. (10-2) and (10-3) andusing therules ofvector algebra, wehave 10-5] THE INERTIA TENSOR 437 N T=Z imkvi k=l N=Z)sm..(<»><:1.)-0»><5.)1==1 =ﬁ‘imkw '[IrX(wX1%)]k=1 =-Q-w-L. (10-152) Therefore Tcanbeexpressed intheform T==1,~w-l-w. (10-153) Equation (10-153) expressed interms ofcomponents along any setof axesisthen '%Ia::1:(-'3: + “l”'%Izz(-92 +Ixywzwy +Iyzwuwz +Izzwzwx =T- (10-154) This istheequation ofafamily ofquadric surfaces inw-space, each sur- facethelocus ofangular velocities forwhich thekinetic energy hasacon- stant value T.IfEq.(10-153) iswritten interms ofcomponents along principal axesx’,y’,z’, Héwéz +iltwtz +ii20122=T, (10-155) then weseethat these surfaces areellipsoids, since themoments ofinertia arenecessarily positive. Ifwedeﬁne avector 1’=%..., (10-155) where aisaconstant, then Eq.(10-153) canbewritten as 1-1-r=a2. (10-157) Thisistheequation oftheinertia ellipsoid. Theconstant adetermines the sizeoftheellipsoid. Itiscustomary toseta=1inwhatever units are being used, forexample, a=1cm-erg-sec. Inthiscase, wenote that the sizeoftheellipsoid (but notitsshape) depends ontheunits being used. Theinertia ellipsoid ofabody, likeitsinertia tensor, isrelative toa particular origin about which moments arecomputed. Thesixcoefficients of.the quadratic form ontheleftofEq.(10-157) arethecomponents of theinertia tensor: I,,,x2 -1-I,,,,y2 +I,,z2 -1-2I,,,xy -1-2I,,,yz +212221: =a2, (10-158) 438 TENSOR ALGEBRA. INERTIA AND STRESS TENSORS [cn.u*.‘ 10 sothat theinertia tensor isuniquely characterized bythecorresponding inertia ellipsoid. This gives usanother convenient geometrical way of picturing theinertia tensor. Bycomparing Eq.(10-157) with Eq.(10-143), weseethat theradius toanypoint ontheinertia ellipsoid is r=a1:'1', (10-150) where I,isthemoment ofinertia about anaxisparallel tor.Inparticular, theprincipal moments ofinertia arerelated byEq.(10-159) tothesemi- principal axes oftheinertia ellipsoid. Weseethat ifthere isdouble de- generacy, theinertia ellipsoid isanellipsoid ofrevolution. Iftheprincipal moments ofinertia areallequal, theellipsoid ofinertia isasphere. Foranysymmetric tensor T,Wecanform aquadratic equation ofthe form (10-157) which deﬁnes aquadric surface that uniquely charac- terizes T.The principal axes ofTaretheprincipal axes ofitsassociated quadric surface. Iftheeigenvalues ofTareallpositive, thesurface isan ellipsoid. Otherwise, itwillbeahyperboloid oracylinder. Ifallthe eigenvalues arenegative, wewould need towrite —a2 fortheright member ofthequadratic equation inorder todeﬁne arealsurface. 10-6 Thestress tensor. Letusrepresent anysmall surface element in acontinuous medium byavector dSwhose magnitude dSisequal tothe area ofthesurface element andwhose direction isperpendicular tothe surface element. Tospecify thesense ofdS,wewilldistinguish between thetwosides ofthesurface element, calling onetheback andtheother thefront. Thesense ofdSisthen from theback tothefront. Wemay then describe thestate ofstress ofthemedium atanypoint Qbyspecify- ingtheforce P(dS) exerted across any surface element dSatQbythe matter attheback onthematter atthefront ofdS. Weunderstand, of course, that thesurface element dSisinﬁnitesimal. That is,allstatements wemake areintended tobecorrect inthelimit when allelements dS—>0. Forasuﬂiciently small surface element, theforce Pmay depend onthe area andorientation ofthesurface element, butnotonitsshape. Thus P isindeed afunction only ofthevector dSatanyparticular point Qinthe medium. Wewillshow that P(dS) isalinear function ofdS. Wemay therefore represent thefunction P(dS) byatensor P,thestress tensor:* P(dS) =P-dS. (10-160) *Thereader iscautioned that many authors deﬁne thestress tensor with the opposite sign from thedeﬁnition adopted here, sothat atension isapositive stress andapressure, anegative stress. Thelatter convention isalmost universal inengineering practice, whereas thedeﬁnition adopted hereismore common in works ontheoretical physics. 10-6] THEsmnss TENSOR 439 P(dS2) /Z 452 ’§ '\\ \-P(dS1) \\dS1 \\ \\ ~:"_~ 1>(_as, -dS2) —(dS1 +dS2) FIG. 10-4. Atriangular prism inacontinuous medium. Toshow thatP(dS) isalinear vector function, wenoteﬁrstthatifdS issmall enough sothatthestate ofstress ofthemedium doesnotchange over thesurface element, then theforce Pwillbeproportional tothe area dSsolong astheorientation ofthesurface iskept ﬁxed. Thus for apositive constant c, P(cdS) =cP(dS). (10-161) Ifthedirection ofdSisreversed, theback and front ofthesurface ele- ment areinterchanged, andtherefore byNewton’s third law, P(—dS) = —P(dS), sothat Eq.(10-161) holds alsoifcisnegative. Now, given any twovectors dS,, dS2,letusimagine atriangular prism inthemedium with twosides dS1,dS2, asinFig. 10-4. Iftheendfaces areperpendicular to thesides, then thethird sideis—dS1 —dS2, asshown. Ifthelength of theprism ismade much greater than thecross-sectional dimensions, we may neglect theforces ontheendfaces, andthetotal force ontheprism is dF=P(dS1) +P(dS2) +P(—dS1 —dS2). (10-162) Ifthedensity isp,theacceleration oftheprism isgiven byNewton’s lawofmotion: pdVa =dF. (10-163) Now ifwereduce alllinear dimensions oftheprism byafactor a,theareas dS,-aremultiplied by042;hence byEq.(10-161), dFismultiplied bya2, 440 TENSOR ALGEBRA. INERTIA ANDsrnnss TENSORS [CHAR 10 anddVismultiplied bya3,sothat apdVa(a) =dF, (10-164) where a(a) istheacceleration ofaprism atimes smaller. Now asa—->0, theacceleration should notbecome inﬁnite; hence weconclude that dF=O, (10-165) from which, byEqs. (10-162) and(10-161), P(dS1) *1“P(dS2) =P(dS1 +dS2). (10-166) Equations (10-161) and(10-166) show that thefunction P(dS) islinear. Note that Eqs. (10-166) and(10-162) imply that there isnonetforce on theprism ifthestress function P(dS) isthesame atallfaces. Any net force canonly result from differences inthestress atdifferent points of themedium; such differences reduce tozero as01——>0. Byconsidering small square prisms, andrecognizing that theangular acceleration must notbecome inﬁnite asthesizeshrinks tozero, wecan show byavery similar argument (see Problem 32)that Pmust bea symmetric tensor. Thestresses ateach point Qinamedium aretherefore given byspecifying sixcomponents ofthesymmetric stress tensor P. Ifthemedium isanideal ﬂuid whose onlystress isapressure pinall directions, thestress tensor isevidently just I P=pl. (10-167) Note that wedidnotprove inChapter 8that inanideal ﬂuid, that is, onewhich cansupport noshearing stress, thepressure isthesame inall directions. This wasproved only inChapter 5foraﬂuid inequilibrium. This logical defect cannow beremedied. (See Problem 33.) According tothedeﬁnition ofP,thetotal force duetothestress across anysurface Sisthevector sum oftheforces onitselements: F=f/1»-ds. (10-10s) S IfSistheclosed surface surrounding avolume Vofthemedium, andif wetake ntobetheconventional outward normal unit vector, then the total force exerted onthevolume Vbythematter outside itis F=-as PdS, (10-109) andbythegeneralized Gauss’ theorem, [seediscussion below Eq.(5-178)], F=-[ffv-mv. (10-170) v 10-6] THEsrmsss TENSOR 441 Since Visanyvolume inthemedium, theforce density duetostress is f,=—V-P. (10-171) Inagreement with anearlier discussion, weseethat this force density arises only from differences instress atdifferent points inthemedium. Equation (10-171) may alsobederived bysumming theforces onasmall rectangular volume element. Theequation ofmotion (8—138) maynowbegeneralized toapply toany continuous medium: 6p£1d%—|—V-P=f. (10-172) This equation may alsoberewritten intheform (8—139): 6v 1 f This equation, together with theequation ofcontinuity (8—127), de- termines themotion ofthemedium when thebody force density fandthe stress tensor Paregiven. The stress Patanypoint Qmay beafunction ofthedensity andtemperature, oftherelative positions and velocities oftheelements near Q,andperhaps also oftheprevious history ofthe medium, which may beasolid (elastic orplastic) oraﬂuid (ideal or viscous). ' From Eqs. (10-172) and (10-173) wecanderive conservation equa- tions analogous tothose derived inSection 8-8. Theconservation equation forenergy analogous toEq.(8—149) is,forexample, 2%(2,212av)=v-(1-v-1»)av.‘ (10-174) The further manipulations oftheenergy equation carried outinSec- tion 8-8cannot allbecarried through inthesame way forEq.(10-174) because ofthedifference inform between thestress term here andthe pressure term inEq. (8-149), asthereader may verify. The energy changes associated with changes involume andshape ofanelement ina continuous medium areingeneral more complicated than those associated with expansion andcontraction ofanideal ﬂuid. Inaviscous ﬂuid, thestress tensor Pwillbeexpected todepend onthe velocity gradients intheﬂuid. This isconsistent with thedimensional arguments inSection 8-14, where wesaw that the term V-Pin Eq.(10-172) must consist ofthecoefficient ofviscosity 1;multiplied by some combination ofsecond derivatives ofthevelocity components with respect tox,y,andz.Iftheﬂuid isisotropic, asweshall assume, then therelation between Pandthevelocity gradients must notdepend onthe orientation ofthecoordinate system. Wecanguarantee that this will 442 TENSOR ALGEBRA. INERTIA ANDsrnnss TENSORS [CHAP. 10 besobyexpressing therelation inavector form that does notrefer ex- plicitly tocomponents. Thedyad 61/ % 6261/ % 62(10-175)§ﬁ%@z Bx 6x 6x v,,=%%§."161/ % 62 hasasitscomponents thenine possible derivatives ofthecomponents of vwith respect tox,y,andz.Hence wemust trytorelate PtoVv. The dyad (10-175) isnotsymmetric, butwecanseparate itintoa sym- metric and anantisymmetric part, asinEqs. (10-64) through (10-66): Vv=(Vv), +(Vv),,, (10-176) (Vv), =%Vv —l—2-(Vv)', (10-177) (Vv),, =%Vv—2(vv)‘. (10-178) The antisymmetric part isrelated, asinEqs. (10-62) and (10-63), toa vector a1=QVXV, (10-179) such that foranyvector dr, A (vv),.- dr=.5><dr. (10-180) Ifdristhevector from agiven point Qtoanynearby point Q’,weseethat thetensor (Vv),, selects outthose parts ofthevelocity differences between QandQ’which correspond toa(rigid) rotation oftheﬂuid around Qwith angular velocity w.This isinagreement with thediscussion ofEq.(8-133), which isidentical with Eq.(S-179). Since noviscous forces willbeasso- ciated with apure rotation oftheﬂuid, theviscous forces must beexpres- sible interms ofthetensor (Vv),,. Since Pisalsosymmetric, Wearetempted towrite simply P=C(Vv),, (10-181) where Cisaconstant. Inthesimple case depicted inFig. 8-10, theonly nonzero component ofVvis(iv,/6y, andEqs. (10-181) and(10-177) then give . I . ‘E 70 %C’.ayA0 1>=20% 00, (10-1s2) 0 00 10-6] THEsmnss TENSOR 443 andtheviscous force across dS=jdSwillbe dF=1>-<zs=2c%l;dsi, (10-188) inagreement with Eq.(8-243) ifC=-21;. Anegative sign isclearly needed, since theviscous force opposes thevelocity gradient. However, Eq. (10-181) isnotthemost general linear relation between Pand Vv that isindependent ofthecoordinate system. Forwecanfurther de- compose (Vv), into aconstant tensor andatraceless symmetric tensor inthefollowing way: _ (Vv). =(Vv). +WY)», (10-184) (Vv), =§,=Tr(Vv),'l =31,-V-VI, (10-185) (Vv),,, -—-=(Vv), —§V-vI. (10-186) This decomposition isindependent ofthecoordinate system, since we have shown that thetrace isaninvariant scalar quantity. Weseeby Eq.(8—116) that thetensor (Vv), measures therate ofexpansion orcon- traction oftheﬂuid. The tensor (Vv),,, with ﬁve independent com- ponents, speciﬁes theway inwhich theﬂuid isbeing sheared. Weare therefore freetoset P=—217(VV)t, -—%17’V -VI, (10-187) with acoefﬁcient 1;which characterizes theviscous resistance toshear, andacoefficient 1)’which characterizes aviscous resistance, ifany, toex- pansion andcontraction. Thelastterm corresponds toauniform pressure (ortension) inalldirections atthegiven point. Thecoefﬁcient 17’issmall andnotvery well determined experimentally foractual ﬂuids. According tothekinetic theory ofgases, 11’iszero foranideal gas. Totheviscous stress duetovelocity gradients, given byformula. (10-187), must beadded ahydrostatic pressure which may also bepresent andwhich depends on thedensity, temperature, andcomposition oftheﬂuid. Ifwelump the lastterm inEq. (10-187) together with thehydrostatic pressure into a total pressure p,thenthecomplete stress tensor is p 1>=p1-1;[Vv+(vv)‘ -gv-v1]. (10-188) Thereader may readily write thisoutinterms ofcomponents. Formula (10-188) isthemost general expression forthestress inan isotropic ﬂuid inwhich there isahydrostatic pressure, plus viscous forces proportional tothevelocity gradient. Itispossible toimagine that the stress might alsocontain nonlinear terms inthevelocity gradients, oreven high-order derivatives ofthevelocity, butsuch terms could beexpected1 1 I 1 444 TENSOR ALGEBRA. INERTIA AND srnnss TENSORS [cn.».1>. 10 tobesmall incomparison with thelinear terms. Experimentally, the viscous stresses inﬂuids aregiven very accurately inmost cases by formula (10-188). Wehave decomposed thetensor Vv, with nine independent com- ponents, intoasum ofthree tensors with one,three, andﬁveindependent components, each alinear combination ofthecomponents ofVv. A similar decomposition isclearly possible foranytensor. The reader may wellaskwhether anyfurther decomposition ispossible. This isaproblem ingroup theory. Westate without proof theresult. Neither ananti- symmetric tensor norasymmetric traceless tensor canbefurther de- composed inamanner independent ofthecoordinate system. Thereader canconvince himself that thisisplausible byalittle experimentation. Letusnow consider anelastic solid. Letthesolid beinitially inan unstrained position, andleteach point inthesolid bedesignated byits position vector rrelative toanyconvenient origin. Now letthesolid be strained bymoving each point rtoanew position given bythevector r+p(r) relative tothesame origin. Wewilldesignate thecomponents ofrby(x,y,z)andofpby(E,17,I).Ifpwere independent ofr,themo- tion would beauniform displacement without deformation. Hence the strain atanypoint may bespeciﬁed bythegradient dyad <"‘_E<’_"<2£ 6x6x6:1: _<2‘FL‘K.Vp-ayayay (10-189) ‘Ea_"‘Pi 62dz62 ' Now Vpcanagain bedecomposed into anantisymmetric part which corresponds toarigid rotation about thepoint r+pand into asym- metric part which describes thedeformation ofthesolid intheneighbor- hood ofeach point: S=av»+%(vp>’- (HH90) The symmetric part canbefurther decomposed into aconstant tensor describing avolume compression orexpansion andasymmetric traceless tensor which describes theshear: s.=§v-P1=25% 1, (10-191) 5,,=2-vp+%(Vp)t -av-P1. (10-192) Ifthesolid isisotropic, then thisisthemost general possible decomposi- PROBLEMS 445 tion. Furthermore, ifHooke’s lawholds, thestress should beproportional tothestrain: P=—~§aV -pl—b$,¢- (10-193) The constants aandbareevidently related tothebulk modulus andthe shear modulus. Ifthesolid isnotisotropic, asforexample, acrystal, then therelation between PandSmay depend onthechoice ofaxes, and must therefore bewritten: R‘em» I-IO P1," = 551213121. (10-194) Since PandShave sixindependent components each, there arethirty-six constants c,-,-1,1. Byusing thefactthat there isanelastic potential energy which isafunction ofthestrain, itcanbeshown that inthemost general case there aretwenty-one independent constants c,-11,1. PROBLEMS 1.Theproduct cTofatensor bya.scalar hasbeen used inthetextwithout formal deﬁnition. Remedy thisdefect bysupplying asuitable deﬁnition and proving that thisproduct hastheexpected algebraic properties. 2.Show that thecentrifugal force inEq.(7-37) isalinear function ofthe position vector roftheparticle, and ﬁnd anexpression forthecorresponding tensor indyadic form. Write outthematrix ofitscoefﬁcients. 3.Deﬁne time derivatives d'I'/dt and d’T/dt relative toﬁxed and rotating ‘coordinate systems, aswasdone inChapter 7forderivatives ofvectors. Prove that _ d'I d’T E=-dT+a1><T—TXw, where thecross product ofavector with atensor isdeﬁned intheobvious way. 4.Write outtherelations between thecoefficients a;,-corresponding tothe relations (10-80). Write down another relation between theunitvectors, in- volving atriple cross product, and write outthecorresponding relations be- tween thecoefﬁcients. 5.Transform thetensor T=AB-1-BA, where A=5i—3j+2k, B=5j+10k, into acoordinate system rotated 45°about thez-axis, using Eq.(10-74). Trans- form thevectors AandB,using Eq.(10-71), andshow that theresults agree. 446 TENSOR ALGEBRA. INERTIA ANDsrnnss TENSORS [CHAP. 10 6.Write down andprove twoadditional relations likethose inEqs. (10-91) through (10-93), involving algebraic properties which arepreserved bytrans- formations ofcoordinates. 7.Prove Eqs. (10-91) and(10-92). 8.Write down thematrix fortheorthogonal tensor Awhich produces a rotation byanangle aabout the2-axis. Decompose Ainto asymmetric and anantisymmetric tensor asinEq.(10-66). What isthegeometrical interpre- tation ofthisdecomposition? 9.Prove that Det(T)[Eq. (l0—95)] isthesame inallcoordinate systems. 10.(a)Prove that thetensor given byformula (10-100) hastheproperty given byEq.(10-23). (b)Prove bydirect calculation that thistensor isrepre- sented bythesame matrix inallcoordinate systems. 11.Prove bydirect calculation that thequantity M(T)deﬁned byEq.(10-139) hasthesame value after thecoordinate transformation (10-74). 12.Diagonalize thetensor inProblem 5.(That is,ﬁnd itseigenvalues and thecorresponding principal axes.) 13.Diagonalize thetensor 7 \/6. —~/8 T=~/6 2-5\/§ - —\/8 -5\/E —3 (Hint: Thesecular equation canbefactored; theroots areallintegers.) 14.What aretheprincipal axesandcorresponding eigenvalues ofthetensor inProblem 2?Interpret physically. 15.Verify thestatements made inthelastparagraph ofSection 10-4 regard- ingtheprincipal axistransformation found intheworked-out example. 16.Prove that iftwotensors SandThave asetofprincipal axes incommon, then S-T=T-S.(The converse isalso true.) 17.Prove that ifatensor Tsatisﬁes analgebraic equation a,,T;‘—l—-~-—I—a2T2—|—a1T-l—a()'| =0, where ‘T"’means T-T---T(nfactors), then itseigenvalues must satisfy the same equation. The null tensor Oisdeﬁned intheobvious way. 18.Usetheresult ofProblem 17toshow that theeigenvalues ofthetensor Arepresenting a180° rotation about some axiscanonly be=1=1. [Hint: Consider theresult ofapplying Atwice.] Show that theroots cannot allbe—|-1. Then show that —1must beadouble root. [Hint: UseEqs. (10-137) and (10-81).] Can you guess thecorresponding eigenvectors? This entire problem istobe answered byusing general arguments, without writing down thematrix forA. 19.Show that theeigenvalues ofanorthogonal tensor [Eq. (10—87)] arecom- plex (orreal) numbers ofunit magnitude. [Hint:LetCbeaneigenvector (pos- sibly complex) ofT,andconsider thequantity (T-C)-(T-C*).] Hence show that oneeigenvalue must be=l=1,andtheother twoareoftheform exp(=1=ioz), forsome angle oz. PROBLEMS ‘ 447 20.Write outthecomponents oftheorthogonal tensor Acorresponding toa rotation byanangle 0about thez-axis. Find itseigenvalues. Find andinterpret theeigenvectors corresponding totherealeigenvalue. 21.Find thecomponents ofthetensor corresponding toarotation byan angle 0about thez-axis, followed byarotation byanangle 1/1about they-axis. Find itseigenvalues. (Hint: According toProblem 19,oneeigenvalue is:l=1; hence youcanfactor thesecular equation.) Show that theresult implies that this transformation isequivalent toasimple rotation about some axis. (You arenotasked toﬁndtheaxis.) Find theangle ofrotation bycomparing your result with theeigenvalues found inProblem 20. 22.Show that theeigenvalues ofanantisymmetric tensor arepure imaginary (orzero). Hence show that anantisymmetric tensor must have onezero eigen- value and two conjugate imaginary eigenvalues. Find theeigenvectors corre- sponding tothezero eigenvalue forthetensor (10-62). 23.Find theinertia tensor ofastraight rodoflength l,mass m,about its center. Usethisresult toﬁndtheinertia tensor about thecentroid ofanequi- lateral pyramid constructed outofsixuniform rods. Show that thistensor can bewritten down immediately from symmetry considerations, given theresult ofProblem 17,Chapter 5. _ 24.Translate tothecenter ofmass Gtheinertia tensor calculated about the origin forthethree disks inFig. 10-2. Verify thestatement made inthetext regarding thedouble degeneracy oflg. 25.Calculate themoment ofinertia ofacircular cone about aslant height. [Hint: Calculate theinertia tensor about theapex relative toprincipal axes, anduseEq.(10-143) .] 26.Formulate and prove themost comprehensive theorem you can with regard totheinertia tensor ofaplane lamina. What canyou sayabout the principal axes andprincipal moments ofinertia? 27.Find, bywhatever method requires theleast algebraic labor, theinertia tensor ofauniform rectangular block ofmass M,dimensions aXbXc,about asetofaxes through itscenter, ofwhich thez-axis isparallel toside c,andthe y-axis isadiagonal oftherectangle aXb. 28.(a)Auniform sphere ofmass M,radius a,-hastwopoint masses iM, %M, located on itssurface andseparated byanangular dis- tance of45°. Find theprincipal axes and principal moments ofinertia about thecenter ofthesphere. (b)Find theinertia tensor about parallel axes through thecenter of mass. Arethey stillprincipal axes? 29.(a)Find theinertia tensor ofaplane rectangle ofmass M,dimensions aXb.(b) Usethisresult toﬁndtheinertia tensor about thecenter ofmass ofthehouse ofcards shown inFig. 10-5. Each card hasmass M,dimen- sions aXb(a<b).Useprincipal axes. FIG. 10-5. Ahouse ofcards. 448 TENSOR ALGEBRA. INERTIA ANDsrnnss TENSORS [cnA1>. 10 30.Find theequation fortheellipsoid ofinertia ofauniform rectangular block ofdimensions lXwXh. 31.Find theequation fortheellipsoid ofinertia ofanobject intheshape ofanellipsoid whose equation is §+ﬁ+i_( Z2 w2 h2--' 32.Prove that thestress tensor Pissymmetric. 33.Prove that ifthere isnoshear onanysurface element atsome point, then thestress tensor Patthat point isaconstant tensor [Eq. (10-23)]. 34.Derive Eq.(10-171) bycalculating thenetforce onarectangular volume element. 35.Derive from Eq.(10-173) theequation ‘3%’,"—)+ v-<pw+ P)=r. which expresses theconservation oflinear momentum. Show from thisequation that themomentum current tensor (pvv —|-P)represents theﬂow ofmomentum, andinterpret physically thetwoterms inthistensor. 36.Derive from Eq. (10-172) alawofconservation ofangular momentum inaform analogous toEq.(8-148). 37.Write theequations ofmotion (10-173) incylindrical components fora moving viscous ﬂuid. Usethese equations, together with suitable assumptions, toderive Poiseui11e’s law(8-252) forsteady viscous ﬂowinapipe. Write the stress tensor forthis case, incylindrical coordinates, asafunction ofr,z,<p. 38.Write outthecomponents ofthestress tensor Pinaviscous ﬂuid. 39.(a)Show that therate ofproduction ofkinetic energy perunit volume duetostresses inamoving medium is Q=—v-(V-P). ~ (b)Show that therate atwhich work isdone bythestresses onthemedium, perunit volume, is dW —d?‘ ——V ' 'V). (Hint: Calculate thework done across thesurface ofanyvolume Vanduse Gauss’ theorem.) (c)Using these results, calculate therate atwhich energy isdissipated per unit volume byviscous stresses inamoving ﬂuid. Write itoutinterms of components. 40.Find therelation between theconstants aandbinEq.(10-193) andthe bulk modulus Band shear modulus ndeﬁned byEqs. (5—116) and (5—118). [SetupSandPforthesituations used indeﬁning Bandn.] 41.Asolid issubject toastress consisting ofapure tension -rperunit area inonedirection. Find thestrain Sinterms of1,a,andb.Using this, andthe PROBLEMS o 449 result ofProblem 40,express Young’s modulus Y[Eq. (5—114)] interms ofB andn.[Hint:Use symmetry todetermine theform of5.] *42. Find themost general linear relation between SandPforanonisotropic elastic substance which possesses cylindrical symmetry relative toaspeciﬁed direction. *43. (a)Assume that inanonisotropic elastic solid, there isanelastic potential energy Vperunit volume, which isaquadratic function ofthestrain com- ponents. Show that there are21constants required tospecify V. (b)Show that ifthestrain inasolid inequilibrium isincreased by65,thework done perunit volume against thestresses (exclusive ofanywork done against body forces) is ‘ aw--iP-~6S.-- — u 1- 1',i=1 [Hint: Calculate thework done onavolume element bythestresses onits surface anduseGauss’ theorem.] (c)Combine results (a)and (b)toshow that Pisalinear function ofSin- volving 21independent constants, ingeneral. m. CHAPTER 11 THE ROTATION OFARIGID BODY 11-1 Motion ofarigid body inspace. The motion ofarigid body in space isdetermined byEqs. (5-4) and(5-5): g=F, (11-1) dLE_N, (11-2) where ' P=MV, (11-3) L=I-w, (11-4) FandNarethetotal force onthebody andthetotal torque about a suitable point O,Visthevelocity ofthecenter ofmass, andIandware theinertia tensor andtheangular velocity about thepoint 0.Foranun- constrained body moving inspace, thepoint 0istobetaken asthecenter ofmass. Ifthebody isconstrained byexternal supports torotate about aﬁxed point, that point istobetaken asthepoint O.Ifthepoint Ois constrained tomove insome fashion, thereader may supply theappro- priate equation ofmotion. (See Chapter 7,Problem 3.) Equations (11-2) and (11-4) fortherotation ofarigid body bear a formal analogy toEqs. (11-1) and(11-3) forthemotion ofapoint mass M.There are,however, three differences which spoil theanalogy. Inthe ﬁrstplace, Eq.(11-4) involves atensor I,whereas Eq.(11-3) involves a scalar M;thus Pisalways parallel toV,while Lisnotingeneral parallel tow.Amore serious difference isthefactthat theinertia tensor Iisnot constant with reference toaxes ﬁxed inspace, butchanges asthebody rotates, whereas Misconstant (inNewtonian mechanics). Finally, and perhaps most serious, isthefactthat nosymmetrical setofthree coordi- nates analogous toX,Y,Zexist with which todescribe theorientation of abody inspace. This point wasmade inSection 5-1, anditissuggested that thereader review thelastparagraph inthat section. Forthese rea- sons, wecannot proceed tosolve theproblem ofrotation ofarigid body byanalogy with themethods ofChapter 3. There aretwogeneral approaches totheproblem. Weshall ﬁrst, in Sections 11-2 and 11-3, trytoobtain asmuch information aspossible 450 J 11-2] EULER'S EQUATIONS orMOTION ronARIGID BODY 451 from thevector equations (11-2), (11-4) without introducing asetof coordinates todescribe theorientation ofthebody. Weshall then, in Sections 11-4and11-5, useLagrange’s equations todetermine themotion interms ofasetofangular coordinates suggested byEuler. 11-2 Euler’s equations ofmotion forarigid body. Thediﬂiculty that Ichanges asthebody rotates may beavoided byreferring Eq.(11-2) to asetofaxes ﬁxed inthebody. Ifwelet“d’/dt” denote thetime derivative with reference toaxes ﬁxed inthebody, then byEq.(7-22), Eq.(11-2) becomesI . %+wXL=N. (11-5) Since Iisconstant relative tobody axes, wemay substitute from Eq.(11-4) toobtain |-%+~><(|-~)=N. (1141) (Recall that d’w/dt =dw/dt.) Itismost convenient tochoose asbody axes theprincipal axes, e1,e2,e3ofthebody. Then Eq.(11-6) becomes 11°31 -|'(Ia—I2)waw2 =N1, 124,2 "l"(I1*Islwiwa =N2, (11-7) Iadls +(I2"I1)w2w1 =N3- These areEuler’s equations forthemotion ofarigid body. Ifonepoint inthebody isheld ﬁxed, that point istobetaken astheorigin forthebody axes, andthemoments ofinertia andtorques arerelative tothat point. Ifthebody isunconstrained, thecenter ofmass istobetaken asorigin forthebody axes. Inorder toderive theenergy theorem from Euler’s equations, wemulti- plyEq.(11-6) bywt w-I-%=w-N. (11-8) Since Iissymmetric, theleftmember is don do Id’ _g _ cu-I-E-=2?-I-w—§a(w-I-w)_dt, where Tisgiven byEq.(10-153). Here wehave used thefact that d/dt, d’/dt have thesame meaning when applied toascalar quantity. 452 THEROTATION orAmen) BODY [cn.u>. 11 Comparing Eqs. (11-8) and(11-9), weobtain theenergy theorem: %=w-N, (11-10) inanalogy with theorem (3—133) forthemotion ofaparticle. From Eqs. (11-7) wenote immediately that abody cannot spin with constant angular velocity w,except about aprincipal axis, unless external torques areapplied. Ifdw/dt =0,Eq.(ll-6) becomes wX(I-w)=SN. (11-11) Theleftmember iszero only ifI-asisparallel tow,that is,ifwisalong a principal axis ofthebody. Ifawheel istospin freely without exerting forces and torques onitsbearings, then itmust benotonly statically balanced, i.e.,with itscenter ofmass ontheaxis ofrotation, butalso dynamically balanced, i.e.,theaxis ofrotation must beaprincipal axis oftheinertia tensor, asanyautomobile mechanic knows. Inorder tosolve Eqs. (11-7) for(v(t), wewould need toknow thecom- ponents oftorque along the(rotating) principal axes, anuncommon situ- ation, except forthecaseN=0.Wenowconsider afreely rotating sym- metrical body, with noapplied torque. Letthesymmetry axisofthe body beea,sothatI1=I2.Then thethird ofEqs. (11-7) is I3¢.;)3 =0, and(03i_s_constant. Theﬁrsttwoequations may bewritten 051+§O)3(.02 =0, (1)2 —Bw3w1 =0, where 3= (11_14) Equations (11-13) areapair ofcoupled linear ﬁrst-order equations in wl,0:2.Letuslook forasolution bysetting .1 601 -= A161”, (.02 = A291“. Wereadily verify that formulas (11-15) satisfy Eqs. (11-13) provided that p=:l=iﬂw3, (11-16) and A2==FiA1. (11-17) 11-2] EULER’s EQUATIONS orMOTION FOR ARIGID BODY 453 Wehave foimd acomplex conjugate pair ofsolutions, wl=e*"”"", (-12==r=ie*""“=*‘, (11-18) andthese may besuperposed with arbitrary constant multipliers toform therealsolution: (.01=Acos(Bw3t+10), (.02=Asin(/Swat+a).(11-19) The angular velocity vector wtherefore precesses inacircle ofradius A about thee3-axis, with angular velocity Be->3. Theprecession isinthesame sense as0:3ifI3>I1,andintheopposite sense otherwise. The magni- tude ofwis w=[013+A211”, (11-20) andisconstant, aresult which canalsobeproved bydirect calculation of d(w2)/dt from Eq.(11-7). The constants (03,A,0aredetermined bythe initial conditions. There arethree arbitrary constants, since Euler’s equations arethree ﬁrst-order differential equations. Since anuncon- strained rotating rigid body hasthree rotational degrees offreedom, we should expect atotal ofsixarbitrary constants tobedetermined bythe initial conditions. What aretheother three‘? Theinstantaneous axisofrotation, determined bythevector w,traces outacone inthebody (thebodycane) asitprecesses around theaxisof symmetry. Thehalf-angle abofthebody cone isgiven by l tanorb=Cg- (11-21) Alternatively, ifthebody isinitially rotating with angular velocity w about anaxismaking anangle oq,with thesymmetry axis, then thecon- stants (1)3andAaregiven by ' A=wsin0:1,, (.03=wcosab. (11-22) .Inorder toﬁndthemotion inspace, weneed tolocate theas-axis with respect toadirection ﬁxed inspace. Wecould dothisbytracing outstep bystep themotion relative tospace axes, allowing thebody torotate with constant angular velocity about anaxisinthebody cone which precesses with angular velocity B403. Itiseasier tolocate wrelative to L,since .byEq.(11-2) Lisconstant ifN=0.The angle a,between co andLisgiven by cu-L as-I-w 2T COSCI; —-5? —-T —Z Since byEq.(11-10) Tisconstant, theangle a,isconstant. Theaxisof 1 454 THEROTATION orAmorn BODY [cn.u>. 11 rotation therefore traces outacone inspace, thespace cone. The space cone hasahalf-angle oz,given byEq.(11-23) anditsaxisisthedirection oftheangular momentum vector L.The line ofcontact between the space cone andthebody cone atanyinstant istheinstantaneous axisof rotation. Since thisaxisinthebody isinstantaneously atrest, thebody cone rolls without slipping around thespace cone. This gives acomplete description ofthemotion (seeFig. 11-1). Wecanexpress a_,interms oftheconstants w,ab.Wehave |=(9191 -I"ezezlli +eaesla =I11 +G3€3(I3 _ Bysubstituting from Eqs. (11-14), (11-22), and(11-24) intoEqs. (11-4), (10-153), and(11-23), andwith w=wn,weobtain: 2T=w2I1[1 +5cos2ab], (11-25) L='(0I1[l1 +BCOS a;,e3], 2 cos ac 1 _#' -[1+(26+B2)cosab]/2 Note thata,depends ononlyabandnotonw.Itisclearfrom Eq.(11-26) thatthespace coneliesinside thebody coneifB>0andoutside if[3<0 (seeFig.11-1). Thisisclear alsoif,when thebody conerollsonthespace cone, theprecession oftheaxisofrotation istohave thesense given by Eq.(11-19). [The reader should check this, remembering that Eq.(11-19) describes themotion oftheaxisrelative tothebody.] Next, consider thecase when theinertia tensor isnondegenerate. We shall number theprincipal axes sothat I3>I2>I1. Itwas shown above that abody may rotate freely about aprincipal axis. Letusstudy small deviations from thissteady rotation. Ifwisnotalong aprincipal axis, then itcannot remain constant. Letusassume that coliesvery close toaprincipal axis, saytoe3,sothat (.03>>(.01and603>>C02. Then if N=0,weseefrom thethird ofEqs. (11-7) that0:3isconstant toﬁrst order in(.01andL02’.Theﬁrsttwoequations then become apairofcoupled linear equations inwl,C02,which wesolve asinthepreceding example toobtain W1=AlI2(Ia _I2)l1/2c°$(l9<-031+ 9), _ (11-28) wz=AlI1(Ia —I1)l1/2$1I1(l3¢°ai+ 9), Where Aand0arearbitrary constants and _ _ 12,= '. (11-29) 11-3] Po1Nso'r’s SOLUTION FORAFREELY ROTATING BODY 455 1 L Q es e/-' ~2: 91 \\\ / // / 4-|qf-_;_2'-'L:-----___-___-_-_\\\ \ \ \ \ FIG. 11-1. Free rotation ofasymmetrical body. The vector oitherefore moves counterclockwise (looking down from the positive e3-axis) inasmall ellipse about thee3-axis. Inasimilar manner, wecanshow that if0)isnearly parallel tothee1-axis, itmoves clockwise inasmall ellipse about that axis, andthat ifcoisnearly parallel tothe e2-axis, thesolution isofanexponential character. Inthelatter case, of course, thecomponents (.01and(.03willnotremain small, andtheapproxi- mation that (02isconstant willhold only during theinitial part ofthe motion. Weconclude that rotation about theaxes ofmaximum and minimum moments ofinertia isstable, while rotation about theinter- mediate axisisunstable. This result isreadily demonstrated bytossing atennis racket intheairandattempting tomake itspin about anyprin- cipal axis. The general solution ofEqs. (11-7) forw,when N=0,can also inprinciple beobtained. Weshall solve theproblem inthenext section byadifferent method. 11-3 Poinsot’s solution forafreely rotating body. Ifthere areno torques, N=0,then Eqs. (11-2) and(11-10) yield four integrals ‘ofthe equations ofmotion: L=I-co=aconstant, (11-30) T=Q-w-I-w=aconstant. (11-31) 456 THE ROTATION OFARIGID BODY [CHAP. 11 Poinsot* hasobtained ageometrical representation ofthemotion based onthese constants, andutilizing theinertia ellipsoid. Letusimagine the inertia ellipsoid (10-157) rigidly fastened tothebody androtating with it.Ifweletrbethevector from theorigin tothepoint where theaxisof rotation intersects theinertia ellipsoid atanyinstant, rr-5w, (11-32) then comparison ofEqs. (11-31) and(10-157) shows that a/20,2T_F (11-33) Thenormal totheellipsoid atthepoint risparallel tothevector V(r-|-1)=2I1x1e1 +212m, +213m, =23L,(11-34) where :01,:02,maarethecomponents ofralong theprincipal axes. The tangent plane totheellipsoid atthepoint 1'istherefore perpendicular to theconstant vector L(seeFig.11-2). Letlbetheperpendicular distance from theorigin tothistangent plane: . .. 1/2 Z=% =2$0 =ﬁg;-)— =aconstant. (11-35) Thetangent plane istherefore ﬁxed inspace (relative totheorigin O)and iscalled theinvariable plane. Itsposition isdetermined bytheinitial con- ditions. Moreover, since thepoint ofcontact between theellipsoid and theplane liesontheinstantaneous axisofrotation, theellipsoid rolls on theplane without slipping. The angular velocity atanyinstant hasthe magnitude ‘ 1/2 w=(22% r. (11-36) This gives acomplete description ofthemotion. \ Astheinertia ellipsoid rolls ontheinvariable plane, with itscenter ﬁxed attheorigin, thepoint ofcontact traces outacurve called the polhode ontheinertia ellipsoid, andacurve called theherpolhode onthe invariable plane. This isillustrated inFig. 11-2. Thepolhode isaclosed curve ontheinertia ellipsoid, deﬁned asthelocus.of points rwhere the tangent planes lieaﬁxed distance Zfrom thecenter oftheellipsoid. In Fig. 11-3 areshown various polhodes onanondegenerate inertia ellipsoid. Note that thetopological features ofthediagram areinagreement with *Poinsot, Theorie Nouvelle delaRotation desCorps, 1834. z 11-3] 1>o1Nso'r’s SOLUTION FORAFREELY ROTATING BODY 457 Inertia ellipsoid riable plane FIG. 11-2. Theinertia ellipsoid rolls ontheinvariable plane. "=1 \ $2 13 ~F10. 11-3. Polhodes onanondegenerate inertia ellipsoid. theconclusions attheendofthepreceding section. Ingeneral, theherpol- hode isnotclosed butﬁllsanannular ringintheinvariable plane. Inthecase ofasymmetrical body itcanbeshown (Problem 7)that the polhodes arecircles about thesymmetry axis and theherpolhodes are circles intheinvariable plane. Inthat case, randtherefore, byEq.(11-36), w(but notw!)areconstant during themotion. Poinsot’s description of themotionin this case agrees with that inthepreceding section. The ,./l’ K 458 THEROTATION orARIGID BODY [cn.u>. 11 polhode andherpolhode aretheintersections ofthebody andthespace cones with theinertia ellipsoid andtheinvariable plane, respectively. 11-4 Euler’s angles. The results inSections 11-2 and 11-3 regarding themotion ofarigid body were obtained without theuseofanycoordi- nates todescribe theorientation ofthebody. Inorder toproceed further with thediscussion, itisnecessary tointroduce asuitable setofcoordi- nates. Wechoose asetofaxes ﬁxed inthebody, which aremost con- veniently taken astheprincipal axes, with origin atthecenter ofmass, orattheﬁxed point ifoneexists. These axes willbelabeled with sub- scripts 1,2,3,asbefore. Ifthere isanaxisofsymmetry, itwillbenum- bered 3;otherwise, theaxes may benumbered inanyorder. Weneed three coordinates tospecify theorientation ofthebody axes with respect toaﬁxed setofspace axes x,y,z.Therelation between thetwosetsof axes could bespeciﬁed bygiving thecoefﬁcients ofthetransformation from coordinates x,y,ztox1,x2,x3.There arenine coeﬂicients butonly three ofthem areindependent, aswehave seen, andtotrytousethree of thecoefficients ascoordinates isnotconvenient. Aswaspointed outin Section 5-1, there isnosymmetric setofcoordinates analogous tox,y,z with which todescribe theorientation ofabody. Among thevarious coordinate systems that have been introduced forthispurpose, oneof themost useful isduetoEuler. InFig.11-4, theEuler angles 0,¢,upareshown. These areusedtospec- ifytheposition ofthebody axes 1,2,3relative tothespace axes x,y,z. Thebody axes 1,2,3areshown asheavy lines; thespace axes x,y,zare lighter. The angle 0istheangle between the3-axis and the2-axis. Since the3-axis isthus singled outforspecial treatment, ifthebody has anaxis ofsymmetry, itshould betaken asthe3-axis. Likewise, ifthe external torques possess anaxis ofsymmetry inspace, that axis should betaken asthez-axis. Theintersection ofthe1,2-plane with themy-plane Z 2 3.: " Z FIG. 11-4. Euler’s angles. 11-4] EULEn’s ANGLES 459 iscalled thelineofnodes, labeled Zinthediagram. The angle ¢ismeas- ured inthexy-plane from thex-axis tothelineofnodes, asshown. The angle atismeasured inthe1,2-plane from thelineofnodes tothe1-axis. Weareassuming that both setsofaxes x,y,zand1,2,3areright-handed. Itwillbeconvenient alsotointroduce athird (right-handed) setofaxes, 5,11,§,ofwhich Eisthelineofnodes, §'coincides with thebody axis3, and1;isinthe1,2-plane.T ' , Inorder toexpress theangular velocity vector wintermsof Euler’s angles,,we ﬁrst prove that angular velocities may beadded like vec- tors, inthesense ofthefollowing theorem: Given aprimed coordinate system rotating with angular velocity 0:1with respect toanunprimed system, andastarred coordinate system rotating withangular velocity (-02relative totheprimed system, theangular velocity ofthestarred system relative tothe unprimed system iswl+<02. (11-37) Toprove thistheorem, letAbeanyvector atrestinthestarred system: d*A l‘ -I -0. (11-38) Then bytheorem (7-22), itsvelocity relative totheprimed system is I id?=(D2XA. (11-39) Now applying theorem (7-22) again, weﬁndthevelocity ofArelative to theunprimed system: %=%+a,><A=(...,+@2)><A. (11-40) Aﬁnal comparison with theorem (7-22) shows that (wl+(02)isthe angular velocity ofthestarred system relative totheunprimed one. Now consider Figure 11-4 andsuppose that thebody ismoving sothat 0,¢,1/1arechanging with time. If0alone changes, while ¢,11/areﬁxed, thebody rotates around thelineofnodes with angular velocity 985. If45 alone changes, thebody rotates around thez-axis with angular velocity 43k. If1/1alone changes, thebody rotates around its3-axis with angular velocity 1//e3. Now ifweconsider aprimed coordinate system rotating with 1'The reader iscautioned that thenotation forEuler’s angles, aswell asthe convention astoaxes from which they aremeasured, andeven theuseofright- handed coordinate axes, arenotstandardized intheliterature. Itistherefore necessary tonote carefully how each author deﬁnes theangles. Theconventions adopted here arevery common, butnotuniversal. -460 THE ROTATION orARIGID BODY [CI-IAP. 11 angular velocity ¢'>kabout thez-axis, andletthe5,17,;-system rotate with angular velocity 96grelative tothis primed system, then bytheorem (11-37), theangular velocity ofthe£,n,§-system is$65+43k. The axes 1,2,3rotate with angular velocity tearelative to£,'q,§‘, hence theangular velocity ofthebody is - w=0e;+dk—l-tea. (11-41) Wehave, from Fig. 11-4, therelations e;=e1cos1I/ —e2sin1//, e,,=e1sinat—|—e2cosib, (11-42) ef"' e3; and k=e;cos9 +e,,sin0 =e1sin0sin1/1—|-e2sin0cosup+e3cos0. (11-43) Wemay therefore express winterms ofitscomponents along theprincipal axes: wl=9008111 +dsin 0sin¢, (4)2=-0sin¢ +d>sin 00081]/, (11-44) (.03=it+<1»cos0. Thekinetic energy isnow given byEq.(10-153): T='2'I1@I +tlzwi +2Iaw§- (11-45) Thekinetic energy isarather complicated expression involving 0,43,it,0, andgt.Note that 0,¢,31/arenotorthogonal coordinates, i.e.,cross terms involving 19¢and 1,0113appear inT.Inthecase ofasymmetrical body (I1=I2),theexpression forTsimpliﬁes totheform: » T=%I102 +%I1<i>2 sin20+§~I3(¢ +<13cos0)2. (11-46) The generalized forces Q9,Q,,,Qaa.reeasily shown tobethetorques about the5-,2-,and3-axes. ' We arenow inaposition toWrite down Lagrange’s equations forthe rotation ofarigid body subject togiven torques. Ifthetorques arede- rivable from apotential energy V(0,4>,1l/), then there will beanenergy integral. IfVisindependent of4»,then inspection ofEqs. (11-44) shows that ¢willbeanignorable coordinate. Unfortunately, thisisnotenough toenable ustogive ageneral solution oftheproblem. However, fora symmetrical body, ifVisindependent of¢also, weseefrom Eq.(11-46) 11-5} THE SYMMETRICAL TOP 461 that both ¢anditareignorable. Wehave then three constants ofthe motion, enough tosolve theproblem. This casewillbesolved inthe next section. Afewother special cases areknown forwhich theproblem canbesolved,* butforthegeneral problem ofthemotion ofanunsym- metrical body under theaction ofexternal torques, asforthemany-body problem, there arenogenerally applicable methods ofsolution, except by numerical integration oftheequations ofmotion. 11-5 The symmetrical top. The symmetrical top, represented in Fig. 11-5, isabody forwhich I1=I2.Itpivots around aﬁxed point O that liesontheaxisofsymmetry adistance lfrom thecenter ofmass G which alsoliesontheaxisofsymmetry. Theonly external forces arethe forces ofconstraint atOandtheforce ofgravity. Therefore, byEq.(11-46), theLagrangian function is L=%I102 +2111132 sin”0+*2l—I3(¢ +<13cos0)2—mglcos0. (11-47) The coordinates 1/»and ¢areignorable, and wehave therefore three integrals ofthemotion: M-@__ dt-aw-0, (11-48) 221_IE_ dt-ad)-0, (11-49) dE 6LW-—-E_0, (11-50) where I W=Ia(¢+¢'>COS0), (11-51) p4,=I1¢'>sinz0I3cos0(1//+43cos0), (11-52) E= Q-I192 + %I1¢-J2 SIII2 6 —l—%I3(¢ +<13cos6)2+mglcos0. (11-53) WeuseEqs. (11-51) and(11-52) toeliminate pt,<1»from Eq.(11-53): _ 2(rt—rtCOS0)’ Ll _E-$110 + 2I1sin20 +213-+ mglcos0. (1154) *See, forexample, E.J.Routh, TheAdvanced Part ofaTreatise ontheDy- namics ofaSystem ofRigid Bodies, 6th‘ed.London: Macmillan, 1905. (Also New York: Dover, 1955.)1 1 1 462 THE ROTATION orARIGID BODY [CHAP. 11 Z 3 \\\\2 .:1»4 ,¢’~ 4"Q--&- ll "18 ¢‘P 1 I E FIG. 11-5. Coordinates forthesymmetrical top. Wecannowsolve theproblem bytheenergy method. Ifweset E’=E-L5, (11-55)213 _ 92 ‘V’=(%I-1l - +mglcos0, (11-56) then 2, 1/2 9= [E’—‘V’(6)]} 1 (11-57) and0isgiven, inprinciple, bycomputing theintegral ' ' ea I1'2 andsolving for0(t). The constant 00istheinitial value of0.Once 0(t) isknown, Eqs. (11-51) and (11-52) canbesolved foritand<13andinte- grated togive 1//(t), 4>(t). Comparison ofEqs. (11-44) and(11-51) shows that P-I’=Iawa, (11-59) ‘sothat w3isaconstant ofthemotion. If0:3=0,then Eq.(11-56) re- duces essentially totheformula (9—137) foraspherical pendulum, asit 11-5] THE SYMMETRICAL TOP 463 IV) @ Q___-O1/2 1r 0 FIG. 11-6. Eﬁective potential energy forthesymmetrical top. should. InFig. 11-6, ‘V’(0) isplotted versus 0forwasé0.The ‘torque’ associated with the‘potential energy’ ‘V’(0) is ‘ ' _ (3‘V’ _ . (p-pcos0)(p —pcos0)‘N’——W —mgls1n0 — ¢ "'I1Sinafo “’ -(11-60) Inspection ofEq.(11-60) shows that, ingeneral (ifp,;épt),the‘torque’ ‘N’ispositive for0é0andnegative for0é1r,andhasonezerobetween 0and1r.Hence ‘V’hasoneminimmn, asshown inFig.11-6, atapoint 00 satisfying theequation "W111 $31490—(IN—Pt0039o)(P¢ *11¢COS90)=0-(11-61) IfE’=‘V’(00), theaxisofthetopprecesses uniformly atanangle 01, with thevertical, andwith angular velocity 2 _P4»—Pw¢0$9o_ $0— I1sin?00 (11-62) Solving Eq. (11-61) for(p,,—ptcos60), and using Eq. (11-59), we obtain sin20,, 41112111 "2 (1)4, -p¢,0OS 99) =‘£131.03 H [1:|:(I. -—12$ COS 00) (11-63) Weseethat if00<1r/2, there isaminimum spin angular velocity below which thetopcannot precess uniformly attheangle 00: comm == cos00>1l2 ' (11-64) a . 464 THE ROTATION orARIGID BODY [CHAP. 11 For(.03>wmin; there aretworoots (11-63) andhence twopossible values of(131,,aslow andafastprecession, both inthesame direction asthespin angular velocity 0:3. For(.03>>wmin, thefast andslow precessions occur atangular velocities LIsWe 4,0_I1cos00 (11-65) and A"W1,$0 '_' Iawa Itistheslow precession which isordinarily observed with arapidly spin- ning top. For00>rr/2(tophanging with itsaxisbelow thehorizontal), there isonepositive andonenegative value for<50. (Towhat dothese motions ofuniform precession reduce when m3—>0?) Study ofFig. 11-6 shows usthat themore general motion involves a nutation oroscillation oftheaxis ofthetopinthe0-direction asitpre- cesses. The axis oscillates between angles 01and 02which satisfy the equation _ g2E’=%h +mglcos0, (11-67) where 72¢,pi,andE’aredetermined from theinitial conditions. Ifwe multiply Eq.(11-67) bysinz0,itbecomes acubic equation incos0.We seefrom Fig.11-6thatthere must betworealroots cos61,cos02between -1and+1. Thethird root forcos6must lieoutside thephysical range ——1to+1. Infact, inspection ofEq. (11-67) willshow that thethird root isgreater than +1. (Inthecase ofuniform precession discussed inthepreceding paragraph, thetwo physical roots coincide, cos01= cos02=cos02.) Ifinitially 0=0,then theinitial value cos01ofcos0 satisﬁes Eq.(11-67); knowing oneroot ofacubic equation, wemay factor theequation andﬁnd allthree roots. During nutation, theprecession velocity varies according toEq.(11-52): ._p,,,—p,,cos6_ ¢— I1sin?0 (11-68) If|p¢|<lppl,wecandeﬁne anangle 02asfollows: cos02,=%-- (11-69) For0>02,qihasthesame signas(.03,andfor0<03,ithasanopposite sign. Thederivative with respect to0oftheright member ofEq.(11-67) isnegative at0=03;hence weseefrom Fig. 11-6 that 03<02,where 02isthelargest angle satisfying Eq.(11-67). Infact, 03<00.If03<01 11-5] THE SYMMETRICAL TOP 465 Z z z 3 3 | 0391 | 01 3 | .01=03ii /1‘.,-+ \\ 10»Mini!“ \0( 0‘AWLN2 '4'4's:1o1b!" <QAIQ* ’ (8) (b) (C) FIG. 11-7. Locus oftopaxis (3)onunit sphere. (orif|p4,|>|p¢.|andp,,,pthave thesame sign), then 43hasthesame sign as0:3throughout thenutation, andthetopaxis traces outacurve like that shown inFig.11-7(a).If03>01,43changes signduring thenutation andthetopaxismoves asinFig.l1—7(b). Itisclear that ifthetopisset inmotion initially above thehorizontal plane with <13opposite insign to (.03,themotion necessarily willbelikethatshown inFig.11-7(b). Animportant special caseoccurs when thetop,spinning about itsaxis with angular velocity w,-.;,isheldwith itsaxisinitially atrestatanangle 01andthenreleased. Initially, wehave - 0=91, 9=0, (i)=0, ¢=(.03. Wesubstitute inEqs. (11-51), (11-52), and(11-53) toﬁnd pa=I3w3, p,,=I3w2, cos01, E’=mglcos01. (11-71) Inthiscase, weseethat 03=01,andthemotion isasshown inFig.11-7(c). Anelementary discussion ofthiscase, based ontheconservation ofangu- larmomentum, wasgiven’ inSection 4-2. Now Eq.(11-56) becomes I§w§ (cos01—cos(9)2 ] ‘V, =T1 +0!COS 0I where 2Imgl<1= (11-73) awe _ The turning points forthenutation aretheroots ofEq.(11-67), which becomes inthiscase, ifwemultiply bysinz0, (cos01—cos(9)2—a(cos 01—cos0)(1 —cosz 0)=0. (11-74) 466 THE ROTATION orARIGID BODY [CHAP. 11 Theroots are 1 cos0=cos01, 1 (11-75) K cos0=5;[14(1-42cos01+4112)‘/2]. The angle 02isgiven bythesecond formula, using theminus sign inthe bracketed expression. The plus sign gives aroot forcos6greater than +1. Letusconsider thecase ofarapidly spinning top, that is,when oz<<1.Wethen have cos62écos01—asin”01. _ (11-76) The angle 02isonly slightly greater than 01,andtheamplitude ofnuta- tion isproportional toa.Ifweset 02 = 00+G, 01 = 00 — (1, andsubstitute inEq.(11-76), weﬁndthat, toﬁrst order inaanda, aé§asin01. (11-78) Wenowset 0='-00-I-6501-I-G-l-5, andsubstitute inEq.(11-72), which becomes, tosecond order inaand6, 2‘V’-v(o.,)+A%1.155’. (11-so) The ﬁrst term isconstant, andthesecond leads toharmonic oscillations in6with afrequency wo=%;(03. (11-81) Thenutation isgiven by _ 0i01+a—acoswot. (11-82) Wesubstitute inEq.(11-68) toobtain <13toﬁrst order ina: . I4>='=T:-% [1—coswot]. (11-83) Theaverage angular velocity ofprecession is . L 13030 ___ mgl _(¢)av-—-—-I1Sin01_-ﬁg (11-84) 11-5] THESYMMETRICAL TOP 467 or>% IV) a<% 0 1r}2 ii’ 0 FIG. 11-8. Effective potential energy when p¢=12¢. Thetopaxistherefore precesses very slowly andnutates very rapidly with very small amplitude. Inpractice, thefrictional torques which wehave neglected usually damp outthenutation fairly quickly, leaving only the uniform precession. Asaﬁnal example, consider thecase when thetopisinitially spinning with itssymmetry axisvertical. Inthiscase, solongasthe3-andz-axes coincide, thelineofnodes isindeterminate. Weseefrom Fig.11-4 that theangle It+¢isdetermined astheangle between thex-and1-axes, although 1//,¢>separately areindeterminate. Hence wehave initially Pr=130/’ -|-03)=laws, (11-35) In=Ia(¢+11>)=Pa (11-36) Equation (11-56) inthiscase becomes 22 _ 2 ‘V’=-kla [——-MSing‘): 0)+acos0]1 (11-87) where aisgiven byEq.(11-73). This, ofcourse, isjustaspecial case of Eq.(11-72). InFig. 11-8 weplot ‘V’forthecase when p,=pg. The form ofthecurve depends upon thevalue ofoz.Weseethat arapidly spinning top(a<2-)canspinstably about thevertical axis; ifdisturbed, itwillexhibit asmall nutation about thevertical axis. Aslowly spinning top (a>%)cannot spin stably about avertical axis, but will execute a large nutation between 01=0and02given byEq.(11-75).Inthiscase, 5cos02'=i-1. (11-ss) 468 THEROTATION orARIGID BODY [cnAI>. 11 The minimum spin angular velocity below which thetopcannot spin stably about avertical axisoccurs when a=2,or,byEq.(11-73), 0......=[iv <11-89>I3 Note that thisformula agrees with Eq.(11-64). Ifinitially 0:3>wmin, atopwillspin with itsaxisvertical, butwhen friction reduces 0:3below wmin, itwillbegin towobble. Alloftheabove conclusions about thebehavior ofasymmetrical top under various initial conditions caneasily beveriﬁed experimentally with atoporwith agyroscope. PROBLEMS 1.Usetheresult ofProblem 3,Chapter 10,toderive Eq.(11-6) directly from theequation %(|-~)=N. 2.(a)Assume that theearth isauniform rigid ellipsoid ofrevolution, look upitsequatorial andpolar diameters, andcalculate theangular velocity of precession oftheNorth Poleontheearth’s surface assuming thatthepolar axis (i.e., theaxisofrotation) deviates slightly from theaxisofsymmetry. (An irregular precession ofroughly thissortisobserved with anamplitude ofafew feet, andaperiod of427days.) (b)Assume that theearth isarigid sphere andthat amountain ofmass 10‘9 times themass oftheearth isadded atapoint 45°from thepolar axis. Describe theresulting motion ofthepole. How long does thepole take tomove 1000 miles? (c)Forarigid ellipsoidal earth, asinpart (a),how massive a“mountain” must beplaced ontheequator inorder tomake thepolar precession unstable? The earth is,ofcourse, notofuniform density, butismore dense near its center. Even more important, theearth isnotrigid, butbehaves asanelastic spheroid forshort times, andcandeform plastically over long times. Theresults inthis problem aretherefore only suggestive and donotcorrespond tothe actual motion oftheearth. Forexample, theobserved precession period of427 days islonger than would becalculated forarigid earth. When plastic deforma- tion istaken into account, anappreciable wandering ofthepole canresult even foranellipsoidal earth with amuch smaller “mountain” than that cal- culated inpart (c).* *Anexcellent short discussion oftherotation oftheearth, treated asan elastic andplastic ellipsoid, willbefound inanarticle byD.R.Inglis, Review ofModern Physics, vol.29,p.9(1957). 1 PROBLEMS 469 3.Show thattheaxisofrotation ofafreely rotating symmetrical rigid body precesses inspace with anangular velocity 00,,=([3—|-S602 a1,)w3, where thenotation isthat used inSection 11-2. 4.Show that iftheonly torque onasymmetrical rigid body isabout theaxis ofsymmetry, then (00%+0,2)isconstant, where co;andwzareangular velocity components along axes perpendicular tothesymmetry axis. IfN3(t)isgiven, show how tosolve forw1,(.02,and(03. 5.Asymmetrical rigid body moving freely inspace ispowered withjetengines symmetrically placed with respect tothe3-axis ofthebody, which supply a constant torque N3about thesymmetry axis. Find thegeneral solution forthe angular velocity vector asafunction oftime, relative tobody axes, anddescribe how theangular velocity vector moves relative tothebody. 6.(a)Consider acharged sphere whose mass mand charge eareboth dis- tributed inaspherically symmetrical way. Show that ifthisbody rotates ina uniform magnetic ﬁeld B,thetorque onitis N=% LXB (gaussian units), where gisanumerical constant, which isoneifthemass density iseverywhere proportional tothecharge density. (b)Write anequation ofmotion forthebody, andshow thatbyintroducing asuitably rotating coordinate system, youcaneliminate themagnetic torque. (c)Compare thisresult with Larmor’s theorem (Chapter 7).Why isno assumption needed here regarding thestrength ofthemagnetic ﬁeld? (d)Describe themotion. What points inthebody areatrestintherotating coordinate system? 7.Prove (without using theresults ofSection 1-2) that iftwo principal moments ofinertia areequal, thepolhode andtheherpolhode areboth circles. 8.(a)Obtain equations, interms ofprincipal coordinates :01,2:2,2:3,fortwo quadric surfaces whose intersection isthepolhode. Your equations should contain theparameters I1,I2,I3,l. (b)Find theequation fortheprojection ofthepolhode onany coordinate plane andshow that thepolhodes areclosed curves around themajor andminor poles oftheellipsoid, butthat they areofhyperbolic type near theintermediate axis, asshown inFig.11-3. (c)Find theradii ofthecircles ontheinvariable plane which bound the herpolhode. 9.Find thematrix (a,-,-) which transforms thecomponents ofavector from space axes tobody axes. Express a;,-interms ofEuler’s angles. [Hint: The transformation can bemade upofthree consecutive rotations byangles 0,45,1/1, about suitable axes, andtaken inproper order.] 10.Write outtheHamiltonian function interms of0,ik,¢,po,p,;.,p¢fora freely rotating unsymmetrical rigid body. Express thecoefﬁcients interms oftheparameters I1,I3,(I2-—I1). 470 THE ROTATION orARIGID BODY [CHAP. 11 11.UseLagrange’s equations totreat thefreerotation ofanunsymmetrical rigid body near oneofitsprincipal axes, andshow that your results agree with thelastparagraph ofSection 11-2. 12.SetupLagrange’s equations forasymmetrical top, theend ofwhose axis slides without friction onasmooth table. Discuss carefully thediﬁerences inthemotions between thiscaseandthecase when theendofthetopaxispivots about aﬁxed point. 13.Agyroscope isconstructed ofadisk ofradius a,mass M,fastened rigidly atthecenter ofanaxle oflength (3a/2), mass (2M/7), negligible cross section, andmounted inside twoperpendicular rings, each ofradius (3a/2), mass (M/3). The axle rotates infrictionless bearings attheintersection points oftherings. One ofthese intersection points pivots without friction about aﬁxed point O. SetuptheLagrangian function and discuss thekinds ofmotion which may occur (under theaction ofgravity). 14.Discuss thefree rotation ofasymmetrical rigid body, using theLa- grangian method. Find theangular velocity foruniform precession and the frequency ofsmall nutations about thisuniform precession. Describe themo- tion andshow that your results agree with thesolutions found inSection 11-2 andinProblem 3. 15.Atopconsists ofadisk ofmass M,radius r,mounted atthecenter ofa cylindrical axle oflength l,radius a,where a<<l,andnegligible mass. Theend oftheaxlerests onatable, asshown inFig.11-9. Thecoeﬂicient offriction is;l..Thetopissetspinning about itssymmetry axiswith avery great angular velocity 4030,andreleased withitsaxisatanangle 01with thevertical. Assume that(.03isgreat enough compared with allother motions ofthetopsothatthe edge oftheaxleincontact with thetable slides onthetable inadirection per- pendicular tothetopaxis, with thesense determined bywg.Write theequa- tions ofmotion forthetop. Assume that thenutation issmall enough tobe neglected, andthat thefriction isnottoogreat, sothat thetopprecesses slowly atanangle 90which changes slowly duetothefriction with thetable. Show “r ll “N =1?l Fro. 11-9. Asimple top. PROBLEMS 471 that thetopaxis willatﬁrst risetoavertical position, andﬁndapproximately thetime required andthenumber ofcomplete revolutions ofprecession during this time. Describe theentire motion ofthetoprelative tothetable during thisprocess. How long willitremain vertical before beginning towobble‘? 16.Obtain atoygyroscope, andmake thenecessary measurements inorder to predict therate atwhich itwillprecess, when spinning atitstopspeed, ifits axis pivots about aﬁxed point atanangle of45°with thevertical. Calculate theamplitude ofnutation iftheaxisisheldatanangle of45°andreleased. Per- form theexperiment, andcompare themeasured rateofprecession with the predicted rate. 17.Aplanet consists ofauniform sphere ofradius a,mass M,girdled atits equator byaring ofmass m.The planet moves (inaplane) about astar of mass M’. SetuptheLagrangian function, using ascoordinates thepolar co- ordinates r,ozintheplane oftheorbit, andEuler’s angles 0,4:,1,0,relative to space axes ofwhich thez-axis isperpendicular totheplane oftheorbit, andthe :1:-axis isparallel totheaxis from which orismeasured. You may assume that r>>a,and usetheresult ofProblem 13,Chapter 6.Find theignorable co- ordinates, andshow that theperiod ofrotation oftheplanet isconstant. 18.Assume that theplanet ofProblem 17revolves inacircle ofradius rabout thestar, although thisdoes notquite satisfy theequations ofmotion. Assume that theperiod ofrevolution isshort incomparison with anyprecession ofthe axisofrotation, sothat instudying therotation itispermissible toaverage over theangle oz.Show that uniform (slow) precession ofthepolar axismay occur if theaxis istilted atanangle 00from thenormal totheorbital plane, andﬁnd theangular velocity ofprecession interms ofthemasses M,m,M’,theradii a,r,theangle 60,andtheangular velocity ofrotation. Show that ifthedayis much shorter than theyear, theabove assumption regarding theperiod ofrevo- lution andtherateofprecession isvalid. Find thefrequency ofsmall nutations about this uniform precession and show-that when theday ismuch shorter than theyear, itcorresponds tothefree precession whose angular velocity is given inProblem 3. 19.Find themasses M,mrequired togive theplanet inProblem 17thesame principal moments ofinertia asauniform ellipsoid ofthesame mass andshape astheearth. Show that, with theapproximations made inProblem 18,ifthe sunandmoon lieintheearth’s orbital plane (they dovery nearly), theeffect ofboth sunandmoon ontheearth’s rotation canbetaken into account simply byadding theprecession angular velocities that would becaused byeach separately. Theequator makes anangle of23.5° with theorbital plane. Find theresulting total period ofprecession. (The measured value is26,000 years.) *20. Write Lagrangian equations ofmotion fortherigid body inProblem 5. Carry thesolution asfarasyoucan. (Make useoftheresults ofProblem 5if you wish.) Show that you canobtain asecond order differential equation in- volving 0alone. Can you ﬁnd any particular solutions, orapproximate solutions, ofthisequation forspecial cases? Describethe corresponding motions. (Note that thisproblem, totheextent that itcanbesolved, gives themotion ofthe body inspace, incontrast toProblem 5,where wefound theangular velocity relative tothebody.) 472 THEROTATION orAmorn BODY [cmu>. 11 21.Anelectron may forsome purposes beregarded asaspinning charged sphere likethat considered inProblem 6,with gvery nearly equal to2.Show that ifgwere exactly 2,and theelectron spin angular momentum isinitially parallel toitslinear velocity, then astheelectron moves through anymagnetic ﬁeld, itsspinangular momentum would always remain parallel toitsvelocity. 22.Anearth satellite consists ofaspherical shell ofmass 20kgm, diameter 1m.Itisdirectionally stabilized byagyro consisting ofa4-kgm disk, 20cm indiameter, mounted onanaxle ofnegligible mass whose frictionless bearings arefastened attheopposite ends ofadiameter oftheshell. Theshell isinitially notrotating, while thegyro rotates atangular velocity wo. Aone-milligram dust grain traveling perpendicular tothe gyro axis with avelocity of 3'>< 104m/sec buries itself intheshell atoneendoftheaxis. What must be therotation frequency ofthegyro inorder that thegyro axis shall thereafter remain within 0.1degree ofitsinitial position? Anaccuracy oftwosigniﬁcant ﬁgures intheresult willbesatisfactory. 23.Agyrocompass isasymmetrical rigid body mounted sothat itsaxis is constrained tomove inahorizontal plane attheearth’s surface. Choose a suitable pair ofcoordinate angles and setuptheLagrangian function ifthe gyrocompass isataﬁxed point ontheearth’s surface ofcolatitude 00.Neglect friction. Show that theangular velocity component <03along thesymmetry axisremains constant, andthatifw3>(I1/I3)w@ sin00,where woistheangular velocity oftheearth, then thesymmetry axisoscillates inthehorizontal plane about anorth-south axis. Find thefrequency ofsmall oscillations. Inanactual gyrocompass, therotor must bedriven tomake upforfrictional torques about thesymmetry axis, while frictional torques inthehorizontal plane damp the oscillations ofthesymmetry axis, which comes torestinanorth-south line. CHAPTER 12 THEORY OFSMALL VIBRATIONS Animportant andfrequently recurring problem istodetermine whether agiven motion ofadynamical system isstable, andifitis,todetermine thecharacter ofsmall vibrations about thegiven motion. Thesimplest problem ofthis kind isthat ofthestability ofapoint ofequilibrium, which weshall discuss ﬁrst. Inthiscase, wecanusethemachinery of tensor algebra developed inChapter 10togive anelegant method ofsolu- tion forthesmall oscillations. Amore general problem occurs when we aregiven anyparticular solution totheequations ofmotion. Wemay then askwhether that solution isstable, inthesense that every solution which starts from initial conditions near enough tothose ofthegiven solution willremain near that solution. This problem willbediscussed in Section 12-6. Methods ofsolution will begiven forthespecial case of steady motion. 12-1 Condition forstability near anequilibrium configuration. Letus consider amechanical system described bygeneralized coordinates x1,...,xf,and subject toforces derivable from apotential energy V(x1, ...,xf)independent oftime. Ifthesystem issubject tocon- straints, wewill suppose thecoordinates chosen such that 2:1,...,x; areunconstrained. Thecoordinate system istobeﬁxed intime, therefore thekinetic energy hastheform T= -M;;,:i:1:i:;,. (12-1) a-"M- I-1N)|-~ Lagrange’s equations then become ’a . ’1aM..... avZa(Mu¢$z)— Z '§"T'xlxm_l'5_‘=0; k=1»-~-»f-1-1 l,m=1 $7‘ wk (12_2) These equations have asolution corresponding toanequilibrium con- ﬁguration forwhich thecoordinates allremain constant ifthey canbe solved when allvelocity-dependent terms aresetequal tozero. The system can therefore beinequilibrium inany conﬁguration forwhich thegeneralized forces vanish: 6VE-O, lc-1,...,f. (123) 473 474 -rnnonr orSMALL VIBRATIONS [cH1u>. 12 These fequations aretobesolved fortheequilibrium points, ifany, of thesystem. Thequestion ofstability iseasily answered inthiscase. IfV(:z:1, ...,xf) isaminimum foranequilibrium conﬁguration :09,...,as?relative toall nearby conﬁgurations 1:?+61:1,...,:0?+6z;, then this isastable conﬁguration. Thetotal energy E=T+V (12-4) isconstant. Let E=I/($9,...,15;’)+an (12-5) betheenergy corresponding toany initial conditions xi’+6:01),..., 2:)’—|-62:)’;:i:Y,...,dz)’near equilibrium. Then if6:51’,...,5:0)’;riff,...,:31? aresmall enough, wecanmake 5Eassmall asweplease. Since Tisnever negative, themotion isrestricted byEq.(12-4) toaregion intheconﬁgura- tion space forwhich V(a:1,...,9”)sI/($9,...,:c,9)+an (12-6) Since Visaminimum at(mg,...,:v?),if5Eissuﬁiciently small, the motion isrestricted toasmall region near :09,...,mt’. Furthermore, since T38E, (12-7) thevelocities £1,...,at;arelimited tosmall values. Therefore theequi- librium isstable inthesense that motions atsmall velocities near the equilibrium conﬁguration remain near theequilibrium conﬁguration. Conversely, ifVisnotaminimum near 2:1,...,2:9,then itisplausible that theequilibrium isunstable, because insome direction away from xlf,...,xi-),Vwilldecrease. Ifwecanchoose thecoordinates sothat x1, say, corresponds tothat direction, and sothat m1isorthogonal tothe other coordinates, then Eq.(12-2) for2:1is d . ’10M1... ..__ aVE(M11$1) —[mgl 5W ilizﬂvm ——F1 (12-8) Forsmall enough velocities suchthatquadratic terms inthevelocities are negligible, thisbecomes M1151=— <12-9) But aswemove away from equilibrium inthe:01-direction, 6V/6x1 be- comes negative, and:01hasapositive acceleration away from theequi- librium point. InSection 12-3 weshall present amore rigorous proof that theequilibrium isunstable ifVisnotaminimum there. 12-2] LINEARIZED EQUATIONS orMOTION 475 Here thetest foraminimum point should berecalled. If20?,...,act} isanequilibrium conﬁguration forwhich Eq. (12-3) holds, then itisa minimum ofV(:c1, ...,22;)relative tonearby conﬁgurations, provided that allthedeterminants inthefollowing sequence arepositive: 62V 62V ‘ii a’V _2 8a:6:1: BI?§>°» at*2>0,..., . 6:v26a:1 31% (E3; where thederivatives areevaluated at$2,...,a:?.*62V 61116131’ >0, H 6:0? (12-10) 12-2 Linearized equations ofmotion near anequilibrium configuration. Wewish now tostudy themotion ofasystem intheneighborhood ofan equilibrium conﬁguration. The coordinates willbechosen sothat the equilibrium conﬁguration liesattheorigin 2:1----=as;=0.The potential energy Vistobeexpanded inaTaylor series in1:1,...,CE]. The constant term V(0, ...,0)may beomitted asitdoes notenter into theequations ofmotion. The linear terms areabsent, inview of Eqs. (12-3). Ifourstudy isrestricted tosmall values of2:1,...,av,-,we may neglect cubic andhigher-order terms in1:1,...,it/,sothat fV=Z1K.m.w.. lc.l=1 azv)K=-— H axkaxl :|:1=...=2:j=0where(12-11) (12-12) Since thecoordinate system isstationary, thekinetic energy is R‘LM\0-NI-I§ T= - Hilyqﬁilr. (12-13) Ingeneral, thecoeﬁicients M1,1may befunctions ofthecoordinates, butsince thevelocities aretobesmall, tosecond order inx1,...,xf; :i:1,...,dc;wemay take M1,;tobethevalues ofthecoefficients at x1=---=:v;=0. Equations (12-11) and (12-13) canbewritten inasuggestive way by introducing inthef-dimensional conﬁguration space aconﬁguration vector xwith components 2:1,...,a:;: x=(a:1,...,:c;). (12-14) *W.F.Osgood, Advanced Calculus, New York: MacMi1lan, 1925, p.179. 476 THEORY orSMALL VIBRATIONS [cHA1>. 12 The coefﬁcients K1,;andMklbecome thecomponents oftensors K11" "K1! K= E 1 /1'--Kn (12-15) Mn...Mu M=5 - _ M/1''~M11 These tensors aresymmetric, orcanbetaken assuch, since byEq.(12-12) Km=Km, (1245) andinthedeﬁning equation (12—13) only thesum %(M1,;+M11¢)isde- ﬁned asthecoeﬂicient of£1,221 =:icl:i:,,. Therefore, wemay require that Mk; =Mlk. (12-17) Thekinetic andpotential energies may now bewritten as T=ﬁt~M-11, (12-18) V=<}x-K-x. (12-19) TheLagrange equations (9-79) maybewritten as M-ir'+K-x=0. (12-20) This equation bears aformal resemblance toEq.(2-84) forthesimple harmonic oscillator. Ifwewrite Eq. (12—20) interms ofcomponents, weobtain adirect generalization ofEqs. (4—135) and (4—136) fortwo coupled harmonic oscillators. Wemay solve Eq. (12-20) bythesame method used tosolve Eqs. (4—135) and(4—-136). Wetry x=ca", (12-21) where C=(C1,...,Cf)isaconstant vector whose components C1,...,6‘,- may becomplex. Wesubstitute inEq.(12-20) anddivide bye1”‘: p2M-c+K-c=0. (12-22) Ifwewrite thisinterms ofcomponents, Weobtain fZ(p2Mk; +K100, =0,lo=1,f. (12-23) l=1 12-3] NORMAL MODES orvrsrwrron 477 IfC1,...,C’;arenotallzero, thedeterminant ofthecoefficients must vanish: P2M11 +K11 ‘''P2M1J -l"K115 =0. (12-24) P211411 +K11'''P211411 +Kr! This isanequation oforder finp2whose fsolutions, pf=-w,g, give thefnormal frequencies ofoscillation. Wemay then substitute anyp? inEqs. (12-23) andsolve forthecomponents C1;ofthevector C,-(except foranarbitrary factor). The solution may then beobtained asasuper- position ofnormal vibrations, just asinSection 4-10 fortwo coupled oscillators. Inthenext section weshall consider analternative way, utilizing themethods oftensor algebra developed inChapter 10,ofde- termining thesame solution. 12-3 Normal modes ofvibration. Ifthecoordinates x1,...,:0;are orthogonal, thetensor Mwillbeindiagonal form: Mk; =-Mk 81,1. (12-25) Ifthecoordinates arenotorthogonal, wecandiagonalize Mbythemethod ofSection 10-4, generalized tofdimensions. (We willusethesame method below todiagonalize thepotential energy.) Letussuppose that thishasbeen done, andthat thecoordinates :01,...,:0;arethecom- ponents ofxalong theprincipal axes ofM,sothat Eq.(12-25) holds. (If1:1,...,ac,»arerectangular coordinates ofasetofparticles, Mkisthe mass oftheparticle whose coordinate iswk.) Wenow deﬁne anew vector ywith coordinates y1,...,yfgiven by yk=(Mk xk)l/2; k=1;---1f~ Note that theconﬁguration ofthesystem isspeciﬁed now byavector y inanew vector space related tothe2:-space byastretch orcompression along each axis, asgiven byEq.(12-26). The kinetic energy interms ofyis IT=es’-w=Zat <12-21> k=1 Clearly, theexpression forthekinetic energy does notchange ifwerotate they-coordinate system, which isourreason forintroducing thevector y. Thepotential energy isgiven by V=%Y'W'Y= kzyk?/1, (12-28) EMlet-IE 478 THEORY orsmnn VIBRATIONS [cniun 12 where KWk;= (12-29) Theequations ofmotion are Y—|-W-y=O. (12-30) The tensor Wissymmetric, and cantherefore bediagonalized bythe method given inSection 10-4. Lete,~beaneigenvector ofWcorrespond- ingtotheeigenvalue W,-: W'8;=Wjej. (12-31) Leta;,~bethecomponents ofe,-inthey-coordinate system: 61'=(a1,-, ...,0],"), ‘=1,. ..,f. (12-32) Then wemay write Eq.(12-31) interms ofcomponents inaform corre- sponding toEqs. (10-108): t(Wm —W15z¢z)¢lz1 =0, k=1,---,f- (12-33) l=1 Again, thecondition foranonzero solution is W11 —W1 W12 W11 W21 W22 —W1" " W2! =()_ (12-34) Wm Wm ---W/1"W1" This isanalgebraic equation oforder ftobesolved forthejroots W,-. Note that itisthesame asEq.(12-24) ifp2=—W,~ andwedivide theleft side ofEq. (12-24) byM1-M2 ---Mf,remembering that Mk; isnow given byEq.(12-25). Each root W,istobesubstituted inEq.(12-33), which may then besolved fortheratios a1,-:a2,-:---:a_;,~. The aljcan then bedetermined sothat e,-isaunit vector: -M» P-'Q:59=1. (12-35) Theproofs given inSection 10-4 canbeextended tospaces ofanynumber ofdimensions, soweknow that theroots W;arereal, andtherefore the 12-3] NORMAL MODES orVIBRATION 479 coefﬁcients a;,-arealsoreal. Moreover, theunit vectors e,~,elareorthog- onal* forW;aéW). Wehave therefore e_.,--e,=6,-,, (12-36) or fZa,,~a,, =a,-,. (12-37) l=1 Inthecase ofdegeneracy, when twoormore roots W,-areequal, We canstillchoose thea;,-sothat thecorresponding e,-areorthogonal. The situation isprecisely analogous tothat described inSection 10-4, except that forf>3itcannot bevisualized geometrically. The proof of lemma (10-125) canbegeneralized tomultiple degeneracies inspaces of anynumber ofdimensions. Now letthecomponents oftheconﬁguration vector yalong e1,...,e,- beq1,...,qj: I y=Zq,-e,= <12-38) j=1 Interms ofcomponents intheoriginal y-coordinate system, f 1/k=Zawq» (12~39),'=1 Conversely, bydotting e,into Eq.(12-38) and using Eqs. (12-32) and (12-36), weobtain: f qr=2akr'!/k- (1240)k=1 These equations areanalogous toEqs. (10-67) and(10-69). The potential energy inthe coordinate system q1,...,Qf,which diagonalizes W,is IV=E%W,-q,*. (12-41) j=l Since Visaminimum attheorigin y=0,theeigenvalues W1,...,W; must allbepositive; otherwise forsome values ofq1,...,qf,Vwould be negative. IfVwere notaminimum, some oftheeigenvalues W,-would benegative. (The special case W,=0may ormay notcorrespond to *Itiscustomary tousetheterm “orthogonal” rather than “perpendicular” inabstract vector algebra when thevectors have only analgebraic, and not necessarily ageometric, signiﬁcance. 480 THEORY OFSMALL VIBRATIONS lomr. 12 aminimum, depending onhigher-order terms which wehave neglected.) Letusset W,=41?. (12-42) Thekinetic energy (12-27) inthiscase is up-Ita.'u,t\° T= (12-43) Inview ofEqs. (12-41) and (12-43), theLagrange equations separate intoequations foreach coordinate q,-: q,+w?q,- =0, j=1,...,f. (12-44) The coordinates q,arecalled thenormal coordinates. The solution is q,-=A;cosco,-t—|-B,~sinw,-t, j=1,...,f, (12-45) where A,-,B,-arearbitrary constants. Wemay write thesolution in terms oftheoriginal coordinates, using Eqs. (12-26) and(12-39): xk=Mil” ‘Za;,,~(A,~ coswit+B;sinw,-t). (12-46) i=1 Thecoeﬂicients are IA,-=q,-(0)=Za,,,M,£'%,,(o) (12-47) k=1 and IB,~=Q,-—1q,(0) =Z¢.,;1a,,,-M,§'22,(o). (12-48) k=1 Wetherefore have thecomplete solution forsmall vibrations about a point ofstable equilibrium. When thenumber ofdegrees offreedom islarge, solving Eq. (12-34) may beaformidable jobwhich, ingeneral, canbedone only numerically fornumerical values ofthecoefficients. However, insome cases wemay know some oftheroots beforehand (often weknow that certain normal frequencies arezero), orfrom symmetry considerations wemay know that certain roots areequal. Any such information helps infactoring Eq. (12-34). IfVisnotaminimum at2:1=---=x,»=0,andsome oftheco- efﬁcients W,arenegative, then weobtain exponential-type solutions. This proves that themotion isunstable inthis case, since thesolution (except forvery special initial conditions) willcontain terms which in- crease exponentially with time, atleast until thelinear approximation 12-4] FORCED VIBRATIONS 481 wehave made intheequations ofmotion isnolonger valid. The case when some W;iszero willnotbediscussed indetail here. Inthelinear approximation Wearemaking, thecorresponding q,-isconstant inthat case, and this corresponds towhat was called neutral equilibrium in Chapter 2.Themotion willproceed atconstant q,-until q,-islarge enough sothat nonlinear terms inq,-must beconsidered. Itmay benoted that inﬁnding thenormal coordinates wehave found atransformation from coordinates x1,...,:c,=toq1,...,q;which simul- taneously diagonalizes two tensors Mand K,ormore correctly, which simultaneously diagonalizes twoquadratic forms, Tand V.Unless two tensors have thesame principal axes, itisofcourse impossible simul- taneously todiagonalize them byarotation ofthecoordinate system. However, ifthecoordinate system isallowed tostretch orcompress along chosen axes, asinthetransformation (12-26), then wecanbring two quadratic expressions todiagonal form simultaneously (provided that at least oneispositive ornegative deﬁnite). Weﬁrstﬁndtheprincipal axes oftheﬁrst tensor. Bystretching and compressing along theprincipal axes, wecanreduce this toaconstant tensor (provided that theeigen- values areallpositive orallnegative). Inthecase above, wereduced Mto‘Iwiththetransformation (12-26). Since allaxesareprincipal axes foraconstant tensor, theprincipal axes ofthesecond tensor, asmodiﬁed bythestretching ofcoordinates, willreduce bothtensors todiagonal form. The reader willﬁnditinstructive togive ageometrical interpretation of thisprocedure, inthecase oftensors intwoorthree dimensions, by representing each tensor byitsassociated quadric curve orsurface, just astheinertia tensor wasrepresented inSection 10-5 bytheinertia ellip- soid. When wearedealing with vectors andtensors inphysical space, weordinarily donotconsider nonuniform stretching ofaxes because thisdistorts thegeometry ofthespace. When wedeal with anabstract vector space, wemay consider anytransformation which isconvenient forthealgebraic purpose athand. 12-4 Forced vibrations. Wenow wish todqermine themotion ofthe system considered inthepreceding section when itissubject toprescribed external forces F1(t), ...,F;(t) acting onthecoordinates x1,...,xf.We willagain restrict ourconsideration tomotions which remain close enough totheequilibrium conﬁguration sothat only linear terms inx1,...,:cf need tobeincluded intheequations ofmotion. Ifweintroduce the vector F(t) =(F1,...,Ff), (12-49) wemay write theequations ofmotion intheabbreviated form, M-i+K-x=F(t). (12-50) 482 THEORY orSMALL VIBRATIONS [CHAR 12 where wehave simply added theforces F(t) toEq.(12-20). Note that Eq.(12-50) may beobtained from theLagrangian function L=T—V—V’, (12-51) where TandVaregiven byEqs. (12-11) and(12-13), and i fV’=-Z2,.F,,(¢). (12-52) k=1 Again suppose that thecoordinates x1,...,2:,»arechosen tobeorthog- onal sothat Misdiagonal. Ifthecoordinates ac),arenotinitially orthog- onal, and arotation ofthecoordinate system isperformed toprincipal axes ofM,then thecomponents F;,(t) must besubject tothesame trans- formation asthecoordinates wk.Since weshall follow thisprocess through inthecasewhere wediagonalize thetensor K,weshall notfollow itthrough indetail forM,butsimply assume that, ifnecessary, ithasbeen carried outandthat Misdiagonal. Wetransform now tothenormal coordinates found inthepreceding section [Eqs. (12-26), (12-39), and(12-40)]: 18;,= Mil/211),,-q,-, (12-53) j—1 f4)=Z)M,%’%.,-21.. (12-54> k=1 The generalized forces Q,(t) associated with F;,(t) areobtained byusing Eq.(9-30). Q1'(t) =2!:M1T1/2<1kjFk(t)- (12-55)k=1 Theinverse transformation is I .F10)=ZfjM1%’”a.-<2.-<1). <12-56>i=1 The reader may also verify Eqs. (12-55) bysubstituting Eqs. (12-53) in Eq.(12-52) andcalculating Q.=_215. (12-57) .1 aq1'Innormal coordinates, _ V’=—iq1Q,-(t), <12-58) J'=1 4 12-4] FORCED VIBRATIONS 483 sothat theequations ofmotion are 41-+wig)-Q,-<0, 1'=1,.--.1". <12-59> Each ofthese equations isidentical inform with Eq.(2-86) fortheun- damped forced harmonic oscillator (b=0).Therefore thenormal modes behave likeindependent forced oscillators, andthesolution canbeob- tained bythemethods described inChapter 2. Itistempting totry"togeneralize ourresults tothecase when linear damping forces arealsopresent. Wecaneasily write down theappropriate equations. Inthegeneral case when thecoordinates arenotorthogonal andthere isfrictional coupling between coordinates, theequations ofmo- tion willbe f Z(M11551 -1-Bkziiz —|-K1111) =0, 7°=1»---11', (12450)z1 or,invector form, M-i+B-x-I-K-x=0. ~ (12-61) Unfortunately, asthereader may perhaps convince himself with some experimentation, itisgenerally notpossible simultaneously todiagonalize three tensors M,B,Kwith anylinear transformation ofcoordinates, even ifstretching isallowed, Themethod ofthepreceding section therefore failsinthiscase, andthere arenonormal coordinates. Thesituation is notimproved byassuming that 2:1,...,2:;areorthogonal sothat Mis diagonal, oreven byassuming that there isnofrictional coupling sothat Bisdiagonal. Ifweapply thetransformations (12-53) and(12-54) which diagonalize Tand V,thecoordinates q,-areingeneral still coupled by frictional forces: _ f2-+Zbaa.+w?q,-=0. (1242) r=l where "fP'1~s bjr Z _l/2Ml—ll2akjalrBl§l- Note that thematrix b,-,isnotdiagonal even ifBk;is.There isaspecial case which sometimes occurs when thefrictional forces areproportional tothemasses, sothat B=2'YM. Themethod ofSection 12-3 then works, since inthey-coordinate system inwhich M—+1,wehave B-—>2"/‘I andthenormal coordinates q,-along theprincipal axes ofWsatisfy the separated equations: q,+2vq,-+a§q,-=0. ~(12-64) 484 THEORY 01-‘SMALL VIBRATIONS [cnx1=. 12 Itmay, ofcourse, alsohappen that inthey-space inwhich Mbecomes ‘I thetransformed tensors BandKhave thesame principal axes, butthis would beanunlikely accident. When thedamping forces arevery small, aperturbation method similar tothat which willbedeveloped inthe next section canbeapplied toﬁnd anapproximate solution interms of damped normal modes. Except inthese special cases, theproblem ofdamped vibrations can behandled only bydirect substitution ofatrial solution like(12-21) in theequations ofmotion (12-60). The secular equation analogous to Eq.(12-24) isthen oforder 2finp.Each root allows asolution forthe vector C.Ifthere arecomplex roots, they occur inconjugate pairs, p,p*, with corresponding conjugate vectors C,C*. The twosolutions (12-21) canthen becombined toyield arealsolution which willbedamped and oscillatory, andcanbecalled anormal mode. Ifall2fsolutions arecom- bined with appropriate arbitrary constants, thegeneral solution toEqs. (12-60) canbewritten. Itisclear onphysical grounds, since thefrictional forces reduce theenergy ofthesystem, that therealparts ofallroots p must benegative ifV(x) hasaminimum atx=0;themathematical proof ofthisstatement isadiﬂicult exercise inalgebra. 12-5 Perturbation theory. Itmayhappen thatthepotential energy is given by V=V°+V’, (12-65) where V°(x1, ...,2:,-)isapotential energy forwhich wecansolve the problem ofsmall vibrations about aminimum point 2:1=---=2:;=0, and where V’(x1, ...,xf)isvery small forsmall values ofx1,...,xf. WewillcallV°theunperturbed potential energy, and V’theperturba- tion. Weexpect that thesolutions forthepotential energy Vwillapproxi- mate those fortheunperturbed problem. Inthissection weshall develop anapproximate method ofsolution based onthisidea. Weshall assume that V’isstationary atx1=---=:0;=0,sothat (<11) =0. (12.66) axk £1=...=:|:f=(] Ifthisisnot’thecase, itisnotdiﬂicult toﬁndapproximately thevalues ofxlf,...,ac?forwhich Visstationary. Weleave this asanexercise. The origin ofcoordinates should then beshifted slightly to2:9,...,2?. This willalter slightly thequadratic terms inV°,butthese small changes canbeincluded inV’.Inanycase, wewilltherefore have anexpansion 12-5] PERTURBATION THEORY 485 ofVaround theequilibrium point oftheform (12-65), with V0 =Z %K]9lxkZ1, (12-67) 2,1 V’ = Kjcﬂkibj, (12-68) Ev(-NI-I where thecoefficients K;',1aresmall. The precise criteria that must be satisﬁed inorder that K1,;canbeconsidered small willbedeveloped asWe proceed. Weﬁrst transform tothenormal coordinates qtil,...,q?fortheun- perturbed problem. Wethen have fV”=Z4W?<q?>”, (1249) j=1 V’- 5-.q?q?, <12-70> {MTM5'3F Wir= —1’2 _’2akﬂ1zrK1§l, (12-71) where W?aretheroot ofthesecular determinant (12-34) fortheun- perturbed problem, and where again weassume, forsimplicity, that 001,...,'x;areorthogonal coordinates. The coefﬁcients W}, aretobe treated assmall. The superscript “O”willremind usthat thevariables q?arenormal coordinates fortheunperturbed problem. Theequations ofmotion forqlf,...,q?are ' r4?+W;-’q§-’+ ZW;-.49=0,1=1.--.,1.<12-72> r=1 Weseethat thediagonal element ofW’adds tothecoefﬁcient ofqg,while theoff-diagonal elements couple theunperturbed normal modes. We expect that ifW’issmall, there willbeanormal mode oftheperturbed problem close toeach normal mode oftheunperturbed problem, that is, asolution with frequency w,-near (0?=(W;-))1’ 2andforwhich q?islarge while theremaining qg,r25j,aresmall. However, ifthetensor W°has degenerate eigenvalues, sothat two ormore oftheunperturbed fre- quencies areequal (orperhaps nearly equal), then weexpect that even a small amount ofcoupling canradically change themotion, asinthecase oftwocoupled oscillators that weworked outinChapter 4.This insight willhelp indeveloping aperturbation method. 486 THEORY orSMALL VIBRATIONS [CHAP. 12 Ifwetrytoﬁndanormal mode ofoscillation bysubstituting q?=0,-21"‘, 3'=1,...,f, (12-73) inEqs. (12-72), weobtain g (1)2+W§?)o,- +ijW;~.0. =0. (12-74) r=1 Now letusassume that themode which weseek isclose tosome un- perturbed mode, sayj=1.Wethen set P2 = _WII1 C1=1+C’1, (12-75) Cjzogv j=2r"'7f2 where, ifW{,C{,...,C}arezero, Eq.(12-73) represents asolution ofthe unperturbed problem. Hence fortheperturbed problem, weassume that W{,C{,...,C}aresmall. Wesubstitute Eqs. (12-75) inEqs. (12-74), andcollect second-order terms ontheright-hand side: —W'1 +W11=—iW116’? +W’1C"1, (12-76) r-1 f 1-1 (12-77) When weneglect second-order terms, theﬁrstequation gives W{: W’;-W’11. (12-78) Therefore 41--22-W?+W11. (12-79) Equations (12-77), ifweneglect theright members, yield thecoeﬂicients C','~: i ; W,-1 '___— _]—2,...,f. Thecoeﬁicient CIisnotdetermined; thiscorresponds tothefactthat the normal mode (12-73) may have anarbitrary amplitude (and phase), although itmust, ofcourse, benear theamplitude (and phase) C?=1 which waschosen inEqs. (12-75) fortheunperturbed solution. Itwillbe convenient torequire that C1,...,C;bethecoefficients ofaunit vector: IZ0?=1. (12-s1) j=1 12-5] PERTURBATION THEoRY _ 487 Wethen obtain thefollowing equation forC{: "1." Q./\ 0'1=— 0;-)2. (12-s2) Toﬁrstorder insmall quantities, C'1-0. (12-83) Bysubstituting inEqs. (12-73), multiplying byanaribitrary constant §—Ae“’, and superposing thecomplex conjugate solution, weobtain the ﬁrst-order approximations totheperturbed normal mode: qi-A<=0S(w1t +0), (12-84) 0.AW"1 -q,= cos(w1t+0), _7=2,...,f, where 0:1isgiven byEq.(12-79). Weseethat theﬁrst-order effect ofthe perturbation istoshift (cfbythediagonal perturbation coefficient W{1 andtoexcite theother unperturbed modes weakly with anamplitude that isproportional totheperturbation coupling coefﬁcients W,’-1andin- versely proportional tothedifferences inimpertmbed normal frequencies (squared). Thisisaphysically reasonable result. Wecannow formulate more precisely therequirement that W’besmall. Inourderivation, wehave assumed that W1<<(W?-W21, j=2,...,f, (12-85) 09-<<1. (12-se) Equations (12-78) and(12-80) show that thisisjustiﬁed if W;-1<<|W2-W21, j=1,...,;. (12-s7) This isthecondition forthevalidity offormulas (12-79) and(12-84). First-order approximations totheremaining modes areobtained from these formulas byinterchanging thesubscript ~11)with anyother. The astute reader willnote that thecondition (12-87) inthediagonal co- efﬁcient W{1 isnecessary only because wehave neglected thelast term in Eqs. (12-77). Equations (12-77) areeasily solved forC}even ifthelastterm ontheright isincluded. This allows ustoremove therestriction onthesizeof thediagonal coefficients ifwewish. This isalsoobvious because wecanalways include any diagonal coefficient inV0[Eq. (12-69)]. The normal coordinates 488 THEORY orSMALL VIBRATIONS [CHAP. 12 fortheunperturbed problem arestillthesame; only the(squared) frequencies arealtered byadding additional diagonal terms. However, inthetransformation tonormal coordinates, diagonal and off-diagonal terms become intermixed, so that unless allterms inV’(701,..._.1c;) aresmall theoff-diagonal terms of V’((19,...,q?)areunlikely alltobesmall. Conditions (12-87) clearly cannot besatisﬁed ifthere isadegeneracy— if,forexample, W?=W3= Inthat case, asmentioned earlier, we expect that even with very small coupling oftheunperturbed modes, anyperturbed mode with w2near WYwillshow appreciable excitation of allthree unperturbed modes. Wetherefore set 112=-W9-W’, (12-ss) andassume that only C4,...,C;aresmall, while C1,C2,C3may allbeof order 1.Wesubstitute inEqs. (12-74) andtranspose second-order terms totheright members: f (W11 ‘TW’)C1 + W’12C2 -l" Wiscs =— Wiron r=4 ' 1 W21C1 +(W22 —W’)C2 —|- W2aUa =—ZW21C'n r=4 fW110. +W402 +(W13-W')0.~.=—ZW4.0.. ' r=4 (12-89) and 8 f(W?-W?)0,~+ZW;-,0,=-Z‘W;~,o,+W’C',-, j=4,...,f. 1'=1 r=4(12-90) Ifweneglect theright members, then Eqs. (12-89) become astandard three-dimensional eigenvalue problem fortheeigenvalue W’and the associated eigenvector (C1,C2,C3). There willbethree solutions corre- sponding tothree perturbed normal modes with frequencies (.02= I/V?-1-W’near thedegenerate unperturbed frequency. Ingeneral, the three roots W’willbedifferent, andsotheperturbed modes willnolonger bedegenerate. The remaining coefficients C4,...,C’;canbefound toa ﬁrst-order approximation from Eqs. (12-90) byneglecting theright mem- bers. Wemay also require that Cbeaunit vector [Eq. (12-81)]. In analogy with Eq. (12-83), this willmean that toﬁrst order thethree- dimensional vector (C1,C2,C3)should beaunitvector. Thecase ofdouble 12-5] PERTURBATION THEORY 489 ormultiple degeneracy ofany order must betreated inthesame way. Clearly, ifthedegeneracy isofhigh order, theﬁrst-order perturbation equations [Eqs. (12-89) with right members zero] may bealmost as diﬂicult tosolve astheexact equations (12-74). When fZ4,wemay have more than onedegenerate normal frequency; inthat case, theabove method canbeapplied separately toeach group ofdegenerate unper- turbed modes toﬁndtheperturbed modes. Incases ofapproximate degeneracy (WY iW2iWg) when condi- tions (12-87) failforagroup ofneighboring unperturbed modes, the method oftheprevious paragraph canalsobeapplied. Equations (12-89) areslightly modiﬁed bytheaddition ofsmallterms, likeW2-W§’,in thediagonal coefficients. Thereader canreadily formulate theprocedure forhimself. When theﬁrst-order approximate solution hasbeen found, the ap- proximate values ofthecoefficients W{,C}may besubstituted inthe right members ofEqs. (12-76), (12-77) [orEqs. (12-89), (12-90)]. The resulting equations arethen solved toﬁndasecond-order approximation. If,forexample, wesubstitute Eqs. (12-80) inEq.(12-76), weobtain the second-order approximation tothefrequency correction: W1 W11 +N2W9_W9 (1291) where wehave used thefactthat W’isasymmetric tensor. Weseethat thesecond-order frequency shift inmode 1contains acontribution dueto coupling with each oftheother modes. The modes tend torepel one another insecond order; that is,each higher frequency mode (W2 >W?) reduces thefrequency ofmode 1,andeach lower frequency mode increases it.The same result wasobserved inthesolution totheproblem oftwo coupled oscillators inChapter 4.The procedure canbecarried outina straightforward way tosuccessively higher-order approximations, butthe labor involved rapidly increases. Toanyorder ofapproximation, wemayintroduce normal coordinates q1,...,qffortheperturbed problem bysetting f1?=Z(1.-at (12-92) r=1 where C,-,,j=1,...,f,arethecoefficients fortherthperturbed normal mode found toanyorder ofapproximation bytheperturbation theory. Thevectors C,=(C1,, ...,Ch)are,tothegiven order ofapproximation, orthogonal unit vectors (orcanbemade so),asweshall seepresently. 490 THEORY orSMALL VIBRATIONS [CHAP. 12 Wemay therefore solve Eqs. (12-92) for fq.=Z0,-.151 (12-92)j=1 FromEqs.(12-92) and(12-73), ifp2=-W2-W;=-ofisthe approximate value forthefrequency, then theapproximate solution for q,must be q,éA,cosw,t-1-B,sinw,t. (12-94) Comparison with Eqs. (12-45) shows that theq,are(approximate) normal coordinates. The Lagrangian, tothegiven order ofapproximation, must therefore be f L-2(ed?-4651?). (12-95)T: . asmay also beveriﬁed bystraightforward substitution ofEqs. (12-92) inEqs. (12-69), (12-70), and (12-43), toany order ofapproximation inC,-,. Alternatively, wemay note that Eqs. (12-74) arejust theequa- tion wewould obtain ifwewere tolook foraneigenvector CofW= W°’-S; W’corresponding totheeigenvalue —p2. Hence theapproximate solutions wehave obtained forEqs. (12-74) arealsoapproximate solu- tions totheproblem ofdiagonalizing W.Equations (12-92) must there- foredeﬁne approximate normal coordinates fortheperturbed motion. 12-6 Small vibrations about steady motion." Letamechanical system bedescribed bycoordinates 2:1,...,ca),and byaLagrangian function L(x1, ...,xf;2&1,...,rt);t). Ifasolution :c‘f(t), ...,x)’(t) isknown, wemay look forsolutions close totheknown solution bydeﬁning new coordinates y1,...,y): 2,,=22(1)+y,,, lc=1,...,1. (12-96) Wesubstitute intheLagrangian L,andexpand inpowers ofy1,...yf; ()1,...,3);. Since :v‘f(t), ...,x?(t) satisfy theequations ofmotion, the reader canreadily show that nolinear terms iny1,...,yf;()1,...,1); occur inL.Terms inLindependent ofy1,...,y);()1,...,1);donot affect theequations ofmotion andmay beignored. Ifweassume that 1/1,...,y);1);,...,1))aresmall, and that wemay neglect cubic and higher powers ofsmall quantities, Lbecomes aquadratic function ofthe newvariables. Theequations ofmotion willthen belinear iny1,...,y,-; 1);,...,1);;1);,...,1);.However, thecoefficients intheequations will, ingeneral, befunctions ofthetime t,andthemethods Wehave sofarde- veloped willnotsuffice tosolve them. Todevelop methods forsolving equations with time-varying coefficients isbeyond thescope ofthisbook. 12-6] SMALL VIBRATIONS ABOUT STEADY MOTION 491 Wewilltherefore consider only cases when thecoeﬂicients inthelinearized equations turn outtobeconstant. s Wecanguarantee that thecoefﬁcients willbeconstant byrestricting ourselves tosteady motions. Suppose that some ofthecoordinates ml,...,ac;areignorable, i.e.,donotappear intheLagrangian function. Wewillalsoassume that Ldoes notdepend explicitly ont.Wedeﬁne a steady motion asoneinwhich allofthenonignorable coordinates are constant. This deﬁnition evidently depends upon thesystem ofco- ordinates chosen. Weshould perhaps deﬁne steady motion asmotion for which, insome coordinate system, thenonignorable coordinates areall constant. Wehave seen inSection 9-10 that ignorable coordinates areparticularly easy tohandle interms oftheHamiltonian equations ofmotion. Letus therefore introduce coordinates x1,...,xf,and corresponding momenta pl,...,pf,ofwhich xb,-+1, ...,ac;areignorable. The Hamiltonian function is H=H(@1, ---,111v;P1,---»PMPN+1, ---,P!)- (12437) Inview ofEqs. (9-198), themomenta ply,-+1, ...,pfareallconstant. Wehave therefore todeal only with 2Nequations (9—198), which for steady motion reduce to 6H 6H%-O, E-0, k-1,...,N. (1298) Forgiven values p?v+1, ...,pf},wearetoﬁnd thesolutions, (ifany, (of these equations forx1,...,:cN;p1, ...,pN.Anysuch solution 11:1,...,xN; plf,...,pg;deﬁnes asteady motion. The ignorable coordinates will all have constant velocities given by ,;g=(@), _7'=N+1,...,f, (12-99) 31):‘0 where thesubscript wonimplies that thederivative istobeevaluated at :v‘1’,...,x%,;p‘1’,...,p§.’. Given asteady motion, letuschoose theorigin ofthecoordinate system sothat ac?—---—-1%=pg 'pg;=O.Inorder tolook for motions near thissteady motion wehold pN+1, ...,pfﬁxed andexpand Hinpowers ofx1,...,xN,pl,...,pN,which weregard assmall. We may omit anyterms which donotdepend onx1,...,pN. Linear terms areabsent because ofEqs. (12-98). Ifweneglect cubic terms insmall quantities, Hbecomes aquadratic function ofx1,...,xN,pl,...,pN, with constant coeﬁicients. Itmay bethat Hseparates into apositive 492 THEORY orSMALL VIBRATIONS [CHAI-'. 12 deﬁnite “kinetic energy,” ‘T’(p1, ...,pN), and a“potential energy,” ‘V’(x1,...,:vN). Inthat case, themethods ofthepreceding sections are applicable. Inorder toapply these methods, wemust express the“kinetic energy,” ‘T’,interms of£1,...,abN,which may bedone bysolving for pl,...,pNthelinear equations -_é‘l’ _ _ x,._apk, k-1,...,N. (12100) Theproblem isthus reconverted toLagrangian form, with ‘L’(:01,...,:cN; £1,...,akN)=‘T’—‘V’. Note, however, that wecannot obtain the correct ‘L’simply bysubstituting 23?from Eq.(12~99) intheoriginal L. The transition toHamiltonian form isnecessary inorder tobeable to eliminate the ignorable coordinates from the problem byregarding pN+1, ...,pfasgiven constants. The “potential energy” ‘V’willcon- tain terms involving pN_|_1, ...,p,»from the original kinetic energy, andthese willappear with opposite signin‘L’. If‘V’(ac1,..., xlv) has aminimum atx1- =xN=0,then x1,...,xN,ifthey aresmall enough, undergo stable oscillations. We may then saythat thegiven steady motion isstable inthesense that for nearby motions (with thesame pN+1, ...,pf),thecoordinates oscillate about their steady values. These oscillations canbedescribed bynormal coordinates, which may befound bythemethod ofSection 12-3. Ifthecoordinate system weuseisamoving one,orifmagnetic forces arepresent and must bedescribed byavelocity-dependent potential (9-166), orif,asoften happens, theignorable coordinates arenotorthog- onal tothenonignorable coordinates, then cross product terms xkp; will appear inH.Thus, ingeneral, thequadratic terms inHhave theform F‘:P’1sH= %a1¢zPkPz +bklwkllz +écmm), (12-101) where wemay aswell assume that am=am, 011»=61.z- (12-102) The coeﬁicients am,bu,ck;arefunctions oftheconstants pN+1, ...,pf andoftheparticular steady motion whose stability isinquestion. We nolonger have aseparation into kinetic and potential energies, and the methods ofthepreceding sections cannolonger beapplied. Itmay be that Hasgiven byEq. (12—101) ispositive (ornegative) deﬁnite in :01,...,:vN,p1,...,pN,inwhich case wemay besure that thesteady motion isstable. For0:1,...,xN,pl,...,pNmust remain onasurface ofconstant H,andifHispositive deﬁnite, thissurface willbean“ellip- soid” inthe2N-dimensional phase space. 12-6] SMALL VIBRATIONS ABOUT STEADY MOTION 493 Inanycase, wemay. study thesmall vibrations about steady motion bysolving thelinearized equations given bytheHamiltonian function (12-101): N it=2(akzpz —|-bllcxl)> z= 1N (12-103) Z§k='"Z(bk1p1-I-Ck1.’£1), I0=1,...,N. l=1 Wecould return toaLagrangian formulation involving Nsecond-order equations inx1,...,aw,butitisjust aseasy todeal directly with Eqs. (12-103). Letuslook foranormal mode inwhich allquantities have thesame time dependence: xv),=Xke”, pk=P,,@"’. (12-104) Wesubstitute inEqs. (12-103) andobtain the2Nlinear equations zv Z[(5116 —P51¢z)Xz —|-11111131] =0, B1 NZ[c;,;X;+(bk;+pa,.,)P,1 =0,k=1,...,1v. (12-105) I-=1 Thedeterminant ofthecoefﬁcients must vanish: bu"‘I7 b21 an l11N 512 522—P'''(121 '''<12N 611 612 '''bu+P''°b1N =0- (12-106) 021 62-2 ‘''1721 b2N 0 u - OI - - O- I O - 6N1 6N2 bN1 "'b1v1v-I-P Wenow note that ifpisanyroot ofthisequation, sois—p. First, letus setp=—p’ inthedeterminant. Now, interchange theupper Nrows andthelower Nrows. Next, interchange theleftNcolumns with the right Ncolumns. Finally, interchange rows and columns, i.e., rotate about themain diagonal. None ofthese operations changes thevalue ofthedeterminant (except possibly itssign, which does notmatter). Wenow have thesame equation forp’that weoriginally hadforp,so that ifpisaroot, soisp’.Weseetherefore that when weexpand the determinant (12-106), only even powers ofpappear, and Wehave an algebraic equation ofdegree Ninp2.Iftheroots areallnegative, asthey 494 THEORY orSMALL VIBRATIONS [cnxrn 12 willbeifHinEq.(12-101) ispositive deﬁnite, then thenormal modes areallstable. Each root p2=-1»? gives two values p=iiw,-. We substitute p=iwjinEqs. (12-105) and solve forX1,,P1,,which will, ingeneral, becomplex. There is,ofcourse, anarbitrary- constant which may bechosen inany convenient way. The solutions forp=—iw,- willbeXi‘),Pf,-. Wesubstitute inEqs. (12-104), multiply byanarbitrary constant A,-em and superpose thetwo complex conjugate solutions to obtain therealsolution forthenormal mode j: wk=A,-Ck, COS(wji +ﬁkj—|-0;), (12-1079) Pk=AiDki 00$(wit+1%;+91), where Xki=éckjewkj,_ (12-10s) PM=%Dk1@m°’} andA,-and0,-areanarbitrary amplitude andphase. Thegeneral solution isnow asuperposition ofnormal modes: N £131,=2 A,-C1,,‘ COS(co,-t —|-ﬁhj—|-9,-), J'=1 N (12-109) pk=Z AjDkj COS (co,-t -|-'Ykj —|-0,‘). i=1 Note that wecannot represent theabove result interms ofnormal coordinates q1,...,qNlinearly related to2:1,...,xNonaccount ofthe phase differences Bk,-,'Y;,,~which arise because ofthecross terms inco- ordinates and momenta. Itispossible toﬁnd alinear transformation connecting the2Nvariables 001,...,xN;pl,...,pNwith asetofnormal coordinates and momenta q1,...,qN;pl,...,pN,each ofwhich oscil- lates atthecorresponding normal frequency. Such transformations be- long tothetheory ofcanonical transformations ofHamiltonian dynamics, andarebeyond thescope ofthisbook. Ifany-root p2ofEq.(12-106) ispositive orcomplex, thecorresponding normal mode isunstable, and the coordinates and momenta move ex- ponentially away from their steady values. Aroot p2=0would corre- spond toneutral stability. One common case inwhich roots p2=O arise occurs when anignorable coordinate has been included among the.731,...,xN. Ifx_,-isignorable, andisincluded inx1,...,xN,then p,-isconstant andmay take anyvalue. Ifwetake p,-slightly different from p?fortheinitially given steady motion, there isanew steady 12-6] SMALL VIBRATIONS ABOUT STEADY MOTION 495 motion with 02,-constant andslightly different from This new motion isgiven by xj=A;+B,~aZ,~t, allother ac),=A,,, (12-110) where A).may beslightly different from $9,.This motion corresponds to anormal mode with p2=0.Insome cases, thealgebra required toignore explicitly acoordinate xiistooformidable, andwemay prefer toinclude ac,- among thenonignorable coordinates. This increases thedegree ofthe secular equation (12-106) byone, butsince theextra root isp2=0,we know that p2willfactor outandtheremaining equation willhave the same degree asifwehadignored x,-.Itmay alsobethat we.have chosen acoordinate system inwhich some ignorable coordinate :v,-does notap- pear. Arootp2=0ofEq.(12-106) willstilloccur. Since thecoordinates actually used willbefunctions oftheignorable one(among others), in thecorresponding normal mode, several orallofthecoordinates may exhibit constant velocities. These may befound bysubstituting inthe equations ofmotion (12-103). The case ofdegeneracy, when amultiple root forp2occurs, ismore complicated forEqs. (12-103) than forEqs. (12-20), where theforce isderivable from apotential energy depending only onx1,...,xN. We cannolonger make useofthediagonalization theory forasymmetric tensor toshow thatforamultiple rootp2,Eqs. (12-105) have acorre- sponding multiplicity ofindependent solutions, aswedidfortheanalogous equations (12-23) or(12-33). Sometimes Eqs. (12-105) may have only oneindependent solution, even when p2isamultiple root ofEq.(12-106), andwemust look forother forms ofsolution than (12-104). Wewillnot carry out the algebraic details here,* but the result isthat when Eqs. (12-105) donotyield enough independent solutions forX1,,Pkfor amultiple root p2,X1,,P),should bereplaced inEqs. (12-104) bypoly- nomials intofdegree (n—1),where nisthemultiplicity oftheroot p2. The resulting expressions must besubstituted inEqs. (12-103), which then give 2Nn relations between the2Nn coeﬁicients inthe2Npoly- nomials. These relations canbeshown toleave justnarbitrary coeﬁicients, sothat thecorrect number ofarbitrary constants areavailable. Alter- natively, wecanslightly alter thecoefficients intheHamiltonian (12-101) sothat thedegeneracy inp2isremoved, ﬁndthesolution, andthen ﬁnd itslimiting form asthecoefficients approach their original values. (See *Forafurther discussion ofproblems ofsmall vibrations, seeE.J.Routh, Dynamics ofaSystem ofRigid Bodies, Advanced Part. New York: Dover Pub- lications, 1955. Chapter 6.Alesscomplete butmore elegant treatment using matrix methods isgiven inR.Bellman, Stability Theory ofDiﬁerential Equations, New York: McGraw-Hill, 1953. 496 THEORY orSMALL VIBRATIONS [CHAP. 12 Problem 24,Chapter 2.)When powers oftappear inthesolution, itis clear that thesolution isnotstable even when p2isrealandnegative, but represents anoscillation whose amplitude after along time willincrease assome power oft.Hence degeneracy generally implies instability inthe caseofEqs. (12-103). Itwillbefound that multiple roots ofEq.(12-106) ordinarily mark theboundary between real and complex solutions for p2inthesense that asmall change insome coefﬁcient ah),blot;orck; willsplit thedegeneracy andlead ontheonehand totworeal, oronthe other hand totwocomplex, roots forp2,depending onthesense ofthe change. Thesituation isclosely analogous mathematically totheproblem ofthedamped harmonic oscillator, where adouble rootforpinEq.(2—125) marks thedividing linebetween theoverdamped andunderdamped cases andleads tosolutions linear int.Inthepresent case, there isnodamp- ing; Eq. (12-106) contains only even powers ofp,andifcomplex con- jugate roots forp2occur, then thecorresponding four roots phave the form :l:'Y:1;iw,and some ofthesolutions grow exponentially. Thus multiple roots canmark theboundary between stable andunstable cases. Inthecase ofEqs. (12-20), where theforces arenotvelocity-dependent, theboundary between stability and instability occurs only when some root forp2iszero; degenerate negative roots forp2always correspond to stable solutions. Weshould further remark that even when theroots p2areallnegative anddistinct, sothat thesolutions ofEqs. (12-103) areallstable, we cannot guarantee that theexact solutions ofthenonlinear equations aris- ingfrom thecomplete Hamiltonian (12-97) arestable. Forvibrations about anequilibrium point when theforces arederivable from apotential energy, wewere able toprove that stable solutions ofthelinearized equa- tions areobtained only around aipotential minimum, and that inthat case, thesolutions areabsolutely stable iftheamplitude issmall enough. Wesawabove that iftheHamiltonian Hispositive (ornegative) deﬁnite near thesteady motion, then wecanalso show that thesolutions are stable. But thesolutions ofEqs. (12-103) may allbestable even when Hisnotpositive ornegative deﬁnite. Inthat case, allwecansayis that, foraslong atime asweplease, ifwestart with sufﬁciently small amplitudes, thesolutions oftheexact problem given bytheHamiltonian (12-97) will‘approximate those ofthelinearized problem given bythe Hamiltonian (12-101). This istrue because thenonlinear terms can be made assmall aswelike bymaking theamplitude sufficiently small, andthen their effect onthemotion canbeappreciable, ifatall,only if they areintegrated over avery long time. Nevertheless, ifweneglect thenonlinear terms, weareprevented from asserting complete stability foralltime. Cases areindeed known where thelinearized solutions are stable, and yetnomatter how near thesteady-state motion webegin, 12-7] BETATRON OSCILLATIONS INANACCELERATOR ,497 theexact solution eventually deviates from thesteady-state motion by alarge amount. Toﬁnd thecriteria that determine ultimate stability inthegeneral case isperhaps theoutstanding unsolved problem of classical mechanics. 12-7 Betatron oscillations inanaccelerator. Inacircular particle ac- celerator, forexample acyclotron, betatron, orsynchrotron, charged particles revolve inamagnetic guide ﬁeld which holds them within a circular vacuum chamber asthey areaccelerated. Since theparticles revolve many times while they arebeing accelerated, itisessential that theorbits bestable. Since theparticle gains only asmall energy incre- ment ateach revolution, itispermissible tostudy ﬁrst thestability of theorbits atconstant energy, and then toconsider separately theac- celeration process itself. We will beconcerned here only with the stability problem atconstant energy E.Letusassume that themag- netic ﬁeld issymmetrical about avertical axis, sothat wemay write, using cylindrical polar coordinates (Fig. 3-22), B(p; ‘P:Z)=Bz(pr z)k +Br(p: Themagnetic ﬁeldinasynchrotron orbetatron isalsoafunction oftime, increasing astheenergy increases, butsince wearetreating Easconstant, wealso take Basconstant. Wewillsuppose that inthemedian plane z=0,theﬁeld isentirely vertical: B(p: ‘P1 = Aparticle ofappropriate energy Emay travel inacircle ofconstant radius p=a(E). Wecallthisorbit theequilibrium orbit. Weareinterested in thestability ofthis orbit; that is,wewant toknow whether particles near this orbit execute small vibrations about it.Such vibrations are called betatrzm oscillations because thetheory wasﬁrstworked outforthe betatron. Thevector potential (Section 9-8) foramagnetic ﬁeld with symmetry about thez-axis canbetaken tobeentirely inthe<p-direction: A=A,,(p, z)m, (12-113) sothat 6 6 m0 k8 BA=55(pA,,,) —117"-* (12-114) 498 THEORY orSMALL VIBRATIONS [cn.u>. 12 Weseefrom Eq.(12-114) that Aisgiven by _1 PA..<p.Z)=p/0pB.<p. 2)dp. <12-115) since A,,must vanish atp=0because oftheambiguity inthedirection ofm.Note that 21rpA,, isthemagnetic ﬂuxthrough acircle ofradius p. The Lagrangian function isgiven byEqs. (9—154) and(9-166): L-211012+P2¢2+22>+§p¢A.</1,2). <12-no where e,marethecharge andmass oftheparticle tobeaccelerated. If thevelocity oftheparticle iscomparable with thespeed oflight, that is, ifthekinetic energy iscomparable with orgreater than mcz, therelativistic form fortheLagrangian should beused (Problem 23,Chapter 9).The momenta are __ all — m 'P»-6,,—P, p.-we+§P111» <12-117) p,=m2. TheHamiltonian function isgiven byEq.(9-196) or(9—200): ~ 2 2 _ A2 H=5:,+5;,+lp”252;?” *1- (12-118) Weseethat zpisignorable andp.,may betaken tobeagiven constant. The Hamiltonian function then hastheform H=‘T’+‘V’, (12-119)with 2 2 1; z 1 - -2T=7%l*,ni =5m(p2+2), (12-120) ‘V’= . (12_121) The problem reduces toanequivalent problem ofstatic equilibrium. Thesteady motions aregiven bythesolutions forp,2oftheequations aw" .I=§p¢B. =0, (12-122) 6‘V’ .. W =—p‘P (m¢ +2Ba) =0: 12-7] BETATRON OSCILLATIONS INANACCELERATOR 499 where wehave used Eqs. (12-117) and (12-114). The ﬁrst equation above issatisﬁed inthemedian plane z=0,and usually nowhere else (p¢;é0).Thesecond equation gives ¢=-;n‘-’;B.0(p>. (12-124) which isequivalent toEq. (3-299). Wemay solve Eq.(12-124) forp, given qb,oralternatively, for(Z:with p=a,theradius oftheequilibrium orbit. Note that theenergy -§_<'ma2¢2 ofaparticle executing thesteady motion islessthan thetotal energy Hoftheparticle whose motion weare studying, butthedifference isofsecond order insmall quantities for vibrations near thesteady motion. According tothedevelopment inthe preceding section, thetwoparticles should bechosen tohave thesame pa. Wenow set p=a+:0, (12-125) where aistheradius oftheorbit forthesteady motion. Next, byexpand- inginpowers ofre,z,at,2,weobtain ﬂ"=%m@P+¥) uamm q‘V’=§mw2(1 —n)x2 +1}mw2nz2, (12-127) where wehave set . B.@=¢=-iﬁ@, (mna a6B,),=— -—-—— 1 12-129 n <B= ap ==0.n=a ( ) andhave used Eqs. (12-114), (12-117), (12-124), and 8B 8B_1=._1>, _ ap 62 (12 130) which follows from thefactthat VXB=0, (12-131) I aswecanseefrom Eq.(9—162). The quantity niscalled theﬁeld index. Weseeimmediately that themotion isstable only if O<n<1. (12-132) Inacyclotron, the‘ﬁeld isnearly constant atthecenter, sothat n<<1, andthen falls rapidly near theoutside edge ofthemagnet. Inabetatron orsynchrotron, themagnetic ﬁeld hasaconstant value ofnandincreases 500 THEORY OFSMALL VIBRATIONS [crnun 12 inmagnitude astheparticles areaccelerated soastokeep aconstant. We seefrom Eq.(12-129) that thevalue ofndoes notchange asB,isincreased provided theshape ofthemagnetic ﬁeld asafunction ofradius does notchange, that is,provided BB,/62 increases inproportion toB,. Since thevariables x,zareseparated in‘T’and ‘V’,wecanimmedi- ately write down thebetatron oscillation frequencies. Itisconvenient toexpress them interms ofthenumbers ofbetatron oscillations per revolution, 1/,and11,: V1=(£5= “Tn)1/21(12-133) (-92"=2;=n1/2_ Ifthere areimperfections intheaccelerator, sothat B,isnotinde- pendent of(0,thedifference between B,anditsaverage value gives rise toaperiodic force acting onthecoordinate x.Theresulting perturbation oftheorbit canbetreated bysolving thecorresponding forced harmonic oscillator equations. IfB,isnotzero everywhere inthemedian plane (z=0),vertical forces actwhich drive thevertical betatron oscillations. Ingeneral, such imperfections alsolead tovariations intheﬁeld index n, sothatn=n(¢), andnforthesteady motion becomes aperiodic func- tion oftime. Inalternating gradient accelerators, theﬁeld index nis deliberately made tovary periodically inazimuth (0.Thesolution of thisproblem istoocomplex forinclusion here. 12-8 Stability ofLagrange’s three bodies. Aparticular solution of theproblem ofthree bodies moving under their mutual gravitational attractions wasdiscovered byLagrange. This solution isasteady motion inwhich thethree masses remain atthecorners ofanequilateral triangle asthey revolve around their common center ofmass. Wewish toin- vestigate thestability ofthissteady motion. This problem isanexample ofarather general class ofproblems incelestial mechanics concerned with thestability ofparticular solutions oftheequations ofmotion. When theparticular solution isasteady motion, theproblem canbe treated bythemethod ofSection 12-6. Wewillsimplify theproblem byconsidering only motions conﬁned to asingle plane. There arethen sixcoordinates, two foreach particle. A little study shows that there arethree ignorable coordinates. Two ofthem represent rigid translations ofthethree particles, andmay betaken as thecartesian coordinates ofthecenter ofmass. The corresponding con- stant momenta arethecomponents ofthetotal linear momentum. The third ignorable coordinate will represent arigid rotation ofthethree particles about thecenter ofmass. The corresponding constant mo- 12-8] STABILITY orLAGRANGE’S THREE BODIES 501 mentum isthetotal angular momentum. The remaining three non- ignorable coordinates willspecify therelative positions ofthethree parti- cleswith respect toeach other. These must beconstant inasteady motion; therefore any steady motion must bearigid translation and rotation ofthesystem ofthree bodies. Wemay, forexample, choose thecoordinates asinFig. 12-1. Here, Xand Yarecoordinates ofthe center ofmass, variation ofawith theremaining coordinates held ﬁxed represents arotation oftheentire system about thecenter ofmass, and rl,r2,6determine theshape andsizeofthetriangle formed bythemasses ml,m2,m3. Weexpect toﬁndthree normal modes ofvibration ofrl,r2,0 about their steady values. Ifweshould happen tooverlook anyignorable coordinate, wewill ﬁnd only those steady motions inwhich that co- ordinate isconstant. The ignorable coordinate willthen reveal itself as azero root (p2=0)ofthesecular equation (12-106). According toEq.(4-127), thekinetic energy willseparate into apart depending onXand Y,andapart depending onrl,r2,0,anda.The potential energy depends only onrl,1'2,and0.The coordinates X,Yare therefore orthogonal torl,r2,0,and oz,andthecenter-of-mass motion separates outoftheproblem. Thecenter-of-mass energy T......=+lM<X’ +Y2)=$4(pf:+pt),.M=ml+m2+ma, (12-134) isconstant, andmay beomitted from theHamiltonian. The center of mass moves with constant velocity, and wemay study separately the motion relative tothecenter ofmass. The ignorable coordinate ozis evidently notorthogonal to0,since theangular velocity ofmlinvolves "11 X T1 ma 0.111.5” i *1 ms Y FIG. 12-1. Coordinates forthethree-body problem. 502 THEORY orSMALL VIBRATIONS [c11».1>. 12 (9—|-11)andthisappears squared inthekinetic energy. This isnotan accidental result ofourchoice ofcoordinates, butaninherent consequence ofthefactthat rotation ofthesystem asawhole inﬂuences the“internal” motion described byrl,r2,0. Itwould beastraightforward algebraic exercise tosetupthekinetic energy interms ofrl,r2,6,oz,ﬁndthemomenta p,lp,2, pl,pa,setupthe Hamiltonian, ﬁnd thesteady motions, andcarry through theprocedure ofSection 12-6 toﬁndthesecular equation (12-106), which would bea third-order equation inp2whose roots determine thecharacter ofsmall deviations from steady motion. This procedure, however, isextremely tedious, asthereader may verify. Itturns out, interestingly enough, that alesslaborious way ofﬁnding theactual solution istoabandon the Hamilton-Lagrange formalism, and tosetuptheequations ofmotion from ﬁrst principles. Wewillstillneed theresults oftheabove general considerations asaguide tothesolution. The algebra isstillsufficiently involved sothat there isarather high probability ofalgebraic mistakes. Itistherefore desirable toreplace 0,abythecoordinates al,0:2shown in Fig. 12-2, soastointroduce analgebraic symmetry between particles mlandm2. This reduces theamount ofalgebra needed andprovides a check ontheresults, inthat ourformulas must exhibit theproper sym- metry between subscripts “1”and“2”. Neither alnora2isignorable now, and weseethat theignorable coordinate nolonger appears explicitly. Wecould notmake useoftheignorable property anyway, since weare notgoing towrite theequations inHamiltonian form. Oursecular equa- tion willturn outtobeoffourth order inp2,butweknow that oneroot willbep2=0,and canbefactored out. Weshow also inFig. 12-2 several auxiliary variables r3,01,02,03which willbeneeded. Wewillwrite theequations ofmotion ofmlinterms ofcomponents directed radially away from m3and perpendicular totheradius rl.In applying Newton’s laws ofmotion directly, wemust refer allaccelera- tions toacoordinate system atrest. Theacceleration ofmlisitsacceler- ation relative tom3plus theacceleration ofm3. The latter acceleration canbefound byapplying Newton’s lawofmotion tom3,which isat- ml Ta '1 TIL T1 02> 2 I11 T2 34% ma Fro. 12-2. Alternative coordinates forthethree-body problem. 12-8] STABILITY OFLAGRANGE’s THREE BODIES 503 tracted bymlandm2. Theforce onmlisthegravitational attraction of m2andm3. Wehave therefore, intheradial direction, .. .2 m1G WLQG mlm G mmGm1(T1—T1d1+i2+i2cos9a =—W£_—%cos01- T1 7'2 7'1 7'3 (12-135) The corresponding equation forthemotion ofml,perpendicular torl,is ml <T1&1 +2’!"1&1 '_ ' sin 03) Z — Lmfg S111 01. T2 Ts Two similar equations canbewritten forthemotion ofm2. The four equations canbeconveniently rewritten intheform: 2 (ml+m3)G +m2G m2G' l:1*7'1él1'l“'__ii‘ "iCOS03+_i‘COS61=0,2 2 2T1 7'2 T3 F2-r2&§+ +@¥cos03+#cos02 =0, T2 T1 TsG G (12-137) T1121 +2i‘1é¢1 '—12% Sill 03+m% S111 61=0, 7'2 1'3 .. .. mG. mG.r2a2+2r2a2+-#2—s1n03 —-%_"SlI192= O. T1 T3 The algebraic symmetry between subscripts “I”and“2”isexhibited in theabove equations. (Note that 03=al—042andchanges, sign ifwe interchange thetwoparticles.) The auxiliary variables 0l,02,03,13 may beexpressed interms ofrl,r2,al,0:2byusing thesineandcosine laws forthetriangle. Wenow seewhy itiseasier touseNewton’s laws directly here. We canexpress theacceleration ofm3very simply interms ofthegravitational forces onm3. IntheLagrangian formulation ofthecorresponding equa- tions forrl,r2,ozl,0:2(oroz,0),theterms which represent theacceleration ofm3have tobeexpressed kinematically, i.e.,interms ofthecoordinates, velocities, and accelerations ofmlandm2,because they arederived by differentiation ofthekinetic energy Tingeneralized coordinates. This is very complicated, and involves explicitly theposition ofthecenter of mass relative toml,m2,m3,which wedonotneed inthepresent formula- tion. The resulting equations areequivalent toEqs. (12-137), butcon- siderably more complicated inform. One reason forthesimplicity of Eqs. (12-137) is,ofcourse, theuseoftheauxiliary variables 0l,02,03,1'3; 504 THEORY orSMALL VVIBRATIONS [CHAP. 12 intheLagrangian formulation, anysuch auxiliary variables would have tobedifferentiated with respect torl,r2,ozl,0:2inorder towrite down theequations ofmotion. Weﬁrst look forsteady motions. Weknow from ourpreliminary dis- cussion that asteady motion canonly bearigid rotation about thecenter ofmass (plus auniform translation). Wetherefore take rl,r2,r3,0l,02,03 tobeconstant, andset 0:2=wt, al=wt+03. (12-138) Ifwesubstitute intothelastofEqs. (12-137), weobtain -Esin03=-15sin02. (12-139) 1'1 ‘ Ts From thelawofsines, T1_ Ta _ sin02—sin03’ (12140) wethen have, unless sin03=sin02=0, r?=T3. (12-141) Inthesame way, from thethird ofEqs. (12-137), weﬁndr2=r3,unless sin03=sinBl=O.Wemay therefore set 1'1='I'2=T3=a, 01:02=03= (12-142) The only possible steady motion, unless themasses lieinastraight line, isoneinwhich thethree masses lieatthecorners ofanequilateral triangle. Wemust still verify that theﬁrst two ofEqs. (12-137) aresatisﬁed. This isthecase if MG<02=-F, (12-143) where Misthetotal mass. This istheparticular solution ofthethree- body problem found byLagrange. The case when thethree masses lie inastraight lineisleftasanexercise. Wenow seek solutions formotions near thesteady motion. Letusset T1 Z a+-T1; T2 Z a+$21 (¥1=¢0t+%7T+€1, 0£2=(.0t-I-G2, where xl,2:2,el,e2arefour new independent variables which wewill 12-8] STABILITY OFLAGRANGE/S THREE BODIES L505 regard assmall. Wesubstitute inEqs. (12-137), retaining only linear terms. Weﬁrstcalculate 03=- 61"—G2, and, toﬁrstorder, from thelawofcosines, 7%=112[1+”%:-Q +\/§(el—@2)]- (12-146) Now, from thelawofsines, sinal=1-:sin03; (12-147) hence, toﬁrstorder, 1 1 —- 01=§—§(€1 —£2) (12-148) and, similarly, 1 1 -02=1'--01--2)+-l/51 <12-149)3 2 2 a Wenote asacheck that Bl+02+03=1r.Weareready tosubstitute inEqs. (12-137), which, toﬁrstorder, become 22m1 -1'Qms —IitmzI21-'-2a(.0é1 -'(0) 9mG' 3\/3mG—%%w2—W7L(@1_62)=0, _1,,;2_2,,,,é2_ (w2+ G)x2 9m1G 77!/1G ’ -$5-"i'I1—‘j,T-(61-'62)=0) 1131+2w531 —5%’? ($1—I2)—% (51'—52)=0, “E2-l"201532 —ﬁle, ($1—$2)+ (611—'52)=O-4a3 4a2 (12-150) Note thatthesecond andfo1n'th equations maybeobtained from theﬁrst andthird byinterchanging thesubscripts “1”and"2"andreversing the signs ofal,0:2,andw;this symmetry follows from thechoice ofcoordi- nates inFig. 12-2. Anormal mode istofound bysetting i $1=X161“, $2=X261, , G1=E16pt, G2='-Egept. I 506 THEORY OFSMALL VIBRATIONS [CHAP. 12 Forconvenience Weset p=(G/a3)"2P. (12-152) Substituting Eqs. (12—151) inEqs. (12-150) and using Eq. (12-143), weobtain X(P2—3M+gm2)-}£l—%m272 -(¥ mg+2M1'”P) E1+%-§mgE2=0, -%m1%+(P2-3M+§m1)€?--3l[-ME, V +(¥m,-2M1'”P) E2=0, -I-(P2 -'2m2)E1+gm2E2= 0; 3\/5 X1 1/2 3\/5 X2"Tm17+(2M P+T“‘ 7 +gm1E1 +(P2-Zm1)E2 =0. (12-153) Thesecular determinant is (P2 —3M+2mg) (—2mg) —(2M‘/ZP +%;' mg) mg) (—2ml) (Pa —3M+2011) — ml) —(2M1”P —%L§ ml)ew <~+~> (re — ml) (2M'/2P +% m1) ml) (P2 —2m1)=0. (12-154) Weknow from previous considerations that thismust beafourth-degree equation inP2,oneroot ofwhich isP2=0.This fact, together with the symmetries inthearray ofcoefficients, encourages ustotrytomanipulate theabove determinant tosimplify itsexpansion and tobring outex- plicitly thefactor P2. Weaddthesecond column totheﬁrst, andthe third tothefourth, then subtract theﬁrst rowfrom thesecond, andthe third from thefourth: 12-8] STABILITY OFLAGRANGE'S THREE BODIES 507 (P2-3M) (-2111,) -(2M‘”P +¥111,) —(2M”2P) 0 [P2-3M+2(m,+mo][2M'”P -%§(m,-1113)] 0 =0 (2M‘”P) (¥m,) (P2-21",) P“ 0 [2M””P +$0111 -mp] -[P2-2(m,+ma] 0 ‘ (12-155) Wefactor Pfrom thelast column, then multiply thelast column by 3M1/2/2,andsubtract from theﬁrst column. Wecanthen factor Pfrom theﬁrstcolumn, toobtain P (-%1",) -(2M‘”P +¥'11,) -(2M"’) 9 3\/§P, 0 [P2-3M+Z(m,+my] [2M"”P -T(m,-1112)] 0 =0'cwem <»~a> P o[2M"’P +$(m,-mo] -[P-2(m,+ma] o (12-156) Thefactor P2isnowinevidence. Tosimplify theexpansion ofthede- terminant, wemultiply thethird row by2PM "U2and subtract from theﬁrstrow: 0 * * -(2M‘” +2P’M'””) 20 [P2-3M+9(m,+1112)] [2M‘”P -¥(M1-m;|)] 0P4 ‘£0 11/2 2M U if 8 0[2M"’P +¥(m,-mo] -[P-2(m,+mﬂ] 0 (12-157) The stars indicate terms that wedonotwrite here, since they willnot appear intheresult. Wecannowexpand inminors oftheﬁrst column, and expand theresulting three-rowed determinant inminors ofthelast column, with theﬁnal result: P2(M +P2)[P4 +MP2+a1(m1m2 +'m2m3 +m3m1)] =0.(12-158) Note, asacheck onthealgebra, that thethree masses enter symmetrically 508 THEORY OFSMALL VIBRATIONS [CHAP. 12 inthisequation, asthey must. Itisafortunate accident that anaddi- tional factor appears explicitly, sothat wehave only tosolve aquadratic equation forP2. Wehave, ﬁnally, four roots: 2___ 2___ _P‘O’ P_M’ (12-159) P2=_%M =|=%lM2 "27(m1m2 +mzms +msm1)l1/2- Asweknow, thezero root results from thefact that there isanaddi- tional ignorable coordinate. The root P2=——M yields astable oscilla- tory mode. The lasttwo roots forP2willboth bereal and negative provided that (mi—|-m2+ms)2 >27(m1m2 —|-mzma —|-msm1)- (12-160) Ifthisinequality isreversed, thelasttworoots arecomplex, andwehave four complex values ofP.Ofthese roots twogiverisetodamped andtwo toantidamped oscillatory solutions. Intheintermediate case when the two members oftheinequality (12-160) areequal, itcanbeshown that theamplitude ofoscillation grows linearly intime. Hence theLa- grangian motion ofthree bodies isunstable when condition (12—160) is notsatisﬁed. Ifoneofthebodies, saym1,ismuch smaller than theother two, wehave therestricted problem ofthree bodies studied inSection 7-6,andthecondition forstability reduces to _ (mg+m3)2>27m2m3, (12-161) or,ifm3isthelargest, m3>24.96m2. (12—162) Ifmgisthesunandm2istheplanet Jupiter, this condition issatisﬁed; hence, ifweneglect theeffects ofallother planets, there arestable steady motions inwhich asmall body revolves around thesunwith thesame period asJupiter andatthecorner ofanequilateral triangle relative to thesunandJupiter. (There aretwosuch positions.) The Trojan aster- oidsareagroup ofbodies with thesame period asJupiter which appear tobeinthisposition. Since Eq.(12-161) isalsosatisﬁed bytheearth- moon system, thecorresponding steady motion ofanartiﬁcial satellite intheearth-moon system isstable. Consideration ofmotions perpendic- ulartotheplane ofsteady motion does notalter these conclusions. How- ever, inview oftheremarks attheendofSection 12-6, ourconclusions about thestability arevalid only forlimited periods oftime. Weleave asanexercise thesolution ofEqs. (12-153), todetermine theratios ofthevariables and hence theoscillation pattern foreach normal mode (see Problem 30). Insolving Eqs. (12—-153), itmay be PROBLEMS ~ 509 helpful tosubject them tothesame series ofmanipulations which led from thedeterminant (12-154) tothedeterminant (12-157). Note that adding onerowofthedeterminant toanother corresponds toadding the corresponding equations. Adding two columns corresponds togrouping thecorresponding variables, that is,tointroducing anew variable which isalinear combination ofthetwooriginal ones. PROBLEMS 1.Find thetransformation tonormal coordinates forthetwocoupled oscil- lators shown inFig.4-10. 2.Solve Problem 26,Chapter 4,bytransforming from $1,rmtonormal co- ordinates bythemethod ofSection 12-3. 3.Amass mmoving inspace issubject toaforce whose potential energy is V=V0exp[(5222 +5y2—|-822—-8yz—26;/a —82a)/a2], where theconstants V0and aarepositive. Show that Vhasoneminimum point. Find thenormal frequencies ofvibration about theminimum. 4.Amass mishung from aﬁxed support byaspring ofconstant kwhose relaxed length isl=2mg/Ic. Asecond equal mass ishung from theﬁrstmass byanidentical spring. Find thesixnormal coordinates andthecorresponding frequencies forsmall vibrations ofthis system from itsequilibrium position. Each spring exerts aforce only along thelinejoining itstwoends, butmay pivot freely inanydirection atitsends. 5.Anionofmass m,charge q,isheldbyalinear attractive force F=—kr toapoint A,where risthedistance from theiontothepoint A.Anidentical ionissimilarly bound toasecond point Badistance lfrom A.The twoions move (inthree-dimensional space) under theaction ofthese forces and their mutual electrostatic repulsion. Find thenormal modes ofvibration, andwrite down themost general solution forsmall vibrations about theequilibrium point. 6.The mass mginFig. 4-10 issubject toaforce F2=Bsinwt.The sys- temisatrestatt=O.Find themotion bythemethod ofnormal coordinates, using theresult ofProblem 1. 7.ThepairofionsinProblem 5issubject toaplane polarized electromagnetic wave incident perpendicular totheline E whose electric ﬁeld E0coswtis directed at45°tothelineE. Find thesteady-state motion. 8.Themass inProblem 3issubject toaforce F,=F,=F,=Be_'“. Find aparticular solution. 9.Set upthe tensors M,B,KforProblem 25, Chapter 4,show that all three canbesimultaneously diagonalized, andsolve theproblem bythemethod ofnormal coordinates. 10.The masses m1and mginFig. 4-10 aresubject tofrictional forces —'Ym1a':1, —’Ym2:t2, respectively. Find thegeneral solution. 510 THEORY orSMALL VIBRATIONS [CHAP. 12 ll.Assume that V°(ac1, ...,xf)hasaminimum at:01=---=av;=0,and that V’(x1,...,xf)issmall, butthat Eq. (12-66) does notnecessarily hold. Find approximate expressions toﬁrstorder inV’anditsderivatives atx1=--- =ac,»=0,forthecoordinates 1:9,...,:v?ofthenew equilibrium point for V=V0+~V’.Iftheexpansion ofV0about :01.=---=2:;=0isgiven by Eq.(12-67) (plus higher-order terms), andifthequadratic terms intheexpan- sion ofVaretobe v=Z‘,e<K2.+Kiri)?/kl/I; 1=,z e where yk=ark—mg,ﬁnd approximate ﬁrst order expressions forthecoeffi- cients K,:,. 12.Find thesecond-order approximations forthecoefﬁcients C15,C1’,which aregiven toﬁrst order byEqs. (12-80) and (12-83). 13.Find thethird-order approximation tothefrequency correction given to second order byEq.(12-91). 14.Formulate theequations tobesolved toobtain aﬁrs't—order approxima- tion inthecase when conditions (12-87) failforagroup offour nearby modes, i.e.thecase ofapproximate degeneracy. 15.Atriple pendulum isformed bysuspending amass Mbyastring of length lfrom aﬁxed support. Amass mishung from Mbyastring oflength Z, andfrom thissecond mass athird mass mishung byathird string oflength l. Themasses swing inasingle vertical plane. Setuptheequations forsmall vibrations ofthesystem, using ascoordinates theangles 01,02,03made byeach string with thevertical. Show thatifM>>>m,thenormal coordinates canbe found ifterms oforder (m/M)1/2 areneglected. Find theapproximate normal frequencies toorder m/M. [Hint: Transform Ktoaconstant tensor, anddiag- onalize M.] 16.InFig. 12-3, thefour masses move only along ahorizontal straight line under theaction offour identical springs ofconstant k,andaweak spring of constant k’<<Ic.Find, toﬁrst order ink’,anapproximate solution forthe normal modes ofvibration. 17.Find theapproximate solution toProblem 16forthecase when the masses areallequal. Does theapproximate result suggest away tosolve the problem exactly? 18.Auniform ellipsoid ofrevolution ofmass M,whose axis ofsymmetry is twothirds aslong asitsequatorial diameter, ismodiﬁed byplacing masses m, 2m,3m,m,2m,3minsequence around itsequator atpoints 60°apart. Two masses, each 4m,areplaced atopposite ends ofadiameter, making anangle of 45°with theaxis and atthelongitude ofthemasses m.Ifm<<M,ﬁnd, to ﬁrst order inm/M, thenew principal axes. [Hint: The perturbation procedure 7 monmoom /cmk k’ k It 1 "'2 ma m4 % FIG. 12-3. Four coupled harmonic oscillators. PROBLEMS 511 developed inSection 12-5 fordiagonalizing thetensor Wmay beapplied to diagonalize approximately anysymmetric tensor |°—I—I’iftheeigenvectors ofl° areknown andI’issmall.] 19.Apply theperturbation method totheproblem ofthestring with variable density considered inthelastparagraph ofSection 9-9, assuming that a<<0'0. Find thelowest normal frequency tosecond order ina,andwrite outthecor- responding solution u(x,t)toﬁrst order ina. 20.Formulate aﬁrst-order perturbation method ofsolving Eqs. (12-60), treating thefriction asasmall perturbation, andassuming thesolution without friction isalready known. Show why, even inﬁrst order, onecannot introduce normal coordinates which include theeffects offriction. 21.Two charges +Ze are.located atﬁxed points adistance 2aapart. An electron ofmass m,charge —e,moves intheﬁeld ofthese charges. Find the steady motions andthesmall vibrations about thesteady motions. *22. Two charges +Ze and —Ze arelocated attheﬁxed points z=aand z=—a. Anelectron ofmass m,charge -—e,moves intheﬁeld ofthese charges. Sketch agraph ofzvs.r,where risthedistance from thez-axis, showing the values of2,rforwhich there aresteady motions. Investigate thestability of these steady motions. q 23.Amass mslides without friction onasmooth horizontal table. Itistied toaweightless string oftotal length lwhich passes through ahole inthetable andistied atitslower endtoamass Mwhich hangs below thetable. Setup theHamiltonian function using ascoordinates thepolar coordinates r,ozofthe mass mrelative tothehole, andthespherical angles 0,goofthemass Mrelative tothehole. Find thesteady motions andthenormal frequencies ofsmall vibra- tions about asteady motion. 24.Assume that thethree masses inFig.’4—16 arefreetomove inaplane but areconstrained toremain inastraight line relative tooneanother. Choose your coordinates sothat asmany aspossible willbeignorable, ﬁnd thesteady motions, and ﬁnd thenormal modes ofvibration about them. 25.Asymmetrical rigid body ismounted inweightless, frictionless gimbal rings. Ahairspring isattached tooneoftherings soastoexert arestoring torque —k¢ about thez-axis, where ¢istheEuler angle. Find thesteady mo- tions andinvestigate thecharacter ofsmall vibrations about them. 26.InProblem 13,Chapter 11,ahairspring isconnected between thedisk axle andtherings which exerts arestoring torque —k1I/’ ,where 1]/’istherela- tive angle ofrotation between disk andrings. The “gyroscope” moves freely inspace with noexternal forces. Find thesteady motions andinvestigate the small vibrations about them. . 27.Two masses mareconnected byarigid weightless rodoflength 2l.One mass isconnected with theorigin byaspring ofconstant Ic,theother bya spring ofconstant 2k.The relaxed length ofboth springs iszero. The masses move inasingle plane. Choose ascoordinates thepolar coordinates r,0ofthe center ofmass relative totheorigin, andtheangle ozwhich therodmakes with theradius from theorigin tothecenter ofmass, taking a=0when thestronger spring isstretched least. Find thesteady motions and theconditions under which they arestable. 512 THEORY orSMALL VIBRATIONS [CHAP. 12 *28. InProblem 23,asecond mass mslides without friction onthetable, and isconnected totheﬁrst mass byarigid weightless rodoflength a.(Assume thearrangement isingeniously contrived sothat therodand string donot become entangled.) Assume also that M=2m. Useasanadditional coordi- nate theangle Bbetween therodandthestring. Find thesteady motions, and determine which arestable. Find thenormal vibrations about thestable steady motions. 29.Thevector potential duetoamagnetic dipole ofmagnetic moment /4is, inspherical coordinates relative tothedipole axis, nsin0 A=TM m. ' Find thesteady motions foracharged particle moving insuch aﬁeld, and show that they areunstable. 30.Find thesolution ofEqs. (12-153) forX1,X2,E1,E2,when P2=—M, anddescribe thecorresponding oscillation. [See thehint inthelastparagraph ofSection 12-8.] *31. Analyze thecase which was omitted inSection 12-8 when thethree bodies m1,m2,m3lieinastraight line. Show that there arethree possible steady motions, oneforeach mass lying between theother two. [Hint: You willneed Descartes’ ruleofsigns.] Show that motions near each ofthese steady motions areunstable. Compare your results with those-of Section 7-6and Problem 17ofChapter 7. u 32.Find thesolution ofEqs. (12-153) forthedouble root P2=—§M, when theinequality (12-160) becomes anequality. Show that inthis case, Eqs. (12-150) have asecond solution inwhich X1,X2,E1,E2arecertain linear functions oft,sayX1=X1—|-X1’t, etc. (You cansimplify thealgebra alittle byassuming that oneoftheadditive constants, sayX1’,iszero. This isallow- able, since Xfcanalways bemade zero bysubtracting from thesecond solution asuitable multiple oftheﬁrst solution you found inwhich X1isconstant. The linearity oftheequations permits linear superposition ofsolutions.) 33.Find thesolution ofEqs. (12-153) forX1,X2,E1,E2,for_therootP2=0, and show that itcorresponds toanew steady motion near thechosen one. Since your solution hasonly onearbitrary constant, there must beanother solution ofEqs. (12-150) corresponding toP2=0.Guess itsform, andverify bysubstitution. *34. Show that ifmotions ofLagrange’s three bodies outoftheplane ofthe steady motion areconsidered, atleast oneofthethree additional coordinates isignorable. Choose astwononignorable coordinates thedistances q1=21—23 and q2=.22-.23,where 24istheperpendicular distance ofmifrom theplane ofsteady motion. Setupthelinearized equations ofmotion bythemethod used inSection 12-8. Solve forthecorresponding normal vibrations andshow that theresult canbeinterpreted ascorresponding simply toasmall change in theorientation oftheplane ofsteady motion. BIBLIOGRAPHY 515 BIBLIOGRAPHY The following isalist,bynomeans complete, ofbooks related tothe subject matter ofthistext which thereader may ﬁndhelpful. ELEMENTARY MECHANICS TExTs 1.CAMPBELL, J.W., AnIntroduction toMechanics. New York: Pitman, 1947. 2.MILLIKAN, R.A.,RQLLER, D.,AND WATSON, E.C.,Mechanics, Molecular Physics, Heat, andSound. Boston: Ginn andCo., 1937. INTERMEDIATE MECHANICS TExTs 3.BECKER, R.A.,Introduction toTheoretical Mechanics. New York: McGraw-Hill, 1954. 4.LINDSAY, ROBERT Bnocn, Physical Mechanics, 2nd ed. New York: D.Van Nostrand, 1950. 5.MACMILLAN, WILLIAM D.,Theoretical Mechanics. New York: McGraw- Hill. Vol. 1:Statics andDynamics ofaParticle, 1927. Vol. 3:Dynamics ofRigid Bodies, 1936. I 6.Oscoon, WILLIAM F.,Mechanics. New York: Macmillan Co.,1937. 7.ScoTT, MERIT, Mechanics, Statics and Dynamics. New York: McGraw- Hill, 1949. 8.STEPHENSON, REGINALD J.,Mechanics andProperties ofMatter. New York: John Wiley &Sons, 1952.‘ 9.SYNGE, JOHN L.,ANDGRIFFITH, BYRON A.,Principles ofMechanics, 3rd ed.New York: McGraw-Hill, 1959. ADVANCED MECHANICS TExTs 10.CoRBIN, H.C.,ANDSTEIILE, PHILIP, Ulassical Mechanics. New York: John Wiley &Sons, 1950. 11.GoL1>sTEIN, HERBERT, Classical Mechanics. Reading, Mass.: Addison- Wesley, 1950. 12.LAMB, HORACE, Hydrodynamics, 6thed. Cambridge: Cambridge Uni- versity Press, 1932. (New York: Dover ‘Publications, 1945.) 13.LANDAU, L.D.,and LIFSHITZ, E.M., Mechanics. London: Pergamon Press, 1960. (Reading, Mass.: Addison-Wesley, 1960.) 14.LANDAU, L.D.,andLrrsnrrz, E.M.,Fluid Mechanics. London: Pergamon Press, 1959. (Reading, Mass.: Addison-Wesley, 1959.) 15.LANDAU, L.D.,and LIFSHITZ, E.M., Theory ofElasticity. London: Pergamon Press, 1959. (Reading, Mass.: Addison-Wesley, 1959.) 16.LORD RAYLEIGH, TheTheory ofSound (2vols.), 2nded. London: Mac- millan, 1894—96. (New York: Dover Publications, 1945.) 17.ROUTH, EDWARD Jonrz, Dynamics ofaSystem ofRigid Bodies, Advanced Part, 6thed. London: Macmillan Co., 1905. (New York: Dover Publications, 1955.) 516 BIBLIOGRAPHY 18.SLATER, J01-IN C.,ANDFRANK, NATHANIEL H.,Mechanics. New York: McGraw-Hill, 1947. 19.WEBSTER, ARTHUR GoRDoN, TheDynamics ofParticles andofRigid, Elastic, andFluid Bodies. Leipzig: B.G.Teubner, 1904. 20.WHITTAKER, E.T.,ATreatise ontheAnalytical Dynamics ofParticles and Rigid Bodies, 4th ed. Cambridge: Cambridge University Press, 1937. (New York: Dover Publications, 1944.) 21.WINTNER, AUREL, TheAnalytical Foundations ofCelestial Mechanics. Princeton: Princeton University Press, 1941. TExTs ONELEcTRIoITY ANDMAGNETISM 22.FOWLER, R.G.,Introduction toElectric Theory. Reading, Mass.: Addison- Wesley, 1953. . 23.FRANK, N.H.,Introduction toElectricity andOptics, 2nded.New York: McGraw-Hill, 1950. 24.HARNWELL, GAYLoRD P.,Principles ofElectricity andMagnetism, 2nded. New York: McGraw-Hill, 1949. 25.PAGE, L.,ANDADAMs, N.I.,Principles ofElectricity. New York: D.Van Nostrand, 1931. 26.SLATER, JoIIN C.,ANDFRANK, NATHANIEL H.,Electromagnetism. New York: McGraw-Hill, 1947. Woarcs ONRELATIVITY ANDQUANTUM MEcHANIcs 27.EINsTEIN, ALBERT, ANDINFELD, LEOPOLD, TheEvolution ofPhysics. New York: Simon &Schuster, 1938. (Anexcellent popular account.) 28.BEHGMANN, PETER G.,AnIntroduction totheTheory ofRelativity. New York: Prentice-Hall, 1946. 29.BoIIM, DAVID, Quantum Theory. New York: Prentice-Hall, 1951. 30.BoRN, MAX, Atomic Physics, tr.by,John Dougall, 4thed.New York: Hafner, 1946. 31.HEISENBERG, WERNER, ThePhysical Principles oftheQuantum Theory, tr. byCarlEckart andFrank C.Hoyt. Chicago: University ofChicago Press, 1930. (New York: Dover Publications, 1949.) 32.LANDAU, L.D.,andLIFSHITZ, E.M.,Quantum Mechanics—N on-relativistic Theory. London:Pergamon Press, 1958. (Reading,Mass.:Addison-Wesley, 1958.) 33.LINDsAY, ROBERT BRUCE, AND MARGENAU, HENRY, Foundations of Physics. .New York: John'Wiley &Sons, 1936. 34.ToLMAN, RIcnARD C.,Relativity, Thermodynamics, andCosmology. Ox- ford: Oxford University Press, 1934. _ TExTs ANDTREATIsEs oNMATIIEMATIoAL Torres 35.BELLMAN, RICHARD, Stability Theory ofDiﬁerential Equations. NewYork: McGraw-Hill, 1953. 36.CHURCHILL, RUEL V.,Fourier Series andBoundary Value Problems. New York: McGraw-Hill, 1941. A BIBLIOGRAPHY 517 37.Covaam‘, Rrcnsnn, Diﬁerential andIntegral Calculus, tr.byE.F.Mc- Shane. London: Blackie &Son, 1934. 38.HOPF, L.,Introduction totheDiﬁerential Equations ofPhysics, tr.by Walter Nef. New York: Dover Publications, 1948. 39.JACKSON, DUNHAM. Fourier Series andOrthogonal Polynomials. Menasha, Wisconsin: George Banta Publishing Co., 1941. 40.VONKARMAN, T.,ANDB101‘, M.A.,Mathematical Methods inEngineering. New York: McGraw-Hill, 1940. 41.KAPLAN, W.,Advanced Calculus. Reading, Mass.: Addison-Wesley, 1952. 42.KELLOGG, OLIVER D.,Foundations ofPotential Theory. Berlin: J.Springer, 1929. ~43.KNEBELMAN, M.S.,ANDTHOMAS, T.Y.,Principles ofCollege Algebra. New York: Prentice-Hall, 1942. 44.LEIGHTON, WALTER, AnIntroduction totheTheory ofDiﬁerential Equa- tions. New York: McGraw-Hill, 1952. ’45.LEVY, H.,ANDBaooorr, E.A.,Numerical Solutions ofDifferential Equa- tions, New York: Dover Publications, 1950. 46.MILNE, W.E.,Numerical Calculus. Princeton: Princeton University Press, 1949. 47.Osooon, WILLIAM F.,Introduction toCalculus. New York: Macmillan, 1922. 48.Oseoon, WILLIAM F.,Advanced Calculus. New York: Macmillan, 1925. 49.Osooon, WILLIAM F.,AND Gmmsrnrn, WILHAM C.,Plane andSolid Analytic Geometry. New York: Macmillan, 1938. _ 50.Pnmcn, B.0.,Elements oftheTheory oftheNewtonian Potential-Function, 3rded.Boston: Ginn &Co., 1902. 51.PEIRCE, B.0.,AShort Table ofIntegrals, 3rded.Boston: Ginn &Co., 1929. 52.PHILLIPS, H.B.,Vector Analysis. New York: John Wiley &Sons, 1933. 53.WHITTAKEB, E.T.,AND Ronmson, G.,TheCalculus ofObservations. New York: VanNostrand, 1924. 54.WILLs, A.P.,Vector Analysis, with anIntroduction toTensor Analysis. New York: Prentice-Hall, 1931. 55.WILSON, EDWIN B.,Advanced Calculus. Boston: Ginn &Co.,1912. 56.WYLIE, D.R.,Jn., Advanced Engineering Mathematics. New York: McGraw-Hill, 1951. ANSWERS TO ODD-NUMBERED PROBLEMS ANSWERS TOODD-NUMBERED PROBLEMS CHAPTER 1 1.4.06 X1042 dyne; 9.22 X10‘3 dyne. 5.(b)umg/ (sin0—ucos0). 7.t=(vo/9)[(sin 0+pcos0)'1+(sin2 0——112cos”0)“1'2]. 9.2.20><102"tons. 11.1.4X1011sunmasses. CHAPTER 2 1.:4:=200t—2000(1 —e“/2°)(3 —e“/2°), (a:inft,tinsec), on=200ft/sec. Assumption Findependent ofv. 3.(a)v=vo—|- +tan_1 1 w=(to+_.§’—°)o—to+?1“—;—‘lan"m m 2 —-I£s£1n[l+(L:$-ti“) ]»(wherea: =Oatt =to).2m1r 5.(b)t,=m(1 —-e_°"'°)/(ab), 2:,=[m/(a2b)][l —e_°"'° —av0e_'"° 7.t=[»2,‘""’ -(1-n)bi1"“-"’; x={»E?""’-I»%.""’-<1—n>b¢1‘2'"”“-"’}/<2 —ml»; t,=t§,‘""’/(1 -mi,(n<1); rt,=tE,2""/(2 -mi,(n<2). 9.mii=k/1:3,xe[¢%+(ls/m)t2]1l2. 11.(a)2:2=€—|— cos <2‘/gt-1- 00)- (b)2:é.‘i%+ J? cos<\/;lt+ 00/2)‘: (interpret). 002 b15.x=-%ln(1+—";?) +1—'—Llncosl:.‘l;g(t0 -0].(0<t<:0), m bog m bg -'=%l.I1 1-i-ﬁg —?lI100Bh -1-n-(l—t0) v(l>t()), where to=Vm/bg ta.n_1 (\/b/mg vo). 521 ANSWERS TO ODD-NUMBERED PROBLIJMS :1:=(1312 —|-tv9MG/2)2'3. <<=><21»/a>"°. <21r/3><m3b‘/4a’>"°. (b)mm=0,=|;\/2 a;at=i=\/2 a,co=(v0/3ma2)1l2. <0)a>[1+<4vo/9m»%>1". a>[1+<1v0/s6mv?>>1"‘. :1:=a,(7/2)"2a. of=C1em+ 0261*»Y1=we/m)+o/2m>211/2 —<1»/2m). 2=[(76/m) -l-.(b/2m)2]1/2 —|-(5/2'"); (b)ac=Ae’/‘ cos(w1t+ 0),‘Y=b/2m, an=[(10/m) ——7211/2. :1:=:c0e“Y‘[cos w1t—|- ('Y/001) sinco1t], it=:EQ(1 —|-"Yl)6_'Yt, it=$()('Y1 —'Y2)"1('Y1e“’Y2‘ —'Y26"71‘). (a)lc=4.9X104kgm/sec“2, b=7.07 X104kgm/sec"1. (b)0.076 sec. av=(F0/ma2)[1 —(1+at+%a2t2)e"“]. :4:=(00/wo) sinwot,t_§31r/2w0; at=(B/m)(cog—m2)'1[cos(wt—l— 0)+cosorsinwgt—l— (co/we) sinOtcoswot] _—l—(v0/wo) sinwot,a=(31rw/2am) —|-0,tZ31r/2wg. :1:=(F0/lo) +Ae—“/‘ cos(w1t—|— 0). (a)x=Oift <to,x=(po/lo 6t)[1 —cosw0(t —t0)],ifto$tﬁt0+6t, at=(2120/k 8t)sin(ﬁwo8t)sinw0(t—to—1}8t),ift>to+6t. (a)mwﬁzc =(%A—|-514-B)e""’°"3 cos(%\/2 wot) —|-(%\/2 A—|-%\/2 B)e_"’°‘/3 sin(§\/2 wot) —%Acoswot—£13cos3w0t —%B sinSwot. x=Fg[w1e_“t —w1e_"cosw1t —(‘Y-a)e_" sincolt],(t>0)m.[<v—oz+as -' CHAPTER 3 (a)k[lnctn(s/1;)+1-~/51;(b)——k1r/\/3. (b)a—f=—l(ﬁ)1I2 h g__]:<I)1/2 h of 2f(f+h)'6h_2 h1+1.’ an 1(a)” fan 1(1)” f af2/ f+h’6h 2h j+h’ f h 1%_%.) (<2_A_.»_1@A» A.) (P6r ash+ 31 anm+ <9» n6<r+p k' ANSWERS TOODD-NUMBERED PROBLEMS 523 (a)5-sin_1 - "0 (b)Angle ofelevation should beincreased by 5mew 3 mg cot? ao-—1’ where asisangle ofelevation with noairresistance. (a)5bx4y2 —6abxyz3; (c)—LIF,dy—/:F,dy—f:F,dz. I I I F=—kz2h/p -—2lczln(p/a)k; V=I022ln(p/a). F.=—ae”<rr3 —if)-razor“+T53).F.=—ye2<1~r3 +T53). F2=——2e2(rf3 +rig). 0=(I0/m)1/2, co.=2(k/m)1/2. krz E—|-(E2—w2L2)1/2 cos(2wt—|-20:0), w=(Ic/m)1'2, tan(6—00)=(wL)“1[E ——(E2—w2L2)1/2] tan(wt+0:0), (60=angle ataphelion). (This isLissajous’ ﬁgure with co,=w,,,i.e.an ellipse.) ‘ (a)F=(1—|—ar)Ke-""/1'2; (d)L2=—mKa(1 —|-aa)e“"“‘, E=(1—oza)Ke-°‘“/2a; (e)1-,=21r[—K(1 +aa)e"°“‘/ma3]“1/2, -r,-=21r[—-K(1 +aa—a2a2)e"'“/ma3]"1/2. [Stable circular motion isnotpossible ifaaZQ-(1—|-\/5).] (c)Ellipse precesses 21r(l —oz)/oz radians perrevolution, insame di- rection as9ifoz<1,inopposite direction ifoz>1,Where 0:2= 1+(mK’/L2). (b)Opposite direction, 1.2X10-7 gm-m—3. N=(6/5)nmMG(R2/r3)m cos0sin0, (N),v =(3/5)17mMG(R2/r3) sinozcosoz(LXk)/L, 0),,=21617(R2/r2) cosoz,opposite todirection ofrevolution. 0.6cosadegrees perrevolution. (21"‘1/Y1){[2T2/(T2 "if'1)]"2 —1], v2=(21rT1/Y1)(T1/T2)‘/2i[211/(T2 +T1)l1'2 —1}- Venus: -5700 mi/hr; Mars: 6700 mi/hr.91 Perigee atpoint ofmaximum 9;as=gR2r2/41-2; e=(A—1)/()\+ 1),X=ratio ofmaximum tominimum 9.(There aremany other possible answers.) 1 1—e2cos2 0 2 (3,) 7-'5 = ; =—'47l' T/LT/T2. <0)mo=to=0,¢o=—<qB/2m) 4[(qB/2rrw)2 —(qt/mp%>11'*.(<1)w.=2[(qB/2nw)2 —<qa/2mp?.>11/2. 524 ANSWERS TOODD-NUMBERED PROBLEMS CHAPTER 4 3.cos‘1[1 -0.29am?/(ml +mm]. 5.M1=18,100 kgm, M2=1320 kgm. 7.r=ro[1+(51r)“1(a2/r%)(wo —w)Yo]2, Yo=length ofpresent year. 195miles. Less ifmoon were included. 13-(1"l"'Y)P1F =(P11—P21)608191 =l=["/P11 +P202 -" (1)11 —pg1)2 Sin2 (7111/2, 'Y=mg/m1. Q=pi[1_(ml/m3)$in2!,4+(ml/m4)5i1'12l73]_ 2m1 sin? (193—|-04) 25.$1=$2=Ar"cos(w1t—|— 0),v=b1/2m1, of=-1”+(kl+I53)/ml; A and2:1=—:c2 =Ae_7‘ cos(w2t—|— 0), ti=-v2+(ki—to/mt15. CHAPTER 5 3.wgIc2(1 +;42)“2/41rpga turns. ~ 5.0=9o+(No/b)t +[o:No/(Izwg —|-b2wo)][b sinwot—Iwocoswot]. 9-9=l44r2(h +h’)/1'2l[1 +2h'5/(h -'h')]- 11.ma=0,y@ =—4a/91r;Io,, =Io,==Io,=31ra4o/8, Io.=31ra4o/4, Io,=(8112-32)a4¢/1081,16, =(8112-64)a4¢/2161. - 13.30yards. 17.(a),lsin (a/\/3); (b)(5/36)Ml2. 19.2\/2 kgm-wt, acting atapoint onthird side extended 0.75 mbeyond thecorner. Equilibrant direction is135° toleftof31kgm-wt force. 21.(a)Fo=(0,-6lb,—14lb) atcenter, F.,=(0,—-3lb,-8lb)atany front corner, —F, atadjacent rear corner. (:c-axis outward, y-axis horizontal toright, z-axis vertical, origin atcenter ofcube.) (b)F1=(0,0,2lb) atcenter, F2=(0,—-6lb, ——16lb) atcenter of front face. (There areother correct answers.) (c)Fo[part (a)] atthe point (65/116 ft,0,0),N=(0,9/58 lb-ft, 21/58 lb-ft). 23.A=(100W/Y)e1°°""/Y. 25.(a)sinh(wa/C’) =irwl/C,13 =—(C/w) cosh(wa/C). 27.2Csinha =W,2Csinh[(wo/C) +a]=W+wl, 3=—(0/w) cosh[(wo/0) +al, where y=B+(C/w) cosh[(wx/C) d:oz],(+ifx>0,—ifx<0). 29.-—pL4/192Y(b2 —|-a2)——pL2/8n. 31.p=po—Bln[1—(gpod/B)], where disdepth; about 2%. ANSWERS ToODD-NUMBERED PROBLEMS 525 CHAPTER 6 E=‘(MG/r2)(r/T)! (T2G’): =_*(MGr/a3): (Ts0'): 9=(MG/T)» (TZ11),=(MG/2¢l3)(3a2 —T2),(TSI1)- Tiliq-E 92 =—-41:)-r2Gp arbitrary constants determined bydr Azp dr ’ ' °° 2 M=/E) dr,p~0ur+w. _M2G [1(r2—|-az) _ a4 a2 :l P_4ara4 n r2 2(r2+ a2)2 _1'2+a2' T_AMG1‘ [(13 +a2)2 (r2—|-az) 3 1'2] _2a2R a4 In r2 —2_E5' (11)9—9o=(MG/¢12)(2 —\/5); (b)9—00=—(i)(MG/¢12)- (a)(MG/r)—|-(MGaz/4r3)(1 ——3cos20); _ (b)g,=--(MG/1'2) -—(3MG'a2/4r4)(1 —~3cosz 0), gs=(3MGa2/4r4) sin20. (a)21rcrG, toward sheet. (b)Itishalftheﬁeldoutside theshell. CHAPTER 7 (b)ma* =—bv* —bgt. S 2pwv cos0,w=angular velocity ofearth’s rotation, 0=colatitude; approximately 0.0003 lb-in_2-mile“. (a)2mwgt sin0,eastward. (b)(8w2h3/9g)1/2 sin0. w=(la/m)1/2; inrotating system, mmoves with angular velocity -—2w incircle ofarbitrary radius with arbitrary center. 2.3X106radians/sec. Increase wifelectron circles inpositive sense relative toB;otherwise, decrease w. 1,21 1 4(<1-2R)2"2QM”) [s+Z';+-1.?-l - CHAPTER 8 (b)u=Asin(mrx/2l) cos(mrct/2l) +Bsin(mrrc/2t) sin(mrct/2l), n=1,3,5,.... _4t .rra: 1rct 1.31r:c 31rct 1.51r:c 51rct _ 23u E,m(s1nlcos l———§sm Icos I+25s1n Icos l 5u=Acoswt(coslax—ctnklsinkz),k=w/c. ' 526 F‘ GO 11 17 19 23 25 27 1 3 5 11 15 17 19 21.ANSWERS TO ODD-NUMBERED PROBLEMS 3: (52%=Ae-""2" sin(nrra:/l) cos{[(n21r2c2/l2) -—(b2/4a2)]1/2t —|-0}, =1,2,3,... 6u bc -r ='-T$)z=0;9("7) =E)?-_‘_—;f(§),E =-1!- u=f(a:——ct)—|-g(a:—|-ct),where f(£) =g(£) =function obtained by joining bystraight lines thepoints f=(—1)"l/20, £=(n+i-)l. H=(RT/M) 111(P/Po), visasolution oftﬁ—v2+(2RT/M) ln(vS/voSo) =2gh, 2??=I>0voSo/v»5'- v=—(a/r2)n. —(l1rA/wpoL,,) sin(l1r:v/L,) cos(m1ry/L,,) sin(k,z—-wt), —(m1rA/wpoL,,) cos(lrra:/L,) sin(m-rry/L,,) sin(lczz—wt). v,=(k,A/wpo) cos(l-rrx/L.,) cos(m-rry/Ly) cos(k,z—wt). Acos(lrrx/L.,) cos(m-rry/L,,) cos(k,z—|-wt). v=(31/2pl3)(l2 —422), dp/dz =12171/pl3, where :visdistance from plane midway between walls. Iv,= v,,= CHAPTER 9I T=}ma2(w2 cos”I+122sin”I)exp(2wcoszI+2usin?I), Q,=asinI(F.sinI—F;cosI)exp(wcos”I+usin”I), =acosI(F, cosI+F9sinI)exp(wcosz I+usinzI); =—2ms2 sin2§',Qw=—ms2 cos2 I. 2 2coT=%m<r+h>(§+%->12 =Lj'J"—")f. Pr»= h.Q.» Qt (c)‘Q,’=‘F.’=mrwz sin20+2mrw¢ sing0, =r‘F9’ =mr2w2 sin0cos0+2mr2w¢ sin0cos0, =rsin0‘F,’ =—2mrr‘w sing0—2mr2w9 sin0cos0. 0-*2=[29/HUI +l2)l[1 =b(1—P-)1/2], / I4=[ml/(mi —|-m2)l[4l1l2/(Z1 —|-l2)2l- (b)0é[aw2/ (g—lw2)] coswt. cos61=2E/3mgR ifpg<§mR2E +8E3/27mg2; otherwise thestring willnotcollapse. (b)¢2=21—2(m+ M)g/(M2), (Oz=(m+M)g sin20/(m —|-2Msin20)lcos0, where cos0=1—(z/2l). (b)U=-%’r"w2(w2 +2/2)—1"w(r27 —21¢) =—-%mw2r2 sin20-—rnwrzqb sin20.‘Q0’ ‘Q¢' ANSWERS TO ODD-NUMBERED PROBLEMS 22 q 2 q 2 q 21/2 29.H=c[mc +(p,,—-2A,) —I—<p,,—ZA,,) —|—<p,-—;A,>jl —|—q¢ 2 2 2 2 2 2 31 H=I7z+ Pv+P1+Pr_|_ P0 + Po ' 2(m1 +M2) 2p 2ar2 2]l.1'2Sl112 0 —(ml+msgz— 2 2_ _Pa P m1m2G V’"“(ml+'”2)"Z’ V’_2)n2+ 2,.r2;in20 _1' CHAPTER 10 5.Til=-40,T52=15,T13=3s\/2, T52=10,T53=15\/2, T53=40. 13.Ti=4.,Tt=10,T5=-s,e'1=(e1+e2)/\/2, 62=(-61 +62+es)/\/§, 6%=(61—-62+263)/\/5 21.Eigenvalues: +1,e*i°‘;_ cosa=§(cosy —\—cos0—|—coswkcos0). 23.|=(5/36)Ml2l, M=6m. P 25.0.15mhzsin”a(6+tan”a). Ma‘+b* 2 Mab3—a3b 27. In =-:3(—ﬁ+_b2-+6 SI," =E -i-Jkbz 2 M 20.252 2 M 2 2 I“=0,I,,f,=ﬁ(m+c):I,,=0,I,,=ﬁ(a 29.(8.)1.=..;.2/.2, 12=M2/12, 13=m(a2—|— b2)/12; e1lla,e2llb. <1»)I1=emf+—‘QBmbz.I2=emf+-§)nit I3=‘H-maz —|-imbz; (e3vertical). 31.('w2+h2):c2 —|-(h2—|-l2)y2 +(l2+w2)z2 =50.2/M. 37-P=[Po-(Aw/l)11 +(1'AP/2l)(hk —|-141)- ac.(O)2n[(Vv),,|2—|— 17'(V-v)2 2 =e»1+»'>Z(‘i’-‘) -en—2»'>Z‘-15%. 87'," ..62. 6x,-| 'l>J 6v. 2 6v.- at),-' 2 3 2_ __ __ T -.= ,--_ +11(an)+2»6%am»Where!Iélr, 21 (e3indirection oftension); Y=9nB/ (n+3B). ANSWERS TO ODD-NUMBERED PROBLEMS CHAPTER 11 we=Na(t +to)/Is. wi=wio00$[010+to)2l —w2oSiI1l'1(i+ i0)2l, wg=w1osin[a(t+to)2]+w2ocos[a(t—|-to)2], 11=Na(1s —11)/(21311), to=wao1s/Ns- cosy cos¢ —cos0sin¢ sing]: ' cos.1,sin¢+cos0cos¢sinit sin1/1sin0 —sin wkcos¢——cos0sin¢cosy —sin ¢sin¢+cos0cos¢cosit cositsin0 sin0sin¢ —sin 0cos4: cos0 L=i;Ma292 —|-iMa2d§2 sin20—|-ﬁ;Ma2(\,h —|- cos(9)2 +%Ma2(\//' +qicos0)2—-ﬁMag cos0, _ where 1//,it’refer todisk andrings, respectively. Angular velocities wg,wéofdisk andrings areseparately constant. Precession andnutation afortopinSection 11-5, butwithpireplaced bypi—|-pg’,I1by-Z-Maz, I3by§Ma2, (.03by(.03—|- Top rises intime té(r2w3o01)/(pgl) after 01/(21rp.) revolutions ofpre- cession; center ofmass moves with angular velocity (gl/r3wso) incircle of radius (gl2/uw§o)1'3, 90°outofphase with precession. Topwobbles after l*(1‘2wso)/ (2MI<1)- L=£1192+=lI1d>2 sin’0+%Is(~l/ +<13cos(9)2+lr(M+ m)(r2+ rzdz) —(M +m)M’G/r —mM'Ga2/(4r3)[1 —3sinz0cos2(¢ —a)]. M=6.0X1024 kgm, m=1.6X1022 kgm, a=6400 km., 24000 years. (21r)"1[(wo COS0o)(I3w3 —I1wo COS00)/I1]1’2. CHAPTER 12 $1=K2[m1SAw2l"1/2q1 —-‘HMIS/Aw2l‘1/2q2, I2=%[m23/Aw2]_1/2q1 —|-K2[m2S Aw2l'1/292, S=1:Awz+§(w§o —wgo), inthenotation ofSection 4-10. av_V_ V_O0?=ﬁe 725$-e "(1s+\/1s),27n%e "(12.-\/73). I1=ll+A1<=0$(w1t+ 91)—|-A200$(w2t+ 92), Z2=a—|-A1cos(w1t +01)—A2cos(w2t—I— 02), y;=A3cos(w3t +03)—|-A4co(w4t —|-94), 1/2=—As008(wst+ 93)+A4608(w4#+ 94), 21=A5cos(w5t+05)—|-Aocos(wot—|-05), 22=—A5 cos(wst+05)+A6cos(wot+06), °~’i=‘"3=Q’?=7’/mi‘"2=k(l+6<l)/(l+2a), ‘"2=‘"5=kl/(l+ 2“), where 2:1,y1,21,andZtiyg, 22aremeasured from A,Brespectively, the at-axis isparallel toAB, andaisthepositive root ofha(l+2a)2 — q2=0. I ANSWERS TO ODD-NUMBERED PROBLEMS 529 Both ions oscillate inphase parallel toelectric ﬁeld with amplitude (qEo/M)/(wﬁ —w2)- $2=Al,/A0, where A0=lKl;),,,|, andALisA0with K9,,replaced by 82V’ A5,, 63V° <37’/3$l)0, Kin =(5;;(Tl)o+ 0' IWI / I 11IW5;W51+Zi“"1_ +2 . t _ ,=2Wl’—W9,,1-2(W‘2 —W?)<W‘1’ —W9) ‘ =§(1+%),tz =i@[1_<\/§+1>g],2 W1 2 (03 -‘= :z:1= w1= x1= a:2= w2= :c1= wa= :v1= :c2= w4=1 4 \/5+1‘ \/%{%[1+<\/§—1>§]- 2:4=(1—|-\/§)A cos(w1t+ 0),x2=x3=—2A cos(w1t+ 9), [k(3+\/5)/2ml"2; —w4=[(1+\/5)+20¢’/k\/5)lA COB(wzl+9), _x3= will+(2k'/k)/ (3+\/5)]; :04=2Acos(wgt+0),:63=1:3=(1+\/5)A cos(w3t+ 0), [k(3—V5)/2m1"’; [2+(1+\/3)(k’/In/3)lA cos(w4t+0), -$3=[(1+\/5)-2(Ic'/lax/5)]A cos(w4t+0), wall+(2k’/k)/(3 —\/5)1-__x4 =—[2 —2(1—|-\/5)(lc'/lax/§)]A cos(wgt+0), Y ‘ Bysymmetry, modes 1and3willbeexactly asgiven above foranyk’; hence tworoots wl,003areknown, andthesecular equation canbefactored. w%=1L2_L|:1_ 8a_|_64a02 » . 4Z200 3100 9-n'2a 64a2 1 1+1%,=3;,,___ 1'20‘—no?—4)’ .1/2. 8 ‘M=Acos(w1t—I—0) 2 "'”°,-=s,s,1, _ '" Steady motions: z=0,r=rq,9=wo. (Cylindrical coordinates with +Ze atz==|=a,r=0.)Normal vibrations: r=TQ+ Acos(w1t+ 0), z=0;r=0,z=Acos(w2t—l— 0),ifr>x/§a, otherwise unstable. 22Z¢2 1'34'2 2_21% 4 1% 1“°=7.§1+;ﬁ '°"-°’°v+ 5+ , wg=wgﬁ——2/i—|—1.- la2 a2 530 ANSWERS TOODD-NUMBERED PROBLEMS 23.Steady motions: ' r=To,0A=00,112=Ma/(mm cos00),452=9/[(1—To)cos00]- . 3M Z4-:2=<p2(A ;|=B),A= —~vcos2 00)+1+3cos2 00, 3M Z B2 =A2 +E-W COS2 00*E —3COS2 [Lower mode unstable ifcos200<‘Q;andr0<il(secz 00—3)]. 25.Steady motions: 0=00,dz=0,w3=constant. Normal vibrations:e=00+A)»;->3cos(wt—l—B), ¢=(00/sin90)Asin(wt+B)1 we=>\%§+ (0)3/$i112 90),A=Is/11; 0=00—|-A—|-(w§Bt/Magsin 00), ¢=B,(unstable mode with 0:2=0). ‘27.Steady motions: r=ro,oz=1r,a-mode always unstable; r=ro,oz=0,possible onlyifl<3r0,both modes always stable. 33.X1=X2=0,E1=E2=A. Guessa:1=:z:2=B, €1=€2=dd; checks ifdz=—3wB/2a. INDEX OF SYMBOLS | INDEX OFSYMBOLS Thefollowing listisnotintended tobecomplete, butincludes important symbols and those which might give rise toambiguity. Ingeneral, standard mathematical symbols, andsymbols used inaspecialized sense occurring only once, areomitted. Tofacilitate reference, thepage onwhich thesymbol ﬁrst occurs islisted immediately after thedeﬁnition ofthe symbol. When useofasymbol inaparticular sense isrestricted tooneor twosections orchapters, thisisindicated bychapter orsection numbers inparentheses following thedeﬁnition. Scalar quantities aredesignated inthetextbyitalics. Vector quantities aredesignated byboldface letters beginning inChapter 3.Anitalic letter isused forthemagnitude ofthevector represented bythesame letter inboldface. Anitalic letter with subscripts isused todenote com- ponents ofthevector represented bythesame letter inboldface. In Chapter 2,roman letters areused forcomplex quantitie. Tensors are represented bysans-serif boldface capitals, beginning inChapter 10.The same letter initalics with adouble subscript designates atensor com- ponent, andwith aprime orsingle subscript, aneigenvalue. Adotovera letter indicates differentiation with respect totime. Single quotes are used tomark quantities associated with ﬁctitious forces which arise in moving coordinate systems. LATIN LETTERS b>I=>I1>amplitude ofvibration, 33 area, 220(Chapter 5) constant coefficient, 60 i AA’ plane perpendicular tobeam, 239(Section 5-10) Avector potential, 390(Sections 9—8, 12—7) acceleration, 4,89 constrained generalized coordinate, 373(Section 9-4) distance from focus todirectrix inparabola, 130(Chapter 3) asemimajor axisofellipse orhyperbola, 129(Chapter 3) Bbulk modulus, 234(Chapters 5,8) Bconstant coefficient, 45 Bmagnetic induction, 139 533__§ Q93 534 b b QQQ O c curl D Dz: det div dS,dS do" dﬂ d*/dt E ‘E, Ea: E0 E e e F,F F,F’ F0 F0 F F7; F7; Fink f f f f f,fINDEX OF SYMBOLS frictional force constant, 28 semiminor axisofellipse, 129(Chapter 3) acurve inspace, 84 arbitrary constant, 42 number ofconstraints, 372(Chapter 9) phase velocity ofwave, 296(Chapter 8) speed oflight, 27 curlof,98 constant coeﬂicient, 298 component ofelectric displacement, 27(Chapter 2) determinant ofatensor, 420 divergence of97 element ofsurface, 97,438 scattering cross section, 137 element ofsolid angle, 263 time differentiation relative tostarred coordinates, 272 total energy, 31 . ﬁctitious total energy inmoving coordinate system, 286 electric ﬁeldcomponent, 26 amplitude ofharmonic electric ﬁeld intensity, 26(Chapter 2) electric ﬁeld intensity, 139 magnitude ofelementary electronic charge, 26 unitvector (usually with subscript denoting direction oraxis) 410 force, 7,74 fociofellipse, hyperbola, orparabola, 128, 129(Chapters 3,4) realamplitude ofharmonic force F,51 complex amplitude ofharmonic force F,51 total external force, 156 total external force onparticle k,155 total internal force onparticle lc,155 force exerted byparticle Zonparticle k,156 arbitrary function, 302(Sections 8-3, 8—10) frictional force, 17(Chapters 1,9) number ofdegrees offreedom, 373 body force density, 246(Chapters 5,8) force perunit length, 237, 295(Sections 5-9, 8-1) 1° QNCOQQ 8 Se grad H h,h’ h h h I Iz I 11 i a~wwl>'~1-...\l kz k12 k k L,L L L L0,L0 l lINDEX OF SYMBOLS gravitational constant, 10 center ofmass, orcenter ofgravity, 212 gravitational potential, 260 acceleration ofgravity, 10,228 arbitrary function, 302(Section 8-3) gravitational ﬁeld intensity, 259 effective acceleration ofgravity, 279 gradient of,95 Hamiltonian function, 397 distance from center ofmass toaxisandtocenter ofoscillation, 213(Chapter 5) arbitrary vector function, 335(Section 8—10)535 position vector of0*relative toO,269(Chapter 7) unit vector radially outfrom z-axis, 92 ﬂuid current, 331(Chapter 8) moment ofinertia about z-axis, 207 inertia tensor, 409 \/T1, 44 unit vector parallel tox-axis, 71 impulse, 214 unit vector parallel toy-axis, 71 central force constant, 125 angular wave number, 301(Chapter 8) spring constant, 32 radius ofgyration ofbeam, 243(Section 5-10) radius ofgyration about z-axis, 207 negative acceleration ratio, 5(Chapter 1) unit vector parallel toz—axis, 71 wave vector, 336(Section 8—10) angular momentum, 101, 103 Lagrangian fimction, 367 length, 244 angular momentum about point O,101, 103 length, 12 unit vector indirection ofincreasing 0,polar spherical coordinates, 93coordinates, 90, 536 ssﬁﬁﬁ N N,N N No,No n n 11 O OO’ “U §v F°l=v?v»=:»e©©©©"*l-u'e"e~e-u'd;°“u'"u'"QINDEX OF SYMBOLS Mach number, 344(Sections 8-13, 8-14) mass, usually total mass ofabody orsystem ofparticles, 10 molecular weight, 250(Section 5-11) mass, usually ofaparticle, 6 unit vector indirection ofincreasing <p,92 bending moment, 239(Section 5-10) torque, 102, 103 total number ofparticles, 155(Chapters 4,9) torque about point O,79,81 shear modulus, 235(Chapter 5) unit vector along radius, inpolar coordinates, 90,inspherical coordinates, 93 ' unit vector normal tosurface, outward normal from closed sur- face, 97 point inspace, usually theorigin, 88 linethrough centroid ofbeam cross section, 241(Section 5-10) points inspace, 226 power, 303(Chapter 8) power perunitarea, 336(Section 8—10) component ofdipole moment perunitvolume, 27(Chapter 2) total linear momentum, 156 stress tensor, 438 complex coefficient inexponential time dependence (e"'), 44 generalized momentum (usually with subscript), 361 pressure, 246(Chapters 5,8,10)V linear momentum, 8,100 excess pressure, 312(Chapter 8) energy absorbed ininelastic collision, 176(Chapter 4) generalized force (usually with subscript) 363 point inspace, 158(Chapter 4) source density, 324(Chapter 8) electric charge, 77 generalized coordinate (usually with subscript), 355 gasconstant, 250(Section 5-11) Reynolds number, 349(Section 8-14) center ofmass coordinate vector, 158 r,r’ S HHS 1's T1; T2 Re 00 'i'ﬁ'ﬁ'ﬁmum?/2?!) To T1 T2 t tr U U u U M, ‘V, V V V v,v v v VwINDEX OF SYMBOLS distances from fociofellipse orhyperbola, 129(Chapter 3) radial distance from z-axis, 206(Chapter 5) radius, radial distance from origin, 17 position vector, 79 relative coordinate, 179 standard point, 112 turning points ofr-motion, 128(Chapter 3)' realpart of,51 shearing force, 239(Section 5-10) surface, surface area, 97(except inChapter 5) strain tensor, 444 distance, 74 impact parameter, 135(Chapter 3) absolute temperature, 250(Section 5-11) kinetic energy, 22 period ofrevolution, 18(Chapter 1) period ofperiodic force, 60(Chapter 2) , velocity independent terms inkinetic energy, 358(Chapter 9) terms inkinetic energy linear invelocities, 359(Chapter 9) terms inkinetic energy quadratic invelocities, 359(Chapter 9) time, 4 trace, 419 function ofx,y,zinseparation ofvariables, 337(Chapter 8) velocity dependent _potential, 388(Section 9-8) height ofstring above horizontal axis, 294(Sections 8-1to8-5 9-9) potential energy perunit mass, 328(Sections 8-8to8—10) 1/rincentral force orbit, 123(Chapter 3) effective potential energy, 123 potential energy, 31 volume, 97 center ofmass velocity, 181 velocity, 4,89 ﬂuid velocity, 314(Chapter 8) relative velocity, 181(Chapter 4) wind velocity, 111537 538 W W WI w X X 113 we X X0 "<1"<,"<1 fl Z Z 2 Oi (1 B 'Y 7 '71,'72 71,'Y2 A Aw2 5 51% 6m 5i 5V 78VINDEX OF SYMBOLS work, 74 weight ofbeam, 244(Section 5-10) load onbeam, 244(Section 5-10) weight perunit length, 237(Chapter 5) function ofasinseparation ofvariables, 297(Chapter 8) x-coordinate ofcenter ofmass, 158 rectangular coordinate, 4 coordinate ofstandard point, 31 complex number whose realpart isx,51 complex amplitude ofharmonically oscillating coordinate av,51 function ofyinseparation ofvariables, 338(Chapter 8) y-coordinate ofcenter ofmass, 158 Young’s modulus, 234(Chapter 5) rectangular coordinate, 4 function ofzinseparation ofvariables, 338(Chapter 8) z-coordinate ofcenter ofmass, 158 1 rectangular coordinate, 4 GREEK LETTERS angular acceleration, 208(Section 5-12) asymptote angle ofhyperbola, 130(Chapter 3) phase angle forforced oscillations, 52(Chapter 2) damping coefﬁcient, 47 ratio ofspeciﬁc heats, 333(Section 8-10) damping coefficients foroverdamped oscillator, 49(Chapter 2) damping coefﬁcients forcoupled oscillators, 196(Chapter 4) increment of,83 wf—cogforcoupled oscillators, 190 increment invirtual displacement, 157(Chapters 4,9) Kronecker symbol, 416 mass ofvolume element ofﬂuid, 317(Chapter 8) small time increment, 58 increment inpotential energy, 363(Chapter 9) volume element, 314(Chapter 8) 6 6 '7 "7 =|>Q<z><=z>¢bqsq>@@ '71 172,173,I74 K K A I4 I4 F‘: V 5 E P P 0' 1 T,-r '6-'9~'S'SINDEX orSYMBOLS 539 eccentricity ofellipse orhyperbola, 129 0 dielectric constant, 27(Chapter 2) coefficient ofviscosity, 19,346 phase ofwave, 302(Section 8-3) function oftinseparation ofvariables, 297(Chapter 8) scattering angle inone-body collision problem, 135 angle between string andhorizontal, 237(Section 5-9,Chapter 8) angle between twovectors, 73(Chapter 3) angle ofrotation about axis, 206 angle ofshear, 234(Sections 5-8, 5-10) phase angle, 26 polar angle, polar coordinates, 90,spherical coordinates, 93 Euler angle, 458 scattering angle inlaboratory coordinates, 173 angles ofprojection ofparticles mg,m3,m4incollision, 173,176 coupling constant, 191(Chapter 4) amplitude ofoscillation ofpendulum, 211(Chapter 5) wavelength, 301(Section 8-3) coefﬁcient ofsliding friction, 17 reduced mass, 180 coeﬂicient ofstatic friction, 17 frequency (cycles/sec orrev/sec), 142 phase ofwave, 301(Chapter 8) lineofnodes, 458(Chapter 11) density, 5,204 4 radial distance from z-axis, 92(Chapter 3andSection 12-7) linear density ofstring, 295(Sections 8-1to8-5, 9-9) period, 107 tension, 15,237 angle ofbending ofbeam, 242 (Section 5-10) azimuth angle, 92 - A electric potential, 140 ' velocity potential, 332(Section 8-9) s 540 ¢ IP 9 co co,w we W0 “>1 4°10,Q20 ml; (‘,2 i>c><’.’’2wt:d<-oo F IINDEX OF SYMBOLS Euler angle, 458 Euler angle, 458 angular velocity, 281 angular frequency (radians/sec), 26 angular velocity, 208, 273 cut-off angular frequency, 308 natural (angular) frequency ofundamped oscillator, natural (angular) frequency ofdamped oscillator, 47(Chapter 2) natural (angular) frequencies ofoscillators without coupling, 189 (Chapter 4) natural (angular) frequencies ofcoupled oscillators, 190(Chap- ter4) OTHER SYMBOLS nulltensor, 446 nullvector, 75 unit tensor, 409 del,symbolic operator, 96 dotproduct, 73 cross product, 75 average value, 53 Y SUBSCRIPTS ﬁnal value (after collision), 172(Chapter 4) initial value (before collision), 172(Chapter 4) g,0,0’,Q,etc.designate values at,orrelative topoint G,O,O’,Q,etc., 79 ,-,,-,1,,1,,,,,etc. designate quantity associated with particle 2',j,etc., i, j,=1,2,...,155(Chapters 4,5) ,-,,-,1,,1,etc. designate vector andtensor components, 410 ,-,1,,1,,,,,,,designate quantity associated with corresponding mode ofvibra- tion orfrequency ofoscillation, 59 ,,,,ma, maximum value of,110, 19 min minimum value of,19 ,,,,,,,,,1,,,,,,etc.assubscript tovector symbol, designate corresponding 1component ofvector, 4;ingeneral, designate aquantity associ- ated with thex-,y-,z-,r-,etc., coordinate oraxis, 106 INDEX OF SYMBOLS ,,component indirection ofn,77 0initial orstandard value, 13541 0,1,2,etc.designate value attime to,t1,t2,etc., 21;orvalues atpoints 0,1,2,etc., 112; orquantities associated with particle number 1,2,etc., 15;orused simply tonumber asetofquantities, 42 SUPERSCRIPTS °external, 155 linternal, 155 ‘transpose ofatensor, 413 ’dimensionless variables, 343(Sections 8-13, 8-14) ’relative toprimed coordinate system, 216 *complex conjugate, 45 *relative tomoving coordinate system, 269 INDEX 1 91INDEX Acceleration, 4 Angular wave number, 301 centripetal, 91,276 Anomalous dispersion, 56 components, incylindrical coordi- Antinode, 339 nates, 92 Antisymmetric tensor, 413 inpolar coordinates, 91 Aperiodic orbit, 124 inrectangular coordinates, 4,89, Applied force (see:Force) Approximate solutions, 337 inspherical coordinates, 94 Arbitrary constant, 25,42, coriolis, 91,276 155 ofgravity, 10,279 Arbitrary function, 302 normal andtangential, 148 Archimedes’ principle,'248 ratio of,5 Area, ofellipse, 129 ADAMS, J.C.,134 oforbit, 125 Addition, oftensors (see: Tensor) from Pappus’ theorem, 2,414, 444 44,104, 21 ofvectors (see:Vector) swept outbyradius vector, 124,133 Adiabatic bulk modulus, 333 Associative law, 69 Adiabatic relation, 323 Asteroid, 508 Air,motion of,280 Astronomical bodies, motio Airresistance, 35,110, 111 165ff,171, 175nof, Alpha particle, 138 ' Asymptote ofhyperbola, 130 Angle ofrepose, 17 Atmosphere, 251 Angle ofscattering (see: Scattering Atom, 57,105, 167, 176, 188, 391 angle) Bohr theory, 135 Angular acceleration, 208 jellymodel, 57 Angular frequency, 51,124 inmagnetic ﬁeld, 283 Angular momentum, 101ﬂ',120, planetary model, 57 158ﬂ',361 Atomic collision, 175, 176r conservation of,120, 166f,168f, Atomic particles, motion of,165ff 170, 203, 205, 326, 383 Atwood’s machine, 14,375 internal, 187 Axial vector, 417 orbital, 167, 188 ofrigid body, 203, 206 Baseball bat,215 rotational, 167 Bead, sliding onahoop, 38 spin, 167, 188 sliding onawire, 3707 vector, 103 Beam, equilibrium of,239ff Angular momentum integral, 121 Bending moment, 239 Angular momentum theorem, 102,103, Bernoulli’s theorem, 329 160,205 Beta-ray spectrometer, 142 Angular position, 206, 208 Betatron, 142, 497, 499 Angular velocity, 208 Betatron oscillations, 497ff addition of,459 Blow (see: Impulsive force) interms ofEuler’s angles, 460 Body cone, 453 vector, 273 Body force, 246, 321, 345 545 546 INDEX BOHR, N.,135 Boundary condition, forairinabox, 339 foropen-ended pipe, 341 forstring, 296, 298, 350 Bounded orbit, 124, 134 BRAHE, TYCHO, 132 'Bulk modulus, 234, 249, 323, 329, 333, 445 Cable (see: String) Calculus ofvariations, 391 Catenary, 238 Cavity, normal vibrations in,198, 341 'Celestial motions (see: Astronomical bodies) Center ofgravity, 228, 257f relative toapoint, 258 (seealso: Center ofmass) Center ofmass, 158, 179,215ff motion of,158, 185, 187 ofhemisphere, 220 relative todifferent coordinate systems, 216 velocity of,185 Center ofmass coordinate system, 182 Center ofoscillation, 213 Center ofpercussion, 215 Central force (see: Force) Centrifugal force (see: Force) Centripetal acceleration (see: Acceler- ation) Centripetal force, 17 Centroid, 220 ofbeam cross section, 243 from Pappus’ theorems, 221 (seealso: Center ofmass) CHADWICK, J.,175 Characteristic value (see: Eigenvalue) Circular pipe, 340, 346 Clock, 210 Closed orbit, 124, 134 Cloud chamber, 142 , Coefficient offriction (see: Friction) Coeﬂicient ofrestitution, 178 Coeﬂicient ofviscosity (see: Viscosity)Collision, 135, 171ff,182ff elastic, 172ff,184 endoergic andexoergic, 176 ﬁrst andsecond kind, 176 inelastic, 176if Comet, 134 Commutative law, 69 Complex number, 45,46,51 Component, ofatensor (see:Tensor) oftorque, 81 ofavector (see: Vector) Component force, 16,17,77ff Compound pendulum, 212f Compressibility, 317, 345 Compression, 233, 245, 328 Compton effect, 201 Conﬁguration, 356, 399 Conﬁguration space, 399 Conic section, 128ff Conservation, ofangular momentum (see: Angular momentum) ofenergy (see:Energy) oflinear momentum (see:Linear momentum) Conservation laws, 165ff,383 forﬂuid motion, 323ff forrigid body, 203 Conservative force (see: Force) Constant ofthemotion, 33,105f,123, 290, 381ff existence ofinthree-body problem, 290 Constant tensor, 409 Constraint, 203, 355, 368, 370ii’ equations of,373, 379 force of,374 holonomic, 370f,372 moving, 374, 387 nonholonomic, 372 Continuity, equation of,318, 323, 330 Continuous medium, 3,5,294, 441 Conveyor belt, 169 Coordinate, dimensionless, 343, 348 generalized, 354if ignorable, 381 21 \ 11 lorthogonal, 359, 360 relative, 179 Coordinate system, 9,270, 399 center ofmass, 182 curvilinear, 93,367 cylindrical, 92,98,220 laboratory, 182 left-handed, 417 moving, 269f,354,357,369,374,397 parabolic, 147, 401 polar, 90f,93,102 rectangular, 4,88f,91,220 rotating, 271ff,357, 359, 369, 401, 404, 445 spherical, 93,99,220 translation of,269f Coordinate transformation, 414if Coriolis acceleration (see: Acceleration) Coriolis force (see: Force) Coriolis’ theorem, 276, 283 Couple, 228f Coupled electric circuits, 197 Coupled harmonic oscillators (see: Harmonic oscillator) Covariant equations, 270 Cream separator, 277 Critical damping, 50 Cross product, 75 Cross section (see: Scattering) oftube ofﬂow, 331 Curl, 98,113f,148 ofvelocity, 320 Curve, 88f Curvilinear coordinates, 93,367 Cut-off frequency, forstring of particles, 308, 309‘ forwave inpipe, 342 Cyclone, 280 Cyclotron, 142,285,497,499 Cylinder, rolling down anincline, 370f rolling onacylinder, 376ff Cylindrical coordinates (see: Coordinate system) Damped oscillator (see: Harmonic oscillator)INDEX 547 Damping (see: Force) over-, under-, andcritical, 50 Decomposition ofatensor, 444 Deﬁnition, 1 Deformation, elastic, 41,231, 233f plastic, 41,231 (seealso:Strain) . Degeneracy, 421, 422, 425ff,488, 495 approximate, 489 Degenerate eigenvalue, 422 Degenerate principal moment of inertia, 432 Degree offreedom, 372 Delsymbol (V), 96ff,318 Density ofmass, 204, 313, 333 Derivative (see: Differentiation) Determinant ofatensor,,420, 427 Diagonal form, 421 Diagonalization ofasymmetric tensor, 421if,478ff infdimensions, 481 Dielectric constant, 27,56 Dielectric medium, 55 Diﬁerence equations, 307 Differential equation, ordinary, 22ff, 104 existence theorem, 23-f extraneous solutions introduced bydifferentiation, 144 general solution, 42,50 homogeneous, inhomogeneous, 42,59 independent solutions, 43 linear, 41ff numerical methods ofsolution, 24,105, 185 order of,41 particular solution, 43,50 simultaneous, 104,190 partial, 296, 297, 302 general solution, 299, 302, 341 numerical methods, 297, 307 separation ofvariables, 297 Differentiation, ofavector, 82f incurvilinear coordinates, 93,95 total andpartial, 314, 361 Dimensionless coordinates, 343, 3481 548 INDEX Dimensions, 12 Dipole moment, 27 Directional derivative, 95 Directrix, 130 Discrete string (see: Vibrating string, made upofparticles) Disk, moment ofinertia of,224 rolling onatable, 371 Dispersion, 56,309, 342 Distortion ofabeam, 241if Distortion ofawave shape, 343 Distributive law, 69 . Divergence ofavector function, 97, 99,315 Divergence theorem (see: Gauss’ divergence theorem) Dotproduct, 73,412 Dyad, 407 Dyad product, 407 Dyadic, 408ff Dynamic balance, 452 Dynamics, 3,5 Earth, gravitational ﬁeldof,267 interior of,245 laws ofmotion on,278f rotation of,171,468 Earth satellite, 152,508 Eccentricity ofellipse, 129, 131 Effective potential energy (see: Potential energy) Eigenvalue, 422ﬂ' Eigenvector, 422if EINSTEIN, A.,2,270 Elastic collision (see:Collision) * Elastic limit, 41 Elastic solid, 198, 335, 444f Electric circuit, coupled, 197f oscillating, 41 Electric current, 319 Electric ﬁeld intensity, 26,55,139, 261, 389 Electric potential, 140, 261, 390 Electromagnetic energy, 168 Electromagnetic ﬁeld, 139ff,168,389 Electromagnetic force (see: Force) Electromagnetic theory, 7,8,140, 390Electromagnetic vibrations, 337 Electromagnetic wave (see:Wave) Electron, 26,55 spinof,472 Electrostatic force 1(see: Force) Elementary particle (see: Particle) Ellipse, 128f semi-major axis, 132 ' Ellipsoid ofinertia, 437f,456 Elliptic integral, 211 Elliptic orbit, 133 Endoergic collision, 176 Energy, 168, 383 absorption inadielectric medium, 55,56 conservation of,31,105, 167f,203, 327f,382f foracontinuous medium, 441 foraﬂuid, 327ﬂ",331 from Lagrange’s equations, 383, 389 foraparticle, 31,105, 140 forarigid body, 203,451 forasystem ofparticles, 163, 167f constant ofthemotion (see: Energy integral) electromagnetic, 168 ofexpansion andcompression, 328 ﬂow of,303, 336 internal, 176, 185f kinetic (see: Kinetic energy) potential (see: Potential energy) relativistic formula, 175 inthree-body problem, 286,289, 293 - Energy density, kinetic, 327 potential, 328 Energy integral, 33f,105, 115, 382f Energy theorem, 22,101, 163, 327,329,452 Engineer, 24,41,231 Equation ofcontinuity (see: Continuity) Equation ofmotion, ofacontinuous medium, 441 ‘ofaﬂuid, 322 INDEX 54$) ingeneralized coordinates, 354f, 367,397 inamoving coordinate system, 270, 277 ‘ ofaparticle, 13,21,100 ofarigid body, 205,207,225,450 451, 460 ontherotating earth, 279 ofasystem ofparticles, 155 ofavibrating string, 295, 306, 395 Equation ofstate (see: Fluid) Equilibrant, 228 Equilibrium, 34,226, 473E ofa.beam, 239E conﬁguration of,473E ofaﬂuid, 245E neutral, 34 point of,34,289, 380 ofarigid body, 226 stable, 34,474f ofastring orcable, 235E unstable, 34,65,474, 480 Equilibrium orbit, 497 Equivalent systems offorces, 227E Escape velocity, 38 EULER, L.,313,322 Euler’s angles, 458E Euler’s equations ofmotion foraﬂuid, 313, 322 Euler’s equations ofmotion fora.rigid body, 451 Euler’s theorem, 382 Exact science, 1 Exoergic collision, 176 External force (see:Force) Falling body, 14,35E Fictitious force (see:Force) Field index, 499 Field theory ofgravitation, 259E Flagpole, 232 Flexible strings andcables, 235E Fluid, 245,313ff,440 conservation laws for,323ff current of,331 equation ofmotion of,313,322 equation ofstate of,249fequilibrium of,245ff expansion of,314E,328 ﬂow of,318f homogeneous, 322, 328 similar problems in,343 ideal, 321, 328, 345, 440 incompressible, 317, 328, 332, 345 irrotational ﬂow of,320, 331 kinematics of,313E potential energy in,328 steady ﬂowof,329E viscous, 329, 345E,441E Flux ofgravitational ﬁeld intensity, 263 Focus, ofconic section, 129 Force, 7 applied, 25f central, 118f,120E,164, 283 centrifugal, 18,122,277, 367, 369, 401, 404 centripetal, 17 component, 77 conservative, 30f,115, 117, 162, 163 ofconstraint, 203,374f coriolis, 277, 279, 280,283, 369, 40.1, 404 damping, 28f deﬁnition of,8 depending onposition, 30,105, 112 depending ontime, 25,105 depending onvelocity, 28,105 electromagnetic, 139, 389f,391 potential for,390 equivalent systems of,227,229f‘ external, 155E,178, 187 ﬁctitious, 122,271,277,367 frictional (see:Friction) ' generalized, 363ff impulsive, 57f,213(seealso: Impulse) internal, 155E,178, 187, 203, 233, 326 inverse square law, 37f,120, 125ff reduction ofasystem of,230 resultant, 227, 230 units of,11 550 INDEX Force density, 246, 321 duetopressure, 322 duetostresses, 441 Forced vibrations, 481E (seealso: Harmonic oscillator) Forced harmonic oscillator (see: Harmonic oscillator) Foucault pendulum, 280ff Fourier integral, 61 Fourier series, 61,299, 300, 392 Frame ofreference, 270 Friction, 28,30,164, 167, 177, 329, 368, 374, 389, 391 coefficient of,17 incoupled oscillators, 195 drysliding, 17,28 lubricated surfaces, 28 static, 17 GALILEO, 2 Gauss’ divergence theorem, 97,248, 319, 330, 400 Gaussian units, 139 General solution (see:Differential equation) Generalized coordinates, 354E,374 kinetic energy in,358 potential energy in,363 forvibrating string, 392 Generalized force, 363ff ' Generalized momentum, 361, 391 Generalized velocity, 357 Gradient, 95 * Gravitation, 7,8,10,18,120, 125, 133,167,25711,391 constant of,10,125, 257 ﬂuid inequilibrium under, 250 Gravitational ﬁeld equations, 262ff Gravitational ﬁeld intensity, 259 ﬂuxof,263 Gravitational potential, 260 Gravitational units offorce, 11 Gravity, acceleration of,10,213, 259 effective, 279 Green’s function, 62 Group theory, 444 Gyrocompass, 291, 472Gyroscope, 161, 468 HAMILTON, W.R.,3,399 Hamiltonian function, 397 Hamilton’s equations, 396E Harmonic oscillator, 32,39E,398, 400 coupled, 188E,307, 396, 476ff with applied force, 196, 481ff with damping, 196, 483f normal coordinates for,395, 480, 482, 485, 489 normal mode, 192E,310, 395, 477ff perturbation theory for,484ff types ofcoupling, 196f damped, 39,47ff critically, over-, under-, 50 energy of,48 forced, 40,50E,59E free, 39 isotropic, 108 power delivered byapplied force, 53, 54 intwoorthree dimensions, 105, 106E Harmonic wave (see: Wave) Heat,163,323,329,389 ﬂow of,323 HEISENBERG, W., 2 Hemisphere, center ofmass of,220 Herpolhode, 456 Holonomic constraint, 370f,372 Homogeneous differential equation (see:Differential equation) Homogeneous ﬂuid (see: Fluid) Hooke’s law, 41,234, 236, 249, 445 Hydrogen molecule ion,115 Hyperbola, 128, 129f Hyperbolic orbit, 135E Ideal ﬂuid (see: Fluid) Ignorable coordinate, 381, 398, 491, 492,4941 Impact parameter, 135 Impulse, 21,57f,100, 213 Incident particle, 182 Incompressible ﬂuid (see:Fluid) INDEX 551 Increment ofavector, 83 Index ofrefraction, 56 Inelastic collision (see: Collision) Inertia, moment of(see: Moment of inertia) Inertia ellipsoid, 437f,456 Inertia tensor, 409, 410, 430ff Inhomogeneous diEerential equation (see:Differential equation) Initial conditions, 23,33,42,104, 172,296 Initial instant, 23,104 Inner product, 73 Integral ofadiEerential equation (see: Constant ofthemotion; Energy integral; Angular momentum integral) Integration, ofavector (see: Vector) over avolume, 219f Internal angular momentum, 187 Internalicoordinate, 185 Internal energy, 176, 185 ' Internal force‘ (see: Force) Internal linear momentum, 186 Internal motion, 185ff Internal velocity, 185 Intrinsic energy andangular momentum, 188 Invariable plane, 456 Invariance oftheequations ofmotion, 270 Invariant, 270, 419 ofatensor, 427 Inverse square law, 120, 125 Ionosphere, 26 Irrotational ﬂow, 320, 331f Isotropic ﬂuid, 441, 443 . Isotropic harmonic oscillator, 108 Isotropic solid,"444 Jacobian determinant, 356 Jelly model ofatom, 57 JOULE, J.P.,168 Jupiter, 508 KEPLER, J.,132 Kepler’s laws, 133Kinematics, 4,87 ofﬂuids, 313ff inaplane, 88E inthree dimensions, 91E Kinetic energy, 22 ingeneralized coordinates, 358 ofaﬂuid, 327 internal, 181 , inN-body problem, 186 ofaparticle, 22,100 l relativistic formula, 175 ofrotation, 208, 437, 460 ofasystem ofparticles, 163 intwo-body problem, 181 Laboratory coordinate system, 182 LAGRANGE, J.L.,3,354,500 Lagrange’s equations, 3,365ff,368, 373, 476 foravibrating string, 391ff Lagrange’s solution ofthethree-body problem, 500ff stability of,504ff,508 Lagrangian equations ofmotion ofa ﬂuid, 313 Lagrangian function, 367, 375,388 relativistic, 404 forvibrating string, 395 Laminar ﬂow, 346 Laplace’s equation, 264, 332 Larmor’s theorem, 283f Left-handed coordinate system, 417 Legendre polynomials, 267 LEVERRIER, U.J.J.,134 Light, velocity of,27,165, 175, 176, 312 . Line integral, 84E Line ofaction, 226 Line ofnodes, 459 Linear combination, 44 . Linear diEerential equation (see: Differential equation) Linear momentum, 7,21,100, 156, 158, 325f,361 conservation of,156, 157, 165f, 168f,203, 205, 325f,383 density of,319, 325 552 INDEX internal, 186 measurement of,142 potential, 391 relativistic formula for,175 inthetwo-body problem, 181 vector, 100 Linear momentum theorem, 21,100, 156, 326 Linear oscillator (sec: Harmonic oscillator) Linear vector function, 407, 411, 412 (seealso: Tensor) Linear vector operator, 407,408,412 (seealso: Tensor) Linearized equations ofmotion, 475E Lines offorce, 263 Liouville’s theorem, 399f Lissajous ﬁgure, 107 Logarithmic derivative, 48 Longitudinal wave, 335 mks units, 11,139 Mach number, 344f Macroscopic body, 165,166 Magnet, 140 Magnetic ﬁeld, 77,141f,283,389 Magnetic force (see:Force, electro— magnetic) Magnetron, 145 Magnitude ofavector, 68 Major axis (see: Ellipse; Hyperbola) Many-body problem (see:N-body problem) Mass, 6,11 center of(see:Center ofmass) conservation of,323 ﬂow of,318f inasound wave, 336 rest, 175 unit, 6 _ Mass spectrometer, 142 Matrix, 410,416 product of,412 sum of,412 MAXWELL, J.C.,2 Median plane, 497 Mercury, 134, 165Minimum, testfor,475 Minor axis (see: Ellipse) mks units, 11,139 Mode ofpropagation inapipe, 342 Mode ofvibration (see: Normal mode) Modulus ofelasticity (see:Bulk; Shear; Young’s modulus) Molecule, 166,176,iss,391 Moment, ofaforce, 79E ofavector, 80f Moment ofinertia, 207, 221ff,430 ofdisk, 224 ofring, 224 ofsphere, 224 Momentum (see: Linear; Angular; Generalized momentum) potential, 391 Moon, 18,171,285 Moving constraint (see: Constraint) Moving coordinate system (see: Coordinate system) Moving origin ofcoordinates, 296f Multiplication, ofvectors (see:Vector) oftensors (see:Tensor) N-body problem, 155,185E Neptune, 143 Neutral equilibrium (see:Equilibrium) Neutral layer, 242 Neutron, 175 NEWTON, ISAAC, 2,7,10,18,126, 132,133,262,333 Newton’s laws ofmotion, 3,7E, 165E,270,354,368 (seealso: Equation ofmotion) Newton’s third law,7E,140,156,157, 165s,203,233,326,329 strong form, 140, 160, 167 weak form, 140, 157, 164, 178, 179 Node, 339 Nonholonomic constraint, 372 Normal coordinates, 480, 483, 489, 494 forvibrating string, 395 Normal frequency ofvibration, 298, 310, 340 INDEX 553 Normal mode ofvibration, 41,192, 477E ofcoupled oscillators, 192E,310, 477E ofﬂuid inabox, 337E ofvibrating string, 298E,310, 395 Nuclear collision, 176, 177 Nuclear reaction, 177 Nuclear theory oftheatom, 138 Nucleus, 176, 177, 188 radius of,138 Null tensor, 446 Null vector, 75 Numerical methods ofsolution (see: DiEerential equations) Nutation, 464 Octopole moment, 267 Open-ended pipe, 340 Orbit, aperiodic, 124 bounded, 124,134 foracentral force, 123E,127E closed, 124,134 foraninverse square lawforce, 126E,133E,135E precessing, 134, 151 Orbital energy, linear momentum, angular momentum, 188 Organ pipe, 340 Orthogonal coordinates, 359, 360 Orthogonal tensor, 418 Orthogonal transformation, 418 Orthogonal vectors, 479 (seealso: Perpendicular vectors) Oscillations (see:Harmonic oscillator; Normal mode ofvibration; Small vibrations) Oscillator (see: Harmonic Oscillator) Outer product, 75 Overdamping, 50 Pappus’ theorems, 221 Parabola, 110,128,130 Parabolic coordinates, 147, 401 Parallel axistheorem, 222 forinertia tensor, 431 Parallel vectors, 75Parametric representation ofacurve, 86,88,89,91 Partial diEerential equation (see: DiEerential equation) Particle, 3,4,166, 188 elementary, 165, 167 string of(see: String) system of,3,155E,185E Particle accelerator, 497 alternating gradient, 500 Pascal’s law, 247 Pendulum, compound, 212f Foucault, 280E simple, 208E spherical, 384E Perfect gas,250 Perihelion, 131 Period ofrevolution, 125, 133 Periodic force, 60 Perpendicular axistheorem, 223 Perpendicular vectors, 75 Perturbation theory, 484E degenerate case, 488f ﬁrst-order, 486f second-order, 489 Phase, 45,51 ofawave, 301 Phase space, 399, 400 Phase velocity, 301 Pipe, normal vibrations in,198, 340 viscous ﬂow in,346E wave propagation in,341E Plane lamina, 223 Plane wave, 334 Planet, 10,126, 132f,165, 168 Planetary model oftheatom, 57 Plastic ﬂow, 41,231 Pluto, 134 Poinsot’s solution forarotating body, 455E Poiseuille’s law,348 Poisson's equation, 264 Polar coordinates (see: Coordinate system) Polar vector, 417 Polarization, 55 Polhode, 456 554 INDEX Potential, electric, 140, 261, 390 gravitational, 260 scalar andvector, 390 velocity, 332 velocity-dependent, 389E Potential energy, 31,112,162 effective, 123, 126f,385, 388, 463, 492 inaﬂuid, 328 ingeneralized coordinates, 363 ofaparticle, 31f,33f,105,112E rotational, 208 ofasystem ofparticles, 162, 164 ofavibrating string, 393, 394 Potential momentum, 391 Power, 22,53f,303, 336 Power factor, 53 Precession, ofearth’s axis, 471 ofanelliptical orbit, 134, 151, 152 ofaFoucault pendulum, 280, 282 ofagyroscope, 161,465 oftheperihelion ofMercury, 134, 152, 165 ofarotating body, 453 ofatop,463, 464 Pressure, 246E,322, 333 asapotential-energy density, 327 Pressure-velocity relation inasound wave, 335 Prevailing westerly winds, 280 Primitive term, 2 Principal axes, 422 ofarigid body, 432 Principal moments ofinertia, 432 Products ofinertia, 430 Projectile, 108E’ Projection, 70 Propagation ofawave (see: Wave) Pseudovector, 417 Quadrupole moment, 267 I Quantum mechanics, 2,3,9,41,57, 135, 139, 165, 168, 171, I75, 181, 198,337, 383, 398 Radio wave (see:Wave, electro- magnetic)Radius ofcurvature, 148 Radius ofgyration, 207 ofbeam cross section, 243 Range ofprojectile, 110 Recoil, 177 . Rectangular coordinates (see: Coordinate system) Reduced mass, 180 Reduction ofasystem offorces, 230 Reﬂection ofawave, 304 Relative coordinate, 179, 182 Relativity, 2,3,9,165, 383 andthedeﬁnition ofmass, 6,8 energy, momentum formulas, 175 general theory, 10,135, 165,271, 283, 368 Lagrangian function in,404 Newtonian principle of,9,270 special theory, 9,142, 153, 270 Relaxation methods, 237 Repose, angle of,17 Residual nucleus, 177 Resonance, 40,53,54,193 Rest mass, 175 Restitution, coeﬂicient of,178 Restricted three-body problem (see: Three-body problem) Resultant, 16,77,227, 230 Reynolds number, 349 Right-hand 1'ule, 75,77 Rigid body, 3,203ff,355, 370, 450 coordinates for,204, 205f,450, 458E * equations ofmotion for(see: Equation ofmotion) rotation of,166, 167, 171 about anaxis, 206E free,452ff,455E inthree dimensions, 451 E,455 E, 460, 461E energy of,167, 177,436E,451 Rocket, 170 Rolling cylinder, 371, 376E Rotating coordinate system (see: Coordinate system) Rotation, virtual, 160 — V Y ? INDEX 555 RUTHERFORD, E.,57,135, 138, 181 Rutherford scattering cross section, 138, 181ﬂ',184 Saddlepoint, 288 Satellite, 126, 165 Scalar, 69 Scalar point function, 84 Scalar potential, 390 Scalar product, 73 Scalar triple product, 76 Scattering, 135ﬁr,175,181ff angle of,135, 173, 175, 183 cross section, 136, 138, 175, 184 SCHROEDINGER, E.,2 Secular equation, 190, 196, 423, 477, 478, 493 Separation ofvariables, 297, 337, 338, 341 Shear modulus, 235, 241, 445 Shearing force, 239 Shearing strain, 234,443,444 Shearing stress, 233,321,443 Similar problems, 343, 349 Simple harmonic oscillator (see: Harmonic oscillator) Simultaneous linear differential equations, 104, 190 Singular point, 287 Singular solutions, 42 Small vibrations, 198,473ff,490 about steady motion, 491if (seealso: Harmonic oscillator) Solar system, 165, 167 Solid, conservation laws in,329 elastic, 444 Sound wave (see:Wave) Space cone, 454 Sphere, moment ofinertia of,224 Spherical coordinates (see: Coordinate system) Spherical pendulum, 384ff Spherical shell, gravitational ﬁeld of, 261 f Spherical symmetry, 217, 262, 432 Spherical wave, 336Spin (see: Angular momentum) Stability, 289f,473ff,496,508 Stable equilibrium (see:Equilibrium) Standard point, 31,32,112 Standing waves, 305,338 Star, 165 State ofa.mechanical system, 396 Statically indeterminate structure, 232 Statics, 3,225E ofbeams, 239ff ofﬂuids, 245ﬁ ofstrings andcables, 235ff ofstructures, 231f Statistical mechanics, 398,400 Steady ﬂow, 329ff,346ff Steady motion, 491 Steady state, 53 Stokes’ theorem, 98,113 Strain, 233ff,241f inasolid, 444f inastressed ﬂuid, 249 Streamline, 330 Stress, 233ﬁ,242,326 inanelastic solid, 444f inaﬂuid, 245, 441if Stress tensor, 438if inanelastic solid, 444f inaviscous ﬂuid, 441if String, equilibrium of,235if vibrating (see: Vibrating string) Strong form ofNewton’s third law(see: Newton’s third law) Structure, 231f Sun, mass of,20,133 Superposition, 59if,192, 196, 299, 477, 494 Symmetric tensor (see:Tensor) Symmetrical top, 161,461ff - Symmetry, 217, 383, 432 Synchrotron, 497, 499 System ofparticles (see:Particle) Target particle, 182 Taylor series, 30,40,41 Temperature ofaﬂuid, 250, 323 Tension, 233, 295 556 mnnx Tensor, 233, 406, 408E,416 antisymmetric, 413,414,444 associated quadric surface, 438 components of,410 constant, 409 ' determinant of,420, 427 diagonal form, 421 diagonalization of,421E dotproduct of,412 eigenvalue of,422ff eigenvector of,422E invariant scalars of,427 null, 446 orthogonal, 418 principal axes of,422 product byascalar, 409, 445 sumof,408,412 symmetric, 413, 414, 421, 422, 438, 444 trace of,419,427 transpose of,413f unit, 409 Terminal velocity, 36,37,111 Terrestrial motions, 165E THOMSON, J.J.,142 Three-body problem, 285,500E restricted, 285E Tides, 167,171 Time derivative, inarotating coordi- natesystem, 272ff,445 ofavector, 82f Top, 161,461E Tornado, 280 Torque, 79,102, 120, 159’f, 207, 208, 225E,239, 450, 452 component of,81 Torsion, 239 Total time derivative, 314, 361 Trace, 419, 427 Trade winds, 280 Transformation ofcoordinates, 414E orthogonal, 418 Transient, 53,62 Translation, ofcoordinate system, 269 virtual, 157 ' Transmission line,41,198,312 Transpose ofatensor, 413Traveling wave (see: Wave) Triple vector product, 76 Tube ofﬂow, 330 Turbulent ﬂow, 349 Turning point, 33,37,126, 127, 128, 132 Two-body problem, 120,178E, 185 Underdamped oscillator, 50 Unit mass, 6 Unit tensor, 409 Unit vector, 90,92,94 Units, 11E cgs,11,139 English, 11 gaussian, 139 mks,11,139 Unstable equilibrium (see: Equilibrium) Uranus, 134 Varignon’s theorem, 227 Vector, 68'E - addition of,16,69,72 algebraic and"geometric deﬁnition of,68,71,73 axial, 417 with complex components, 424 component of,70,271E cross product of,75 curlof,98,113f,148 diEerentiation of,82f,93,95 divergence of,97 dotproduct of,73 equality of,68 infdimensions, 475 free, sliding, ﬁxed, 68,226 inner product of,73 integration of,146 lineintegral of,84E magnitude of,68,72 moment of,80f multiplication byascalar, 69,71 null, 75 » outer product of,75 parallel, 75 perpendicular, 75 INDEX 557 polar, 417 projection of,70 resolution intocomponents, 16,79 scalar product of,73 subtraction of,73 triple products of,76 vector product of,75 Vector analysis, 95E Vector angular momentum (see: Angular momentum) Vector angular velocity (see: Angular velocity) Vector identity, 96 Vector moment, 81 -Vector point function, 84 Vector potential, 390 Vector torque, 81 Velocity, 4 components, incylindrical coordinates, 92 inpolar coordinates, 90 inrectangular coordinates, 89,91 inspherical coordinates, 94 along acurve, 148 generalized, 357 internal, 185 ofsound, 312, 333 insound wave, 335, 340 (seealso: Escape velocity; Terminal velocity) Velocity-dependent potential, 389 Velocity potential function, 332 Vibrating membrane, 198 Vibrating string, 198,294E general solution, 299, 302, 395 Lagrange’s equations for,391E made upofparticles, 305E,394_ normal mode, 2_98, 310, 395 withvariable density, 396 (seealso: Wave) Vibration, ofﬂuid inabox, 337E theory of,198,473E,490, 491E (see also: Normal mode; Harmonic oscillator, vibrating string) Virtual displacement, 157, 362 Virtual rotation, 160 Virtual translation, 157Viscosity, 321,345E,441E coefficient of,19,346,443 Viscous ﬂow, 348,441E Viscous force density, 349 Viscous stress tensor, 441E Vortex, 320, 326 Wave, electromagnetic, 26,56,312, 343 harmonic, 300f,336 longitudinal, 335 inapipe, 341E plane, 334 propagation of,310f reﬂection of,304 inasolid, 335 sound, 312,332E power in,336 speed of,333 velocity in,335, 340 spherical, 336 standing, 305, 338 onastring, 300E onastring ofparticles, 309 onatransmission line,312 transverse, 335 traveling, 301f,308, 334ff,338, 342 Wave equation, 296,311f,333,334, 336 Wave mechanics, 337, 399 (seealso: Quantum mechanics) Wave number, 301 Wave vector, 336 Wavelength, 301 Weak coupling, 192f Weak form ofNewton’s third law(see: Newton’s third law) Wind, 111,280 Work, 22,74,101,112 incompression, 328 done byforce ofconstraint, 374 expressed asalineintegral, 85,101 Young’s modulus, 234,236,242,449 YUKAWA, H.,150 |_|IF|H|u__‘_IH_____M___a___|_‘b-"_____‘_|__“|| ______ a"ii §5~!_1d__‘"u_T_"_____._-:5“ |_.r--- _ _ FDR ADVANCED UNDERGRADUATE LEVEL COURSES THE TF1-VLDR MANUAL OF ADVANCED UNDERGRADUATE EXPERIMENTS INPHYSICS ﬁ';1u.'J;rn."-1'4! Fugthi".lm1-"Hcrrrc .lz?-*w.'r'f:firJI1 HfI'.I'l;','.~:l1t'-'1 Trr.lcIu‘r,~1 5-'30 pp,2_’}.§{Uri-ll (I555) Willi-= itsprim.-ir_\' l'un|-tiun istous:-.=isL |l‘.'1l‘ll4'l.'3 inplanning lliiioriitiil-y 4'|'1ll.l"-‘355, 5{_|_||]|||'|[__._; g]|[1|||_;1 i_:.['||] '_[1‘\-|,\|"',' ||ﬂ|‘-[I]! 1-_-f[‘l"('l'.\l‘|' l\¢H'l'l\, I'll‘!'|‘\'v.'li !i'-AL§1‘[I'.‘TZl[ glliilif in thrir lnlrurnLory worl-:. PRlNClP|.E$ OF ELECTRICITY #4-ND MAG-HETISM BYlfl-J-{Lilli-.?{lZ\' .’\I.l'1FGH, f..‘arr1rg2'r {nmrutr ofTwlnnilngy, =\.\'n Iﬁw-ziiariaz W.l‘u<.m, I.I:'.,-if. Reswa-roll Lﬂburrlfory 430pp,ISOflius (I960) .-‘tU'.'&i,-lJl'l<'l'l-i for|.~m1r.~:r~s inr~l|_*|'l.ri<-ity and nmgnvtislu \\'l1lI'll nmi-:1-s usenfveetrlr l1fJl.{I.[l*'}ﬂ unrl 1|.~n-.-a .'\1n;wriun I-urn ntsinalisciissing thesnurcv ofIn:-gnutiv llulrln. Ilinly ii‘.k1wn'l<-rlgu nl|':ll*|‘|llllh isuusumull. l}|'sig1u~1'l prim:irily forutwo-semester |_-ours:-, |.lu~l-oi.-l-1 isntliiplulilv Ina||11£'.-at-muster culirew. 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