Osgood mechanics032869mbp

TEXTCROSS WITHIN THE BOOKONLY crc o> oe oo a<ti<OU_160351 3fh za<Q 5-4cr OSMANIA UNIVERSITY LIBRARY Mo.$f)Accession No. bookshould bereturned onorbefore thedate lastmarked below MECHANICS THEMACMILLAN COMPANY NEWYORK BOSTON CHICAGO DALLAS ATLANTA SANFRANCISCO MACMILLAN &CO., LIMITED LONDON BOMBAY CALCUTTA MELBOURNE THEMACMILLAN COMPANY OFCANADA, LIMITED TORONTO MECHANICS BY WILLIAM FOGG OSGOOD, PH.D., LL.D. PERKINS^ PROFESSOR OFMATHEMATICS, EMERITUS INHARVARD UNIVERSITY NEWYORK THEMACMILLAN COMPANY 1949 COPYRIGHT, 1937, BYTHEMACMILLAN COMPANY. Allrights reserved nopart ofthisbookmay be reproduced inanyform without permission inwriting from thepublisher, except byareviewer whowishes toquote brief passages inconnection with areview written forinclusion inmagazine ornewspaper. Published June, 1937. Reprints! Nov. 1946. Reprinted, May, 1948. Reprinted November, 1949 STUPANDELECTROTYPED BYJ.S.GUSHING CO. PRINTED INTHEUNITED STATES OFAMERICA PREFACE Mechanics isanatural science, and likeanynatural science requires foritscomprehension theobservation andknowledge ofa vastfund ofindividual cases. Arid sothesolution ofproblemsis ofprime importance throughoutallthestudy ofthissubject. ButMechanics isnotanempirical subject inthesense inwhich physics andchemistry, when dealing with theborder region oftUe human knowledgeoftheday areempirical. The latter take cognizance ofagreatnumber ofisolated facts, which itisnotas yetpossible toarrange under afewlaws, orpostulates. Thelaws ofMechanics, likethelaws ofGeometry, sofarasfirstapproxima- tions go thelaws thatexplain themotion ofthegolfballorthe gyroscope ortheskidding automobile, andwhich make possible thecalculation oflunar tables andthepredictionofeclipses these laws areknown, and willboasnew aridimportant two thousand years hence, asintherecent pastofscience when first theyemerged intothelight ofday. Here, then, istheproblem oftraining thestudent inMechanics toprovide himwithavastfund ofcasematerial andtodevelop inhimthehabits ofthought which referanewproblem back tothe fewfundamental laws ofthesubject. The physicistiskeenly alive tothe firstrequirement and tries tomeet itbothbysimple laboratory experiments andbyproblemsinthepart ofageneral course onphysics which isespecially devoted to"Mechanics." The interest ofthemathematician toooften begins with virtual velocities andd'Alembert's Principle, andthevariational principles, ofwhich Hamilton's Principleisthemost important. Both arc right, inthesense thatthey aredping nothing that iswrong ;but each takes suchafragmentary view ofthewhole subject, that his work isineffectual. Theworld inwhich theboyand girlhave lived isthetrue laboratory ofelementary mechanics. The tennis ball, thegolf ball,theshellontheriver;theautomobile good oldModel T, initsday,and thehome-made autos andmotor boats which vi PREFACE youngsters construct andwillcontinue toconstruct theamateur printing press ;thegamesinwhich themechanics ofthebodyisa part ;allthese things gotoprovide thestudent with richlaboratory experience before hebegins asystematic study ofmechanics. It isthisexperience onwhich theteacher ofMechanics candraw, and draw, anddraw again. TheCambridge Tripos offifty years andmore agohasbeen discredited inrecent years, andthecriticism wasnotwithout foundation. Itwasamethod which turned outproblem solvers sosaid itsopponents. But itturned outaClerk Maxwell and it vitally influenced thetraining ofthewhole group ofEnglish physicists, whose workbecame soillustrious. Inhisinteresting autobiography, From EmigranttoInventor, Pupin acknowledges in nouncertain terms thedebt heowes tojust this training, andto Arthur Gordon Webster, through whom hefirstcame toknow thismethod amethod which Benjamin Osgood Peirce also prized highly inhiswork asaphysicist. And sowemake no apologies foravailing ourselves tothefullest extent ofthatwhich theoldTripos Papers contributed totraining inMechanics. But wedonotstop there. Afterall, itisthelaws ofMechanics, their comprehension, their passing over intothefleshandblood ofour scientific thought, andthemathematical technique andtheory, that isourultimate goal. Toattain tothisgoalthemathematical theory, absurdly simple asitisatthestart,must besystematically inculcated intothestudent from thebeginning. Inthisrespect thephysicistsfailus.Because themathematics issimple, they donotthink itimportant toinsist on it.Anyway togetan answer isgoodenough forthem. Butadayofreckoning comes. Thephysicist ofto-dayisindesperate need ofmathematics, and atbest allhecandoistogrope, trying onemathematical expedient after another andholding tonooneofthese longenough totest itmathematically. Nor ishetobeblamed. Itistheold(and most useful) method oftrialanderror heisemploying, andmust continue toemploy forthepresent. Isthewriter onMechanics, percontra, toaccept thechallenge ofpreparing thephysicist tosolve these problems? That istoo large atask. Rather,itisthewisdom ofPasteur who said : "Fortune favors theprepared mind" thatmay wellbeaguide for usnowand inthefuture. What canbedone, andwhatwehavo attempted inthepresent work,istounite abroad anddeepknowv PREFACE vii edgeofthemost elementary physical phenomena inthe field of Mechanics with thebestmathematical methods ofthepresent day, treating with completeness, clarity, and rigor thebeginnings ofthesubject ;inscope notrestricted, indetail notinvolved, in spiritscientific. Thebook isadapted totheneeds ofafirstcourse inMechanics, given forsophomores, andculminating inathorough study ofthe dynamicsofarigidbody intwodimensions. This coursemaybe followed byahalf-course orafullcourse which begins with the kinematics andkinetics ofarigidbody inthree dimensions and proceedstoLagrange's Equations andthevariational principles. Soimportant areHamilton's Equations and their solution by means ofJacobi's Equation, that thissubject hasalsobeen in- cluded. Itappears that there isaspecial need fortreatingthis theory, foralthoughitisexceedingly simple, thecurrent text- books areunsatisfactory. They assume anundefined knowledge ofthetheoryofpartialdifferential equations ofthefirst order, but theydonotshowhowthetheoryisapplied. Asamatter offact, notheoryofthese equations atallisrequired forunderstanding thesolution justmentioned. What isneeded isthefact that Hamilton's Equations areinvariant ofacontact transformation. Asimple proofisgiven inChapter XIV, inwhich themethod mostimportant forthephysicist, namely, themethod ofseparating thevariables,issetforth withnoinvolved preliminaries. But even thisproofmaybeomitted orpostponed, andthestudent may strike inatoncewithChapter XV. Theconcept ofthevector isessential throughout Mechanics, but intricate vector analysisiswholly unnecessary. Acertain minute amount ofthelatter ishowever helpful, andhasbeen set forth inAppendix A. Appendix Dcontains adefinitive formulation ofaclass of problems which ismost importantinphysics, andshows how d'Alembert's Principle andLagrange's Equations apply. Itties together thevarious detailed studies ofthetextand gives the reader acomprehensive view ofthesubject asawhole. Thebook isdesigned asacareful andthorough introduction to Mechanics, butnotofcourse, inthis briefcompass, asatreatise. With theprinciples ofMechanics once firmly established and clearly illustrated bynumerous examples thestudent iswell equipped forfurther study inthecurrent text-books, ofwhichmay viii PREFACE bementioned :Routh :AnElementary Treatise onRigidDynamics and alsoAdvanced Dynamics, bythesame author; particularly valuable for itsmany problems. Webster, Dynamics good material, andexcellent forthestudent who iswelltrained inthe rudiments, buthard readingforthebeginner through poor pre- sentation andlacunae inthetheory ;Appell, Mecanique rationelley vols. iand iiacharming book, which thestudent mayopen atanychapterforsupplementary reading andexamples. Jeans, Mechanics, may alsobementioned forsupplementary exercises; asatext itisunnecessarily hard mathematically fortheSopho- more, and itdoes notgofarenough physically fortheupper- classmari. Itisunnecessary toemphasize theimportanceof further study bytheproblem method ofmore advanced and difficult exercises, such asarefound inthese books. Buttogo further inincorporating these problems into thepresent work would increase itssizeunduly. Itisnotmerely aformal tribute, butoneofdeep appreciation, which Iwish topay toTheMacmillan Company and toThe Norwood Press fortheir hearty cooperationinallthemany dif- ficult details ofthetypography. Good compositionisadistinct aidinsetting forth thethought which theformulas aredesigned toexpress. Itsbeautyisitsownreward. Tohisteacher, Benjamin Osgood Peirce, who firstblazed the trail inhiscourse, Mathematics4,given atHarvard inthemiddle oftheeighties theAuthor wishes toacknowledge hisprofound gratitude. Outofthese beginnings thebook hasgrown, developed through theAuthor's courses atHarvard, extending overmore than forty years, andoutofcourses given later atTheNational UniversityofPeking. Mayitprove ahelp tothebeginner in hisfirstapproach tothesubject ofMechanics. WILLIAM FOGG OSGOOD May 1937 CONTENTS CHAPTER I STATICS OFAPARTICLEPAGE 1.Parallelogram ofForces 1 2.Analytic Treatment byTrigonometry 3 3.Equilibrium. TheTriangle ofForces. Addition ofVectors .4 4.ThePolygon ofForces 7 5.Friction 9 6.Solution ofaTrigonometric Equation. Problem 12 Exercises onChapterI 15 CHAPTER II STATICS OFARIGID BODY 1.Parallel Forces inaPlane 21 2.Analytic Formulation;nForces 23 3.Centre ofGravity 26 4.Moment ofaForce 28 5.CouplesinaPlane 29 6.Resultant ofForces inaPlane. Equilibrium 32 7.Couples inSpace 34 8.Resultant ofForces inSpace. Equilibrium 36 9.Moment ofaVector. Couples 37 10.Vector Representation ofResultant Force andCouple. Resul- tant Axis. Wrench 38 11.Moment ofaVector about aLine 40 12.Equilibrium 41 13.Centre ofGravity ofnParticles 42 14.Three Forces 43 Exercises onChapter II 46 CHAPTER III MOTION OFAPARTICLE 1.Rectilinear Motion 49 2.Newton's Laws ofMotion 50 3.Absolute Units ofForce 55 4.Elastic Strings 58 ix x CONTENTS PAOB 5.AProblem ofMotion 60 6.Continuation; theTime 63 7.Simple Harmonic Motion 64 8.Motion under theAttraction ofGravitation 69 9.WorkDonebyaVariable Force 72 10.Kinetic Energy andWork 75 11.ChangeofUnits inPhysics 76 12.TheCheck ofDimensions 79 13.Motion inaResisting Medium 81 14.Graph oftheResistance ;84 15.Motion inaPlane andinSpace 86 16.Vector Acceleration 90 17.Newton's Second Law 92 18.Motion ofaProjectile 93 19.Constrained Motion 95 20.Simple Pendulum Motion 97 21.Motion onaSmooth Curve 99 22.Centrifugal Force 101 23.TheCentrifugal OilCup 105 24.TheCentrifugal Field ofForce 106 25.Central Force 108 26.TheTwoBody Problem 114 27.TheInverse Problem toDetermine theForce 114 28.Kepler's Laws 115 29.OntheNotion ofMass 118 CHAPTER IV DYNAMICS OFARIGID BODY 1.Motion oftheCentre ofGravity 120 2.Applications 123 3.TheEquation ofMoments 126 4.Rotation about aFixed Axisunder Gravity 127 5.TheCompound Pendulum 130 6.Continuation. Discussion ofthePoint ofSupport.... 132 7.Kater's Pendulum 133 8.Atwood's Machine 134 9.TheGeneral Case ofRotation about aPoint 136 10.Moments ofInertia 137 11.TheTorsion Pendulum 139 12.Rotation ofaPlane Lamina, NoPoint Fixed 139 13.Examples141 CONTENTS xi PAGE 14.Billiard Ball,Struck Full 143 15.Continuation. TheSubsequent Motion........ 145 16.Further Examples 146 Exercises onChapter IV 151 CHAPTER V KINEMATICS INTWODIMENSIONS 1.TheRolling Wheel' 154 2.TheInstantaneous Centre 155 3.Rotation about theInstantaneous Centre 157 4.TheCentrodes 159 5.Continuation. Proof oftheFundamental Theorem.... 162 6.TheDancing TeaCup 165 7.TheKinetic Energy ofaRigid System 166 8.Motion ofSpace withOnePoint Fixed 168 9.Vector Angular Velocity 170 10.Moving Axes. Proof oftheTheorem of8 172 11.Space Centrode andBody Centrode 174 12.Motion ofSpace. General Case 175 13.TheRuled Surfaces 176 14.Relative Velocities 177 15.Proof oftheTheorem of12 179 16.Lissajou's Curves 182 17.Continuation. TheGeneral Case. TheCommensurable Case. Periodicity 186 Professor Sabine's Tracings ofLissajou's Curves between pages 190-191 CHAPTER VI ROTATION 1.Moments ofInertia 191 2.Principal Axes ofaCentral Quadric 194 3.Continuation. Determination oftheAxes 196 4.Moment ofMomentum. Moment ofaLocalized Vector . .197 5.TheFundamental Theorem ofMoments 199 6.Vector Form fortheMotion oftheCentre ofMass.... 201 7.TheInvariable LineandPlane 201 8.Transformation of 202 9.Moments about theCentre ofMass 204 10.Moments about anArbitrary Point 205 xii CONTENTS PAGE 11.Moments about theInstantaneous Centre 207 12.Evaluation ofaforaRigid System ;OnePoint Fixed... 208 13.Euler's Dynamical Equations 210 14.Motion about aFixed Point 212 15.Euler's Geometrical Equations 214 16.Continuation. TheDirection Cosines oftheMoving Axes . .216 17.TheGyroscope 217 18.TheTop 220 19.Continuation. Discussion oftheMotion 222 20.Intrinsic Treatment oftheGyroscope 225 21.TheRelations Connecting v,F,and K 229 22.Discussion oftheIntrinsic Equations 231 23.Billiard Ball 237 24.Cartwheels 241 25.R&ume* 245 CHAPTER VII WORK ANDENERGY 1.Work 248 2.Continuation :Curved Paths 250 3.Field ofForce. Force Function. Potential 253 4.Conservation ofEnergy 256 5.Vanishing oftheInternal Work foraRigid System.... 258 6.Kinetic Energy ofaRigidBody 260 7.Final Definition ofWork 261 8.WorkDonebyaMoving Stairway 264 9.Other Cases inWhich theInternal Work Vanishes .... 266 10.Work andEnergyforaRigidBody 266 CHAPTER VIII IMPACT 1.ImpactofParticles 270 2.Continuation. Oblique Impact 274 3.Rigid Bodies 277 4.Proof oftheTheorem 279 5.Tennis Ball,Returned withaLawford 282 CHAPTER IX RELATIVE MOTION ANDMOVING AXES 1.Relative Velocity 285 2.Linear Velocity inTerms ofAngular Velocity 285 CONTENTS xiii PAGE 3.Acceleration 287 4.TheDynamical Equations 290 5.TheCentrifugal Field 291 6.Foucault Pendulum 292 CHAPTER X LAGRANGE'S EQUATIONS ANDVIRTUAL VELOCITIES 1.TheProblem 297 2.Lagrange's Equations intheSimplest Case 299 3.Continuation. Particle onaFixed orMoving Surface . . .303 4.TheSpherical Pendulum 306 5.Geodesies 308 6.Lemma 310 7.Lagrange's EquationsintheGeneral Case 312 8.Discussion oftheEquations. Holonomic andNon-Holonomic Systems 313 9.Continuation. TheForces 315 10.Conclusion. Lagrange's Multipliers 316 11.Virtual Velocities andVirtual Work 318 12.ComputationofQr 320 13.Virtual Velocities, anAidintheChoice ofthe TTT 321 14.OntheNumber moftheQT 322 15.Forces ofConstraint 325 16.Euler's Equations, Deduced from Lagrange's Equations. . .325 17.Solution ofLagrange's Equations 326 18.Equilibrium 330 19.Small Oscillations 333 Exercises onChapterX 336 CHAPTER XI HAMILTON'S CANONICAL EQUATIONS 1.TheProblem 338 2.AGeneral Theorem 339 3.Proof ofHamilton's Equations 342 CHAPTER XII D'ALEMBERT'S PRINCIPLE 1.TheProblem 345 2.Lagrange's EquationsforaSystemofParticles, Deduced from d'Alembert's Principle 348 3.The SixEquationsforaSystemofParticles, Deduced from d'Alembert's Principle 349 riv CONTENTS PAGE 4.Lagrange's EquationsintheGeneral Case, andd'Alembert's Principle 350 5.Application:Euler's Dynamical Equations 352 6.Examples 353 CHAPTER XIII HAMILTON'S PRINCIPLE ANDTHEPRINCIPLE OF LEAST ACTION 1.Definition of5 356 2.TheIntegral ofRational Mechanics 360 3.ApplicationtotheIntegral ofKinetic Energy 362 4.Virtual Work 364 5.TheFundamental Equation 364 6.TheVariational Principle 370 7.Hamilton's Principle 370 8.Lagrange's Principle ofLeast Action 372 9.Jacobi's Principle ofLeast Action 377 10.Critique oftheMethods. Retrospect andProspect.... 379 11.Applications 379 12.Hamilton's Integral aMinimum inaRestricted Region. . .381 13.Jacobi's Integral aMinimum inaRestricted Region.... 386 CHAPTER XIV CONTACT TRANSFORMATIONS 1.Purpose oftheChapter 389 2.Integral Invariants 392 3.ConsequencesoftheTheorem 395 4.Transformation ofHamilton's Equations byContact Trans- formations 400 5.Particular Contact Transformations 403 6.Theft-Relations 407 CHAPTER XV SOLUTION OFHAMILTON'S EQUATIONS 1.TheProblem andItsTreatment 410 2.Reduction totheEquilibrium Problem 413 3.Example. Simple Harmonic Motion 415 4.H,Independent of t.Reduction totheForm, H'=Pi . . .420 5.Examples. Projectile invacuo 424 CONTENTS xv PAGE 6.Comparison oftheTwoMethods 429 7.Cyclic Coordinates 430 8.Continuation. TheGeneral Case 433 9.Examples. TheTwo-Body Problem 434 10.Continuation. TheTop 438 11.Perturbations. Variation ofConstants 440 12.Continuation. ASecond Method 444 APPENDIX A.Vector Analysis 447 B.The Differential Equation:(du/dt)z=f(u) 456 C.Characteristics ofJacobi's Equation 466 D.TheGeneral Problem ofRational Mechanics 476 INDEX . . . . 491 MECHANICS CHAPTER I STATICS OFAPARTICLE 1.Parallelogram ofForces. Byaforceismeant apush ora pull.Astretched elastic band exerts aforce.Aspiral spring, likethose used intheupholstered seats ofautomobiles, when compressed byaload, exerts aforce. Theearth exerts aforce ofattraction onafalling raindrop. The effect ofaforce acting atagiven point, 0,depends not merely onthemagnitude, orintensity, oftheforce, butalsoon thedirection inwhich itacts. Layoffaright linefrom Ointhedirection oftheforce, andmpke thelength of^^ thelineproportional totheintensityoftheforce;forpJG1 example,ifFis10Ibs., thelength maybetaken as 10in.,or10cm., ormore generally, tentimes thelength which represents theunit force. Then thisdirected right line, orvector, gives acomplete geometric picture oftheforce. Thus ifabarrel offlour issuspended byarope (andisatrest), theattraction of gravity thepulloftheearth willberepresented byavector pointing downward and oflength W,theweight ofthebarrel. Ontheother hand, theforcewhich therope exerts onthebarrel willberepresented byanequal andopposite vector, pointing upward. For, action and reaction areequal and opposite. When two forces actatapoint, they are equivalenttoasingle force, which isfound as follows. Lay offfrom thepoint thetwovec- tors,PandQ,which represent thegiven forces, andconstruct theparallelogram, ofwhich the Fia2 rightlinesegments determined byPandQare twoadjacentsides. Thediagonal oftheparal- lelogram drawn from determines avector, R,which represents 1 2 MECHANICS thecombined effect ofPand Q.This force, R,iscalled the resultant ofPandQ,andthefigure justdescribed isknown asthe parallelogram offerees. Example1.Two forces of20pounds eachmake anangle of 60witheach other. Tofindtheir resultant. Here,itisobvious from thegeometry ofthe figure that theparallelogramisarhombus, and that_the length ofthediagonal inquestionis 20N/3=34.64. Hence theresultant isaforce of 34.64 pounds,itsline ofaction bisecting theangle between the given forces. Example2.Two forces of7pounds and9pounds actata point andmake anangle of70witheach other. Tofind their resultant. Graphical Solution. Draw theforces toscale, constructing the angle bymeans ofaprotractor. Then complete theparallelo- gram andmeasure thediagonal. Find itsdirection with the protractor. Example3.Apicture weighing 15Ibs.hangs from anail in thewallbyawire, thetwosegments ofwhich make angles of30 with thehorizon. Find thetension inthewire. Here, theresultant, 15,ofthetwounknown tensions, TandT9 isgiven, andtheangles areknown. Itisevident from thefigure thatTalsohasthevalue 15.Sotheanswer is :15Ibs. Decomposition ofForces. Conversely, agiven force canbede- composed along anytwodirections whatever. Allthat isneeded is,toconstruct theparallelogram, ofwhich thegiven force is thediagonal andwhose sides liealong thegiven lines. FBlTKf, FIG.4 FIG.5 If,inparticular, the lines areperpendiculartoeach other, thecomponentswillevidently be : Fcos<p, Fsin<p. STATICS OFAPARTICLE 3 EXERCISES 1.Two forces of5Ibs.and12Ibs.make aright angle with each other. Show that theresultant force is13Ibs.andthat it makes anangleof2237'with thelargerforce. 2.Forces of5Ibs.and7Ibs.make anangle of100with each other. Determine theresultant force graphically. 3.Iftheforces inQuestion1make anangle of60with each other, findtheresultant. Give first agraphical solution. Then obtain ananalytical solution, using however notrigonometry beyond atable ofnatural sines, cosines, andtangents. 4.Iftwoforces of12Ibs.and16Ibs.havearesultant of20Ibs., what anglemust theymake witheachotherandwith theresultant? 5.Aforce of100 Ibs.acts north. Resolve itintoaneasterly andanorth-westerly component. 6.Aforce of50Ibs.acts east north-east. Resolve itintoan easterly andanortherly component. Ans. 46.20 Ibs.;19.14 Ibs. 7.Aforce of12Ibs.acts inagiven direction. Resolve itinto twoforces thatmake angles of30and40with itsline ofaction. Only agraphical solution isrequired. 2.Analytic Treatment byTrigonometry. The problemof finding theresultant calls forthedetermination ofoneside ofa triangle when theother two sides and theincluded angle arc known; and also offinding theremaining angles. The first problemissolved bytheLaw ofCosines inTrigonometry: (1)c2=a2+62-2abcosC. Here, a=P,b=Q, c=R, C=180-w andhence c (2)722=P2+Q2+2PQcos co. FIG. Example. Forces of5Ibs.and8Ibs.make anangle of120 witheach other. Find their resultant. Here, R2=25+64-2X5X8X=49; R=7Ibs. Tocomplete thesolution and find theremaining angles we canusetheLaw ofSines : 4 MECHANICS ~ I, (3) 1 (4)sinA sinB sinC Thus sm^>sm There isnodifficulty hereabout thesignwhen theadjacent angleisused, since sin(180 to)=sin co. Inthenumerical example above, Equation (4)becomes : 8 7 sin<p|\/3 Thus sin<p=4-V3,cos<p=|, <t>=8147'. The third angleisfound from thefactthat thesum ofthe angles ofatriangleistworight angles: A+B+C=180. Tosumup,then :Compute theresultant bytheLaw ofCosines andcomplete thesolution bytheLaw ofSines. EXERCISES Giveboth agraphical andananalytical solution each time. 1.Forces of2Ibs.and3Ibs.actatright angles toeach other. Find their resultant inmagnitude anddirection.* 2.Forces of4Ibs.and5Ibs.make anangle of70with each other. Find their resultant. 3.Equilibrium. The Triangle ofForces. Addition ofVec- tors. Inorder that three forces beinequilibrium,itisclearly necessary andsufficient thatanyoneofthem beequalandopposite totheresultant oftheother two. Thecondition canbeexpressed conveniently byaidoftheidea oftheaddition ofvectors. First ofall,twovectors aredefined asequaliftheyhave the same magnitude, direction, and sense, nomatter where inthe plane (orinspace) theymaylie. *Observe that inthiscase itiseasier todetermine theangle from itstangent. Square roots should becomputed from aTable ofSquare Roots. Huntington's Four-Place Tables areconvenient, andareadequate fortheordinary cases that arise inpractice. Butcases notinfrequently arise inwhich more elaborate tables areneeded, andBarlow's willbefound useful. STATICS OFAPARTICLE 5 Vector Addition. LetAandBbeanytwovectors. Construct Awithanypoint, 0,asitsinitial point. Then, with theterminal pointofAasitsinitial point, construct B.The vector, C,whose initial pointisthe initial point ofAandwhose terminal point istheterminal pointofB,isdefined asthevector sum, or,simply, thesumofAandB : C=A+B. Itisobvious that B+A=A+B. Anynumber ofvectors canbeadded byapplying thedefinition successively. Itiseasily seenthat (A+B)+C=A+(B+C). Consequently thesum A!+A2+---+An isindependent oftheorder inwhich theterms areadded. Foraccuracy andcompletenessitisnecessary tointroduce the nilvector. Suppose, forexample, thatAandBareequal and opposite. Then theirsum isnotavector inanysense asyet considered, fortheterminal point coincides with theinitial point. When thissituation occurs, wesaythatwehaveanilvector, and denote itby: A+B=0. Wewrite, furthermore,*B=-A. Equilibrium. The condition, necessary andsufficient, that three forces beinequilibriumisthat their vector sumbe0.Geo- metrically this isequivalent tosaying that thevectors which represent theforces canbedrawn sothat the fig- urewillcloseandformatriangle. From theLaw ofSineswehave : P Q_E sinp sinq sine FIG.8" p+q+e=180. *Itisnotnecessary forthepresent togofurther intovector analysis than the above definitions imply. Later, thetwoforms ofproduct willbeneeded, and thestudent may beinterested even atthisstage inreading Chapter XIII ofthe author's Advanced Calculus, orAppendix A. 6 MECHANICS FIG.9Since sin(180 A)=sin4,wecanstate theresult inthe following form. Letthree forces, P,Q,andE,acting onapar- ticle, beinequilibrium. Denote theangles between the forces, asindicated, byp,q,e.Then Equa- tion (1)represents anecessary condition forequilib- rium. Conversely, thiscondition issufficient. Wethus obtain aconvenient solution inallcases except theoneinwhich themagnitudes oftheforces, butnoangles, aregiven. Here, theLaw ofCosines* gives one angle, andthen asecond angle canbe computed bytheLaw ofSines. Example1.Forces of4,5,and6areinequilibrium. Find theangles between them. First, solve theproblem graphically, measuring the angles. Next, apply theLaw ofCosines : 42=52+62_2X5X6Xcosp, cos<p=f, <p=4123'. Asecond angleisnowcomputed bytheLaw ofSines : 5 4 sin i 5V7 16'sin<p 5546'. Thethird angleis8251'. Example2.A40 Ib.weight rests onasmooth horizontal cylinder and iskeptfrom slipping byacord thatpasses over the cylinder and carries a10Ib.weight atitsother end. Find theposition ofequilibrium. Thecord isassumed weightless, and since it passes over asmooth surface, thetension in itisthesame atallpoints. The surface of thecylinderissmooth, hence itsreaction is normal toitssurface. Let 6betheunknown angle that theradius drawn totheweight FIG.10 *Ifwehadalargenumber ofnumerical problems tosolve, itwould paytouse themore elaborate theorems ofTrigonometry (e.g.Law ofTangents). But for ordinary household purposes themore familiar law isenough. STATICS OFAPARTICLE makes with the vertical. Then from inspection ofthe figure weseethat 10 40 orsin6sin90' sin6= ,=1429'. EXERCISES Find 1.Forces of7,8,and9pounds keep aparticle atrest, theangles theymake withoneanother. 2.Forces of51.42, 63.81, and71.93 grs.keep aparticle atrest. What angle dothe firsttwoforces make witheach other? Find theother angles. 3.Aweightless string passes overtwosmooth pegs atthesame levelandcarries weights ofPandPatitsends. Inthemiddle, there isknotted aweight W.What angle dothesegments ofthestring p make with thevertical, when the systemisatrest? w Ans. sin6-^ 4.Aboat isprevented from drifting down stream bytworopes tiedtothebow oftheboat,andtostakes atopposite points onthe banks. Oneropeis125 ft.long; theother, 150ft.;andthe stream is200 ft.broad. Ifthetension intheshorter ropeis 20Ibs.,what isthetension intheother rope? 6.Twosmooth inclined planes, back to back, meet along ahorizontal straight line,andmake angles of30arid45with thehorizon. Aweight of10Ibs.placed onthe firstplaneisheldbyacordthat passes over thetopoftheplanes and carries aweight W,resting ontheother plane andattached to theendofthecord. Findwhat valueWmust have. 4.The Polygon ofForces. From the case ofthree forces thegeneralization to thecase ofnforces acting atapoint pre- sentsnodifficulty. Addtheforces geometri- cally,i.e.bythevector law.The vector sumrepresents theresultant ofallnforces. Thus, inthefigure, theresultant isgivenby FIG.13FIG.12 8 MECHANICS thevector whose initial pointisthepoint andwhoso terminal pointisP. Thecondition forequilibriumisclearly thattheresultant bea nilvector, orthat thebroken line close andform apolygon butnotnecessarily apolygoninthesense ofelementary geometry, since itssidesmay intersect, as inFigure14. This condition willobviously beful- filled ifandonlyifthesum oftheprojectionsoftheforces along each oftwo lines that intersect iszero.* Analytically, the resultant canberepresented asfollows. LetaCartesian system ofcoordinates beassumed, and letthe components oftheforceFkalong theaxes beXkandYk.If, now, thecomponents oftheresultant aredenoted byXandY, wehave : v_v ,v, ,v AA! -\-A2~r*~rAn, Y=Yt+F2++Yn. Suchsums arewritten as (1) X*, or Xk> or Z*> t=l I depending onhow elaborate thenotation should betoinsure clearness. The forces willbeinequilibrium if,andonly if,theresultant force isnil,andthis willbethecase if (2) 2)Xk=0,2Yk=0. t-i t=i EXERCISE A4Ib.weightisacted onbythree forces, allofwhich lieinthe same vertical plane:aforce of10Ibs.making anangle of30with thevertical, andforces of8Ibs.and12Ibs.ontheother side of thevertical andmaking anglesof20with theupward vertical and15with thedownward vertical respectively. Find theforce that willkeep thesystem atrest. Space ofThree Dimensions. Ifmore thantwoforces actata point, theyneed not lieinaplane. Buttheycanbeadded two *Cf.Osgood andGraustein, Analytic Geometry, pp.1-6. STATICS OFAPARTICLE 9 atatimebytheparallelogram law, the firsttwothus being replaced byasingle force their resultant and this force in turncompounded with thethird force; etc.Thebroken line thatrepresents theaddition ofthevectors nolongerliesinaplane, butbecomes askewbroken lineinspace, andthepolygonofforces becomes askew polygon. Thecomponents oftheresultant force along thethree axesare : nn^n (3)X=2^X^ Y=2^Yk, Z=2^Zk. Thecondition forequilibriumis : Inallofthese formulas, Xk,Yk,Zkarealgebraic quantities, being positive when thecomponent hasthesense ofthepositive axis ofcoordinates, andnegative when thosense istheopposite. Insolving problems inequilibriumitisfrequently simpler tosingle outthecomponents thathaveonesense along thelineinquestion andequate their sum, each being taken aspositive, tothesum ofthecomponentsintheopposite direction, each ofthese being taken aspositive,also. Themethod willbeillustrated bythe examplesinfriction ofthenextparagraph. 5.Friction. Letabrick beplaced onatable;letastring be fastened tothebrick, and letthestring bepulled horizontally with aforceFjust sufficient tomove thebrick. Then thelaw of physicsisthat A F=pR, R\ rJ-p^ whereR(here, theweight ofthebrick)isp15 thenormal* pressure ofthetable onthe brick,aridju(the coefficient offriction)isaconstant forthetwo surfaces incontact. Thus,ifasecond brick were placed ontop ofthefirst,Rwould bedoubled, andsowould F. Wecanstate thelawoffriction generally bysaying: When twosurfaces areincontact andone isjustonthepoint ofslipping *Normal means, atright angles tothesurface inquestion. Thenormal toa surface atapoint isthelineperpendicular tothetangent plane ofthesurface at thepoint inquestion. 10 MECHANICS over theother, thetangential forceFduetofriction ispropor- tional tothenormal pressure Rbetween thesurfaces, or where/*,thecoefficient offriction,isindependent ofFandR, anddepends onlyonthesubstances incontact, butnotonthe area ofthesurfaces which touch each other. Formetals onmetals nusuallyliesbetween 0.15and0.25 inthecase ofstatical fric- tion. Forsliding friction/*isabout 0.15;cf.Rankine, Applied Mechanics. Asimple experiment often performedinthelaboratoryfor determining /xisthefollowing. Letoneofthesurfaces berepre- sented byaninclined plane, theangle of which canbevaried. Lettheother surface berepresented byarider, orsmall block of thesubstance inquestion, placed onthe plane. Iftheplaneisgraduallytilted from ahorizontal position, therider willnotslip foratime. Finally, apositionwillbereached forwhich therider just slips. Thisangle oftheplaneisknown as theangle offriction and isusually denoted byX.Letusshow that ntan X. Resolve theforce ofgravity, W,into itstwocomponents along theplane andnormal totheplane. These are : WsinX,WcosX. Andnow theforces acting uptheplane (i.e.thecomponents directed uptheplane) must equal theforcesdown theplane, or F=WsinX; andtheforces normal tothe.plane andupward must equal the forces normal totheplane anddownward, or R=WcosX. Hence F sinX, x-=an^RcosX But F=/J2. Consequently ntanX. STATICS OFAPARTICLE 11 Example. A50Ib.weightisplaced onarough inclined plane, angle ofelevation, 30.Acord attached totheweight passes overasmooth pulley atthetopoftheplaneandcarries aweightW atitslower end. Forwhat values ofWwillthesystem bein equilibriumifjut= -J? Here, X<30,and so,ifWisvery small, the50Ib.weight will slipdown theplane. SupposeWisjustlargeenough toprevent slipping. Then friction actsuptheplane, andtheforces which produce equilibrium arethose indicated. Hence 'F+W 50sin30W 50cos30 Fio.17 Fia.18 F+W=50sin30=25, R=50cos30=25V3, and F=%R. Itfollows, then, that W=6~ 25=17.8 Ibs.o If,now,Wisslightly increased, the50Ib.weight willobviously stillbeinequilibrium, and this willcontinue tobethecase until the50Ib.weightisjustonthepoint ofslipping uptheplane. This willoccurwhenW=32.2 Ibs.asthestudent cannowprove forhimself. Consequently, thevalues ofWforwhich there is equilibrium arethose forwhich 17.8^Wg32.2. EXERCISES 1.IfthecylinderofExample 2, 3,isrough, /z= ,findthe totalrange ofequilibrium. Ans. 449'gg2344'. 2.Consider theinclined planes ofExercise 5, 3.Iftheone onwhich the10Ib.weight rests isrough, /z=T^,findtherange ofvalues forWthat willyield equilibrium. 12 MECHANICS 3.Prove theformula H=tanX bymeans ofthetriangle offorces,i.e.theLaw ofSines, 3,(1). 6.Solution ofaTrigonometric Equation. Problem. A50Ib. weight rests onarough horizontal plane, ju=-.Acord is fastened totheweight, passes overasmooth pulley 2ft.above theplane, and carries aweight of25Ibs.which hangs freely at itsother end.Tofind allthepositions ofequilibrium. Resolving theforces horizontally andvertically, wefind : 25cos6=R, 25sin+R=50. Hence, eliminating R,weobtain theequation: FIG. 19 (1)6 cos6+sin=2. Thisequationisoftheform : (2) acos+bsin=c, and issolved asfollows.*Divide through byVa2+b2 :50 cosacos+sinasin6= Va2+62' or (5)cos(0 a)= Va2+b2 *Thestudent should observe carefully thetrigonometric technique setforth inthisparagraph, notmerely because equations ofthistype areimportant inthem- selves, butbecause thepractical value ofaworking knowledge oftrigonometry isnotconfined tosolving numerical triangles. Offargreater scope andimportance inpractice arethepurely analytical reductions toother trigonometric foi-ms, and thesolution oftrigonometric equations. That isone ofthereasons why the harder examples attheend ofthechapter arevaluable. They notonly give needed practice informulating mathematically physical data;they require also theability tohandle analytical trigonometry according tothedemands ofpractice. STATICS OFAPARTICLE 13 Theangleaismost easily determined from theequation: (6) Thus aisseen tobeoneoftwoangles which one isrendered clearbyplotting thepoint ontheunit circle : z2+2/2=1, whose coordinates are =__JL__ 7b__ x' y~ Theangle from thepositive axis ofxtotheradius drawn tothis pointisa.Thuswehave agraphical determination ofa.Itis notnecessary tocompute thecoordinates accurately, butmerely toobserve inwhich quadrant thepoint lies,soastoknow which root oftheequation fortanatotake. Thus ifb>0,amust be anangle ofthe firstorsecond quadrant. Finally,ifc/Va2+b2 isnumerically greater than1,theequation hasnosolution. Indefining a,itwould, ofcourse, Jiave answered justaswell if sinaandcosahadbeen interchanged, and ifeither orboth the ratios in(4)hadbeen replaced bytheir negative values.* Returning now tothenumerical equation above, weseethat tana= , sina>0, a=928'; cos(6-9280- -928'=7048'. Since 6intheproblem before usmust beanangle ofthe first quadrant, thelower signisimpossible, and =80 16'. Wehave determined thepointoftheplane atwhich allthe friction iscalled intoplayandthe50Ib.weightisjustonthe point ofslipping. Forother positions, Fwillnotequal pR. Such aposition willbeoneofequilibriumiftheamount offriction actually called into play, orF,islessthan theamount thatcould *Equation (2)might alsohave been solved bytransposing onetermfrom the left- totheright-hand sideandsquaring. Onusing thePythagorean Identity: sin2-fcos2=1, weshould beledtoaquadratic equation inthesineorcosine. Thisequation will, ingeneral, havefourrootsbetween and360,andthree ofthemmust beexcluded. Moreover, theactual computation bythismethod ismore laborious. 14 MECHANICS becalled intoplay, orpR. Itseems plausible thatsuch pointslie totheright ofthecritical point ;butthisconclusion isnotim- mediately justified, for,although theamount offriction required, 01F=25cosB, islessforalarger 0, still, theamount available, or n50-25 sin Mfi-- g- , isalso less.Wemust prove, therefore, that F<nR or OK ^50-25sin25cos<-5-- b This willobviously besoif 6cos6<2-sin0, 8016'<<90, orif 6cos+sin<2, rif 6 1 2 orif. cos(0-9287 )<-=' As0,starting with thevalue 8016',increases, 928'also increases, andconsequently cos(0928') decreases. Conse- quently ourguess isborne outbythefacts, andthe50Ib.weight willbeinequilibrium atallpoints onthetable within acircle of radius .343ft.,oralittle over4in.,whose centre isdirectly under thepulley. EXERCISES 1.Solve thesame problemiftheplaneisinclined atanangle of15with thehorizon, andthevertical plane through theweight andthepulleyisatright angles totherough plane. Ana. 4333'g0g 6924'. 2.Atwhat angle should theplane betilted, inorder that the region ofequilibrium may justextend indefinitely down theplane? 3.Find theangle ofthethird quadrant determined bythe equation: n. M3sin6-2cos=1. STATICS OFAPARTICLE 15 4.Show thatthere areinalleightways ofsolving Equation (2), given bysetting theright-hand sides ofEquations (4)equal to cosa,+sinaand sina, cosa,where the signs are independent ofeach other. 5.Evaluate theintegral: dx s-.acosx+bsinx 6.Solve theequation: 2cos2 <p4cos<psin<p3sin2 <p=5. Suggestion. Introduce thedouble angle, 2<p. EXERCISES ONCHAPTER I* 1.Arope runs through ablock, towhich another ropeis attached. The tension inthe first ropeis 120Ibs.,andtheangleitincludes is70. What isthetension inthesecond rope? 2.Amanweighing 160 Ibs. islying ina hammock. Therope athisheadmakes an angle of30with thehorizon, andtherope athisfeet,anangle of15. Find thetensions inthetworopes. 3.Aload offurniture isbeing moved. Therope that binds itpasses overtheround ofachair. Thetension ononesideofthe round is40Ibs.andontheotherside, 50Ibs.;andtheangleis 100. What force doestheround have towithstand ? 4.Acanal boat isbeing towed byahawser pulled byhorses onthebank. Thetension inthehawser is400 Ibs.and itmakes 4 anangle of15with thebank. What is theeffective pullontheboat inthedirec- tionofthecanal? 5.Acrane supports aweight ofaton asshown inthe figure. What arcthe forces inthehorizontal andintheoblique FlG -21member? 6.Three smooth pulleys canbesetatpleasure onahorizon- talcircular wire. Three strings, knotted together, passover the *Thestudent should begin eachtimebydrawing anadequate figure, illustrating thephysical objects involved, andheshould putintheforces with colored inkor pencil. Abottle ofredink,used sparingly, contributes tremendously toclear thinking. 16 MECHANICS pulleys andcarry weights of7,8,and9Ibs.attheir freeends. Howmust thepulleys beset,inorder that theknotmaybeat restatthecentre ofthecircle ? 7.Atelegraph poleattwocross- roads supports acable, thetension inwhich isaton.Thecable liesin ahorizontal plane and isturned through aright angle atthe pole. The poleiskeptfrom tipping byaFIG.22 FIG.23stayfrom itstoptotheground, thestay making anangle of45with theverti- cal.What isthetension inthestay? 8.The figure suggests astake ofa circus tent, with atension of500 Ibs. tobeheld. What isthetension in thestay,ifthestake could turn freely? 9.Twomen areraising aweight of150 Ibs.byarope that passes overtwosmooth pulleys and is knotted atA.How hard arethey pulling? 10.If,inthepreceding question,in- stead ofbeing knotted atA,thetwo ropes themenhave hold ofpassed over pulleys atAandwere vertical above A, howhardwould thementhenhave to pull? 11.AweightWisplaced inasmooth hemispherical bowl;a string, attached totheweight, passes over theedge ofthebowl and carries aweightPatitsother end. Find theposition of equilibrium. 12.Solve thesameproblem foraparabolic bowl, therimbeing atthelevel ofthefocus. 13.Oneendofastringismade fasttoapegatA.The string passes overasmooth pegatJ?,atthesame level asA,andcarries a weightPatitsfreeend.Asmooth heavy bead, ofweight W,can slideonthe string. Find theposition ofequilibrium andthe pressure onthepegatB. 14.Abead weighingWIbs.can slideonasmooth vertical circle ofradius a.Tothebead isattached astring that passesFIG.24 STATICS OFAPARTICLE 17 overasmooth pegsituated atadistance %aabove thecentre of thecircle, andhasattached toitsother endaweight P.Find allthepositionsofequilibrium. 16.A50Ib.weight restsonasmooth inclined plane (angle with thehorizontal, 20)and iskeptfrom slipping byacordwhich passes over asmooth peg1ft.above thetopoftheplane, and which carries aweight of25Ibs.atitsother end. Find the position ofequilibrium. 16.Aheavy bead canslideonasmooth wire intheform ofa parabola with vertical axisandvertex atthehighest point, A string attached tothebead passes overasmooth pegatthefocus oftheparabola andcarries aweight atitsother end.Show that ingeneral there isonlyonepositionofequilibrium ;butsome- times allpositions arepositions ofequilibrium. 17.Aweightless bead*canslideonasmooth wire intheform ofanellipse whose planeisvertical. Astringisknotted tothe beadandpasses overtwosmooth pegs atthe foci,which areat thesame horizontal height. Weights ofPandQfiroattached tothetwoends ofthestring. Find thepositionsofequilibrium. 18.Aninextensible flexible string has itsendsmade fastattwo points and carries aweightless smooth bead. Another stringis fastened tothebeadanddrawn taut. Show thatevery position ofthebead isoneofequilibrium,ifthesecond stringisproperly directed. 19.Give amechanical proof, based onthepreceding question, thatthefocal radii ofanellipse make equal angles withthetangent. 20.Abead ofweightPcan slide onasmooth, vertical rod. Tothebead isattached aninextensible string oflength 2a, carrying atitsmiddle point aweightWandhaving itsother end made fasttoapegatahorizontal distance afrom therod.Show thatthepositionofequilibriumisgivenbytheequations: Ptan<p=(P+W}tan6, sin+sin<p=1, where0,<paretheangles thesegments ofthestring make with thevertical. *Questions ofthistypemaybeobjected toontheground thataforcemust act onmass, andsothere isnosense inspeaking offorces which actonamasslcss ring. But iftheringhasminute mass, thedifficulty isremoved. Theproblem maybe thought of,then, asreferring toaheavy bead, whose weight isjustsupported by avertical string. Since theweight ofthebeadnowhasnoinfluence ontheposition ofequilibrium, themass ofthebeadmaybetaken asvery small, and so,physically negligible. 18 MECHANICS 21.If,inthepreceding question, P=Wyshow that=2155', V=3849'. Determine these angles whenW=2P. 22.Aflexible inextensible stringintheform ofaloop60in. longislaidovertwosmooth pegs20in.apart and carries two smooth beads ofweightPandW.Find theposition ofequi- librium,ifthebeads cannot come together. Ans. Wsin6=Psin^>, cos6+cos(f>=3cos6cosp; cos4e- |cos3e- I(i-~)cos2e 2 hence 9cos46-6cos36-8\cos28+6Xcos6-X=0, where_(W2-P2 )X~TP 23.Show thatif,inthepreceding question, P=5and TF=10,=258',andfindthereaction onthepeg. 24.Oneendofaninextensible string ain.longismade fast to apegAandattheother end isknotted aweight W.Asecond string, attached toW,passes over asmooth pegatB,distant bin.fromAandatthesame level, andcarries aweightPatits other end. Find theposition ofequilibrium. IfP=Wjhow farbelow thelevel ofthepegswillthe first^-25.*Observe thebraces that stiffen the frame ofarailroad car. Formulate a reasonable problem suggested bywhat you sec,and solve it. 26.Abridge ofsimple typeissuggested bythefigure. In designing suchastructure, thestiffness ofthemembers atapoint A B J_ FIG.26 *Thefollowing fourproblems aregiven only inoutline, andthestudent thus hastheopportunity offilling inreasonable numerical dataandformulating aclean- cutquestion. Itisnotnecessary thatherespond toalltheproblems; buthe should demand ofhimself thathedevelop anumber ofthemandsupplement these byothers oflikekindwhich hefinds ofhisown initiative ineveryday life. For, imagination isoneofthehighest oftheintellectual gifts, andtoomuch effort cannot bespent indeveloping it. STATICS OFAPARTICLE 19 where thesecome togetherisnottobeutilized, buttheframe is plannedasifthemembers were allpivotedthere. Draw such a bridge toscaleand findwhat thetensions andthrusts willbeif itistosupport aweight of20tons ateach ofthepoints A,B. Make areasonable assumption about theweight oftheroad bed, butneglect theweightofthe tierods, etc.j100lbs< 27.Thetension ineach ofthetraces attached to_j_Jo,o 100Ibs.awhiffle-tree 3ft.longis100Ibs.Thedistance from 1(j thering tothewhifflc-tree is10in.What istheten- sion inthechains ? Fm -27 28.Have youever seen afunicular asmall passenger car, hauled upasteep mountain byacable? How isthetension in thocable related totheweight ofthecar?When thedirection ofthecable ischanged byafriction pin,orroller, overwhich the cable passes, what isthepressure onthepin? FRICTION 29.Consider theinclined planes ofQuestion 5, 3. Ifboth .irerough and/z=-faforeach,what istherange ofvalues forW consistent withequilibrium? 30.Aweightless bead can slideonarough horizontal wire, ju=0.1.Acord isattached tothebeadand carries aweight at itsother end, thusforming asimple pendulum. Through what angle canthependulum swing without causing thebead toslip? 31.Awater main 5ft.indiameter isfilled withwater toa depth of1ft.Amouse tumbles inandswims tothenearest point onthewall. Ifthecoefficient offriction between herfeetandthe pipeis -J-,cansheclamber up,orwillshebedrowned? 32.Aheavy bead isplaced onarough verticalcircle, the coefficient offriction beingf.Iftheangle between theradius drawn tothebeadandthevertical is16,findwhether thebead will slipwhen released. 33.Aropeisfastened toaweight that restsonarough hori- zontal plane, andpulled until theweight justmoves. Find the tension intherope,andshow that itwillbeleastwhen therope makes, with thehorizontal, theangle offriction. 34.Thesame question foraninclined plane. 20 MECHANICS 35.A50Ib.weightisplaced onarough inclined plane, M=t, angle ofinclination, 10.Astring tiedtotheweight passes over asmooth pegatthesame level astheweight andcarries aweight of7Ibs.atitslower end.When thesystemisreleased fromrest, will itslip? 36.Aweightisplaced onarough inclined plane and isattached toacord, theother endofwhich ismade fasttoapegintheplane. Find allpositions ofequilibrium. 37. Iftheparabolic wiredescribed inQuestion 16isrough, and theweights arePandW,find allpositions ofequilibrium. Ans.WhenP^W,thelimiting positionisgivenbyoneor theother oftheequations: i-P~W1 ?-W~P1tan2~"P+W' /z'tan2"W+P' n' Find theother positions ofequilibrium, and discuss the caseP=W. 38.Cast iron rings weighing1Ib.each can slideonarough horizontal rod,M= -J.Astring 6ft.longisattached toeach of these beads and carries asmooth bead weighing 5Ibs.How far apart canthetwobeads ontherodbeplaced,ifthesystemisto remain atrestwhen released ? 39.Anelastic string 6ft.long,obeying Hooke's Law,isstretched toalength of6ft.6in.byaforce of20Ibs.Theends ofthe string aremade fastattwopoints 6ft.apart andonthesame level.Aweight of4Ibs. isattached tothemid-point ofthestring andcarefully lowered. Find theposition ofequilibrium, neglect- ingtheweight ofthestring. Ans. isgivenbytheequation: cot=120 (1-cos0). 40.Solve thepreceding equation for0,toone-tenth ofadegree. Ans. =14|. 41.Themast ofaderrick is40ft.high,andastayisfastened ^20toablock ofstone weighing 4tonsand resting onapave- ment, M=fTheboom is 35ft.long,and itsend isdis- oo tant20ft.from thetopoftheriG.&Q.11-mast. Isitpossible toraise a5tonweight, without thederrick's being pulled over, thedis- tance from thestone tothederrick being 120ft.? CHAPTER II STATICS OFARIGID BODY 1.Parallel Forces inaPlane. Lottwoparallel forces,Pand Q,actonabody atAandB,and letthem have thesame sense. Introduce twoequal andopposite forces, Sand S',atAandBas shown inthefigure, and,com- pounding them withPandQ respectively, carry theresulting forces back tothepointDin which their lines ofaction meet. These latter forces arenowseen tohavearesultant, (1) R=P+Q, parallel tothegiven forces and having thesame sense, itsline ofaction dividing thelineAB intotwosegments, ACandCB. Letthelengths ofthesegments bedenoted asfollows: AC=a,CB=6,AB=c,DC=h. From similar trianglesitisseenthatfrI(329 HenceP Sh a'Qh AS' 6' aP=bQ. a+6=c.(2) Moreover, (3) Tosumup,then :Theoriginal forces,PandQ,havearesultant determined bytheequations (1), (2),and(3). Example. The familiar gravity balance, inwhich onearm, a, fromwhich theweightPtobedetermined issuspended,isshort, andtheother arm, b, fromwhich theriderQhangsislong,isa FIG.30 case inpoint. 21 22 MECHANICS Q 1OppositeForces. IfPandQareoppositeindirection, and unequal (Q>P,say), theyalsohave aresultant. Introduce a forceE(Equilibriant) parallel toPandQandhaving thesense ofP,determining itsothatQwillbeequal andopposite tothe resultant ofPandE.Then Q=P+E, cP-bE, a=b+c. ThusPandQareseen tohave aresultant, FlG -31 (4) R=Q-P, having thesense ofQ,itslineofaction cuttingACproducedin thepointBdetermined bytheequations: (5) aP=bQ, (6) a=b+c. IfPandQareequal, theyform acouple and, asweshallshow later, cannot bebalanced byasingle force;i.e.theyhave no resultant (force). Example. Consider apair ofnutcrackers. The forces that actononeofthemembers arei)P,thepullofthehinge ;if)Q, thepressure ofthenut; and Hi)the forceEthehand exerts, balancing the resultant, R,ofPandQ. Wehave heremade useoftheso-called Principle oftheTransmissibilily ofForce, which saysthat theeffect ofaforceona bodyisthesame, nomatter atwhat pointinitsline itacts. Thus aservice truck willtowamired caraseffectively (butnomore effectively) when thetow-ropeislong, aswhen itisshort, provided that ineach case theropeisparallel tothe road bed. Moreover,itisnotnecessary tothink ofthepoint ofapplication aslying inthematerial body. Itmight bethecentre ofaring. Forwecanalways imagine arigid weightless truss attached tothe body andextending tothedesired point. Butwealways think ofabody,i.e.mass, onwhich thesystem offorces inquestionacts.FIG.32 STATICS OFARIGID BODY 23 EXERCISES 1.A10tontruck passes over abridge that is450 ft.long. When thetruck isone-third oftheway over,howmuch ofthe load goes tooneend ofthebridge, andhowmuch totheother end? Ans.6ftons tothenearer end. 2.Aweight of200 Ibs. istoberaised byalever 6ft.long, the fulcrum being atoneendofthelever, andtheweight distant 9in. from thefulcrum. What force attheother end isneeded,ifthe weight ofthelever isnegligible? 3.Acoolie carries twobaskets ofpottery byapole6ft.long. Ifonebasket weighs 50pounds andtheother, 70pounds, how far arctheends ofthepolefrom hisshoulder? 2.Analytic Formulation;nForces. Suppose thatnparallel forces act.Then two,which arenotequal andopposite, canbe replaced bytheir resultant, andthis inturncombined withathird oneofthegiven forces, until thenumber hasbeenreduced totwo. These will ingeneral have aresultant, but,inparticular, mayform acouple orbeinequilibrium. Thus theproblem could besolved piecemealinanygiven case. Anexplicit analytic solution canbeobtained asfollows. Begin withn=2anddenote theforces byPlandP2.Moreover,letP1 aridP2betaken asalgebraic quan- tities, being positiveifthey actin onedirection;negative,ifthey actp\ intheopposite direction.-- 1- AT 1 TVI*XssQ 3J=1 ** *2Nextdraw alineperpendicular* Fl(J33 tothe lines ofaction ofP1and P2,andregard this line asthescale of(positive andnegative) numbers,liketheaxis ofx.Letxlfx2bethecoordinates ofthe pointsinwhichP1?P2cutthe line.Weproceed toprove the following theorem. TheforcesPlandP2havearesultant, (l) R=P,+P provided Pl+P2^0.Itslineofaction hasthecoordinate: *Anoblique direction could beused, butintheabsence ofanyneed forsucha generalization, theorthogonal direction ismore concrete. 24 MECHANICS Suppose, first, thatPtandP2areboth positive. Then, by 1, R=P,+P2 , wherea=xx1? o=x2x, provided Xj<x2(algebraically). Hence (x-x,)Pl=(x2-x)P2, andfrom thisequation, therelation(2)follows atonce. Thecasethatx2<x lisdealt with inasimilar manner, asisalso thecasethatPlandP2areboth negative. Next, suppose P1andP2have opposite senses, but Pl+P,*0. Let P1<0,P2>0, |P1 |<P2, where |x |means thenumerical orabsolute value ofx.Thus |-3 |=3, |3 |=3.Moreover, letxl<x2.Then, by 1, PlandP2have aresultant, 12=-?!+P2, andthecoordinate, x,corresponding toitisobtained asfollows : a=xxltb=xxtl andhence, from1,(5): (x-x{)(-Pt)=(x-x2)P2. Soagainwearrive atthesame formulas, (1)and (2),asthe solution oftheproblem. Itremains merely totreat theremaining cases inlikemanner. The final result willalways beexpressed byformulas (1)and (2). Wearenowready toproceed tothegeneral case. THEOREM 1.Letnparallel forces,Plt ,Pn,act.They will havearesultant,R=P!^----+Pn, providedthissum^0,and itslineofaction willcorrespondtox, where /o\ ;=_P\Xli"*''"TPnXn (3) x- STATICS OFARIGID BODY 25 Theproof canbegiven bythemethod ofmathematical induc- tion. Thetheorem isknown tobetrue forn=2.Suppose itwerenottrue forallvalues ofn.Letmbethesmallest value ofnforwhich itisfalse.Wenowproceed todeduce acon- tradiction. Suppose, then, thatPD- ,Pmisasystemofparallel forces, forwhich thetheorem isfalse, althoughitistrue forn2, 3, ,m-1.Byhypothesis, Pi+'+Pm*0. Now,itispossible tofindm 1oftheP,'swhose sum isnot : Pi+'+Pm-,*0, letussay. Thesem Iforces have,byhypothesis, aresultant : R'=Pl+--+Pw-lf and itsxhasthevalue : _//I3*ir* '" 'T~-*wi#w~ix~ />!+-TP^;' since thetheorem holdsbyhypothesisforallvalues ofn<m. Next, combine thisforce withPw.Since R'+Pm=1\+---+Pm*0, thetwoforces have aresultant, R=R'+Pm=P1+--+Pw, and itslineofaction isgiven bytheequation _R'x'+PmxmPlgl++Pwa,X#'+PMA+--+P But this result contradicts theassumption that thetheorem is false forn=m.Hence thetheorem istrue forallvalues ofn. Couples. Let (4) P,+---+Pn=0, Pn*0. Thei1 Pl+-+Pn-!^0, andtheforcesPlf ,Pn_1have aresultant, ft'=P4-...4-P ^ir n^ *n1> whose lineofaction isgivenbytheequation: -/=P!%lH~''"TPn-i ^n~i 26 MECHANICS If,inparticular, x'=xn,thisresultant, 72',willhave thesame lineofaction asPn;andsince R'+Pn=0, orRf=-Pn, thenforces willbeinequilibrium. Wethenhave : _PIX\I''~T1n-iXn-iXn r> yfn (5) Pl*l+'+PnX n=0. Andconversely,ifthiscondition holds, wecanretrace ourstops andinfer equilibrium. But,ingeneral,x'^xn.Hence _^P,XlH-----hPn-i gn-i~ (6) Pl*l+''+PnX n^Q. Wethenhave acouple. And conversely,if(4)and(6)hold,we canretrace ourstepsandinfer thatwehave acouple. Wehave thusproved thefollowing theorem. THEOREM 2.Thenparallel forcesPlt ,Pnformacouple if, arulonlyifPI+...+P.=0. Equilibrium. The case ofequilibrium includes notonly the caseabove considered (Pn^0),butalsothecase inwhich all nforces vanish. Wethushave thefollowing theorem. THEOREM 3.Thenparallel forcesPly ,Pnareinequilibrium if,andonly ifPI+...+Pn= ^ PiXl+----hPnXn=0. 3.Centre ofGravity. Letnparticles,ofmasses Wi, ,mn, befastened toarigid rod,theweight ofwhichmaybeneglected, and letthem beacted onbytheforce ofgravity.Iftherod is supported atasuitable point, (?,and isatrest, there willbeno tendency toturn inanydirection. This pointiscalled thecentre ofgravity ofthenparticles, and itspositionisdetermined bythe equation: mlxl+-+mnxn - ml+-+mn STATICS OFARIGID BODY 27 Iftheparticleslieanywhere inaplane, being rigidly connected byatrusswork ofweightless rods,and ifwedenote thecoordinates ofmkby(xk,yk) 1thecentre ofgravityisdefined inasimilar manner (secbelow) and itscoordinates, (x, ?/),aregiven byEquations (1)and /9xfl_*KIy\H----+ >nyn (2) y~ n^+.-.+m.' For,lettheplaneoftheparticles bevertical, theaxis ofxbeing horizontal. Then thesystemisacted onbynparallel forces, whose lines ofaction cuttheaxisofxatright angles inthepoints %!,'', xnyandtheir resultant isdetermined inposition byEqua- tion(1).Onrotating theplane through aright angleandrepeat- ingthereasoning, Equation (2)isobtained. Thecentre ofgravity ofanymaterial system, made upofpar- ticles andline, surface, andvolume distributions,isdefined asa point, (7,such that,iftheparts ofthesystem berigidly connected byweightless rods,and ifGbesupported, there willbenotendency ofthesystem torotate, nomatter how itbeoriented. Wehave proved thoexistence ofsuch apointinthecase ofnparticles lyingonaline. Fornparticles inaplanewehaveassumed that acentre ofgravity exists and liesintheplane, andthenwehave computeditscoordinates. Weshall provelater thatnparticles always haveacentre ofgravity, andthat itscoordinates aregiven byEquations (1), (2),and . .ml++mn Inthecase ofacontinuous distribution ofmatter,likeatri- angular lamina orasolid hemisphere, themethods oftheCalculus lead tothesolution. Itisthedefinite integral, defined asthe limit ofasum, that ishereemployed, andDuhamel's Principle isessential intheformulation. Inthesimpler cases, simple integrals suffice. Buteven insome ofthese cases, surface and volume integrals simplify thecomputation. Thefollowing centres ofgravity aregiven forreference. fl)Solid hemisphere: x=fa. b)Hemispherical surface : x a. c)Solid cone : x=fh. d)Conical surface : xfA. e)Triangle:Intersection ofthemedians. 28 MECHANICS 4.Moment ofaForce. LetFbeaforce lying inagiven plane, and letbeapoint oftheplane. Bythemoment ofFabout ismeant theproduct oftheforcebythedistance from ofits lineofaction, orhF.Amoment mayfurthermore bedefined as analgebraic quantity, being taken aspositive when ittends toturn thebody inone direction (chosen arbitrarily asthepositive direction), andnegative intheother case. Finally,if lieson thelineofaction oftheforce, themoment isdefined as0. LetaforceFactatapoint (x,y),and letthecomponents ofF along theaxesbedenoted byX,Y.Then themoment (taken algebraically) ofFabout theorigin, 0,is : (1) xY-yX. Proof. Lettheequation ofthelineofaction ofFbewritten inHesse's Normal Form : xcosa+ysina=h. / iFIG.34O FIG.35 Suppose, first, thatthemoment ispositive. Then itwillbe hF=x(Fcosa)+y(Fsina). Here,2iraisthecomplement of6,Fig.34 : Hence andsincecosa=sin6,a=-+2rr. sina=cos0, X=Fcos6, Y=Fsin6, theproofiscomplete. STATICS OFARIGID BODY 29 If,however, themoment isnegative,a.and 6willbeconnected bytherelation, Fig.35 : ,-a--- Bt-+|. Hence cosa= sin9,sina=cos 6. Themoment willnowberepressed as hF=x(Fcosa)+y(Fsina), andthuswearrive atthesame expression, (1),asbefore. Thesame istrue foranilmoment. Hence theformula (1)holds inallcases. From(1)weprove atonce thatthemoment oftheresultant of twoforces acting atapointisthesum ofthemoments ofthetwo given forces. Letthelatter beFj,F2,with themoments xY1yX land xY2yX 2. Thecomponents oftheresultant force areseen totake theform : Xl+X2andYl+F2,and itsmoment is From thisexpression thetruth ofthetheorem isatonce obvious. Finally, themoment ofaforceabout anarbitrary point, (x ,?/), isseentobe : (2) (x-x)Y-(y-y,}X. Thephysical meaning ofthemoment ofaforce about apoint isameasure oftheturning effect oftheforce. Suppose thebody were pivoted at0.Then thetendency toturnabout 0,dueto theforce F,isexpressed quantitatively bythemomenta Anda setofforces augment orreduce oneanother intheir combined turningeffect according tothemagnitude andsense ofthesum ofthemoments oftheindividual forces. From thispoint ofview amoment isoften described inphysics andengineering asatorque. 6.Couples inaPlane. Acouple hasalready been defined as asystem oftwoequal andopposite parallel forces. Acouple cannot bebalanced byasingle force, but isanindependent me- chanical entity ;theproofisgiven below. Bythemoment ofacouple, taken numerically,ismeant the product ofeither forcebythedistance between thelines ofaction oftheforces. 30 MECHANICS THEOREM. Two couples having thesamemoment andsense are equivalent. Supposefirstthattheforces oftheonecouple areparallel tothe forces oftheother couple. Then, byproper choice oftheaxis ofx,wecanrepresent thecouples asindicated, where JP.+P.-O. 0<P,; P.I , Il .P3+P4=0,0<P3; p *\" P, x^xt+h,0<h; Fxo.36J*4=*3+J, <l> Now consider thesystem offour forces,PlfP2,P3tP4. These areinequilibrium. For, P,+P2-P3-P4= and Pr _|_p-r _P ,._ /:>/r 1^lI*2**/2*3"^S*4**/4 (P -4-PW P It(P -\-P}r4-P7= fl^JTj-fX2/^2-I i V.^3I-*4/*'4'^3^ VJ> Hence the firstcoupleisbalanced bythenegative ofthesecond couple, andthus thetheorem isprovedforthecasethat allthe forces areparallel. Iftheforces ofthetwocouples areoblique toeach other,let OAandOBbetwo lines atright angles totheforces ofthe first couple andtothose ofthesecond couple respectively. Layofftwoequal distances, OA=handOB=h,on these lines. Thenbythetheorem just proved the firstcouple canberepre- sented asindicated bytheforcesP and P.* Furthermore, thesecond couple, reversed insense, canberep- resented bytheforcesQandQ.Let thelines ofaction ofPatAandQatBmeet inC,andcarry these forces forward sothat each acts atC.Then thefour *Itmight seem thatthere aretwocases tobeconsidered, forcannot thevectors that represent theforces ofthe first couple beopposite insense? True. But thenwecanbegin with thesecond couple. Itsforces willberepresented bythe QandQofthediagram ;andtheforces ofthefirstcouple, reversed insense, will nowappear asPandP. STATICS OFARIGID BODY 31 A:(i,o)forces obviously areinequilibrium, forallfourareequal inmagni- tude, sincebyhypothesis themoments ofthetwogiven couples areequal ;andtheforces make equal angles with theindefinite lineOC,insuchamanner thattheresultant ofonepairisequal andoppositetothat oftheotherpair, thelines ofaction ofthese resultants coinciding. Tosum up,then, theeffect ofacouple inagiven planeisthe same, nomatter whore itsforcesact,andnomatter how large or small theforcesmay be,provided only that themoment ofthe coupleispreserved both inmagnitude andinsense. Composition ofCouples. From theforegoingitappears thattwo couplesinaplane canbecompounded intoasingle couple, whose moment isthesumofthemoments oftheconstituent couples ;allmo- ments being taken algebraically. For,assume asystem ofCartesian axes intheplane, andmark the pointA :(1,0). The firstcouple canberealized byaforceP1atA parallel totheaxisofy(and either positive ornegative) andanequal and opposite force Piatthe w origin, acting along theaxis ofy. Themoment ofthiscouple, taken algebraically,isobviously P,. Dealing with thesecond couple inasimilar manner, wenow have astheresult two forces,P1andP2,atAparallel totheaxis ofy',andtwoequal andopposite forces at along theaxis of y.These forces constitute aresultant couple, whose moment is thesum ofthemoments ofthegiven couple. This laststatement isatfault inoneparticular.Itmayhappen thatthesecond coupleisequal andopposite tothefirst,andthen theresultant forces both vanish. Inorder that thiscasemaynot cause anexception, weextend thenotion ofcouple toinclude a nilcouple:i.e.acouple whose forces areboth zero, orwhose forces lieinthesame straightline;andwedefine itsmoment tobe0. Wearethus ledtothefollowing theorem. THEOREM. //ncouplesactinaplane, theircombinedeffect is equivalenttoasingle couple, whose moment isthesumofthemoments ofthegiven couples. 32 MECHANICS Remark. Themoment ofacoupleisequal tothesum ofthe moments ofitsforces about anarbitrary point oftheplane. This isseen directly geometrically from thedefinition ofamoment. In particular,letapoint bechosen atpleasure. Thecouple can berealized bytwoforces, oneofwhich passes through 0.The moment ofthecoupleisthen equal tothemoment oftheother force about 0. 6.Resultant ofForces inaPlane. Equilibrium. Letany forces actinaplane. Then they areequivalent f)toasingle force, orii)toasingle couple ;or,finally, Hi)they areinequi- librium. Let beanarbitrary point oftheplane. Beginning with theforceFDletusintroduce attwo forces equalandopposite toFj.Thetwoforces checked form acouple, and theremaining *force istheoriginal forceFx,transferred to thepoint 0. Proceeding inthismanner witheach ofthe ,' remaining forces, F2, ,Fn,wearrive at anewsystem offorces and couples equiv- alent totheoriginal system offorces and consisting ofthose n forces,allacting at0,plusncouples..These nforces areequiva- lent toasingle force, R,at0;orareinequilibrium. And the ncouples areequivalent toasingle couple, orareinequilibrium. Ingeneral, theresultant force, R,willnotvanish, nor willthe resultant couple disappear. The latter can, inparticular, be realized asaforce equal andopposite toRandacting at0,anda second force equaltoR,buthaving adifferent line ofaction. Thus theresultant ofallnforces ishereasingle force. Inciden- tallywehaveshown that anon-vanishing couple cannotbe balanced byanon-vanishing force;for,theeffect ofsuchaforce andsuch acoupleisaforce equal tothegiven force, buttrans- ferred toanew lineofaction, parallel totheoldline. Itmayhappen thattheresultant force vanishes, buttheresult- antcouple does not. Forequilibrium,itisnecessary and suffi- cient thatboth theresultant forceandtheresultant couple vanish. This condition can,with thehelp oftheRemark attheclose of 5,beexpressed inthefollowing form. EQUILIBRIUM. Asystem ofnforces inaplane willbeinequi- libriumif,andonly if STATICS OFARIGID BODY 33 i)theyaresuchaswould keepaparticleatrestiftheyallacted at apoint; and ii)thesum ofthemoments oftheforces about apoint (onepoint isenoughj and itmay bechosen anywhere) oftheplaneiszero. Analytically, thecondition canbeformulated asfollows. Let thepoint about which moments aretobetaken, bechosen as theorigin, and lettheforceFractatthepoint (xr,yr).Then Xr=0, r=l 2)(XrYr- yrXr)=0. FIG.40Example. Aladder rests against awall, the coefficient of friction forboth ladder andwallbeing thesame, M- Iftheladder isjustonthepoint ofslipping when inclined at anangle of60with thehorizontal, what isthe value ofju? Since allthefriction iscalled into play, the forces areasindicated inthefigure,RandSbeing unknown, andJJLalsounknown. Condition?)tellsusthatthesum oftheverti- calcomponents upward must equal thesumofthe vertical components downward, or R+S=W. Furthermore, thesum ofthehorizontal components totheright must equal thesum ofthehorizontal components tothe left,or S=nR. Finally, themoments about apoint, 0,oftheplane must balance. Itisconvenient tochoose asapoint through which a number ofunknown forces pass; forexample, oneend ofthe ladder, saytheupper end. Thus 2acos60R=2asin60R+acos60TF, or R=V3/ HenceW 2(1-S 2(1-MV3) 34 MECHANICS Wecannoweliminate RandS.Theresulting equationis Thus /i=2-V3 =0.27. Theother root, being negative, hasnophysical meaning. EXERCISES 1.Ifintheexample justdiscussed thewall issmooth, butthe floor isrough, and if/*= ,find allpositionsofequilibrium. 2.Ifintheexample ofthetext/*could beasgreat as1,show that allpositions would bepositions ofequilibrium. 3.Aladder 12ft.longandweighing 30Ibs.rests atanangle of60with thehorizontal against asmooth wall, thefloor being rough, /u=.Amanweighing 160Ibs.goesuptheladder. How farwillhegetbefore theladder slips? 4.Inthelastquestion, howrough must thefloorbetoenable theman toreach thetop? 6.Show thatanecessary andsufficient condition forequilibrium isthatthesum ofthemoments about each ofthree points, A,B, andC,notlyinginaline, shall vanish foreach point separately. 7.Couples inSpace. THEOREM I.Acouplemay betransferred toaparallel plane withoutalteringitseffect }provided merely that its moment andsense arepreserved. Itissufficient toconsider twocouples inparallel planes, whose moments areequal andopposite, andtoshow that their forces are inequilibrium. Construct acube withtwoofitsfaces intheplanes of.thecouples. Then onecouple can berepresented bytheforces marked PandPjandthereversed couple, bytheforcesQandQ(P=Q). Consider theresultant ofPat AandQatB. Itisaforce of p+Q(=2P) parallel toPand having thesame sense, andpassing through thecentre, 0,ofthecube. Turn next toPatCand QatD.The resultant ofthese forces isobviously equal and STATICS OFARIGID BODY 35 opposite totheresultant just considered andhaving thesame line ofaction. Thefour forces are,then,inequilibrium. This completes theproof. Example. Inacertain type ofauto(Buick 45-6-23) thelastbolt intheengine headwassonear thecowl thataflatwrench could notbeused. Thegarage manimmediately bentaflat wrench through aright angle, applied oneendofthe wrench tothenutand, passing ascrew driver through theopening intheother end,turned thenut. Thus the applied couple was transferred from thehorizontal plane through thescrew driver totheplaneofthe riut.IQ ' Vector Representation ofCouples. Acouple canberepresented byavector asfollows. Construct avector perpendicular tothe plane ofthecouple and oflength equal tothemoment ofthe couple. Asregards thesense ofthevector, either convention is permissible. Letusthink ofourselves asstanding upright onthe plane ofthecouple andlooking down ontheplane.Ifweare ontheproper side oftheplane, weshall seethecouple tending to produce rotation intheclock-wise sense. Andnowthedirection from ourfeettoourheadmaybetaken asthepositive sense of thevectoror,equally well, theoppositedirection. THEOREM II.Thecombined effect oftwocouplesisthesame as thatofasingle couple represented bythevector obtained byadding geometricallythetwo vectors which represent respectively thegiven couples. Thetheorem hasalready been provedforthecase that the planes ofthegiven couples areparallel orcoincident. Ifthey intersect, layoffaline seg- ment ofunit length, AB, ontheir lineofintersection, andtake theforces ofthe couples sothatthey actat FlQ43AandBperpendicularly to the lineAB.Then itis easily seenthattheresultant ofthetwoforces atAandtheresult- antofthetwoforces atBform anewcouple. Finally, thevector representations ofthese three couples are three vectors perpendicular respectively tothethree planes of thecouples, equal inlength totheforces ofthecouples, andso\R QX^/ 36 MECHANICS oriented astogivethesame figure yielded bythree oftheforces, properly chosen, onlyturned through 90. 8.Resultant ofForces inSpace. Equilibrium. Letanyn forces actonabodyinspace. Letthem berepresented bythe vectors F,, ,Fn.Let beanarbitrary pointofspace. Intro- duce attwoforces that areequal andoppositetotheforceFk. Then thenforcesFlf ,Fnathave aresultant : (1) R=F!++Fn, acting at0,orareinequilibrium. And theremaining forces, combined suitably inpairs, yieldncouples, Clt ,Cn,whose resultant couple, C,is : (2) C=C,+---+Cn, or,inparticular, vanishes;thecouples being then inequilibrium. Ingeneral,neitherRnorCwillvanish. Thus thengiven forces reduce toaforceandacouple. Theplane oftheresultant couple, C,will ingeneral beoblique tothelineofaction oftheresultant force, andhence thevector Coblique tothevector R.Let C=C,+C2, where C^iscollinear with R,andC2isperpendicular toR.The couple represented byC2canberealized bytwoforces inaplane containing theresultant force,R;and itsforces canbecombined withR,thus yielding asingle force R,whose lineofaction, how- ever, hasbeen displaced. This leaves only thecouple Cj.We have, therefore, obtained thefollowing theorem. THEOREM. Any system offorces inspaceisingeneral equiva- lent toasingle force whose lineofaction isuniquely determined, and toasingle couple, whose planeisperpendiculartothelineof action oftheresultant force. Inparticular,theresultant forcemay vanish, ortheresultant couple may vanish, orbothmay vanish. Equilibrium. Thegiven forces aresaid tobeinequilibriumif andonlyifboth theresultant force andtheresultant couple vanish. Forcompletenessitisnecessary toshow that theresultant force, R,together with itslineofaction, andtheresultant couple, Ct,areuniquely determined. For itisconceivable thatadiffer- STATICS OFARIGID BODY 37 entchoice, O',ofthepointOmight have ledtoadifferent result. Now, thevectorRisuniquely determined by(1),andsoisthe same ineach case;butCdepends onthechoice of0',andsoCl might conceivably bedifferent fromCJ,though eachwould be collinear withR.Thisis,however, notthecase. For, reverseR inthesecond case,and alsothecouple C[.Then thereversed force andcouple must balance the first force and couple. But thissituation leads toacontradiction, asthereader willatonce perceive. 9.Moment ofaVector. Couples. Given aforce, F,acting along aline//,andanypoint inspace. Bythevector moment ofFwith respectto(orabout)ismeant thevector product*/\F (1) M-rXF, where risavector drawn from toapointofL. Itisavector atright angles totheplane ofand L,and itslengthisnumerically equal tothe moment ofFabout inthat plane. Itssense depends onwhether weareusing aright-handed oraleft-handed system. Referred toCartesian axesFIG.44 (2) (3)M=xaybz c X Y Z L=(y-V)Z- (z-e)Y M=(z-c)X-(x-a)Z N=(x-a)Y-(y- P:(x,y,z) r=r'- rj, r'=xi+yj+zk, FIG.45 *Thestudent should read 3ofAppendix A.This, together with themere definitions thathavegone before, isallofVector Analysis which hewillneed for thepresent. 38 MECHANICS Vector Representation ofaCouple. Letacouple consist oftwo forces, F!andF2: F,+F2=0, acting respectively along two linesLandL2.The vector C which represents thecoupleisseenfrom thedefinition ofthe vector product tobe : (4) C=rXPlf where rrepresents anyvector drawn from apoint ofL2toapoint ofLj.Let beanypoint ofspace. Then thesum ofthevector moments ofFxandF2with respect to yields thevector couple: (5) C=r,XP!+r2XF2, where r^r2areanyvectors drawn from toL1andL2respec- tively. For hence rXF!=T!XF!-r2XFj=T!XFj+r2XF2. 10.Vector Representation ofResultant Force and Couple. Resultant Axis. Wrench. Let Pi, ,Fnbeanysystem of forces inspace. LetPbeanypoint ofspace, and letequal and opposite forces, F*and FA,k=1, ,n,beapplied atP. Consider thenforces F^- - ,Fnwhich actatP.Their re- sultant isR=F!+-+Fn. Theremaining forces yield ncouples, consisting each ofF*at PA', (%k, Vk,Zk)and F&atP:(x,y,z).Let rk,rbethevec- torsdrawn from theorigin ofcoordinates (chosen arbitrarily) to PkandPrespectively. Then the Jfc-th couple, C*,isrepresented bytheequation: C*=r*XF*-rXF*. Wearethus ledtothefollowing theorem. THEOREM. Thegiven forcesFlf ,Fnareequivalenttoasingle force, (6) R=F,++Fn, actingatP;and toacouple, (7) C=5)r*XF*-rXR. fc-i STATICS OFARIGID BODY 39 Resultant Axis. The resultant axis isthelocus ofpoints P, forwhichCliesalongR;i.e. iscollinear withR.Thecondition forthis isobviously thevanishing ofthevector product: (8) RXC=0, R*0. Let (9) iXF^Li (10) Thus (11)L=R=Fj+Zk. Then thecondition (8)becomes, byvirtue of(7): RX(Li+Mj+tfk)=RX(rXR), or: (12) This istheequation oftheresultant axis invector form. To reduce toordinary Cartesian form, equate thecoefficients ofi,j, krespectively. Thuswefind : Y+zZ=YN-ZM (13) -Z(xXzZ)=ZL-XN zZ)=XM-YL Oneofthese equations maybecome illusory through thevanish- ingofallthecoefficients;butsometwoalways define intersecting planes, fortherank ofthedeterminant is2,sinceR>;and between thethree equations there exists anidentical relation. Let(,ij,f)bethecoordinates ofthenearest point ofthelineto theorigin. Then Hence (15) $YN-ZM >?=ZL-XN ZM- YL ft2 40 MECHANICS Thuswehave found onepoint oftt*eresultant axis,andthe direction oftheaxis isthat ofR.The resultant coupleisgiven by(7),where (16)r=i+Tjj+fk. Wrench. Awrench isdefined astwo forces, acting atarbitrary points; moreover, neither force shall vanish, and their lines of action shallbeskew. Lettheforces beF*,acting at(xk,y^Zk),k=1,2.Thereader willdowell tocompute theresultant force, axis,and couple. Suppose, inparticular, thatFxisaunit force along thepositive axisofZ,andF2isaforce of2,parallel totheaxisofyandacting atthepoint (1,0,0). EXERCISE LetF!andF2betwoforces, thesum ofwhose moments about apointis0.Show thatFt,F2,and lieinaplane. 11.Moment ofaVector about aLine. Letaline, L,anda vector, F,begiven. LetL'bethelineofF,and let0,Ofbethe points ofLandL'nearest together. Letrbethevector from to0',and let |r |=h.Letabe aunit vector along L.Assume co- z ordinate axes asshown. Then rL^r Bythemoment ofFaboutLismeant : FlG -46 M=hYa, where a=k.ThemomentMcanbeexpressedininvariant form asfollows. Since r=hi, wehave : rXF=- k(rXF)=hY. Hence (1) M={a-(rXF))a. Moregenerally,rmaybeanyvector drawn from apointofL toapoint ofL'.Wehave thus arrived atthefollowing result. STATICS OFARIGID BODY 41 Themoment ofavectorFabout alineLisgiven bytheformula: M={a- (rXF)}a, where risany vector drawn fromapoint ofLtoapoint oftheline ofF,andaisaunit vector collinear withLandhavingthesign attributed toL. Inparticular, themoments ofFabout thethree axes arere- spectively: yZ-zY,zX-xZ,xY-yX. EXERCISE Aforce of12kgs.actsatthepoint ( 1,3, 2),and itsdirec- tion cosines arc(3, 4, 12). Find itsmoment about the principal diagonal oftheunit cube;i.e.the linethrough the origin, making equal angles with thepositiveaxes. 12.Equilibrium. In8wehavo obtained anecessary and sufficient condition fortheequilibrium ofnforces, F,, ,Fn, interms ofthevanishing oftheresultant forceandtheresultant couple. Bymeans ofEquations (6)and (7)of10wecanformu- late these conditions analytically. Thefirst, namely, R=0, gives: andnowthesecond, namely, C=0,reduces Equation (7)tothe vanishing ofthefirsttermontheright, or (2)2(ykZk-zkYk)=0,2(zkXk-xkZk)=0, This lastcondition, which wasobtained from thevanishing of acouple, admits two further interpretationsinterms ofthe vanishing ofvector moments, namely: i)Thesum ofthevector moments ofthegiven forces with respect toanarbitrary pointofspaceis0. ii)Thesum ofthevector moments ofthegivenforces about anarbitrary lineofspaceis0. Thecondition ii)isequivalent tothefollowing: iif )Thesum ofthevector moments ofthegivenforces about each ofthree particular non-complanarlines is0. 42 MECHANICS Necessary and Sufficient Conditions. Itisimportant forclear- ness toanalyse these conditions further, astowhether they are necessary orsufficient orboth. Condition i),regarded asanecessary condition,isbroadest when istaken asanypoint ofspace. ButConditionii)issufficient ifitholds forthelines through justoneparticular point 0,the condition (1)beingfulfilled. Condition ii),regarded asanecessary condition,isbroadest when theline istaken asany lineinspace. Butasasufficient condition, though true asformulated, itislessgeneral than (1)andCondition ii'),which may, therefore, betaken asthe broadest formulation ofthesufficient condition. EXERCISES 1.Show thatCondition i)issufficient forequilibrium. 2.Show thatCondition ii)issufficient forequilibrium. 13.Centre ofGravity ofnParticles. Letthenparticles ^i, ,wnbeacted onbygravity. Thusnparallel forcesarise, andsince theyhave thesame sense, theyhave aresultant not 0. Lettheaxis ofzbevertical and directed downward. Then theresultant isaforce directed downward andofmagnitude n (1) R=Wjg++mng=g5)mk, theresultant axisbeing vertical. Furthermore, Xk=0,Yk=0, Zk=mkg.Thus, 10,(11): n n (2)L=gVmkyk,M=-gVmkxk,N=0. t-l *-i Thenearest point oftheresultant axistotheorigin hasthecoordi- nates givenby(15), 10 : (3) *=*=5-, .,_**> .-f-0. Ifanypoint ofthis line issustained, thesystem ofparticles (thought ofasrigidly connected) willbesupported, and the system willremain atrest. Inparticular, onepoint onthis line hasthecoordinates : STATICS OFARIGID BODY 43 If,secondly, weallow gravity toactparallel totheaxis ofx, theresultant axisnowbecomes parallel tothat axis,andthenearest point totheoriginisfound byadvancing theletters cyclically inEquations (3).Again thepoint whose coordinates aregiven by(4)liesonthis axis. And, similarly, when gravity acts parallel totheaxis ofy.Itseems plausible, then, that ifthispoint be supported, thesystem willbeatrest,nomatter inwhat direction gravity acts. Thisis,infact, thecase. Toprove thestatement, letthepointPof10betaken as(x,y,0).ThenC=0.For mkamkpmkyij k Xij 2 Sniky since thecoefficient ofeach oftheunit vectorsi,j,kisseen at once tovanish, nomatter what valuesa,0,7may have. Thus theexistence ofacentre ofgravity fornparticlesis established. Itisapoint such that, nomatter howthesystem beoriented, theresultant couple duetogravityisnil. 14.Three Forces. Ifthree non-vanishing forces, acting on arigidbody, areinequilibrium, theylieinaplane andeither pass through apoint orareparallel. Proof.Iftwoforces inspace areinequilibrium, theymust be equal andopposite, andhave thesame lineofaction;orelseeach must vanish. Exclude thelatter case astrivial. Take vector moments about anarbitrary point, O,inthelineofaction ofone oftheforces. Then thevector moment oftheother forcemust vanish by12.Thus thesecond force either vanishes orpasses through ;i.e.through every point ofthelineofaction ofthe first force. Finally, theymust beequalaridopposite. Inthecase ofthree forces, nooneofwhich vanishes, andno twoofwhich have thesame lineofaction, take vector moments about apointinthelineofaction ofthe first force, butofno other force. Thesum ofthesecond andthird vector moments about must bezero. Hence thesecond andthird forces liein aplane through 0.They are, therefore, equivalent toasingle force acoupleisimpossible, since itcould notbebalanced bythe first force. Thus the firstforce reduces totheresultant, reversed insense, ofthesecond andthird forces, andthetheorem isproved. 44 MECHANICS ATrigonometric Theorem. Thefollowing trigonometric theorem isuseful inmany problems oftheequilibrium ofabody acted on bythree forces. Letalinebedrawn from thevertex ofatriangle, dividing theopposite sideintotwosegments oflengthsmand n, andmaking angles6and<pwith these sides. Then (m+n)cotif/mcot6ncot<p, where\l/istheangle this linemakes with the segment n. Theproofisimmediate. Project thesides ofthetriangle onthis line,produced: (m+n)cos\(/=acos 6cos<pt andthenapply thelawofsines : m b n FIG.47 sin i sin0' sn sn<p Example1.Auniform rodoflength 2aisheldbyastring oflength 21attached tooneend oftherodarid toapegina smooth vertical wall, theother end oftherod resting against thewall. Find allthepositions ofequilibrium. The three forces ofW,T,andRmust pass through apoint, andthismust bethemid-point ofthestring. Hence, applying theabove trigo- nometric theorem toeither ofthetriangles ABO orABC,wehave : (1)2tan6=tan<p. Asecond relation isobtained from purely geometrical consider- ations, namely:* (2)Icos=2acos<p. Itremains tosolve these equations. Squaring (1)andreducing, wehave : 4sec26=3+sec2 <p, or: 4cos2 <pcos2=1+3 cos2 <p *Itwould bepossible tousethegeometric relation /sin=asin<p. Butthefurther computation ofthesolution would belesssimple. STATICS OFARIGID BODY 45 Combining with (2),weget 4Z2cos2 <p=4a2cos2 t 1+3cos2 <p Since cos<pcannot vanish,itfollows that cos<p= Butaand Iarenotunrestricted, for<cos?<1.Hence andso a<I<2a, or,thestring must belonger than therod,butnottwice aslong. Furthermore, there arealways two positions ofequilibrium,in which therod isvertical, regardless ofIand a. Remark. What thetrigonometric theorem hasdone forusis toeliminate theforces. Withoutit,weshould havebeen obliged towritedown twoorthree equations involving TandR,andthen eliminate these unknowns, with which wehave noconcern so farastheposition ofequilibrium goes. Example2.Suppose that, inthe lastexample, thewall is rough. Then thereis,inaddition, anupward force offriction, F/xft,making four forces inall, when therod isjuston thepoint ofslipping down thewall. Butthe forcesRandFcanbecompounded intoasingle forceSmaking anangle \with thenormal to thewall, and sotheproblemisreduced toa three-force problem. Applying thetrigonometric theorem tothetriangle ABCwefind : 2acot<p=acot6atan\, (3) 2cot<p=cot- /z. Itisbetter here totakethegeometric relation intheform : (4) Isin8=asin(p. From(3)wenowhave : esc2=4cot2 <p+4/icot <p+M2+1. Hence 72 Z2sin2=- A-- : :- :- ;r-rr= 2sin2 4cot2 99+4/icot <p+M2+1 46 MECHANICS This lastequation canbegiven theform : Z2 4cos2 <p+4/icos<psin<p+(1+/j2 )sin2 <p= This equation,inturn, could bereduced toaquartic insin<p orcos<p;butsuch procedure would bebadtechnique. Rather, let 2cos2 <p=1+cos2y?, 2cos<psinv?=sin2<p, 2sin2p=1cos2<p. Theequationisthusreduced toanequation oftheform : Acos2<p+Bsin2^=C, andnowcanbesolved bythemethod ofChapter I,6. EXERCISE 1.Complete thestudy ofExample 2,i)computing A,B,C, andii)finding when therod isjustonthepoint ofslipping up. (K2- J-5-M2 ; ii)thesame equation with thesignofMreversed. 2.Ifa=1,I=If, p,=0.1, find allpositions ofequilibrium. EXERCISES ONCHAPTER II* 1.Show that,inatackle and fallwhich hasnpulleysineach block, thepower, P,exerted is2n+1times thetension intherope. 2.What force applied horizontally tothe hubofawheel (atrest) willjustcause thewheel tosurmount anobstacle ofheight h? 3.Twoheavy beads ofthesame weight can slideonarough horizontal rod.Tothebead isattached astring that carries asmooth heavy bead.How farapart canthebeads onthe rodbeplacedifthey aretoremain atrestwhen FIG.50 released ? 4.Agateisraised onitshinges anddoesnot fallback.How rough arethehinges? *Begin eachproblem bydrawing afigure showing theforces, andthelengths andangles which enter. STATICS OFARIGID BODY 47 5.There hasbeenalight fallofsnow onthegate.Acat weighing 5Ibs.walks along thetopofthegate, andthegate drops. The disconcerted catspringsoff. Itisobserved from hertracks inthesnow that shereached apoint 2ft.from the end ofthegate. The distance between thehinges is2^- ft., andthecentre ofgravity ofthegateis5ft.from thevertical linethrough thehinges.Ifthegateweighs 100Ibs.,what is thevalue of/z? 6.Arodrests inasmooth hemispherical bowl, oneendinside thebowlandtherimofthebowl incontact with therod. Find thepositionofequilibrium. a-fV32r2+a2 Ans. cos6= ,or where theradius ofthebowl isr,thedistance ofthecentre ofgravity oftherodfrom itslower end isa,andtheinclina- tion oftherodtothehorizon, 0,provided a<2r. 7.Auniform rodrestswithoneendonarough floorandthe other endonasmooth plane inclined tothehorizon atananglea. Find allpositions ofequilibrium. 8.Thesameproblem where both floorandplane arerough. 9.Apicture hangs onawall. Formulate theproblem of equilibrium when thewall issmooth, andsolve it. 10.Thesame question where thewall isrough. 11.Asmooth rodrests with oneendagainst avertical wall, apegdistant hfrom thewall supporting therod. Find the position ofequilibrium. [ftAns. cos=\*a 12.Thesame problem where thewall isrough, thepegbeing smooth. Find allpositions ofequilibrium. 13.Abarrel islyingonitsside.Aboard islaidonthebarrel, with itslower end resting onthe floor. Find allpositions of equilibrium. (Barrel, floor,andboard areallrough.) 14.Aplank 8ft.longisstood upagainst acarpenter's work- bench, which is2ft.8in.high. The coefficient offriction between either thefloor orthebench andtheplankis .Iftheplank makes anangle of15with thevertical,will itslipdown when letgo? 48 MECHANICS 15.Asmooth uniform rod rests inatest-tube. Find the positionofequilibrium. Ans. Thesolution isgivenbytheequations: 2tan=cot^>,rsin^+r=2acos 0. 16.Auniform rod2ft.long rests with oneendonarough table. Totheother endoftherod isattached astring1ft.long, made fast toapeg2ft.above thetable. Find allpositionsof equilibrium. Ans. Onesystemoflimiting positionsisgiven forju=2 bytheequations: cot<p=2+2cot0, 2cos6+cos<p=2. Solve these equations bymeans oftheMethod ofSuccessive Approximations. 17.Awater tower is100 ft.highand100 ft.indiameter. Find approximately thetension intheplates nearthebase. 18.Water isgradually poured intoatumbler. Show thatthe centre ofgravity oftheglassandthewater islowest when itisin thesurface ofthewater. 19.Ifoneattempts topulloutatwo-handled drawer byone handle, what isthecondition thatthedrawer willstick fast? CHAPTER III MOTION OFAPARTICLE 1.Rectilinear Motion.* Tubesimplest case ofmotion ofmat- terunder theaction offorce isHhat inwhich arigidbodymoves without rotation, each point ofthebody describing aright line, andtheforces that actbeing resolved along that line. Consider, forexample, atrain ofcars,andneglect therotation ofthewheels and axles. The train ismoved bythedraw-bar pull oftheloco- motive, andthemotion isresisted bythefriction ofthetracks andthewind pressure. Obviously,itisonly thecomponents of theforces parallel tothetracks that count, andtheproblem of Dynamics, orKinetics, asitismore specifically called,isto determine therelation between theforces andthemotion; or, ifone will: Given theforces, tofind thedistance traversed asafunction ofthetime. Amore conventional example, coming nearer topossible experi- mentation inthelaboratory, would bethat ofablock ofiron *Thestudent must notfeelobliged tofinish thischapter before going on.What isneeded isathorough drill inthetreatment oftheearly problems bythepresent methods, forthese arethegeneral methods ofMechanics, toinculcate which is aprime object ofthisbook. Elementary text-books inPhysics sometimes write down three equations: *=^at2 , v at,t?2=2as, andgiveanunconscionable number ofproblems tobesolved bythisdevice. The pedagogy ofthisprocedure istotally wrong, since itreplaces ideas byarule of thumb, andeven thisrule isbadly chosen, since itdisguises, instead ofrevealing, themechanical intuition. Now, afeeling forMechanics isthegreat object to beobtained, andthehabits ofthought which promote such intuition are,fortu- nately, cultivated byjustthesame mathematical treatment which applies inthe more advanced parts ofMechanics. Itisahappy circumstance thathere there isnoconflict, buttheclosest union, between thephysics ofthesubject andthe mathematical analysis.Athorough study of1-12through working eachproblem bythepresent general methods ismost important. Moreover, 22should here beincluded with, of course, thedefinition ofvector acceleration given in16andthestatement of Newton's Second Law in 17.Thestudent should thenturn toChapter IV,the most revealing chapter inthewhole elementary part ofthebook, andstudy itic alldetail. Theremaining sections ofthepresent chapter should beread casually atanearly stage, soasnottoimpede progress. Ultimately, they areimportant ; butthey aremost useful when thestudent comes torecognize their importance through hisexperience gathered from thelaterworkabove referred to. 49 50 MECHANICS placed onatableanddrawn alongbycords, soapplied that the block doesnotrotate andthateach pointofitdescribes aright line withvarying velocity. Itisclear thatablock ofplatinum having thesame mass, i.e.containing thesameamount ofmatter,ifacted onbythe same forces, would move justliketheblock ofiron,ifthetwowere started sidebysidefrom restorwith thesame initial velocities. Wecanconceive physical substances ofstillgreater density, and thesame would betrue.Oncompressing thegiven amount of matter intosmaller andever smaller volume, weareledtothe idea ofaparticle, ormaterial point,i.e.ageometrical point, to which theproperty ofmass isattached. This conception has theadvantage thatsuch aparticle would move exactly asthe actual body does ifacted onbythesame forces; butweneed saynothing about rotation, since thisideadoesnotenterwhen weconsider only particles. Moreover, there isnodoubt about where theforces areapplied theymust beapplied attheone point, theparticle. 2.Newton's Laws ofMotion. SirIsaacNewton (1642-1727), whowasoneofthechief founders oftheCalculus, stated three lawsgoverning themotion ofabody. FIRST LAW. Abody atrestremains atrestandabodyinmotion moves inastraightlinewithunchanging velocity, unless some external forceactsonit. SECOND LAW. The rateofchange ofthemomentum ofabody isproportionaltotheresultant external forcethatactsonthebody. THIRD LAW. Action andreaction areequalandopposite. Themeaning oftheFirstLaw isclear enough,ifwerestrict ourselves forthepresent tobodies and particlesasdescribed and moving in 1.*TheThird Law, too,isself-explanatory. Con- sider, forexample, two particlesofunequal mass, connected by aspring, themass ofwhich isnegligible. Then thepull (orpush) ofthespring ontheoneparticleisequal andoppositetoitspull (orpush) ontheotherparticle. TheSecond Law isexpressedinterms ofmomentum, andthe momentum ofaparticleisdefined astheproduct ofitsmassby *We might consider, furthermore, such material distributions aslaminae, i.e. material surfaces; andalso wires, ormaterial curves. Finally, rigidcombinations ofallthese bodies. MOTION OFAPARTICLE 51 itsvelocity, ormv. Here, visnotanessentially positive quan- tity themere speed. Wemust think oftheposition ofthe body asdescribed byasuitable coordinate,s.The lattermay bethedistance actually traversed bytheparticle; orwemay think ofthepathoftheparticle astheaxis ofx,and sasthe coordinate oftheparticle. The velocity, v,willthenbedefined asds/dt: and ispositive when sisincreasing ;negative, when sisdecreasing. TheSecond Lawcannowbestated intheform : (2) t. Here, /denotes theresultant force, and ispositive when ittends toincrease s;negative, when ittends todecrease s. Ordinarily, misconstant always, inthecase ofthebodies citedabove andso d(mv)_dv ~dT~m 'dt Thequantity dv/dtisdefined astheacceleration, and isoften represented bya : Itispositive when visincreasing, negative when visdecreasing. Newton's Second Lawcannowbestated intheform :Themass times theacceleration isproportionaltotheforce: (4) maoc/. From theproportion wenowpass toanequation: (5) ma=X/, where Xisaphysical constant. Thevalue ofXdepends onthe units used. Ifthese aretheEnglish units, thepound being the unitofmass, thefoottheunit oflength, thesecond theunitoftime, and thepound thegravitational unit offorce, then Xhasthe value 32(or,more precisely, 32.2), andNewton's Second Law of Motion becomes here : A,) m 52 MECHANICS Inthedecimal system, thegramme being theunit ofmass, the centimetre theunit oflength, thesecond theunit oftime,andthe gramme thegravitational unit offorce, X=981,andNewton's Second Law ofMotion becomes here : A2) mf t=981/. In 3weshall discuss theabsolute units. Inparticular, the units ofmass, length, andtime having been chosen arbitrarily, asinPhysics, theso-called "absolute unit offorce"isthat unit which makes X=1inNewton's Equation, sothathere : A\ dv . A,) m di=f- Three Forms fortheAcceleration. The acceleration isdefined asdv/dt, andsince v=ds/dt, wehave : x/2/!\ U* (6) a=^ Athirdform isobtained bystarting with theequation: (7\dv_dsdv ('dt~ dids andthen replacing ds/dtbyitsvalue,v.Thus (8) .-.*. These three forms fortheacceleration : dvdzs dv connect thethree letterss,t,vinpairs inallpossible ways. Which form itisbetter touseinagiven case, willbecome clear from practiceinsolving problems. Example1.Afreight train weighing 200tons isdrawn by alocomotive that exerts adraw- jtPP-I/==8000 tbarpull of9tons. 5tons ofthis s force areexpended inovercoming frictional resistances. Howmuch speed willthetrain have acquired attheendofaminute,ifit starts from rest? MOTION OFAPARTICLE 63 Herewehavem-200X2000=400,000 Ibs., /=9X2000-5X2000=8000 Ibs.* andhence Equation A,)becomes : 400,000^=32X8000, or = dt 25 Integrating with respect tot,wefind : V== Tjngt~f~C . Since v=when t=0,wemusthaveC0,andhence Attheendofaminute,t=60,andso ^=MX60=38 -4ft -Persec. Toreduce feetpersecond tomiles perhour itisconvenient tonotice that30miles anhour isequivalent to44ft.asecond, asthestudent canreadily verify; orroughly, 2miles anhour corresponds to3ft.asecond. Hence thespeed inthepresent case isabout two-thirds of38.4, or26miles anhour. Example 2.Astone issent gliding over theicewithaninitial velocity of30ft.asec. Ifthecoefficient offriction between the stoneandtheiceis-fa,how farwillthestone go? Here, theonly force thatwetakeaccount ofistheretarding force offriction, andthisamounts toone-tenth ofapound offorce forevery pound ofmass there is m inthestone. Hence,ifthere are 4 \10-q mpounds ofmass inthestones theforce willbe^m lbs.,f andFlG *52 since ittends todecreases,itistobetaken asnegative: *Thestudent must distinguish carefully between thetwomeanings oftheword pound, namely (a)amass, and (6)aforce twototally different physical objects.Thus apound oflead isacertain quantity ofmatter. Ifitishungupbyastring, thetension inthestring isapound offorce. tThestudent should notice thatmisneither amass noraforce, butanumber, like alltheother letters ofAlgebra, theCalculus, andPhysics. 54 MECHANICS Nowwhatwewant isarelation between vands,fortheques- tion is:How far (s=?),when thestone stops (v=0)?Sowe usethevalue (8)ofaandthusobtain theequation: dv 16 V ds=~-5> or vdv=/-ds. rr"216 .nHenceT>= ^-s+C. Todetermine Cwehave thedata that,when s=0,v=30. Since inparticular theequation must hold forthese values, ^!=+C, C=450, andso v2=900-^s. When thestone stops,v=0,andwehave =900-3s,s=141 ft. EXERCISES* 1.Ifthetrain ofExample1wasmoving attherate of4m. anhourwhenwebegan totake notice, how fastwould itbemov- inghalfaminute later? Giveacomplete solution, beginning with drawing thefigure. Ans. About 17m.anh. 2.Asmall boy seesaslideontheiceahead, andruns for it. Hereaches itwith aspeed of8miles anhourand slides 15feet. How rough arehisshoes? Ans. M=.15. 3.Show that,ifthecoefficient offriction between asprinter's shoes andthetrack isTV>n^^cs^possible record inahundred- yarddashcannot belessthan 15seconds. 4.Anelectric carweighing 12tons getsupaspeed of15miles anhour in10seconds. Find theaverage force that actsonit, *Itisimportant that thestudent should work these exercises bythemethod setforth inth4etext,beginning each timebydrawing afigure andmarking (t)the force, bymeans ofadirected right line, orvector, drawn preferably inredink; and(t'i)thecoordinate used, assorx,etc.Heshould nottrytoadapt suchformulas ofElementary Physics as v=at, a=a2 , vz=2as topresent purposes. For,thoobject ofthese simple exercises istoprepare theway forapplications inwhich theforce ianotconstant, andheretheformulas just cited donothold. MOTION OFAPARTICLE 55 i.e.theconstant forcewhich would produce thesame velocity in thesame time. 6.Inthepreceding problem, assume that thegiven speed is acquired after running 200 feet. Find thetime required and theaverage force. 6.Atrainweighing 500tonsandrunning attherate of30miles anhour isbrought torestbythebrakes after running 600 feet. While itisbeing stopped itpasses overabridge. Find theforce withwhich thebridge pullsonitsanchorage. Ans. 25.2 tons. 7.Anelectric car isstarting onanicytrack. Thewheels skidand ittakes thecar15seconds togetupaspeed oftwomiles anhour. Compute thecoefficient offriction between thewheels andthetrack. 3.Absolute Units ofForce. The units interms ofwhich we measure mass, space, time,andforce arcarbitrary, aswaspointed outin 2.Ifwechange one ofthem, wethereby change the value ofXinNewton's Second Law. Consequently, bychanging theunit offorce properly, theunits ofmass, space, andtime being heldfast,wecanmake X=1.Hence thedefinition above given: DEFINITION. The absolute unit offorce_isthat unitwhich makes X=1inNewton's_ Second Law ofMotionij* (1) moT^f. Inorder todetermine experimentally theabsolute unit offorce, wemay allow abody tofallfreely andobserve how far itgoes in aknown time. Itisaphysical lawthat theforce withwhich gravity attracts anybodyisproportional tothemass ofthatbody. Letthenumber gbethenumber ofabsolute units offorce with *Wehave already metaprecisely similar question twice intheCalculus. In differentiating thefunction sinxweobtain theformula Dxsinx=cosx onlywhenwemeasure angles inradians. Otherwise theformula reads: DxsinxXcosx. Inparticular, iftheunit isadegree, X=Tr/180. Wemay, therefore, define aradian asfollows :Theabsolute unit ofangle (theradian) isthatunitwhich makes X=1 intheabove equation. Again, indifferentiating thelogarithm, wefound; X This multiplier reduces tounity when wetakea=e.Hence thedefinition : Theabsolute (natural) base oflogarithms isthatbasewhich makes themultiplier logo eintheabove equation equal tounity. 56 MECHANICS which gravity attracts theunit ofmass. Then theforce, measured inabsolute units, withwhich gravity attracts abody ofmunits ofmass willbemg.Newton's Second LawA3)gives forthiscase : dv , dv ds s=%gt*+K,K=0, andwehave thelaw forfreely falling bodies deduced directly from Newton's Second Law ofMotion, thehypothesis being merely that theforce ofgravityisconstant. Substituting in thelastequation theobserved values s=S,t=T,weget: 28 9= ?*' IfweuseEnglish units formass, space, and time, ghas, to two significant figures, thevalue 32,i.e.theabsolute unit of force inthissystem, apoundal,isequal nearly tohalfanounce. Ifweusec.g.s. units, granges from 978to983atdifferent parts oftheearth, andhasinCambridge thevalue 980.Theabsolute unit offorce inthissystemiscalled thedyne. Since gisequal totheacceleration withwhich abodyfalls freely under theattraction ofgravity, giscalled theacceleration ofgravity. But this isnotourdefinition ofg;itisatheorem about gthat follows fromNewton's Second Law ofMotion. The student cannow readily prove thefollowing theorem, which isoften taken asthedefinition oftheabsolute unit of force inelementary physics:Theabsolute unit offorce isthat force which, acting ontheunit ofmass fortheunit oftime, gener- atestheunit ofvelocity. Incidentally wehave obtained twooftheequations forafreely falling body: v=gt,s=%gt2 . Thethird isfound bysetting a=vdv/ds andintegrating: dv 2gs. MOTION OFAPARTICLE 57 Example. Abodyisprojected down arough inclined plane withaninitial velocity ofVQfeetpersecond. Determine the motion completely. The forces which actare :thecomponent ofgravity, mgsin7 absolute units, down theplane, andtheforce offriction, pR= nmgcos7uptheplane. Hence ma=mgsin7nmgcos7, dv -IT=gsm7 cos7. Integrating thisequation, we get v=g(sin7 /zcos7)t+C, o= +<?> =g(sin7-cos7)+v . Asecond integration gives B)s=\g(sin7-^cos7)P+VQt, theconstant ofintegration herebeing0. Tofind vinterms ofswemay eliminate tbetween A)and B).Orwecanbegin byusing formula (8), 2,fortheacceler- ation : dvf. , v-r=g(sin7- /xcos7),do %v2=g(sin7Mcos7)s+K,^= +X, t;2=20(sin7 /zcos7)s+#o- EXERCISES 1.If,intheexample discussed inthetext, thebodyispro- jected uptheplane, findhow faritwillgoup. 2.Determine thetime ittakes thebodyinQuestion1to reach thehighest point. 3.Obtain theusual formulas forthemotion ofabody pro- jected vertically: v2=2gs+vl or=2gs+vl ; v=gt+VQ or=-gt+VQ; 8=ot*+vt or=-tf* +M. 58 MECHANICS 4.Onthesurface ofthemoon apound weighs only one-sixth asmuch asonthesurface oftheearth. Ifamouse canjump up1footonthesurface oftheearth, howhigh could shejump onthesurface ofthemoon? Compare thetime she isintheair inthetwocases. 6.Ablock ofironweighing 100pounds rests onasmooth table.Acord, attached totheiron, runsover asmooth pulley attheedge ofthetableand carries aweight of15pounds, which hangs vertically. Thesystemisreleased with theiron 10feet from thepulley. How longwill itbebefore theironreaches the pulley, andhow fast will itbemoving? Ans. 2.19 sec.;9.1 ft.asec. 6.Solve thesame problem ontheassumption thatthetable is rough, n=^,andthat thepulley exerts aconstant retarding force of4ounces. 7.Regarding thebiglocomotive exhibited attheWorld's Fair in1905bytheBaltimore andOhio Railroad theScientific American said :"Previous tosending theengine toSt.Louis, the engine wastested atSchenectady, where shetooka63-car train weighing 3150 tonsupaone-per-cent. grade." Findhow long itwould take theengine todevelop aspeed of15m.perh.inthesame trainonthelevel, starting from rest, thedraw-bar pullbeing assumed tobethesame asonthegrade. 8.IfSirIsaac Newton registered 170pounds onaspring balance inanelevator atrest,andif,when theelevator was moving, heweighed only 169pounds, what inference would he drawabout themotion oftheelevator? 9.What doesamanwhose weightis180pounds weigh inan elevator that isdescending withanacceleration of2feetper second persecond ? 4.Elastic Strings. When anelastic stringisstretched bya moderate amount, thetension, T,inthestring isproportional tothestretching, i.e.tothedifference, s,between thestretched andtheunstretched length ofthestring: (1) Tocs, orT=ks, where fcisaphysical constant, whose value depends bothonthe particular string andontheunits used. MOTION OFAPARTICLE 59 Suppose, forexample, thatastring isstretched 6in.byaforce of12Ibs.;todetermine k.Ifwemeasure theforce ingravitar tional units,i.e.pounds, then T=12 when s=. Hence, substituting these values inequation (1),wehave: 12=&, or k=24, (2) T=24s. Ifwehadchosen tomeasure theforce inabsolute units,i.e. poundals, then, since ittakes (nearly) 32ofthese units tomake apound, thegiven force of12pounds would beexpressed as (nearly) 12X32,orprecisely 120,poundals. Hence, substitut- ingthepresent value oftheforce in(1),which, toavoid con- fusion, wewillnowwrite intheform : T=k's, wehave : 120=k'\or k'=240, (3) T'=240s. When thestringisstretched 1in.,s=^andthetension asgiven by(2)isT=2,i.e.2pounds. Formula(3),onthe other hand, gives 20,or64(nearly) asthevalue oftheten- sion, expressedinterms ofpoundals, and this isright; for it takes 64half-ounces tomake 2pounds, andsoweshould have T'=20.* Thelawofstrings stated above isfamiliar tothestudent inthe form ofHooke's Law: rr\ __. I' where Iisthenatural, orunstretched, length ofthestring, and lf ,thestretched length; thecoefficient Ebeing Young's Mod- ulus. Foragiven string, E/l=kisconstant, andV I=sis variable. *Itiseasy tocheck ananswer inanynumerical case. Thestudent hasonly toaskhimself thequestion: "Have Iexpressed myforce inpounds, orhave I expressed itinterms ofhalf-ounces?" Just asfivedollars isexpressed bythe number 5whenweusethedollar astheunit, butbythenumber 500whenwe usethecent, so,generally, thesmaller theunit, thelarger thenumber which expresses agiven quantity. 60 MECHANICS EXERCISES 1.Anclastic stringisstretched 2in.byaforce of.3Ibs. Find thetension (a)inpounds; (b)inpoundals, when itisstretched sft. Ans.(a)T=18s;(6)T=180s. 2.When thestring ofQuestion1isstretched 4in.,what is thetension (a)interms ofgravitational units; (b)interms of absolute units? Ans.(a)6pounds; (6)192poundals. 3.Anelastic stringisstretched 1cm.byaforce of100 grs. Find thetension(a)ingrs. ;(6)indynes, when itisstretched scm. Ans.(a)100s; (b)98,000s. 4.Oneendofanelastic string 3ft.longisfastened toapeg atA,anda2-pound weightisattached totheother end.The weightisgradually lowered tillitisjustsupported bythestring, and itisfound thatthelength ofthestring hasthusbeen doubled. Find thetension inthestring when itisstretched sft. Ans. fsIbs.;^spoundals. 5.AProblem ofMotion. Oneend ofthestring considered inthetext of4isfastened toapegatapoint ofasmooth horizontal table;aweight of3Ibs. isattached totheother end ofthestring andreleased from restonthetable with thestring .stretched one foot.How fast willtheweight bemoving when thestring becomes slack? Theweight evidently describes astraightlinefrom thestarting point, A,toward thepeg0,andwewish toknow itsvelocity when ithasreached apoint B,onefootfromA. The solution isbased onNewton's Second Law ofMotion. Itisconvenient here totake asthecoordinate, notthedistance APthattheparticle hastravelled at --- <-'- -janyinstant, but itsdistance sfrom B. T-, KA The force which acts isthetensionriG.54 i.iofthestring; measured inabsolute units itis240rs. Since ittends todecreases,itisnegative. Hence Newton's Lawbecomes : (1) (2)___fJ2atit)Tointegrate thisequation, replace -^byitsvalue v-=- : MOTION OFAPARTICLE 61 Hence vdv= (3)Ivdv=8gIsds, Todetermine C,observe thatinitially,i.e.when theparticle wasreleased atA,v=and s=I.Hence =- 4(7+C, C=40, and (3)becomes (4)t;2=80(1-s2 ). Wehavenowdetermined thevelocity oftheparticle atan arbitrary point ofitspath, andthus areinaposition tofind its velocity attheonepoint specified inthequestion proposed, namely, atB.Here, s=0,and v2Uo=8g=8X32, v\s=sQ=16(ft.persec.) EXERCISES* 1.Theweight intheproblem justdiscussed isprojected from Balong thetable inthedirection ofOBproduced withavelocity of8ft.persec. Findhow faritwillgobefore itbegins toreturn. Ans. Newton's equationisthesame asbefore, and the integral, (3),isthesame;butinitiallys=and v=8. HenceC=32,andtheanswer is6inches. 2.If,intheexample worked inthetext, thetable isrough andthecoefficient offriction, /i,hasthevalue^,how fast will thebody bemoving when itreaches B? Ans. Newton's equation nowbecomes : 3^=-24g8+i-3g, andtheanswer is :4Vl5=15.49 ft.persec. 3.Solve theproblem ofQuestion 1,forarough table, M=T- Ans. The required distance isthepositive root ofthe equation 16s2+s4=0,ors=.4698ft.,orabout 5fin. *Inthefollowing exercises andexamples, itwillbeconvenient totakethevalue ofgasexactly 32when English units areused. Begin each exercise bydrawing a figure showing thecoordinate used, andmark theforces inredink. 62 MECHANICS 4.Find where theweight inQuestion 2willcome torest ifthestring, afterbecoming slack, doesnotgetintheway. 6.The2Ib.weight ofQuestion 4, 4,isreleased from rest atapointBdirectly under thepeg.4andatadistance of3ft. fromA;thestring thus being taut, butnotstretched. Find how faritwill fallbefore itbegins torise. Ans. 6ft. 6.If,inthelastquestion, theweightisdropped from the pegatA,findhow faritdescends before itbegins torise. Ans.Toadistance of6+3\/3=11.196 ft.below A. 7.Iftheweight inthelasttwoquestionsiscarried toapoint 7ft.belowAandreleased, show that itwill risetoadistance of 5ft.belowAbefore beginning tofall. 8.If,inthelastquestion, theweightisreleased from apoint 10ft.below A,show;that itwill risetoaheight of1ft.and10in. below A. 9.The string oftheexample studied inthetext of4is placed onasmooth inclined plane making anangle of30with thehorizon, andoneend ismade fasttoapegatAintheplane. Ifaweight of1Ibs.beattached totheother endofthestring andreleased from restatA,findhow fardown theplaneitwill slide. Assume theunstretched length ofthestring tobe4ft. 10.Thesame questioniftheplaneisrough, /*=^V3. 11.Acylindrical sparbuoy (specific gravity ^)isanchored sothat itisjustsubmerged athigh water. Ifthecable should break athigh tide,show that thesparwould jump entirely out ofthewater. Assume that thebuoyancy ofthewater isalways justequal totheweight ofwater displaced. 12.Aparticle ofmass 2Ibs. liesonarough horizontal table, and isfastened toapostbyanelastic band whose unstretched lengthis10inches. The coefficient offriction is-,andtheband isdoubled inlength byhangingitvertically with theweight at itslower end. Iftheparticle bedrawn outtoadistance of 15inches from thepostandthen projected directly away from thepostwithaninitial velocity of5ft.asec., findwhere itwill stop forgood. MOTION OFAPARTICLE 63 6.Continuation;theTime. Thetime required bythebody whose motion wasstudied in5toreach thepointBcanbe found asfollows. From equation (4)wehave : (5)v=-^=V8gVl s2 . Since sdecreases as tincreases, ds/dtisnegative, andthelower sign holds. Replacing V8gbyitsvalue, 16,weseethat (6) This differential equationisreadily solved byseparatingthe variables,i.e.bytransforming theequation sothat only the variable soccurs ononeside ofthenewequation, andonlyton theother;thus (7) I6dt=- Hence 16*=-f ,ds=-sin-1s+C. =-f,ds=-siJVl-s2 Ifwemeasure thetimefrom theinstant when thebodywas released atA,then t= arid s=1arethe initial values which determine C : =-sin-11+C, C=~ Thus l&t=-sin-1s. The right-hand side ofthisequation hasthevalue cos""a . Hence wehave, asthefinal result,* (8) 16t=cos"1 s, or s=cos16$. *Inevaluating theabove integral wemight equally wellhave used theformula Vl- 2 Weshould thenhavehad : 16*=cos-18-C'. Substituting theinitial values t-0,*=1inthisequation, wefind : =cos'11-C", or C"-0, andthefinal result isthesame asbefore. 64 MECHANICS This equation gives thetime ittakes thebody toreach an arbitrary point ofitspath. Inparticular, thetimefromAto Bisfound byputtings=: (9)W=cos-1= |,t= J2=.09818 sec. EXERCISES 1.Show that ifthebody,inthecase just discussed, hadbeen released from restatanyother distance from thepeg,thestring being stretched, thetime tothepoint atwhich thestring becomes slackwould havebeen thesame. 2.Show that ittakes thebody twice aslong tocover the first halfofitstotalpath asitdoes tocover theremainder. Find thetime required tocover theentire path inthecase ofthefollowing exercises attheclose of 5. 3.Exercise 1. Ans.^=.09818.OA 4.Exercise 5. Ans. t=\^K / / -',total time,TrA/^=.9618 sec.*04Jv6s s2 ^ 5.Exercise 6. Ans. t=\ ^+sin-1 77==? 6.Exercise 7. Ans. .9618 sec. 7.Exercise 9. 8.Exercise 10. 9.Exercise 8. 7.Simple Harmonic Motion. Thesimplest andmostimportant ease ofoscillatory motion which occurs innature isthatknown asSimple, Harmonic Motion. Itisillustrated with the least amount oftechnical detail bythefollowing example, orbythe firstExercise below. Example. Ahole isbored through thecentre oftheearth, a stone isinserted, theairisexhausted, andthestone isreleased from restatthesurface oftheearth. Todetermine themotion. Theearth ishereconsidered asahomogeneous sphere, atrest inspace. Itsattraction, F,on ^ thestone diminishes asthestone nears thecen- FIG.55 tre,and itcanbeshown tobeproportional,at MOTION OFAPARTICLE 65 anypoint ofthe hole, tothedistance ofthestone from the centre: ~ ,Focr, orF=kr. Todetermine theconstantfc,observe that, atthesurface, r=R(theradius oftheearth), and,ifwemeasure Finabsolute units,F=wgr,wheremdenotes themass ofthestone. Hence mg=kR or k=~^,/ andF=^rR Asthecoordinate ofthestonewewilltake itsdistance, r, from thecentre oftheearth. Then Newton's Second Law gives us: /i\d*r mg (1) m^=-Rr- For,when rLspositive, theforce tends todecreaser,andsois negative. When risnegative, theforce tends toincrease ralge- braically, andsoispositive. Hence(1)isright inallcases. Inorder tointegrate Equation (1),which canbewritten in theform : (2) <*L=-L r wdp Rr> weemploy thedevice ofmultiplying through by2dr/dt: cydrd^r =_2gdr dtdt*~ RT dt d/rfr\2 The left-hand side thus becomes -77 (-n )Hence each side at\dt/ canbeintegrated with respect tot:* *Thismethod canbeapplied toanydifferential equation oftheform : Multiply through by2dy/dx: a4? <&ccte2 Theleft-hand sidethusbecomes (~ JHence Integrating, wehave 66 MECHANICS --^.Cr^~Rr dtCd ( Jdt\dt or _ - - dt~R-R Todetermine(7,observe that initially,i.e.when thestonewas atAyr=Randthevelocity, dr/dt,=0.Hence 0=~|ft2 +c, orC=!#2 . 7 it Thus finally: <3>(I)'- 1<'-"> Atthecentre oftheearth,r=0,and (dr/dt)2=gR.Ifwe taketheradius oftheearth as4000 miles, thenR=4000X5280, g=32,andthevelocityisabout 26,000ft.asec., orapproxi- mately 5miles asecond. Thestone keeps onwith diminishing speed andcomes torest foraninstant when r=J?,i.e. itjustreaches theother side oftheearth, andthen falls back. Thus itoscillates throughout thewhole length ofthehole, reaching thesurface attheendof each excursion, andcontinuing thismotion forever. The result isnotunreasonable, forthere isnodamping ofany sort, no friction orairresistance. TheTime. Tofind thetimeweproceed asin 6.From Equation (3)itfollows that Hence, separating thevariables, wehave : '" dr dt=-\F====^. 9VR*-r or t= ^\|cctfh1~+C. I Initially,t=and r=72;thus (7=0,and (4)t=\cos-1~, or r=Rcos(^VB)' * <7 /t \A// MOTION OFAPARTICLE 67 ThetimefromAto isfound byputtingr=: *-' Oncomputing thevalue ofthisexpression itisseen tobe21min. and16sec.ThetimefromAtoBistwice theabove. Hence thetime ofacomplete excursion, fromAtoBandback toAis Thistime isknown astheperiod oftheoscillation.* TheGeneral Case. Simple Harmonic Motion isalways dom- inated bythedifferential equation A\**X_ %~ '~dfi~ ' where thecoordinate xcharacterizes thedisplacement from the position ofnoforce. This equation canbeintegrated asinthe special caseabove, and itisfound that B) where hdenotes thevalue ofxwhich corresponds totheextreme displacement. The velocity when x=isnumerically nh,and thus isproportional both tonandtoh.Asecond integration gives C) x=hcosntj provided thetime ismeasured from aninstant when x=h. Theperiod, T,isinversely proportional ton : ^ 27T andtheamplitudeis2h.Thus theperiodisindependentof theamplitude. Themotion represented byEquation C)isknown asSimple Harmonic Motion. Thegraph ofthefunction isobtained from *Inthe firstequation (4)theprincipal value oftheanti-cosine holds during the firstpassage ofthestone fromAtoB.Thesecond equation (4)holds with- outrestriction. 68 MECHANICS thegraph ofthecosine curve byplotting thelatter toonescale ontheaxisoft,andtoanother scaleontheaxis ofx. FIG.56 EXERCISES 1.Two strings liketheonedescribed inthetext of4are fastened, oneend ofeach, totwopegs,AandB,onasmooth horizontal table, thedistance ABbeing double thelength of either string, andtheother end ofeach stringismade fastto a3Ib.weight, which isplaced at0,themid-pointofAB.Thus each stringistaut, butnotstretched. Theweight beingmoved toapointCbetween andAandthen released from rest,show that itoscillates withsimple harmonic motion. Find thevelocity withwhich itpasses andtheperiod oftheoscillation. Itis assumed thatthestring which isslack innowise interferes with orinfluences themotion. Ans. The differential equation which dominates themotion d2x isffi=256z, where xdenotes thedisplacement ofthe at 3Ib.weight ;hence themotion issimple harmonic motion. Therequired velocityisnumerically 16h,where hdenotes themaximum displacement. The periodis27T/16= .3927 sec. 2.Work thesame problem fortwo stringsliketheone of Question 4, 4,anda2Ib.weight. 3.Show that themotion ofExample 7, 5,issimple harmonic motion, andfindtheperiod. 4.Ifastraight holewerebored through theearth fromBoston toLondon, asmooth tube containing aletter inserted, the air exhausted from thetube, andtheletter released atBoston, how longwould ittaketheletter toreachLondon? MOTION OFAPARTICLE 69 6.Ifintheproblem ofQuestion 9, 5,theweight were re- leased withthestring taut,butnotstretched, anddirected straight down theplane, show that theweight would execute simple harmonic motion. Determine theamplitude andtheperiod. 6.Work theproblem ofthetext forthemoon;cf.thedata in 8. 7.Asteel wire ofonesquare millimeter cross-section ishung upinBunker HillMonument, andaweight of25kilogrammesis fastened tothelower end ofthewireand carefully brought to rest. Theweightisthen given aslight vertical displacement. Determine theperiod oftheoscillation. Given thattheforce required todouble thelength ofthewire is21,000 kilogrammes, andthat thelength ofthowire is210 feet. Ans.Alittle over halfasecond. 8.Anumber ofironweights areattached tooneendofalong round wooden spar, sothat,when lefttoitself, thespar floats vertically inwater. Aten-kilogramme weight having become accidentally detached, thesparisseen tooscillate withaperiod of4seconds. The radius ofthesparis10centimetres. Find thesum oftheweights ofthesparandattached iron. Through what distance docsthespar oscillate ? Ans.(a)About 125kilogrammes ;(6)0.64metre. 8.Motion under theAttraction ofGravitation. Problem. To findthevelocity which astone acquiresinfalling totheearth from interstellar space. Assume theearth tobeatrestandconsider onlythe \A force which theearth exerts. Letthestone bere- j leased from restatA,and letrbeitsdistance from thecentre oftheearth atanysubsequent instant. Then the force, F,acting on itis,bythelaw of gravitation, inversely proportional tor: ElA*-? SinceF=mgwhen r=JR,theradius oftheearth, X ,rmgR*mq=-F^ and v=^ 70 MECHANICS Newton's Second Law ofMotion heretakesontheform : dzr_mgR2m ~dT*~ ~7*~' Hence Tointegrate thisequation, weemploy themethod of7and multiply by2dr/dt: drd2r2gR2dr d/dr\*=2gR*dr dtdt2r2dt'rdi\dt) r2dt Integrating with respect totwefind : Initially dr/dt=andT=Z;hence o.' +c,c~ Since dr/dtisnumerically equal tothevelocity, thevelocity Vatthesurface oftheearth isgivenbytheequation: IfIisvery great, thelastterm intheparenthesisissmall, and so,nomatter how greatIis,Vcannever quite equalV2gR. Here g=32,R=4000X5280, andhence thevelocityinques- tion isabout 36,000 feet, or7miles, asecond. This solution neglects theretardingeffect oftheatmosphere; butastheatmosphereisvery rare ataheight of50miles from theearth's surface, theresult isreliable down toapoint com- paratively neartheearth. Inqrder tofindthetime itwould take thestone tofall,con- sider theequation derived from (2): Hence 2 ,andMOTION OFAPARTICLE 71 Vlrrdr Turning toPeirce's Tables, No.169,wefind : dr Vlr-r2 =Vlr-r2+~sin- Thus t= fji Initiallyt=and r=I: Finally, then, fr-H+ss-sm- Forpurposes ofcomputation, abetter form ofthisequation isthefollowing: (3) EXERCISES* 1.Iftheearth hadnoatmosphere, withwhat velocity would astone have tobeprojected from theearth's surface, inorder nottocomeback? 2.Ifthemoon were stoppedinitscourse, how longwould ittake ittofalltotheearth? Regard theearth asstationary. Ans. 4days, 18hrs., 10min. *Inworking these exercises, thefollowing datamaybeused : Radius ofthemoon,^that oftheearth. Mass ofmoon, ^Tthat ofearth. Mean distance ofmoon from earth, 237,000 miles. Acceleration ofgravity onthesurface ofthemoon, thatonthesurface ofthe earth. Diameter ofsun,860,000 miles. Mass ofsun,333,000 that oftheearth. Mean distance ofearth from sun,93,000,000 miles. Acceleration ofgravity onthesurface ofthesun,905 ft.persec.persec. 72 MECHANICS 3.Solve thepreceding problem accurately, assuming that the earth andthemoon arereleased from rest ininterstellar space attheir present mean distance apart. Theircommon centre of gravitywillthenremain stationary. 4.Thesameproblem fortheearthandthesun. 6.Iftheearth andthemoon were held atrestattheir present mean distance apart, withwhat velocity would aprojectile have tobeshotfrom thesurface ofthemoon, inorder toreach the earth? 6.Iftheearth andthemoon were held atrestattheir present mean distance apart, andastone were placed between them at thepointofnoforceandthen slightly displaced toward theearth, withwhat velocity would itreach theearth ? 7.Ifaholewere bored through thecentre ofthemoon, as- sumed spherical, homogeneous, andatrest ininterstellar space, andastone dropped in,howlongwould ittakethestone toreach theother side? 8.Show that iftwospheres, eachonefoot indiameter andof density equal totheearth's mean density (specific gravity, 5.6) were placed with their surfaces ofaninchapart andwere acted onbynoother forces than theirmutual attractions, theywould come togetherinabout fiveminutes andahalf. Given thatthe spheres attract asifalltheirmass were concentrated attheir centres. 9.WorkDone byaVariable Force. Ifaforce, F,constant inmagnitude andalways acting along afixed lineAB inthe same sense, beapplied toaparticle,* and iftheparticle bedis- placed along thelineinthedirection oftheforce, thework done bytheforceontheparticleisdefined inelementary physics as F I W=Fl, Ap Bwhere Idenotes thedistance through whichpeoC3 theparticle hasbeen displaced. Suppose, however, that theforce isvariable, butvarying con- tinuously andalways acting along thesame fixed line.How shall theworknowbedefined ? *Or,more generally, tooneandthesame pointPofarigid ordeformable material body. MOTION OFAPARTICLE 73 Letacoordinate beassumed ontheline;i.e.think oftheline astheaxis ofx.Lettheparticle bedisplaced from A:x=a toB:x=6,and leta<b.LetF,tobegin with, always act inthedirection ofthepositive sense along theaxis. Then F=f(x), where f(x)denotes apositive continuous function ofx. Divide theinterval (a,b)upintonparts bythepoints xl9 xz, ,zn_i,and letXQ=a,xn=b.Then,if xt+i Xk=Azfc, thework, ATF*,donebytheforce indisplacing theparticle through the fc-th interval ought, inorder tocorrespond tothegeneral physical conception ofwork, toliebetween thequantities FiAx and Fi'Az, where FJandF'k'denote respectively thesmallest andthelargest values off(x)inthisinterval.* Wehave, then : (1) FiAx^ATF t^Fi'te. Onwriting outthedouble inequality (1)fork=0,I, , n 1andadding thenrelations thus resulting together, wefind thatW=2AWk liesbetween thetwosums : (2) F'^x+F(Ax++n_,Az, (3) F'Jte+F('&x++K'^Az. Each ofthesesums suggests thesum (4) /(* )Az+f(Xl)A*!+-+/(*_,) Axn, whose limit isthedefinite integral, ft (5)lim[/(x )A*+/(x,)Ax++/(*._0 Ax]=f/(*) As. n-oo Ja ThatWisinfactequal tothisintegral: h (6) W= Jf(x)dx, a follows fromDuhamePs Theorem. *Thisstatement ispure physics. Itisthephysical axiom onwhich thegeneral- ization ofthedefinition ofwork isbased. More precisely, itisoneoftwophysical axioms, theother being thatthetotalwork,W,forthecomplete interval isthesum ofthepartial works, ATT*, forthesubintervals. 74 MECHANICS IftheforceFacts inthedirection opposite tothat inwhich thepointofapplicationismoved, weextend thedefinition and saythatnegative work isdone. ForthecasethatFisconstant, thework isnowdefined asfollows : (7) W=F(b-a). Here,Fistobetaken asanegative number equal numerically totheintensityoftheforce. Thus (7)isseen tohold inwhichever direction theforce acts, provided that a<6.Will (7) still hold if6<a? Itwill. There areinallfour possible cases : i)++ ii) Hi)H h Incasesi)andii)theforce overcomes resistance, and positive work isdone. Incases in)andiv)theforce isovercome, and negative work isdone. Hence (7)holds inallcases. Itisnoweasy toseehow thedefinition ofwork should be laiddownwhenFvaries inanycontinuous manner. Theconsider- ations areprecisely similar tothose which ledtoEquation (6), andthatsame equation isthefinal result inthis,themostgeneral, case : & W=Cf(x) dx. a Example. Tofindthework done instretching awire. Letthe natural (orunstretched) length ofthewirebeI,thestretched length, V.Then the tension, T,is __, r-^-iT > igivenbyHooke's Law : O AP B FIG.59 T=\~~ where Xisindependent ofIand V,and isknown asYoung's Mod- ulus. Letthewire,initsnaturalstate,liealong the lineOA,and letit,when stretched,liealong OB,OPbeing anarbitrary inter- mediate position. Letxbemeasured from A,and letx=hat B.Then T=\~ ,TT7 /\x,xr,and W=I\jdx= jIxdx=-==- MOTION OFAPARTICLE 75 This istheworkdoneonthewirebytheforce that stretches it. Ifthewire contracts, thework donebythewireonthebody to which itsendPisattached willbe /(->!)*<"Xa2 21 21 EXERCISES 1.Intheproblem of7compute thework donebytheearth ontheparticle when thelatter reaches thecentre. 2.Aparticle ofmassmmoves down aninclined plane. Show thatthework doneonitbythecomponent ofgravity down the planeisthesame asthework donebygravity ontheparticle when itdescends vertically adistance equal tothechange in levelwhich theparticle undergoes. 3.Aparticleisattracted toward apoint byaforce which isinversely proportional tothesquare ofthedistance from 0. Howmuch work isdoneontheparticle when itmoves from a distance atoadistance balong aright linethrough 01 4.Iftheearth andthemoon were stopped intheir courses andallowed tocome together bytheirmutual attraction, how much work would theearth havedoneonthemoon when they meet? 5.Find thework donebythesunonameteor when thelatter moves along astraightlinepassing through thecentre ofthe sun,fromaninitial distance Rtoafinal distance r. 10.Kinetic Energy andWork. Letaparticle ofmassm describe aright linewith velocity v=ds/dt. Itskinetic energy isdefined asthequantity: mv2 2' Lettheparticle move under theaction ofanyforceFwhich varies continuously: F=/(s). Then Newton's Second Law canbewritten intheform : dvf/^ fi5-/(). Hence mvdv=f(s)ds. 76 MECHANICS Integrate thisequation between the limits aand6,denoting thecorresponding values ofvbyvland v2: 6 /mvdv=If(s) ds. Theleft-hand sidehasthevalue : mv2Pa ~2~n Theright-hand sideis,bydefinition, theworkWdone onthe particle bytheforce F.Hence (1) i andweinfer theresult : THEOREM. Thechange inthekinetic energy ofaparticleis equaltotheworkdoneonitbytheforce which actsonit. Thistheorem expresses,inthisthesimplest case imaginable, thePrinciple ofWork andEnergyinMechanics. Bymeans of itafirst integral oftheequation arising from Newton's Second Law canbefound inthecase ofaparticle, when theforce is known asafunction oftheposition, andthestudent willdowell togoback over theforegoing problems and exercises, and ex- amine their solution from thisnewpoint ofview;e.g.Equation (4)in5,Equation (3)in7,andEquation (2)in8are,save forthefactor m/2, theEquation ofEnergy, as(1)isoften called. EXERCISES Work theExercises of5,7,8,sofaraspossible, bythe Method ofWork andEnergy. 11.Change ofUnits inPhysics.* Tomeasure aquantity istodetermine howmany times acertain amount ofthat sub- stance, chosen arbitrarily asthe unityiscontained inagiven *Theintroduction ofthisparagraph andthenext atthisstage seems torequire justification. Ifthesetwopurely physical subjects aresufficiently important tobe taken uphere, thenwhy not,atthebeginning ofthischapter, the firsttimethey areneeded ?But ifthey aremerely forreference, whybreak theunity, coherence, ofthepresentation byplacing them here rather than attheendofthechapter? TheAuthor feelsthat this isabout thetimewhen thebeginner inMechanics should turn hisattention systematically tothese subjects, foruntil hehassomeknowledge oftheproblems studied inthischapter, hocanhardly beexpected torecognize the importance ofChange ofUnits and oftheCheck ofDimensions. MOTION OFAPARTICLE 77 amount ofthesubstance.* Thus tomeasure thelength ofright lines istofindhowmany times aright linechosen arbitrarily as theunit oflength afoot oracentimetre oracubit iscon- tained inagiven right-line segment. Thenumber, s,thus result- ingiscalled thelength ofthe line. Itdepends ontwothings theparticularlineandtheunit chosen. Ifadifferent unit of length bechosen, thesame line willhave adifferent number, s',assigned toit,and itslength thenbecomess'.fNow, forall lines,s'willbeproportionaltos: (1)s'ocs or s'=cs, where cisaconstant depending ontheunits. Itisdetermined inanygiven casebysubstituting particular values forsands', known tocorrespond. Thus ifwewish totransform from feet toyards, consider inparticular alinewhich isayard long. Here, s'willequal1and swillequal 3,so 1=3c, c=i, and,t (2)'=$8. Example.Ifayardistheunit oflength, aminute theunit oftime, atontheunit ofmass, andakilogramme theunit of force, findXinNewton's Second Law. Wewillstart withNewton's EquationintheEnglish units : ^2<? (3)*<y=32/, *Theword substance heremaybetoonarrow initsconnotations, forwewant aword that willinclude every measurable quantity, from thelength ofalight- wave tothewheat crop oftheworld. Such awoid obviously does not exist, andsoweagree tousesubstance inthissense asaterminus tcchnicus. tItwould seem paradoxical tosaythat thesame linehasalength of6when thefoot istheunit,andalength of2when theyard istheunit. But itmust bo remembered that thelength isafunction oftwovariables, theunitbeing oneof them. Theattempt issometimes made tomeet theapparent difficulty bysaying "3 ft.=1yd." But thismakes confusion worse confounded; for3=1isnot true, while ontheother hand totrytointroduce "concrete numbers," like3ft., 10Ibs.,5sees., intomathematics, isnotfeasible. Totrytochange units inthis way leads toblunders andwrong numerical results. There isonlyonekind of number inelementary mathematics. Toattempt toqualify itasabstract, isto qualify thatwhich isunique. Thedenominate attribute (3ft.,10Ibs., etc.) ispart ofthephysical thing conceived;itdoesnotpertain tothemathematical counterpart, which ispurely arithmetical. tCompare thisequation with theattempted form ofstatement mentioned in the lastfootnote :"1yd.=3ft." Itwould seem tofollow from that state- ment that'yds.=3sft.But'=.What acheerful prospect forgetting therightanswer bythatmethod 1 78 MECHANICS andwrite thetransformed equation intheform : Then theproblemistodetermine X'.Here, from (2): Next,mf=km, i_j.v9000 lc m'm I-ftXZOOO,lc- 20Q(), m- 2QO() Similarly, i'_ ?=$ 60'J2.20' Thus ,d2s'_602d2sm"^- 2000*X3m^ XT=__L_\'/*j220A;- The left-hand sides ofthese equations areequal by (4).On equating theright-hand sidesanddividing by(3)wefind : 602X' 2000X32.20X32'X'=422.4.* Ondropping theprimes, Newton's Second Law, written inthe new units, appears intheform : EXERCISES 1.Iftheunits oflength, time,andmass arerespectively amile, aday,andaton,compute theabsolute unit offorce inpounds. 2.Iftheacceleration ofgravityis981 inthe c.g.s. system, compute gintheEnglish system. 3.Iftheacceleration ofgravityis32.2 intheEnglish system, compute ginthec.g.s. system. *More precisely, theresult should botabulated as : X'=4.2X102 , since thedata, namely, X-32,arecorrect only totwosignificant figures. MOTION OFAPARTICLE 79 4.Iftheunit offorce beapound, theunit oftimeasecond, andtheunit oflength afoot, explain what ismeant bytheabsolute unitofmass,andshow that itisequal (nearly) to32Ibs. 6.Formulate and solve thesame probleminthedecimal system. 6.Iftheunitofmass isapound, theunit oflength, afoot,and theunit offorce, apound, findtheabsolute unit oftime. Arts. .176 sees. 12.TheCheck ofDimensions. The physical quantities that enter inMechanics canbeexpressedinterms oftheunits of Mass[Af],Length [L],andTime[7"].Thus velocityisofthe dimension length/time, orL/T=LT~l .Acceleration hasthe dimension LT~2 ,andforce, thedimension ML/T~*. When, anequationiswritten inliteral form, as eachtermmust have thesame dimension. For,suchanequation remains true,nomatter what theunits ofmass, length, andtime may be;and iftwoterms had different dimensions inanyone ofthefundamental quantities (mass, length, time), achange ofunitswould lead toanewequation notingeneral equivalent totheoldone. This principle affords auseful check oncomputation. Thus, ifanellipseisgivenbytheequation: allthequantities x,y,a,bareofdimension oneinlength, orL. Thedimension ofitsareamust beL2 ;and itis,forA=irab. Thevolume oftheellipsoidofrevolution corresponding torota- tionabout theaxisofxshould beofdimension L3 ,and itis : V=7ra62 . This principle affords auseful check onputtinginorleaving outg,when problems areformulated literally. Thus inthe Example of3,ifwehadforgotten ourginwriting down the right-hand side, thecheck ofdimensions would immediately haveshown uptheoversight. For, theleft-hand member isof dimension ML/T~~*\ hence every termontherightmust have 80 MECHANICS thissame dimension. Itdoes, inthecorrect equation ofthetext. Itis,ofcourse, onlywhen allthequantities which enter arein literal form, that thecheck canbeused. Ifsome arereplaced bynumbers, thecheck doesnotapply. Observe that incomputing thedimension ofaderivative,like d2s/dt2 ,wemay think ofthelatter asaquotient, thenumerator being adifference, andhence ofthedimension ofthedependent variable, while thedenominator isthought ofasapower. EXERCISES Determine thedimension ofeach ofthefollowing quantities: 1.Kinetic energy. Ans.ML2T~2 . 2.Work. Ans.ML2T~2 . 3.Moment ofinertia. Ans.ML2 . 4.Momentum. Ans.MLT~l . 6.Couples. Ans.ML2T'2 . 6.Volume density. Ans.ML~Z . 7.Surface density.Ans.ML*2 . 8.Line density. Ans.ML"1 . 9.Theacceleration ofgravity. Ans.LT~2 . 10.Thewind resistance canoften beassumed proportional tothesquare ofthevelocity.Ifitiswritten ascv2 ,what is thedimension ofc? Ans.ML~l . 11.InQuestion 10,what istheanswer when thewind re- sistance istaken persquare foot ofsurface exposed? 12.Check thedimensions ineachequation occurringin 3. 13.In4,Equation (3),thecheck fails. Explain why. 14.What arethedimensions ofYoung's Modulus? 15.IntheExample treated in8,wewished tofindthevelocity ofthestone atthecentre oftheearth inmiles persecond. But ifwesubstituted forR,intheformula (dr/df)2=gR,thevalue ofRinmiles(i.e.4000), weobtained awrong answer, eventhough thedimensions ofboth sides ofthisequation arethesame, namely, L2/T~2 .Explain why, andshowhowFormula (2),which is onehundred percentliteral, canbeused toyield acorrect result, whenR=4000. 16.Examine each equation in8astowhether theCheck ofDimensions isapplicable. MOTION OFAPARTICLE 81 13.Motion inaResisting Medium. When abody moves through the airorthrough thewater, these media opposere- sistance, themagnitude ofwhich depends onthevelocity, but doesnotfollow anysimple mathematical law. Forlowvelocities upto5or10miles perhour, theresistance Rcanbeexpressed approximately bytheformula : (1) R=av, where aisaconstant depending both onthemedium andonthe sizeandshape ofthebody, butnotonitsmass. Forhigher velocities uptothevelocityofsound (1082ft.asec.) theformula (2) R=cv* gives asufficient approximationformanyofthecases that arise inpractice. Weshallspeakofother formulas inthenext para- graph. Problem 1.Aman isrowinginstillwater attherate of3miles anhour, when heships hisoars. Determine thesubsequent motion oftheboat. HereNewton's Second Law gives us : /o\dv (3)m^=-av. TT j*mdvHence at= ,av /A\ AmiVQ (4)t=- log-^ where VQistheinitial velocity, nearj^^Bft. asec. Tosolve (4)forv,observe that a* i *>o=log--, orm v v Hence _at (5)9 v=VQe . Itmight appear from(5)that theboatwould never come to rest,butwould move moreandmore slowly, since _at lime~=0. =00 Wewarn thestudent, however, against such aconclusion. For theapproximation weareusing,R=av,holds only foralimited 82 MECHANICS time, andeven forthat time isatbestanapproximation. It willprobably notbemany minutes before theboat isdrifting sidewise, andthevalue ofaforthisaspectoftheboatwould bequite different,ifindeed theapproximation K=avcould beused atall. Todetermine thedistance travelled, wehavefrom(3): dv mv~r=av,ds andconsequently: (6)=o-^. Hence, even iftheabove lawofresistance helduptothelimit, theboatwould nottravel aninfinite distance, butwould ap- proach apoint distant feetfrom thestarting point, thedistance traversed thus being proportional totheinitial momentum. Finally, togetarelation between sandt,integrate (5): ds -? />7\ffiVn/1 ~m\ (7)8=-(l-em ). From this result isalsoevident that theboat willnever cover adistance ofSft.while theabove approximationlasts. EXERCISE Ifthemanandtheboat together weigh 300 Ibs.and ifasteady force of3Ibs. isjust sufficient tomaintain aspeedof3miles anhour instillwater, show thatwhen theboat hasgone 20ft., thespeed hasfallen offbyalittle lessthanamileanhour. Problem 2.Adrop ofrain fallsfrom acloud withaninitial velocity ofvft.asec. Determine themotion. Weassume that thedropisalready ofitsfinal size, not gathering further moisture asitproceeds, and take asthe lawofresistance : R=ct;2 . MOTION OFAPARTICLE 83 Theforces which actarei)theforce ofgravity, mg,downward, andif)theresistance oftheair,cv2 ,upward. Asthecoordinate oftheparticle wewilltake thedistance AP,Figure 60,which ithasfallen. Then, Newton's Second Lawbecomes : dv, A m-jj=mg cv2 . TT dvmg cv2 Hence v~r= ,as m cv*mvdv mgcv*' s=-log(mg-cv2 )+<7, andthus finally FIG.60 /ON Wim<7 00 (8)s=5-log- -t' v72cm#cv2 Solving forvwehave mgcv2 (9) ^gg-^-^gvy When sincreases indefinitely, the lastterm approaches as itslimit, arid_hencethevelocityvcannever exceed (orquite equal) vVmg/cft.asec. This isknown asthelimiting velocity. Itisindependent oftheheight andalsooftheinitial velocity, and ispractically attained bytherain asitfalls, foraraindropis notmoving sensibly faster when itreaches theground than itwas atthetopofahigh building. EXERCISES 1.Work Problem2,taking asthecoordinate oftheraindrop itsheight above theground. 2.Find thetime interms ofthevelocity andthevelocity in terms ofthetime inProblem 2. 3.Show that,ifacharge ofshotbefired vertically upward, itwillreturn with avelocity about 3times that ofraindrops 84 MECHANICS ofthesame size;andthat ifitbefired directly downward from aballoon twomiles high, thevelocitywillnotbeappreciably greater. 4.Determine theheight towhich theshot will riseinQuestion 3,andshow thatthetime tothehighest pointis where vistheinitial velocity. 14.Graph oftheResistance. The resistance which theat- mosphere orwater opposes toabody ofagiven sizeandshape caninmany cases bedetermined experimentally with areason- abledegree ofprecision andthus thegraph oftheresistance : JLcanbeplotted. Themathematical problem ^'2then presentsitself ofrepresenting thecurve with sufficient accuracy bymeans ofasimple function ofv.Intheproblem ofvertical motion intheatmos- phere, Problem2, 13, dV ^f/\m-^=mg f(v), according asthebodyisgoing uporcoming down,sbeing meas- ured positively downward. Now ifweapproximate tof(v)by means ofaquadratic polynomialorafractional linear function, or wecanintegrate theresulting equation readily. And itisobvi- ousthatwecansoapproximate, atleast, forarestricted range ofvalues for v. Another case ofinterest isthat inwhich theresistance ofthe medium istheonly force that acts, asinProblem 1: dvff,m_-/(). Aconvenient approximationforthepurposes ofintegrationis /()=avb . MOTION OFAPARTICLE 85 Hereaand baremerely arbitrary constants, enabling ustoim- posetwo arbitrary conditions onthecurve, forexample, to make itgothrough twogiven points, andaretobedetermined soastoyield agood approximation tothephysical law. Some- times thesimple values 6=1,2,3canbeusedwith advantage. Butwemust notconfuse these approximate formulas with simi- larly appearing formulas that represent exact physical laws. Thus, ingeometry, theareas ofsimilar surfaces andthevolumes ofsimilar solids areproportional tothesquares orcubes ofcor- respondinglinear dimensions. This lawexpresses afact that holds tothefinest degree ofaccuracyofwhich physical measure- ments haveshown themselves tobecapable andwithnorestric- tionwhatever onthe size ofthebodies. But thelawR=av2 orR=cv*ceases tohold,i.e.tointerpret nature within thelimits ofprecision ofphysical measurements, when vtranscends certain restrictedlimits, andthestudent must becareful tobear this factinmind. EXERCISES Work outtherelations between vands,andthose between vandI,iftheonly force actingistheresistance ofthemedium, which isrepresented bytheformula : 1.R=a+bv+cv\ 2.R=~-~V --3.R=av*. 7+dv 4.Show that itwould befeasible mathematically tousethe formulas ofQuestions1and2inthecase ofthefalling raindrop. 5.Atrain weighing 300tons, inclusive ofthelocomotive, can justbekeptinmotion onalevel track byaforce of3pounds totheton.Thelocomotive isable tomaintain aspeedof60 miles anhour, thehorse power developed being reckoned as1300. Assuming that the frictional resistances arcthesame athigh speeds asatlowonesandthat theresistance oftheair ispro- portional tothesquare ofthevelocity, findbyhowmuch the speed ofthetrain willhave droppedoffinrunning halfamile ifthesteam iscutoffwith thetrain atfullspeed. 6.Amanandaparachute weigh 150pounds. How large must theparachute bethat themanmay trust himself toitat anyheight,if25ft.asec. isasafevelocity withwhich toreach theground? Given thattheresistance oftheairisasthesquare 86 MECHANICS ofthevelocity and isequal to2pounds persquare foot ofoppos- ingsurface foravelocityof30ft.asec. Ans. About 12ft.indiameter. 7.Atobogganslide ofconstant slopeisaquarter ofamile longandhasafallof200 ft.Assuming that the coefficient offriction isTQ,that theresistance oftheair isproportional tothesquare ofthevelocity and isequal to2pounds persquare foot ofopposing surface foravelocity of30ft.asec.,andthat aloaded toboggan weighs 300pounds andpresents asurface of3sq.ft.totheresistance oftheair;findthevelocity acquired during thedescent andthetime required toreach thebottom. Find thelimit ofvelocity that could beacquired byatobog- ganunder thegiven conditions ifthe hillwere ofinfinite length. Ans. (a)68ft.asec.;(b)30sees.;(c)74ft.asec. 8.Theropes ofanelevator break andtheelevator fallswith- outobstruction till itenters anairchamber atthebottom of theshaft. Theelevator weighs 2tonsand itfallsfrom aheight of50ft.The cross-section ofthewell is6X6ft.and itsdepth is12ft. Ifnoairescaped from thewell,how farwould the elevator sink in?What would bethemaximum weight ofa man of170pounds? Given that thepressure andthevolume ofairwhen compressed without gain orlossofheat follow the law : pvl-4l=const., andthat theatmospheric pressureis14pounds tothosquare inch. 9.Intheearly days ofmodern ballistics theresistance of theatmosphere toacommon ballwasdetermined asfollows. Anumber ofparallel vertical screens were setupatequal dis- tances, theballwasshotthrough them (with apracticallyhori- zontal trajectory), andthetime recorded (through thebreaking ofanelectriccircuit) atwhich itcuteach screen. Explain the theory oftheexperiment, andshowhowpoints onthograph of theresistance asafunction ofthevelocity could beobtained. 16.Motion inaPlane andinSpace. Vector Velocity. When apointPmoves inaplane orinspace, itsposition atanyinstant canberepresented byitsCartesian coordinates : (1) *-/(*), 2/ MOTION OFAPARTICLE 87 where thefunctions arecontinuous, together withany deriva- tivesweshallhave occasion touse. The velocity ofPhasbeen defined asds/dt. For, hitherto, wehave regarded thepath asgiven, and itwasaquestion merely ofthespeed andsense ofdescription ofthepath. Butnowwe need more.Weneed toputinto evidence thedirection and sense ofthemotion, andsoweextend theidea, defining velocity more broadly asavector. Layoffonthetangent tothepath, inthesense ofthemotion, adirected linesegment whose length isthespeed ofthepoint, and letthevector thusdetermined be defined asthevector velocity ofthepoint P. Composition andResolution ofVelocities. Amouse runs across thefloor ofafreight car.Todetermine thevelocity ofthemouse inspace,ifthevelocity ofthecar isu, andthevelocity ofthemouse relative tothecar isv. Letthemouse startfrom apointP ononeside ofthecarandrunacross thefloor inastraight linewithconstant velocity, v,relative tothecar. LetQ bethopoint shehasreached attheend oftseconds. Thenp~_ Letthevelocity, u,ofthetrain beconstant, and let bethe initial positionofP.Then OP=ut. InFigure 62,thelineOArepresents thevector velocity uof thetrain, andABrepresents thevector velocity vofthemouse relative tothefreight car. Their geometric, orvector, sum is represented by07?.From similar triangles itappears that thepath ofthemouse in spaceistheright linethrough and5, andthathervelocityinspaceisthevector OB,oru+v. Thus hervelocityinspacemay bede- scribed, from analogy with theparallelo- gram offorces, astheresultant ofthetwocomponent velocities, ualong thedirection ofOAandvalong thedirection OCthrough Oparallel toA.ut FIG.62 FIG.63 88 MECHANICS Similarly, anyvector velocity mayberesolved intotwocom- ponent velocities along anytwo directions complanar with the given velocity ;Fig. 63. Theextension tospaceisobvious. Any three non-complanar vector velocities canbecomposed intoasingle velocity bythe parallelepiped law.And conversely anygiven vector velocity canbedecomposed into three component vector velocities along anythreenon-complanar directions. TheGeneral Case. Returning now tothegeneral case ofmotion inaplaneorinspace, wemaydefine theaveragevector velocity for theAtseconds succeeding agiven instant asthevector*(PP') divided byA,orthevector (PQ): When Atapproaches asitslimit, thelength ofthis vector, namely, thechordPP,divided byAt,approaches thespeed of thepoint atP;or, r~P**'rAsn n hm-=hm=Dts=v,numerically.A=OAt AJ-=OAt Moreover thedirection ofthevariable vector (PQ) approaches afixed direction asitslimit. And sothevariable vector (PQ) approaches afixed vector, v,asitslimit, or lim--V-=lim(PQ)=v. A*=At AZ= This vector, v,isdefined asthevectorvelocity ofthepoint P. Cartesian Coordinates. Toprove that theabove limit actually exists, consider thecomponents of(PP') and(PQ) along the. axes. These arc : iA#AyAz Ax, Ay,A* and -,-,- The lastthree variables approachlimits : lim-7=Dix, lim--7=D ty, lim~=Dtz.A-0At Af-0At A/=*0At Hence (PQ) approaches alimit, v,andthecomponentsofvalong theaxesare : *When itisnot feasible torepresent vectors bybold face type, the ( ) notation maybeused, as:(PP') or,later, (a). Thestudent should draw the figure which represents thevectors (PP') and(PQ)> MOTION OFAPARTICLE 89 dx__dy __dz Vx~Tt'Vv~ ~dt'Vz" ' These equations admit thefollowing physical interpretation. Consider theprojections, L,Af,TV,ofthepointPontheaxes of coordinates. The velocities withwhich these points aremoving along theaxes areprecisely dx/dt, dy/dt, and dz/dt. And sowe cansay:Theprojections ofthevectorvelocity valongtheaxes are equal respectivelytothevelocities oftheprojections. Finally, observe that, just astheaverage vector velocity ap- proaches theactual vector velocity asitslimit, sotheprojections oftheaverage vector velocity approach theprojections ofthe actual vector velocity astheir limits. Remark. Thestudent may raise thequestion:Ifvx,vy,and vzarethecomponents ofthevector velocity, v,arethey not, therefore, themselves vectors, andshould they notbewritten assuch, vz,vy,vz?Yes, this iscorrect. But itdoes notcon- flictwith theother view ofvx,vy,and vzasdirected linesegments ontheaxes ofx,y,and z.For,asystem ofvectors whose direc- tion (butnotsense)isfixed, constitute asystem ofone-dimen- sional vectors, andthese areequivalent todirected linesegments, since thetwosystems stand inaone-to-one relation toeach other. One-dimensional vectors canberepresented arithmeti- callybytheordinary realnumbers, positive, negative, and zero. EXERCISES 1.Show that,ifpolar coordinates intheplane areused, the component velocities along andorthogonal totheradius vector arerespectively: dr dB 2.Apointmoves onthesurface ofasphere. Show that dB .Ad<p where 6and<pdenote respectively theco-latitude andthelongi- tude. 3.Apointmoves inspace. Show that dr dB wherer,6,<parethespherical coordinates ofthepoint. 90 MECHANICS 16.Vector Acceleration. Letapoint describe apath, as in 15.Bythevector changeinitsvelocityismeant thevector (1) Av=v'-v, cf.Fig. 65,p.96.Theaveragevector acceleration isdefined asthe vectorAv AT WhenAapproaches 0,theaverage vector acceleration approaches alimiting value, and thislimiting vector isdefined asthevector acceleration ofthepoint: /\rAv (a)=hm Ar=oAt Cartesian Coordinates. Thecomponents ofthevector acceler- ation along theCartesian axes,atx,a^,andaz,arcreadily com- puted. For, thecomponents ofthevector (1)along theaxes arerespectively: v'xvx v'yvy v'zvz AJ'AJ'AZ Asinthecase ofvelocities, thecomponents ofthelimiting vector andthelimits approached bythecomponentsofthevariable vector arerespectively equal.* Hence ,.At>z~ ,.Avy~ ..Avz~ax=hm =Dtv x,av=hm =Dtv v,az=hm-=Dtv t, A/=*** A/=&t A/=At or: d'2z Osculating Plane andPrincipal Normal.] Letavector rbe drawn from anarbitrary fixed point ofspace tothevariable pointPthat istracing outthecurve(1),15.Then dr.v--a=*- Letsbethearc,measured inthesense ofthemotion;and let .ds *Thistheorem istrue ofanyvector which approaches alimit, asthestudent canreadily verify. fCf.theAuthor's Advanced Calculus, p.304, 8. MOTION OFAPARTICLE 91 Then r'isaunit vector lying along thetangent and directed inthesense ofthemotion. Furthermore, =r ds isavector drawn along theprincipal normal, toward thecon- cave side oftheprojection ofthecurve ontheosculating plane, and itslengthisthecurvature, K,atP. Ontheother hand, theacceleration dv, dsdr= and*-__ Hence EXERCISES 1.Apoint describes acircle with constant velocity. Show that thevector acceleration isnormal tothepathanddirected toward thecentre ofthecircle, andthat itsmagnitudeis "2 2 , or orr. 2.Show that,when apointisdescribing anarbitrary plane path, thecomponents ofthevector acceleration along thetangent andnormal are :<j2s v* at=w'an=? whoro pdenotes theradius ofcurvature, andthecomponent<xn isdirected toward theconcave side ofthecurve. 3.Apoint describes acycloid, therolling circlemoving forward with constant velocity. Show that theacceleration isconstant inmagnitude andalways directed toward thecentre ofthe circle. 4.Prove byvector methods that, inthecase ofmotion in where ar,adenote thecomponents oftheacceleration along andperpendicular totheradius vector. Use thesystem of-ordinary complex numbers, a+bi,where i=V1,andset .'r=re6\ 5.Obtain thesame results bygeometric methods. 92 MECHANICS 17.Newton's Second Law. Letaparticle move under the action ofany forces, and letFbetheir resultant. Let (a)be itsvector acceleration. Then Newton's Second Law ofMotion asserts that themass times thevector acceleration isproportionalto thevector force ,or,iftheabsolute unit offorce isadopted, (1) m(a)=F. InCartesian coordinates thelawbecomes : W>-J7Z=X, (2) td?z If,inparticular, X,Y,Zarecontinuous functions ofx,y,z, dx/dt, dy/dt, dz/dt, and tyitthen follows from thetheory ofdifferential equations that thepathisuniquely determined bythe initial conditions;i.e. iftheparticleisprojected from a point (z ,2/o>ZG)withavelocity whose components along theaxes are(u^v,w),thepathiscompletely determined. Thisremark isstriking when oneconsiders that thecorresponding theorem isnottrue ifonedetermines themotion bymeans oftheprinciple ofWork andEnergy;cf.theAuthor's Advanced Calculus, p.351, Singular Solutions. The essential point here isthatEqua- tions (2)never admit asingular solution, whereas theequations ofWork andEnergy do. Inthemore general cases itisalsoseenthatthepathisuniquely determined bythe initial conditions. This statement iscon- firmed inthecase ofeach oftheexamples considered below. Forageneral treatment,cf.Appendix A. Osculating Plane. The force, F,alwaysliesintheosculating plane ofthepath. For,from 16,andEquation (1)above, Hence wecanresolve Fintoacomponent Talong thepathanda component Nalong theprincipal normal, andweshallthenhave : y"72o tYllfim=T =N where p=l//c. MOTION OFAPARTICLE 93 EXERCISE Show that(r'xr")F=0. Hence, inCartesian coordinates, -(z'x")Y+(x'y")Z=0, where andy'z"-z'y",etc. /dxfrff1 fJU ~;. JUdsetc. 18.Motion ofaProjectile. Problem. Tofindthepathofa projectile acted ononlybytheforce ofgravity. Thedegree ofaccuracy oftheapproximation tothetruemotion obtained inthefollowing solution depends ontheprojectile and onthevelocity withwhich itmoves. Foracannon ball itis crude, though suggestive, whereas forthe16Ib.shot, used in putting theshot,itisdecidedly good. Hitherto wehaveknown thepath ofthebody; herewedo not.Thepathwillobviously beaplane curve, andsoNewton's Second Law ofMotion becomes : m dt2 whereX,Yarethecomponents oftheresultant force along the axes,measured inabsolute units. Inthepresent caseX=0,Y=mg,andwehave (2) FIG.64 Ifwesuppose thebody projected from with velocity VQat anangle awith thehorizontal, theintegration ofthese equations ives:dx dt-jj=(J=VQcosa, x=v1cosa; ^=vQsina- 0tf, y=V/sina- 94 MECHANICS Eliminatingtweget: OX" (3) y=xtana--^- 5-2^cos2a Thecurve hasamaximum atthepointA :(z,,t/J, t$sinacosa#?,sin2a/M_ U/% U *> ff' *>-~2T~ Transforming toasetofparallel axesthrough A,wehave : x=x1+a?!, y=y'+ylt V' *22COS2n,&UQLUo Cc This curve isaparabola with itsvertex atA.The height ofitsdirectrix aboveAisv\cos2 a/2</, andhence theheight above ofthedirectrix oftheparabola represented by(3)is vlsin2a .v%cos2a_v% ~~2g~+2i~" 27' The result isindependent oftheangle ofelevation a,and so itappears that alltheparabolas traced outbyprojectiles leaving with thesame velocity have their directrices atthesamelevel, thedistance ofthis levelabove being theheight towhich the projectile would rise ifshotperpendicularly upward. EXERCISES 1.Show thattherange onthehorizontal is R= sin2a, j/ andthatthemaximum rangeRisattained whena=45 : ~ g' Theheight ofthedirectrix above ishalf thislatter range. 2.Aprojectileislaunched with avelocity ofVQft.asec.and istohitamark atthesame levelandwithin range. Show that there aretwopossible angles ofelevation andthatone isasmuch greater than45astheother isless. 3.Find therange onaplane inclined atanangle j3tothe horizon andshow thatthemaximum rangeis *~ ~g1+sin MOTION OFAPARTICLE 95 4.Asmall boycanthrow astone 100 ft.onthe level. He isontopofahouse 40ft.high. Show thathecanthrow the stone 134 ft.from thehouse. Neglect theheight ofhishand above thelevels inquestion. 6.The best collegiate record forputting theshot was, at onetime, 46ft.andtheamateur andworld's record was49ft. Gin. Ifaman puts theshot46ft.andtheshot leaves hishand at aheight of6ft.3in.above theground, findthevelocity with which helaunchesit,assuming that theangle ofelevation ais themost advantageous one. Am. v=35.87. 6.Howmuch better record cantheman ofthepreceding question make than ashorter man ofequal strength andskill, theshotleaving thelatter's hand ataheight of5ft.3in.? 7.Show that itispossible tohitamarkB :(x6, 2/&),provided 8.Arevolver cangiveabullet amuzzle velocity of200 ft. asec. Isitpossible tohitthevaneonachurchspire aquarter ofamileaway, theheight ofthespire being 100 ft.? 9.Ithasbeenassumed that thepath oftheprojectileisa piano curve. Prove thisassumption tobecorrect byusingall three Equations (2), 17. 19.Constrained Motion. Letaparticle beconstrained to move inagiven curve,likeasmooth bead that slides onawire. Consider first thecase ofaplane curve. Letthecomponent oftheresultant ofalltheforces along thetangent beTandalong thenormal beN.Then Newton's Second Law ofMotion, 17, gives thefollowing equations: (1)mv2=N. P Theproof given in17wasbased onvector analysis. Wewill giveone fortheplane casewithout theuseofvector methods. Geometric Proof. Compute thecomponents ofthevector acceleration along thetangent andalong thenormal. Let <p 96 MECHANICS betheangle which thetangent hasturned through inpassing fromPtoP' .Then thecomponent ofAvalong thetangent willbe vfcostp v= (t;+Av)cos<p v =Avcos<p v(1 cos<p). Bythedefinition ofcurvature, K=lim~r, p=lim PP'PP' Now, thecomponent oftheaverage ac- celeration along thetangentis vfcos& vAv 1cos (p TT=-r-COS<f> V - At At At LetAtapproach0.Then<papproaches 0,andthelimit ofthe firsttermontherightis /,.AZA/V \~ flimMlimcos^J=Dtv. Toevaluate thelimit ofthesecond term, write 1cosy?__1cos<pjp_As At<p AsAt The first factor approaches 0,andthesecond andthird factors remain finite, since each approaches alimit. Hence thelimit oftherighthand side is0. Wehave proved, then, that ,. vfcos<p v~ lim =Dtv, andthus the firstofEquations (1)isestablished. Toobtain thesecond ofEquations (1),consider thecomponent oftheaverage acceleration along thenormal, or Thiscanbewritten asvsm <p At ;sin<p(?_As v_ <pAsAt' MOTION OFAPARTICLE 97 where sisassumed toincrease with t.The limit ofthisproduct isseen tobe : i 2 vX1X-Xv=-,P P andthisproves thetheorem. Thecomponent Nmeasures thereaction ofthecurve. Itis thecentripetal forceduetothemotion. The foregoing analysis yields the first ofEquations (1)for twisted curves. EXERCISE Usethepresent geometric method toobtain theformulas : 1d where ar,otedenote respectively thecomponents ofthevector acceleration along andperpendicular totheradius vector. 20.Simple Pendulum Motion. Consider thesimple pendu- lum. Here ^sm-772=-mgsin0,at andsince s=10, This differential equationischaracteristic forSimple Pendulum Motion. Wecanobtain afirstintegral bythemethod of7: 2g.de ~~T: 77^ ~T~sinu-jr.dtdt2I at = cos+C, where aistheinitial angle ;hence (2)^=^(cos0-cos). Thevelocityinthepath atthelowest point is Itimes theangular velocity for=0,or V20Z (1 cosa),and isthesame thatwould havebeen acquired ifthebobhadfallen freely under theforce ofgravity through the MECHANICS same difference inlevel. Equation (2)isvirtually theIntegral ofEnergy. Ifweattempt toobtain thetimebyintegrating Equation (2), weareledtotheequation: de 20JVcos cosa This integral cannot beexpressedinterms ofthefunctions at present atourdisposal. ItisanElliptic Integral.* When0, however,issmall, sin6differs from 6byonlyasmall percentage ofeither quantity, andhence wemay expect toobtain agood approximation totheactual motion ifwereplace sin6in(1)by: (3)_g-~ This latter equationisofthetypeofthedifferential equation ofSimple Harmonic Motion, 7,A),n2-having here thevalue g/l. Hence, when asimple pendulum swings through asmall amplitude,itsmotion isapproximately harmonic and itsperiod isapproximately 9 The Tautochrone. Aquestion that interested themathema- ticians oftheeighteenth century was this :Inwhat curve should apendulum swing inorder thattheperiod ofoscillation may bo absolutely independent oftheamplitude? Itturns outthat thecycloid has thisproperty. For, thedifferential equationof motion is dzs FIG.67where $ismeasured from thelowest ' point, andsince s=4asinr, , d*s gwehave -j-z=-~s.at24a *Cf.theauthor's Advanced Calculus, Chapter IX,page 195,where thisintegral isreduced tothenormal form. MOTION OFAPARTICLE 99 This isthe differential equation ofSimple Harmonic Motion, 7,A),andhence theperiodoftheoscillation, 9 isindependent oftheamplitude. Acycloidal pendulum maybeconstructed bycausing thecord ofthependulum towindontheevolute ofthepath. The resist- ances duetothestiffness ofthecord asitwinds upandunwinds would thusbeslight ;butintimetheywould becomeappreciable. 21.Motion onaSmooth Curve. Letabead slide ona smooth wireunder theforce ofgravity. Consider firsttheplane case. Choosing theaxes asindicated, wehave : (1) Hencedt*dx ds _9 2~g~ds~dt Integrating thisequation with respect to/,wefind : Ifwesuppose thebead tostartfrom restatA,then =2gx Q+C,\A:(x Q,y) FIG.68 (2) Butthevelocity thatabody falling freely from restadistance of xxattains isexpressed byprecisely thesame formula. Inthemore general case that thebead passes thepointA withavelocityvwehave : (3)eg=2gx<>+C, t>2- eg=2g(x-z). Thus itisseenthatthevelocity atPisthesame thatthebead would have acquired atthesecond level ifithadbeen projected vertically from thefirstwith velocity. 100 MECHANICS Thetheorem also asserts that thechange inkinetic energyis equal totheworkdoneonthebead;cf. 10. Ifthebead starts from restatA,itwillcontinue toslide till itreaches theendofthewire orcomes toapoint A'atthesame v*>^Llevel asAInthelatter case itwill ingen- eraljust risetothepointA1andthen retrace itspathback toA.But ifthetangent tothe FIG.69curve atA'ishorizontal, thebeadmay approach A1asalimiting position without everreachingit. EXERCISES 1.Abead slides onasmooth vertical circle. Itisprojected from thelowest point withavelocity equal tothatwhich itwould acquire infalling from restfrom thehighest point. Show that itwillapproach thehighest point asalimit which itwillnever reach. 2.From thegeneral theorem (2)deduce the first integral (2)ofthedifferential equation (1), 20. Space Curves. Thesame treatment applies tospaceofthree dimensions. Itisinteresting, however, togiveasolution based onCartesian coordinates. Choose theaxisofxasbefore positive downward. Thenwehave : (4).D ~dT*="%+R*> d*y_D~*~"' whereRx,Rv,Rtarethecomponents ofthereaction Rofthewire along theaxes. SinceRisnormal tothecurve, wehave : < s-S+*-8+*!- Tointegrate Equations (4)multiply through respectively by dx/dty dy/dt, dz/dt and add.Wethus find, with the aid of(5): MOTION OFAPARTICLE 101 But 2/efo\2 .(dy\2/dz\*"2=(&)+U)+b) Hence (6)reduces to (7) ^md(v2 )=mgdx. Onintegrating thisequation, wefind : /o\ mvZ mva / \ (8)-7j--- 2^="V(*- *<>) This isprecisely theEquation ofEnergy.Itcould have been written down atthestartfrom thePrinciple ofWork andEnergy. Itisthegeneralization of(2)forspace curves. EXERCISE Abead slides onasmooth wire intheform ofahelix, axis vertical. Determine thereaction ofthewire inmagnitude and direction. 22.Centrifugal Force. When aparticle ofmassmdescribes acircle with constant velocity, theacceleration isdirected toward thecentre, and itsmagnitudeis The force which holds theparticleinitspath is, therefore, normal tothepathanddirected inward. Itsmagnitudeis ,rmv2 9FIG.70N=-=mco2r. r Why, then, theterm" centri/u^aif force" theforce that "flees thecentre"? Theexplanationisaconfusion ofideas. Ifthe mass isheld initspathbyastring fastened toapegatthecentre, 0,doesnotthestring tugat inthedirection OPaway from the centre and isnotthisforce exerted bytheparticle initsattempt, ortendency, toflyaway from thecentre? Theanswer tothe first question is,ofcourse, "Yes." Now oneofthestandard methods ofthesophistsistobegin with aquestion onanon- controversial point, conceded without opposition intheir favor, andthen toconfuse theissue intheir second question "and isnotthisforce exerted bytheparticle?" 102 MECHANICS Matter cannot exertforce, foraforceisapush orapull,and matter canneither push norpull ;itisinert. The particle does notpullonthestring, thestring pullsontheparticle. Buteven thisstatement willbeaccepted only half-heartedly,ifatall, bypeople whohave notyetgrasped thebasic idea ofthescience ofMechanics thestudy ofthemotion ofmatter under the action offorces. What comes first isamaterial system solid bodies, particles, laminae andmaterial surfaces, wires, anycombi- nation ofthese things, including even deformable media (hydro- dynamics, elasticity) andthen thissystemisacted onby forces. ISOLATE THESYSTEM Themanwho firstuttered these words deserves amonumentum aere. Inthepresent case there aretwosystems, each ofwhich canbeisolated :(1)theparticle ;(2)whatever thepegisattached to think ofasmooth table, theparticle going round andround inahorizontal circle andbeing held initspathbyastring whose other end isattached toapegatapoint ofthetable. Inthe case ofthe firstsystem, theforce that acts isthepullofthestring toward thecentre, and thisforce iswhat isnow-a-days described as"centripetal" force the force that"seeks the centre." The second system hasnothing todowith the particle. In particular, thissystem maybethetable. Inthat case, thefloor, aswellasgravity, exerts certain forces, andunder theaction of alltheforces, thetable stays atrest. The force ofthestring, varyingindirection, causes theforces ofthefloor tovary. Andnow, after allissaidanddone, comes therejoinder: "But theparticle didpullonthestring, forotherwise thestring would nothave pulled onthepeg." There isnoanswer tothese people. Some ofthem aregood citizens. They vote theticket ofthe party that isresponsible fortheprosperityofthecountry ;they belong totheonly truechurch;they subscribe totheRedCross drive buttheyhavenoplaceintheTemple ofScience;they profaneit. Example1.Abullet weighing1oz. isshot intoasling, con- sisting ofastring 5ft.longwithoneendfastened at0,theother endcarrying aleather cup.Ifthevelocity ofthebullet is600 ft. asec.,howstrong must thestring be,nottobreak? MOTION OFAPARTICLE 103 Thetension inthestring willbe o 5 ^ mv2 I FIG.71 wherem=^,v=600, r=5;or 6002 16X54500; 4500what? pounds? No, fortheforce ismeasured inabsolute units, orpoundals, andso,togettheanswer inpounds, wemust divide by32.The tension, then, that thestring must beable towithstand is141Ibs. Example2.Arailroad train rounds acurve of1000 ft.radius at30m.anhour.How high should theouter railberaised, fciftheflanges ofthewheels arenotto press against either track? Standard gauge, 4ft.8|in. Ifaplumb bob ishungupinacar, anddoesnotoscillate, then itshould be atright angles totheaxles ofthewheels. Itwilldescribe itscircular pathinspace under theaction oftwo forces, namely, gravity, mg,downward, andthetension, \ T,ofthe string. Letthestring make mg 7ananglea.with thevertical. Then the vertical component ofTjustbalances gravity, forthere isnovertical motion ofthebob. Hence Tcos a.~mg. Thehorizontal component ofTyields thenormal forceNwhich keeps thebobinitscircular path, or Hence v2442 Since thedistance between therails is4ft.8in.,itfollows that theouter railmust beraised 3.42 in. 104 MECHANICS EXERCISES 1.Aparticle weighing 4oz. isattached toastring which passes through asmall hole, 0,inasmooth table and carries aweightWatitsother end. Ifthe firstweightisprojected along thetablefrom apointPatadistance of2ft.from with avelocity of50ft.asecond inadirection atright angles toOP, thestring being tautandthepartbelow thetable vertical, how greatmustWbe,that the4oz.weight may describe acircular path? Ans. 9Ibs.12oz. 2.Aboyonabicycle rounds acorner onacurve of60ft. radius attherate of10m.anhour,andhisbicycle slipsoutfrom under him.What isthegreatest value p,could havehad? Ans. Notquite |. 3.Aconical pendulumislikeasimple pendulum, onlyitis projected sothat itmoves inahorizontal circle instead ofina vertical one.Show that Zo>2=gseca. 4.Iftheearth were gradually tostop rotating, howmuch would Bunker HillMonument beoutofplumb? Given, that theheight ofthemonument is225 ft.andthelatitude ofCharles- town is4222'. Ans. About 4in. 5.Anocean liner of80,000 tons issteaming eastontheequator attherate of30knots anhour. Ifsheputsabout andsteams west atthesame rate,what istheincrease inherapparent weight? 6.Iftheearth were held inhercourse bysteel wires attached tothesurface onthesidetoward thesunandevenly distributed asregards across-section byaplane atright angles tothem, show thattheywould have tobeasclose together asblades of grass. Itisassumed that their other ends areguided near the earth's surface. 7.Show that asteel wireoneend ofwhich ismade fast to thesunandwhich rotates inaplane with constant velocity, making onerotation inayear, could justabout reach tothe earth without breaking. Neglect theheat ofthesunand all forces ofgravitation. 8.Asteel wire 1sq.mm. incross-section, breaking strength 70kgs.,isstrung round theearth along theequator. Show that, iftheearth gradually stopped rotating, thewirewould snap. MOTION OFAPARTICLE 105 9.What isthesmallest latitude such thatthewire described inthepreceding question,ifstrung round theearth onthat par- allel,would notbreak ? 10.Iftheearth hadasatellite close by,how often would thelatter riseandsetinaday? Ans. About 18times. 11.Aboyswings abucket ofwater around inavertical circle without spilling any. Does notthebucket exert apullonthe boy'shand ? Explain thesituation byisolating asuitable system, namely: i)thebucket ofwater;ii)theboy. 23.TheCentrifugal OilCup.Adevice once used fordeter- mining thespeedofalocomotive consisted ofacylindrical cup containingoilandcaused torotate about itsaxis,which was vertical, withanangular velocity proportional tothespeed of thetrain. Letussechow itworked. Suppose the oiltoberotatinglikearigid body, withnocross currents orother internal disturbances. What willbetheform ofthefreesurface ?Imagine asmall par- ticle floating onthe oil. Itwillbeacted onbytheforce ofgravity, mg,downward andthebuoyancy, B,ofthe oilnormal to thesurface. The resultant ofthese two forces must just yield the centripetal forceNrequired tokeep theparticlein itspath.Now N=mco2x. FIG.73 Ontheother hand, theslope ofthecurve isdetermined bythe factthatthetangentisnormal toB.Thus BcosT=mg, Bsinr=N. Hence U'X or dx" gX' Itfollows, then, that 0)v-$*. Thus itappears thatthefreesurface isaparaboloid ofrevolution 106 MECHANICS ToGraduate theCup. Itiseasily shown that thevolume of asegment ofaparaboloid ofrevolution isalways halfthevolume ofthecircumscribing cylinder. If,then,wemark thelevel of the oilwhen itisatrest,theheight, h,towhich itrisesabove this levelwhen itisinmotion will justequal thedepth, h,ofthe lowest point ofthesurface below thispoint. From (1)itfollows, then, that ifadenotes theradius ofthecup, or EXERCISES 1.Atomato can4in.indiameter isfilled with water and sealed up. Itisplaced onarevolving tableandcaused torotate about itsaxis,which isvertical, attherate of30rotations asec. Find thepressure onthetopofthecan. Ans. Theweight ofacolumn ofwater 4ft.high (nearly) andstanding ontopofthecan. 2.How greatisthetendency ofthecantoripalong theseam? 24.TheCentrifugal Field ofForce. Itispossible toview the mechanical situation inthe oilcupfrom astatical standpoint. Imagine very tiny insects crawling slowly round onthosurface ofthe oil.Tothem the oiland allthey could secofthewalls andtopofthecupwould appear stationary, andtheywould refer theirmotion totherotating space asifitwere atrest. Wecanreproduce thesituation, sofarasstatical problems areconcerned, inaspace that isactually atrestbycreating a field offorce,inwhich theforcewhich actsonaparticleofmassm distant rfrom afixed vertical axis istheresultant oftheforce ofgravity, mg,vertical anddownward, andaforcemcoV directed awayfrom theaxis,where coisaconstant. Thus themagnitude of theforcewould be +(mco2 r)2= and itwould make anangle <pwith thedownward vertical, where wVtan <p= MOTION OFAPARTICLE 107 Tobring themechanical situation nearer toourhuman intui- tion,wemight think ofalarge round cup,500 ft.across atthe top,constructed with theflooringintheform oftheparaboloid inquestion and rotating with thesuitable angular velocity. There would beasmall opening atthevertex, through which observers could enter and leave. Theview ofallsurrounding objects would becutoff,andthemechanical construction would besonearly perfect that,whenwewere inside thecup,weshould notperceive itsmotion. Suppose, forexample, that theslope oftheflooralong therimwere45. Then, since uPx tanT= , g itfollows that ~ 32' co=T4T(nearly), or.36. Thetime, T,ofacomplete revolution isgivenbytheequation: 27T- jT, or T==17sees. 0> Thus thecupwould make nearly four revolutions aminute. Since co2 /gr=^^,theintensity ofthefieldwould be (0.004r)2 , andupon therimofthecup, thiswould amount tomgV2, or 41percent greater than gravity onthefixed surface oftheearth roughly, two-fifths more.Amovie actress whowasmain- taining herweight inHollywood, wouldtipthescales at, well,howmuch ? What wehave said applies, however, only tobodies that are atrestinthe field.When abody moves,stillother forces enter, andthese willbeconsidered inthechapter onRelative Motion. Nevertheless, wecandescribe themotion ofaprojectile directly, since itwould beaparabolainthefixed spacewestarted with. Imagine atennis court laidoutwith itscentre atthelowest point ofthebowl. Lobtheballfrom theback linetotheback line,andwatch theslice ! 108 MECHANICS Onemay reasonably inquire concerning theengineering prob- lems oftheconstruction. There willbeatendencyofthecup toburst toflyapart, duetothe"centrifugal force." Can it beheld together byreinforcing itwith steel bands round the outer rim, orwillthese have alltheycandotohold themselves together? Itturns outthat only one-seventieth ofthebreak- ingstrength willbeneeded tohold theband together, thus leav- ingsixty-nine seventieth^ forreinforcing. Butsince attherimthe"centrifugal force" isasgreat asthe force ofgravity, anyunbalanced load willcause thecuptotug onitsanchorage unmercifully. Ahundred menweigh approxi- mately 8tons,and iftheywerebunched atapoint oftherim, thereaction ontheanchorage would be8tons. Thestudent willfind itinteresting tocompute thereaction incasearacing carwere driven along therimat100miles anhour. 25.Central Force. Letaparticle beacted onbyaforce directed toward afixed point, O,anddepending onlyonthe distance from O,notonthedirection. Newton's Second Law ofMotion, 17,thenbecomes : (1> d.(r*\ = rdt\ dt)' whereRisacontinuous function ofr. Law ofAreas. Thesecond equation admits afirst integral: (2)r^==h ' This equation admits astriking interpretation. Consider the area, A,swept outbytheradius vector drawn from tothepar- ticle. Then A=/r*dd, FIG.74 dt~~ dt MOTION OFAPARTICLE 109 andhence (3) A=ft(-g, or,equal areas areswept outinequal times. Wehave tacitly assumed that h5*0.Ifh=0,then(2) reduces todd=0,andthepathisastraightlinethrough 0. Work andEnergy. Thekinetic energy oftheparticleis nw^_m(WWW2" 2W+r < Byvirtue of(2)thisbecomes : =i 2~ 22 2 Ontheother hand, thework,cf.Chap. VII, 3. r (6) TF=Cfidr. Hence I _~ J?1_2/ 7>~ mh*J This isadifferential equationofthe first order connecting rand6,and itsintegral gives theform ofthepath. TheLaw ofNature. Newton discovered theLaw ofUniversal Gravitation, which says thatanytwo particlesintheuniverse attract each other with aforce proportional totheir masses and inversely proportional tothesquare ofthedistance between them. Thislaw isoften referred toastheLaw ofNature. Inthepresent case, then, theparticleisattracted toward withaforce proportional to1/r2 ,andso (8) ll^,=- Thus (9) ^ TheLaw ofEnergy, asexpressedintheform ofEquation (7), herebecomes : 110 MECHANICS where X=m/x,andCisaconstant depending ontheinitial condi- tions. Theform ofthisequation suggests asimplification consisting insubstituting forritsreciprocal: (11) u= J- Thus (10)becomes : /io\ i 9 2/4.~ (12) ^5+= - S;+C'. Thisequation admits further reduction. Write : Since theleft-hand sidecannever benegative, theright-hand sidecanbewritten as52 ,andBitselfmaybechosen aseither oneofthesquare roots. Finally, set *-u~? Then (13)goesover into : dr2 (14)*+^-B". Thegeneral integral ofthis differential equation canbewritten intheform : (15) x=Bcos(0-7), where yistheconstant ofintegration. When5=0,thetruth ofthisstatement isobvious, forthen (14)reduces to +*- de*+x u' andtheonly solution ofthis differential equationis* x=0. IfS2*0,then (14)yields: d8= *Wehave hereanexample ofadifferential equation ofthe first order, handed tousbyphysics, whose general integral doesnotdepend onanarbitrary constant, butconsists ofaunique function ofalone. MOTION OFAPARTICLE 111 where, however, thetwo signs arenotnecessarily thesame. Butinallcases thislastequation leads to(15).* Wesetouttointegrate Equation (12),andwehave arrived attheresult : (16) u=~+Bcos(e-7). This equation canbethrown into familiar formbytakingB asthenegative radical andsetting where enow istheconstant ofintegration. Thus (16)yields: (\7\ r= - 1}M1-ecos(0-7) The Orbit. Thepath oftheparticleisgiven byEquation (17). This istheequation ofaconic referred toafocus aspoleand having theeccentricitye. TheCase e<1.Ife<1,theconic isanellipse, andthe length ofthetransverse axis is M(1-e*) Denoting thelength ofthesemi-axes byaand6,wehave : h* . A2 -Md-e2 )' MVf=T' Thedistance between thefoci is (19)c= Thearea oftheellipseis (20)TTOfc= Th tion: A=pr. Hence (21) T2=47r2-- *Itisworth thestudent's while tofollow through these multiple-valued func- tions, thathemay secure afirmer holdontheCalculus, eventhough the final result Equation (15) issimple.(1-e2 )1 Theperiodic timeTisconnected with theareaAbythe rela- 112 MECHANICS Determination oftheConstants ofIntegration. Letthebody be projected from thepoint (r,0)=(a,0)withaninitial velocity VQinadirection making anangle ftwith theprime direction =o.Todetermine theorbit. Wewillmention firstageneral formula. Let\l/betheangle from theradius vector produced tothetangent. Then since each siderepresents thecomponentveofthevector velocity, v,perpendicular totheradius vector. Byvirtue of(2)this becomes : (22) h=vrsin^, andthis istheformula wehadinmind. Todetermine theconstants in(17), then, write theequation intheform : /f>O\_r*/'-j //j \\ Hence (24)-=-^(1 ecos7), ecos7=1ah* an Furthermore, du fie.= -sin(Q-y).dO h'2 Since du__1dr__idr__vcos\f/ d6 r2dO hdt h' wehaveinitially: /rt/\ M^ ^nCOSp. Vi\ilCOSp (26) T^sin7=,-. esin7= hih M From(22), (27) h=VQasinj8, cos2 ft=1^- U i/ i -2y^ Squaring thesecond equation in(24)and (26),andadding, wefindbytheaidof(27): (28) e MOTION OFAPARTICLE 113 Theevaluation isnowcomplete. Bymeans of(27), hisdeter- mined;(28)then gives e,and(24)and(26)yield 7. From (28)weinfer that and thisequation contains theinteresting result that theorbit willbethefollowing conic : i) ellipse,ifv%<; ii) parabola,ifv\= ; Hi) hyperbola,ifv%>, irrespectiveofthedirection, 0,inwhich thebodyislaunched. Formotion inacircle, e=0.From (24) (30)-=4h2=MO. an2 Moreover, from (26)weseethat=7r/2,andsoweinferfrom (27)that A2=via*. Hence, bytheaidof(30), (31)v*= Jj. Conversely, conditions (30)and (31) are sufficient, that the path beacircle. Forfrom (27) follows that cos2 /3=0,and (29)gives 6=0.The result checks with thefactthatthenumer- icalvalue ofR,orw/*/a2 ,isequal tothecentripetal force, or EXERCISES 1.Show that if then2, , r2-rr=hand u=-,dt r' 114 MECHANICS 2.Ithasbeenassumed thattheorbit isaplane curve. Prove thistobethecasebymeans ofaconstraint, consisting ofasmooth plane through 0,thepoint ofprojection, andthetangent tothe path atthat point. UseNewton's Equations, 17, (2),and show thattheforce oftheconstraint is0. 26.TheTwoBody Problem. Iftwo particles ofmasses w, m'jattracting each other according tothelawofnature, and acted onbynoother forces, beprojected inanymanner, their centre ofgravity, G,willdescribe aright line, with constant velocity, orremain permanently atrest;cf.Chapter IV,1.Con- sider thelatter case. LetGbethefixed point 0,and letthe distances oftheparticles from ber,r'.Then theforce oftheir mutual attraction is /=K (r+r')2> whereKisthegravitational constant. Ontheother hand, mr=m'r'. Hence ,_m+m' r ~m'r' andso '\2fm \2 , = (- --,)m'.\m+ml Thus theparticlemisattracted toward with theforce that would beexerted byamassMfixed at0,andsotheorbit ofm isdetermined bythework of 25.Inparticular,ifmdescribes anellipse,mfwilldescribe asimilar ellipse with thesame focus, being turned through anangle of180. 27.The Inverse Problem toDetermine theForce. Let aparticle move inaplane according totheLaw ofAreas. Then r^=h T dtft' andthecomponent oftheforce perpendicular totheradius vector, 0,isnil.Hence theparticleisacted onbyacentral force, R, either attractive orrepulsive. From Exercise1,25,wehave : MOTION OFAPARTICLE 115 Example. Letthepathbeanellipse (or,more generally, any conic) with thecentre offorce atafocus. Then 1 ecos(07),u= , p=const.,P d*u . 1 W*+U= p> and mh*lK~ pr* The forceis,therefore, anattractive force, inversely proportional tothesquare ofthedistance from thecentre, when rliesbetween itsextreme values forthis ellipse. Butanarbitrary rangeof values, <a<7<0,canbeincluded insuchanellipse, andso theresult isgeneral. EXERCISE Show that ifthepathisanellipse with thecentre offorce at thecentre, theforce isproportional tothedistance from the centre. 28.Kepler's Laws. From observations made byTycho Brahe, Kepler deduced thelaws which govern themotion of theplanets. 1.The planets describe plane curves about thesunaccording tothelawofareas; 2.Thecurves areellipses with thesunatafocus; 3.Thesquares oftheperiodic times ofrevolution areproportional tothecubes ofthemajor axes oftheellipses. NewtonysInferences. From Kepler's lawsNewton drew the following inferences. Consider aparticular planet. From the firstlaw itfollows that theforce acting onitisacentral force, since thecomponent atright angles totheradius vector isnil. From thesecond law,combined with the first, itfollows from 27that theforce isinversely proportional tothesquare ofthe distance from thecentre, or 116 MECHANICS Ithasbeenshown in25,(21)that T2=47T2-,M whereTdenotes theperiodic time, andaisthesemi-axis major. Forasecond planet, Kepler's third lawgives, then, that p!=p,orthat/*isthesame foralltheplanets. Tosum up,then,Newton inferred that theplanets areat- tracted toward thesunwithaforce proportional totheir masses andinversely proportional tothesquare oftheir distances from thesun. From here itisbutastep totheLaw ofUniversal Gravita- tion. Ifthesunattracts theplanets, somust, bytheprincipal ofaction andreaction, theplanets attract thesun. LetMdenote themass ofthesun,thought ofasatrest.* Then Thus theLaw ofUniversal Gravitation isevolved :Anytwo bodies (particles)intheuniverse attract each other withaforce proportional totheir masses and inversely proportional tothe square ofthedistance between them, or win The factorKiscalled thegravitationalconstant. Itsvalue inc.g.s. units is#=6.5X 10-*; cf.Appell, l.c.,pp.390-405. EXERCISES 1.Show thatthe first oftheequations (1), 25 : dt* dt2 *Foramore detailed treatment cf.Appell, Mecanique rationnette, vol. 1,3ded., 1909, 229etseq. MOTION OFAPARTICLE 117 onmaking thetransformation (11): . r*r dtandemploying (2): goesover intotheequation: Hence obtain (16): u=~+Bcos(B 7). 2.Prove that h v -,P where pdenotes thedistance from tothetangent tothepath. 3.Show that the earth's orbit, assumed circular, would become parabolicifhalfthesun'smassweresuddenly annihilated, thesunbeing assumed tobeatrest. 4.Asmooth tube revolves around oneend inafixed plane with constant angular velocity. Aparticleisfree tomove in thetube. Determine themotion. 5.If,inthepreceding question, anelastic stringismade fasttotheparticle andattached totheendofthetube, deter- mine themotion. 6.Aparticleisattracted toward afixed centre with aforce proportional tothedistance. Show that thepathisaplane curve, andthat itcanberepresented bytheequations: x=Acos(nt+a), y=Bsin(nt+a). Isitanellipse? 7.Show that acomet describing aparabolic path cannot remain within theearth's orbit, assumed circular, formore than (2\-1-thpart ofayear, ornearly 76days. 8.Ashell isdescribing anelliptical orbit under acentral attractive force. Prove that,ifitexplodes,allthepieceswill meet again atthesamemoment;andthat after halftheinterval between theexplosion andthecollision, each piecewillbemoving 118 MECHANICS with thesame velocity asattheinstant ofexplosion, butinthe oppositedirection. 9.Show thataparticle, moving under theaction ofacentral force, cannot havemore thantwoapsidal distances;cf .Appendix B. 10.Find thelaw offorcewhen aparticle describes acircle, thecentre offorce being situated onthecircumference. Ans. Theinverse fifthpower. 11.Iftwospheres, eachonefoot indiameter and ofdensity equal tothemeandensityoftheearth(5.6)were released from rest ininterstellar space with their surfaces-^inches apart, how longwould ittakethem tocome together? How great would theerror beiftheirmutual attraction were taken asconstant? 12.Acannon ball isfired vertically upward from theEquator with amuzzle velocityof1500 ft.asec.How farwest ofthe cannon would itfall,iftheearth hadnoatmosphere? 13.Show thataparticle acted onbyacentral repulsive force varying according totheinverse square, will ingeneral describe abranch ofahyperbola with thecentre offorce atthat focus which liesontheconvex side ofthebranch. What istheexcep- tional case? 29.OntheNotion ofMass. Matter isinert. Itcannot exert aforce;itcannot push orpull.Ityields toforce, acquiring velocityinthedirection inwhich theforce acts wearethink- ingofaparticle. Byvirtue ofitsinertness itpossesses mass, whichmaybedescribed asthequantity ofmatter which abody contains. Mass ismeasured bytheeffect which force produces onthe motion ofabody.Weassume that forcemaybemeasured by aspring balance. Ifaforce, constant inmagnitude and direc- tion,beapplied toabody initially atrest, thebodywillacquire acertain velocityinagiven time. Ifthesame force beapplied toanother body, and ifthesecond body acquire thesame veloc- ityinthesame time, thetwobodies shall besaid tohave the same mass. Thus different substances canbecompared asto their masses andonadopting anarbitrary mass astheunit in thecase ofonesubstance, theunitcanbedetermined inthe case ofother substances. MOTION OFAPARTICLE 119 Itwasproved experimentally byNewton that theforces with which gravity attracts twomasses equal according totheabove definition, areequal. And soone isledtoinfer thephysical lawthat theweight ofabodyisproportional toitsmass. This law affords aconvenient means ofmeasuring masses, namely, byweighing. Inabstract dynamics, however (toquote from Maxwell), matter isconsidered under noother aspect than thatunder which itcanhave itsmotion changed bytheapplication offorce. Hence anytwobodies areofequal mass ifequal forces applied tothese bodies produce,inequal times, equal changes ofvelocity. This istheonly definition ofequal masses which canbeadmitted in dynamics, and itisapplicable toallmaterial bodies, whatever theymaybemade of.* InEngineeringithasbecome customary todefine masses as equal when their weights areequal. Wehave hereaquestion ofasense ofvalues, andMaxwell hasgoneonrecord asdeclaring unequivocally fortheinertia property. Touseweight todefine mass islikesaying thattwolengths areequalwhen therodsby which wemeasure them have thesame weight. Just asspace andtimestand above massand force, so,initselementary impor- tance, theinertia property towers above thelawofgravitation. *Maxwell, Matter andMotion, Art.XLVI. CHAPTER IV DYNAMICS OFARIGID BODY 1.Motion oftheCentre ofGravity. Letasystem ofparticles beacted onbyanyforces whatever. The lattermaybedivided intotwo classes :i)theinternal forces;ii)theexternal forces. By i)wemean that theparticlemzexerts onm1aforceF12 which may have anymagnitude andany direction whatever, orinparticular notbepresent atall,F12=0.Theparticlemlexerts aforce onw2,which isdenoted byF2l.Andnowwe assume thephysical lawthat action and reaction areequal and opposite;i.e.that thevector F21 >2/2)isequal andopposite tothevec- torF12,or Fi2+F21=0. FIG.75 Forconvenience wewillthink of theparticles andforces aslyingin aplane. Thetransition tospace ofthree dimensions isimmediate. Denote thecomponents ofavector forceFalong theaxes of coordinates byX,Y.Then *i2+X21=0, F12+721=0. Suppose there arethree particles. Then Newton's Second Law ofMotion gives forthefirst ofthem theequations: j.__=X1+Xn+X1 There areinallthree such pairs ofequations, those inxbeing the following: d*x d2x d2x ~dfi=-^3+ 120 DYNAMICS OFARIGID BODY 121 Onadding these three equations together, thecomponents Xuontheright, arising from theinternal forces, annul onean- other inpairs, andonly thesum oftheXiremains : aX\ .aXn ,CiX-i -*r. *rr.m*~dP+m*~W+m*^W=Xl+X>2+*3* Inasimilar manner weinfer,bywriting down thethree equa- tions inyandadding, that m* "eft2^"^"*2^ft2^"^"*3~eft^~~**^~*?~"~*3' Precisely thesame reasoning shows thatif,instead ofthree, wehaveanynumber, n,ofparticles, theinternal forces annul oneanother inpairs, andthusweobtain theresult : Coordinates oftheCentre ofMass. The left-hand sides of these equations admit asimple interpretationinterms ofthe motion ofthecentre ofmass ofthesystem. The coordinates, (x yy),ofthecentre ofmass aregiven bytheequations: (2)x= V=mnxn2 mn mnyn m1++wn Ifwedenote thetotalmassbyM,then !rnkxk= Hence wehave : d*xka'Xk_TUT- V~M~M ~jfLi 2* fcl*l/ H/w j^ andthusEquations (1)canbewritten intheform : (3) 122 MECHANICS These equations areprecisely Newton's Second Law ofMotion foraparticle ofmass Af,acted onbythegiven externalforces, each transferred tothe particle. Wecan state theresult as follows. THEOREM. Thecentre ofmass ofanysystem ofparticles moves asifallthemass were concentrated thereand alltheexternal forces acted there. Inthecase ofparticlesinspace, there isathird equation, (3)being superseded nowby (4)*=^,. Remark. There isonedetail inthestatement ofthetheorem that requires explicit consideration. Wehave written down tbedifferential equationsofthemotion, butwehave notinte- grated them. Ifwedonotstart theparticle ofmassMincoin- cidence with the initial position ofthecentre ofmass,itobvi- ously cannot describe thesame path. More thanthis,wemust giveitthesame initial velocity (i.e.vector velocity).Isthis enough toinsure itsalways remainingincoincidence with the centre ofmass? Theanswer tothisquestionisacategorical Yes;cf.Chapter III, 17andAppendix B. Generalized Theorem. Wehave proved thetheorem ofthe motion ofthecentre ofmass forasystemofparticles. Inthe case ofarigid body,wecanthink ofthebody asdivided upinto alargenumber ofcells, each ofsmallmaximum diameter; the mass ofeach cellasthen concentrated atoneofitspoints, and thenparticles thus resulting asconnected bymasslcss rods, after themanner ofatruss.* Tothisauxiliary system ofpar- ticles thetheorem asabove developed applies. Aridnow itis intuitionally evident, orplausible, that thesystem ofparticles willmove inamanner closely similar tothat oftherigid body, when the cells aretaken very small. One istempted tosay *Itisoften necessary touseatruss, atsome ofwhose vertices there areno masses. Wemay think ofminute masses attached atthese points andacted on bygravity orbynoexternal forces atall.The effect ofthese small masses isto modify slightly thevalue ofMinEquations (4).Andnow itfollows from the theory ofdifferential equations that theintegrals of(4)arethereby alsomodified only slightly. Hence thephysical assumption ismade, thatEquations (4)hold evenwhen there arenomasses atthevertices inquestion. DYNAMICS OFARIGID BODY 123 that themotion oftheactual bodyisthelimit approached by themotion ofthesystemofparticles asngrows largeandthe cells small. And thisis,infact, true. But this isnotamathe- matical inference farfrom it itisanew physical postulate. Wethusextend thetheorem andelevate ittoaPrinciple.* PRINCIPLE OFTHEMOTION OFTHECENTRE OFMASS. The centre ofmass ofanymaterial system whatsoever moves asifallthe mass were concentrated there,and alltheexternal forces acted there : A) 2.Applications. The Glass ofWater. Suppose aglass of water isthrown outofathird-story window. Asthewater falls, ittakes onmost irregular forms, breaking first into large pieces, and these into smaller ones. The forces that actaregravity andtheresistance oftheatmosphere, thelatter spread out all over thesurfaces ofthepieces. Andnow thePrincipleofthe lastparagraph tells usthat thecentre ofgravity moves asif allthemass were concentrated there and allthese forces trans- ferred bodily (i.e.asvectors) tothat point. TheFalling Chain. Letachain hang atrest, thelower end justtouching atablo, and letitbereleased. Todetermine the pressure, F,onthetable. Weidealize thechain asauniform flexible string, oflength/ anddensity p(hence ofmassM=pi),andthink ofitasim- pinging always atthesame fixed point, 0,ofthetable. Let s bethedistance thechain hasfallen and letxbetheheight of thecentre ofgravity above the table. Then thePrinciple of theMotion oftheCentre ofMass gives theequation: *A"Principle" inMechanics iswelldescribed inthewords ofProfessor Koop- nrian (of.theAuthor's Advanced Calculus, p.430): "According totheusage of thepresent daytheword principle inphysics haslost itsmetaphysical implication, andnowdenotes aphysical truth ofacertain importance and generality. Like allphysical truths, itrests ultimately onexperiment ;butwhether itistaken asa physical law, orappears asaconsequence ofphysical laws already laiddown, doesnotmatter." 124 MECHANICS (1) Now,~a' s-ld*s I G* 'OplIdt*' Moreover, from thelaws offreely falling bodies, Fia.76 Onsubstituting those values in(1),we obtain : (2) Hence (3)OP(s- 1)+ F- gpl. orF=gps+pv2 , F=3gps. Thismeans thatthepressure ofthechain onthetable isalways just three times theweight ofthat part ofthechain which has already come torestonthetable. Itappears, then, thatFismade upoftwo parts, i)thepres- sure gpsonthetable, ofthat part ofthechain already atrest; andii)apressure (4) P=pv*, duetotheimpact ofthechain against thetable. AStream ofWater, Impinging onaWall. Suppose ahose is turned onawall (oraconvict!).Todetermine thepressure. Weidealize themotion bythinking ofthestream ashitting thewall atright angles, thewater spattering inalldirections along thewallandthus giving upallitsvelocity intheline of motion ofthestream. Dynamically, this isprecisely thesame case asthat ofthechain falling onthetable, sofarastheimpactisconcerned, andhence thepressureisgivenby(4): FIG.77PA==ov * DYNAMICS OFARIGID BODY 125 Example. Afireengineisable tosend a2in.stream toa vertical height of200 ft.Find thepressureifthestream is played directly onadoor. Ans. 541 Ibs. TheCrew ontheRiver. Thecrew isoutforpractice. Ob- serve thecut-water oftheshellanddescribe how itmoves, and whyitmoves asitdoes. What system doyoudecide toisolate ? theshell? ortheshell, oars,andcrew? EXERCISES 1.Ifamanwere placed onaperfectly smooth table, how could hegetoff? 2.Ifashellwere firedfrom agunonthemoon andexploded initsflight, what could yo.usayabout themotion ofthepieces? 3.Agooseisnailed upinanairtightboxwhich restsonplat- form scales. Thegoose fliesup. Willthescales register more or lessorthesame? 4.Apailfilled with water isplaced onsome scales. Acork isheldsubmerged byastring tied tothebottom ofthepail. The string breaks. Dothescales register more orlessorthe same? 6.Aman, standing inthestern ofarowboat atrest,walks forward totheprow. What canyousayabout themotion of theboat? 6.When theman stops attheprow oftheboat, boatandman willbemoving forward withasmall velocity. Explain why. 7.Auniform flexible heavy stringislaidover asmooth cylinder, axis horizontal, andkept from slipping byholding oneend,A,fast, thepart ofthestring fromAuptothecylinder being vertical. Thepart ofthestring ontheother side ofthe cylinder is,ofcourse, alsovertical,itslower end, /?,being below thelevel ofA,andthewhole stringliesinavertical plane per- pendicular totheaxis ofthecylinder. The stringisreleased from rest. Determine themotion, there being asmooth guard which prevents thestring from leaving theupper side ofthe cylinder. 8.If,inQuestion 7,thedifference inlevel between Aand Ris2ft.,and ifthedistance fromAuptothecylinderis8ft., 126 MECHANICS compute thevelocity ofthestring when theupper endreaches thecylinder, correct tothreesignificant figures. 9.Findhowlongittakes theupper endofthestring toreach thecylinder. 10.The sporting editor ofaleading newspaper recentlyre- ported anew stroke which acertain coach haddeveloped, the advantage ofwhich wasthat itgaveanevenmotion totheshell andavoided thejerkiness 01theold-fashioned strokes. Examine thisnews item. 3.TheEquation ofMoments. Recall theformula forthe moment ofaforceFabout theorigin, namely, (1) xY-yX. Consider asystemofparticles acted onbyany external forces whatever, and interacting ononeanother by forces that areequal andopposite, but arenowassumed each time tolieinthelinojoining thetwopar- ticles inquestion. Moreover, theparticles shall lieinafixed plane. Begin with thecase of three particles, asin1,and writedown thesixequations that express Newton's Second Law ofMotion fortheseparticles.* Next, form theexpression:FIG.78 FIG.79 mlr andcomputeitsvalue from theequationsinquestion, namely, (x,Yl-VlX,)+(x,Y12-VlX12)'+(x,Yn-y,Xn). Theparentheses represent respectively themoments ofFDF12, F13about theorigin. Now, dothesame thingfortheparticlera2,and finally, forra3. Onadding these three equations together,itisseen that the *Itisimportant thatthestudent dothis,anddoitneatly, andnotmerely gaze atthethree equations printed in 1andtrytoimagine thethree notprinted. Heshould write outthe fullequation derived below from these, neatly onasingle line,andthen write theother twounder thisone. DYNAMICS OFARIGID BODY 127 moments oftheinternal forces about theorigin destroy one another, andthere remains ontheright-hand sideonly thesum ofthemoments oftheapplied forces. Ifthere aren>3particles, mltw2, ,mn,theprocedureis thesame, andwearethus ledtothe THEOREM OFMOMENTS: B) mkd*Xk -yA). Werefrain from writing down thecorresponding theorem in three dimensions, because weshall have noneed ofitforthe present. 4.Rotation about aFixed Axis under Gravity. Let the system ofparticles of3berigidly connected, and letonepoint, O,ofthetrass-work beatrest, sothat thesystem rotates about asapivot. Forexample, take thecase ofauniform rod,one endofwhich isheldfast,andwhich isreleased from restunder gravity. Divide therod intonequal parts, x andconcentrate themass ofeach part,fordefi- niteness, atitsmost remote point. Wethus have asystemofnparticles, andweconnect them rigidly byamassless truss-work asshown inthefigure.* Wearenowready tocompute each side of Equation B)fortheauxiliary system ofnpar- ticles. Let rbethedistance from toany point fixed intherod.Then (1) x=rcos0, y=rsin0, where varies with thetime, t,but risconstant with respect to t.HenceFia.80 (2)dx . _dO -rr= TSill-7T,dt dt$-- Weobserve next that, inallgenerality, bymere differentia- tion,i.e.purely mathematically, (3)<L(d-i 4*\-d*y_ di\x ~diydt)~xd? *Cf.thefootnote, I. 128 MECHANICS andweproceed tocompute theparenthesis bymeans ofEqua- tions (1)and(2).Wefind : dy dx ~dO Inthepresent casewehave : d(dyk dxk\_d*8 dt\Xk dtyk~dt)-~rt~dT forrkdoesnotchange with thetime,andsodrk/dt=0.Hence Thesumwhich here appearsisthemoment ofinertia* ofthe system about : Thus theleft-hand side oftheEquation ofMoments reduces to theexpression: Turning now totheright-hand side ofB)weseethat the kthparticle, m/t,yields amoment about equal tothequantitymkgrksin0,andsothesum inquestion becomes ]-mkgrksin0, or-(2)mkrk)gsin0. t t But mkrk=Mhy where histhedistance from tothecentre ofgravity, (?,ofthe system ofparticles. Hence, finally, (6) I^2=-MghsmO. This issubstantially theequation ofSimple Pendulum Motion, Chapter III, 20 : ^\ d26 g. . *Moments ofinertia forsuch bodies asinterest ushere aretreated inthe Author's Introduction totheCalculus, p.323. DYNAMICS OFARIGID BODY 129 Hence thesystem ofnparticles oscillates likeasimple pendulum oflength l=m or (82)I= j,where 7=MW, kdenoting theradius ofgyration. More precisely, whatwemean bythelaststatement isthis. Letasimple pendulum besupported at0,letitslength bek*/h, and letitbeplaced alongside therod,thebobbeing atapoint distant Zfrom O. Ifnowbothbereleased from restatthesame instant, theywill oscillate sidebyside,though nottouching each other. TheActual Rod. Asngrows larger and larger, themassless rodweighted with thenparticles comes nearer andnearer to theactual rod,dynamically. This isnotamathematical state- ment. Itexpresses ourfeeling from physics forthesituation ourintuition. And sowhonwesaythatthemotion oftheactual rod isthelimit approached bythemotion oftheauxiliary rod, wearestating anewphysical postulate. The resultis,that the actual rodoscillates likeasimple pendulum oflength fc_2-iL2-?/ h# 3 EXERCISES Apply themethod setforth inthetext, introducing each time anauxiliary setofparticles, andproceeding tothe limit. Do nottryshort cutsbyattempting touseinpart theresult ofthe exercise worked inthetext. 1.Arod10ft.longandweighing 30Ibs.carries a20Ib.weight atoneendanda30Ib.weight attheother. Itissupported atitsmiddle point. Find thelength oftheequivalent simple pendulum. Ans. 30ft. 2.Equal masses arefixed atthevertices ofanequilateral triangle andthelatter issupported atoneofthevertices. Ifit beallowed tooscillate inavertical plane, findthelength ofthe equivalent simple pendulum. 130 MECHANICS 3.Arigiduniform circular wire*6in.indiameter andweigh- ing12Ibs.hasa4Ib.weight fastened atoneofitspoints and isfree tooscillate about itscentre initsown plane. Find the length oftheequivalent simple pendulum. 4.Equal particles areplaced atthevertices ofaregular hexa- gonandconnected rigidly byaweightless truss. Thesystem ispivotedatone oftheparticles andallowed tooscillate ina vertical plane under gravity. Find thelength oftheequivalent simple pendulum. 6.Generalize tothecase ofnequal particles placed atthe vertices ofaregular n-gon. 6.TheCompound Pendulum. Consider anarbitrary lamina, orplane plate ofvariable density. Let itbesupported ata point aridallowed tooscillate freely initsown plane, assumed vertical, under gravity. This isessentially themost general compound pendulum. Todetermine themotion. Divide thelamina upinanyconvenient manner into small pieces andconcentrate themass ofeach piece atoneofitspoints. Connect these particles with one another andwith thesupport at byatruss-work. The auxiliary sys- temcanbedealt withbythePrin- cipleofMoments. Set %k ^kCOS6k, Thenyk=rksin0*. FIG.81Now,draw alineinthelamina, forexample, thelinethrough and thecentre ofgravity, G,ofthe particles, anddenote theangleitmakes with theaxisofxby0. Then where ctkvaries withfc,but isconstant asregards thetime. Hence d0k=de d2Bk=d*0 dt" dt' dP~~ dt* *Byawire isalways meant amaterial curve. DYNAMICS OFARIGID BODY 131 Thus theleft-hand side oftheEquation ofMoments, 3, becomes n\ V1& Tdze (1)?m*r*^=/^> where 7denotes themoment ofinertia ofthesystem ofparticles about 0. Theright-hand sideofB), 3,canbewritten (2) 2)"mkgyk=-02)mky*' The lastsum hasthevalue My,whore thecoordinates ofG aredenoted by(x,y).Letthedistance from toGbeh.Then y=hsin and(2)becomes (3)-M0/isin 6. Onequating (1)and (3)toeachother, wehave d~n (4)/5y[=-JfffAsinfl. This istheEquation ofSimple Pendulum Motion, and Itappears, then, thattheauxiliary system ofparticles oscillates likeasimple pendulum. Asweallowntoincrease without limit, themaximum diameter ofthe little pieces approaching 0,itseems plausible thatthemotion willapproximate moreandmoreclosely tothat oftheactual compound pendulum, and this consider- ation leads ustolaydown thephysical law, orpostulate, that Equation (4)holds forthecompound pendulum, where 7and hnow refer tothelatter body. Remark. Wehave thought ofthemass ofthecompound pendulumastwo-dimensional, orlyinginaplane. But this is obviously anunnecessary restriction. Conceive ablock of granite, blasted from thequarry asirregular andjagged as you please. Mount itontwoknife-edges, soitcanswing about ahorizontal axis.Now thisblock willobviously oscillate exactly asaplane lamina perpendicular totheaxiswould,ifthemass oftheactual block were projected parallel totheaxisonaplane atright angles totheaxis. 132 MECHANICS Theabove "obviously"isnot tobetaken mathematically, but isanew physical law, orpostulate. Itistrue thatwhen wecome totreat thegeneral case ofmotion inthree dimensions, thispostulatewillbemergedinmore general ones. EXERCISES Find thelength oftheequivalent simple pendulum when the compound pendulumisoneofthefollowing. 1.Auniform circular disc, free torotate initsown plane about apointinitscircumference. Ans. I=fr. 2.Acircular wire, about apoint ofthewire. Ans. I=2r. 3.Question 1,when theaxis istangent tothedisc. Ans. I=fr. 4.Question 2,when theaxis istangent tothewire. Ans. I= -Jr. 6.Arectangular lamina, about aside. 6.Asquare lamina, about avertex. 7.Atriangle, about avertex. 6.Continuation. Discussion ofthePoint ofSupport. Let 7=Mfc2 bethemoment ofinertia ofthecompound pendulum about a parallel axisthrough thecentre ofgravity,(?.Bythetheorem of10themoment ofinertia about theactual axis willbe : andthelength oftheequivalent simple pendulumissoonfrom (5), 5,tobe : (6)I=^A2 - The question arises: What other points ofsupport, (i.e. what other parallel axes), yield thesame period ofoscillation? Clearly they arethose, andonly those, whose distance, x, from satisfies theequation, ,_*+* x (7) x*-Ix+Jk2=0, DYNAMICS OFARIGID BODY 133 where kand Iaregiven, andwhere, more- over, (6)istrue, or (8) h*-Ih+k2=0. One root ofEquation (7)isxl=h. Theother isseen tobe _i j,_*2 ^-I-ft_~ Wecanstate theresult asatheorem. FIG.82 THEOREM. Thelocus ofthepoints 0,forwhich thetime ofoscil- lation isthesameyconsists oftwoconcentric circles with their centre atG,their radii being , ,k2 hand -r- fi EXERCISES 1.Draw twoconcentric circles about G,ofradii hand k*/h. Show that thelength, Z,oftheequivalent simple pendulum cor- responding toanaxisthrough apoint ononeofthese circles isobtained bydrawing alinefrom through G,andterminating itwhore itmeets theother circle. Thistheorem isduetoHuygens. 2.Show that thelocus ofthepoints ofsupport, forwhich thetime ofoscillation isleast, form acircle withGascentre and ofradius k. 7.Kater's Pendulum. Theexperimentfordetermining the value ofg,theacceleration ofgravity, bymeans ofasimple pendulum andtheformula T= 9 isfamiliar toallstudents ofphysics andmathematics. The chief error intheresult arises from theerror indeterminingI. Thebob isnotsensibly aparticle andthestring stretches. Toattain greater accuracy, Kater made use ofHuygens's Theorem, 6,Ex.1, constructing acompound pendulum that could bereversed. Itconsists essentially ofamassive rod, orbar,provided withtwo sets ofadjustable knife-edges. These edgeslieintwo parallel lines, andthecentre ofgravity, (?,is 134 MECHANICS situated intheir plane, atunequal distances, hand A',from them. The knife-edges arenow soadjusted experimentally that the period when thependulum oscillates about theonepairisthe same aswhen itisreversed andallowed tooscillate about the other pair. Since I=h+h', thedetermination ofthelength oftheequivalent simple pendulum cannowbemade with great accuracy bymeasuring thedistance between theknife-edges. Indeed, theaccuracy inthus deter- mining gisnow sogreat thatvery small errors, likethose due tothebuoyancy ofthe air,thechanges inthependulum dueto changesintemperature, andthegive ofthesupports have to beconsidered. Foranelaborate and interesting account,cf. Routh, Elementary Rigid Dynamics, 98etseq. 8.Atwood's Machine. AnAtwood's Machine consists ofa pulleyfreetorotate about ahorizontal axis,andastring passing over thepulley andcarrying weights,MandM+m,atitstwo ends. Itmaybeused tomeasure theacceleration ofgravity. Ourproblemistodetermine themotion ofthesystem. The "system" which wechoose toisolate isthecomplete system ofpulley and weights, themass ofthestring being assumed negligible. This isnotarigid system, butstill,ifwereplace thepulley byasystem ofparticles rigidly connected, theinternal forces ofthecomplete auxiliary systemwillsatisfy thehypothesis* of 3,andthus theEquation ofMoments willhold. Fortheauxiliary system ofparticles due tothewheel the contribution totheleft-hand side oftheEquation ofMoments, B), 3,becomes asinthecase ofthecompound pendulum: where Idenotes themoment ofinertia ofthissystem about the axis,andBistheangle through which thewheel hasrotated. *Consider ashort interval oftime intheduration ofthemotion. Inthe auxiliary system, leteach vertical segment ofthestring befastened toaparticle near thepoint oftaiigericy ofthestring intheactual case. Then itisplausible physically thatthemotion oftheauxiliary system during thisshort interval differs butslightly from that oftheactual system. Hence wemayassume thattheforce ofthestring always actsatthepoints oftangency with thewheel, andneglect the restofthestring which isincontact withthewheel. Butthis isanewphysical law. DYNAMICS OFARIGID BODY 135 Lettheradius ofthewheel (more precisely, ofthegroove in which thestring lies)bea.Observe, too,that yl=const.+a0,=const. aO. Thus theremaining contributions tothe left- hand sideofB), 3,willbe (2) (M+m)a2-^+Ma2-^ Theright-hand sideofB)reduces to (3) (M+m)gaMga=mga. Thus B)becomes : dt2 This, fortheauxiliary system ofparticles. Andnowweassume, physically, thatthelimitapproached bythemotion oftheauxiliary systemisthemotion oftheactual system ;i.e.thatEquation (4) holds fortheactual system. Letsdenote thedistance theweight andriderhave descended. Then s=aO,andfrom (4)itfollows that (5)mga' dt2I+(2M+m)a2 Onintegrating thisequation wehave,inparticular, that (6) s= I+(2M+m)a2 Corresponding values ofsand tcanbeobserved experimentally. Thus Equation (6)isequivalent toalinear equation inthetwo unknowns, //a2andg: ~ II0. IfMisheld fastandmisgiven different values,itisclear thatthe coefficient ofgwilltakeondifferent values, andsoweshallhave twoindependent linear equations fordetermining theunknown physical constants, 7/a2andg. 136 MECHANICS EXERCISES Inworking these exercises usethemethod, notthe result, of thetext. Begin eachtimebydrawing afigure. 1.Suppose that thewheel isauniform circular discweighing 10Ibs.,andthat5Ib.weights arefastened tothetwoends of thestring. What willbetheacceleration duetoa1oz.rider? 2.Work thecase inwhich thewheel isahoop,i.e.auniform circular wire, themasses ofthespokes being negligible; and show that theacceleration oftherider does notdepend onthe radius, butonlyonthemass ofthehoop, andMandm. 3.Determine thetensions inthestring inthegeneral case. 4.Find thereaction ontheaxis. 6.Prove theassertion inthetextabout thecoefficient ofg's taking ondifferent values whenmisvaried. 6.*Howrough must thestring beinthegeneral case,inorder nottoslip? 9.TheGeneral Case ofRotation about aPoint. Consider anarbitrary rigidbodyintwodimensions, acted onbyanyforces initsplane, and free torotate about apoint 0,i.e.about an axisthrough perpendicular totheplane. Then,Isay, its motion isdetermined bythePrinciple ofMoments, B) /-jT2=5JMoments about 0. The Principleisrendered plausible bydividing theactual distribution into small pieces, asintheexample ofthecorn- pound pendulum andtheAtwood's machine, andobserving that thePrincipleistrue fortheauxiliary system. The limit ap- proached bythemotion oftheauxiliary systemisthemotion defined byEquation B)ofthepresent paragraph. Andthuswe areledtolaydown thephysical postulate that this isthemotion oftheactual system. Equation B),then,isanindependent physical law,made plausible bythemathematical considerations setforth above, butnotfollowing mathematically from them. The Effect ofGravity. Whenever gravity acts, thecontribu- tion ofthis force totheright-hand side ofEquation B)can *Thisproblem ismore difficult than theothers, and isessentially aproblem in theCalculus;cf.theauthor's Advanced Calculus, Chapter 14, 8. DYNAMICS OFARIGID BODY 137 always bewritten asthemoment ofasingle force, that force being theattraction ofgravity onasingle particle ofmass equal tothemass oftheentire bodyandsituated atthecentre ofgravity ofthebody. This istrue inthemost general case ofmotion, when nopoint ofthebodyispermanently atrest. Here, again, wehave anewphysical postulate. EXERCISES 1.Aturn table consisting ofauniform circular disc isfree torotate without friction about itscentre. Amanwalks along therimofthetable. Find theratio oftheangle turned through bythetable totheangle described bytheman,ifmanandtable startfrom rest. 2.Thesame problem when themanwalks inalong aradius ofthetable, thesystem notbeing, however, initially atrest. 10.Moments ofInertia. Themoment ofinertia ofthe simpler andmore importanjb distributions ofmatter aredeter- mined bythemethods oftheIntegral Calculus;cf.forexample theauthor's Introduction totheCalculus, p.323,andtheAdvanced Calculus, pp.58,79,88. Ml2 1.Auniform*rodoflengthIabout oneend :5o TI/T 2 2.Arodoflength 2aabout itsmidpoint:$ 3.Acircular discabout itscentre :5& Mr2 4.Acircular discabout adiameter :~T~~' 5.Asquare about itscentre : fMa2 . 6.Asquare about aside;cf.Example1. 7.Ascalene triangle about aside : , where hdenotes thealtitude. o A i i *r *8.Asphere about adiameter : 9.Acubeabout alinethrough thecentre parallel toanedge ; cf.Example5. *Itwillhenceforth beunderstood that thedistribution isuniform unless the contrary isstated. 138 MECHANICS AGENERAL THEOREM. Themoment ofinertia ofany distribu- tionofmatter whatever, about anarbitrary axis,isequaltothemo- ment ofinertia about aparallel axis through thecentre ofgravity, increased byMh2: where hdenotes thedistance between theaxes. Wewillbeginbyproving thetheorem forasystem ofparticles. Letthe firstaxisbetaken astheaxisofzinasystem ofCartesian coordinates, (x,y,z) ;and letthesecond axisbetheaxis ofz' inasystemofparallel axes. Then /=2mk(x,?+2/*2 ), 7=2w*W+yi2 ). Since x=x'+x, y=y'+y, itfollows that )=2k. 2xmkxi+2y <kyi- The lasttwoterms vanish because 0'isthecentre ofgravity, andhence 2mkx't=o,2 Itremains merely tointerpret theterms that areleft,and thusthetheorem isproved forasystemofparticles. Ifwehave abody consisting ofacontinuous distribution of matter, wedivide itupintosmall pieces, concentrate themass ofeachpiece atitscentre ofgravity, form theabove sums, and take their limits. We shall have asbefore 2mkXr t=0, Smkyi=0,andhence lim2 *(**2+2/t2 )=Km n-ooj7 n=oc or since these limits arebydefinition themoments ofinertia forthe continuous distribution. DYNAMICS OFARIGID BODY 139 Example. Tofind themoment ofinertia ofauniform cir- cular discabout apointinitscircumference. Here, 7=%Mr2 andh=r.Hence T Q,.,/=fMr2 . 11.TheTorsion Pendulum. Letarodbeclamped atits mid-point toasteel wireandsuspended, therodhorizontal and thewire vertical. Lettherodbedisplaced slightly initshori- zontal plane, thewireremaining vertical, aridthen released. To determine themotion. The forces acting ontherodamount toacouple, duetothe torsion ofthewire, andthemoment ofthecoupleispropor- tional totheangle through which therod isdisplaced such isthelaw ofelasticity. Thus thePrinciple ofMoments, 9, yields inthiscasethedifferential equation, Ma2 where /= ^isthemoment ofinertia oftherod,andKisthe constant ofthewire. Equation (1)istheequation ofSimple Harmonic Motion, andthustheperiod ofoscillation, (2) T=2* isthesame, nomatter what the initial displacement mayhave been, provided merely that thedistortion ofthewire isnotso great astoimpair thephysical lawabove stated, andprovided dampingisneglected. 12.Rotation ofaPlane Lamina, NoPoint Fixed. Letarigid plane lamina beacted onbyany forces initsplane, and let it move initsplane. Todetermine themotion. The centre ofgravity willmove asifallthemass were con- centrated there and alltheforces were transferred tothatpoint ; 1.Itremains toconsider therotation. PRINCIPLE OFMOMENTS. Thelamina rotates asifthecentre ofgravity were heldfastand thesame forces actedonthelamina as those applied intheactual casey (1)^"77/2=SMoments about(?, 140 MECHANICS where Idenotes themoment ofinertia about thecentre ofgravity, G; 6istheanglethatalinefixed inthelamina makes withalinefixed intheplane, and theright-hand side isthesum ofthemoments of theforces about G. Proof. Consider firstasystem ofparticles rigidly connected. Let(x,y)beaxes fixed intheplane, and(,77)parallel axeswhose originisatG.Then (2) x=+a, y= 77+y, and theomitted terms vanishing forthereason that >k*=0, ]mkrjk=0, andhence, too, Remembering that dt\dt weseethatEquation B), 3,herebecomes : Because xk=&+x, yk= rik+y, theright-handsideofEquation (4)becomes : 2fen-rjkXk)+7 Since DYNAMICS OFARIGID BODY 141 itfollows that Onsubtracting thisequation from(4),there remains : (5) * Inthisequationiscontained theproof ofthetheorem fora system ofnparticles. For, theleft-hand sidereduces tothe left-hand side of(1),since thedistance ofthepoint (&, ?/*)from thecentre ofgravity, G,doesnotchange witht;andtheright- hand side expresses precisely thesum ofthemoments ofthe applied forces about G. Finally, wepass toacontinuous distribution ofmatter in theusual way, laying down anewphysical postulate totheeffect thatEquation (1)shall hold forallrigid distributions ofmatter inaplane. 13.Examples. Ahoop*rollsdown arough inclined plane without slipping. Determine themotion. The forces are: theforce ofgravity andthereaction ofthe plane. Letthe latter force beresolved intoanormal com- ponent, R,andthetangential force offric- tion, F,acting uptheplane. Then, for themotion ofthecentre ofgravity, weshall have: -.FIQ84 Thesecond equation forthemotion ofthecentre ofgravity merelytellsusthat (2) R=Mgcosa, afact thatwecould have guessed, since thecentre ofgravity always remains atthesame distance from theplane. However, letusformulate thesecond equation, andprove ourguess right. Letydenote thedistance ofthecentre ofgravity from theplane. *Apipe, thethickness ofwhich isnegligible, when placed ontheplane with its axishorizontal, would move inthesameway. Thetwoproblems aredynamically identical. 142 MECHANICS Then Buty=a,theradius ofthehoop, andsotheleft-hand side of thisequationis0. Turning now totherotation ofthehoop,wewritedown Equa- tion (1)oftheTheorem, 12 : (4)/-^=aF, I=Ma2 . Since there isnoslipping, (5)5=a0, where,forconvenience, wetake as6theangle that theradius drawn tothepoint ofcontact with theplane atthestart has turned through,5being also atthestart. Equations (1)and(4)cannowbewritten intheform : Ma-JTJ=MgsinaF, (6) Ma*<^=aF. Oneliminating Fbetween these equations, wefind : or (8) ^= |sina . Hence itappears that thecentre ofthehoopmoves down the plane with just halftheacceleration itwould have iftheplane weresmooth. Equation (2)appears tohave played nopart inthesolution. Butwehaveassumed that there isnoslipping, andsoFcannot begreater thanpR: (9) F^R. Toascertain what this condition means forthe coefficient offriction, ju,andthesteepness oftheplane, a,solve Equations (6)forFandsubstitute : F^ys*na 2' DYNAMICS OFARIGID BODY 143 Mgsina .. 2-5 ^M^ cosa, (10) tana^2M. Hence itappears thatamaynotexceed tan"1 2ju. EXERCISES 1.Show that,ifthehoop bereleased fromrest, gt. at* . v=~sma, s=~-sma, v2=0ssina. 2.Show furthermore that at . at2 .= TJ-sina, ^=T-sina,<&& 4<z W2*Lgjnaa 3.Solve theproblem studied inthetext forasphere. Show that d*6 5g. d*s 50. ^5=^ama, ^J^Bina. 4.Prove thatthespherewillslipunless tanagJJLI. 6.Make acomplete study ofadisc, orsolid cylinder. 14.Billiard Ball, Struck Full.Abilliard ball isstruck full bythecue.Todetermine themotion. The forces are: theforce ofgravity, acting downward atthe centre ofgravity, andthereaction ofthe billiard table, which yields avertical component, R,andahorizontal component, F. Letsbethespace described bythecentre oftheball,and6,the angle through which theballhasturned.* The Principle oftheMotion oftheCentre ofGravity, 1, yields theequations: (1)__ dt*F R=Mg FIG.85 *Itisofprime importance thatthestudent begin eachnewproblem, ashere, bydrawing afigure showing theforces andthecoordinates used insetting upthe differential equations ofthemotion. Itiswell, too,tonote atthesame timeany auxiliary relations, asinthepresent instance, F=pR. 144 MECHANICS ThePrinciple ofRotation about theCentre ofMass, 12, yields theequation: tv\ rd*enw T(2) 7=^ I Finally, solongasthere isslipping, (3) F=MB. From Equations (1), (2),and(3)itappears that /A\ &S (4.) is=- r' d<2~ 2a Theintegrals ofthese equations areasfollows : ,g. fv=v-ngt,s=vt- \ t>2=vl- and Thus astheballadvances,itscentre moves more andmore slowly, while thespeed ofrotation steadily increases.Finally, pure rolling will set in.This takes placewhen thevelocityof thepoint oftheball incontact with thetable isnil.Now, the velocity ofthispoint oftheball ismade upoftwo velocities, namely, i)thevelocity oftranslation, orthevelocity thepoint would have iftheballwere notrotating,i.e.v,asgiven by(50; andii)thevelocity duetorotation, orthevelocity thepoint would have iftheballwere spinning about itscentre, thought of asatrest. The latter isavelocity ofao>inthedirection opposite tothemotion ofthecentre, and isgiven by(52).Thus theve- locity forward ofthepoint oftheballincontact with thetable is (6) vow. Slipping continues solong asthisexpressionispositive, and ceases when itvanishes : (7)v oo>=0. DYNAMICS OFARIGID BODY 145 Thetime isgivenbytheequation or Thecorresponding value ofsisseentobe : Theangle through which theballturns is Finally, (11) 1= |>0,!=T2' 7 la EXERCISES 1.Solve thesame problemincase thetable isslightly tipped andtheball isprojected straight down theplane. 2.Work thelastproblem with themodification that theball isprojected straight uptheplane. 15.Continuation. TheSubsequent Motion. Attheend of thestage ofthemotion just discussed, theballhasboth amo- tion oftranslation andone ofrotation, thepoint ofthe ball incontact with thetable being atrest. Iffromnowontheforce exerted bythetable ontheball consists solely ofanupward component Randatangential component F,thelatter force will vanish, andtheball willcontinue torollwithout slipping. For, suppose thetable isrough enough toprevent slipping. Then s=oQ,andsince equations (1)and(2)stillhold,wehave : HenceFvanishes, andtheangular and linear accelerations are both0,too. But inpractice theball willslow up.How isthistobeac- countedfor,iftheresistance oftheairisnegligible? Theanswer is,thatthereaction ofthetable isnotmerely aforce, withcom- ponents RandF.but, inaddition, acouplej themoment ofwhich 146 MECHANICS wewilldenote byC.This couple hasnoinfluence onthemotion ofthecentre ofgravity; thusEquations (1), 14,remain as before. ButEquation (2)nowbecomesO (13) FIG.86Furthermore, (14) Hence dP 7Ma' dt* 7Ma*' la SinceCissmall, theballslowsupgradually. EXERCISES 1.Ifthecentre oftheballwasmoving initially attherate of 6ft.asec.and iftheballstops after rolling 18ft.,show that C=IMa. 2.Ifthe initial velocityofthecentre was VQand iftheball rolled Ift.,show thatCisproportional totheinitial kinetic energy andinversely proportional tothedistance rolled. 16.Further Examples, i)HOOP ONROUGH STEEPLY IN- CLINED PLANE. Suppose, intheExample studied inthetext of13,thatadoesexceed tan"1 2/*.What willthemotion then be,thehoop being released from rest? Equations (1), (2),and (4)willbeasbefore. Butnow(5)is replaced bytheequation: (i) p=& allthefriction now being called into play. Oneliminating P andR,wefind : (2)dzs-=<7(sin<*- -== cosa. dt2a The integrals ofthese differential equations canbewritten down atonce. Inparticular,itisseen that theratio ofsto6 isconstant,ifthehoop starts from rest : sa(sinaucosa) ,, ,% -,-=- -- '-=o(tan acotX- 1).pcosa DYNAMICS OFARIGID BODY 147 FIG.87The lastparenthesis hasthevalue 1when tana=2/z,and is>1when aislarger. Thus themotion isoneinwhich acir- cleofradius , _ r=a(tanacotX 1) andcentre atthecentre ofthehooprolls without slipping onalineparallel tothe plane andbeneath it.Wehave herean illustration ofthegeneral theorem thatany motion ofalamina initsownplane canbe realized bytherolling without slipping ofacurve drawn inthe lamina onacurve drawn intheplane ;cf.Chapter V,4. ii)LADDER SLIDING DOWN ASMOOTH WALL. First, draw a figure representing theforces andthecoordinates. The three equations ofmotion thusbecome : (3) FIG.88"77/2=a^s*n"~a^cos^^= U/t" With these three Dynamical Equations areassociated two Geometrical Equations: (4) x=acos0, y=asin 6. These fiveequations determine thefiveunknown functionsx,y, 0,R,S,thetimebeing theindependent variable;orthey deter- mine five ofthevariables x,y,0,R, /S,tasfunctions ofthe sixth. Eliminate72,Sbetween the firstthree equations: <5>'.->-^Macos~Afgracos 6. From theGeometrical Equations follows : dx .dO-~~=acos6-jr,dt dt d2x 148 MECHANICS Combining these with(5)andreducing weobtain : This differential equation canbeintegrated bythedevice of multiplying through by2d0/dt andthen integrating each side with respect tot: 30 dO - dt Since d itfollows that fdO\230. (dt)=-2Hs The constant ofintegration, C,isdetermined bythe initial conditions. Iftheladder isreleased fromrest,making ananglea with thehorizontal, then dB/dt=and 6=ainitially, andso =- jjsina+C.Zd Hence, finally, *> '-(-..-*.) Tofindwhere theladder will leave thewall. This questionis answered bycomputing Randsettingit=: Ed2x -/>d2*/de\2R=M--trr=MasmO-jrz Macos6[-77),at1at2\at/ (8) R=fMgcos6(3sin6-2sina). HenceR=when 3sin62sina=0. Let ftbetheroot ofthisequation: ft=sin-1 (fsina). Observe thatcos8cannot vanish when g6< DYNAMICS OFARIGID BODY 149 The intuitional evidence isherecomplete:theladder leaves thewalland slides along with thelower endincontact with the floor. Butsuppose apersonisunwilling totrust hisintuition andsays: "Ah, youhave notproven your point inmerely showing thatR=foracertain value of0.The ladder might stillremain incontact with thewall,Rincreasing astheladder continues toslide." The logic ofthisobjectionisvalid. The objection canbemetasfollows. Think oftheupper end oftheladder asprovided witharing that slides onasmooth vertical rod.Then theladder willnot leave thewall.How aboutRinthiscase? Formula(8)now holds cleardown tothefloor;butR<when 6<sin"1(|sina), andsothevertical rodhastopullontheladder instead ofpush- ing. Thisproves thatourintuition was correct. TheTime. From Equation (7)itappears that (9)Vsina sin6 This integral cannot beevaluated interms oftheelementary functions. Onmaking thesubstitution x=sin6, theintegral goesover intoanElliptic Integral oftheFirst Kind, andcanbetreated bywell-known methods; cf.theAuthor's Advanced CakuluSj Chapter IX. iii)COINONSMOOTH TABLE. Acoin isreleased from rest with onepoint oftherimtouching asmooth horizontal table. Todetermine themotion. The forces acting are: Gravity, Mg,down, andthereaction, R,ofthetable upward. Thus thecentre ofgravity ofthecoin descends inarightline. Let itsheight above thetable bede- notedbyy.Then thefurther Dynamical Equations become : (10) =-aRcosO. at* TheGeometrical Equationis : (11) yosinfl. 150 MECHANICS Oneliminating Randywefind : (12) (fc2+a2cos20)^-a2sin cos (9(~f)2=-agcos 0. (Zf \ttf/ This differential equation comes under ageneral class, namely, those inwhich one ofthevariables fails toappear explicitly. Thegeneral plan ofsolution insuch cases istointroduce anew variable, And thiscanbedone here. But inthepresent case there isa short cut,due tothespecial form ofthedifferential equation. Itisobserved that, onmultiplying theequation through by 2dQ/dt ytheleft-hand sidebecomes thederivative ofacertain function with respect to tysothattheequation takes ontheform : (13) Onintegrating each side ofthisequation with respect tot,w find: (F+o2cos2 0) =-2agsin9+C. \Gfv/ Todetermine Cmake useofthe initial conditions, dO/dt= and=a.Thus (14) (A;2+a2cos2^{^)=2a0(sina-sin0). Theangular velocity, w,ofthecoinwhen itfalls flatonthe table isgivenbytheequation: _2agsina Wl~ Buthere isanassumption, namely, thatthepoint ofthecoin initiallyincontact with thetable remains incontact till=0. This isplausible enough physically; butinthis gunws,isthere notanappreciable admixture ofunimaginativeness and the question which themoron sofrequently asks:"Why shouldn't it?" Theangular velocity dd/dt ofthecoin issteadily increas- ing, asweseeboth intuitionally andfrom (14).Mayitnot increase tosuchanextent that thelowest pointinthecoinmay DYNAMICS OFARIGID BODY 151 kickupandleave thetable before thecentre comes cleardown? Themoron certainly cannot answer thisobjection byphysical intuition. Itishere thatmathematics sitsasjudge over thesituation. Replace theactual problem byoneequivalent during theearly stage ofthemotion, and seewhether thisstage laststhrough to theend. Letthelowest pointofthecoinbeprovided with a ringthat slides onasmooth horizontal rod.Then thecoin will fallasweguessed. Compute now thereaction, R.The test is:DoesRremain positive throughout themotion? Weleave ittothestudent tofindout. EXERCISES ONCHAPTER IV 1.Ahomogeneous solid cylinderisplaced onarough inclined plane and released from rest. Will itslipasitrolls, orwill it rollwithout slipping? Ans. Itwillslipiftheangle ofinclination oftheplane isgreater thantan"1 (|M)- 2.Thesame problem forahomogeneous spherical shell (material surface). 3.Abilliard ball issetspinning about ahorizontal axisand isreleased, justtouching thecloth ofthe billiard table. How farwill itgobefore pure rolling sets in? 4.Acircular dischasastringwound round itscircumference. The freeendofthostringisfastened toapeg,A,andthedisc is released from rest inavertical plane with itscentre below the level ofA,and thestring tautand vertical. Show that the centre ofthedisc willdescend inavertical right linewithtwo- thirds theacceleration ofgravity. 6.The disc ofthepreceding problemislaid flatonasmooth horizontal table;thestringiscarried overasmooth pulley atthe edge ofthetable, andaweight equal totheweight ofthedisc isattached totheendofthestring. Thesystemisreleased from rest, thestring being tautandtheweight hanging straight down. Show that theacceleration oftheweightisthree-fourths that ofgravity. 6.Find thetension ofthestring inthelastquestion. 7.Solve theproblemofQuestion 3with themodification thatthetable isinclined atanangleatothehorizon. 152 MECHANICS Discuss infullthecasethat therotation oftheball isinsuch asense that theballmoves down theplane faster than itwould ifithadnotbeen rotating. 8.Study theproblem ofthelastquestion when therotation isintheopposite sense. 9.Abilliard ball isplaced onabilliard table inclined tothe horizontal atanangle a,and isstruck fullbythecue,sothat itstarts offstraight down theplane without anyinitial rotation. Study themotion. 10.Thesame problem when theball issostruck that itstarts straight uptheplane. 11.Ifamanwere placed onaperfectly smooth table, how could heturnround ? .12.Aplank canrotate about oneend,onasmooth horizontal table.Aman, starting from theother end,walks toward the pivot. Determine themotion. 13.Asmooth tube, theweight ofwhichmay beneglected, canturn freely about oneend.Arod isplaced inthetubeand thesystemisreleased from restwith therodhorizontal. Deter- mine themotion. 14.Aspindle consists oftwoequal discs connected rigidly withanaxle,which isasolid cylinder. The spindleisplaced onarough horizontal table, andastringiswound round theaxle and carried over asmooth pulley above theedge ofthetable. Aweightisattached tothelower endofthestring andthesystem isreleased from rest. Determine themotion. Consider firstthecase inwhich thestring leaves theaxlefrom thetop ;then, thecasethat thestring leaves theaxlefrom the bottom. Ineach case, thesegment ofthestring between the axleandthepulleyshallbehorizontal andatright angles tothe axis,andthepartbelow thepulley, vertical. 15.The centre ofgravity ofafour-wheeled freight car is 5ft.above thetrack andmidway between theaxles, which are 8ft.apart. The coefficient offriction between thewheels (when they arelocked) andthetrack is^-.Ifthecar isrunning atthe rate of30m.anh.,inhowshort adistance can itbestopped by applying thebrakes totherearwheels only? How far,ifthe brakes areapplied tothefronfr wheels only? DYNAMICS OFARIGID BODY 153 16.Auniform rod issuspended inahorizontal position by two vertical strings attached toitsends. One stringiscut. Find theinitial tension intheother one. 17.Ahoopishunguponapegand released. Findwhether itwill slip. 18.Auniform circular disc, ofradius 1ft.andweight 10Ibs., canrotate freely about itscentre, itsplane being vertical. There isaparticle weighing1Ib.fixed intherim,andafineinextensible weightless string, wound round therimofthedisc, hasaweight ofPIbs.fastened toit.Thesystemisreleased from restwith the 1Ib.weight atthelowest point andtheother weight hanging freely atthesame level.How greatmayPbe,ifthe 1Ib.weight isnottobepulled over thetop? 19.Abilliard ball rolls inapunch bowl. Determine the motion. 20.Asolid sphereisplaced ontopofarough cylinder ofrevolu- tion, axis horizontal, and slightly displaced, under theaction of gravity. Findwhere itwillleave thecylinder. 21.Auniform rod isreleased from rest, inclined atanangle, with itslower end incontact with arough horizontal plane. Will itslipatthestart? Determine themotion. 22.Apacking box issliding overanicysidewalk. Itcomes tobareground. Will ittipup? CHAPTER V KINEMATICS INTWODIMENSIONS 1.TheRolling Wheel. When awheel rollsoveralevel road without slipping, thenature ofthemotion isparticularly acces- sible toourintuition, forthepoints ofthewheel lowdownmove slowly, thepoint incontact with theground actually being at rest fortheinstant, and itismuch asifthewhole wheel were pivoted atthispoint androtating about itasanaxis. Thisis, infact, precisely thecase, thevelocity ofeach pointofthewheel attheinstant being thesame asifthewheel were rotating per- manently about that point. Ifthewheel isskidding,itisnotsoeasy toseethat asimilar situation exists, andyetitdoes. Nomatter howthewheel is moving, provided itisrotating atall,there isateach instant a definite point (farawayitmay be),about which thewheel rotates atthatinstant. This pointiscalled theinstantaneous centre. Toprove thisassertion, wewillbeginbygiving ageneral formu- lation oftheproblem ofthemotion ofanyplane lamina inits plane. Itmakes theproblem more concrete tothink ofanactual lamina, likeadiscoratriangle orafinite surface, S.Butwe arereally dealing with themotion ofthewhole plane, thought ofasrigid. Themotion may bedescribed mathematically asfollows. Draw apair ofCartesian axes inthemoving plane;i.e.think X FIG.90 ofthisplane asasheet ofpaper, anddraw the(,?/)-axes inred inkonthepaper. Assume further asystem ofaxes fixed in 154 KINEMATICS INTWODIMENSIONS 155 space the(x,/)-axes. Then the(,^-coordinates ofanarbi- trary pointPofthemoving plane areconnected with the(x,y)- coordinates ofthesame pointbytherelations : x=XQ+cos677sin0, y 2/0+?sin+ycs0- The position ofthemoving planeisknown when onepoint, as0',isknown andtheorientation, asgiven by 0,isknown. Themotion may, therefore, becompletely described bystating howXQ, 2/0) vary with thetime;i.e.bysaying what func- tions x,2/o> are f : Woshallassume attheoutset that these functions arecontinuous andpossess continuous derivatives ofthe first order. Later,it willbodosirable torestrict them further byrequiring thatthey have continuous derivatives ofthesecond order. EXERCISE Express andrjinterms ofxandyyi)geometrically, byreading theresult offfrom thefigure ;ii)analytically, bysolving Equa- tionsA)for ,77.Theformulas are : ff=(x-x)cos+(y- 2/0)sin0, Irj=(x XQ)sin+(yyQ)cos 0. 2.TheInstantaneous Centre. LetPbeapoint fixed inthe moving plane mark itwith adotofredinkonthesheet of paper. Lotthecoordinates ofPbe(x,y).Then they aredeter- mined asfunctions oftbyEquations A),(, rj)being thecoordi- nates ofPwith reference tothemoving axes. Ofcourse, and TJareconstants with respect tothetime, fortheredinkdotdoes notmove inthopaperitmoves inspace. The vector velocity, v,ofPinspace canbedetermined by moans ofitscomponents along theaxes ofxandyywhich arefixed inspace, Chapter III, 15 : -^ -,-^- 156 MECHANICS These derivatives canbecomputed interms oftheknown func- tions (1),namely, x,yQ,0,and oftheir derivatives, bymeans ofEquations A).Thus (2)dx___dxQ, dy_dy*,f dt dt Theparentheses that here enter areseenfrom Equations A) tohave thevalues : Hence-(y- dx_ dt~X-XQ. ~dt dy__dy, ,_ dt~~dt+(XdB These equations express thecomponents ofthevector velocity vofthepointPalong theaxes fixed inspace,interms ofthe coordinates (x,y)ofPandtheknown functions (1). New Notation. Since derivatives with respect tothetime occur frequentlyinthework which follows, theNewtonian nota- tionwith thedot isexpedient: (4)._dxX~ ' x= dt*'etc. Thus theformula forthecomponents ofthevelocity assumes tho finalform : B)*=z-(y- 2/0)0> y= 2/0+(*-#o)* TheInstantaneous Centre. Wenow inquire what pointor points (ifany) ofthebody areatrestatagiveninstant. A pointis"atrest" ifitsvelocityis0.Hence thecondition is, that x=andy=0,or : (5)=x-(y- j/)6, =y+(*-zn)6. KINEMATICS INTWODIMENSIONS 157 These equations yield aunique solution fortheunknown x andywhen, andonlywhen, 6^ : C) +Xn -- -- A THEOREM. Atanyinstant atwhich d6/dt=6isnot0,there is oneandonlyonepoint ofthebody atrest. This pointiscalled theinstantaneouscentre, and itscoordinates^ (xu!/i)iaregiven byEquations C). If6=0,nopoint ofthebodyisatrest, orelse allpoints are; there isnever asingle point atrest, totheexclusion ofall others. When 6=0,XQandyQnotboth vanishing,allpoints ofthe body aremoving inthesame direction with thesame speed, andwehave amotion oftranslation. EXERCISES 1.Show that thecoordinates (j, 77j)oftheinstantaneous centre, referred tothemoving axes, arcgivenbytheequations: ^osin6?yncos& " a xQcos+2/0sin6 1/1= ^ 2.Acircle rollsonalinewithout slipping. Show that the point ofcontact isatrest. 3.Abilliard ball isstruck fullbythecue. Find theinstan- taneous centre during thesubsequent motion. 3.Rotation about the Instantaneous Centre. The very name"instantaneous centre7' implies thatthemotion ofthebody isone ofrotation about that point. Letusmake this state- ment precise. Suppose thebodyisrotating about theorigin, 0,withangular velocity 6=o>.What willbethevector velocity ofanarbitrary point P:(x, y)?Theanswer isgiven byEquations B),where 0*istaken atO,andthusx=yQ==yQ=0.Hence 158 MECHANICS (1) f*"""* Iy=xu. The result checks, forthese arethecomponents ofavector at right angles with theradius vector rdrawn from toPandhav- ingthesense oftheincreasing angle 6,itslength being Vx2+y2 ft=ru, <6. If6<0,itssense isreversed. Itistheform ofEquations (1)that isimportant. Wesay thatanymotion ofthepoints ofthe(x,2/)-plane such that, at agiven instant, thevelocity ofeach pointisgivenby(1),isone ofrotation oftheplane asarigid body about 0.The velocities ofthepointsintheactual motion before andafter theinstant in question maybedifferent from those ofthe rigidbody that is rotating permanently about 0.But forashort space oftime before andafter theinstant, thediscrepancywillbesmall because ofthecontinuity ofthemotion, andattheoneinstant, theveloci- tieswill alltally exactly. Ifthepoint about which thebodyispermanently rotating hadbeen thepoint (a,6)instead ofthe origin, Equations (1) would havebeen thefollowing: (2)f*=-&-&)*, 1y=(x-a)6. Wearenowready tostateandprove thefollowing theorem. THEOREM. Themotion oftheactual bodyatanarbitrary instant t,atwhich 6?*0,isoneofrotation about theinstantaneous centre. Toprove thetheorem wehave toshow that, attheinstant2, where (xlty^aregivenbyEquations C), 2,and(#,y}aregiven byEquations B)ofthesame paragraph. Todothis, eliminate XQand2/between Equations B)andC).Thiscanbedonemost conveniently bywriting Equations C)intheform(5)of2: =*-(Vi-2/oH =2/0+(x l-x)6, KINEMATICS INTWODIMENSIONS 159 andthen, inthisform, subtracting them respectively fromEqua- tions B).The result isEquations (3)ofthisparagraph, andthe theorem isproved. Translation and Rotation. From theforegoing result anew theorem about themotion oftheplane canbederived atonce. LetAbeanarbitrary point, and letitsvector velocity bedenoted byV.Impress oneach point oftheplane, asitmoves under thegiven law,avector velocity equal andopposite toV.Then Aisreduced torest,andthenewmotion isoneofrotation about Awith thesame angular velocity asbefore. Wethushave the THEOREM. The field ofvector velocities isthevectorsum ofthe fields consisting i)ofthetranslationfielddue tothevectorvelocity ofanarbitrary point,A;andii)oftherotationalfield withAas centre. Inother words, thegiven motion consists ofrotation about an arbitrary point, A,plusthetranslation ofA. Thetheorem also follows immediately from Equations B),if wetakethepoint 0'atA. 4.The Centrodes. Wearefamiliar with themotion ofa circular discwhen itrollswithout slipping onaright lineora curve awheel rolling ontheground. Con- sider,more generally, themotion ofalamina when anarbitrary curve drawn initrolls without slipping onanarbitrary curve fixed inspace. Womaythink ofabrass cylinder, orcam, ascutwith itsfacecorresponding to the first curve, andattached tothebody; asecond such cam, with itsface corresponding tothesecond curve, being fixed inspace. Andnow the firstcam isallowed torollwithout slipping onthesecond cam. Thus agreat varietyofmotions ofthelamina canberealized, andnowtheremarkable fact isthat allmotions canbegenerated inthisway, with the single exception ofthe translations, provided that thefunctions(1)of 1have continuous deriva- tives ofthesecond order, andthespace centrode istraced out bytheinstantaneous centre withnon-vanishing velocity. Anecessary condition forthetruth ofthisstatement isevident from intuition, namely:thepoint ofcontact ofthetwocams 160 MECHANICS must betheinstantaneous centre oftheactual motion. This fact suggests theproof thefaces ofthecams,i.e.thecurves, must bethelocioftheinstantaneous centres inthebodyandinspace. Definition. The locus oftheinstantaneous centre inthebody iscalled thebody centrode, andthelocus oftheinstantaneous centre inspaceiscalled thespace centrode. THEOREM. Anymotion ofarigid lamina which isnottransla- tioncan begenerated bytherolling ofthebody centrode (without slipping) onthespace centrodc, providedthespace centrode istraced outbytheinstantaneous centre with non-vanishing velocity;the functions (1)of1having continuous derivatives ofthesecond order. Before wocanprove thetheorem, wemustmake clear toour- selves how toformulate mathematically therolling ofonecurve without slipping onasecond curve. Astheindependent variable, the*timo most naturally suggests itself; but itisbetter atthe outset nottochooseit,buttotake, rather, avariable Xwhich merely corresponds tothofactthat, foranarbitrary (i.e.variable) value ofX,thecurves meet ina(variable) point P.Andnow weshalldemand further : i)that thecurves betangent toeach other atP ; ii)that thearcoftheonecurve corresponding toanytwo different values ofX,namely, XjandX2;andthearcoftheother curve corresponding tothesame values ofX,have thesame length. Thus, inparticular, both curves may bemoving amore general casethan theonethat interests ushere. Lettheequationoftheonecurve, C,referred toasystem of Cartesian axes, (x,y),be : (1) *=ff(A),=A(X), where thefunctions #(X), h(\) arecontinuous together with their first derivatives, andthelatter donotvanish simultane- ously: (2) <<7'(X)2+/i'(X)2 . Letthesecond curve, F,referred toasecond systemofCartesian axes, (, r;),berepresented bysimilar equations, (3) (4) KINEMATICS INTWODIMENSIONS 161 Thecoordinates ofanypoint oftheplane, referred totheone setofaxes, areconnected with thecoordinates ofthesame point, referred totheother setofaxes,bytheequations: x=x+%cos6TIsin0, y= 2/o+sin9+ rjcos 0. Andnowwerequire thatXQ,yQy6befunctions ofXwhich have continuous firstderivatives :(5) (6) *o=/00, 2/o=* 00,= where /'(X), ^'(X), ^'(X) existandarecontinuous. SinceCandFalways meet inapoint P,whose coordinates areexpressed bytheequations (1)and(3),itfollows thatEqua- tions (5)willhold identicallyinXifthevalues ofx}yfrom(1), andthose of,77from(3),besubstituted therein. Thevector vwhose components are _dx__dy Vx~ d\'Vy~~ d\ istangent toCatPand itslengthis Thevector uwhose components arc istangent toFatPand itslengthis Therequirements i)andii)demand that these twovectors beidentical. This condition isboth necessary and sufficient. Theanalytical formulation ofthocondition isasfollows : (7)vx=u$cosBUysin0, vy=u%sinB+u^cos B. Wenowhave allthematerial out ofwhich toconstruct the proof. From Equations (5)it follow? that FIG.92 162 MECHANICS dx The first lineinthese equationsisnothing more orlessthan the first ofEquations (7),andthelatter equations wehave set outtoprove. Hence thesecond linemust vanish,iftheequation istobetrue,andso,bytheaidof(5),weobtain thefirstofEqua- tions (8): <8) Thesecond equationisobtained inasimilar manner from the second oftheabove equations. Equations (8)represent anewform ofnecessary and suffi- cient condition forthefulfilment ofConditionsi)andii). 5.Continuation. Proof oftheFundamental Theorem. It isnow easy toprove thetheorem of 4.Thetwo curves, C andF,arehere thespace centrode andthebody centrode, and wewillnowtake asourparameter X,thetime t.Equations (1), 4,thus represent thecoordinates, xlandyltoftheinstanta- neous centre inspace, and inEquations (8),the (x,y)arethe coordinates ofthissame point, (xl9yj.The other quantities that enter into (8)arethefunctions(6)thatdetermine theposi- tion ofthemoving body; andX=t.Thus Equations (8)go over intothefollowing: f*o- (2/i- 2/o)*=0, 12/o+(xl-z)6=0. Butthese areprecisely Equations (5)of2,which determine the instantaneous centre. Equations (8)arethusshown tobetrue. KINEMATICS INTWODIMENSIONS 163 Discussion oftheResult. From Equations (8) 4,itappears thatanecessary condition forthetruth ofthetheoremis,that thecoordinates ofapointPofCsatisfy theequations: (10)X=Xn-dy/d0 dX/dX' ,dxaIdB y= 2/o+-3r/3T- Butthese conditions arenot sufficient, since thefunctions xand ythus defined willnotingeneral admit derivatives. Tomeet this latter requirement wedemand, therefore, that thefunctions (6)possess, furthermore, continuous second deriv- atives. But this isnotenough, even ifthecase thatxandy reduce toconstants isexcluded (rotation about afixed point). Itis,however, sufficient whenweaddthehypothesis of(2), 4,andsodemand that dy/de\~dx/dx/'d de benotboth (dO/d\ being, ofcourse, 5^0). Inother words, theequations (12) td0d2x_d^cteo _,^!^o = dXdX2dX2dX"*"dX2dXU| dX'dX2""" dX*"d\"~ dX2dX= shall never hold simultaneously. This excluded case includes thecase inwhich anordinary cusp occurs; but italsoincludes more complicated singularities. If,inparticular, thefunctions (6)areanalyticintheneighbor- hood ofapoint, X=X,and ifthecase ofpermanent rotation about afixed point beexcluded, thecurveCwillatmost have acuspandotherwise besmooth intheneighborhood ofthepoint ; andthesame willbetrue ofP. Acceleration ofthePoint ofContact. Letthepoint (XQ,y), atagiven instant, t,betaken atthepoint ofcontact ofCandT. Then itfollows from (10) or,more simply, from (8) since x=xQandy=yQ>that dX 164 MECHANICS Lettheorigin, furthermore, betaken atthispoint (x ,y),and letCbetangent tothex-axis here. Now, thederivatives ofx andyin(10)cannot both vanish. Oncomputing them itis seenthattheyreduce respectively to d\2~d\' d\2d\ Thesecond, dy/d\, hasthevalue0,sinceCistangent totheaxis ofxattheorigin. Hence weinfer that IfXisthetime, t,these derivatives become thecomponents along theaxes oftheacceleration ofthepointofcontact, thought ofasapointfixed inthemoving body. From (13)itappears that this acceleration isnever0,but isavector orthogonal tothe centrodes attheir point ofcontact. Thereader canverifythis result inthecase ofthecycloid. Example. Abilliard ball isprojected along asmooth hori- zontal table withaninitial spinabout thehorizontal diameter which isperpendiculartothelineofmotion ofthecentre. Deter- mine thetwocentrodes. Take thepath described bythecentre oftheballastheaxisofx, andthecentre oftheballas(x ,y).Then Equations (10)give: FIG.93x=XQ, y=--. Hence thespace centrode isahorizontal straight lineatadistance c/cobelow thecentre ofthe ball,andtheinstantaneous centre isalways beneath thecentre ofthe ball. Thismeans that the ball rollswithout slipping onarightlinedistant c/wbelow the centre. Hence thebody centrode isacircle ofradius c/coabout thecentre oftheball. EXERCISE Abilliard ball isstruck fullbythe cue. Determine the space centrode andthebody centrode during thestage ofslipping ; cf.Chapter IV,14. KINEMATICS INTWODIMENSIONS 165 Thecoordinates being chosen asintheExample, theequations ofthespace centrode are : SC~~*Crt~~Cv"""" _2a_V~ 52acl 5/ t' where adenotes theradius oftheball,and ctheinitial velocity ofitscentre. Thetime that elapses during thestage ofslipping is2c/7/A<7 seconds. Thespace centrode meets thebilliard table at theangle Theequations ofthebody centrode, referred tosuitable polar coordinates, are : 2ac 12a P=- t5' 6.TheDancing TeaCup.When anempty teacupisset down onasaucer, thecupsometimes willdance foralongtime before coming torest.Two features ofthisphenomenon attract attention; first, that theenergy, obviously slight,isnotearlier dissipated bydamping, andsecondly, thatwecanhearanoise inwhich solittle energyisinvolved. Thesecond point canbe disposed ofeasily because ofthephysical factthat theenergy ofsound waves issurprisingly small. Toexamine the first critically weneedmore lightonthenature ofthemotion. The results which wohave obtained inthis chapter furnish the clue. The following exampleishighly sug- gestive. Consider themotion ofalamina,inwhich thebody centrode isarightlinemaking a small (variable) angle with thehorizontal. Forthespace centrode takeacurve suggested bythe figure. Such acurve canbedefined suggestively asfollows. Begin with thecurve (1) y=sin- 166 MECHANICS Thiscurve gives satisfactorily thepart ofthefigure nottoonear thelines y=1,but itistangent tothese lines, whereas it should have cusps onthem. The desired modification issimple. Forexample, toconvert thecurve y=f( X)=X* from onewhich istangent totheaxis ofxintoonewhich hasa cuspontheaxis,itisenough toreplace f(x)by[/(z)]*: Apply thisideatothecurve(1). Itwill suffice toset (2) y asthereader caneasily verify. Now allow thebody centrodo thetangent line todescend according toareasonable law.Wehave hereapicture ofwhat goesonastheteacupdances. The lineoscillates through smaller andsmaller angles asitspoint ofintersection with theaxis ofx descends. Tyndall,* inhispopular lectures, showed anexperiment with acoalshovel illustrating thesame phenomenon. The all-metal shovel washeated near itscentre ofgravity and laidacross two thin lead plates clamped inavise, with their edges horizontal. Astheshovel boremore heavily ononeoftheplates, thelatter expanded with theheat, throwing theshovel onto theother plate. Then theprocess was reversed. Thus vibrations like those oftheteacuparose, anddieddown. 7.TheKinetic Energy ofaRigid System. The kinetic en- ergy ofanysystem ofparticlesisdefined as T=i2)m^\ vk2=xk2+yi2+zt2 . Werestrict ourselves totwodimensions, andthus vi?=xj?+yj?. Suppose, now, that the particles arerigidly connected. In Equations A), 1,letthepoint (x ,y)betaken atthecentre ofgravity, (x,y).Thus Equations (2), 2,become : *Tyndall, Heat Considered asaMode ofMotion, Lecture IV. KINEMATICS INTWODIMENSIONS 167 k=x(&sin+*;*cos0)0, Vk=+(&cos- 77*sin0)0. Onsquaring andadding, multiplying bymk,andthenadding with respect tok,wefind : For,each oftheremaining terms involves asafactor oneofthe quantities 2) andeach ofthese is0,since thecentre ofgravityisattheorigin ofthe(,7y)-axes. Hence itfollows that (1) T=|MF2+i/fl2 , whereVdenotes thevelocity ofthecentre ofgravity andIisthe moment ofinertia about thecentre ofgravity,12being theangular velocity. Second Proof. The resultmay alsobeobtained bymeans of theinstantaneous centre, 0.Forthemotion, sofarasthe velocities that enter into thedefinition ofTareconcerned,is oneofrotation about 0.Hence (2) T=i/'ft2 , where /'denotes themoment ofinertia about 0.Now, /'=I+Mh\ where histhedistance from tothecentre ofgravity, and (3) V=Aft. Onsubstituting thisvalue of/'in(2)andthenmaking u$eof(3), Ttakes ontheform(1),andthiscompletes theproof. Generalization. Themost general rigid bodies withwhich we areconcerned aremade upofparticles andmaterial distribu- tions spread outcontinuously along curves, over surfaces, and throughout regions ofspace. When such abody rotates about an.axis, thekinetic energy isdefined byEquation (1).We canstate theresult intheform : The kinetic energy ofanyrigid material system which isrotating aboutanaxis,isgiven bytheformula: T 168 MECHANICS Remark. Theformula holds even forthemost general case of motion ofany rigid distribution ofmatter inspace. For, such motion ishelical,i.e.duetothecomposition oftwovector fields ofvelocity, i)afield corresponding torotation about anaxis; andii)afield oftranslation along that axis;cf .12below. EXERCISES 1.Aball rollsdown arough plane without slipping. Deter- mine thekinetic energyinterms ofthevelocity ofitscentre. 2.Aladder slidesdown awall, thelower end sliding onthe floor. Find thekinetic energy interms oftheangular velocity. 3.Auniform lamina intheform ofanellipseisrotating in itsplane about afocus. Compute thekinetic energy. 4.Ahomogeneous cube isrotating about oneedge. Determine thekinetic energy. 8.Motion ofSpace withOne Point Fixed. Consider any motion ofrigid space, onepoint, 0,being fixed.Weshallshow that there isaninstantaneous axis,i.e.alinethrough 0,the velocity ofeach point ofwhich is0;andthat thevelocities of allthepoints ofthemoving space, considered atanarbitrary instant, form avector fieldwhich coincides with thevector field arising from thepermanent rotation ofspace about this axis. Wegive firstageometrical proof which appeals strongly to the intuition. The refinements which acritical examination ofthedetails calls forarebest given through anew proof by vector methods. LetQbeapoint ofthefixed space, distinct from O. Ifits velocityis0,then thevelocity ofevery point oftheindefinite right linethrough andQis0,since avariable rightline isevi- dently atrest iftwoofitspoints areatrest. If,ontheother hand,Qismoving, passasphere, with centre at0,through Qandconsider the field ofvector velocities cor- responding tothepoints ofthissphere. Thevectors areevidently alltangent tothesphere, andtheyvary continuously, together with their first derivatives, forwearenotconcerned with dis- continuous motions. Pass agreat circle, (7,through Qperpendicular tothevector velocity ofQ.LetPbeapoint ofCnear Q.Then thevector KINEMATICS INTWODIMENSIONS 169 velocity ofPwillalsobeatright angles totheplane ofCandon thesame side ofCasthevector atQ.For, since thevector velocity ofQisatright angles tothechord QP,thevector velocity ofPmust lieintheplane through Pperpendicular toQP.But italso liesinthetangent plane tothesphere atP.Andnow I say,there,must beapointAofC(and hence twopoints) whose velocity is0.For, otherwise,allthevectors that represent the velocities ofthepoints ofCwould bedirected toward thesame sideofC.Inparticular, then, thepoint Q'diametrically opposite Qwould have such avector velocity. But thatwould mean that themid-point ofthediameter Q'Q,i.e.thecentre ofthe sphere,isnotatrest.From thiscontradiction follows thetruth oftheassertion thatthere isapointAofCwhich isatrest. Hence thewhole indefinite linethrough andAisatrest,andtheexist- ence ofaninstantaneous axis, 7,isestablished. Rotation about theInstantaneous Axis. Itremains toprove that thevector field oftheactual velocities coincides with the field ofthevector velocities due toarotation about I.Con- sider anarbitrary point, P,noton/.ThenPcannot beat rest, unless allspaceisatrest. For,ifthree points, notinaline, ofmoving space areatrest,allpoints must beatrest. Pass aplane, M,through Pandthe axis. Then thevector velocity ofPmust beperpendicular toM.For letQbeanypoint of/. SinceQisatrest, thevector velocityofPmust lieinaplane through Pperpendicular toQP. Consider next thecircle, C,through Pwith/asitsaxis. The vector velocity ofPistangent toC.For itisperpendicular to anylinejoining Pwith apoint of/.Moreover, thevector velocities ofallpoints ofCareofthesame length. Forother- wisetwopointsofCwould beapproaching each other, orreced- ingfrom each other.* Lastly, themagnitude ofthevector velocity ofPispropor- tional toitsdistance from /.LetMbetheplane determined by7andP.Consider two points,P1andP2,inMbutnoton 7,distant A,andh2respectively from 7.Letuhlbethemagni- tude ofthevector velocity ofPt.Then coA2isthemagnitude of thevector velocity ofP2.Forotherwise P2would issue from therigid planeM.f Thiscompletes theproof. *Exercise 4below. tExerciso 5 170 MECHANICS EXERCISES 1.Givearigorous analytic proof that iftwopoints ofamoving straight lineareatrest,every point oftheline isatrest. 2.ApointQismovinginanymanner, andasecond point, P, issomoving that, atagiven instant,itisneither approaching Q norreceding from Q.Give arigorous analytic proof that the vector velocity ofPisorthogonal tothe lineQP,ifthevector velocity ofQisorthogonal tothat line. 3.Ifthree points ofspace areatrest,and ifthese points do not lieonaline,allspaceisatrest. Prove rigorously analyt- ically. 4.Proveanalytically thestatement ofthetextwhich refers tothisExercise. 5.Thesame forthisstatement. 9.Vector Angular Velocity. Letspace rotate asarigidbody about afixed axis, L,with angular velocityo>.LetPbean arbitrary point fixed inthemoving space. Then thevelocity ofPwillberepresented by avector vperpendicular totheplane deter- mined byPandthe lineL,and oflength hw,where histhedistance ofPfrom L. Let beanypointofL.Layofffrom alongLavector oflength wanddenote this vector by (co). Letrbethevector drawn fromOtoP.Then thevector velocity vofPisrepresented bythe vector product of(w)andr:FIG.95 (1) Xr; cf.Appendix A. Letasystem ofCartesian axes(or,y,z)beassumed with as origin, and leti,j,kbeunitvectors along these axes. Write (2) Then (3)co2k. iJk xyz KINEMATICS INTWODIMENSIONS 171 Thecomponents ofvalong theaxesarethusseen tobe : vx= Vy=XUg ZO) Vg=(4) Composition ofAngular Velocities. Consider tworotations about axeswhich pass through 0.Letthem berepresented bythe vectors(o>)and (a/)-Anarbitrary pointPofspace hasavector velocity vgivenby(1): v=()Xr, duetothefirst rotation, andavector velocityv' : V=(')Xr, duetothesecond rotation. Letthese vectors, vandv',beadded. Then athird vector field results oneinwhich tothepointPisassigned thevector v+v'. Itisnotobvious that this third vector field canbe realized byamotion ofrigid space far less, then, that itis precisely the field ofvelocities duetotheangular velocity repre- sented bythevector (5) (0)=()+(0- That this isinfacttrue that istheLaw oftheComposition of AngularVelocities. Theproofisimmediate. Wehave : (6) v+V=()Xr+(')xr. Now, thevector, orouter, productisdistributive : (7) {()+(')}Xr=()Xr+(')Xr. Hence (8) v+v'={()+(a/)}Xr=(12)Xr, andwearethrough. The result canbeformulated asthe fol- lowing theorem. THEOREM. Angularvelocities can becompounded bytheLaw ofVector Addition. EXERCISE Prove thelawofcomposition forangular velocities bymeans ofEquations (4). 172 MECHANICS 10.Moving Axes. Proof oftheTheorem of 8.Letspace bemoving asarigidbody withonepoint, 0,fixed. Leti,j,k bethree mutually orthogonal unitvectors drawn from andfixed inspace, and leta,0,7beasecond setofsuch vectors fixed in thebody. Thescheme oftheir direction cosines shall bethe following: aj87 rjf (1) Thus'l '2 a= with similar expressions for/3,7,where thedirection cosines areanyfunctions ofthetime, t,continuous with their first(and forlater purposes their second oreventhird) derivatives, and satisfying thefamiliar identities;cf.Appendix A.Observe that (2)187= aa=0,etc. 07+7/3=0,etc. Wearenow inaposition toprove analytically theexistence of aninstantaneous axis. LetPbeanarbitrary point fixed inthe body, and letrbethevector drawn from thefixed point toP. Then (3) r=fa+vp+f7. SincePisfixed inthebody, ,77,fareconstant with respect tothetime,andso (4) t=$a+40+fy. Anecessary and sufficient condition thatPbeatrestis,that theprojections offonthree non-complanar axes allvanish. Hence, inparticular, thecondition thatPbeatrestcanbeex- pressed intheform : (5) at=0, fit=0, 7*=0. Applying thiscondition tothevector(4),wefindthethree ordinary equations: KINEMATICS INTWODIMENSIONS 173 +fay= (6) tfa+ f07= =0 Let (7) a=yp, b=ay, c=pa. From(2)itfollows that a=pj y b=7, c=aft. Equations (6)arcnow seen toadmit theparticular solution: =a, 77=b, f=c. These cannot allbeunless thebodyisatrest, since thevanishing oftheabove scalar products would mean that pa=0, ya= ; andofcourse aa=0.Thus thevector awould beatrest,and likewise, each oftheother vectors, Pand 7. Thegeneral solution oftheequations (6)isgivenbytheequa- tions : (8)=Xa, ij=X6, f=Xc,-oo<X<oo . These points, andthese only, areatrest. They form theinstan- taneous axis, and itremains toshow that the*latter deserves itsname. Instantaneous Axis. Letavector(co)bedefined asfollows: (9)o>=7/3, a,=ay, co$=fta ; (10) (CO)=C00!+^0+W^7. Then(co)iscollinear with theinstantaneous axis,whose equa- tions (8)cannowbewritten intheform : (11) 1=JL=L. CO^ COr, CO^ Wehave seen that thevector velocity vofanarbitrary point fixed inthebodyisgiven by(4).Thecomponents ofvalong the(, 77,f)-axes canbewritten intheform : V{=at=%aa+yap+fay =yt=%ya+rjyp+yy 174 MECHANICS Hence ff=f From (12) itfollows that (13) V=COfW, CO^=(CO)XT, r andsoweseethattheactual vector velocity vofPisthesame asthevector velocity whichPwould have ifrigid space were rotating about theinstantaneous axiswithangular velocityco. Thus theactual field ofvector velocities ofthepointsPcoin- cides with thefield ofvector velocities duetorotation about the instantaneous axis represented bythevector angular velocity (co),andtheproof iscomplete. 11.Space Centrode andBody Centrode. The locus ofthe instantaneous axis infixed spaceiscalled thespace centrode, and itslocus inthemoving space, thebody centrode. The actual motion consists oftherolling oftheonecone (thebody centrode) without slipping ontheother cone (thespace centrode). Toprove thisstatement consider thepath traced outbya specified point intheinstantaneous axis. Take, forinstance, theterminal point ofthevector(co),theinitial point being at0. Thelocus ofthispointisacertain curveConthespace centrode : (co)=coxi+coj+cosk, andacertain curveTofthebody centrode : Itissufficient toshow that these curves aretangent andthat corresponding arcs areequal. This will surely bethecase if d(u>)/dt forCisequal tod(co)/cft forF.Now, the firstvector hasthevalue Thevalue ofthesecond vector is : +a*,/?+co$ KINEMATICS INTWODIMENSIONS 175 The last linevanishes because itrepresents thevelocityof thepoint (o>)fixed inthebody, thispoint lying ontheinstan- taneous axis. The first line isthevector(co). This completes theproof. EXERCISE Treat themotion oftheplane byanalogous vector methods. Let _ bethevector drawn from thefixed tothemoving origin, and letp,a-beunitvectors drawn along thepositive axes ofandrj. Letrbethevector from thefixed origin toanarbitrary point P. Then r=f+p+r?<7. Thevector velocity inspace ofapointPfixed intheplaneis given bythevector equation: *=f+fP+^- Theinstantaneous centre isgivenbysetting t=0. Ontheother hand, p=e", a=C+D'. Thecomplete treatment cannowbeworked outwithout diffi- culty. 12.Motion ofSpace. General Case. Let rigid space be moving inanymanner, subject totheordinary assumptions about continuity. Reduce apointAtorestbyimpressing on allspace amotion oftranslation whose vector isequal andoppo- sitetothevector velocity ofA.Thevector field ofthevelocities inthe original motion iscompounded bytheparallelogram lawofvector addition outofthetwovector fieldsi)oftranslation andif)ofrotation about theinstantaneous axis, 7. Letthevector that represents thetranslation beresolved into two vectors, one, T,collinear with7,the other, A,atright angles to7.Thevelocity~H ofanypoint, P,distant hfrom theaxis, is, inthecase ofpure rotation, hu;itsdirec- tion isatright angles totheplane through Pandtheaxis,and its sense isadefinite oneofthetwopossible senses. Hence itisseen 176 MECHANICS that itispossible tofindapoint, B,whose vector velocity dueto therotation isequal andopposite tothevector velocity A.(Draw alinefrom apoint oftheaxis, perpendicular to/andA,and measure offonit,intheproper direction, adistance h=A/u.) AllpointsinthelineLthrough Bparallel to/will alsobeat rest. Itthusappears that theoriginal motion isoneofrotation aboutLcompounded bythelawofvector addition with amotion oftranslation parallel toLtiadrepresented bythevector T. This vector fieldis,ingeneral, thesame asthat ofthevector velocities ofthepointsofanutwhich moves along afixedmachine screw (orofthepoints ofamachine screw which moves through afixed nut). Thetwoexceptional cases arethose ofrotation, corresponding toapitch ofthethreads, and translation, the limiting case, asthepitchbecomes infinite. 13.TheRuled Surfaces. Wehave seen in 12that the vector field ofvelocities,inthegeneral case ofthemotion ofrigid space,isthesum oftwovector fields one, rotation about an axis,L;theother, translation parallel toL.The locus ofLin spaceisaruled surface S,thespace centrode, andthelocus ofL inthemoving spaceisalsoaruled surface, S,thebody centrode. From analogy with therolling cones weshould anticipate Jhe theorem governing thepresent case. THEOREM. ThesurfaceistangenttoSalong L,and itrolls and slides onS. Anintuitional proof canbegiven asfollows. First ofall,it isclearfrom thevery definition ofLthatSslides onSalong L. Soitisnecessary toprove only thetangency ofthetwosurfaces. LetLbethelineLattime t= ,and letPbeapoint ofL . Pass aplane through Pperpendicular toL ,cutting Sinthe curve Cjand letPbethepoint inwhichLattime t=tQ+At cuts C.LetQbethepoint fixed inS,which willcoincide with Pattime tQ+A.Thevector velocity ofQattheinstant t thasacomponent, c,parallel toLandacomponent huatright angles totheplane through LandQ.Obviously hisinfinitesimal with AJ.IntimeMthepointQwill, then, have been displaced, save astoaninfinitesimal ofhigher order, parallel toLbya distance cAt.But itwillhave reached P. Theproofisnow clear. Theplane through LandQmakes aninfinitesimal angle with thetangent plane toSatPQbecause KINEMATICS INTWODIMENSIONS 177 itcontains apointQofSinfinitely near toP,butnotonL . Theplane through LandPmakes aninfinitesimal angle with thetangent plane toSatPQbecause itcontains apointPofS infinitely near toP,butnotonZ/ .And thesetwoplanes make aninfinitesimal angle with each other, because whenQisdis- placed parallel toLbyadistance cA, itsdistance fromPisan infinitesimal ofhigher order than thedistance ofPfromP . Instead ofdeveloping thedetails needed tomake theintuitive proof rigorous, wewilltreat thewhole question byvector methods. First, however, adigression onrelative velocities. 14.Relative Velocities. LetapointPmove inanymanner inspace, and letitsmotion bereferred toasystem ofmoving axes. Consider, first, thecase that themoving axeshave afixed origin, 0.Letasystem ofaxes (x,y,z),fixed inspace, with origin atbechosen;letthemoving axesbedenoted by(, 77,f), andreferred tothefixed axesbythescheme ofdirection cosines of 10.Letrbethevector drawn from toP : 0)r=f+4/9+[7. Then or (3) v=vr+v, where theterms ontheright have thefollowing meanings. The vector, ,.. d,drj_ ,df <4> ^-dr+s'-1-*" represents thevelocity ofPrelative tothemoving axes;i.e. what itsabsolute velocity would beifthe(, t;,f)-axes were fixed andPmoved relative tothem justasitactually doesmove. Secondly, thevector (5)ve=**+it+r7 represents thevelocity inspace ofthat point fixed inthebody, which attheinstant tcoincides with P.Tosaythesame thing inother words :Letusconsider thepointPatanarbitrary instant oftime,t=t.LetQbethepoint fixed inthebody, 178 MECHANICS which atthisoneinstant coincides withP.Then veisthevector velocity ofQ.Itisthe vitesse d'entrainement, thevelocity with which thepointQisbeing transported bythebody attheinstant t. The analytic expressionforveweknow allabout. Invector form itis : (6)ve=()Xr or (7) Itscomponents along theaxes,ifwewrite v'=vejare : (8) Thuswehave asthefinal solution ofourproblem this: The components ofthevector velocity ofPalongtheaxes of%,ry,fare : (9)=+ General Case. Lettheaxes of(x,y,z)befixed inspace. Let(, TJ,f)bethemoving axes,whoseorigin, 0',hasthecoor- dinates (XQ,i/o,2).Then (10)r=r+r'. Hence (11) v=v+v'. Here, FIG.97 (12)dxQ. and v'isgiven by(9).The v'of(11) is,ofcourse, nottheV of(8). KINEMATICS INTWODIMENSIONS EXERCISE179 Denoting thecomponentsofvalong the(,17,f)-axes by >v*yvl>show thatthecomponents ofvalong these axes are : (13) Here,dy 'dt+^~ 4+ =7*0. 16.Proof oftheTheorem of 12.Letasystem ofCartesian axes fixed inspace, (x,y,2),with origin inbeassumed. Let O' :(x ,y^ZQ)beapoint fixed inthebody, themotion of0' beingknown : (1) *o=/(0, 2/o=v(0, *o=lKO. Finally,letPbeanypoint fixed inthebody. Then (2)r=rc+r', cf. 14,(10)with thespecialization thathere (3)dt Then thecomponents oftheabsolute velocity ofP(i.e. itsvelocity infixed space) along theaxes of(,77,f)aregivenbytheformulas of 14,13 : (4) Wecanformulate theproblem asfollows :Tofindapoint ^(i> i/Dfi)fixed inthemoving space whose absolute vector velocityiscollinear with thevector(o>),oris : (5) *,=(<>). Here, (co)isthevector angular velocity ofthemoving space, whose rotation isdefined bythedirection cosines of 10. 180 MECHANICS Byvirtue of(4)thevector equation three ordinary equations: (6)(5)isequivalent tothe Since (7)-=0, afurther necessary condition is : (8) utat+co^f Wecandispose atonce ofthecase o>= ;forthen thespace inwhich 0'isatrest,isstationary, andsothemotion ofthegiven spaceistranslation (unlessitbeatrest). Thus alllines parallel tothevector that represents thetranslation areaxes such as weseek. If o)^0,weobtain from (8)aunique determination ofk. Onsubstituting thisvalue in(6),twoofthese equations, suitably chosen, determine uniquely two ofthethreeunknownslfrjl}ft aslinear functions ofthethird, andthen theremaining equation (6)istruebecause of(8). i= i> i7i=61, fi=ci beaparticular solution of(6),then anarbitrary solution, i'JuTi>willsatisfy theequations: (1- fli)+(fi-ciK= -(fi-c^cu*= Hence (9) andthus &,T^,ftisseen tobeanypoint ofthe linethrough (a,, bi,Cj)collinear with(<*>). This linewedefine asL.These con- ditions aresufficient aswellasnecessary. KINEMATICS INTWODIMENSIONS 181 Thelocus ofLinspaceistheruled surface S;itslocus inthe body (i.e.themoving space)isthesurface S.These surfaces have the lineLincommon. Wewish toshow that they are tangent along L,andthatSslides overSinthedirection ofL. The lastfact isclearfrom thedefinition ofL. Thepoint (&, rjlfft)isnotuniquely determined bythetime, butmaybeanypoint ofL.Wewill, forourpurposes, select it asfollows. Let Z/beaparticular L,and letPbeanarbitrary point ofL,once chosen andthen held fast. Pass aplaneM through Porthogonal toL .Then (xlfyltzjshallbetheinter- section ofthevariable lineLwithM,and itslocus shallbedenoted byC.Thepoint (ft,ylyfjshall bethepoint ofSwhich coin- cideswith (xu yi,2i)attime t=t.Itslocus inSshallbedenoted byF.Thiscurve canberepresentedintheform : F: ti=F(t), 77!=$(0, fi=*(0- Itstangent vector atanarbitrary pointis ka+*?i*+#, dt+dt1*+ dt7' provided thisvector 7*0. Toshow thattwosurfaces which intersect atapointPare tangentitissufficient toshowi)thattheyhaveacommon tangent vector,t;andii)thatatangent vectortjtotheonesurface and atangent vector t2totheother surface, neither collincar witht, arecomplanar witht,allthree vectors, emanating fromP . The surfaces Sand2satisfy i)because they areboth tangent toL.Secondly, consider thevector^drawn from tothepoint ofintersection ofCandFattime t t.Itsderivative isavector tangent toC,providedit5^0. Ontheother hand, consider thepoint (|t,77^J\)ofF,forwhich t=t.Letthevector drawn from 0'tothispoint bedenoted byr[.Then *i=TO+i{, where risgivenby(1),and 182 MECHANICS Hence^-^tt-L*'!* j-*j,y dt" dta^'Mft^dty 4-fo+^d+7/^+^7. This last line isprecisely thevector velocityofthat point fixed inS,which attheinstant inquestion,t= t,coincides with (x\>V\yzi)-This vector, t,letuscallit,liesalongLbecause of (5), unless itbe0. Theother vector ontheright of(11)isthevector (10);i.e.. avector t2tangent toTat(, rjlyf,)or(xl9yltzj.Equation (11)thussaysthat ti=t,+1. Now, thevector drjdt=tx^ willnot liealong L.Hence t.2 willnot, either. Consequently Conditionii)issatisfied, aridthe surfaces SandFaretangent along L.Thecase t= isincluded; itdoesnotlead toanexception. EXERCISE Letacylinder ofrevolution rolland slideonasecond cylinder which isfixed, the first cylinder always being tangent along an element, and there being noslipping oblique totheelement. Choose thepoint (x ,i/ ,z)intheaxis ofthemoving cylinder, anddiscuss thewhole problem bythemethod ofthisparagraph. 16.Lissajou's Curves. Inonedimension, orwithonedegree offreedom, themost important periodic motion isSimple Har- monic Motion. Itcanberepresented analyticallyintheform : (1) x=acos(nt+7), where (2) T=- 'n istheperiod, where aistheamplitude, andwhere 7isdetermined bythephase. Intwodimensions, orwithtwodegrees offreedom, animpor- tant case ofoscillatory motion about afixed pointisthat in which theprojectionsofthemoving point ontwo fixed axes KINEMATICS INTWODIMENSIONS 183 atright angles toeach other, execute, eachbyitself, simple har- monic motion : Ix=acos(nt+7) \u) 1 Iy=bcos(mi+e) Itispossible togeneralize atonce tondimensions : (4) xk=akcos(nkt+yk),fc= Z, ,n. Letusstudyfirst thetwo-dimensional rase, beginning with some simple examples. Wemay set7= if,asusually hap- pens, theinstant from which thetime ismeasured isunimpor- tant. Example1:m=n.Dynamically, this casecanborealized approximately bythesmall oscillations ofaspherical pendulum. Let7=0, <p=nt. Then mt+e= <f>+, (x=acos <p y=Acos<pBsin<p A=bcose,B=bsin c. Assume that neither anorbvanishes, since otherwise weshould bethrown backonrightlinemotion along oneoftheaxes.We willtakea>0,b>0. Ingeneral, B^0.Thepath ofthemoving pointisthen anellipse with itscentre attheorigin. For, (6)cos<p= ^,sin<p=-^x-^y. Onsquaring andadding wefind : (7) B2x2+(Ax-ay)2=a2B2 , andthisequation represents acentral conic which doesnotreach toinfinity,i.e.anellipse. 184 MECHANICS Theaxescanbefound bythemethods ofanalytic geometry, orcomputed directly bymaking thefunction COS2^__2AB cos<psin^+B*sin2^ Wehave omitted thespecial case :B=0.Here,e=or TT,andthemotion isrectilinear, along theline : Inallcases, thepathisconfined within therectangle: x=a, y=b, and continually touches allfour sides, sometimes being a diagonal, but,ingeneral, anellipse inscribed intherectangle. Example2.m=n+h,where hissmall. Ifwewrite the equationsintheform : x=acosnt (9) y=bcos(n+A/+e) then,fortheduration oftimeT2w/n, ht+ isnearly constant, andthepathis FIG.98 nearly anellipse which, however, does notquiteclose. Andnow, inthenext in- terval oftime, thepath againwillbeanear-ellipse, but ina slightly different orientation itspoints oftangency with the circumscribing rectangle willbeslightly advanced orretarded, depending onwhether hispositive ornegative. Thus asuccession ofnear-ellipseswillbedescribed,allinscribed inthesame fixed rectangle x=a,y=6.Themotion can berealized approximately experimentally asfollows. Blackburn's Pendulum. Bythis ismeant themechanical sys- temthat consists ofanordinary pendulum, theupper end of *Their directions aredetermined, ineither way,bytheformula : cos2*= or cos2y . 62sin2e 2abcos c where ydenotes theangle from theaxis ofxtoanaxisoftheconic. Thelength* oftheaxes arefound tobe : where A8=a4-f2a*&* cos2+b*. KINEMATICS INTWODIMENSIONS 185 which ismade fastatthemid-point ofaninextensible string whose twoends arefastened atthesame level. When thebob oscillates inthevertical plane through thesupports, thesecond string remains atrest,andwehave simple pendulum motion, thelength ofthependulum being Z,thelength ofthe firststring. Secondly, letthebob oscillate inavertical plane atright angles tothelinethrough thepoints ofsupport, andmid-way between these. Again, wehave simple pendulum motion;butthelength isnow I'=I+d,where ddenotes thesaginthesecond string. Forsmall oscillations, thecoordinates ofthebob willevidently begiven approximately byEquations (3),andbysuitably choos- ingIand d,wecanrealize anarbitrary choice ofmand n. TheSand Tunnel.* Ifthebob ofthependulumisatunnel ofsmall opening,filled with finesand, thesand, asitissues from thetunnel, willtrace outacurve onthefloorwhich shows ad- mirably thewhole phenomenonofthe Lissajou's Curves. In particular,ifthesecond stringisdrawn astaut asisfeasible, sothatdissmall, thetwoperiodswillbenearly, butnotquite, equal ;and itispossible toobserve thenear-ellipses steadily advancing, flashing through near-right lines (the diagonals of thefixed rectangle). Example3.m 2n.Begin with thecase7=0*e=0,andset <pnt: (10) x=acos<p, y=bcos2<p. Hence (11) */=I**-6 andthecurve isanarcofaparabola, passing through thevertices (a,6),(a, 6)ofthecircumscribing rectangle andtangent to theopposite side atthemid-point. Thesandpendulum may bereleased from restatthepoint (a,6),and itthen traces re- peatedly theparabolicarc. Inthegeneral case, (12) x=acos<p, y=Acos2<pBsin2<p ; A=bcos6,B=bsin e. *Thisexperiment should beshown inthecourse. Itisnotnecessary tohave aphysical laboratory. Atunnel canbebought attheFiveandTen,andstring isstillavailable, even inthisageofcellophane andgummed paper. 186 MECHANICS When eissmall, thecurve runs along near totheparabola;cf. Fig. 100. Itissymmetricintheaxisofy,since^and<p' <p+TT givex'=x,y'=y.Itistangent once toeach ofthesides x=a,x=aofthecircumscribing rectangle, and twice toeach oftheother two. sides. When ehasincreased to?r/2, A=and (13) x=acos<p, y= bsin2p or FIG.99This curve isobtained atoncebyaffine transformations from thecurve (15) yz which isreadily plotted. When chasreached thevalueTT,wehave again anarcofa parabola theformer arc,turned upside down. As econtinues toincrease, thenewcurves arethemirrored images oftheoldin theaxis ofx,for e'=e+TTreverses thesigns ofAandB.All these curves except thearcs ofparabolas arequartics, inscribed in thefixed rectangle, andhaving symmetryintheaxisofy. Example4.m=2n+h,where hissmall. Here, fx=acosnt (16)_ Iy=bcos[2n*+ht+e] and forasingle excursion, Misnearly0. Thus thenewcurve runs along close toan oldcurve forasuitable fixedc,butastime elapses, thesuitable eadvances, too. FIG.100 Thestudent canreadily trace these curves with thesand tunnel. Ifhedoes hisbest tomake d= J,there willbeenough discrepancy toprovide forasmall h. 17.Continuation. TheGeneral Case. TheCommensurable Case. Periodicity. Letmandnbecommensurable, where pandqarenatural numbers prime toeach other. Then n=ap,m=aq. KINEMATICS INTWODIMENSIONS 187 (x=a< ,y=b(Let<p=at,7=0.Then =acosp<p, y=6cos(?+e). These functions areperiodic with theprimitive periods 2ir/p and27r/<7, and evidently have thecommon period2ir.The smallest positive value ofcoforwhich acosp(<p+co)=acosp^> 6cos{(/(p+co)+e}=fecos{</p+e} isco=27T. For,iftoistobeaperiod ofthe firstfunction, then x2?r CO=A P And ifcoistobeaperiod ofthesecond function, then 27T CO=M Hence X/*.-=-, Xg= ,, andthesmallest values ofX,ninnatural numbers which satisfy thisequation arcX=p,n=q. From theperiodicity ofthefunctions itappears thatthecurve isclosed, arid thus, as tincreases, thecurve istraced out re- peatedly. Foranon-specialized value ofe,thecurve istan- gent toeach ofthesides x=a, aofthecircumscribing rec- tangle ptimes, corresponding tothesolutions oftheequations cosp<p 1,1;andqtimes toeach ofthesides y 6,6. Alinex=x', a<x'<a,cuts thecurve in2ppoints ;a liney=y'j b<y'<6,in2qpoints. These curves are allalgebraic, and rational, orunicursal. For,onsetting=tan-J-p,thevariables xandyappear, by deMoivre's Theorem, asrational functions of .The curves are allsymmetricintheaxis ofy. TheIncommensurable Case. Aperiodic.Ifontheother hand n/misincommensurable, thecurve never closes. Itcourses every region contained within therectangle.IfPbeanarbi- trary point oftherectangle, thecurve willnotingeneral pass through P;but itwillcome indefinitely near toPnotmerely once, butinfinitely often;possibly, occasionally passing through P. 188 MECHANICS Theproof canbegiven asfollows. Consider acircle andthe angle <patthecentre. Let<f>==2wabeanangle which is incommensurable with2?r;i.e. letabeirrational. Then the points ofthe circle which correspond to,2,3, (denote thembyPt,P2, )are alldistinct. Hence theymust have at leastonepoint ofcondensation, P.Butfrom thisfollows that every point must beapointofcondensation. For, letPnand Pmbetwo points near P.Then thepoint corresponding to nwmust benear thepoint corresponding to <p=0.Hav- ingthusobtained anarcofarbitrarily small length, wehave but totake multiples ofit,i.e.toconstruct thepoints P*(n-m)i k=1,2,3, ,tocome arbitrarily near toanypoint onthe circumference. Turning now totheequationsofthecurve, let m <p=nt,=a.n Then x=acos<f>, 1 y=bcos(cup+17). Let agx'^a,and let<p=<p'bearoot oftheequation xfacos<p. Thecurve cutsthelinex=x'inthepoints forwhich y=bcos{<*(<?'+2kw)+r?}, 6cos{(- <?'+2kw)+y}. Andnow, since theangles 2kair lead topoints onthecircle which areeverywhere dense, thecorresponding values ofthecosine factor arealsoeverywhere dense between 1and+1. Itisofinterest tostudy themultiple points ofthecurve. These occurwhen t) <f>'=kw+1 ^,I^0; it) ^=)br+^~a'fc?0; provided (19) -n*(1+ka)ir. When theinequality (19) holds, there isaone-to-one cor- respondence between thevalues of<pandthepoints ofthecurve, KINEMATICS INTWODIMENSIONS 189 provided themultiple points (which arealways double points with distinct tangents) arecounted multiply. If,however, (20) 77=(1+kQa)7r, then a<p+ f\=a(<p+ fc7r)+i7r. Set (21)= <f>+kQw. Then theEquations (18)become : x=a!cos <22> I= where a!=aora,andlikewise &'=6or 6.Let ^<oo. Then there isaone-to-one correspondence between thevalues of6aridthepoints onthecurve. Thepointforwhich 9=: x=a', y=&', isanend-point ofthecurve. Itissimple, noother branch going throughit.Thedouble points correspond tothevalues 0'=kw+->0,a where /CTT-->0,I^; a. orwhere -far+->0, k^0. andn-Dimensions. Inthecase ofmotion with three degrees offreedom, theequations canbereduced totheform : (23)xacos <p y=bcos(ay+??) z=ccos($v+f) The casethat a,/3arebothcommensurable canbediscussed as before. The curve closes, themotion isperiodic. When a andftareboth irrational, andtheir ratio isalso irrational,itcan 190 MECHANICS happen that thecurve courses every region, however small, of theparallelepiped: a^xga, 6^2/^6, c^z^c, andhasnomultiple points, thecorrespondence between the points ofthecurve and thevalues of<pwhen oo<<p<oo being one-to-one without exception Whether theformer prop- ertyispresent for allsuch values ofaand0,provided further- more thator,/3,and /3/aarenotconnected byalinear non-homo- geneous equation with integral coefficients, andthat77, J"arenot specialized,Icannot say,thoughIsurmise ittobe.The latter property, however, canbeestablished. Thesame statements hold inthegeneral case, xk=akcos(oLk<p+rik), k=1,-- ,n. If;mayhappen,inadynamical system withndegrees offree- domandcoordinates qlt ,qn,that only asub-set, <ft, , qm,1gm<n,execute aLissajou's motion. Thus aBlack- burn's Pendulum suspendedinamoving elevator willhave its projection onahorizontal plane executing aLissajou's motion, whereas thevertical motion isnotperiodic atall. The lateProfessor Wallace Clement Sabine drew mechanically some very beautiful curves, which areherereproduced inhalf-tone. Istillhave thehalf-tone which Dr.Sabine gave me. SofarasI havebeen abletoascertain, thecurves werenever published. The figures hereshown weremade from lantern slides inpossession of theJefferson Physical Laboratory, and itisthrough thecourtesy of theLaboratory that Ihavebeenenabled toreproduce them here. CHAPTER VI ROTATION 1.Moments ofInertia. Themoment ofinertia ofnparticles, w,-,with respect toanaxis isdefined asthesum : (1) /=i><r<, <=i whore rdenotes thedistance ofwt-from theaxis;cf.Chapter IV, 10. Let beanarbitrary point ofspace, and letCartesian axes with asorigin beassumed. Letthemoments ofinertia about theaxesbedenoted asfollows : (2)A= Theproducts ofinertia aredefined asthesums : (3) -0=5miViz<>E= These definitions areextended intheusualwaybythemethods ofthecalculus tocontinuous distributions. Interms oftheabove sixconstants itispossible toexpress themoment ofinertia about anarbitraryaxisthrough 0.Let thedirection cosines oftheaxisbea,0,yand letP :(x,y,z) beanarbitrary pointinspace. Then r2=p2_<^ or r2=x2+tf+z2-(ax+py+yz)2 . Since op of+P+72=1,FlG - thelastexpression forr2canbewritten intheform : (x2+ 2/2+*2)(2+P+72 )-(ox+fry+yz)*. 191 192 MECHANICS Hence -2yazx- Thus /=a25)m;(^2+z>2 )+25Jmt-(*i2+a*2 )+etc. or: (4) 7=4a2+502+CV-2D07~2#y-2Fa/3. This isthedesired result. Themeaning oftheformula can beillustrated bytheEllipsoid ofInertia. Consider thequadric surface, (5) Ax2+By2+Cz*-2Dyz-2Ezx-2Fxy=1. Itisknown astheEllipsoid ofInertia, and itsuse isasfollows. Letanarbitrarylinethrough with thedirection cosines,0,7 meet thesurface inthepoint (X,Y,Z),and letpbethelength ofthesegment oftheaxisincluded between thecentre ofthe ellipsoid and itssurface. Then X=ap, Y=0p,Z=yp. SinceX,Y,Zsatisfy (5),itfollows that (6) p2(^la2+Bp+C72-2D/37-2#ya-2Fa/3)=1. Oncombining (4)and (6)wefind : (7) P2/=1, /=^ and/isseen tobethesquare ofthereciprocal ofp.From this propertyitappears that theEllipsoid ofInertia isinvariant of thechoice ofthecoordinate axes. Ifallnparticles mzlieonaline,Equation (5)nolonger repre- sents anellipsoid. Lettheaxis ofzbetaken along this line. ThenA=Band alltheother coefficients vanish. Thus (4)be- comes7=4('+ *). Here,A^exceptinthesingle casethat i=1andmlliesat the origin. Inallcases but this one, thequadric surface (5) still exists, being thecylinder ofrevolution (8) A(*2+2/2 )=1, andthetheorem embodied in(7)isstilltrue. ROTATION 193 Suppose, conversely, that(5)fails torepresent atrue(i.e. non-degenerate) ellipsoid. Ifallthe coefficients A,B, , Fare0,thesystem ofparticles evidently reduces toasingle particle situated at0.Inallother cases, (5)represents acentral quadric surface, S. Ifthis isnotatrueellipsoid, then there is aline, L,which meets 8atinfinity;i.e.which doesnotmeetS inanyproper point, but issuch thatasuitably chosen variable lineL'always meets S,thepointsofintersection recedingin- definitely asLfapproaches L.Themoment ofinertia about L'isgivenby(7)andapproaches asL'approaches L.Hence themoment ofinertia aboutLis0.But ifthemoment ofinertia ofasystem ofparticles about agiven axis is0,itisobvious that alltheparticles must lieonthis axis. Wesee,then, that (5)represents atrueellipsoidinallcases except theone inwhich theparticleslieonaline,andthat (7) holds inallthelatter cases, too,except theoneinwhich the system reduces toasingle particle situated at0. Parallel Axes.We recall, finally, thetheorem relative topar- allelaxes;Chapter IV, 10 : THEOREM. Themoment ofinertia, I,about any axis, L,is equaltothemoment ofinertia, 7,about aparallel axis,L,through thecentre ofgravity, plus thetotalmass times thesquare ofthedis- tance, h,between theaxes: I=J+MW. EXERCISE Show that themoment ofinertia about any line, L,inspace isgivenbytheformula : 7={A+M(y\+*?)}a2+{B+M(z\+x$\ /32 +{C+M(x\+tf)}72-2(Z>+MVlzJ0y -2(E+MZ.X,) ya-2(F+Mx.y,) aft where theorigin ofcoordinates isatthecentre ofgravity and x\>y\jz\arc^iecoordinates ofanypointonL,andwhere a,fty arethedirection cosines ofL. Foranarbitrary system ofaxes, replace xltylyzlrespectively by *i~*> y\-y> *i- *> 194 MECHANICS where #, ?/,2arethecoordinates ofthecentre ofgravity, and xuVnz\arethecoordinates ofanypoint onL allreferred to thenew axes. 2.Principal Axes ofaCentral Quadric. Letaquadric surface begivenbytheequation: (1) Ax2+By*+Cz*+2/)7/*+2Ezx+2Fxy=1, where thecoefficients arearbitiary subject tothesole restriction thatthey shall not allvanish. Theproblem is,sotorotate the axesthatthenewequation contains only thesquare terms. Let (2)F(x, y,z)=Ax2+Eif+Cz*+2Dyz+2Ezx+2Fxy, (3) *(*,P,s)=x*+y*+z*. Consider thevalue ofthefunction F(x, y,z)onthesurface ofthesphere (4)'x*+ ?/2+z2=a2 , or *(x, y,z)=a2 . Since F(x, y,z)iscontinuous andthesphereisaclosed surface, thefunction must attain amaximum value there, and alsoa minimum. Lettheaxesbesorotated that themaximum value isassumed ontheaxis ofz}inthepoint (0,0,f),where f=a.Wethink ofEquations (2)and (3)nowasreferring tothenew axes. Inaccordance with theMethod ofLagrange*weform the function F+\$, theindependent variables being x,y,z,with Xasaparameter ; andwethen soteach ofthofirst partial derivatives equal to : (5) b\+X*,=0,F2+X$2=0, t\+X$3=0. These three equations, combined with (4),form anecessary condition onthefourunknowns x,y>ZjXforamaximum : 'Ax+Fy+Ez+\x= (6) Fx+By+Dz+\y= Ex+Dy+Cz+\z= Butweknow thatthepoint (0,0,f),f^0,yields amaximum. HenceD=0,E=0, *Lagrange's Multipliers, Advanced Calculus, Chapter VII, 5. ROTATION 195 andthenewF(x, y,z)hastheform : .F(x, y,z)=Ax*+2Fxy+By*+Cz\ Ifthecoefficient oftheterm inxydoesnotvanish,itcanbe made todosobyasuitable rotation oftheaxesabout theaxis ofz;cf.Analytic Geometry, Chap. XII,2.Thus F(x,y,z)is reduced finally byatmosttworotations (thesemaybecombined intoasingle rotation, butthat isunessential) tothedesired form : (7) F(x, y,z)=Ax*+By*+Cz\ Here, A,B,Cmaybeanythree numbers, positive, negative, or0, except thatwehave excluded astrivial thecase that allthree vanish. The original equation (1)willobviously repre- sentanellipsoidifandonlyifthenew coefficients A,B,Cin(7) areallpositive. Wehave thus established thefollowing theorem. THEOREM. An arbitrary homogeneous quadratic function F(x, y,z)canbereduced byasuitable rotation oftheaxes ofcoordi- nates toasum ofsquares. Thenewcoefficients ofxr ,y',z1may beanynumbers, positive, negative, or0. EXERCISE Show, bythemethod ofmathematical induction andLa- grange's Multipliers, thatanarbitrary homogeneous quadratic function innvariables, canbereduced toasum ofsquares byasuitable rotation. Byarotation ismeant alinear transformation : x{=anx1++ainxn x'n=ani#i++annxn such that, foranytwocorresponding points (xlt ,xn)and (x( 9 ,x'n),therelation holds : r'2_L ...4.r'2r2J_ ...4.r2 *l\ \*n~*\\ \An) andthedeterminant ofthetransformation, A=San ann, which necessarily hasthevalue1,isequal to+1. 196 MECHANICS Itiseasy towritedown theconditions thatmust holdbetween thecoefficients ofthetransformation, butthese conditions are notneeded forourpresent purpose. Obviously, theresult of anytworotations isarotation. 3.Continuation. Determination oftheAxes. Inthe fore- going paragraph wehave been content toshow theexistence of atleast onerotation, whereby thegiven function isreduced toasum ofsquares. Wehave notcomputed thevalues ofthe new coefficients, norhavewedetermined thelengths ofthoaxes. Now, anyrotation oftheaxes carries thesecond function, $,over into itself : *'(*', */',z')s*(*', ',z')=*(x, y,z). Thefunction : fl=F+\3>, goesover intothefunction : F'+X*', where Xremains unchanged. Now, thecondition : an_ 00~ '~ 'a?~ ' isequivalent tothecondition : since thedeterminant ofthe linear transformation does not vanish. Hence Equations (6)ofthepreceding paragraph will beofthesameform forthetransformed functions. When F'(x', y',*')=A'x'*+B'y'*+C'z">, theequationfordetermining Xreduces tothefollowing: (A'+X)(B'+X)(C"+X)=0. Thus thethree roots ofthedeterminant ofEquations (6), A+\ FE (8) As FB+\D EDC+ ROTATION 197 arcseen tobethenegatives ofthecoefficients A',B'tC',andso theaxes ofthequadric arefound. Iftheroots ofthedeter- minant(8)aredenoted by\19X2,X3,thelengths ofthesemi- axes are IncaseaX=0,thequadric reduces toacylinder, ormorespe- cially, totwoplanes. Allthree X'swillvanish ifandonlyifthe original F(x, y,z)vanishes identically. When thoXthave once been determined, Equations (6)give theequations oftheaxes ofthequadric. Ingeneral, thethree \iare distinct, andEquations (6)then represent arightline foreachX. 4.Moment ofMomentum. Moment ofaLocalized Vector. LetAbeavector whose initial point, P,isgiven, and let be anypoint ofspace. Letrbethevector drawn from toP. Bythemoment ofAwith respectto ismeant thevector, orouter product:* (1) rXA. Wehavemet thisidea inStatics, where themoment ofaforce F,acting atapoint P,with respecttoapoint wasdefined as thevectorM=rXF. Themoment ofmomentum ofaparticle withrespecttoapointis defined asthevector| (2)ff=rXmv, whore risthevector drawn from thepoint to theparticle, andvisthevector velocity ofthe FIG.102 particle. Themoment ofmomentum ofasystem ofparticles with respect toapointisdefined asthevector n (3)ff=J)rkXmkvk, 1=1 *Cf.Appendix A. fContrary tothousual notation ofwriting vectors inboldface, asa,x, i, etc.,orbyparentheses, as(co), itseems hereexpedient todenote thevector moment ofmomentum bya,thevector momentum byp,andthevector angular velocity byo>. 198 MECHANICS where rkisdrawn from thepoint inquestion tonik,andv^istin vector velocityofm,k. Inthecase ofacontinuous distribution ofmatter theextensioi ofthedefinition ismade intheusualwaybydefinite integrals. InCartesian form ahasthevalue, forasingle particle: (4)J1 y- dy 'dtm dtm dtdx dz dz di dx dz dx theorigin being at0.Andso,forasystemofparticles,th( components ofaalong theaxes are : (5) Rate ofChange of<r.Since (&} *L(^y._ di dt\dt dt itisseenfromEquations (5)that (7)dxk dz dx dzx-y-^~v These equations, invector form, become : (8) ljt=^mkTkXa*' where a^denotes thevector acceleration ofthefc-th particle. ROTATION 199 The result, Equation (8),could have been obtained atonce from(3).Ifwedifferentiate Equation (2),wefind : da^,dv.di^, _=mrx__+w ._*xv . Now, -77=vand vXv=0. at Hence do-.ydv dt=mrX dt=r*a> where adenotes thevector acceleration. Similarly, from (3) wederive(8). 5.TheFundamental Theorem ofMoments. InChap. IV, 3,itwasshown that, inthecase ofanysystem ofparticles ina plane such that theinternal forces between anytwo particles areequal andopposite and liealong thelinothrough theparticles, themoments oftheinternal forces annul each other, andthe equation ofrotation becomes : (1) I;mk(xk^-ykd ji*)=2(XkYk fc= 1 Thetheorem and itsproof canbegeneralized atonce tospace ofthree dimensions. Newton's Second Law ofMotion isex- pressed fortheparticlemkbytheequations d*Xk_v_i_ '<~W~Xk+ J =Yk+XY where F*/denotes theinternal force which isexerted onthe particle w*from theparticle my. Multiplying thethird ofthese equations byy^thesecond byzkandadding, andobserving that themoments oftheinternal force cancel inpairs, since the forcesF/jfcandF*/areequal andopposite andhave thesame line ofaction, the first ofthefollowing three equations isobtained. Theothertwoarededuced inasimilar manner. 200 MECHANICS (2) These equations express theFundamental Theorem ofMo- ments. Invector form itis : (3) S-i>XF, or: ^7=2)(Moments oftheApplied Forces about 0). Equation (3)canbededuced more simply byvector methods. Write Newton's Second Law inthevector form : (4) mkak=F*+2)Fw. Next, form thevector product, (5) mkTkXa*=rfcXF*+2)r*XF*/, andadd. Thesumonthe left isequal tod<r/dtby 4,(8).On theright, thevector moments oftheinternal forces cancel in pairs, andthere remains theright-hand sideof(3). FUNDAMENTAL THEOREM OFMOMENTS. The rate ofchange ofthevector moment ofmomentum ofanysystem ofparticlesis equaltothevector moment oftheapplied forces, providedthat the internal forces between each pair ofparticlesareequalandopposite andinthelinethroughtheparticles: or,inCartesian form tEquations (2). The foregoing result applies tothemost general systemof particles, subject merely tointernal forces ofthevery general nature indicated. Bytheusual physical postulateofcontinuity weextend thetheorem tothecase ofcontinuous distributions ROTATION 201 ofmatter, ortoanymaterial point set.Forexample, oursolar systemisacase inpoint, andwewillspeak ofitindetail in 7. 6.Vector Form fortheMotion oftheCentre ofMass. Let beanarbitrary fixed pointinspace, and letfbethevector drawn from tothecentre ofgravity, (?,ofamaterial system. Let Fi, ,Fnbetheforces that act;i.e.theapplied, orexternal, forces. Then thePrinciple oftheMotion oftheCentre ofMass isexpressed bytheequation: (1) M-., where v=df/dt. Equation (1)ismerely thevector form of Equations A),Chapter IV,1.Itcanbederived byvector methods, byadding Equations (4), 5,andobserving that Mi=2)=1 Letpdenote themomentum, p=Mv. Equation (1)nowtakes ontheform : (2) Thuswehave foranysystem ofparticles, rigid ordeformable, andeven forrigid bodies and fluids, thetwoequations ofmomen- tum : THEEQUATION OFLINEAR MOMENTUM : A)\ THEEQUATION OFMOMENT OFMOMENTUM : 7.TheInvariable LineandPlane. Incasenoexternal forces act, <a-" 202 MECHANICS andthevector crremains constant. The linethrough collinear with <Tiscalled theinvariable linewith respectto0,andaplane perpendicular toit,theinvariable plane withrespectto0. The solarsystemisacase inpoint,ifwemay neglect anyforce thestarsmay exert.Wemay consider theactual distribution ofmatter and velocities, and then, onchoosing afixed point, 0,thecorresponding value ofawillbeconstant. Orwemay replace thesunandeach planet byanequal mass concentrated atitscentre ofgravity, andconsider thissystem. Again, thevector a-corresponding toagiven pointwillbe constant, andobviously nearly equal totheformer a. Letuschoose oneofthese cases arbitrarily anddiscuss itfurther. The vector adepends onthechoice of0.Canwenormalize thischoice? The centre ofmass ofthesolar systemisnotat rest,andso,sinceweareneglecting any force exerted bythe stars*, themomentum ofthesystem, p=Mv,isconstant and 5^0.The direction ofthisvector, porv,doesnotdepend on thechoice of0.Thepoint 0'canbesochosen that a'iscollinear with p. Weshallshow inthenextparagraph, Equation (5),that (2) a=a'+MrXV, where a,a'arereferred to0,0'respectively.Ifaisnotalready collinear withp,let <rberesolved intotwocomponents; one, collinear withp,theother,cr,atright angles. Wewish, then, sotodetermine rthat (3) MrQXv=(7,<TO*0. Since o-andvarcperpendicular toeach other, thiscanbedone. ThepointOfwillbeanypointofalinecollinear with p.This isknown astheinvariable lineofthesolar system. Forafurther discussion,cf.Routh, Rigid Dynamics, Vol.I,p.242. 8.Transformation of <r.Let beapoint fixed inspace, pand letO'beasecond point, moving orfixed. Let Pbetheposition ofaparticle ofthesystem. Then o r CD fr=r'+r; Fia.103 IV=v'+V, where v'expresses thevelocity ofPrelative to0',andVisthe velocity of0' . ROTATION 203 Forasingle particle, themoment ofmomentum with respect to isthevector <r=rXrav=mr'X(v'+v)+mrXv, or (2)<r=mr'Xv;+mrXv+mr'Xv . The firsttermontheright hasthevalue a'r=r'Xwv', ortherelative moment ofmomentum, referred to0'asamoving point. Forasystem ofparticles weinfer that *xv* or (3)<r=<r'r+MrXv+Mi'Xv , where o>istherelative moment ofmomentum referred to0'as amoving point ;visthevelocity ofthecentre ofmass;and f' isthevector drawn from 0'tothecentre ofmass. Thesecond termontheright, MrXv=rXMv, canbeinterpreted asthemoment ofmomentum, relative toO, ofthetotalmomentum, Mv, ofthesystem, thought ofasamass,M,concentrated atO1andmoving with thevelocityv. Thethird term, Mr'XV=f'XMv, isthemoment ofmomentum, relative to0',ofthetotal mass,Mjconcentrated atthecentre ofgravity andmoving with ve- locity V . If,inparticular, 0'betaken at(?,then f'=0,v=v,and (4)<r=<r'r+MrXv, where o>denotes therelative moment ofmomentum, referred toGasamoving point, and MrXv=fXMv isthemoment ofmomentum, Mv, ofthetotal mass, concentrated atGandmoving with thevelocityofG,referred to0. 204 MECHANICS Let cr'denote thevalue ofareferred tothepoint 0'asafixed point;i.e. Xvk. Then </=5Jm^iXvi+]mkriXv or </=v'r+Mr'Xv . Thus Equation (3)goesover into : (5) a=</+MrXv. Thisamounts tosetting v=in(3). 9.Moments about theCentre ofMass. Wehave theFunda- mental EquationofMoments, 5 : (I)' S Andwehave theEquation ofTransformation, 8,(4): (2) a=*'r+MfXv, where o>istherelative moment ofmomentum, referred toGas amoving point. Differentiate this lastequation, observing that since df/dt=viscither orelsecollinear with v.Thuswefind : /ONda_dff'rMd? (6)dt~ Ht+MXTt Ontheother hand, r=r'+f, wherer,faredrawn from;r'from 0.Thus (4) 5)r*XF=gr;XF*+fX5)F*. t k Ic Substituting inEquation (1)thevalues found inEquations (3) and(4),weobtain theresult : ROTATION 205 Recall theEquation oftheMotion oftheCentre ofMass, 6: (6) " From itfollows that MrX^=fX F*. fc Thus these terms cancel in(5)andthere remains : Inthisequationisembodied theresult whichmaybedescribed asthe PRINCIPLE OFMOMENTS WITH RESPECT TOTHECENTRE OF MASS. The rateofchange oftherelative vector moment ofmomen- tum, referredtothecentre ofmassGregarded asamoving point, isequaltothesumofthevector moments oftheapplied forces with respecttoG: EXERCISE Show thatarigidbodyisdynamically equivalent, ingeneral, toapair ofequal masses ontheaxis ofor,asecond paironthe axis ofy,andathird pairontheaxis ofz,each pairbeing situ- ated symmetrically with respect tothe origin, and allsixdis- tances from theorigin being thesame;itbeing assumed thatthe principal axes ofinertia liealong thecoordinate axes. Discuss theexceptional cases. Usetheresults of 12. 10.Moments about anArbitrary Point. Consider themost general transformation, 8,(3): (1) *=<j'T+MrXv+Mr'Xv, anddifferentiate : Ontheother hand, (3) 2r"xF*=2rixF* t I 206 MECHANICS From theEquationofLinear Momentum, 6,(1)follows that (4) MIOX%-2roxF*- G/v^^T Moreover, -jjV-f- V-. For, hence vXv=vXv'+vXv=vXv', and vXv'+v'Xv=0. Substituting, then, intheEquation ofMoment ofMomentum, 6,B): g=2r*XF*, t andreducing, wefind : This equationisgeneral, covering allcases oftaking moments about amoving point 0',relative tothat point. When, however, oneuses theexpression:"taking moments about apoint O'" themeaning ordinarily attached tothese wordsis,thattheequa- tion (6) f=?ri><F* shallbetrue. Hence wemusthave (7)f'X = forevery value of t. Let t=Tbeanarbitrary instant. Let0'beapoint which describes acertain path, (8)r-rftr). Consider thisasthevector roftheforegoing treatment. Then ROTATION 207 Attheinstant t=r, /dv*\ /a2f<A (-5-),.,= (IF;,.; Then Equation (7)istohold forthisvector rattheoneinstant t=T. Thuswehave ingeneral, notasingle curve traced outby0' and(7)considered foravariable point ofthat curve, asinthe case of8,where f'=0, butaone-parameter family of curves, andEquation (7)considered foronepoint ofeach curve; cf.forexample, thenextparagraph. 11.Moments about theInstantaneous Centre. Consider the motion ofalamina,i.e.arigid plane system,initsown plane. LetQbetheinstantaneous centre atagiven instant,t=T. Then<r,referred tothepoint Q,isavector perpendicular tothe plane, and itslengthis Tde 1 dt' where /isthemoment ofinertia ofthelamina about Q. What does itmean to"take moments aboutQ"?From the foregoingitmeans totakemoments about apoint 0'describing acurve TO=*o& T) which attheinstant t=rpasses through Q. There isanunlimited setofsuch curves. Letusselect, inpar- ticular, thecurveCwhich isthepath ofthatpointfixed inthe lamina, which passes through Qattheinstant t=r.Observe that this isanarbitrary choice ofC.This curveCisknown in terms oftherolling ofthebody centrode onthespace centrode. The velocity of0'atQis0,but itsacceleration,ifQisan ordinary point,isnormal tothecentrodes atQanddoesnotvanish; Chapter V, 5.If,then, Equation (7), 10istobesatisfied, the centre ofgravity, G,must lieinthenormal tothecentrodes. In particular, thenormal tothebody centrode must passthrough thecentre ofgravity. Hence thebody centrode must beacircle with thecentre ofgravity atthecentre,ifthecondition istobe permanently satisfied. Theequation ofmoments nowbecomes : I-JJ2~S(Moments about Inst. Centre). 208 MECHANICS Theonly case, then, ofmotion inaplane,inwhich wemay permanently takemoments about theinstantaneous centre, thought ofasapoint fixed inthemoving body,isthat inwhich acircle rollsonanarbitrary curve, thecentre ofgravity being atthecentre ofthecircle; andthelimiting case, namely, that thepointQispermanently atrest. This lastcasecorresponds totheidentical vanishing ofdvjdt. Moments about anArbitrary Point. Consider nowanarbi- trary point 0'fixed inthebody. Let itbeatQattheinstant t=r,and letCbethecurve, TO=r(J,T), which itisdescribing. Takemoments aboutQwith reference tothispoint, 0'.Then da[=d?0 dt dt2' If,furthermore, Equation (7), 10issatisfied, theequation of moments becomes : /-72/) ^77/2"=2}(Moments about Q). Equation (7)heremeans, ingeneral, that theacceleration of0' iscollinear with thelinedetermined byQandG.Inparticular, theequationissatisfied iftheacceleration ofO'isatQ ;orif Qcoincides with G.* EXERCISE Abilliard ball isstruck fullbythecue. Consider themotion while there isslipping. Show that itisnotpossible totake moments about theinstantaneous centre. Find thepoints ofzero acceleration and verify thefactthat itispossible totakemoments about them, explaining carefully whatyoumeanbythese words. Show that thepoints whose acceleration passes through the centre oftheball lieonacircle through thecentre ofthe ball, ofradius one-fifth that oftheball, thecentre being directly above thecentre oftheball. 12.Evaluation of <rforaRigid System; One Point Fixed. Consider arigid system ofparticles with one point, 0,fixed. *Edward V.Huntington hasdiscussed thisquestion, Amer. Math. Monthly, vol.XXI (1914) p.315. ROTATION 209 Themotion isthenoneofrotation about anaxispassing through ]cf .Chapter V, 8.Letthevector angular velocity be denoted byw,and leta,p,7beasystem ofmutually perpen- dicular unit vectors lying along Cartesian axes with theorigin at0.When wewish these axes tobefixed, weshall usethe coordinates (x,yyz)andreplace QJ,j3,7by i,j,k.Inthegeneral case, thecoordinates shallbe,17,f. LetPbeanypoint fixed inthebody, and letrbethevector drawn from toP : a) Thevelocity ofP, (2)r=$+lift+fy. " isexpressed interms ofthevector coasfollows (Chapter V, 9): (3) v=coXr. InCartesian form, (4) v= or (5) Forasingle particle, then, ahasthevalue : (6) a=rXrav, (7) Hence (8)a=my r 210 MECHANICS These formulas lead inturn tothefollowing: <T=m[(rj*+f2 )C0-&CO,,- (9) =m[-rft+(f =m[-fcof-ftw,+(2+ i;2 Forasystemofparticles theybecome : and so,finally, (10)or=- These aretheformulas which givethecomponents ofaalong theaxes of,77,fwhen theoriginisfixed. Itisobviously im- material whether theaxesarefixed ormoving. 13.Euler's Dynamical Equations. Consider thecase ofa rigid body, onepoint ofwhich isfixed. TheEquationof Moments, (1)-rr=2 (Moments about), referred tothispoint, admits asimple expressioninterms ofthe angular velocity, &,ofthebody. Letthe(, r/,f)-axes befixed inthebody, and letPbeapoint which moves according toany law. Let r=<*+T70+f% where risthevector drawn from toP.Thenwehave seen (Chapter V, 14): dr= dt This result applies tothevector : o-= cr$a+(Ty13+<TS7, andthus gives ustheleft-hand sideof(1). ROTATION Ontheright-handsideof(1)let XF*=La+M/3+Ny.211 Thuswehave : (2)=L,dt dt dfft-LJ~ _j."|C0O"TJ WTJ(7A==xV. at Onsubstituting for <T, o^,,a^their values from (10), 12,the equations known asEuler's Dynamical Equations result. In particular,ifthe(, 77,f)-axes arelaidalong theprincipal axes ofinertia, then andEquations (2)assume theform : dp /j- (3) where P= r= When theaxes ofcoordinates donotcoincide with theprin- cipal axes ofinertia, Euler's Equations take thegeneral form : (4)1dt dt dt -(En,-(C-B)w^=L, andtwoothers obtained byadvancing theletters cyclically. Eider's Dynamical Equations alsoapply totherotation of arigidbody about itscentre ofmass; 9.Here, there isno restriction whatsoever onthemotion. 212 MECHANICS 14.Motion about aFixed Point. Letthebodymove under theaction ofnoforces, save thereaction at0.Then Euler's Equations become thefollowing: Afirst integralisobtained bymultiplying theequations re- spectively byp,q,andr,andadding: (2) Ap*+Bq*+Cr*=h. This istheEquationofEnergy, Chapter VII, 5,6. Asecond integralisfound bymultiplying Equations (1)re- spectively byAp,Bq,andCr,andadding: dr (3) A2p2+B2q2+C2r2=I Thisequation corresponds tothefactthat dt=0, andso a=Apa+Bqf$+Cry isconstant. From Equations (2)and (3),two ofthevariables, asp2and #2 ,can,ingeneral, bedetermined interms ofthethird, andthen, onsubstituting inthethird equation (1),adifferential equation forralone isfound. Itisseen that tisexpressedasanelliptic integral ofthe firstkind inr.Thus, p,qtand rarefound as functions oft. Exercise. LetA=3,B=2,C=1;and letp,q,rallhave theinitial value 1.Work outthevalue of tinterms ofthein- tegral. ROTATION 213 TheBody Cone. Onmultiplying (2)by Z,(3)byA,andsub- tracting, wefind : (4)A(l- Ah)p*+B(l- Bh)g2+C(l-Ch)r2=0. Theequationsoftheinstantaneous axisare : (5)-*=?= , 'pqr' whenp,q,raretheabove functions of t.Hence thelocus of theinstantaneous axisinthebodyisthequadric cone : (6)A(l- Ah)?+B(l- Bh) T,2+C(l-Ch)f2=0. More explicitly, letp,q,rsatisfy (2)and (3)andhence (4). Then anypoint (, ry,f)of(5)satisfies(6),andhence lieson thequadric cone. Conversely,let(, 77,f)beapoint ofthe quadric cone, (6).If(, TJ,f)^(0,0,0),determinep,q,r,p bythefourequations P=/*> Q=M, r=/if, Thus (3)issatisfied. And (4)holds, too.Hence (2)istrue. Consequently, (, r/,f)liesonaninstantaneous axis. TheSpace Cone. Poinsot obtained anelegant determination ofthespace centrode. Consider theellipsoid ofinertia, 2.It isasurface fixed inthebody. Letm bethepoint inwhich theraydrawn from and collinear with cocuts S. Then thetangent planeMtoSatmis aplane fixed inspace, thesame forall points, m.Themotion isseen tobeone ofrolling ofthesurface SontheplaneMwithout slipping. pIG^4 Toprove the first statement,itis sufficient toshow thatthetangent planeatmisperpendicular to cr,andthat itsdistance fromOdoes notdepend onm.The equation ofSis : S: 42+5T;2+Cf2=1. Thecoordinates ofmare : PP, PQ, pr, where 214 MECHANICS Hence theequationofMis M : PAp^+pBqv+pCrf=1. The direction components ofitsnormal areAp,Bq,Cr.But these areprecisely theprojectionsof <rontheaxes. HenceM isperpendicular toa.Moreover, thedistance oftheplaneM from is 1=1=Jh P^l*l' andsoisconstant. This completes theproofofthe first state- ment. Toprove thesecond statement; consider somuch ofthebody cone, (6),asliesinS.LetFbethecurve onSwhich marks theintersection ofthese twosurfaces. Then FrollsonMwith- oujbslipping, andthecurve ofcontact, C,canservo asadirectrix ofthespace centrode. Forthebody cone, (6),rolls without slipping onthespace cone, andthecurves F,Caretwocurves onthese cones, which curves aroalways tangent atthepointM oftheinstantaneous axis. Theangular velocity, w,ispropor- tional tothedistance Om;for 15.Euler's Geometrical Equations. Euler introduced asco- ordinates describing thepositionofarigid body, onopoint ofwhich isfixed at0,thethree angles, 6,p,and^represented inthefigure. Between thecomponentsoftheangular velocity wabout theinstantaneous axis, andthese coordinates and their derivatives, exist thefollowing relations : d$,ddV= Sin COS<prr+SHI (p-r. dt at ... <ty, de q= sin sinip-j-+costp-r (it Ctt(1) dt' dt These areknown asEnter' sGeometrical Equations. ROTATION 215 Ageometrical proof canbegiven bycomputing thevector velocities ofcertain suitably chosen pointsintwoways. Begin FIG.105 with thepointCinwhich thepositive axisoffpierces thesurface oftheunit sphere. Aswelookdown onthesphere from above thispoint,itisevident from thefigure that dO dt=psin<p+qcos<p sin6-=pcos<p+qsin<p. (it FIG.106 These equations yield the firsttwoofEquations (1). Toobtain thethird equation, consider themotion ofE. Its velocityismade upofavelocityo>tangent tothearcEA,and two velocities perpendicular tothis arc.Ontheother hand, itsvelocityiscomposed ofthevelocity tptangent tothearcEA; thevelocity \ftcos0,alsotangent toEA;and perpendicular toEA. Hence andthis isthethirdEquation (1). 216 MECHANICS Ifp, <7,rhave onc6"beendetermined asfunctions ofthetime, Equations (1)yield asystem ofthree simultaneous differential equations ofthe firstorder fordetermining 0,<p,$asfunctions ofthetime. Thus intheproblem of14, themotion ofa rigidbody under noforces, oracted onbytheoneforce ofcon- straint that holds thepointOfixed, p,q,rwere determined explicitly asfunctions ofthetime, andthefurther study ofthe problemisbased ontheabove Equations (1). 16.Continuation. TheDirection Cosines oftheMoving Axes. Themoving axesarerelated tothefixed axesbythescheme, Vf y z andthequestion is,toexpress thenine direction cosines interms oftheEulerian angles. Thiscanbedone conveniently byvector methods,ifweeffect thedisplacement onestep atatime. Let i,j,kbeunit vectors along theoriginal axes,and a, /ft,7unit vectors along thedisplaced axes. Letthe first displacement bearotation about theaxis ofzthrough theangle ^,and let i,jgoover into i1;jj.Then it=icos^+jsin^ J!=isin^+jcos^ kk*V|JEL. Next, rotate about theaxis ofjtthrough anangle 6,whereby ixgoes into i2,andkxintok2=7: 12=itcos ktsin J2=Ji k2= ijsin+kjcos 6. Finally, rotate about k2through anangle ^,wherebyi2goes over into i3=aandJ2goes intoj3=: 13=i2cos<p+J2sin <p j3=i2sin<p+j2cosv k,=k,. ROTATION 217 From these equationsitappears that Zj=cos cos<pcos^ sin<psin^ ml=cos6cos(f>sin^+sin^cos^ n!= sin cosv? Z2=cos6sin<pcos^cos<psin^ m2=cos6sinpsin^+cospcos^ n2=sin sinv? Z3=sin cos^ w3=sin6sin^ n3=cos 0. 17.TheGyroscope. Itisnow possible tosetforth insimplest terms theessential characteristics ofthemotion ofarotating rigid body, which isthebasis ofgyroscopic action. Byagyro- scopeismeant arigidbody spinning athigh velocity about an axispassing through thecentre ofgravity, which isatrest,and acted onbyacouple whose representative vector isperpendicular totheaxis. Consider,inparticular, thefollowing motion. LetA=J?, C7*0,and lettheaxis offbecaused torotate with constant angular velocity, c,intheplane^=0.What willbethecouple? Here, d\l//dt=andEuler's Geometrical Equations give PdO de 3~T7* T= where dd/dt=c.Thecomponents LandMareunknown, butN=0.Thethird oftheDynamical Equations becomes : ~ dt=0; hence r=v, and Pisalarge positive constant. Since here d<p. Hence, from thefirsttwoequations, p=csinvtj q=ccos vt. 218 MECHANICS Onsubstituting these values intheDynamical Equations, we find: L=Ccvcosvt,M=Ccvsin vt. Wemay think ofthecouple asmade upofaforceFacting atthepointC :(, 77,f)=(0,0,1)inFig. 105,andanequal and opposite force at0.ThenFwillbetangent tothesphere at C.Let itberesolved intotwocomponents, oneperpendicular tothe(,f)-plane ;theother, inthat plane. The first willhave thevalue L,taken positiveinthesense ofthenegative ry-axis ; thesecond willequalM,taken positiveinthesense ofthe -axis. When t=0,L=Ccv,M=0, andatany later time, theresult isthesame. Thiscanbeseen directly from thenature oftheproblem, since themotion ofthe axisoffintheplane \f/=isuniform, andhence theforcewhich theconstraint exerts willbethesame force relative tothebody atoneinstant asatanyother instant. Itiseasy toverify analytically thetruth ofthelaststatement. For,theforcenormal totheplane \l/= willalways be Lcos<pMsin<p=Ccv; andtheforce inthatplane willalways be Lsinif>+Mcosp=0. This result brings out inthesimplest form imaginable the essential phenomenoningyroscopic action, namely, this: To cause theaxis tomove inaplane with constant angular velocity, acouple must beapplied whose forces actontheaxisinadirection atright anglestothatplane. Finally observe that ifonethinks ofonesolf as standing onthegyroscope andmoving withit,one's eexertedbody along thepositive axis offandfacing inthe direction ofthemotion oftheaxis, theforceLap- plied tothegyroscopewillbedirected toward one's left,andhence thereaction ofthegyroscope onthe FIG.107constraint willbedirected toward theright, thegyro- scope spinning intheclockwise sense asonolooks down onit.Ofcourse,ifthesense oftherotation were reversed, thesense ofthereaction would bereversed also. ROTATION 219 EXERCISES 1.Show that,ifnoassumption regarding6ismade, but^= andr=v,then T A-4d*e .n 4deL=Asinvt-rz+Cvcos vt-T: at* at TM A *d*en <ddM=ACOS vt-T7T~CvSinvt-rr- at1at 2.Iftheaxis presses against arough plane, \l/=0,thetan- gential force being p,times thenormal force, then* dB provided dO/dt>andfurthermore thepoint oftheaxis in contact with theplane moves backward,i.e.inthesense ofthe decreasing6. Ontheother hand, theaxismust have asufficiently large radius sothat therequirement below relating tothemotion of thepointofcontact canbefulfilled. Hence dOc^t, QcA^t 37=ceA and 9= -/YeA at CIJLV whore cdenotes theinitial value ofdO/dt, andinitially=cA/Cp,v. Moreover,OVSS-T8' where v=dO/dtand s=refer tothepointinwhich thesphere (ofradius 1)iscutbytheaxis. 3.Prove that, intheproblem ofthepreceding question, the normal reaction oftheconstraint is 4.Show that,ifa(small) constant couple, ofmoment,acts onthegyroscope, thevector that represents thecouple being atright angles totheplane \l/=anddirected intheproper *Ifwethink ofthematerial axisasacylinder ofsmall radius, there willbea small couple about theaxis,tending toreduce r.Butasthiscouple approaches when theradius ofthecylinder approaches 0,wemay consider theideal case ofan axisthat isamaterial wire ofnilcross section, thecouple nowvanishing. 220 MECHANICS sense, and iftheaxis ofthegyroscope beconstrained tomove intheplane \l/=0,theacceleration of6isconstant : This lastequationistrue,evenwhen evaries with thetime. 6.Prove that, nomatter how 8varies, theaxis ofthegyro- scope always being constrained tomove intheplane ^=0, thereaction ontheconstraining plane \l/=isnumerically itssense being that oftheincreasing \l/when dO/dt>0,butthe opposite when dd/dt<0. 6.Ithasbeenshown thatarigidbodyisequivalent dynami- cally tothree pairs ofparticles situated atthesixextremities of athree dimensional cross;9. Lettheequivalent system move asthegyroscope didinthe text,i.e.with\l/=and dO/dt=c.Consider,inparticular, aninstant, atwhich themoving axes areflashing through the fixed axes;i.e.=<p \l/ 0.Show, byaidoftheexpres- sions fora,0,7,thatthevector acceleration ofeach ofthefour particles onthe-andthef-axes passes through ;but, inthe case ofeach oftheother two particles,isparallel totheaxis of f.Hence explain thereaction ofthegyroscope ontheconstraint. 7.Discuss theproblem ofQuestion 2forthecase that the point ofcontact isallowed toslipforward. Consider also all cases inwhich dB/dt< initially. 18.TheTop. Thetopisarigidbody having anaxis of material symmetry and, inthecase ofafixed peg,supported at apoint oftheaxis. Letthepositive axisoffpassthrough the centre ofgravity, (7,distant hfrom 0. The third ofEuler's Dynamical Equations becomes, since the applied forces gravity andthereaction ofthepeg both pass through theaxisoff, (1)C%=0. Hence r=v(constant). Theequation ofenergy herebecomes : (2) A(p2+2 )+CV>=H-2Mgh cos 6. ROTATION 221 Furthermore, thevertical component ofthevector aisconstant. For, theapplied forces giving avector moment at reduce to gravity, which isvertical, andsoitsvector moment with respect toOishorizontal. Now, thecomponents of o-along themoving axes areAp,Bq,and Cr.Hence thevertical component ofa is(16): Bqn 2 Onsubstituting fornltn2,nztheir values from 16wehave : (3) Apsin6cos<p+Aqsin6sin<p+Cvcos=K. Turning now toEuler's Geometrical Equations, wefind : (4)dt,-dO p= sin cos<p-~+sin <p-jr ... q=sin sin<p cosdO -=r at Onsubstituting these values ofpandqin(2)and(3)wefind : (5) (6)sin2e--=-bi>cos 6=C theconstants aand depending onthe initial conditions ofthe motion;i.e.they areconstants ofintegration; whereas aand bareconstants ofthebody. The third Equation (4)determines<pafter and^have been found from(5)asfunctions oft: (7) <p=vt Icos6 dtdt. Returning now toEquations (5)andeliminating d\l//dt, we obtain : (8) sin26(~)2=sin2e(a-acos6)- (J8-6i>cos0)2 . 222 MECHANICS The result isadifferential equationforthesingle dependent vari- able,6.Itcanbeimprovedinformbythesubstitution (9) u=cos9: (10)2=(1-u*)(a-au)-CJ-bvuY=f(u). Thus f(u)isseen tobeacubic polynomial, which wewill presently discuss indetail. But firstobserve that thesecond Equation (5)gives: d*P""bvu Hence\f/isgiven byaquadrature after uhasoncebeenfound asafunction of t. Retrospect andProspect. Tosum up,then,wehave reduced theproblem tothesolution ofEquation (10) foruasafunction of t.'Equation (9)gives 0;Equation (11) gives \f/;andEqua- tion(7)gives <p.Wemay concentrate, then, onthesolution of Equation (10). 19.Continuation. Discussion oftheMotion. ThePolynomial (1) fM=(1-u*)(a-au)-(0-bmY becomespositively infinite foru=+oo. Itisnegative or foru=+ 1,1.Hence ingeneral thegraph willbeasindi- cated, or </(M), u,<u<u2] fM=/(iO=o. Moreover,1<u^<u2<1,and f(u) hasone root, u'>1.The roots w^u2will, therefore, besimple roots. The differential equation (a)'- comes under theclass discussed inAppendix B.Inparticular, thesolution isafunction (3) u=<*>(0 single-valued andcontinuous for allvalues oftandhaving the period T7 ,where ROTATION 223 (4) or (5) <p(t+ Furthermore,if (6) *!= then Andsimilarly,if //w\ /,\ then /*7'\ /* \ /j I\ \t) ^C*2T)~^(^2 ~T~Ty. Physical Interpretation. Letasphere/Sbeplaced about ascentre, and letPbethepoint ofintersection ofthepositive axis offwith S.LotCbethecurve thatPdescribes onS.The results justobtained show thatCliesbetween thetwo parallels oflatitude corresponding to (8) u=ult u=u2. Forconvenience let tbemeasured from apoint ontheupper parallel, u=u2.Then there arethree cases according asinitially III. <0. CASE I.SinceFIG.109 bvu eft 1-u2 ispositive when uhas itsgreatest value, u2jd^/dt willremain positive, andso^willsteadilyincrease with t.Let^=when t=0.As tincreases toiT,\!/willincrease to 224 MECHANICS where u=^(0, Equation (3).When t=T, \pwillhave in- creased by^,andonecomplete arch ofthecurveCwillhave been described. Thearch issymmetric intheplane ^=-J-^. The rest ofCisobtained byrotating thisarchabout thepolar axisofSthrough angles that aremultiplaof^. CASE II.Here, d\f//dtisatthestart, andhence |8-bvu2=0. Since udecreases,itfollows that inthefurther course ofthe motion < bvu, andso\[/steadily increases. ThecurveChascusps ontheupper parallel oflatitude. CASE III.Here-bvu< atthestart, and itisconceivable that thisrelation should persist forever. Buteven ifthiswere notthecase,itisstillconceivable thatthevalue of\l/whenPreaches thelower parallel oflatitude should belessthan orequal tothe initial value, ^=0.That neither ofthese cases ispossible that thevalue of\l/corre- sponding tothe firstreturn ofPtotheuppercircle ispositive hasbeenshown byHaclarnard.* ThecurveChasdouble points inthis case, but itproceeds with increasingtinthesense ofthe advancing ^,asindicated. Special Cases. There isstillavariety ofspecial cases tobe discussed, one ofwhich isthat inwhich f(u) hasequal roots lying within theinterval : 1<MI=tig<1. Since /(I) inallcases, andsince theremust beathird rootu1^1.Thusuisadouble rootand f(u)=(w-u, where x(u)<0,-1<u<1. *Butt, desSci.math. 1895, p.228. ROTATION 225 Theonly solution ofEquation (2)inthis case, which takes onthevalue u^when t=0,is u=wt. ThecurveCreduces toaparalleloflatitude. When u=1isaroot, various cases can arise. ThepointP may passthrough thenorth polewithavelocity ;oritmay gradually climb, approaching thenorth pole asalimit; orthe topmay permanently rotate about thepolar axis. Similarly, whenu= 1isaroot. There isagreat wealth ofliterature onthegyroscope and thetop.The reader can refer tothe article ontheGyroscope intheEncyclopaedia Britannica; toWebster, Dynamics; to Routh, Elementary Rigid Dynamics; and toAppell, Mecanique rationelle, vol. II. EXERCISE Treat thecase ofatoponasmooth table. Assume that the pegisasurface ofrevolution. The distance, then, from the centre ofgravity tothevertical through thepoint ofcontact with thetable willbeafunction oftheangle ofinclination of theaxis. Assume axes fixed inthebody with theorigin atthecentre ofgravity. Write downi)theequation ofenergy; ii)theequation that saysthatthevertical component of orisconstant. From thispoint ontheprocedureisprecisely asbefore, and theresult isagain adifferential equation ofthetype treated in Appendix B.Discuss allcases, andshow that ingeneral the axis oscillates between two inclinations, both oblique tothe vertical. Begin with thespecial casethat thepegisapoint. Having studied thiscase indetail, proceed tothegeneral caseandstudy itindetail, also. Then derive thespecial case asaparticular caseunder thegeneral case. 20.Intrinsic Treatment oftheGyroscope.* Themost general case ofmotion ofagyroscope reduces tooneinwhich asingle couple actsonthebody, and thiscouple canbebroken upinto *The results ofthisparagraph arccontained inapaper bytheAuthor: "On theGyroscope," Trans. Amer. Math. Soc., vol.23,April, 1922, p.240. 226 MECHANICS twocouples one, represented byavector atright angles to theaxis ofthegyroscope ;theother, byavector collinear with theaxis. Inthemost important applications that arise inprac- tice,thelatter couple vanishes. But inthegeneral case,itgives risetothethird oftheDynamical Equationsintheform : Theformer couple canberealized byasingle forceFper- pendicular totheaxisandacting atthepointPinwhich the positive f-axis cuts theunit sphere, theother force ofthe couple andtheresultant force acting at0.* Definition oftheBending,K.LetCbethecurve described ontheunitsphere byP,and letSbetheconewhich isthelocus oftheaxisofthegyroscope, andofwhichCisthedirectrix. Con- sider therateatwhich thetangent plane toSisturning whenP describes Cwith unit velocity. This quantityshall bedenoted asthebending oftheconeandrepresented bythenumber K. Itisalsotherate atwhich theterminal pointofaunit vector drawn from atright angles tothetangent plane traces out its pathontheunit sphere.Kshall betaken positive when anob- server, walking along C,seesCtothe leftofthetangent plane, andnegative, whenCistohisright. Itiseasy tocomputeK.LetVbetheangle from theparallel oflatitude through Pwith thesense oftheincreasing ^tothe tangent toCwith thesense oftheincreasings.Then itappears formaninfinitesimal treatment that__.v'ds ds Since tanV= ,. .-, orV=tan"1 ,. .-, -77-; -, d\f/sinB $'sin0' where accents denote differentiation with respect tos,andsince ds2-d6*+dj*sin2 0, or/2+V*sin2=1, itfollows that (3)jc=WB"-0'iHsin-(1+/2 )Vcos 0. *The pointOneed notbethecentre ofgravity inthefollowing treatment. Itmaybeanypoint fixed intheaxisofmaterial symmetry. ROTATION 227 From thedefinition itfollows atonce that thebending ofa cone ofrevolution must beconstant. Tofind itsvalue,letthe coordinates besochosen that theequation ofthecone is=a. Then thelength ofthearcofCis 5=^sinaandso\l/'sina=1. From (3)itnow isseenthat (4)K=cot oc. Thecone liestotheright oftheobserver, ashetravels along C. Ifhereverses hissense, thesign ofKwillbechanged. Butboth cases areembraced inthesingle formula(4),thesecond corre- sponding toaconewhose angleisTTa,or againforwhich sisreplaced bys. Conversely,ifKisconstant, Cisacircular cone. For, theequation (3)can,byelimi- nating ^', U-Ax 1fbewritten intheform : ntt (5)K= the signholding whenever ^'<0.Hence IfKisconstant, set K=cota.Then Equation (6)admits one solution,=a,or=7ra;and, asisshown inthe theoryofdifferential equations, this istheonly solution which, atapoints=s,takes onthevalue a,or TTa,andwhose derivative vanishes there. Further Formulas forK.*From(3)itfollows further that -7-77:sinB2-J-T^cos6sin26cos ,~, (7)*= where the signholds whenever\f/'<0. *These results areinserted forcompleteness. They willnotbeused inwhat follows, andthestudent may passonwithout studying them. They arechiefly ofinterest tothestudent ofDifferential Geometry. 228 MECHANICS IfKisknown, orgiven, asafunction ofs,thenEquation (6) determines asafunction ofs,and\l/isthenfound byaquad- rature : (8) Thebending, K,isconnected with thecurvature, K,ofC,re- garded asaspace curve, bytheformula (9) K*=K2+1. Furthermore,cf.Fig.Illbelow: (ijk (10) n=aXt xyz x'y'z' (11) hence (12) Since IK\= 11' Iand=yz"- zx"-xz" KZ'=xy_yX". (13) ^2=x"*+y"*+z"*, formula (9)follows atoncefrom (12)and (13). Moreover, from (12)itfollows that (14) *=-xyz x'y'z' x"y"z" Finally, thetorsion, T,ofCisconnected with /cbytherelation : vl*= T, theresult obtained byProfessor Haskins.* *Fortheproof ofthisformula cf.theAuthor's paper cited above.d Ts( ROTATION 229 21.TheRelations Connecting v,F,and *.Thephysical phe- nomenon which itismost important tobring home toone's intuition istheeffect oftheforceFonthemotion ofthegyro- scope. Any such explanation must take account ofallthree quantities, v,F,and K.Butmany popular explanations claim- ingcorrectly tobe"non-mathematical," butincorrectly tobe accurate intheir mechanics failbecause they areunaware ofK.Thus, forexample, thestatement oftenmade that"when acoupleisapplied toarotating gyroscope, theforces ofthe couple intersecting theaxis ofthegyroscope atright angles, the axis willmove inaplane perpendicular totheplane oftheforces ofthecouple"isfalse. Infact, theaxis willbegin tomove tangentially tothis plane,ifitstarts fromrest,and allinter- mediate cases arepossible, according totheinitial motion ofthe axis. Asimple andaccurate explanation,interms ofv,F,andK, canbegiven asfollows.* First ofall,however, thethird of Euler's Dynamical Equations, which herebecomes : (1) and requires nofurther comment thani)that itisperfectly general, applying tothemotion ofthegyroscope under any forces whatever; andii)that inthecasewhich most interests us,namely that inwhich there isonly theforceF(and thereac- tionat0)wehave :N=0,andsor= *>,aconstant. LetF,then, beresolved,inthetangent plane, intoacomponent Talong thepositive tangent, andacomponent Q,taken positive when directed toward the leftoftheobserver;i.e.Qisposi- tivewhen Kispositive. Then (2) AKV*+Crv=Q, where v=ds/dtand sincreases inthesense ofthemotion ofP, rbeing givenbyEquation (1). *Cf. theAuthor's paper"On theGyroscope" cited above, p.240. 230 MECHANICS Proof. Lettheunit vector from toPbedenoted bya(it isthevector 7ofthecoordinate system) ;lettbeaunitvector along thepositive tangent toCatP;and letnbeaunit vector normal toaand tandsooriented 'awith regard tothem as ftiswith regard \pX to7and a.These areprincipal axes of t\ inertia, andthemoments ofinertia about them are : L=A In=A, Ia=C. Thecomponents oftheangular velocity &about them are : FIG. Ill o>t=0,wn= I), Now,(Thasthevalue : a=InC0nn+ItCO*t+IaC0aa. Hence (3) From thisequation wecancompute-rr : do- dv dn da Itisclear that (4)da dtvt. Furthermore, from thedefinition ofthebending,itappears that /^x dn (5)-j-r=KVt Hence, finally, d<r (6)=Av-n+(Aw*+C)t+C LetthevectorMwhich represents theresultant moment of alltheappliedforces about bewritten intheform : M=Mnn+M tt+Maa. Since da -. ROTATION 231 wehave : (7) Av^-=Mn,AKv*+Crv=Mt,C%=Ma. us dt Turning now tothecase inwhich wearemost interested, namely, that inwhich aforceFactsatPinadirection atright angles toOP :F=-Qn+Tt, weseethatMn=T,M t=Q,andthus Equations (2)are established. Equation (1)isthethird ofEquations (7). Wehave thus obtained Euler's Dynamical Equationsinthe form: dv (8)Av^-=Tds AKV*+Crv=Q 22.Discussion oftheIntrinsic Equations. The first ofEqua- tions(8), 21, AV=r,ds' admits asimple interpretation. Itshows thatthepointPde- scribes thecurveCexactly asasmooth bead ofmassm=A would move along awire intheform ofCifitwere acted onby atangential force T. Thethird equation, C~=N ^ dt*' shows that thecomponent roftheangular velocity &about the axis ofthegyroscope varies exactly asitwould iftheaxiswere permanently atrestandthesame coupleNrelative totheaxis acted. Thesecond equation, A) AKV*+Crv=Q, expresses thesolerelation which holds between thefourvariables K,v,r,andQ.Intheapplications, however, risconstant, r=v, andsotheequation A') AKV*+Cw=Q expresses thesolerelation between K,v,andQ. 232 MECHANICS TheCaseF=0.Letusbegin with thecasethatFvanishes, buttheaxis isnotatrest. Here,Q=0,T=0.Equation A) gives i) AKV+Cr=0, or,onintroducing theradius ofbending, p=1/|K|,andchoos- ingr> : Av p=='Cr' Ifrisapositive constant, r=v>0,then Cv and since visconstant, for A __n FIG.112 ds~' Kisalso constant, andnegative. The axis ofthegyroscopeis describing acone ofsemi-vertical angle a,where cota= |K|, or tana=p, andthesense ofthedescriptionissuch thattheobserver, walking alongCinthe positive sense, hastheconeonhisright. TheCase K=0.Here, thepath ofP isanarcofagreat circle, and & ' FIG.113 Q=Crv, orQ=Cw, nomatter whatTandthemotion ofPalong itspathmay be. Thepressure oftheaxisagainst theconstraint, inanormal direc- tion,istotheright, and isproportional torandtov; or,if risconstant, tov,thecoefficient then being Cv.Thusweobtain anew, andwith theminimum ofeffort, themain result of 17. General Interpretation ofEquation A).Wecannow give a simple physical interpretationtoEquation A): Am*+Crv=Q. The left-hand side isthesum oftwoterms. Thesecond term expresses theforce, Q2=Crv, ROTATION 233 thatwould berequired tocausePtodescribe agreat circle on thesphere;i.e.tomake theaxismove intheplane through tangent toC.This force,Q2,isalways directed toward the left,forQ2>0. The firstterm, accounts forthebending.~ ,9Ql=AKV*, Itsnumerical value, canbeinterpreted asthecentripetal force exerted onaparticle, ofmassm=A,tomake itdescribe acircle ofradius pwith velocityv.Wlien Kispositive, this force ispositive, andsois directed toward theleft;andvice versa. Consider now theforceQfalong thenormal natP,which (combined with thesmooth constraint ofthesurface ofthesphere) would berequired toholdaparticle ofmassm=Ainthepath C. Letthevector abewritten intheform : Thena=xi yj+zk. v=xi+y]+zk=vt, where x'=dx/ds tx=dx/dt, etc. Furthermore, n=aXt=(yzf-zyf )i+(zz'-xzf )j+(**/'-yx')k. Theacceleration, (a),ofPinspace is,ofcourse : ()=zi+ j+^k. Now, thecomponent oftheacceleration along thenormal n totheplane ofaand tisn-(a), which canbewritten inthe form :xyz x'yfz' xyz Since x=vx',itfollows that x=v*x"+vx', etc., andso xyz x'y'z' xyzxy x"y"z" 234 MECHANICS Thusmw2isequal totheforce Q'tangent tothesphere and normal to(7,which would berequired toholdaparticle ofmass w,describing (7,initspath; thecomponent alongtbeing mvdv/ds, and thethird component, along a,being thereaction normal tothesphere,inwhich wearenotinterested. Itisnatural tothink ofthepointpnonthe linethrough P alongnasthecentre ofbending.Ifwedraw theosculating cone ofrevolution through P,this isthepointQinwhich that linemeets theaxis ofthecone. Anobvious interpretationforthisforce ofmw2isthecentripetal force ofaparticle describing acircle ofradiusp,with centre atQ,tangent toCatP,thevelocity beingv. _,.The forceQ2canberealized physically asfollows. Letanelectro-magneticfield offorce begenerated by anorth-pole situated at0,and lettheparticlemcarry acharge, e,ofelectricity. The force exerted onebythe field willbeat right angles tothepath and tangent tothesphere, and, finally, proportionaltothevelocity, v,ofm.Hence ecanbeso chosen that thisforce willbeprecisely equal toQ2=CW. Inthemore general, but lessinteresting, casethat risvariable, thephysical interpretation can stillbeadapted byusing avariable charge.* Summary oftheResults. Tosum up,then,wecansay:The point P,inwhich theaxisofthegyroscope meets theunitsphere about 0,moves likeaparticleofmassm=Aconstrained to lieonthesphere andcarrying acharge ofelectricity,e.The forces that actonmaresupplied bytheelectromagnetic force ofthe field,Q2=Cw,acting oneyandaforceFacting onm, thecomponents ofFalong thetangent andnormal atPbeing TandQlrespectively. The case ofavariable rcanbemetby avariable charge,e. Asregards thephysical realization ofthecondition that the particlelieonthesurface ofthesphere, wemaythink ofamass- lessrodofunit length,freetoturnabout oneendwhich ispivoted at0,andcarrying theparticleattheother end. *Theidea ofusing theabove electro-magnetic field toobtain #2wassuggested tomebymycolleague, Professor Kemble, towhom Ihadjustcommunicated the results ofthetext,down tothispoint. (Note ofJan. 23,1933.) ROTATION 235 EXERCISES 1.Suppose theaxlePofthegyroscopeiscaused tomove in asmooth slotintheform ofameridiancircle, which ismade to rotate inanymanner. The forceFwillthen benormal tothe meridian, ortangent totheparallel oflatitude. Show that d26 . Suggestion:Combine Euler's Geometrical Equations with Euler's Dynamical Equations. 2.Letthecomponents ofFalong themeridian inthesense oftheincreasing6andalong theparalleloflatitude inthesense oftheincreasing ^bedenoted respectively by and^Show that Ifand^areknown asfunctions of6, \f/}t,these equations suffice todetermine thepathofP. 3.Consider small oscillations oftheaxis ofthegyroscope in theneighborhood oftheaxis 6=v/2,$=0.Let Show that theequationsofQuestion1lead totheapproximate equations: 4.Generalize theequationsofQuestion 2tothecasethatA, ,Carealldistinct. 236 MECHANICS 5.Intrinsic Equations. From theequations: Av%=T,ds' AKV*+Cvv=Q, K= thepath canbedetermined ifT,Qareknown asfunctions of sand v. 6. .Ship's Stabilizer. Thegyroscope canbeused toreduce the rolling ofaship.Amassive gyroscopeismounted inacage, orframe,itsaxisbeing fixed with reference totheframe, and vertical. Theframe ismounted ontrunnions, with axis hori- zontal andatright angles tothekeel,and itisprovided with abrake todampenitsoscillations about this axis. Thus the axis ofthegyroscope hastwodegrees offreedom;itcanrotate intheplane through thekeelandthemasts, andthisplane rotates with therollingoftheship.* Isolate thefollowing systems: i)Theship, exclusive ofthegyroscope andframe; ii)Theframe; Hi)Thegyroscope. The rolling oftheshipisgoverned bytheequation: where the firstterm ontherightisduetothedamping ofthe water; thesecond, totherighting moment produced bythe buoyancy ;andthethird, totheforce exerted bythetrunnions. Theframemaybethought ofasrotating about thepoint, 0, regarded asfixed, inwhich theaxis ofthegyroscope cuts the *Apicture andanaccount oftheship's gyroscope isfound inthearticle on the"Gyroscope" intheEncyclopaedia Britannica and inKlein-Sommerfeld, Theorie desKreisels, vol. iv., p.797. Forthediscussion which follows thereader alsoneeds, however, thetheory andpractice ofOscillatory Motion withDamping ; cf.theAuthor's Advanced Calculus, Chap. XV. ROTATION 237 axis ofthetrunnions. LetEuler's Angles besochosen thatthe axis ofthesphere:6=0,^=0,isparallel tothe keel, the plane^=being vertical. Moreover, let bereplaced by$, where Themotion ofthegyroscope about itscentre ofgravity, the point (9,willbegoverned bytheapproximate equations ofQues- tion 3. Finally, themotion oftheframe isgoverned bytheequations called forinQuestion 4above. These equations aremodified by thecondition <p=0,andthen reduced still further bysetting sin&=0,cos#=1.Thus dt where the firsttermontherightisduetothebrake andother damping, andthesecond, togravity, since theframe issocon- structed that itscentre ofgravityisappreciably below O. Oncombining these fiveequations andneglecting. A+A'in comparison with7wefind : These aretheequations which govern themotion. They are discussed atlength inKlein-Sommerfcld,I.e. 23.Billiard Ball. Letabilliard ballbeprojected along the table, withanarbitraryinitial velocity ofthecentre, 0,andan arbitraryinitial velocity ofrotation. Todetermine themotion. Letthe(x,y)-plane oftheaxes fixed inspace behorizontal. Letmoving axes of(, 17,f)bechosen parallel to(#,y,z),but with theorigin atthecentre oftheball. Thepoint oftheballP,incontact with thetable, shall be slipping, andtheangle from thepositive direction oftheaxis ofxortothedirection ofitsmotion shallbe^. 238 MECHANICS Theforces acting are :gravity, orMgtdownward atR=Mg upward atP;andtheforce offriction, nMg, atPinthesense opposite tothat ofslipping. Hence, forthemotion ofthecentre ofgravity, M-jp=pMgcos\f/ CD rf2. dt2 The vector momentuma-,referred tothecentre ofgravity, hasforitscomponents along themoving axes : where Themoment equation, thus gives:fMa2 . d<r (2) Hence (3)/-=nMgasin u-6 7--=nMga cos^ 'if-- =const. 5dtThe angle^isunknown. Eliminate itbycombining Equa- tions (1)and (2): (4) Hence (5)dt* ^y. dt25dt 2a 2a where A,Bareconstants ofintegration depending onthe initial conditions. Theymayhaveanyvalues whatever. ROTATION 239 LetVbethevelocity ofthelowest point ofthe ball. Then (6)Vx=Vcos$=-7- aco, Vv=Vsin^=-+au(. Combining these equations with(5)weget: where2\dt^ A'=- B'=-fB. Equations (1)nowtakeonthefollowing form. Forabbrevi- ation let dtA', Then (8) Hence andconsequentlydu dt dv du dv ^v-7Tu-77=0,dt dty av, where a, /3areconstants notboth 0.Moreover, uand vare notboth 0. Suppose u>0,a>0.Then av av u Theproof ofthis lastequation requires theconsideration ofthe two cases :i) 7*;ii)=0.These formulas aregeneral, holding inallcases inwhichw^0. 240 MECHANICS Itthusappears that (9) cos sn=ft Hence d?x/dt2andd2y/dt* areconstants, andconsequently the centre oftheballdescribes ingeneral aparabola;inparticular, astraightline. The direction, however, inwhich thepointP isslipping,isalways thesame;cf.Equations (9). This result comprises themain interest oftheproblem, solong asthere isslipping. Slippingceases whenV=0,or (10) TheSubsequent Motion. From thisinstant onthemotion is pure rolling weare, ofcourse, neglecting rolling friction and allother damping. For, attheinstant inquestion, V 0,and theangular velocityisrelated tothelinear velocityofthecentre ofgravity asfollows. Letthecentre oftheballbeattheorigin and letitsvelocity bedirected along thepositive axisofx.Then (11)dx Ttt-Oc#c> Jt 0; au( |t-o=0, aco, \t-o=c, wf |t-o y, where 7canhaveanyvalue, positive, negative, or0. Letusconsider themotion which consists inpure rolling and pivoting, andseewhat force atPisnecessary. First,wehave (12)Mw= - dt*-Y~Y> where X,Yarethecomponentsoftheunknown reaction atP. Next, taking moments about thecentre ofgravity, wefind : (13)UOi) fUWf)-w*___.i___._ax dt dt dt dt ROTATION 241 Finally, (14)Tr Ay.---a*,- These seven equations, (12), (13), (14), together with the initial conditions(11), formulate theproblem completely, and determine theseven unknown functions, x,y,co$,w^,o^,X,F, aswewillnowshow. From thesecond equation (13)itappears that Subtracting thisequation from the firstequation (12),wefind : Buttheleft-hand side ofthisequation vanishes because the first equation (14)isanidentity in t.HenceX=0.Similar con- siderations show thatY=0. Onsubstituting these values in(12)and (13), these fiveequa- tions canbesolved subject tothefive initial conditions(11),and theother condition, that initially x=0,y=0.The centre of theballdescribes thepositive axis ofxwith constant velocity,c. Theangular velocity wisalso constant, itscomponents along theaxes being given bytheir initial values (11). Since 7is arbitrary, wmay beanyvector whatever inthe(77,f)-plane, whose component along the r;-axis isc/a. Theforegoing discussion maybeabbreviated bymeans ofthe Principle ofWork andEnergy, Chapter VII. Themotion ofpure rolling with pivoting requires, then, no force tobeexerted bythetable. Itisuniquely determined by the initial conditions, andhence itcoincides with theactual motion ofthebilliard ball. 24.CartWheels. Consider theforewheels ofacart. Ideal- izedtheyformtwoequal discs connected byanaxleabout which each canturn freely. Todetermine themotion onarough inclined Diane. 242 MECHANICS Wewillbegin withastillsimpler case that ofasingle wheel, ordisc,mounted sothat itcanturnand rollfreely, butwillalways have itsplane perpendicular totheplane onwhich itrolls. The frame which guidesitmay bethought ofassmooth. Itsmass canbetaken intoaccount, butwewilldisregard it,inorder not toobscure themain points oftheproblem. Wewillchoose thecoordinates asindicated, theaxisofbeing inthediscandalways parallel totheplane ;theaxis of77,being theaxis ofthedisc,isalso parallel tothe plane. The axisofyliesintheplane and ishorizontal. The axisofxis directed down theplane. Let v?betheangle through which thedischasturned about the axisofrj;letsbethearcde- scribed bythepoint ofcon- tact,P;and letbetheangle from thepositive axisofxtothepositive tangent atP.Letthere- action oftheplane be F=Xa+Yj9+ZT, where a,0,yareunit vectors along themoving axes. Z=Mgcose,where eistheinclination oftheplane, andThen (1)d2xM-;- Xcos6Ysin6+Mgsin 6 =Xsin0+Ycos0 Theangular velocity, hasthevalue Moreover,=-da, 7=0. Takemoments about thecentre ofgravity: (2)dt ROTATION 243 Since andA=C= Hence or (3) Incomputing theright-hand side ofEquation (2),thecouple which keeps theaxis ofthedisc parallel totheplane must be taken intoaccount. Thevector which representsitiscollinear with theaxisof .Hence thecouple mayberealized bythetwo forces :=0, co,,= <p,o> a2 ,#=pfa2 ,wehave : =-Beta.+5^j3+ Thusat r2=-0. r*XF*=j8XFl7+(- 18)X(-F,y)+(-ay)XF; r*XF,=(2F,+aY)a-aX0.or,finally: (4) Equating, then, thevectors (3)and (4)wefind : (5) Finally, thecondition ofrolling without slipping canbewritten intheform : dx ~dt=vcos0,dy= dtvsin^, 244 MECHANICS where andso /n\ dx d<p n dy d<p. (6) -TT=a--cos0,-~=a- sin 0.v'dt dt dt dt Theformulation isnowcomplete. There areseven unknown functions, namely:x,y,0,p,X,Y,Fl9andseven equations to determine them, namely, Equations (1), (5), (6). Tosolve these equations, beginbydetermining6from(5): 3/\ (7) ft=X, 9=\t+M. Next, eliminate Yin(1): cos<9+ sin^=x+Mgsin ecos 6. at1diLJ AndnowXcanbeeliminated by(5),andxyyby(6).Thus Ma-=iMa~+Mgsin ecos0, or (8) where Hence (9) ^=*(sin-slnM) +*o, and fc, M./ . fcsinM\. v?= ^5(cosM-cos0)+^-- Jt+v?o- From (6),xandycannowbefound asfunctions oft;andfinally X,Y,Fucanbedetermined from(1)and(5). Thesystem ofEquations (1), (5), (6)isanexample ofequa- tions called non-holonomic byHertz because some ofthem, namely (6), involve time-derivatives ofthe first order only andcannot bereplaced bygeometric equations between the coordinates. Aninteresting case ofanon-holonomic problemisthat ofa coin rolling onarough table. Itisstudied indetail byAppell, ROTATION 245 Mecanique raiionelle, Vol. I,p.242, ofthe1904 edition, andan explicit solution isobtained interms ofthehypergeometric function. EXERCISES Thestudent shouldfirst, without reference tothebook, repro- duce thetreatment justgiveninthetext, arranging inhismind theprocedure: i)figure, forces, coordinates; ii)motion ofthe centre ofgravity; Hi)moments about thecentre ofgravity; iv)conditions ofconstraint;v)thesolution oftheequations. 1.Solve theproblemofthetwowheels mentioned inthe text. 2.Coin rolling onarough table. Read casually Appell, adopting hissystem ofcoordinates. Then construct independ- ently thesolution, following themethod used intheproblem ofthetext. 3.Theproblem ofthetext,when themass oftheframe is taken intoaccount. Begin with thecasethat thebottom ofthe frame issmooth and itscentre ofmass isatthecentre ofthedisc. 4.Study themotion ofthecentre ofgravity ofthedisctreated inthetext,bymeans oftheexplicit solution ofx,yinterms oft. 25.Resume. Indealing with themotion ofarigid body, there arethetwovector equations: equivalent tosixordinary equations. Itisalways possible totakemoments about thecentre of gravity. The Principle ofWork andEnergy frequently gives auseful integral oftheequations ofmotion. Iftheright-hand side oftheMoment Equationisavector lyinginafixed plane, thecomponent ofanormal tothisplaneis constant, andthusanintegral oftheequations ofmotion is obtained. Sometimes there areconditions which areexpressed byequa- tionsbetween time-derivatives ofthe first order,t= t,butwhich cannot beexpressed byequations between thecoordinates only. 246 MECHANICS The firststepinsolving aproblemistodraw thefigure, mark theforces, andpassinreview each oftheitems justmentioned; reflecting,incase these arenotadequate, onconsiderations of likenature, whichmaybegermane totheproblem. With theforces andthegeometry oftheproblem inmind, nextchoose asuitable Coordinate System.Ifitisdesirable torefer atothecentre ofgravity, aCartesian system with itsorigin there isusually thesolution. These axesmaybefixed inthebody, coinciding with theprincipal axes ofinertia. Ortheir directions maybe fixed inspace. Ortheymaymove inthebody and inspace subject tosome condition peculiar totheproblem inhand. Final Formulation. Itremains towritedown theequations arising from each oftheabove considerations. They must be innumber equal tothenumber ofunknown functions. Besides the differential equations ofthesecond order, thesemay alsoinclude differential equations ofthefirst order, notreducible toequations between thecoordinates. Thesolution ofthese equationsisapurely mathematical prob- lem.Goback frequently over familiar problems and recall the mathematical technique, writing theequations down onpaper, neatly, andcarrying through alldetails ofthesolution. Inthis way, analytical consciousness isdeveloped;itiscomposed of experience andcommon sense. Further Study. There isavastfund ofinteresting problems inRigid Dynamics, ofallorders ofdifficulty, andtwoinvaluable treatises areAppell, Mecanique rationelle, Vols. IandII,and Routh, Rigid Dynamics, Vols.I,II.Routh's exposition ofthe theoryisexecrable, buthislists ofproblems, garnered from the oldCambridge Tripos Papers, arecapital. Theearth isatop,andthestudy oftheprecession andnuta- tion ofthepolar axis isagood subject forthestudent totake upnext. Webster's Dynamics isalso useful intheimportant applica- tions itcontains. The text ishard reading; butthestudent whooncedominates themethod assetforth, forexample, inthe foregoing treatment, canandshould construct hisown solution oftheprobleminhand. Finally, Klein-Sommerfeld, Theorie desKreiselsyinfourvolumes. This isaclassic treatment ofthesubject. The first three vol- ROTATION 247 umes treat thetheory ofthetopbymodern mathematical methods. Thefourth volume, devoted totheapplications inengineering, canbestudied directly through thetheory which wehave de- veloped above, without reference totheearlier volumes. There isadetailed study ofthegyroscopic effect inthecase ofrail- road wheels, theWhitehead torpedo, theship's stabilizer, the stabilityofthebicycle, thegyro-compass, theturbine ofLeval, andalargenumber offurthertopics. CHAPTER VII WORK ANDENERGY 1.Work. InElementary Physics work isdefined asthe product, force bydistance : (1) W=Fl, theunderstanding being thataforce F,constant inmagnitude and direction, actsonaparticle, P,oratapointPfixed ina rigid orelastic body, anddisplaces Padistance Iinthedirection ofthoforce. The definition shallnowbeextended tothecase ofavariable force,stillacting onaparticle oratafixed point ofamaterial body. Let a^xg6 betheinterval ofdisplacement. Let F=f(x) bethe force, where f(x)isacontinuous function. Divide the interval intonpartsbythepoints XQ=a,xl9 ,xn-\,xn=b, Fk andconsider thefc-thsub-interval : -_,. FlG1163*-i^x^xk,Az*=xk-_!. Andnowwedemand that theextended definition ofwork shall besolaiddown that i)thetotalwork shallbeequal tothesumofthepartial works : ii)thework foranyinterval shall liebetween thework cor- responding tothemaximum value oftheforce inthat interval, andthework corresponding totheminimum force : gAW k^ 248 WORK ANDENERGY 249 where FiF Fi' intheinterval inquestion. Now, since f(x)isacontinuous function,ittakes onitsmini- mum value, Fi,intheinterval : andsimilarly,itsmaximum value : FZ=/(*;'), **- HenceWliesbetween thetwosums : Buteach ofthese sums approaches alimit asnincreases, the longest Axjbapproaching 0,andthislimit is-thedefinite integral: Hmi> =fJ Hence therequirements,i.e.physical postulates i)andii)are sufficient todetermine thedefinition ofthework inthiscase :* b (2)W=Jf(x)dx. a Theforegoing definition applies toanegative force, andalsoto thecasethat 6<a;theworknowbeing considered asanalge- braic quantity. Thus ifaforce, instead ofovercoming resistance, isitselfovercome;i.e.yields,itdoesnegative work. Thework which corresponds toavariable displacement, x, where a^x^6,isbydefinition : X (3)W=Jf(x)dx.a Hence (O f-,. *Strictly speaking, wehaveshown that (2) isanecessary condition forthe definition ofworkaccording tothepostulates i)and ii).Itisseenatonce, however, thatconversely Equation (2)affords asufficient condition, also. 250 MECHANICS EXERCISES 1.Show that thework done instretching anelastic stringis proportional tothesquare ofthestretching. 2.Find thework donebythesunonameteor which falls directly into it. 3.Thework corresponding toavariable displacement from xto6,where a^x^6,isbydefinition : & (5)W=ff(x)dx. X What isthevalue ofdW/dxl 2.Continuation :Curved Paths. Suppose theparticle describes acurved pathCinaplane, andthat theforce, F,varies inmag- nitude and direction inanycontinuous manner. What willbe thework done inthiscase? Suppose thepathCisarightlineandtheforce, though oblique tothe line,isconstant inmagnitude anddirection; Fig. 117. Resolve the force into its twocomponents along theline andnormal toit.Surely, we tett>f< j*amust laydown ourdefinition r^--~- i ofwork sothatthework done \ii \ byFisequal tothesum of FlG117theworks Ofthecomponent forces. Now, thework done bythecomponent along thelinehasalready been defined, namely, Flcos^,whereF= |F |istheintensity oftheforce. Itisanessential part oftheidea ofwork that theforce over- comes resistance through distance (orisovercome through dis- tance). Now, thenormal component does neither;itmerely sidles offandsidesteps thewhole question. Itisnatural, there- fore, todefine itasdoing nowork. Thuswearrive atourfinal definition: Thework donebyFintheparticular case inhand shallbe (6) W=Flcos^. Asecond form oftheexpression ontherightisasfollows. Let XandYbethecomponents ofFalong theaxes. LetTbethe angle thatthepathABmakes with thepositive axis ofx.Then WORK ANDENERGY 251 theprojection ofFonABisequal tothesum oftheprojections ofXandYonAB,or Fcos^=XcosT+YsinT. Ontheother hand, x2xl=IcosT, 2/2 2/i=Zsinr. Hence (7) TF=X(x zarOH General Case. IfCbeanyregular curve, divide itinto narcsbythepointss=0,st, ,sn-i,sn=J.LetFbe thevalue ofFatanarbitrary point of the /b-tharc,and let^ibetheangle from thechord (st-i,sk)tothevector F.Then thesum Jb-l FIG.118 where Zfcdenotes thelength ofthe chord, gives usapproximately whatweshould wish tounderstand bythework, inview ofourphysical feeling forthisquantity. The limit ofthissum,when thelongest happroaches 0,shallbe defined asthework, or (8) W Since ~^-=1,As*' itisclear thattheabove limit isthesame as* /n /* limVFkcosfaAsfc IFcos\l/ds.A J Wearethus ledtothefollowing definition ofwork inthecase ofacurved path: iFcos^ds. o(9) W *Cf.theauthor's Advanced Calculus, p.217. Itisimperative thatthestudent learn thoroughly what ismeant byaline integral. 252 MECHANICS Asecond formula forthework isobtained bymeans of(7): < i (10)W=AxCOST+7SUITES=C(x^+Y-J J^as t or (11) W=Cxdx+Ydy. Theextension tothree dimensions isimmediate. The defini- tion (9)applies atoncewithout evenaformal change. Formula (11)isreplaced bythefollowing: (12) W=Cxdx+Ydy+Zdz or (', 6'.c') Cxdx+Ydy +Zdz. (a.b.c) Example. Tofind thework donebygravity onaparticle ofmassmwhich moves fromaninitial point (XQ,yQ,ZQ)toafinal point (xltyltzjalong anarbitrary twisted curve, C. Lettheaxis ofzbevertical andpositive downwards. Then X=0,Y=0,Z=mg ; W=jXdx+Ydy+Zdz=jmgdz=mg(z l-z). C ZQ Hence thework done isequal totheproductoftheforcebythe difference inlevel (taken algebraically), anddepends onlyon the initial and final points, butnotonthepath joining them. EXERCISES 1.Awell ispumped outbyaforcepump which delivers the water atthemouth ofapipewhich isfixed. Show that the work done isequal totheweight ofthewaterinitially inthewell, multiplied bythevertical distance ofthecentre ofgravity be- lowthemouth ofthepipe. 2.Thecomponents oftheforcewhich acts or?aparticle are : X=2x 3y+4z5,Y=zx+8,Z=x+y+z+l2. WORK ANDENERGY 253 Find thework donewhen theparticle describes thearcofthe helix x=cos0, y=sin0, z=70, forwhich ^g2*. 3.IfthecurveCisrepresented parametrically: C:x=/(X), y=*(X),=f(X), XSX^Xi, show thatthework isgivenbytheintegral: 3.Field ofForce. Force Function. Potential. Aparticle intheneighborhood ofthesolar systemisattracted byallthe otherparticles ofthesystem withaforceFthat varies inmagni- tudeanddirection from point topoint. ThusFisavector point- function throughout theregion ofspace justmentioned. Its components along Cartesian axes, namely, X,YyZ,areordinary functions ofthespace coordinates, x,y,z,oftheparticle. In vector form : (1) F=Xi+Yj+Zk. Theexample serves toillustrate thegeneral idea ofafield offorce.Wemay have anelectro-magnetic field, aswhen a straight wire carries acurrent. Ifthenorthpole, P,ofamagnet isbrought intotheneighborhood ofthewire,itwillbeacted on byaforceFatright angles toany linedrawn fromPtothewire andofintensity inversely proportional tothedistance ofPfrom thewire, thesense oftheforce depending onthesense ofthe current. Iftheaxisofzbetaken along thewire, then Z-0, where(r,0,z)arethecylindrical coordinates ofP,andCisa positive ornegative constant. Thus invector form and 254 MECHANICS Force Function. Itmayhappen thatthere isafunction (4) u=<p(x,y,z) such that ,vY_duv_du7__du (5) X~te> W~d~z Such afunction, w,iscalled aforce function. Invector form : Fcanbewritten insymbolic vector form asfollows. Let Vbeasymbolic vector operator, namely: (7) V=i+'+k- ThenVuisdefined as : Hence (9) F=Vu. Gravitational Field. Inthecase ofthe field generated bya single particle ofattracting matter, there isaforce function : (10)= where risthedistance from thegiven fixed particle tothevariable particle, andXisapositive constant. Inthecase ofnparticles, (n)-2Tk> provided theunits areproperly chosen. Ekctro-Magnetic Field. Fortheelectro-magnetic fieldabove described, (12)11=CO. Wemay alsowrite : (13) u=Ctan~l-; v/X9 but thisformula istreacherous, since only certain values ofthe multiple-valued function areadmissible. However, since the wrong values differ from theright ones onlybyadditive con- WORK ANDENERGY 255 stants, wecanusetheformula forpurposes ofdifferentiation, andweshallhave : (14)X=|H=C-=g- vY=f*-C^jhi,Z=0.'dx x2+2/2' ?/ x2+y2 Work. When aparticle describes anarbitrary path inafield offorce, thework doneontheparticle bythe field isgiven by Equation (12) of 2.Ifthere isaforce function, thisformula becomes : i i&u7.du, \ i.e.thechange which uexperiences along thecurve C. Ifthe region inwhichClies issimply connected, orifuisasingle- valued function, then (16) W=u+const. ThusWisindependent ofthepathbywhich theparticle arrived atitsfinal destination, anddepends onlyonthestarting point andtheterminal point: (17) W=u(x, y,z)-u(a, 6,c). Foranyclosed path,W=0. Such afield offorce iscalled conservative. Itistrueconversely that ifthe field represented bythevector (1)isconservative, then there alwaysisaforce function,u.Forthen theintegral: (*.*. (18) u=IXdx+Ydy+Zdz (a.b.c) isindependent ofthepathandsodefines afunction u(x,y,z). Moreover, (19) **,*y,*z. dx dy dz Potential Energy. When afield offorce hasaforce function, u,thenegative ofu,plusaconstant,isdefined asthepotential energy: (20) *=-u+C. Incase, then, apotential <pexists, 256 MECHANICS EXERCISES 1.Show that the field offorce defined bythevector (3)is notconservative. But ifRbeanyregion ofspace such thatan arbitrary closed curve inRcanbedrawn together continuously toapoint notontheaxis,without evermeeting theaxis,though passing outofR,then the field offorce defined inRby(3)is conservative. 2.Ameteor, whichmayberegarded asaparticle,isattracted bythesun(considered atrest)andbyalltherestofthematter inthesolar system. Itmoves from apointAtoapoint B. Show thattheworkdoneonitbythesun is W=Km(-- where rand rxrepresent thedistances ofAandB,respectively, from thesun,andKisthegravitational constant. 4.Conservation ofEnergy. Letaparticle beacted onby any force whatever. Themotion isdetermined byNewton's Second Law : (i)' U/l/ U>C/ U/l/ Multiply these equations respectively bydx/dt, dy/dt, dz/dt, andadd : dzd^z\ _ydx,ydy-dz }m \dt~dt2~r ~dt~dt2^dt~dfi)~ dt dt+dt Theleft-hand sideofthisequation hasthevalue : ~2dtv*' W+~dP+ ~dt2'where v2= Hencemd2Ydx ,y,dyt^dz 2dt dt dt dt Each sideofthisequationisafunction oft,andthetwofunc- tions are,ofcourse, identical invalue.If,then,weintegrate WORK ANDENERGY 257 each sidebetween anytwolimits,<and tlttheresults must tally: f*.vtdt= (J2dtVdtJ\ o 'o Theleft-hand sideofthisequation hasthevalue : Theright-hand side isnothing more orlessthan Cxdx+Ydy+Zdz, taken over thepath oftheparticle ;2,(13). But this ispre- cisely theworkdoneontheparticle bytheforce that acts. Hence Thequantity mv isdefined asthekinetic energy oftheparticle. Wehave, then, inEquation (3)thefollowing theorem. THEOREM. Thechange inthekinetic energy ofaparticleisequal totheworkdoneontheparticle. Ifinstead ofasingle particle wehave asystemofparticles, thesame result istrue. For,from theequationsofmotion of theindividual particles: ,.^ d2xkv (4)mt-^r-Xt,*--= , weinfer that ^(vdxk .dyk,dzk2(Xxk .vdyk,7dzk\+^ Thekinetic energy ofthesystemisdefined as 258 MECHANICS Onintegrating, then, between anylimits <and <1;wehave 89r.-'.- t Theright-hand siderepresents thesum oftheworks doneon theindividual particles,orthetotalwork done onthesystem. The result istheLaw ofWork andEnergyinitsmost general form forasystemofparticles. THEOREM. Thechange inthekinetic energy ofanysystem of particlesisequaltothetotalworkdoneonthesystem. Conservative Systems. Incase the forces areconservative; i.e. ifthere exists aforce function Usuch that theright-hand side ofEquation (5)becomesC/jUg,andso (7) 1\-T=U,-U Thepotential energy, <l>,isdefined as : (8) $=-U+const. Hence (7)canbewritten : (9) 71 !+*t=T+*o- Lettthetotalenergy bedefined as (10) E=T+*. Wehave, then : (11) E,=Em orthetotalenergy remains constant. This istheLaw oftheCon- servation ofEnergy initsmost general form forasystem of particles. 6.Vanishing oftheInternal Work foraRigid System. Con- sider asetofparticles which form arigid system. Letthem be held together bymassless rods connecting them inpairs. Thus theinternal forces withwhich anytwo particles,rat-and m/, reactoneach other areequal aridopposite: (1) Fty+Fn=0, WORK ANDENERGY 259 and liealong thelinejoining theparticles, andfurthermore the distance between theparticlesisconstant;i.e. (2) rl=(Xi-xtf+(yi- 2/y)2+(Zi-ztf isindependent ofthetime, or Ingeneral, however,ifeach particleisconnected bythese rodswith alltheothers, there willberedundant members, so thatthestresses intheindividual rods willbeindeterminate. In that case,letthesuperfluous rodsbesuppressed. Consider thework done ontheparticle mibytherodcon- nectingitwith mj.Itis : /+Y^dyi+Z/dzi andcanbeexpressed bymeans oftheparametertintheform : C(Y{-L-v^Mi-u7^A,// Bythesame token, thework doneonrn,jbymis Since Xu+Xn=0,Ya+Y,-t=0,Za+Za=0, thesum ofthesetwoworks canbewritten intheform : i /(i v(dyi dy\ (dz{dz ij'~~+Yii\dt~~ "~ This lastintegral vanishes. For, theforceFi;iscollinear with thelinesegment connecting wt-withmjfor : Hence theintegrand vanishes identically by(3). 260 MECHANICS Wehave thusobtained theresult that thework donebythe internal forces ofarigid system ofparticlesisnil. Itfollows, then, that thechange inthekinetic energy ofsuch asystemis equal tothework donebytheexternal, orapplied, forces. Look- ingbackward andalsoforward wecannow state thegeneral THEOREM. Thechange inthekinetic energy ofanyrigid system whatever isequaltothework donebytheapplied forces. Forasystem ofparticles theproof hasbeen given. Before wecanextend ittorigid bodies, wemust generalize thedefini- tions ofkinetic energy andwork. 6.Kinetic Energy ofaRigid Body. Consider arigid body. Letthevolume density, p,beacontinuous function. Denote byvthevelocity ofavariable pointPofthebody. Then the kinetiQ energyisdefined as (i) T= extended throughout theregion Tofspace, occupied bythebody. Thevector velocity vofPisthevector sumi)ofthevelocity Valong theaxisofrotation andii)thevelocityv'atright angles tothat axis. Hence (2)v*=V*+r*a>* where rdenotes thedistance ofPfrom the axis,and coisthe angular velocity about theaxis. Substituting thisvalue in(1)we find: Hence (3) i-Tr+TT Letvdenote thevelocityofthecentre ofgravity, G;and lot hbethedistance ofGfrom theaxisofrotation. Then 0t=F2+A2co2 . Moreover, I=7+Mh\ where 7isthemoment ofinertia about aparallelaxisthrough G.Hence (4) r WORK ANDENERGY 261 Inequations (3)and(4)iscontained thefollowing general theorem. THEOREM. The kinetic energy ofarigid body isthesum of thekinetic energy oftranslation alongtheinstantaneous axisandthe kinetic energy ofrotation about theinstantaneous axis. Itcanalsobeexpressed asthesumofthekinetic energy ofaparticle oflikemass, moving with thevelocity ofthecentre ofgravity, and the kinetic energy ofrotation aboutanaxisthroughthecentre ofgravity, paralleltotheinstantaneous axis. OnePoint Fixed. Letapoint ofthebody beatrest. Let the(, 17,f)-axesliealong theprincipal axes ofinertia, being theorigin. Then thecomponents ofthevector velocity vofany point fixed inthebody are : =770^ fa?,/ fw~~w Vt Hence (5) T=%(Aw$2+BuJ+Cco^2 ). Iftheaxes ofcoordinates arenottheprincipal axes ofinertia, then (6)T=i TheGeneral Case. From(4)and (5)weinfer that (7) T=Mv*+(Ap*+Bq*+Cr2 ), where A,B,Carethemoments ofinertia about theprincipal axes ofinertia through thecentre ofgravity, andp,q,rarethe components ofthevector angular velocity walong these axes. 7.Final Definition ofWork. Wehave hitherto assumed that thepointofapplication, P,oftheforce isfixed inthebody. Sup- posePdescribes acurveCeither inthebody orinspace. How shall theworknowbedefined? Take thetime asaparameter. Divide theinterval TO^t^rt intonparts bythepoints=r<tv<-<tn-\<tn=rv LetQkbethepoint fixed inthebody, which attime t=tkwill 262 MECHANICS reachC;letTkbeitspathinspace, and letVkbeitsvelocity in spacewhen itreaches C;cf .Fig. 120. Fortheinterval oftime wemay take theforce asconstant, F=FA,thevalue ofFat theintersection ofI\with C,and letF* actonthepoint Qkthroughout the in- terval Atffc.Then J?kwilldowork equal approximately to (1) Fkvkcos\{/kA^, FIG.119where\l/kistheangle from TktoF&at Pk.IfQkisdisplaced along thetangent toTfcadistance vkktk,theexpression (1)represents thework precisely. Wewillnow define thework as F Km cos or,dropping ther-notation andexpressing theinterval oftime astQg t^/t: FIG.120 (2) cf.Fig. 119. Invector form thework isW=IFvcostdt; (3)! =/Fvttt, ?r where visthevector velocity ofQatP,andFv isthescalar product ofthese vectors. Wehave used thetime astheindependent variable, orthe parameter, interms ofwhich todefine thedisplacement. But theresult isinnowisedependent onthetime inwhich thedis- placement takes place. Any other parameter, X,would have done equally well, provided d\/dtiscontinuous andpositive (or negative) throughout. For j.ds ,.ds, xvat=-TTat=-=-aA. WORK ANDENERGY 263 This formulation ofthedefinition ofwork inthegeneral case isduetoProfessor E.C.Kemble. Example1.Abilliard ball rollsdown arough inclined plane without slipping. Find thework doneby^ theplane. Here,Ciseither thestraightlineorthe circle; each curve Tisacycloid with cusp atPandtangent normal toC;and v=0.HenceW=0.FIG.121 Example2.Thesame, except thattheballslips. The curveCshall betaken along theplane. Thenormal component R=Mgcosadoesnowork; thecomponent along theplane, F= cos does. Let sbethedistance travelled bythecentre oftheball;0,theangle through which the ball has turned. Thecurve Tisatrochoid tangent toCatP.Hence\l/=or*,FIG.122 ds dd and 'cos {(!s)-<*(0i-0o)l- Observe that inthedefinition, Equation (2),vispositive or0. Itwould not, therefore, berightinthisexample towrite _ds d8 v~Tta dt EXERCISES 1.Check theresult inExample 2bydetermining themotion oftheballandcomputing thechangeinkinetic energy. 2.Atrain isrunning attherate of40m.anh.Thebaggage carisempty, andthesmall sonofthebaggage master isdisport- 264 MECHANICS inghimself onthefloor. Herunsforward, then slides. Ifhewas running attherateof6m.anh.when hebegan toslide,and slid 5ft.,howmuch work didhedoonthecar? Compute bythedefinition andcheck yourwork bysolving forthemotion. 3.Aropeisfrozen tothedeck ofaship. The freeend is^ Ihaaled ,over asmooth pulley atP. ,**!^~^~"^ , Ittakes avertical component ofBR=20Ibs.tofreethefrozenpart. Howmuch work isdone ? Take thefrozen part asstraight, andPinthevertical plane throughit. 4.Extend thedefinition ofwork toabody force, F,whereF isacontinuous vector, defined ateach point ofthebody: 5.Show thattheinternal work duetotherope inanAtwood 's machine isnil.Would thisbethecase iftheropestretched? 6.Anumber ofrigid bodies areconnected byinextensible cords thatcanwindandunwind onthem inanymanner without slipping. Show that thesum oftheworks donebythe;cords onthesystem andthesystem onthecords isnil. First, extend thedefinition ofwork soastoinclude thecaseofthework doneon thesystem bythepart ofacordwhich isincontact withabody. 8.Work Done byaMoving Stairway. Consider thework which anescalator, ormoving stairway, doesonaman ashe walks up.The forces that actontheman are/,S,andMg, where R,Sarethecomponents ofthe forcewhich theescalator exerts onhis foot,andMg acts athiscentre of gravity. The curve Tisalways a right linelyingintheinclined plane, and ds dtFIG.124 where sdenotes thedistance theescalator hasmoved since the mancameaboard WORK ANDENERGY 265 The forceRdoesnowork, since for it^=ir/2.Thewhole work isduetoS=Fcos^,and is : Thespeed oftheescalator isconstant;denote itbyc.Thus (2) W And (3)I=c*t, where Iisthedistance theescalator hasmoved while theman isrunning up. Ontheother hand, consider themotion ofthecentre ofgrav- ityoftheman. Lettheaxisofxbetaken uptheplane. Then M--f=SMgsina, (4) Mu^Mu=ISdt sna, where u dx/dt. Itfollows, then, from(2)and(4)that (5) W=c(Mu lMV,Q)+Mgct^sina. Iftheman steps offwith thesame velocity with which he stepped on,u^=w ,then, with thehelp of(3), (6) W=Mglsina. Now h=Isina isthevertical distance bywhich themanwould havebeen raised inthotimehewasontheescalator ifhehadnotrun,butstood still. Hence, finally, (7) W=Mgh. Itmakes nodifference, then, whether theman runs fastor slowly, upordown. Theonething thatcounts ishowlonghe isontheescalator. Thuswhen small boys playontheescalator, running upanddown, thework theescalator does increases in 266 MECHANICS proportion tothetime they areonit,provided they arrive and leave with thesame velocity. 9.Other Cases inWhich the Internal Work Vanishes. i)Two Rigid Bodies, Rolling withoutSlipping. Here, the action and reaction areequal and opposite, though not in general normal tothe surfaces. Moreover, thevector velocity ofthepoint ofcontact, regarded asapoint fixed intheonebody, -- ^p^f-must bethesame asthevectorvelocityofthe ---^[>^ pointofcontact, regarded asapoint fixed in Vi'v2 theother body. Theworks donebythetwoforcesFj,F2on FIG.125 thetwobodies are : *i i W1=CF.V, cosftdt,W2=(*F2v2cosftdt. to ButFl=F2,i\=v2,ft+ft=TT.Hence W,+W2=0. ii)TwoSmooth Rigid Bodies, Rolling and Slipping. Here theforces FlandF2areequal andopposite, andnormal tothe surfaces atthepoint ofcontact. The ve- locitiesVjandv2arenotequalwhen there is slipping ;buttheir projections onthenormal areequal: vlcosft+vzcosft=0, Since furthermore Fl=F2)wehave : W1+W2=0. Fio.126 Wehave already mentioned thecase ofrigid bodies onwhich inextensible massless strings wind andunwind, 7,Exercise 6: andmassless rodswereshown in5todonowork. Thus syj? terns ofrigid bodies connected byinextensible strings and rods, eventhough thepoint ofapplication oftheforce exerted bythe string orrodbevariable, shownointernal work. 10.Work andEnergy foraRigid Body. THEOREM. The change inthekinetic energy ofarigid body, actedonbyanyforces, isequal totheworkdonebytheseforces. WORK ANDENERGY 267 Weprove thetheorem first fortwospecial cases. CASE I.NoRotation. Here, thechange inkinetic energyis m ^i2_-fl^o2w2 2' i.e.thechange inthekinetic energy ofaparticle ofmassM, moving asthecentre ofgravityismoving. Ontheother hand, consider thework donebyone ofthe forces, F : (2) W= Since there isnorotation, v=v,^=^, and (3) w=CFvcos$dt. HenceWisthework doneonaparticle atthecentre ofgravity bythesame force, andthetheorem istrueby4. CASE II.OnePoint Fixed. Here, Eulcr's Dynamical Equa- tions, Chapter VI, 13,determine themotion. Consider aforce Fwhich actsonthebody atP.Letrbethevector drawn from toP.Then thevofthedefinition ofwork, 7,(3)is v=&Xr. Hence (4) Fvcos^=F-v=F(wXr). Ontheotherhand thevector moment ofFabout is From Euler's Equations, I.e.,wehave : Adp.ndq.~dr T ,,, ,,r ^TT/"" '5/" /it~P'^'' Hence (5) $(Ap*+Bq*+ Theleft-hand side isthechangeinkinetic energy. Now (6) Lp+Mq+Nr=M-w=co-(rXF), 268 MECHANICS andso (7) Fvcos^=Lp+Mq+Nr. For itistrue ofanythree vectors that a-(bXc)+c(bXa)=0, since a-(bXc)=a1a Moreover, HencewXr=-(rXo>). F-(wXr)=-(rXF). From (5)itfollows, then, that forasingle force, thechange inkinetic energyisequal tothework done. Forthecase of nforces theproofisnow obvious. Theextension tothecase of body forces and forces spread outcontinuously over surfaces or along curves, presents nodifficulty. Remark. Wehave shown incidentally that thework done onarigidbody withonepointfixed is h Jo>t, whereM=La+Mp+Nj istheresultant couple. TheGeneral Case. Consider firstasingle force, F.Thework itdoes is W=/Fvd*. Here, V=V+V', where visthevelocityofthecentre ofgravity and v'istheve- locity ofthepointQrelative tothecentre ofgravity, asitflashes through P.Hence W= WORK ANDENERGY 269 The first integral hasthevalue Mv* Thesecond integralisequal totheright-hand sideofEquation (5). Thus thetheorem isproved forone force. Foranumber of forces theproofisnowobvious. EXERCISES 1.Aball isplaced onarough fixed sphere ofthesame size andslightly displaced near thehighest point. Find where it willleave thesphere. Letp,haveanyvalue. 2.Aweightless tubecanturn freely about oneend.Asmooth rod isinserted inthetubeandthesystemisreleased from rest with thetube horizontal. How fast will itbeturning when it isvertical ? 3.Acylindrical can isfilled withwater andsealed up. Itis mounted sothat itcanrotate freely about anelement ofthe cylinder. Show that itoscillates likeasimple pendulum, pro- vided thecan issmooth. 4.Inthepreceding problem, theheight ofthecan isequal toitsdiameter, andthecanweighs 5Ibs.Thewater weighs 31Ibs.Find thelength oftheequivalent simple pendulum. 5.Acircular tube, smooth inside, plane vertical,ispartly filled with water. Thetube isheld fastandthewater isdis- placed, then released from rest. Show that itoscillates like asimple pendulum, anddetermine thelength ofthelatter. 6.Abenttube intheform ofanLismounted sothat itcan slide freely onasmooth table. The vertical arm isfilled with water, andthesystemisreleased from rest.How fast will it bemoving where thevertical armhasjustbeenemptied? Assume thetubesmooth inside; and alsotake theweight of thetubewith itsmount equal totheweight ofthewater. 7.ThecanofQuestion 4isallowed torolldown aroughin- clined plane, starting from rest. Find theacceleration ofthe centre ofgravity. CHAPTER VIII IMPACT 1.Impact ofParticles. Lettwoparticles, ofmassesmland W2,bemovinginthesamestraightlinewith velocities Uiand w2,and letthem impinge oneach other. Tofind their velocities after theimpact. Isolate thesystem consisting ofthetwoparticles. Then no j*1 w2 external forces act,and sothe /$|Qxgfr ^f u2 momentum remains unchanged. FIG.128 Hence (1) m^ m^!+m2u2w2w2=0. Asyet,nothing hasbeen saidabout theelasticity ofthepar- ticles. Theextreme cases are: perfect elasticity (liketwo bil- liard balls) and perfect inelasticity (liketwo balls ofputty). Ineach case there isdeformation ofthebodies fornowwe willnolonger think ofparticles, but, say, ofspheres, andthe velocities ultw2,etc.refer totheir centres ofgravity. During thedeformation themutual pressures mount high, andeven ifother (ordinary) forces act, their effect isnegligible, compared with thepressuresinquestion. Inthecase ofperfect inelasticity, there isnotendency toward arestitution ofshape, and so,when themaximum deformation hasbeen reached, the mutual pressures drop tonothing at all.Atthis point, the velocities ofthetwocentres ofgravity arethesame, u(=u2, andhence thiscommon velocity, which wewilldenote by f7,is givenbytheformula : (2) u=miUl "*"m*u*ml+m2 Thus theproblemissolved forperfect inelasticity. Forpartial elasticity,itishelpful topicture theimpact asfollows. The 270 IMPACT 271 motion ofthecentre ofgravityofeach ball isgiven bythe equation: Forthe firststage oftheimpact,i.e.uptothetime ofgreatest deformation,t=T,wehave, onintegrating each side ofeach equation between thelimits andT : T T (4)mlul<-T / i-oJt=T /=IJ /=e/Rdt. The integraliscalled animpulse,* and isdenoted byP : T (5) P=CRdt. o Hence (6)=P =P. m2U Thesecond stage oftheimpact now begins, astheballs are kicked apart bytheir mutual pressures. Onintegrating the equations (3)between thelimitsTand7V ,wehave : T' (7)m^fftdt',CR'dt'. Now itiseasily intelligible physicallyifweassume that, in thecase ofpartial elasticity, thevalue ofR'stands inaconstant ratio tothevalue ofRatcorresponding instants oftime, orthat (8) whenR'=eR, T-t=t'-T. Here thephysical constant eis called thecoefficient ofrestitution. Itliesbetween and 1:tTt' FIG.129 (9) <e<1, *Sometimes spoken ofasanimpulsive force ;butthisnomenclature isunfor- tunate, sincePisnotofthenature ofaforce, which isapush orapull,butrather isexpressed byachange ofmomentum. Moreover, thedimensions ofimpact are ML/T, notML/T*. 272 MECHANICS being inthecase ofperfect inelasticity and 1forperfectelas- ticity. Hence 7" T (10) P'=/Vdtf=eCRdt=eP. Equations (7)thus yield thefollowing: (U) {^-Ilf/=df The four equations, (6)and(11), contain thesolution ofthe problem. Between them,UandPcanbeeliminated, andthe resulting equations canthenbesolved foru[,u2.The result is : Uf\:*vn..4- (12)' U2ml+m,2 e)m1u1+(ra2 TheCasem2=oo. IfinEquations (12)wealloww2toincrease without limit,weobtain theequations: U2=U2. These equations donotprove that,when themassm2isheld fast,orismoving withunchanging velocity u2,thevelocity ofthe massmlafter theimpactwillbegiven bythe firstequation (13), butthey suggestit.The proofisgiven bymeans ofthe first oftheequations (6)and (11), resulting astheydorespectively from the first oftheequations (4)and(7),combined with (10); Uhaving heretheknown value u2. If,inparticular, u2=0,wehave : (14) u{=-,. Perfect Elasticity,e*=1.Equation (14)becomes inthiscase u[=-MI,andtheballrecedes with thesame velocity asthat withwhich itimpinged. Ifthemasses areequal,ml=w2,Equations (12)become : IMPACT 273 and the balls interchange their velocities. This latter phe- nomenon canbeillustrated suggestively bytwo equal ivory ballssuspended sidebysidefrom strings ofequal length, after themanner oftwo pendulums.Ifoneballhangs vertically atrest,andtheother isreleased from anangle with the vertical, thesecond ball willbereduced torestbytheim- pact, andthe first will risetothesame height onitssideofthevertical asthatF fromwhich thesecond ballwas released. Thus thevelocities willbesuccessively interchanged atthelowest point ofthecircular arc. Critique oftheHypothesis (8). Inthishypothesis wehave taken forgranted anamount ofdetail inthephenomenon before usfarinexcess ofwhat thephysicistwilladmit asreasonable inviewing theactual situation, andhemay easily berepelled by sodogmatic anassumptioninacasethatcannot bosubmitted to direct physical experiment andwhich, afterall,isfarlesssimple thanwehave ledthereader tosuppose, since theproblemis essentially one inthe elasticity ofthree-dimensional distribu- tions ofmatter. The objection, however,iseasily met.We may takeEquation (10): P'=eP, asthephysical postulate governing impact. EXERCISES 1.Aball of6Ibs.mass, moving attherate of10m.anh. overtakes aballof4Ibs.massmoving attherate of5m.anh. Determine their velocities after impact, assuming that thecoeffi- cient ofrestitution is . Ans. 7and9.5m.anh. 2.Thesame problem, when theballs aremoving inopposite directions. 3.Aperfectly elastic sphere impinges onasecond perfectly elastic sphereoftwice themass. Find thevelocity ofeach after theimpact. 4.Newton found that the coefficient ofrestitution forglass is^f.Ifaglass marble isdropped from aheight oftwo feet onaglass slab,howhigh will itrise? 274 MECHANICS 5.Inthelastquestion, what willbetheheight ofthesecond rebound? What willbethetotal distance covered bythemarble before itcomes torest?* 6.Find thetime ittakes themarble tocome torest. 7.Intheexperiment with thependulums described inthe text, theimpingingball willnotbequite reduced torest,because notwomaterial substances arequiteelastic.If,forgiven balls, e 0.9,show that theballwhich isatrestshould beabout 11 percentheavier than theother one,inorder toattain complete restforthelatter. What percent larger should itsradius be? 8.If,inthelastquestion, thependulum bobs areofglass. e=^f,findtheratio oftheir diameters. 9.Iftwoperfectly elastic balls impinge oneach other with equal velocities, show thatoneofthem willbebrought torest ifitisthree times asheavy astheother. 10.Determine thecoefficient ofrestitution foratennis ball bydroppingitandcomparing theheight ofriierebound with theheight from.whichitwasdropped. 11.Some pitchers used todeliver aslow ball toBabe Ruth, believing thathecould notmake ahome runsoeasily asonafast ball. Discuss themechanics ofthesituation. J2.Continuation. Oblique Impact. Lettwospheres impinge atanangle, andsuppose them tobeperfectly smooth. Tode- termine thevelocities after the impact. Letthelineofcentres betaken astheaxis ofx.Thedeforma- tion ofeach sphereisslight, and Fia131theforce exerted bytheother sphere, spread outasit^sover avery small areaandacting normally ateach point ofthisarea, willyield aresultant force, K,nearly parallel totheaxis ofx. Forthe firstsphere wehave : *Thephysics ofthesecond part ofthisproblem (and ofthenext) isaltogether phantastic. After afewrebounds wepassbeyond thedomain within which the physical hypothesis ofthetext applies, andthefurther motion becomes apurely mathematical fiction. Itisamusing forthosewhohaveasense ofhumor inscience. Butfortheliteral-minded person, behephysicist ormathematician, itisdangerous. IMPACT 275 (15) whereX=#cos6,F=72sinc, ebeing numerically small andxltylreferring tothecentre of gravity, andforthesecond sphere, (16) rn^X, *, F. Onintegrating (15)weobtain : ,.,,_>. (17)=-/-Yd*, m,^1^=/Yrf*.*-o Jdl*=oJ Andnowwedenote the firstimpulse byP,andlaydown the postulate thatthesecond impulseis : (18) Thus theintegrals of(15)and (16)load totheequations: m^U mlul=P [mtFm^^ (19)1=P which hold forthe firstepoch oftheimpact, theequations for thesecond epoch being, asinthecorresponding case of1,the following: m}u\ m,U=eP{m*v\m}V= 1. 4m2U=eP Im^m2F=0. Forweassume asthere thephysical postulate: (21) P'=eP. The result atwhich wehave arrived isseen tobethefollowing. Thecomponentofthevelocity ofeach sphere perpendicular to thelineofcentres hasbeenunchanged bytheimpact, (22) v[=vl9v'2=v2. 276 MECHANICS Thecomponentsofthevelocity along the line ofcentres are changed precisely asinthecase ofdirect impact, 1,Equa- tions (12): u\= -*-- mi (23) Mj=ml Kinetic Energy. When e lti.e.when thespheres areper- fectly elastic, thetotal kinetic energyisunchanged bytheimpact, forthen , 2_ >\ ,r~~~"~~ ' asisshown bydirect computation from (23),and theequa- tions (22)hold inallcases, whether e=1ore<1. When e=0,i.e.when thespheres aretotally inelastic, aneasy computation shows thatthekinetic energy hasbeen diminished. Theintermediate case, <e<1,istreated inthesame way. Itfollows from direct computation that theleft-hand side of Equation (24)hasthevalue : (m^+w2M2)2+mlmz(u lu2)ze2 m2) and this isatonceshown tobelessthan theright-hand side. Theterms arising from Equations (22)donot, ofcourse, affect theresult. EXERCISES 1.Asmooth ball travelling south-east strikes anequal ball travelling north-east with one-quarter thevelocity, their line ofcentres atthetime ofimpact being eastandwest. Ife=^, findthevelocities oftheballs after impact. 2.Asmooth ball strikes ahorizontal pavement atanangle of45. Find theangle ofrebound ifthecoefficient ofrestitution isf. 3.Show that thekinetic energy oftheballs ofQuestion 1is diminished intheratio of245/272 bytheimpact. 4.Thecorresponding question fortheballofQuestion 2. IMPACT 277 3.Rigid Bodies. Letarigidbody beacted onbyasingle impulse. Bythat ismeant thepostulates about tobelaid down, suggested bythefollowing physical picture. AforceF acts atapoint (x,y)fora short time, mounting high in intensity. Ordinary forces,if present, produceinthis inter- valoftime, ^t^T,only slight results, and intheulti- mate postulates donotappear, sothey arenotconsidered in^FrG 132 thepresent picture. Thethree equations which govern themotion are : -=Y dt**' Onintegrating with respect tothetimewefind (2)dt M(u'-u)=Cxdt, M(v'-v)=CY U T T 7(0'-0)=Ax-x)Ydt- f(y-y)Xdt. Concerning Fwewillassume that thevector changes con- tinuouslyinmagnitude and direction during theinterval of time inquestion, andthat thepoint ofapplication, (x, T/),also moves continuously, remaining nearafixed point (a,b)through- outtheinterval. Let (3) x=a+ ,y=6+ 17. Then,rjareinfinitesimal with T.Let (4) P=Cxdt, Q=CYdt. 278 MECHANICS The lastEquation (2)nowbecomes : (5) /('-0)=(a-*)Q-(b-y)P T T +CtYdt-Cr,Xdt. Weshould liketoinfermathematically thatfrom thehypoth- esisthat theintegrals (4)approach limits whenTapproaches 0, theintegralsinthelast lineof(5)converge toward 0;forthen weshould have theequations: M(u'-u)=P, (6)' 7(12'- $2)=(a-x)Q-(b-y)P, PandQheredenoting thelimiting values oftheintegrals (4). This inference caninfactbedrawn, provided theangle through which thevector Frangesislessthan 180. Equations (6) then hold, and, inparticular,itfollows, oneliminating Pand Qbetween them, that (7) 7(Q;- fl)=M(a-x)(v'-v)-M(b-y)(u'-u) or (8)fc2 (12'-Q)=(a-x)(v'-v)- (b- y)(u'-u). AnExample. Arod isrotating about oneend,and itstrikes anobstruction, which bringsitsuddenly torestwithout any reaction onthesupport. What point oftherodcomes into contact with theobstruction ? {Letthedistance from thestationary end beh,and let Ibethelength oftherod. O k'Let FIG.133 v=C; then ft=-C. Since M/2 u=0,fi'=0,^=0,12'=0, /=^-, wehave : Hence Thepointiscalled thecentre ofpercussion. IMPACT 279 4.Proof oftheTheorem. The proofisgiven bymeans oftheLaw oftheMean, which isasfollows. Letf(x), <p(x) betwo functions which arecontinuous intheclosed interval a^xgb,and let<p(x)notchange sign there. Then ft//*f(x) <p(x)dx=f(x')I<p(x) dx, a< a:'<6. e/(9) Inthepresent casetheaxescanbesochosen that gY. Hence (10) C^Ydt=' f*Ydt, where' isthevalue ofatasuitable point,t= ',intheinterval gg77 .Now, byhypothesis, andT?approach uniformly, i.e.thelargest numerical value that either hasintheinterval ^tgTapproaches 0;andfurthermore, alsobyhypothesis, theintegral ontheright approaches alimit, Q.Hence theinte- gralontheleftapproaches0. Iftherange oftheangle ofFdoesnotexceed 90,theaxes of coordinates canbesochosen that neitherXnorYchanges sign intheinterval ^t^77 ,andthen itcanbeshown asabove thatboth integralsinthesecond lineof(5)approach0. Inthemore general case thatFiscontained merely within anangle lessthanTT,theaxescanbechosen inmoreways than onesothatYwillnotchange sign. If(x,y)refer toonesuch choice and (x',y')toasecond, then (11) wherex'=ax+by yr=ex+dy FIG.134a=cos7, b=sin7, c= sin7, d=cos7. Thesame transformation holds*with respect tothevector F : *Itisinsuch acase asthepresent onethat thescientific importance ofthe proper definition of(-artesian coordinates, laiddown inAnalytic Geometry, ia revealed. That definition begins with directed linesegments onaline,proceeds tothetheorem thatthesum oftheprojections oftwobroken lineshaving thesame 280 MECHANICS fX'=aX+bY (12> Ir-T +fl' andalsowith respect to(,jj): r=of+6, Hence r r T T T (14)A'*" dt=acAxA+6cAx dt+bdCr,Y dt+ad ft-Ydt.0000 The integral ontheleft,andthelasttwointegrals ontheright, approach thelimit with 77 ,ashasbeenshown above. We willshow that this istrue also oftheother integrals, andhence inparticular ofthelastintegralin(5).Todothis, writedown Equation (14) fortwo choices ofaxes (xf ,y')subject tothe above conditions andcharacterized bytwovalues ofy:yland 72,where yl^0,y2^0,yl^72,andsolve theresulting equa- tions forthetwo integrals inquestion. Thedeterminant of theequations, hasthevalue sinylsiny2sin(yz-7^, and sodoes notvanish. Thus theintegrals forwhich weare solving areseen tobelinear functions ofintegrals that areknown toapproach 0,and thiscompletes theproof. The Restriction. Thetheorem isnottruewhenFisrequired merely tovary continuously with tinthein- terval ^t^T,asthefollowing example shows. Let X=Fcos<?jY=Fsin<p; =pcos^, 77=psin\l/. FIG.135 Then extremities, onanarbitrary line, isthesame for'both lines,andendsbydeclaring thecoordinates ofapoint astheprojections ontheaxes ofthevector whose initial pointistheorigin andwhose terminal point isthepoint inquestion. With that definition, Equations (12)and (13)aremerely particular cases ofEquations (11), sinceboth setsofequations express theprojections ofavector onthecoordinate IMPACT 281 LetFandpbeconstants, and let _2wt TT _2wt Then r T P-fx* Ffa*-0, If,now,weset __ /TT r__*_P *> f rpy theintegrals (4),being always 0,eachapproach limits, andso thePandQofFormulas (6)have each thevalue 0.Butthe thirdEquation (6)doesnothold. Butmayitnot stillbesufficient, inorder tosecure thevanish- ingofthelimit oftheintegral T /(*r- todemand thatY^0?That this isnotenough,isshown by modifying theabove example asfollows. Let X=Fcos?,Y=0. The integral then hashalfthevalue ithadbefore; hence, etc. EXERCISES 1.Auniform rodatrest isstruck ablow atoneend, atright angles totherod. About what pointwill itbegin torotate? 2.Apacking box issliding overanasphalt pavement, when itstrikes thecurbstone. Find thespeed atwhich itbegins to rotate. 3.If,inthepreceding question,thepavementisicy,and if thebox, before itreaches thecurb, comes toabare spot, M=1> what isthecondition that itshould nottip? 282 MECHANICS 4.Iftheboxtips, findwhether itwillslide, orrotate about afixed line. 6.Show that greater braking powerisavailable when the brakes areapplied tothewheels oftheforward truck ofarail- road car. 6.Ifallfourwheels ofanautomobile arelocked, compare the pressureoftheforward wheels ontheground with that ofthe rearwheels. 7.Alamina isrotatinginitsownplane about apoint 0,when itissuddenly brought torestbyanobstruction atapointP situated inthelineOGproduced. Show thatOP isequal tothe lengthoftheequivalent simple pendulum, when thelamina issupported at andallowed tooscillate under gravityina vertical plane. 5.Tennis Ball,Returned withaLawford. Consider atennis ball,returned over thenetwith flattrajectory aridrotation such that thelowest point ofthe ball ismoving backward. The ground thus exerts aforward force, andwewillassume that this state ofaffairs holds throughout theimpact. We shall have, then, thefollowing formulation oftheproblem: M(U-M)=QMfa-{/)'==eQ M(V-VQ)=Q M(v,-F)=eQ where ^^^^no F=pR Fid.136 istheimpulse. First ofall, V=0, forthepoint ofgreatest deformation ismarked bythecentre ofgravity oftheball's ceasing todescend. Thuswehave seven equations fortheseven unknowns, u {,vltcot,[/,V,12,Q. Itisnoweasy tosolve. Observe that VQ<0,co<0, u>0. IMPACT 283 Wehave, then : Q=M(-t>), vl=e(-VQ), Thevalue ofc^isnotinteresting. What wedowant toknow is theslope ofthetrajectory attheendoftheimpact ;i.e. v___ e( VQ)_ ___e\_^" u+(1+e)M(-t, )~ 1+(1+e)/i\f where X=(VQ)/U Qisthenumerical value oftheslope before theimpact. Astheballhasadropduetothecut,Xwillbeconsiderably larger numerically than theslopeinthepart ofthetrajectory justpreceding thelasttenfeetorsobefore touching theground. Itmight conceivably have avalue asgreat as .Thevalue of eisabout 0.8.juvaries considerably andmight beashigh as . Thus &=.15, *i asagainst X=.20,orthesteepness oftherebound isonly three- fourths thesteepnessoftheincident path. Notonlydoes theball riseatasmaller angle, butthehorizontal velocityisincreased bynearly 10percent;for u,=w[l+(1+<0/*A]=1.09t* . For thisdiscussion tobecorrect itisessential that theball maintain itsspinthroughout thewhole impact. This explains thenature ofthestroke. Theracquet hasahighupward velocity while theball isontheguts. The ball loses spinduring theflight before theimpact, dueto theairresistance causing thedrop, and this lossmay easily be comparable with thelossduring theimpact. Itwould beinter- esting totakemotion pictures oftheball,showing thetrajectory justbefore andjustafter theimpact. EXERCISES 1.Abilliardball, rotating about ahorizontal axis,fallson apartially elastic table. Find thedirection oftherebound if ju=%and e=.9. 284 MECHANICS 2.Arod,movinginavertical plane, strikes apartially elastic smooth table. Determine thesubsequent motion. 3.Thepreceding question, with thechange that thetable is rough, M=i- 4.Question 2foratable that iswholly inelastic andinfinitely rough. 5.Arigid lamina isoscillating inavertical plane about a point when itstrikes anobstacle atPwhose distance from isequal tothelength oftheequivalent simple pendulum. Show that itwillbebrought torestwithout anyreaction ontheaxis. ForthisreasonPiscalled thecentre ofpercussion. 6.Arigid lamina, atrest,isstruck ablow atapoint 0.Find thepoint about which itwillbegin torotate. 7%Asolidcone,atrest,isstruck ablow atthevertex ina direction atright angles tothe axis. About what line will it begin torotate ? CHAPTER IX RELATIVE MOTION ANDMOVING AXES 1.Relative Velocity. Itissometimes convenient torefer the motion ofasystem tomoving axes. LetObeapointfixed inspace. Let0'beapoint movinginanymanner, likethe centre ofgravity ofamaterial body, orthecentre ofgeometric symmetry ofabody whose centre ofgravityisnotat0';itis apoint whose motion isknown, oronwhich wewish particularly tofocus ourattention. Finally,letPbeanypoint ofthesystem whose motion wearestudying. Then (1)r=r+r', dr_drdi^ dt" dt^dt' or (2) v=v+v'. , i , iFlG -137 The choice ofnotation ishere particularly important boldface letters denote asusual vectors because wehave twoanalyses toemphasize, namely, i)thebreaking up ofthevelocity vintothetwo velocities vandv';andif)the breaking upofv'intothetwovelocities : (3)v'=vr+v., where vr,therelativevelocity, andvethevitesse d'entrainement are presently tobedefined. For thispurpose wemust first recall theresults ofanearlier study. 2.Linear Velocity inTerms ofAngular Velocity. InChap- terV, 8,wehave studied themotion ofasystem referred to moving axes(, 17,f)with fixed origin 0.Here, (4)r=a+iff+fy and +r/4+f7- 285 286 MECHANICS Thisequation represents ananalysisofthevelocity (6) v= ofthepointPintotwo velocities, namely, (7) v=vr+ve, where (%} v= 4-#4-^ istherelative velocity ofPwith respect tothemoving axes;i.e.the absolute velocity whichPwould have ifthe( ,r;,f)-axes were atrestandthepointPmoved relatively tothem just asitdoes : (9)=f(t), ??=<p(t), f=\l/(i). Secondly, (10) Ve=a+rift+f7 isthe vitesse d'entrainement, the Schleppgeschwitidigkeit, the velocity withwhich that pointQfixed inthemoving space and flashing through Pattheoneinstant, t,ismoving inspace. Let (w)bethevector angular velocity ofthemoving axes : (11) (w)=pa+qp+ry, where Then (12) ve=()Xr,FIG.138 or (13) +(&-fp)j8+ (77??-y. The final result isasfollows :Thecomponentsofv,ordr/dt, along theaxes are (14) RELATIVE MOTION ANDMOVING AXES 287 Irepeat:These aretheformulas when themoving axes have their origin, 0',fixed :r=0,v=0,v=v'. 3.Acceleration. Returning now tothepointPof 1and Equations (1)and(2),wedefine itsacceleration asthevector: Hence (16) a=+, or (17) a=a+a'. The firsttermontheright, a,requires nofurther comment. Itismerely theacceleration infixed space oftheknown point 0'. Thesecond term, a',relates totherotation andadmits ofanum- berofimportant evaluations. First Evaluation. The first ofthese isasfollows. Let a'be denoted bya.Then < -T Wemay identify thevariable vector v'with thevariable vector rof 2,Formula (4) ;for,ofcourse, rwasanyvector, moving according toanylawwewish. Now,wehave evaluated the right-hand side of(18)bymeans ofEquations (14). Hence the componentsoftheright-hand side of(18)areobtained bysub- stitutingintheright-hand side of(14) for,rjffrespectively v >v^v - Ontheother hand, write (19) a=a$a+a^ft~f~fl<7- Thuswoarrive atthefinaldetermination ofainterms ofknown functions : __dv**__ (20)dVrt=~ a$jT-+pity qv$ These aretheformulas referred toastheFirst Evaluation. 288 MECHANICS Second Evaluation. TheTheorem ofCoriolis. Another form forthevector acanbeobtained bydifferentiating (5),2: +f+77/3+fy. Thus o | _y*dt!c^l <fte&y + <*+r/j3+f7. This result isduetoCoriolis. The firstandthird linesadmit immediate interpretations. For, istherelative acceleration, ortheacceleration ofPreferred tothe (>Vyf)-axes asfixed. Next, (23) ae==^+ 77d^+fW~ isthe acceleration d'entrainementj ortheSchleppbesMeunigung, theacceleration with which thepoint Q,fixed inthemoving space andcoinciding attheinstant twith P,isbeing carried alonginfixed space. Thevector (23): ae=&+tip+fy, canbecomputed asfollows. Since asisgeometrically, or kinematically, immediately obvious (24) a=r/3-qy, $=py-ra, y=qa- p/3, wehave : (25) -J-} +rp-qy. The lastlinehasthevalue : RELATIVE MOTION ANDMOVING AXES 289 where (co)=pa+q@+ry. Hence /oc\ /\dQ dr\ f.dr >.dp\/dp (26)*-r-'a+-f '+"- -w2 (+i?/3+r7)+(p$+qr>+rf)(pa+g/8+ry), or (27) ae=(0Xr- 2r+(()r) (), where This vector(w')isthevelocity relative tothefixed axes (, 77,f), withwhich theterminal point of(co)ismoving when theinitial pointisat0';itistherelative angular acceleration, referred tothe (> *?>f)-axes asfixed. Finally, thevector a= 'dtdt^ dtdt^ dtdtj canbeexpressedintheform : (30) a=(co)Xvr, or: For,onrecurring toFormula (10) of2andtaking, asthearbi- trary vectorr,thevector vr,which isgiven by(8),theright- hand sideof(10)comes tocoincide withtheright-hand sideof(29). With theaidof(12), thisvector canbewritten intheform ofthe right-handsideof(30),andthiscompletes theproof. Tosumup,then :From(17), (31) a=a+a', where a'=a,andaisgivenby(20).Asecond evaluation ofa isgivenby(21), (32) a=ar+2a t+ae, where arisgivenby(22)andacby(23) ;thelatter, inadiffer- entform,by(26)or(27). Finally, atisgivenby(29)or(30). 290 MECHANICS 4.TheDynamical Equations. From Newton's Second Law of Motion, written intheform : (1) ma=F, itfollows that (2) ma+ma=F, where aistheacceleration ofthemoving origin, 0',and a=ar+2a,+a*. Thevector aristherelative acceleration and isgiven byFor- mula (22), 3.Thevector acistheacceleration d'entrainement and isdefined by(23) ;itisrepresented by(26)or(27). Finally, a*isdefined by(29)and isrepresented by(30). Ifthemotion ofthemoving axes isregarded asknown, then QOis'aknown function ofttand(o>),i.e.p,q,rareknown from (11),2.Equation (2)cannowbewritten intheform : (3) ma r=FmaQ2ma ma e. Onsubstituting fora*itsvalue from (30)and foraeitsvalue from (26), asystemofdifferential equationsisfound fordeter- mining ,77,f: rf \dt dtdt (4) where thefunctions/,<p, \l/canbewritten down explicitly from theabove formulas. More generally, Equation (1)canbethrown into theform required inagiven problem byusing asuitable form fora,a, ar,a^aaspointed outattheend of 3.Each oneofthese accelerations must bestudied intheparticular case. There is nosingle choice ofsufficient importance tojustify writing down thelong formulas. Butthestudent willdowell tomake hi^ own syllabus, writing down thevalue ofarandeachform foi a,,a{. RELATIVE MOTION ANDMOVING AXES 291 EXERCISE Obtain thedynamical equations inexplicit form from La- grange's Equations, Chapter X.Observe that f6?+& fp+tfop)2 where _ __Vy~V'7-*3-ft-+ *-fa+"I-5- 5.The Centrifugal Field. Letspace rotate with constant angular velocity about afixed linethrough 0,the(, rj,f)-axes being fixed inthemoving space. Then thevector angular velocity (o>)isconstant, and a=0.The vector ae,3,(27) reduces to : (1) a.=-*r+(().r)(), and iseasily interpreted. Kinematically,itis,ofcourse, the centripetal acceleration;geometrically,itisavector drawn from thepointPtoward theaxisand oflengtho>2 p,where pisthe distance fromPtotheaxis. Newton's Lawtakes theform : ma=F, where aisgivenby 3,(32),andthus (2)a=+2q-2r-+pfa+9,+rf) 292 MECHANICS Axis off,theAxis ofRotation. Inthis case, r=w,andtheequations reduce tothefollowing: (3)p=q=0, Fia.139Thus themotion along theaxis offisthesame asitwould be ifspace were notrotating. The projection ofthepathon the(,77)-planeisthesame asthepath ofaparticleinfixed, or stationary, space, when acted oni)bytheapplied forceF;ii)by aforce rao>2 />directed away from; andHi)byaforce atright angles to thepath, equalinmagnitude to2ma)V, andsooriented tothevector velocity vasthepositive axisofis,with re- spect tothepositive axisof17. This third force isknown asthe Coriolis force. Inthecase ofthe Centrifugal OilCup,andthecorre- sponding revolving tennis court, Chapter III, 23itwasenough, forproblems instatics, totake intoaccount theu centrifugal force," ortheforceii)above. But forproblemsinmotion, this isnot sufficient. There istheCoriolis force Hi)atright angles tothepath,liketheforce anelectro-magnetic field exerts ona moving chargeofelectricity. 6.Foucault Pendulum. Consider themotion ofapendulum when therotation oftheearth istaken intoaccount. Wemay think, then, oftheearth asrotating about afixed axisthrough thepoles, which wewilltake astheaxisofz,theaxes ofxandy lyingintheplane oftheequator. LetPbeapoint ofthenorthern hemisphere, and letitsdis- tance from theaxisbep.Bythe vertical through Pismeant thelineinwhich aplumb bobhangs atrest, or,moreprecisely, thenormal toalevel surface. Letfbetaken along thevertical, directed upward; let betangent, asshown, tothemeridian RELATIVE MOTION ANDMOVING AXES 293 through thepoint ofsupport ofthependulum; then77willbe tangenttotheparalleloflatitude through thepoint ofsupport, anddirected west. LetXbethelatitude ofP;i.e.theangle thatfmakes with theplane ofthe equator. Theearth rotates about itsaxis from west toeast, and sothe vector angular velocity, (co),is directed downward. Thus p=cocosX, q=0, r=wsinX, 2rr co= 2460-60=.000727. FIG.140 Wecannowwritedown thedifferential equationsthatgovern themotion. These arecontained inthesingle vector equation of4: ma+ma=F. LetPbethepoint ofsupport, and let(f, 1?,f)bethecoordinates ofthependulum ;Z,itslength, First, compute F : F=G+N, where -au,w.c>u istheforceduetogravity, ortheattraction oftheearth;and N=-j^a-Jtffl- j-#7 isthetension ofthestring. Next, aisthecentripetal acceleration, or : a=co2p(sinX+7cosX). Finally, aisgiven bytheformulas (2)of 5.For, although thesewere written down fortheparticularcase a=0,theyapply generally, where aisarbitrary, providedthevector angular velocityofthemoving spaceisconstant. Thuswecanwritedown explicitly thethree equationsofmo- tion. These wereplace byapproximate equationsobtained as 294 MECHANICS follows. Approximate, first, tothefield offorcebythegravity field.field,U=-U=-mg. Next, suppress those terms which contain co2asafactor, orare ofthecorresponding order ofsmall quantities. Thus Finally, ?+T?+f2= =-I+r?+terms ofhigher order. Weintroduce thefurther approximations which consist insup- pressing theterm ind/dt inthesecond equation, and setting N=mg.The firsttwoequations thusbecome : A) Discussion oftheEquations. Multiply the firstequation A) through bydi/dt, thesecond bydq/dt, andadd. The resulting equation, , dtdt*~^dtdt* integrates intotheequationofenergy: 2~ Z2 or,onintroducing polar coordinates, RELATIVE MOTION ANDMOVING AXES 295 Next, multiply Equations A)by jand respectively, and subtract : This integrates into drj d_ ,2r%-17^-cor+C' or (2) fJ='r+Cf where co'=sinX. 4Special Case. Letthependulum beprojected withasmall velocity from thepointofequilibrium. Theninitially r= ; henceC=and dB di="' Itfollows, then, that=w't. Thismeans that,ifthemotion bereferred tomoving axes, so chosen that f'coincides withf,but'makes anangle u'twith, thependulumwillswing inthe(',f')-plane. Itisnoweasy to determine rasafunction oftfrom(1) ;rexecutes simple harmonic motion. TheGeneral Case. Returning now tothegeneral case,let themotion bereferred toamoving plane through (thepoint ofequilibrium ofthependulum), perpendicular tothe f-axis, androtating with constant angular velocity a/about 0.Then (3) <f>=B-u't istheangular coordinate inthenew plane. Equation (2)now becomes : (4)r-l-C, andthis istheequation ofareas initsusual form. Equation (1)goesover into : or (5)^+r^+2'C+rV=- |r2+. 296 MECHANICS Onsuppressing theterm r2'2because ofitssmallness, wefind : +--!-+* But this ispreciselytheequationofenergy corresponding toan attracting central force ofintensity -y^r.Hence themotion I iselliptic with atthecentre;i.e.thependulum, once released, describes afixed ellipse inthemoving plane. Theaxes ofthis ellipserotate inthepositive sense,i.e.theclockwise sense, asone looksdown ontheearth. Butthependulum describes theellipse ineither sense, according totheinitial conditions, thedegenerate case ofthestraightlinelying between thedescriptioninpositive sense andthat innegative sense. IntheFoucault experiment inthePantheon thependulum was slightly displaced from the positionofequilibrium and released from rest relative tothe earth. Itthen described theellipse inthenegative sense. For initially dr/dt was 0,sothat itstarted from theextremityof anaxis (obviously themajor axis) and itsinitial motion rela- tivetothemoving plane wasinthenegative sense ofrotation;i.e. counter clockwise. Attheend oftwenty-four hours,thasthe value :t=24X60X60,andhence =u't=27rsinX. The result checks, forattheequator should be0,andatthe North Pole, 2w. EXERCISE Obtain theequations ofmotion A),directly from Lagrange's Equations, Chapter X. CHAPTER X LAGRANGE'S EQUATIONS ANDVIRTUAL VELOCITIES INTRODUCTION Inthepreceding chapters, thetreatment hasbeen based on Newton's Second Law ofMotion. Work andEnergy have entered asderived concepts. Itistrue that certain general theorems have been established, whereby some oftheforces of constraint have been eliminated, likethetheorem relating to themotion ofthecentre ofgravity, andthetheorem ofrotation of arigidbody. Butinthelastanalysis, when therehavebeen forces ofconstraint which have notannulled oneanother inpairs,the setting upoftheproblem hasinvolved explicitly anyunknown forces ofconstraint, aswell astheknown forces, andtheformer havethenbeeneliminatedanalytically, anew ineachnewproblem. Weturnnow tomethods whereby, incertain important cases, theforces ofconstraint canbeeliminated once forall,sothat theywillnoteven enter insetting uptheequations onwhose solution theproblem depends. Moreover, weintroduce intrinsic coordinates and intrinsic functions. The intrinsic coordinates are aminimum number ofindependent variables whose values locate completely thesystem. They areoften called generalized coordi- nates, andaredenoted byqlt ,qm.The intrinsic functions are thekinetic energy, thework function oritsnegative, thepotential energy, andtheLagrangean function L.These wehave called intrinsic because theydonotdepend onany special coordinate system, oronanyspecial choice oftheq's. Later, weshall consider intermediate cases inwhich thenumber ofq's,though highly re- stricted,isnotaminimum, and inwhich, moreover, theunknown forces, orconstraints, have notbeenwholly eliminated. 1.TheProblem. Amaterial system may bedetermined in itsposition byoneormore coordinates, qlt- ,</,*andthe *Wechoose, ingeneral, thelettermtodenote thenumber ofthe q's.Butwe replace itbyninthese early examples toavoid confusion with themthat refers tothemass oftheparticle. 297 298 MECHANICS time,t.Forexample,letabead ofmassmslide freely ona smooth circular wire, which rotates inahorizontal plane about .oneofitspoints, 0,with constant angular velocity. The angle <pthat the radius drawn from makes with afixed hori- zontal line isgiven explicitly, Let 9betheangle fromOQproduced to theradius,QP 9drawn tothebead. Then thepositionofmisfully determined by6and t.Thus ifwe set=q, x=f(q, t), y=\(/(q, t). Theproblemofdetermining themotion isthat offinding q asafunction of t. More generally,letasmooth wire, carrying abead, move according toanylaw,and letthebead beacted onbyanyforces. Todetermine themotion. Wewilltreat thisprob- lemindetail presently. Asthesecond illustrative example, consider n masses,ml9 ,mnjfastened toaweightless in- extensible flexiblevstring, oneend, 0,ofwhich is heldfast,and letthesystem beslightly displaced from thepositionofequilibrium. Todetermine themj oscillation. Finally, wemaythink ofarigid body, acted on byany forces. Ifthere arenoconstraints,itwill requiresixcoordinates, qltqz, ,qB9todetermine theposition. Thesemay bethethree coordinates ofoneofthepoints ofthebody, asthecentre of gravity, x, ?/,2;andthethree Eulerian angles, 0, tp9 \f/9which determine theorientation ofthe body. Wemay, however, alsointroduce constraints. If onepointisfixed, there arethree degrees offree- dom,andsothree coordinates, ql9</2,qZ9 forex- / ample, theEulerian angles are required. Or, again, thebody might rotate about afixed axis. Then n=1,and ql=q9asingle coordinate would be sufficient. Or, finally, thebody might befree toslide along LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES afixed lineandrotate about it.Heren=2andqlfqarethe coordinates. Each ofthelasttwoexamples may bevaried bycausing the linetomove inanaltogether specified manner. Then, beside ql=q,orq1and q2,thetime, t,would enterexplicitly. In allsuch cases, themotion isdetermined byLagrange's Equations, which, when there isaforce function U,take the form: A\ 1(1T_^-^ _ iA;dtdb dq~,~dq r' r-1, ...,m, where Tdenotes thekinetic energy, and qrdqr/dt.Weturn now totheestablishment ofthese equations, beginning with the simplest cases. 2.Lagrange's Equations intheSimplest Case. Letabead slideonasmooth wirewhose form aswell asposition varies with thetime : (1) x=f(q,f), */=*>(<?,0,*=*(</,*), where thefunctions f(q, t),<p(q, t), \f/(q, t)arecontinuous to- gether withwhatever derivatives weneed touse,andwhere fa 8ydzV 'dq' dq donot allvanish simultaneously. Themotion isdetermined by theequations: " (2) where X,F,Zrefer tothegiven, orapplied forces,i.e.forces other than thereaction, (X,Y,Z),ofthewire,and (3) XX+/*Y+v1=0, where (X, /z,v)arethedirection components ofthetangent to thewire forthereaction ofthewire isnormal tothewire, though otherwise unknown. 300 MECHANICS Multiply these equations through bydx/dq, Sy/9q, 8z/8qre- spectively andadd : 8yPz8z\ _^+ dt*dq)~y> theremaining terms, namely: ydx .ydy7dzATT~ i "o~i*-~^r~i dq dq dq vanishing because of(3),since dx/dq, dy/dq, dz/dq arethedi- rection components ofthetangent tothewire. The left-hand side ofEquation (4)canbetransformed as follows. Wewrite : tYI (6) T=~(x*+y*+z*) 9 where thedotnotation means atime derivative : dx .dq.x= Tt>^Tvctc - From (1) dx _ dt~ dqdt" dV or /\ .dx ..dx (7)^ITqO+Jt' with similar expressions foryand z.Onsubstituting these values in(6),Tbecomes afunction ofq,q,t.Andnowthe left- hand sideofEquation (4)turns outtobeexpressible intheform : <W<LdL-?L Wdtdq dq For, first,wehave : dT f.dx..dy..ai\-z-r=m[x+y-~+z~).dq \dq dq dq/ From (7)itfollows that dx=dx dq~ dq LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 301 with similar expressions fordy/dq and dz/dq. Thus dT Next, differentiate with respect tothetime : .V ddT__/d2xdxd2ydyd2 'dtdq \dt2dq dt2dq dt2 dq, ./.ddx . .ddy.d< Ontheother hand, dT_/ .dx_j_.dyL.dz __m^ Now,,1AN (10)=^-.dq \dq dq dq/ dx_ddx 'dq~ dt~dq' For,from(7), f)'V s)^IF /ftIT UJU (/JU .. C/JU ~dq~ ~difq dqdt' and, ofcourse, ddx d2xdq d2x Jt~dq= d^~dt dtdq' Substituting thevalue given by(11)andthecorresponding values fordy/dq, dz/dqintheright-hand side of(10),wesee thatdT/dqisequal tothelasthalf oftheright-hand sideof(9), andthus theproofiscomplete:theleft-hand side of(4)has thevalue(8).Wearrive, then, atthefinal result : - dt^j 8q~' This isprecisely Lagrange's Equation forthepresent case. Thecasethattheapplied forces have aforce function, U,isof prime importanceinpractice. Here dVvdU 8U~> Y='' ~' andthusQbecomes : = 4. 4.. dxdq dy~dq"*"dzdq 302 MECHANICS Lagrange's Equation nowtakes theform : dtdq dq dq Example. Consider theproblem stated attheopening ofthe paragraph. Here, theapplied forces areabsent, andsoU=const. Furthermore, x=acosut+acos(6+coJ) y=asinut+asin(0+ui) where q= ;and x=acosinut a(0+co)sin(0+co) 2/=acocos to+a(^+co)cos(^+co<) T=^-((^+a?)2+2co(0+co)cos+co2 ) orn/^T1 r=ma2 (^-fco+cocos0),=ma2co(0+co)sin0, ddT3T /d*B , o. 5ay"-8a"ma"(3i+w8i or g=-..dn, This last istheequation ofSimple Pendulum Motion, d*Q g ^ -& }**' Thus thebead oscillates about themoving lineOQasasimple pendulum oflength Z- ^~co2 would oscillate about thevertical. EXERCISES 1.Abead slides onasmooth circular wirewhich isrotating with constant angular velocity about afixed vertical diameter. Show that tl^fi -^=co2sin6cosB+2sin0. at* a LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 303 2.Ifthebead isreleased withnovertical velocity from a point onthelevel ofthecentre ofthe circle, show that itwill notreach thelowest pointif 3.Abead slides onasmooth rodwhich isrotating about oneendinavertical plane withuniform angular velocity. Show thatd2r=w2r+gsinwt. 4.Integrate thedifferential equation ofthepreceding ques- tion. 6.Abead slides onasmooth rod,oneendofwhich isfixed, andtheinclination ofwhich does notchange. Determine the motion,ifthevertical plane through therodrotates withconstant angular velocity. 6.Intheproblem discussed inthetext, determine thereac- tion,N,ofthewire. Ans.N=ma2 [w2cos6+(0+w)2 ]. 7.Asmooth circular wire rotates with constant angular velocity about avertical axiswhich liesintheplane ofthe circle. Abead slides onthewire. Determine themotion. 8.Show thatif,inQuestion 1,theaxis isahorizontal di- ameter, themotion isgiven bytheequation: ~~fw2cos(p+~coscoHsin<p=0, where<pistheangle which theradius drawn tothebeadmakes withtheradius perpendicular totheaxis. 9.Abeadcanslideonasmooth circular wirewhich isexpand- ing,always remaininginafixed plane. Onepoint ofthewire is fixed, andthecentre describes arightlinewith constantvelocity. Determine themotion ofthebead. .,8Ans. 6=a+- t 10.Thesame problemifthecentre isatrestandtheradius increases atanarbitrary rate. 3.Continuation. Particle onaFixed orMoving Surface. Let aparticle, ofmass w,beconstrained tomove onasmooth sur- face,which canvaryinsizeandshape, (1) x=f(q l9q2,0, V 304 MECHANICS where thefunctions ontheright arecontinuous, together with whatever derivatives wewish touse,andtheJacobians donotvanish simultaneously. Themotion isgiven asbefore by Equations (2)of2,where now, however, thereaction, dueto thesurface,isknown indirection completely. Multiply these equations through respectively bySx/dq l9 dy/d<li, dz/d<li> andadd.Ontheright-hand sidethere remains only since dx/dq lfetc.arethedirection components ofacertain line inthesurface, drawn from(x,y,z),and X,Y,Zarethecom- ponents ofaforcenormal tothesurface. Hence *dq,dt*dq,* Thereduction oftheleft-hand side issimilar tothereduction intheearlier case. Itis,however, just aseasy tocarry this reduction through forasystem withndegrees offreedom, and this isdone inthefollowing paragraph. Thusweseethat dtdqtdqll' and, similarly: dtdq% dq% These areLagrange's Equations forthecase oftwovariables; i.e.thecase of(ql9qz). If,inparticular, aforce function, [/, exists, thenLagrange's Equations taketheform : \ d (/2 uL uLJ duL uLuU dtdqi dqldq dtdq2^2 ^2 Example. Aparticleisconstrained tomove, without fric- tion, inaplane which isrotating with constant angular velocity about ahorizontal axis. Determine themotion. LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 305 Lettheaxis ofrotation betaken astheaxis ofx,and letr denote thedistance oftheparticle from the axis. The coordi- nates oftheparticle are(z,y,z), where y=rcos0,z=rsin0,=co, andwetake ql=x,q2=r.Then yrcosw 7*wsinco, z rsin o>2+rwcos ait, aresult thatmayberead offdirectly, without theintervention ofy,z.Furthermore, U=mgz=mgrsinut. Lagrange's Equations nowbecome : dT. dTn dUn-r-r=mx, ~^~=0,-=0;dx dxJdx aresult immediately obvious. Next, dT. dT or=mgsin< ttr* mco2r=ingsinco/, ^2-o>r=-^sm^. Aspecial solution ofthisequationisfound,eitherbythemethod ofthevariation ofconstantsor,more simply, byinspection, tobe : sni r" 2co Hence thegeneral solution is Since =w^,theequation canbewritten intheform : 306 MECHANICS Aparticular case ofinterest isthat inwhichA=B=0. Thiscorresponds totheinitial conditions oflaunching theparticle from apointintheaxiswith avelocity whose projection along theaxis isarbitrary, theprojection normal totheaxisbeing g/2w. Thepathisthenahelix. EXERCISES 1.Acylinder ofrevolution isrotating with constant angular velocity about avertical axis, exterior tothecylinder, theaxis ofthecylinder being always vertical. Aparticleisprojected along theinner surface, which issmooth. Determine themotion. 2.UseLagrange's Equations todetermine themotion ofa particle inaplane, referred topolar coordinates : d*r dd*\ md 4.The Spherical Pendulum. Consider thespherical pendu- lum;i.e.aparticle moving under gravity andconstrained tolie onasmooth sphere. Take ascoordinates thecolatitude, 6,and thelongtitude, <p,thenorth pole being thepoint ofunstable equilibrium.Then fY)f]1T=^- (02+pSin2 0)^jj=_ mfif2=_mgacose. O/TI PT7%TT-=ma2 6,=ma2 <f>2sin6cos0,-=mgasin0, 00 Cv Cv andthefirst ofLagrange's Equations becomes : ma2Sma2 (f>2sin6cos6=mgasin0, or a Proceeding nowtothesecond equation, wehave : Hence LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 307 Thisequation integrates into (2)^sin*0=h, where theconstant hisofdimension 1inthetime, [T~1 ]. Combining Equations (1)and (2),weobtain: - This istheEquation ofSpherical Pendulum Motion. Incase themotion istobestudied forsmall oscillations near thelowest point ofthesphere, itiswell toreplace6byitssupplement, 6' TT 6.Equation (3)thenbecomes : This lastequation reduces totheEquation ofSimple Pendulum Motion when h=0.Any differential equation oftheform : fA\d26Acos0 D. (4)r=A--T-T+Bsm6,at2sm36 whereA>andB>arearbitrary constants, canobviously beinterpreted interms ofspherical pendulum motion. Afirst integral of(3)canbeobtained intheusualmanner : 2A2cosgdfl 2g. dfi dtdP sin3dt"*"a*m dt' Integrating each sidewith respect to0,weobtain : < --;-?-+* Forafurther discussion oftheproblem,cf.Appell, Mecanique rationelk, vol.i,277. EXERCISES 1.Give anapproximate solution forsmall oscillations near thepointofstable equilibrium, 0,using Cartesian coordinates. Here, andtheapproximate pathisanellipse with ascentre. 308 MECHANICS 2.Treat themotion ofaparticle constrained tomove ona smooth surface, 222=a22+2 2/2 , forsmall oscillations near theorigin. 3.Show thatatopwhich isnotspinning moves likeaspherical pendulum. More precisely, wemean thebodyofChapter VI, 18,when v=0. 6.Geodesies. Letaparticle beconstrained tomove ona smooth surface under noappliedforces. Thepathisageodesic.* Letthesurface begivenbytheequations: x=f(u, i;), y=<p(u, 0),z=t(u, v), where these functions arecontinuous together with their first derivatives, andnot alltheJacobians d(u,vy d(u,v)' d(u,v) vanish. Theelement ofarc isgivenbytheformula : ds*=Edu*+2Fdudv+Gdv\ where thecoefficients arceasily computed, Thekinetic energy hasthevalue : T=%(Eu*+2Fui>+ Lagrange's Equations nowbecome, sinceU=: (6)(Eu+Fit)-i(Euw2+2FUuv+Gut-2 )=0, (Fu+Gv)- -(Evu2+2Fvuv+Gvv2 )=0. Ontheother hand, thegeodesies,intheir capacityofbeing theshortest linesonthesurface, aregiven asextremals ofthe integral A , =fVEu'2 L=/VEu'2+2Fu'v'+Gv'2d\ *0 *Byageodesic ismeant alineofminimum length onasurface minimum, at least, ifthepoints itconnects arenottoofarapart ;cf.Advanced Calculus, p.411. LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 309 intheCalculus ofVariations (cf.theAuthor's Advanced Calculus, p.411)bytheequations: Eu'+Fv' Euu'2+2FUu'v'+Guv'2 -= d Fu'+Gv' Evu'*+2FVu'v'+G,v'* _=u(7) d\VEu'^'2Fu'vr+Gv'22V'Eu'2+^'u'i/^W2 Theparameter Xcanbereplaced byanyother parameter, /u: M=/(A), provided that/(X)iscontinuous, together with itsfirstderivative, and/'(X) 9*0.Inparticular, then, thechoice t= /zisapossible one. Butthen, because oftheequation ofenergy, T=h,or : (8) (#w'2+2*W+Gv'*)=h, u'=u,v'=*, i itfollows thatEquations (7)reduce toEquations (6). Since thevelocity along thepathisconstant, theonly force being normal tothepath,tisproportional to5.Infact, (8) saysthat mds * Thus thetransformation oftheparameter from Xto tamounts insubstance toatransformation tos;i.e.Equations (6)are virtually theintrinsic differential equations ofthegeodesic: (9),~(Eu'+Fv')-(E.w'2+2FUu'v'+Guv"*)= as J-(Fuf+Gv')-$(E,w'2+2FVu'v'+G,v'2 )= as .du .dv U*=-J-,V'=-7-rln' ftn EXERCISES 1.Obtain thegeodesies onacylinder ofrevolution. Observe that,when thecylinderisrolled outonaplane, thegeodesies must goover intostraightlines. 2.Thesameproblem foracone ofrevolution. 310 MECHANICS 3.Show that thegeodesies onananchor ring, ortorus, are given bythedifferential equations: d2 . h2sin6 (10)a(b+acosfl)3" where aand&>aaretheconstants oftheanchorring,andAisa constant ofintegration. 4.Show thatif,inthepreceding question, initially6=ir/2, = or,=Athen (IDa2 (6+acos0)2a2 dt(b+acos0)2 6.Lemma. Wehave seen in2that, inthecase ofasingle #, dx (Pydy dqdt2dqd^z dt2dtdq dq Itisimportant torecognize thisequation asapurely analytic identity, irrespective ofanyphysical meaning tobeattached to T. Itsays that,if T=~(x2+y2+z2 ) and *=f(q,0, y=v(<7,0,^=^(<7,0, where these functions arcanyfunctions subject merely tothe ordinary requirements ofcontinuity, thenEquation (1)istrue. Weturnnow tothegeneral case ofnparticles, mwith the coordinates (a?,yt,Zi) 9i=1,2, ,n.Lettheposition of thesystem bedetermined bymparameters, orgeneralized coordi- nates,?!,, qm,andthetime,t: *i=/ifoi,'' ',?m, yi= <Pi(q ly -,?m, 2i=ti(qi,- ,gm, where thefunctions/,-,(p,^t,arecontinuous together with their partial derivatives ofthe firsttwoorders, andwhere therank of thematrix a)ism :A) LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 311 a)dxl dqm dql Thekinetic energy, canbeexpressedinterms ofglt- ,qm,qly ,qm,t,for with similar formulas foryiyz^ Ouraim istoestablish thegeneral fundamental formula cor- responding to(1): i\-^~' -" r=1, The independent variables inthe partial differentiations are (<7i>'" (7m, (7i> >qm,0>andx^y^Ziaregiven by(3).We have : From (3): Hence_ _ dqr 'dqr-etc. Differentiate with respect to^along agiven curve : (A\ <L?L=V (tfxjdxjd^jjdyi d2ZjdzW*^ rVmiV"^20?r^2^r^207 ddxid^dy/i.d[dzj\ 2' 312 MECHANICS Ontheother hand, (5)I= Now, For,from(3),\* dqr* dqrl dqr< ^ dt while ddqrdt' dtdq r Similar relations hold fordiji/dq randdZi/dq r.Onsubstituting these values in(5),itisseenthatthelastsum in(4)hasprecisely thevalue dT/dqr,andthus therelation I.isestablished. 7.Lagrange's Equations intheGeneral Case. Letasystem ofparticles ntiwith thecoordinates(xi,y^Zi)beacted onbyany forces whatever, X,Ft,Zi.ByNewton's Second Law ofMotion (i)v=Xi Letthepositionofthesystem bedetermined, asin6,bym coordinates ql9 ,qmand t: A) where thefunctions/-, ^,-, derivatives ofthe firsttwoorders, thematrixifei,' ,<7 (<7i, ,?w, i(^li''' i(7m, arecontinuous together with their dql LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 313 being ofrankm.Multiply the first ofEquations (1)by8xi/dq r, thesecond bydyt/dq r,thethirdbydzi/dq r,andadd : /0, v (x*x< (2) 2,*<VdP" ft dx <4.v^<j.7^A^^^' r=1, ,ra. The left-hand sidehasthevalue expressed bytheFundamental EquationI.of 6.Lettheright-hand sidebedenoted byQr: (3) " Itthusappears that ddT These areknown asLagrange's Equations. Wehavededuced Lagrange's Equations fromNewton's Second Law ofMotion. They include Newton's Lawasaparticular case. For ifweset t= thenT7becomes : andQ3<,Q3<+i,Qa<+2 arenowthecomponents oftheforce which actsonmt-.Thus Newton's Equationsresult atonce. 8.Discussion ofthe Equations. Holonomic andNon- Holonomic Systems. Wehave before usthemost general case. Norestrictions havebeenmade ontheforces. These may, then, comprise dissipative forces, likethose offriction orairresistance. Ontheother hand, theremaybeoneormore equationsofthe form: (4) where thefunction Fdoesnotdepend ontheinitial conditions. 314 MECHANICS Moreover itmay happen, whether there arerelations ofthe form (4)ornot,thatthe<?/sandtheir time derivatives arebound together byoneormore equations: (5) $(?!,',?m,% ,?m,=0. Anairplane, rising atagiven angle, would beanexample. Thecasewhich ismost important inpractice,isthat inwhich <f>islinear intheqr: (6) AM+--+Amqm+A=0, where theA'sarefunctions of(qlt ,qm,C),independent of theinitial conditions. Itmayhappen thatarelation oftheform (6)isequivalent to oneoftheform(4).Thus if 'ds dO thisrelation isequivalent totheequation: (8)s-ad=c, which isessentially oftheform(4). Ifnorelations (5)or(6)arepresent ;orifsuch relations (5)or (6)asmayhave entered intheformulation oftheproblem are allcapable ofbeing replaced byequations oftheform(4),the systemissaidtobeholonomic. Examples ofnon-holonomic systems aretheCartWheels of24infra, andtheBilliard Ballonthe rough table, rolling andpivoting without slipping, p.240;alsothe coinontherough table, andthebicycle.* Butwhen theBilliard Ballslips, p.237,thesystemisholonomic, fortheunknown reac- tion ofthetable canbecomputed explicitly, asthereader can easily verify, interms ofthevelocityofthepoint ofcontact, andthus itscomponents areexpressed interms ofthetime deriv- atives ofthegeneralized coordinates. Weare stillleaving inabeyance thequestion ofwhether La- grange's Equations admit aunique solution. Ourconditions are necessary forasolution ofthemechanical problem, butnotalways sufficient. Thestudy ofsufficient conditions willbetaken upin 17andinAppendix D. *Appell, Micaniqae rationelle, Vol. II,Chaps. XXI, XXII. LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 315 9.Continuation. TheForces. Thequestion ofholonomic or non-holonomic hastodowith theleft-hand side ofLagrange's Equations,i.e.withconditions ontheqr,qr,twhich donotinvolve theforces orcontain theconstants oftheinitial conditions. The forces appear onthe right, and itistothese thatwe now turn ourattention. Itmayhappen that the total force Xi,YijZicanbedecomposed intotwoforces : (9) x,=xi+xi Y<=Yi+17, zt=z(+z; insuchamanner that theX{,Y^Z\willbeessentially simpler than theX,Yi,ZiyandthattheX*,F*,Z*disappear altogether from Lagrange's Equations. Forexample, theX{,F{,Z\maybe expressibleinterms ofaforce function : v,_tillv,_d(J r/r_dU Xi~ ~Yi" '" whereUisknown explicitly interms ofX*,2/, ,-,t. Asregards thedisappearanceoftheA7,Y*,Z*theproblems discussed in 1-6have afforded ample illustration. These were theso-called forces ofconstraint, andthey didnotappearin Lagrange's Equations. Returning now tothegeneral case,weobserve that itmay happen thattheX*,F*,Z*fulfil thecondition : r1, ,m.When this istrue, theQrontheright ofLa- grange's Equations takeonthesimpler form : This case isimportant inpractice because itenables ustoget ridofsome oralloftheunknown forces oftheproblem arising from constraints;cf . 15.Butevenwhen allofthelatter forces cannot beeliminated inthisway,theirnumber canbereduced to aminimum;andthen themethod ofmultipliers setforth inthe nextparagraph leads tothefinal elimination. 316 MECHANICS 10.Conclusion. Lagrange's Multipliers. Consider asystem, themotion ofwhich isgivenbyLagrange's Equations: ddT_dT^ =Q Itcanhappen thattheQr'scanbesplit intwo : (14) Qr=Qr+&*, r=1,- ,m, insuch amanner thattheQ'areessentially simpler than theQr known functions, forexample whereas theQ*have the property that (15) Q*TT, H+Qiirm=0, where irlf ,wmareanymnumbers which satisfy theequations: (16) disTTi++amsTrm=0, s=1,-,/*< m. Lettherank ofthematrix (17) beu.an ami From Equations (13)itfollows that ddTdT ~ nomatter what numbers the TT,may be.If,inparticular, the Qrand the jr,aresubject tothecondition expressedin(14), (15),and(16),then (18)ddT Multiply the/zequations (16) respectively byarbitrary num- bers, Xi, ,X^,andsubtract from (18): Ofthemnumbers vlt 9vmitispossible tochoose some setofmp,arbitrarily, andthen therestaredetermined by (16). Fordefiniteness, suppose LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 317 (20) Thenir^+i, ,irmarearbitrary, and iru ,TT>aredetermined. Now lettheXj, ,Vbesochosen that (21) This ispossible because thedeterminant ofthese/zlinear equa- tions inX,, ,XMisnot zero. Forthese values ofXEqua- tions (19)reduce tothefollowing mjuequations: (22) Thedetermination oftheX'sby(21)isindependent ofany choice ofthe TT/S. ThenumbersTJ>+I, ,TTOTarewholly arbi- trary. Hence each coefficient in(22)must vanish. Wehave thus established thefollowing THEOREM. WhenQrcanbewritten intheform: (23) w/ie (24)r=1, (25),als*-!+ therankofthematrix (26)+ama7rm=0,=1, am\ Ctl/x''' iwgf /z,</ien i<zspossibletofindpnumbers \19 ,X^swcA (27)dT numbers aredetermined by&oftheEquations (27),and the values thus obtained arethen substituted intheremaining mp equations. 318 MECHANICS Applicationsofthistheorem occur inpracticeinavariety of problemsinwhich theqr,qr,and tareconnected byrelations of theform (6), 7: (28) ai,^++amqm+a,=0,=1, ,/*, where thear,,a,depend ontheqrand,butnotonthe ini- tial conditions. Asamatter offact, inanumber ofsuch cases thecoefficients arain(28)dolead toasystem ofequations (16)which control asetofnumbersTT^ ,wmforwhich an analysis (14)ofQrwith theresulting relation (15)ispossible. Inother problems, however, thearsofequations (16)have noth- ingtodowithanysuch equations as(28),ifindeed thelatter exist, butmayeven themselves depend onqr,aswellasqrand t. Foracomplete discussion cf .Appendix D. Weturnnow toadirect determination oftheQrfrom purely mechanical considerations. 11.Virtual Velocities and Virtual Work. Letasystem of particles rawith thecoordinates (x%1?/t,zt)begiven (i 1, ,ft).Letdxi, dyt, dzibeany3nnumbers, and letmbe carried tothepoint (xi+8x l,yi+fyi, Zi+dZi).Then tha systemissaid toexperience avirtual displacement (5xi, tyi, dZifl theword"virtual" expressing thefact that theactual system maynotbecapableofsuch adisplacement, even approximately. Thus aparticle constrained tomove onacurve orasurface would ingeneral betaken offitsconstraint, andnot lieeven inthetan- gent lineorplane. Ifforces (Xi,Y^Z$actonthesystem, thequantity (1) W*=J(X,dxi+Yidyi+Zidzi) t=i isdefined asthevirtual workduetothevirtual displacement. Itisconvenient inmany applications torestrict thevirtual displacements admitted toconsideration bylinear homogene- ousequations between the5xt-,6?/ t,5zi. Consider, inparticular, asystem ofparticles whose coordinates aregiven byEquations A), 7: ^=fi(qi,- - ,qm,t) A) , *, LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 319 where therank ofthematrix a) ism.Let (2)fattdxl 09i dqm ^dZn dql dqm where dq }, ,3qmaremarbitrary quantities. Ifm<3n,the toi,8yi}dziaresubject tooneormore linear homogeneous equa- tions. Thus onlyalimited number ofthem cannowbechosen arbitrarily, therestbeing thendetermined. Consider theactual displacement (Ax t-,A?/*,Az)which the system experiences intime Atasitdescribes itsnatural path. Itis : AXt=fi(ql+Ag,, ,qm+Agm,t+At) /(qly ,qm,t) (3) A?/t=<pi(ql+At/!, ,qm+A^m,t+A^) pi(qlf ,gm,^) Since the5#rarearbitrary,itispossible tochoose them equal to theAqr,ordqr=Aqr.Itdoesnotfollow, however, thatthecor- responding tot, diji, dziwill differ fromAxi}Ay iyAzbyinfinites- imals ofhigher order with respect toA.Thiswill, infact, be thecase ifthefunctions /t-,<pt,\f/idonotcontain thetime,i.e. ifdfi/dt=0,etc.Butotherwise ingeneral not. Thevirtual work hasthevalue : (4) , where allmofthenumbers 8qrarearbitrary. If,inparticular, theforces canbebroken upasin(9)9: (5)Xi=XI+XI Ft=Y(+F?, Zt=Z\+Zt, 320 MECHANICS sothat (11)holds: (6)jjj(ir*+17* +* r=1, ,m,then(4)takes onthesimpler form : (7) r,- Ineither case, (8) W5=Q.dq,+...+Q m8qm. Consider theactual displacement (Ax, Ayi,AZi) ofthesystem intime Atasitdescribes itsnatural path.If/,-, ^?t-,\f/idonot containt,thevirtual workWswill differ from theactual work, ATT,byaninfinitesimal ofhigher order thanA;otherwise, this willnotingeneral bethecase. 12.Computation ofQr.InEquation (8), 11thedqrare marbitrary numbers. Wemay, then, setdqk 0,k^r; 5qr7*0,andcompute thecorresponding value ofW6.We shall thenhave : Consider, forexample, aparticle that isconstrained tolie onamoving surface. Itscoordinates aresubject tothecondi- tions : (10) x=f(q lyq2,0, y=<p(qi, q0, 2=iKfc, q*t). Avirtual displacement means thatwefixourattention onan arbitrary instant oftime, t,andconsider thesurface represented by(10)forthisvalue of t.Next, consider apoint (x,y,z)of this surface. Then avirtual displacement (&r,dy,dz)ofthis point means anarbitrary displacement inthetangent plane to thesurface atthepointinquestion. Inparticular,ifweset dq2=andtake dq 1^0,then thevirtual displacement takes place along thetangent tothatcurve inthesurface whose coor- dinates arerepresented by(10)when</2and tareheld fast. Now, thenatural path oftheparticle under theforces that actdoesnotingenerallieinthesurface just considered, nor is ittangent tothesurface. Ifthesurface issmooth, thereaction ontheparticlewillbenormal tothesurface, andsothevirtual LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 321 workWsdue tothereaction willbe0.Buttheactual work donebythereaction inAseconds along thenatural pathwill ingeneral beaninfinitesimal ofthesame order asAt. Itisnoweasy toseehow tocompute QlandQzmcase the surface issmooth. Thevirtual work ofthereaction ofthesurface isnil,andsoweneed consider only theother forces. The virtual displacement takes placeinthetangent plane tothesurface, andwecancompute directly thevirtual work corresponding tothesuccessive virtual displacements givenby8ql^0,8q2= and5ft=0,8qz^;cf.further 16infra. 13.Virtual Velocities, anAid intheChoice ofthe irr.In thegeneral theorem of 10there wasnoindication astohow the 7rrmaybochosen. Incertain cases which arise inpractice, themotion being subject toLagrange's Equations: / d3TdT ithappens that there aregeometric orkinematical relations between theqr'softheform (28), 10 : (2) ali(fft+' '+dmsqm+ CL8=0, 8=1,-,/*< 1, where thear8,a8areknown functions ofqlt ,qm,t,*which donotdepend onthe initial conditions, andwhere therank of thematrix an (3) ISfJL. If,now, thepossible virtual displacements corresponding to anarbitrary choice of5ft, ,8qmaresorestricted that (4) als8qi+--+am88qm=0, 5=1, ,/*, itturns outthat thevirtual work ofcertain forces ("forces of constraint ")will vanish. Hence byidentifying these "forces ofconstraint"with theQ*andsetting 8qr=7rr,thehypotheses ofthattheorem arcfulfilled. *Itmayhappen thatsome orallofthese equations maybeintegrated inthe form:F(qi, -,</m ,t)=0,whereFdoes notdepend ontheinitial conditions; but itisnotimportant todistinguish thiscase. 322 MECHANICS Example. Consider thedisc ofChapter VI, 24asfree to rollwithout slipping onarough horizontal plane which ismoving initsownplane according toanygiven law;forexample, rotat- ingabout afixed point with constant velocity. The force which theplane exerts onthediscatthepoint ofcontact willdowork onthedisc. Butthevirtual work ofthis force,when thevirtual displacementisrestricted asabove,isnil.Thusweareledto asuitable setofmultipliers7rr,namely, the5qrthus restricted. 14.OntheNumber moftheqr.Forasystem ofparticles theqr's,ashasalready been pointed out,canalways beidentified with thecoordinates : Here,m=3w,andLagrange's Equations become identical with Newton's Equations. In'theory, then, there isnodifference between thetwosystems. Inpractice, Lagrange's Equations provideinmany cases an elimination offorces inwhich wearcnotinterested. Consider aladder sliding down awall. Ifthewalland floor aresmooth, wemay takem=1,q=0,and alltheforces in which wearenotinterested willbeeliminated. sHere, (2) x=acos0, y=asin0. MgeMKHence FIG.144(3) r=^(02+1)0. Lagrange's Equation becomes : where U=Mgasin 0. Thuswefindastheequation governing themotion : M(a2+ A;2 )-^=-Mga cos or fK\d?0 ag LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 323 Inthisexample, themaximum ofelimination hasbeen at- tained atoneblow. Ifwethink oftheladder asmade upofa hugenumber ofparticles connected byweightless rods, theforces intherodshave been eliminated, and alsotheforces exerted by thefloorandthewall. Suppose, however, that the floorandthewall arerough. Wecan still write down asingle Lagrangean equation, theleft-hand sidebeing asbefore. Butnow W (7) Q=_J=-Mga cos+2acosOp,S+2asin R. dq Wehavenotequations enough tosolve theproblem. The difficulty canbemetbytakingm=3andsetting (8) 0i=x, 02= 27, 03=0- Tisgivenby(1).Andnow Q!=S-M#, Q2=R+S-Mg, (9)1=a(sin+cos0)S+a(/isin-cos6)R. Lagrange;sthree equations become : (10) =a(sin+ncos0)S+aOsin-cos0) The firsttwoofthese aretheequations ofmotion ofthecentre ofgravity ;thethird, theequationofmoments about thecentre ofgravity. What Lagrange's method here hasdone,isfirst toeliminate theinternal forces between theparticles, justaswedidinChapter IV, 1,whenweproved thetheorem about themotion ofthe centre ofgravity; and similarly, whenweproved thetheorem ofmoments, Chapter IV, 3and 9. 324 MECHANICS Let Qr=+Qf, where Qr~s-MB, Q;=/Z+/*S, Q*=a(sin+/xcos0)S+a(p,sin-cos6)R. Wewish tofindthree multipliers, TT^?r2,7r3,such that (11) Ql^+Q*T 2+Q,*T,=0. Thiscaneasily bedone algebraically, with theresult that TTI+/i7T2+a(sin6+/xcos0)7r3= (12),+ 7r2+(Msin cos0)7r3= asolution ofthese equations being: TJ= sin(0+2X),?r2=cos(0+2X),7r3=I/a. Butamechanical derivation iseasy, too. Consider theresult- antoftheforcesRand p,R.Draw aperpendicular toitthrough thelower end oftherod,anddisplace thisendalong this line. Dothesame thing attheupper endoftherod,anddisplace the upper endalong this line. The resultis,that fSql=-asin (9+2\)8q, 'lo) 1 Idq2=acos(6+2X)S#3 Corresponding tosuch adisplacement thevirtual work ofthe "constraints" must vanish. Andnow itismerely aquestion of trigonometry toshow thatourexpectationisfulfilled : (14) Qj>i+Qlfy,+,*?,=0. Equation (11)corresponds toEquation (15) of10,andEqua- tions (12)aretheEquations (16) ofthatparagraph. ButEqua- tions (13),though corresponding toEquations (4), 13,donot have their origin inEquations (2), 13.The latter would arise from differentiating (2). Returning now toEquations (10),weseethat theunknown reactions RandSareeliminated by(14),where 5qlf8q.2,8qssat- isfy (13). Hence (15) Jflfift+(M+Mg) dq,+Mk*0dq z=0, or (16)-MXasin(0+2X)+(My+Mg)acos(0+2X) +Mk* '6=0. LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 325 This equation, combined with Equations (2),leads atonce tothesolution oftheproblem: (17) (/c2+a2cos2X)~~+a2sin2X^~+agcos(0+2X)=0. 15.Forces ofConstraint. Adefinition of"forces ofcon- straint" from thepointofview ofphysics, which shall beboth accurate andcomprehensive, has, sofarastheauthor knows, never been given. They would beincluded insuch forces as theX*, *,Z*of9,which disappear from theQr;I.e.,Equa- tion (11). Arid still again, theQ*of10arisefromunknown forces andareeliminated bythemethod ofmultipliers. Perhaps these two cases arecomprehensiveinRational Me- chanics. Arethere problems inthisscience notincluded here? Ifnot,theasterisk forces could bedefined astheforces ofcon- straint;cf .Appendix D. 16.Euler's Equations, Deduced from Lagrange's Equations. When arigidbody rotates about afixed point, thekinetic energyis (1) T=i(4p2+#?2+O2 ). Letp,q,rbeexpressedinterms ofEuler's angles, Chapter VI, 15: p=6sin(p \}/sin cos<p (2) q6cos<p+ \l/sinQsin<p r<j>-{- \l/cos Thesecond ofLagrange's Equationsisreadily computed: dTAdp.,,dq. dr=Ap~+Bq~~+CV=Cr; v<P O(D V(D(/<p TT-= 8cos<p+^sin6sin <p=g, = 6sinv>-\-$sin cos^>=p, Hence 326 MECHANICS Tocompute <$>,observe that, nomatter what forcesmay act, theycanbereplaced byaforce atandacouple. The latter canberealized bymeans ofthree forces : o ii) Hi)aforceLyacting atthepoint*r= "Ma" "r= T=a. Avirtual displacement80=0,5<p^0, d\l/=gives asthe virtual work N5<p and this isequal to&8<p. Hence <t>=Nandwehave : Cj|-(A-B)pq=N. This isthethird ofEuler's Dynamical Equations. Theother twofollow from thisonebysymmetry, and areobtained by advancing theletterscyclically. EXERCISE Obtain thesixequations ofmotion ofarigidbodybymeans of Lagrange's Equations. 17.Solution ofLagrange's Equations. Wehave seen in 10that iftheQrsatisfy theconditions oftheTheorem ofthat paragraph, then (i) where thematrix : (2)- dtdqrdqry*^=Qr+2)ar8\9,r=1, ,m, isofrankju<m.Inthecases which arisemost frequently in practice, there arenequations oftheform : *Bythe"point r"ismeant theterminal point ofthevector rwhen theinitial point isat0. LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 327 (3) a,isqi+-+amsqm=aa, s=1, ,ju, where thea'sarefunctions ofglt ,qm,t. The kinetic energy Tisapositive definite quadratic form in theql9 ,qm,butnotnecessarily homogeneous: T=T2+T,+T, where andT19Tarehomogeneous ofdegree1or intheqr,orvan- ishidentically. The coefficients arefunctions ofqlf- - ,qm,t. TheformT2,or (4) 2 </*<** *,./ isapositive definite homogeneous quadratic form. ForTcanbe written intheform : /tu, ,juw,(7arefunctions of#i, -,^m,t. Finally,letQ'rbeaknown function ofqr,qr,t. THEOREM. Them+/*Equations: ddT8T/A iH----+o.g=a., s=1, ,/z, determine uniquely them+p,functions (ft, ,</,Xx, ,XM. Proo/. The firstmofthese equations have theform : (5)Airqi++AmrqmarlXj ar/AXM=Br, r=1, ,m,where risafunction ofthe</,.,qr,and .The remaining /*equations give,ondifferentiating: (6) ai,qi -\----+amsqm=Ca, s=1, ,/*, where C.islikewise afunction ofthegr,qr,and <.Thuswehave m+Mlinear equations inthem+nunknowns: qlt ,#m, Xj, ,X^.Their determinant : 328 MECHANICS _ "" an aii Oi,A OmM doesnotvanish. Forotherwise them+nlinear homogeneous equations: uti++AmiZm+an*?!++ai^= lmti+*'+Ammtm+Oml^+'''+Omplfc= = = would admit asolution (,- ,w,77^ ,T/M)nottheidentity. Moreover, not allthefi, ,minthissolution could vanish; forthenweshould have : ++ =0. Buttherank ofthematrix ofthese equations, namely thematrix (2),is/x.Hence allthei)l9 ,^ofthesolution vanish a contradiction. Next, multiply ther-thequation (8)by r,r=1, ,m,and add. Theterms ini^, ,77^drop outbecause ofthelastp.of theequations (8),andsothere results theequation: where not allthej, ,mare0.This isimpossible, since(4) isapositive definite quadratic form. Equations (5)and(6)admit, therefore, asolution : (9)0, ,m; LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 329 Assuming fordefiniteness thatthedeterminant an <v I/*'** Uft.fi. woseethat the firstmoftheEquations (9)admit theintegral given by(3),or : where,inparticular, /,islinear in<fo+i, ,qm-This isapar- ticular integral which isindependent ofthe initial conditions of themechanical problem. Let (11) qa=Ka, a=+1,- ,m. Thesystemofmdifferential equations (9)isnowseen tobeequiv- alent, under the restrictions ofthedynamical problem, tothe system: dq8 f,j\ 1 ' ''' '* --' (12) dKa ,Km,0,tt=/*+1, Here isasystemof2m pdifferential equationsofthefirstorder fordetermining the2m Munknown functions qr,Ka.Their solution yields themdesired functions, qlt--- ,qm.TheX8are nowuniquely determined asfunctions of tandtheinitial condi- tions. Asregards thefreedom oftheinitial conditions, onwhich ofcourse thedetermination oftheconstants ofintegration depends, theinitial values</r,grofqr,qrarerestricted bytheequation fl\sf]\ H I*** Idrnsqm~$.s- Itisimportant here, asinsomany problems ofthekind dis- cussed inthischapter, todistinguish between constants that are connected with thechoice ofcoordinates andconstants that arise from theinitial conditions ofthemechanical problem. Thus inthe problem of('hap. IV, 13,p.141, Fig. 84,smight equally well have beenmeasured from adifferent level, andthen therelation would havebeen : s=ad+c. 330 MECHANICS 18.Equilibrium. Letadynamical system begiven, withn particles m,,themotion being subject toNewton's Law : (1) Theforces aresaidtobeinequilibriumifZit 1, n. (2)=0,Yt=0, 0, Anecessary and sufficient condition forequilibrium is,that (3)=0,=0,=0. Wearenotinterested inthegeneral case, which, inaccord with thedefinition just given, relates toasingle instant oftime, the forces notingeneral beinginequilibrium atanyother instant. Wehave concern rather withapermanent state ofrestofasystem capableofcertain motions which aresubject togeometric condi- tions.Wearethinking primarily ofsuchproblemsinthestatics ofparticles andrigid bodies aswere studied inChaptersIandII; butalso ofmore general problems,likethefollowing:Auni- form circular dischasaparticle attached toitsrim.The disc rests onasmoothellipsoid andarough table which contains twoaxes oftheellipsoid. Find thepositions ofequilibrium. More precisely, thesystem shall becapable ofassuming the positions defined bytheequations: (4)=fi(q\, ,?*) where thefunctions /,^, \f/idonotdepend ont,andwhere the rank ofthematrix (5) sra. Observe that this lastrequirement does notimply thatmhas theleast value forwhich thexiyy^Zicanberepresented byequa- tions oftheform(4),satisfying theabove requirements. Itis LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 331 stillpossible that theq19 ,qmmaybeconnected byrelations oftheform : (6) Ft(ft, ,qm)=0, j=1, ,p<m. Ontheother hand itdoesimply that ifthex^yiy2,allvanish, then this istrue oftheqr,andconversely; andif,furthermore, both theij#,z,andthexiy#,zallvanish, then this istrue oftheqrandtheqr,andconversely. Themotion ofthesystem is,first ofall,subject totheequa- tions : (7) [71,=Qr,r=1, ,M<m, where bydefinition \T\=<L2L-VL. 1lr dtdqrdqr Tothesemaybeadded further equations: (8) alaqi++am*qm=0, a=1, ,/*< m, where therank ofthematrix (9) is(1.Itispossible thatsome orallofthese equations canbe expressedintheform(6),butthis isunimportant. Afirstnecessary andsufficient condition forequilibriumisthat (10) Qr=0, r=1,--- ,m. For,anecessary and sufficient condition forthevanishing ofthe left-hand side of(7)forft=0, ,&,=isthatft=0, - ,qm=0. Ifthere arerelations oftheform(8),itmayhappen that the Qrcanbesplitupasfollows : (11) Qr=Qr+Q?> where (12) Q*ft++Q*m*qm=0, provided thedqraresochosen that (13) 01,30!++Om^qm=0, s=1, ,/*. 332 MECHANICS Under these circumstances anecessary and sufficient condi- tion forequilibriumisthat (14) <?>/!+-+Q'mdqm=0, provided the8qrsatisfy (13). That thecondition isnecessary appears from thefact that (10)istrue,andhence (15) (Q[+<#)&++(<&+Qi)?*= forallvalues ofthedqr.Ifthe5qrsatisfy (13),itfollows that (12)istrue,and (14)now follows. Suppose conversely that (12)and (14)holdwhen the8qrare subject to(13). Then thesystemisinequilibrium. Suppose thestatement false.From (12)and (14)itfollows that (15)holds, provided (13)istrue,andhence from(7)itfollows that (16) i)[r]r5<yr=0, T~.\ provided (13) holds. Lot (17) qr=cr, r=1, ,m, initially. Then not allthecrare0.Now, (18) dqr=cr, r=1, ,m, isasystem ofvalues satisfying (13). For,ondifferentiating (8) with respect to tandthen settingt=/ ,qr=0,these relations follow, namely: 0>lsC\+'''+OmsCm=0,S=1, ,(JL. Consequently (16)holds forthose values ofcr.Now, (19)' T=^A^qaqft. a,ft Hence f)T=Air(ji+'''+Amrqmj v(]r dBT.... ,A.. . . .,..,^-QT=Airq\+--+Amrqm+terms in(ql9- ,qm), andsoinitially m m ^[T] rdqr=^MirCj H+Amrcn)cr=^Aapcacft. -! r=la,/3 LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 333 Buthero isacontradiction, since (19), being apositive definite quadratic form, canvanish onlywhen allthearguments are 0. Wecanstate theresult asa THEOREM. Anecessary and sufficient condition thatadynamical system,themotion ofwhich isgoverned bytheequations: _rfd_T__8T (20)dt'dq rWr~V"r-1'"->m; i7i++Om.(7m=0,=1, ,/*< m, where therank ofthematrix ofthese lastp,equationsis/*,bein equilibrium and atrest, isthat qr=0,r=1, ,m,andthat (21) Qiffi++Qmdq m=0, where. 5q },-- ,dqmaresubjecttothecondition: (22) a\8qi++am5<$tfw=0,=1, -,/*. 7Tisahomogeneous positive definite quadratic form in //,inparticular, (23) Qr=Qr+Qr*, f=1, ,m, and?/#isknown that (24) Q*fy,++Q*ndqm=0, where dql9 ,5gmaresubjectto(22), ttew(21)can6ereplaced by (25) QlSft+-+Q'mdqm=0. 19.Small Oscillations. Two equal masses areknotted toa string, oneendofwhich ismade fast toapegat0.Determine themotion inthecase ofsmall vibrations. Here, T=^(202+^2+2^cosfc,-6 )), U=mga(2 cos+cos<p). Since6,d,<p,<parcsmall, these functions canbere- placed bytheapproximations: .9X FIG.146 const. U=-mga(Q*+^)+ 334 MECHANICS Equations (1)arctypical foranimportant class ofproblems insmall oscillations ofasystem about aposition ofstable equi- librium, theapplied forces being derived from aforce function, (7.LetTandUboth beindependent oft\letqr= for r=1, ,m,betheposition ofequilibrium, and letT,Ubere- placed bytheirapproximate values when qr,qrareallsmall. Then (2) T=Vdraqr<ls r.s r,s=1,- ,ra,where thecoefficients or,=aar,br,=b,rare constants andeach ofthequadratic forms isdefinite. Lagrange's Equations nowtaketheform : (3) Orltfi+'''+Clrmqm=~ (&rl<7l++&rm?m), r=1,--- ,m. *Tointegrate these equations,itisconvenient tointroduce new variables asfollows. Itisatheorem ofalgebra*thatbymeans ofasuitable linear transformation with constant coefficients : (4) qr=riq(++Mmrf,, r=1,- ,ro, thequadratic forms(2)caneachbereduced toasum ofsquares: T=q(*++q'm\ U=-nt2 ft'2- ---nm2^2 . Lagrange's Equations nowbecome : g$-V*r-l,...,. Their integrals take theform : (7) q'r=Crcos(nrt+7r),r=1, ,m. Returning totheoriginal variables qr,wefindasthegeneral solution ofEquations (3)thefollowing: (8)qr=CiMri cos(nj+TI)++Cmfirmcos(nmt+7m), r=1,- ,m, where theCr,yrarethe2mconstants ofintegration. *Bocher, Higher Algebra, Chap. 13. LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 335 Inthis result, complete asitisintheory, there appear, how- ever, thecoefficients/*r,ofthelinear transformation (4).These canbedetermined bythefollowing consideration. LetCr= in(8)when r9*s;and letC8=1.Thuswehave aspecial solution : (9) q?=\rcos(nt+7), where Xr=/irandn=na,7=y8.Substitute q?in(3): (10) (bn-n*a rl)XiH-----h(bm-n2arm)\m=0, r=1,-- ,m. Anecessary condition that (9)beasolutionis,thatthemlinear Equations (10)admit asolution inwhich theX/sarenot all0. Hence thedeterminant ofthese equations must vanish : (ii)n2an =o. 6mm~n2a Ifthenr2are alldistinct, theyform precisely themroots of thisequationinn2 .Moreover, eachn?leads toaunique deter- mination oftheratios oftheX'sthrough themEquations (10), andourproblemissolved. Itmayhappen thatkoftherootsn2of(11) coincide. Inthat case,koftheX'scanbechosen arbitrarily, andsowestillhave k linearly independent solutions (9)corresponding tosuch aroot n2 .More precisely, letn2beamultiple root oforderfcj ;n22 , amultiple root oforder fc2;etc. Letn2beset=n^inEqua- tions(10). Then, oftheunknown \lt ,Xm,itispossible to choose acertain setof^arbitrarily, andthen therest willbe uniquely determined. Let allbutone oftheseA^X'sbeset =1.Thusweget^sets of(Xx,-- ,Xw),andeach setgives a solution ofEquations (3).Moreover these solutions areobvi- ously linearly independent. Proceeding ton22wedetermine inthesamemanner fc2further setsof(Xn. ,Xm),each setgiv- ingasolution of(3) ;these solutions arelikewiselinearly inde- pendent ofoneanother and also oftheearlier solutions. And soon,totheend. Thus inallcases theroots of(11)lead tom linearly independent solutions (9). The variablesq'rareknown asthenormal coordinates ofthe problem. Each isuniquely determined, save astoafactor of 336 MECHANICS proportionality, when theroots of(11) are distinct. But in thecase ofequal roots, aninfinite number ofdifferent choices arepossible. EXERCISE Carry through theexample given atthebeginning ofthepara- graph. Ans.Two sets oflinearly independent solutions arethe following: f0j=cosfat+7j),f2=cosfat+72); i^=A/2cosfat+7i),I02=~^2cosfat+72), where n*=(2-V2)2 n22=(2+\/2)?- Cv C* Thegeneral solution is : =<?!cos(n^+TJ)+C2cosfat+72), tf?=C^A/2 cos(n^+TI)-C2V2cosfat+72). EXERCISES ONCHAPTER X 1.Asmooth wedge restsonatable.Ablock isplaced on thewedge, andthesystemisreleased from rest. Determine themotion. 2.Two billiard balls areplaced oneontopoftheother, on arough table, andreleased fromrest, slightly displaced from the position ofequilibrium. Determine themotion. 3.Auniform rod ispivoted atoneend,and isacted onby gravity. Will itmove likeaspherical pendulum? 4.Auniform rodoflength 2aandmass3mcanturn freely about itsmid-point. Amassmisattached tooneend ofthe rod. Iftherodis_rotatingabout avertical axiswithanangular velocity of\/2ng/a, and soreleased, show that theheavy end willdiptilltherodmakes anangle ofcos~1(Vn2+1ri)with thevertical, andthen riseagain tothehorizontal. 5.Obtain theequations ofthetopfrom Lagrange's Equations. 6.Determine themotion ofatopwhose peg, considered as apoint, slides onasmooth horizontal plane. 7.Thesame question when thesizeofthepegistaken into account. LAGRANGE'S EQUATIONS. VIRTUAL VELOCITIES 337 8.The ladder ofp.322, the initial position being oblique tothelineofintersection ofthewallandthefloor.* 9.Two equal rods arehinged attheir endsand project overasmooth horizontal plane. Determine themotion. SUGGESTION. Take ascoordinates (1)the x,yofthecentre ofgravity ;(2)theinclination 6ofthelinethrough thecentre of gravity andthehinge ;(3)theangleabetween thislineandeither oftherods. Two ofLagrange's Equations control themotion ofthecentre ofgravity. Athird expresses thefact that thetotalmoment ofmomentum with respect tothecentre ofgravity,isconstant. Andfourthly there istheequation ofenergy, f 10.Arough table isrotating about avertical axis. Study themotion ofabilliard ballonthetable, assuming that there is noslipping. 11.Thesameproblem with slipping. 12.Work theproblem of 19,p.333,when theparticles are notrequired tomove inavertical plane. 13.Two equal uniform rods archinged atoneoftheir ends, andtheother endofonerod ispivoted. Find themotion for small oscillations inavertical plane. 14.Thesame problem when therods arenot restricted to lyinginaplane. 16.Auniform rod issupported bytwostrings ofequal length, attached toitsends, their other ends beingmade fast attwo points onthesame level, whose distance apartisequal tothe length oftherod.Asmooth vertical wire passes through a small hole atthemiddle oftherodand bisects thelinejoining thefixed points. Determine themotion. 16.Ifinthepreceding question thewire isabsent, study the small oscillations oftherodabout theposition ofequilibrium. 17.Abead can slideonacircular wire, noexternal forces acting. Determine themotion intwoand inthree dimensions. Begin withthecase ofnofriction. *Routh, Elementary Rigid Dynamics, p.329. tAppell, Mtcanique rationelle, vol. ii,chap. 24, 446.Many other problems ofthepresent kind arefound inthischapter. CHAPTER XI HAMILTON'S CANONICAL EQUATIONS 1.TheProblem. Theproblemofthischapteristhededuc- tionofHamilton's Canonical Equations: dqr_m dpr__Wi... dt" dpr'dt dqr' ' ' ' from Lagrange's Equations: ddL dLn t :r;Q~--^"~=0,r=1, ,m. cftd</ rdqr The transition ispurely analytical, involving nophysical con- cepts whatever, andforthatreason itiswell tosetthetheorem andproof apartinaseparate chapter. Theproblem canbestated asfollows. Westart outwith a Lagrangean System. Such asystemisdefined asamaterial system which canbelocated bymeans ofmgeneralized coordi- nates qly ,qmandwhose motion isdetermined byLagrange's Equations. Ifwesetqr=Kr,these goover intothe2mequa- tions : A) dML_ dtdKrdqr r=1,< ,m,where (1) L=L(q l9' ,tfm, *i,' ,Km,=i(0r, *r, isafunction ofthe2m+1independent variables#/.,*r,<. The function Liscalled theLagrangean Function. Incase there isawork function, Lisgivenbytheequation: (2) L=T+U, 338 HAMILTON'S CANONICAL EQUATIONS 339 where KT=qrand T=T(q r,?r, isthekinetic energy. Inany case, theHessian Determinant, theJacobian : /\ d(L 1?- ,Lm) (6)0(*i,---,O' where shall notvanish. The(2m+l)-dimensional spaceS2m+iofthevariables(qr,*r,t) shall betransformed onthe(2m+l)-dimensional spaceR2m+i ofthevariables (qr,pr,t)bymeans ofthetransformation : /M\ 8Lt (4) Pr= ^-,r=1, ,m, thef/rgoing over individually into themselves. Thesystem of 2mdifferential equations ofthe firstorder inthe2mdependent variables qr,Kr,namely, Equations A),thereby goes over into asystem of2mlikeequations inthe2mdependent variables qr,pr.These lastequations arethefollowing: fr. dqr811 dpr8H - (5)-*=& -*=-*?r-1'-"-^ where II=H(qr,pr,t)isdefined bytheequation: (6) H=I)prKr~L, r=l theKrbeing functions of(qr,pr,defined by(4). This isthetheorem which isthesubject ofthischapter andto theproof ofwhich wenow turn. Theconverse istrueunder suitable restrictions. 2.AGeneral Theorem. LetF(x l9 ,xn)beanyfunction, continuous together with itsderivatives ofthe firsttwo orders, andsuch that itsHessian Determinant, theJacobian : <" Letatransformation, T,bedefined bytheequations: T:r- r 340 MECHANICS LetG(ylt ,yn)bedefined bytherelation : (2) G(yi, ,j/n)=Jxryr-F(x l>-,xm)> r=l where xl9 ,xnarethefunctions ofylt ,yndefined by 1 .Then (3) xr-%,r-!,...,. For, differentiate(2),regarded asanidentityintheindependent variables ylt ,yn: dxs Theright-hand side,bytheequation^ defining T,reduces toxr, andtheproofiscomplete. Furthermore, For,onperformingfirstthetransformation T,then thetransfor- mation T~~l jtheresult istheidentical transformation. Hence d(xu-- ,xn)d(y l9 ,yn)' or: ,G.) 1 3(*i, ,*n) Inparticular, then, /rv d(Gi. ,Gn)r$(ri, ,/n)"T~f^ir= I I These results maybestated inthefollowing theorem. THEOREM I.LetF(x ly ,xn)beafunction satisfyingthe condition: Performthetransformation: dFT 11 r=1 w ^. 2/r- ,r1, ,n. HAMILTON'S CANONICAL EQUATIONS 341 LetG(y lt ,yn)bedefined bytheequation: n (6) G(y ly ,y)=5)xryr-F(x l9 ,xn), r=l where xrisdetermined asafunction of(ylt ,yn)bytheinverse, T~l ,ofT.Then theinverse ofTisrepresented asfollows: T~lx-r=1 n 1 . xr-^,rl, ,n. Moreover, d(Gi,''' >Gn) ,Q ^(2/1, -,2/n) Inparticular, Theidentical relation canbewritten inthesymmetric form: (7) F(x lt ,xn) r-1 where 3F dG andtheHessian Determinants ofF,Gare^0. Wenowproceed toasecond theorem, which isofimportance intheapplications oftheresults ofthischapterinmechanics. THEOREM II.//F,Garedefined asbefore, and ifeachdepends onaparameter, %,therelation (8) J'ft; a?,,..-, x^+Oft; yu ,2/n)=Jxryr T=l beinganidentity, because ofTorT~l ,either inthen+1arguments ( ;#!,, Xn)orinthen+1arguments ( ;yly ,yn Let (;!,, XB)betheindependent variables in(8).Then = yrft& ft' But andtheproofiscomplete. 342 MECHANICS 3.Proof ofHamilton's Equations. We start outwith the Lagrangean Function L(q r,KT,t),which fulfils thecondition : '"'Lwn(if if}9V\Kli )Km/ andmake thetransformation : (2) pr=-, r=1, ,m. TheHamiltonian Function //(qr,pr,f)isthen defined bythe equation: (3) L+H=%prKr. T If,then,wesetxrKr,yr=pr,andregard theqrand tas parameters,alltheconditions ofthetheorems of2willbemet. Itfollows, then, thattheinverse of(2)isgivenbytheequation: andfurthermore that (5; +g^~ t r,--, m. Itisalsotruethat dt dt' although thisrelation isnotimportant forourpresent purposes. TurnnowtoLagrange's Equations: dq, ~dt d<9L_cuu dtd*rdqr The firstofthese, combined with(4),gives: mdqr-mr-\ m \OJ ,~ ,/1,,III. From thesecond, combined with (2)and(5),weinfer that (9) -37-= -r-, T=1, ,m. HAMILTON'S CANONICAL EQUATIONS 343 Butthese areprecisely theHamiltonian Equations (5)of 1: dqr_9H dpr__3H dt~ dpr>dt- dqr> T~A' ' ' which wesetouttoestablish. Themathematical converse issimple. Given Equations (10) with thecondition Equations (4)define atransformation, andthenLisdefined by(3).Then(2)arid (5)follow from thetheorems of 2.And nowthe first ofEquations (10),combined with(4),gives dqr . -= r=l,...,m. Thesecond equation (10),combined with (2)and(5),leads to theequation: ddL_dL _1 ~7i~n ~^ t f1>""" )W" dt8Krdqr Thuswoarrive atLagrange's Equations (7). We see,then, thataknowledge ofthefunctionHissufficient foracomplete mathematical formulation ofthemotion. But what canwesayofthephysical meaningof//inthegeneral case? There isanimportant restricted class ofcases inwhich thedefinition issimple. Suppose there isaforce function U depending onqr,talone, andfurthermore that thekinetic en- ergyTisapositive definite homogeneous quadratic form inthe (/!,--, qm>LetLbedefined bytheequation: (11) L=T+U. Let Kr=qr>Then Thetransformation : 8L*-aT r nowbecomes : 344 MECHANICS Thus ?,,-2,j-2T.r=l rc/"r Hence (12) ff=J)pr*r-L=T-U. T Stillmore specially,ifneither Tnor [7depends ont,then // becomes thetotalenergy ofthesystem. ThatHishereconstant along anygiven pathappears asfollows. Wehave inthegeneral casetherelation : dHm asisseen atoncebydifferentiating: *H=VM^: +T?dH_dpr,BH dt $,d(Irdt^dprdt"*" dt' andthenmaking useofHamilton's Equations. But inthegen- eralcase dPI/dt^0,andsoHisnotconstant along anarbitrary path. If,however, Hdoes notcontaint,then3H/dt=and sincenowdH/dt=0,wehave : (14) H=h. CHAPTER XII D'ALEMBERT'S PRINCIPLE 1.TheProblem. Thegeneral problem ofRational Mechanics, sofarasitrelates toasystemofparticles, canbeformulated : i)interms ofthe3nequations givenbyNewton's Second Law ofMotion : A) miXi=Xi, rmi/i=Yiywz=Zit i=1,-,n ; ii)interms offurther conditions equivalent to3nrelations between the6n+1variables (xi}yi}ziyXifYi,Zi, t).Apostu- lational treatment ofthese conditions willbefound below in Appendix D. Two extreme casesmay bementioned attheoutset. First, each variable Xi,YiyZifmay begiven asanexplicit function oftheXi,yi,Ziand their first derivatives with respect tothe time,and t: X,=*/fa,yt,zifXi,yiyz{,t) YJ=*/fa,yiyZi, i,yi}Zi,t) Thus A)reduces toasystem ofsimultaneous differential equa- tions fordetermining Xi,yiyZiasfunctions ofthetime, andwith thesolution ofthisproblem thedetermination oftheXi,Yi,Zi isgivenbysubstitution. Secondly, attheother extreme, thepath ofeach particle and thevelocity oftheparticle initspathmay begiven. Thus #ijy\>Zibecome known functions oft,andagain theXi,Ft,Zi arefound bysubstitution. Between these twoextremes there isaclass ofproblems in which constraints occur which canbeeliminated byageneral principle due tod'Alembert. We shall notattempt togive ageneral definition of"constraints/' fornosuch definition exists; butwecanformulate arequirement which embraces theordinary cases that arise inpractice. Let8xitdyi,6ztbeany3nquantities 345 346 MECHANICS whatsoever. Then itisseen atoncefromA)that thefollowing equationistrue : (1)2)(m&~x*>*x*+(m*y*~r)fy<+(m<*<-z*>dZi=- -i Thisequationissometimes referred toastheGeneral Equation ofDynamics. Now itmayhappen that theforceX^F,Zicanbebroken upintotwoforces : (2) Xi=XI+X!, Yi=Y(+Yl Zi=Z(+ZI where thetwonew forces, namely, theX'itF,Z\and the X*tF|*,Z*,aresimpler than theoldforthefollowing reasons. i)TheX't,Yi,Z{areeither explicit functions ofthex^?/,Zi, i,yi fZi,torthey involve inaddition arestricted setofun- known functions arising from forces which arenotgiven asfunc- tions ofthese&n+1variables. ii)TheXf,Yf,Zfhave theproperty that (3) 2)XfdXi+Yfdyi+ZfdZi= !--=! forallvalues ofthe6xi, 5t/,dztwhich satisfy the\iequations: n (4) ^/Aiadx i+Biadyi+CieSz^ 0, a=1, ,M, <=1 where thecoefficients aregiven functions ofthe6n+1variables XiyyijZi,Xifj)itZi, t,andtherank ofthematrix : "11"* "-"nl (5)' is/x;andconversely. Bymeans of-Equations (4)the3nquantities X*,Yf,Z*can beexpressedinterms of/xunknowns asfollows. Multiply the a-thEquation (4)byanarbitrary number \a,andsubtract the newequation from(3).Thus (6)2)(ft- i-l a-l D'ALEMBERTS PRINCIPLE 347 Now, inEquations (4),acertain setof3n nofthe&c, By*,dZi canbechosen atpleasure andthen theremaining /*ofthese quantities willbeuniquely determined, foratleast oneAt-rowed determinant from thematrix (5)does notvanish. Itfollows, then, that\, ,X^canbedetermined uniquely from asuitable setofjuequations chosen from the3nequations: H M f- (7)X*2^Aia\a, Yi a= 1 Andnow theremaining 3n/zEquations (7)willbesatisfied bythese values ofthe X's.ForEquation (6)hasbecome an equation inwhich only those terms appear forwhich6x,, 8yi,6z arearbitrary, andhence their coefficients must each vanish. Equations A)cannowbewritten intheform : (8) mix*=X(+5)Aia\a,nnyi=Y(+JfiiaX, where theXt, ,X^havecome tousaslinear combinations of/zsuitably chosen X*,Y*9Z*.Ontheother hand, theyappear inEquations (8)merely as/zunknown functions, which canbe determined from/*ofthese equations andthen eliminated from theremainder. Virtual Work. Toputfirstthings firstwasnever more impor- tantthan inthestatement ofd'Alembert's Principle. The3n quantities 8xi,SyiydZiaretobegin with3narbitrary numbers, andwethenproceed torestrict thembytheequations (4). Never- theless, whatever values theymay have, they determine bydefi- nition avirtual displacement ofthesystem ofpoints (Xi, t/, 2,-), andthequantity: Ws-2)X{dxi+YiBy*+Z,dZi =i isbydefinition the virtual work corresponding tothis virtual displacement. Thus Equation (3)saysthat theforceX*tY*,Z* issuch that itdoesnovirtual workwhen thevirtual displacement issubject totheconditions (4). 348 MECHANICS D'ALEMBERT'S PRINCIPLE FORASYSTEM OFPARTICLES. Given asystem ofparticles,themotion ofwhich isdetermined inpartby Equations A).Thediscovery ofananalysis ofXiyYi,#by(2) andofthemost general virtual displacement 5Xi,dy^ 6zifwhereby the virtual work oftheforce Xf,Yf,Z*vanishes, this virtual displace- ment being expressed by(4); finallytheelimination ofdXi,dyifdzir andX*jYf,Z*,asabove setforth, whereby 3n nequations free from theseunknowns result; this isthespirit and content of d'Alembcrt's Principle. This enunciation ofthePrinciple doesnotrepresentitshistoric origin, butrather itsinterpretation inthescience today;cf . Appendix1). 2.Lagrange's Equations foraSystem ofParticles, Deduced from d'Alembert's Principle. Letthecoordinates x,yi,zofthe systemofparticles considered in 1beexpressible interms ofm parameters andthetime : (Di?m, ,<im, itfm, where (ql} ,qm)isanarbitrary point ofacertain region of the(</!, ,gm)-spacc andtherank ofthematrix ism.Let Thenbythepurely mathematical process ofdifferentiation and substitution Equation (1)of 1yields: dT8T ZdtWr'Wr where and (4) Now, themquantities 6ql9 ,dqmarewholly arbitrary. Hence thecoefficient ofeachterm in(3)must vanish, andsowearrive D'ALEMBERT'S PRINCIPLE 349 atLagrange's Equations intheirmost general form forasystem ofparticles: ,,, d8T8T Here, norestriction whatever isplaced ontheforcesX,Ft, nor isthenumber ofqr'$required tobeaminimum. Inanimportantclass ofcaseswhich arise inpractice, (6)x,=x'<+xi F,=F;+17, Zi=z;+z*, where Hence r=1, ,TW. Equations (5)nowbecome Lagrange's Equations forthisrestricted case. Thecases ofconstraints thatdonovirtual work arehere included. Cf.further Appendix D. 3.TheSixEquations foraSystem ofParticles, Deduced from d'Alembert's Principle. Letthesystem ofparticles of Ibe subject tointernal forces such thattheaction andreaction between anytwoparticles areequal andopposite and inthelinethrough theparticles: X*ij __~ And letanyother forces X'tjY[,Z(act. Let dXi=a+pz t-yiji fyi=b+yxi aZi dZi=C+ayi-pXi where a,6,c,a, /?,7are sixarbitrary quantities. Since the internal forces destroy oneanother inpairs, and likewise, their moments, Equation (1)goesover intothefollowing: 350 MECHANICS =aJ(m,<-X'()+b2(*<#<-F|)+c2(<*<-Zf) 182)(x Now, setanyfiveofthesixquantities a,6,c,a,#,7equal to0, andthesixth equal to1.Thus the sixequations ofmotion, fromwhich theinternal reactions havebeen eliminated, emerge: (1)n n )m*g=2J Invector form these equations appear astheEquation ofLinear Momentum : ^=F andastheEquationofMoment ofMomentum : dt Wehaveused d'Alembert's Principle todeduce asetofnecessary conditions. These arenotingeneral sufficient, because thefore- going choice ofdxif8yiydziisnotingeneral themost general one. 4.Lagrange's EquationsintheGeneral Case, andd'Alembert's Principle. Consider anarbitrary system ofmasses, towhich Lagrange's Equations, onthebasis ofsuitable postulates, apply: /i\ddTdTn i (1)5a--" r-i,..-,*. Ifwesetbywayofabbreviation : (2)ddT_dT_ (2)~ [^ D'ALEMBERT'S PRINCIPLE 351 then (3) 2([71- Qr)Sqr=0, where dql9 ,dqmareanymquantities whatever. Itmay happen thatQrcanbeexpressedintheform : Qr=Q'r+Q], where Q'risforsome reason simpler thanQrandwhere, moreover, (4) Qffy,++Q;s?m=0, provided (5) daidqi+' '+a>am8qm=0,a=1, ,/I, therank ofthematrix : being /*.Byreasoning precisely similar tothatused in1,itis seenthattheQ*canberepresented intheform : M Qr 2La<*r^a, r=1, ,W, where theXcanbeinterpreted physically ascertain linear com- binations ofasuitable setof/*ofthequantities Q*. Moreover, Lagrange's Equations takeontheform : ddTdT,A _ where now theXaarethought ofasunknown functions, which canbedetermined by /*ofthese equations andthen eliminated from theremaining mnequations. Virtual Work. Inallcases theexpression canbeinterpreted asthevirtual workdoneonthesystem bythe forces which correspond totheQr.Inparticular, then, the condition (4)means that thevirtual work oftheforces which lead totheQ*isnil,provided thatthevirtual displacementcor- responds tothecondition expressed byEquations (5). 352 MECHANICS 5.Application:Euler's Dynamical Equations. Consider a rigidbody, onepoint, 0,ofwhich isfixed, andwhich isacted on byany forces. Itsposition may bedescribed geometrically in terms ofEuler's Angles, Chapter VI, 15 : (1) Qi=0, q2=t, q*=<p. Itskinetic energy is,byChapter VII, 6: (2) T= -i(Ap*+Bq*+Cr2 ), where p= \j/sin6cos<p+6sin<p (3) q= \l/sin6sin(p+6cos<p r=^cos+^ Byd'Alembert's Principle, 4 : where allthree 5^rarearbitrary. Let5q { 0,dq2=0, Compute Thevalue isseen atonce tobe : Ontheother hand,Q3canbecomputed asfollows. Denote thevector moment oftheapplied forces, referred to0,as M=La+Mf3+Ny. Then thevirtual work corresponding tothevirtual displacement (5q ly8q2)5g3)=(0,0,dqz)isseen tobe : HenceQ3=N,andwefind : Thus one ofEuler's Dynamical Equationsisobtained, and theother twofollow bysymmetry, through advancing theletters cyclically. D'ALEMBERT'S PRINCIPLE 353 Thereader willsay:"But this isprecisely thesame solution asthatgivenearlier byLagrange's Equations, Chap. X, 16." True, sofarastheanalytic details ofthesolution go ;andthis is usually thecasewith applications ofd'Alembert's Principle.It istheapproach totheproblem through theGeneral Equation ofDynamics, 1,which hereyields (4),andtheconcept anduse ofvirtual work, that brings thetreatment under d'Alembert's Principle. 6.Examples. Consider theproblem oftheladder sliding down asmooth wall;cf.Fig. 88,p.147. Letusregard this problem asthemotion ofalamina, moving initsown plane. AHthegeneralized coordinates ofthelamina wemay take the coordinates ofthecentre ofgravity: ?i=x, q2=y,qz=8. Then (1) T=%M(fr+p)+pfF2 , where kistheradius ofgyration about thecentre ofgravity. Byd'Alembert's Principle, (2) 2([T\r-Qr)8qr=0. r-lV ' Inthepresent case, Q^=S5x, Q26</2=(R-Mg) by, Q35<73=a(Ssin-Rcos6)d6, andthusQrisdetermined. Let Qr=Q;+QM where Q*=s, Ql=R,Q3*=aOSsin9-Rcos8). Now, x,y,6areconnected bytherelations: (3) x=acos0, y=asin 6. If,then,wesubject dx,dy,SOtothecorrespondingrelations: 8x=asin660, 6?y=acos 60, weseethat (4) Qr^i+Ql5<72+Q,*ff,=0. 354 MECHANICS Thus thevirtual work oftheforces Q?,corresponding tosuch adisplacement,isseen tobenil,andsoEquation (2)isreplaced bythesimpler equation: (5) i;(m,-$W==o. r-lV ' Hence M^j(-asin0)66+(^^f+Mg)acos669+MW 66=0. Onreplacing these second derivatives ofxandybytheir values from(3)adifferential equationinthesingle dependent variable isobtained : andthisdetermines themotion. Rough Wall. Suppose, however, thewall isrough; Fig. 145, p.323. Equations (1)and(2)stillhold. Butnow Qi*fc=(S~R) x,Q26q2=(R+S-Mg) 8ff, Cs^3=a[S(sin6+/zcos6)+R(/xsin6-cos0)]66. Let Qr=Q;+Q;, where Qr=S-/, Q*=B+MS, Q*=aS(sin+Mcos0)+aR(n sin cos0). Thevalues of5^D6q2,8qzwhich make thevirtual work ofthe forceQ*vanish : QI^+c;?i+cr?i=o, arefound bymaking thecoefficients ofBandSzero inthis last equation: /z&7i+6q2+a(Msin6cos0)6qz= ...4-a(sin+Mcos0)6q3= Hence f(1+M2 )fyi=a[(1- /i2 )sin-2Mcos0]6q^ I(1+M2 )5^2=a[-2Msin+(1-M2 )cos0]6q3 D'ALEMBERT'S PRINCIPLE 355 Equation (5)nowbecomes : M^jSq,+(M^JL+Mg)Sq,+Mk^dq,=0, and itremains only tosubstitute thevalues of5qlf8q2from(7), andthevalues ofx,yfrom(3),andreduce. The result is : ((1-M2 )a*+(1+M2 )*)JJ+ ag[2/xsin6-(1- /*2 )cos0]. The virtual displacement (dqlt5g2,dq^)which here ledtothe elimination oftheunknown reactions R,Swasnotonewhich inanywiseconformed tothe"constraints"inthesense ofthe floorandthewall. Ifwereplace Rand /z#bytheir resultant anddraw alineLthrough thebottom oftheladder perpendicular toit,andthendothesame thing atthetopoftheladder, thus obtaining alineL2,theabove virtual displacement corresponds toanactual displacement inwhich thebottom oftheladder is moved alongLtandthetopalongL2. CHAPTER XIII HAMILTON'S PRINCIPLE ANDTHEPRINCIPLE OFLEAST ACTION 1.Definition of8.Anewandindependent foundation for Mechanics isgiven byHamilton's Principle and certain other Principles oflike nature. An integral, forwhich Hamilton's Integral: jV+t/)dt, istypical,isextended along thenatural pathofthesystem, and then itsvalue isconsidered foraneighboring, orvaried, path. The Principle asserts that theintegral isaminimum forthe natural path, oratleast that theintegralisstationary forthis path,i.e.that itsvariation vanishes : (T+U)dt=0. Itistothetreatment ofthissubject thatwenow turn. Obvi- ouslywemust beginbydefining what ismeant byavaried path andbyavariation 8. LetF(x l9 ,xn>x[,- ,x'n,u)beafunction ofthe2n+I variables indicated. Here, (xl9 ,xn)shall lieinacertain regionRofthen-dimensional space ofthevariables (xly ,xn); thevariablesx[ 9 ,x'nshall bewholly unrestricted;andu shall lieinthe interval :agugb.The function Fshall becontinuous, together with itspartial derivatives ofthe first andsecond orders.* Let *Asregards assumptions ofcontinuity, welaydown onGOand for allthe requirement thatwhatever arbitrary functions areintroduced shall becontinuous, together withwhatever derivatives wemay wish touse, unless thecontrary is stated. Foranintroductory treatment oftheCalculus ofVariations cf.theauthor's Advanced Calculus, Chap. XVII. 356 HAMILTON'S PRINCIPLE. LEAST ACTION 357 C : Xi=Xi(u), a^ug6, i=1, ,n, beapath lyinginR.Let ,_dxj(u) Xl~ du' Thus apathTinthe(2n+l)-dimensional space ofthearguments ofFisdetermined. Byavaried path, F',ismeant thefollowing. LetCfbeacurve inRdefined bytheequations: C' : Xi=Xi(u, e), a^ug6,f=1, ,n, where a?t(w, 0)=Xi(u), and isconsidered onlyinaregion forwhich |e |issmall. De- note partialdifferentiation with respect toubyd.Let ,e) // \-i '~~ du' ' , , , bechosen asthevalues ofthe#(,-, #.Thecurve F'inthe (2n+l)-dimcnsional spaceiswhat ismeant byavaried curve. The variation of#,-,or&c,-,isdefined bytheequation: Since xl(u,e)isanyfunction thatconforms merely tothegeneral requirementsofcontinuity, weseethat dxi=IH(U),i=1, ,n, isawholly arbitrary function, restricted onlybytheabove re- quirementsofcontinuity. The variation ofx't,orbx( isnot,however, arbitrary, but is defined bytheequation: dude>o Thus Hence .A< -r--oa;=o-jdu du Itisnownatural tolaydown thefurther definition: (4) 358 MECHANICS Definition ofSF.Bythevariation ofF(a;,-,x't,w)ismeant : 8"=().., where Xiand xjontheright-hand sidearesetequal tox(u, and ZI'(M, e).Hence (6)^-|(J>'+I Itisobvious that 8(F+$)=8F+d$; andalsothat where ,,*,,, x'n,u), k=1, ,m. Finally, thedefinition: (7) ddF=ddF, corresponding tothetheorem : /ON *dFd8F (8)a5?-"5T And similarly, 6 b (9)JFSd*=* a a a Thedependent variables x^u), ,xn(u)playar61e inthe foregoing definitions analogous tothat oftheindependent variables inpartial differentiation. Buttheanalogy holds onlyuptoa certain point, andtoassume itbeyond theorems liketheabove formulas which wecanprove, hasledtoconfusion anderror in physics. Variation ofanIntegral. Consider theintegral: b /f F(Xi,u'' >XH9X19 ,Xn,U)du, HAMILTON'S PRINCIPLE. LEAST ACTION 359 taken along thepath F.By8Jismeant thefollowing:Extend theintegral along thepathF'.Thus afunction /(e)isdefined. andnow,bydefinition: (10) ^ Itfollows atonce asatheorem that (ID The integralissaidtobestationary foraparticular pathFif b 8CFdu=0, a nomatter what functions 8x*=tm(u)may be.The condition isreadily obtained incase dxi isrestricted tovanish foru=a and foru=6: (12) dxi|.=in(a)=0; &e<|Ussb=^(6)=0. For: d($F^\dFfid^* d^V^J5XV" ~d$8Xi+dHW<8Xi' Hence 03) If,now, theintegralinquestionistovanish foranarbitrary choice of&C;,itiseasily seen thateach parenthesisintheinte- grand ofthelastintegral must vanish, or : These areknown asEuler's Equations. Itisclear that 6 b (15)djFdu= j8Fdu. 360 MECHANICS The limits ofintegration maybevaried, too. Let a'=^(a, ),6'=*(&,), where ^?(a, 0)=a, \f/(b, 0)=6. Let 6' J(c)=IF[Xi(u,e), x'i(u,c),u\du. a' Then thevariation oftheintegralisdefined asbefore, by(10). Itfollows that 6 b (16)dCFdu =C6Fdu+F(B i9Bf i9b a a where .A*=z(o), A'<=x(a), < EXERCISE Since l/,6) itfollows (under theordinary hypothesesofcontinuity), onlet- tingeapproach 0,thattheright-hand sideapproaches f.du Theleft-hand sideapproaches 8&.Hence ,eM> d 5-j-=-7-6$.awaw Thus Equation (7)isobtained asatheorem, andnonewdefinition isnecessary. Explain theerror. 2.The Integral ofRational Mechanics. Allthat hasgone before merely leadsuptothedefinition ofthevariation ofthe following integral: ti (1)j> F(x lt---,x n,xl,-",x n,()dt, HAMILTON'S PRINCIPLE. LEAST ACTION 361 where thelimits ofintegration maybeconstant orvariable. The answer would seem tobesimple,since tisthevariable ofintegra- tionandhence theindependent variable ineach ofthefunctions Xi=Xi(t). Butthesymbol written down astheintegral (1)istaken inPhysics tomean something totally different. Let (2)t=t(u), ^u beanyfunction ofusuch that, intheclosed interval(0,1), 0<f-f.du Then thesymbol (1)istaken tomean theintegral: (3) /=JF(xlt,*,|f,..-,|?)t'du, where x[=dxi/du, andthe"variation oftheintegral (1)" is understood tobethevariation ofthis lastintegral. Thus (4) dJ=Cd(Ft')du, o where uisthevariable ofintegration, andtheindependent vari- able ineach ofthefunctions Xi=Xi(u),t=t(u). This lastvariation, (4):&/,comes under theearlier definition ofthevariation ofanintegral. Inparticular, Equation (4)may bewritten intheform : 8J= Ineach ofthese integrals thevariable ofintegration may be changed backfromutot,andthus t <t (5) SJ=CdFdt+ foddt.fbFt'du+ CF8tfdu. 362 MECHANICS Thevariation oft,namely St,isanarbitrary function ofu: U=T(M), and dSt T'(U) dt t'(u)' where uistheinverse function defined by(2). Moreover, by 6Fin(5)ismeant thefollowing: m- where /z,'\ t'5x't-xW ,d .., d 8(7)=- ?5. *,= Tu$x<,st=^a. Wesee,then, that ' Now,when 2istheindependent variable, /o\ *-ddXi (8) 5^*=-ir- Thetwoformulas, (7)and (8),show that 5iisnotinvariant oftheindependent variable. Why should itbe? Similarly, thevariation ofanintegralisnotinvariant ofthevariable of integration. Much oftheconfusion intheliterature arises from losing sight ofthis fact. The"variation oftheindependent variable" issupposed tocover this case. Itdoes sowhen and onlywhen itbecomes identical insubstance with theabove analysis. 3.Application totheIntegral ofKinetic Energy. Bydefini- tion, thekinetic energy Hence n (1) &T=5}mi(idXi +yifoji i-i nomatter what theindependent variable and thedependent functions may be. Iftheformer isu}then HAMILTON'S PRINCIPLE. LEAST ACTION 363 (2) 8T= ,[*,5()+frS()+ 6 ByFormula (7)of2andthecorresponding formulas involving yiyZiythisequation becomes : a,w. Variation oftheIntegral: (4) J= Letthenatural pathinspace berepresented parametrically by theequations: Xi=Xi(u), ^u^1, and let Thevariation ofthisintegral has,by 2,(5)thevalue : (5) Bytheaidof(3),f*Tdt=(*8Tdt+ CTdd *o tot The firstterm ontheright canbetransformed byintegration byparts, theintegrand obviously having thevalue : Hence *t <i (7) fdTdt=-Cj?rm(XidXi+Hidyi+2<62^)d< V V<"1 'l +5)wiifofa, +yi%+^fe {)tl~2TlTd^. <.i<oJ 364 MECHANICS Finally, then : /i ^ (8)5CTdt=-/5)mt(ftfa,+ i:sl The variationsdxi,dy^bz^dtare3n+1arbitrary functions, subject merely totheordinary conditions ofcontinuity. If, inparticular, weimpose on&c-,dyiy5zt-thecondition thatthey vanish attheextremities oftheinterval ofintegration,i.e.for t=tQjtlythen *,/t <i (9)Ardt =-f5)mi(ft fa*+*<+2*820*~2J /o 'o~ *o and '' /!n (10)d(*Tdt=-fTm<(ftfa,-+y<fyi+5f- //i-i /O h 4.Virtual Work. Bythevirtual work oftheforces Xi, F,-,Zt-, considered along thenatural path: (1) Xi=Xi(u), yi=y(w), 2.=ti(u), UQ-^U ^uly ismeant thequantity: (2) Ws= where fai, 5i/i, 5^^are3narbitrary functions ofu,subject merely totheordinary conditions ofcontinuity. Thisquantityisoften denoted bydW;but itisnot, ingeneral, thevariation ofany function, and soitisbetter toavoid this confusion, writing 5WonlywhenW8isthevariation ofafunction. 6.TheFundamental Equation. Combining Equation (9)of 3withEquation (2)of4wehave : HAMILTON'S PRINCIPLE. LEAST ACTION 365 (1) Here the3n+1variations &rt-,6?yt,5zt-,5arearbitrary except that the first3nofthese vanish for t tQ,^;andthenforces Xi,Y{,Ziareanyforces whatever. Ifthese arethetotal forces acting ontheparticles, theright-hand side of(1)willvanish since each parenthesis vanishes byNewton's Law,andweshallhave : t\ (2)J(ST+2T<~+Wt)dt=0. h Letthetotal force bebroken upintotwoforces : (3)Xi=xi+xi Y>=y;+Y;, z,=z;+zi Then (4) Ws=Wv+Ws*. Suppose thatW&*vanishes : (5) W8*=X;txt+Yt8y {+Ztdz<=0, t=i when thevariations dX{, 8yi, dziarechosen subject tocertain conditions. Then Equation (2)takes theform : (6) wherenow dxt, 8yi, dzisatisfy these conditions,dtbeing stillwholly arbitrary, and (7) W9=2)XJ t=i These conditions usually taketheform : (8) 366 MECHANICS whereAta,Bta,Ciaarefunctions ofx,-, /<,Zi,xityf,zitt,andthe rank ofthematrix : An Cni (9) *1I/A* * *v>n/A isM.This case includes both theholonomic andthenon-holo- nomic cases. But itmust beobserved that8T isingeneral no longer, ornotyet,thevariation ofafunction.* Generalized Coordinates. Suppose that thecoordinates xt-,yi}Zi ofeachmass micanbeexpressedinterms ofmparameters q.u''' iqan<ithetime : Xi=fi(q\j >qm, f) Vi~ <?i(qu' iq*n, Zi=^i(q^ ,qmjt) where therank ofthematrix(10) ism.Suppose further thatEquation (5)issatisfied when dfis , ,dfi ,OXi==7 OQ'i ~i~"**H O^m ^^j ^Q'm *._^^ . .fa *(11) the5^, ,8qmbeing arbitrary. Let /1O\ /^ ''O(V^i IV/^2/I^ (12) Qr=> (A<-r+Yi-r+/+I*-'**. r=1, ,m. *The definition ofthevariation ofafunction, itwillberecalled, isbased on thedependence ofthelatter oncertain arbitrary functions, whose variations may alsobetaken asarbitrary. These arbitrary functions areanalogous, letusrepeat, totheindependent variables inthecase ofpartial differentiation. And sofurther assumptions (i.e.postulates ordefinitions) areneeded before dTcanagainmean a variation. HAMILTON'S PRINCIPLE. LEAST ACTION 367 Then (13) W?=Q!&++QmSqm. Ontheother hand, (14) T=T(q l9,.,*, ,&,,<) and5T7asgiven by 3,(2),becomes thevariation ofthis latter function, where qi(u), ,qm(u), t(u)aretheindependent functions. Equation (6)ofthepresent paragraph nowtakes the form: where dTmeans what itsays thevariation ofTandwhereWsistheWyof(13). We willdenote thisequation astheFundamental Equation. Itembraces Equation (2)above, fortheXi,t/,zcanalways be taken asm=3ngeneralized coordinates. Thisequationissometimes written intheform : ti X) C(dT+5W) dt+2T8dt=0, where (15) dW=Qldql+--+Qmdqm. Letusseejustwhat thismeans. First ofall,theequation is trueunder thehypotheses which ledtoEquationI.These were, that thepathisthenatural path ofthesystem, given bythe equations: (16) qr=ffr(tO,t=t(u), where(0)=t,t(l)=tltandthevariations 8qr=ir(u),8t= /(w), arearbitrary functions subject merely totheconditions : 7?r(0)=0, r=0,1,...,m; r?r(l)=0,r=1, ,m, 368 MECHANICS andpossibly toafurther restriction : Irjr(u) |<A, |jr(u) |<h, r=0,1, ,m, where hisadefinite positive constant. Furthermore, inEquation X),8W=W$isnotingeneral the variation ofanyfunction ofqlt ,qm,t.TheQrhave definite values ateach pointofthenatural path, andsoaredefinite func- tions ofu;buttheydonotingeneral haveanymeaning ata point (qr,t)notonthenatural path, nordoesdW. Finally, (17)^SfV where ^ \dudu dudu \du The lastterm intheintegral hasthevalue : /27 Andnow themeaning ofEquation X)isthis :Ifqrand t aresetequal tothefunctions (16)which define thenatural path and ifdqr,8tarechosen arbitrarily, subject merely tothegeneral conditions above imposed, Equation X)willbefulfilled. Thus Equation X)expresses anecessary condition forthemotion of thesystem and this inallcases, bethey holonomic ornon- holonomic. Since Equation X)istrue forallvariations 8qr,8t,itstillrepre- sents anecessary condition when these functions aresubject to anyspecial restrictions wemay choose toimpose onthem. For example,itmayhappen that theQrcanbebroken upintotwo functions : Qr=Qr+Qr*,r=1, ,m, such that QiiQi++Qlfym=0, provided that a*i8qi++Oamfyw=0,a=1, ,/i, HAMILTON'S PRINCIPLE. LEAST ACTION 369 where aar=aar(ql9 ,qm,qlf ,qm,0>andtherank ofthematrix : isIJL.Here,dW isreplaced by (18) 5W= + butonlym/iofthe&/r,anddt,cannowbechosen arbitrarily. Inwhat sense is5Tnowa"variation"? Emphatically,inno sense;fornodefinition hasbeen laiddown which reaches out tothis case,and itisonlyfrom adefinition that5Tcanderive itsmeaning. Nevertheless, Equations (17)and (18) continue todefine thevalues oftheterms dT,bWthatappear inEqua- tionX),andthus thisequation continues tohave ameaning, and toholdwhen acertain setofm/*variations 8qr,anddt,are chosenarbitrarily. Force Function. Finally, theremaybeaforce function, U : (19) whereUisafunction ofthe#,-,y^z,and t.Thus theFunda- mental Equation (2)becomes inthiscase : II. where the(Xi, F<,Zi)of(19)isthetotal force acting onm,-, provided(/doesnotdepend on t;otherwise wemust understand byd(Jthevirtual variation ofU,or : Again, theremay beafunction U(q l, ,qm><)such that in(15) SU r1, m. Then &W=BU 370 MECHANICS and theFundamental Equation takes onthesame form, II., provided Udoesnotdepend on t;otherwise, 'U~W*+~'+Wt*" oq\ oqm 6.The Variational Principle. Thevariational principle asex- pressed bytheFundamental EquationI.of5,orevenby Equation II.,does notassert that theintegral ofsome function, orphysical quantity,isaminimum, oreven stationary: 5I(something) =0 or Id(something)=0. Fortheintegrandisnotavariation,inthesense oftheCalculus ofVariations; noraretheforces oftheproblem varied; they areconsidered onlyalong thenatural path ofthesystem.* The Principle expresses anecessary condition forthemotion ofthe system. Inthenon-holonomic case, thecondition cannot be sufficient, since the first-order differential equations have not been incorporated intotheformulation oftheproblem. Weturnnow tocertain further restrictions whereby Hamil- ton's Integral orananalogous integral doesbecome stationary, andinfact,inarestricted region, aminimum. 7.Hamilton's Principle.Ifweset : t(u, c)=U,tQ^U^ tlt then 8ts andtheFundamental EquationI.becomes : (1) Wecannowsuppress theparameter usince thetime isnottobe varied. *Initsleading ideas thistreatment wasgiven byHolder, Gottinger Naehrickten, 1896, p.122. Unfortunately Holder feltimpelled todefer totheprimitive view of variations as" infinitely small quantities" inthesense oflittle zeros, i.e.infinitely small constants orfunctions ofXi,yltz, t.Inthefoot-notes onpp.130,131the "neglect ofinfinitesimals ofhigher order" renders obscure infact, vitiates thetreatment, sofarasclean-cut definitions go.Thewriter cannot but feelthat theinner Holder would have preferred such atreatment asthat ofthetext,but thathedidnothave thecourage tobreak with theunsound traditions ofthe little zeros, forfearoflosing hisclientele. HAMILTON'S PRINCIPLE. LEAST ACTION 371 Suppose thataforce function Uexists, which depends only ontheXij2/,zandt,orontheqrand t: (2)U=U (xify<,ziyf) orU=U(qr,t). Since tishere theindependent variable with respect tovariation 5,wehaveW8=6U inthesense oftheCalculus ofVariations, and (1)becomes : *i (3) C(ST=0. Itmust beremembered, however, that thevariations&r,, fly,-, bZior5qrsatisfy thecondition ofvanishing when t=tandwhen t=t1 . Herewemeet our firstexample ofanintegral, (4) to thevariation ofwhich vanishes : ti (5)*f(T+U)dt=0. h This equation embodies Hamilton's Principle, which wemay formulate asfollows. HAMILTON'S PRINCIPLE. LetTbethekinetic energy ofasystem ofparticles fand letaforce function U=U(q r,t)exist. The natural path ofthesystemisthatforwhich Hamilton's Integral: (6) l(T+U)dt, isstationary: (7)dI(T+U)dt=0. Here tistheindependent variable, andthevariations ofthedependent variables aresuchasvanish when t=tQandwhen t=tv 372 MECHANICS Anecessary and sufficient condition that (7)betrue isafforded byEuler's Equations, 1,which herebecome : ddTdTdU,___ ______ rp.1,,. i*jj dtdq rdqr~ dqr' ' ' But these areprecisely Lagrange's Equationsforthesystem. Incidentally wehave anewproofofLagrange's Equations,in casewemake Hamilton's Principle ourpoint ofdeparture. Wehave proved thePrinciple forsystems ofparticles withm degrees offreedom, and itcanbeestablished incertain more general cases, e.g.forsystems ofrigid bodies; provided eachtime thataforce function exists. Thecase isalsoincluded, inwhich relations oftheform : *>(?!>' ' ' >?,=0, a=1, ,Ml exist;cf .Bolza, Variationsrechnung, p.554.Themost general case isthat ofasystem having aLagrangean Function, orkinetic potential,L.Intheabove cases, L=T+U. When itisnotpossible toestablish itwithout special postulates consider, forexample, themotion ofaperfectfluid orofanelastic bodyitistaken asitself thepostulate governing themotion ofthesystem. ThePrinciple consists, then,inrequiring that theLagrangean Integral: d (8)fldt to bestationary ;orthat i bCldt =0. V Itwillbeshown in14thatHamilton's Integral (8)isactually aminimum forapath lying within asuitably restricted region ; buttheminimum property does notnecessarily hold forun- restricted paths. 8.Lagrange's PrincipleofLeast Action. Ourpointofdeparture istheFundamental Equation II., 5,inwhichUnowdoes not depend on t: HAMILTON'S PRINCIPLE. LEAST ACTION 373 Moreover, Tdoesnotdepend on t: T=T(qi, ,qm,qiy ,?m). Thuswehave : ti (1) C(ST+SU)dt+ZTddt =0. <0 Here each 5represents avariation inthesense oftheCalculus ofVariations, theindependent functions being ql9-- ,</m,t; buttheintegrandisnotthevariation ofsome function, nor is theintegral thevariation ofsome integral. Nevertheless, the equationistruewhen allm+1variations, qlt- ,qm<t,are chosenarbitrarily. Letusexamine more minutely themeaning ofthis laststatement. These variations aredefined byarbitrary functions : (2) 0r(u,e), (u,), such that qr(u,0)=g r(u), t(u,0)=t(u). Moreover : ?r(0,)=<?r(0), ?r(l,=g r(l)j j(0,c)=<(0,0)=<=const.; i(l,0)=^. Butingeneral ^(1, c)^tvThus 5|-o=0; ^|tt=1^0. Inparticular, then, thefunctions (2)mayberestricted byany further conditions which arecompatible merely with thegeneral conditions ofcontinuity. Such acondition istheone that, not merelyforthenatural path corresponding to=0,butalso for allvaried paths: (3) T=U+ft, or,moreexplicitly: (3') T[?r(u, ),^|]=U[?r(u,c)]+A, where feisaconstant. SinceTishereahomogeneous quadratic polynomialinqlf- - - ,qm,itisclear that t(u ye)isobtained bya quadrature when theqr(u,),r=1, ,m,arechosen arbi- trarily. 374 MECHANICS Let beanyfunction. BydFweshallnowmean thefollowing: (4) SF=- u where qr(u,e),t(u,e) arerestricted bytherelation(3),i.e.(3'). And similarly: (5)b\Fdu= -j-iFdu,J 0*J e-O where theintegrand onthe left isformed forthearguments qr(u), etc.,and theintegrand onthe right, forqr(u,c),etc.; Equation (3')stillholding. Thus itfollows, inparticular, that (6.) bT=8U. Although these definitions areinform identical with theearlier ones, where them+1functions (2)were arbitrary, they arein substance distinct, since thesem+1functions arenow related by(3')- Equation (1)nowbecomes, onsuppressing thefactor 2 : r, (7) CdTdt+Tddt=0. Since obviously, under ournew definition of5, 5/TT/'\ xT7tfirnjj^'0(11)=01'I -f-7Ot, andsince 5t'=dbt/du, Equation (7)takes theform: i (8)Id(Tt')du=0. o Hence, finally: (9)lj*Tdt=0. Wearethus ledtothefollowing Principle. LAGRANGE'S PRINCIPLE OFLEAST ACTION. Letasystem of particles have thekinetic energy Tandaforce function U,whereU depends onlyontheposition ofthesystem, notonitsvelocity orthe HAMILTON'S PRINCIPLE. LEAST ACTION 375 time,andwhereTisindependent oft.Thenanecessary andsuffi- cient condition forthenatural path ofthesystem is,that (10) subjecttothehypothesisthat allvaried paths fulfiltherequirement that (11) T=U+h. Inaddition^thevariations ofthecoordinates shall vanish fort=t and t= Jj. The Principle thusformulated presents aLagrangean problem intheCalculus ofVariations with variable endpoints andone auxiliary condition : (12)CTM=o, Themethod ofsolution developed inthat theory* employs Lagrange's Method ofMultipliers. Briefly outlined itisas follows. Set F=T+\v, where Xisafunction oft,and letqr(t),X($),bedetermined by them+1equations /*o\*fl? d&Ff\ t <13>Wr'dtWr^' '-I.'".*. andthesecond equation (12). From Equation (13) itfollows that^+X^_^r^ +X tyr dqrdtLdq r d or These equations, combined with thesecond equation (12), give: (14) X=- -J-. *Cf.Bolza, Variationarechnung, p.586,where thecase isconsidered that there are,inaddition, relations between thecoordinates, notinvolving thetime. 376 MECHANICS Hence theqr(t)aredetermined from theresulting equations, 8T .dU ddT The latter areLagrange's Equations. Incidentally wehave anewdeduction ofthem, based onLagrange's Principle ofLeast Action. Asinthecase ofHamilton's Principle, soherewecangivea direct proof ofLagrange's Principle ofLeast Action bymeans oftheCalculus ofVariations. For, asabove pointed out, the Principleisequivalent totheLagrangean problem represented by(12). Recurring tothecondition (3)weseethatthefunctions ql(u,e), 'm' 9Qm(u, e)maybechosen arbitrarily, andthefunction t(w,e) thendetermined by(3').Ifthefunction t(u, e)thusdetermined besubstituted intheintegral: i (16)fft'du, then tiscompletely eliminated from that integral. For (17) T=%Ar.qrq., AT.=A.r, r.s thecoefficients Ar8ldepending onlyontheql9- ,qm.Now, /=/2r ' to(18) Let (19) where, asusual, q'r=dqr(u)/du. Then or (20) From(20)and(3)itfollows that Tt'= HAMILTON'S PRINCIPLE. LEAST ACTION 377 andthus tiseliminated, theintegral (18)taking theform : f- Wenowhave before usaproblemintheCalculus ofVariations, ofmuch simpler type thesimplest type ofall,considered at theoutset. Itistheintegral (21), formed forthefunctions qr(u),that istobestationary, andthese functions are allarbi- trary. After thisproblem hasbeen solved,tisdetermined from (20), or (22) t-t,=f-j-JLJVu This isJacobi's Principle ofLeast Action, which wewilltreat inthenextparagraph asanindependent Principle. But itis interesting toseehow itcanbederived from theFundamental Equation of 5,andproved asaparticular caseunder Lagrange's Principle ofLeast Action. EXERCISE Show thatEquation (1)under therestrictions named canbe thrown intotheform : <i dT ,dU ddT\,, -x-ho 17TT-) &Qrdt=0. O(lr0qr (ItG(j[r' Hence deduce Lagrangc's Equations. 9.Jacobi's Principle ofLeast Action. Letasystem ofparticles have thekinetic energyTandaforce function U,whereUdepends onlyontheposition ofthesystem, notonitsvelocity orthetime,and where theconditions imposed onthecoordinates donotcontain the timeexplicitly. Then anecessary and sufficient condition forthe natural path ofthesystem is,thattheintegral: (1) fVU+hVTdt bestationary: (2) 378 MECHANICS Thetime isgiven bytheequation: (3) T=U- or (4) where (5)r vir+T =Vrdt. where <h(u stationary:Wecangiveadirect proof asfollows. The integral (1)has thevalue : i (6)JVu+hVSdu, ,qm(u)arearbitrary functions. Itistobe i (7)dIVU+hVSdu=0. Hence Euler's Equations must hold, or : (8) (V(7+hV5) 7(V[J+hVS)=0, T=1, ,m. d/vTT+1Hence . S d/ du\' Equations (9)determine thepath ;thasnotyetentered in the solution. Equations (3)and(5)now determinet;itis givenby(4). Itfollows furthermore that dqr\du/ dqry dq'r dudqr Combining these equations with(4)and(9)wefind : /ii\ .__ .\'J4O^, O^. Qatoqroqroqr HAMILTON'S PRINCIPLE. LEAST ACTION 379 Thuswearrive atLagrange's Equations.Ifweassume them, thenwehave aproof ofJacobi's Principle. Conversely,ifwe assume Jacobi's Principle, wehave anewproof ofLagrange's Equations. 10.Critique oftheMethods. Retrospect andProspect The symboldistreacherous. Itcananddoesmeanmany things, and writers onMechanics arenotcareful tosaywhat theymean byit.Ind'Alembert's Principle the dxt,8yifdZibeganlifeby being 3narbitrary numbers. Intheir youth they were dis- ciplined toconform tocertain linear homogeneous equations. Thus stillanumber ofthem were arbitrary quantities; therest hadnochoice, theywereuniquely determined. Enter, theCalculus ofVariations. Andnow the dxijdy^ dzi, and 8tbecome thevariations offunctions ofaparameter, orinde- pendent variable,u.Fromnowonthese <$'smust bedealt with under thesanctions oftheCalculus ofVariations atleast, ifthofindings ofthatbranch ofmathematics aretobeadopted. TheFuture. Asthephysicist fares forth over theuncharted ocean ofhisever-expanding science, hiscompassisthePrinciples. Heseeks anintegral which inthenewdomain willdoforhimwhat Hamilton's Principle achieved inclassical mechanics. There is mysticism about this integral. Imagination must guide him, andhowilltrymany guesses. Buthewillnotbehelped byan undefined d.Hemustmake acloan-cut postulate defining the integral, andthen laydown aclean-cut definition ofwhat he moans bythovariation. There isnoshort cut.Athorough- going knowledge oftherudiments oftheCalculus ofVariations isasessential inMechanics asperspectiveisinart. 11.Applications. Lotaparticle beacted onbyacentral attracting force inversely proportional tothesquare ofthedis- tance. Then (1) r= where thepoleisatthecentre offorce,and itisassumed thatthe motion takes placeinaplane (cf.Exercise4,below). Then the integral: (2) f\r*0'*du 380 MECHANICS must bemade aminimum. Set F(r,B,r',0')=>(?+h)(r'2+r2*'2 )' ThenA^-^=n dw00' 20 Since dF/dO=0,itfollows that ^F-^/M ..r*0' _ (6)W'-Vr+h Vr'*+r*0'*" Ifc=0,then=const. andthemotion takes placeinarightline. But ifc7*0,may betaken asthevariable ofintegration:*u=0,and (3)becomes : Hence e, . cdr */;rVhr2+r-c2 Change thevariable ofintegration: =1 "" r Then *=+//T i 22vAl~| /it/* C1L Performing theintegration, wefind : _1 ecos(0 y) =K EXERCISES 1.Discuss indetail thecase c=0. 2.Inthegeneral case,determine theconstantse,K,yinterms oftheinitial conditions. *Itistruethattheinterval foruwas (0,1) ;but itmight equally wellhave beenanarbitrary interval :a^u^b. HAMILTON'S PRINCIPLE. LEAST ACTION 381 3.Obtain thetime. 4.Allowing theparticle freemotion inspace, show that its pathisaplane orbit. Suggestion:UseCartesian coordinates. 5.Discuss themotion ofaparticle invacuo under theforce of gravity. Assume thepath tolieinaplane. 6.InQuestion 5,prove that thepathmust lieinaplane. 7.Explain thecase ofmotion inacircle under thesolution giveninthetext. 12.Hamilton's Integral aMinimum inaRestricted Region.* THEOREM. Theintegral f(1) ldt to isaminimum forthenatural path, providedtQand^arenottoofar apart. TheLagrangean function : ,?m, ft, hastheproperties: (3) 2 1,1 isapositive definite quadratic form. Moreover, (4) H+L=JPrqr, T where (5) Pr= J|r-1, ,m, andtheHamiltonian function H(q ly ,?,?!, -,pm, hastheproperties: *Oarathfodory hasgiven aproof ofthistheorem :Riemann-Weber, Partielle Differenlialgleichungen dermathematischen Physik, 8.cd.1930, Vol. I,Chap. V. 382 MECHANICS Anecessary condition thattheintegral (1)beaminimumis,that h 8fL8ILdt=0. Theextremals aregivenbyKuler's equations: which areprecisely Lagrange's equations. Bythetransformation (5),theinverse ofwhich isgiven by (7),Lagrange's equations (8)goover intoHamilton's canonical equations, Chap.XI : dqr^M dpr___Mr-i...m(9)dt~ dpr'dt* dqr' r~1' 'm ' The latter canbesolved bymeans ofJacobi's equation ;cf.Chap. XVandAppendix C : /imW .uf dv dv (10)_+ff(,1,...,, m,_...,_ asfollows. Let (qrQ ,pr,t)beapoint intheneighborhood of which Equations (9)aretobesolved. Asolution of(10): (11) V=S(q l9- ,qm,!,, ,t), canbefound*such that * * (12) -f^ ^(ll' '' , ) *Theexistence theorem inquestion follows atonoofrom thetheory ofcharacter^ istics asapplied toEquation (10). That theory tellsusthutthere exists asolution of(10): V-Stai, -,Qm,0, such that,when t=to,Sreduces toagiven function ^(71, ,Qm) : S(qi,--,qn,to)- <f>(qi, ,qm). Here, <p(q\, ,qm)isanyfunction which, together with itsfirst derivatives, is continuous intheneighborhood ofthepoint (qi, ,qm).Such afunction is: <P(Ql,'''.Qm)=Sr7r, r where the oti, , aremarbitrary constants, orparameters. Thefunction *S thus resulting isthefunction required inthetext. If,aswemay assume, thefunction //(qr,pr, isanalytic inthepoint (qr, pr,Jo),and if,asishere thecase, <f>(qr)isanalytic inthepoint (gr),then the fundamental existence theorem oftheclassical Cauchy Problem, formulated forthe simplest case, applies atonce,andthetheory ofcharacteristics isnotneeded. HAMILTON'S PRINCIPLE. LEAST ACTION 383 inthepoint (qrQ ,<*r,tQ)andfurthermore theequations: /-<o\ SS ~ 8S t (13) pr= Wf, fir-^,r=l,.-.,m aresatisfied the firstset,when (qv,pr,J)aregiven, bythe values ar=ar;andthen thesecond setdetermines /3r. Bymeans ofthisfunction SEquations (9)aresolved. The solution iscontained in(13)and isobtained explicitly bysolving (13)forqr,pr: (. f?r=/r(i, ,m,ft,' ,ftn, Ipr=grfai,''' ,OW,ft,''' ,j8m,t) Properties oftheExtremals. Ifr/r= </r(0represents an extremal, and ifqr=dqr/dt,thenby(5)and (13): 05) ^r=^f, r=l,.-.,n. Moreover : (16) 2)'M'+&= r For, sinceSisasolution of(10),itfollows, bytheaidof(13), that (17) St+H(q l9 ,?,?!, ,p*,0 =0. Onsubstituting thisvalue ofHin(4),andreplacing printhe resulting equation byitsvalue from(13),Equation (16) results. TheFunction E(q r,q'T,qr,t).Consider thefunction V=L(q r,q'r,t), where(</r,</J,t)are2n+1independent variables. Let (qr,qr, beanarbitrary point, anddevelop L'about thispointbyTaylor's Theorem withaRemainder. Wehave : (18) V=L+2)Lir(q'r-qr)+E(q r,&gr,0, r where L,Lqrareformed forthearguments (qriqr,t),and (19) E(q r,Qr, <jr,=i2liri,(q'r-qrM-?.) r,s thecoefficient I^rQ8being thevalue ofL<jr^foramean value of thearguments qr,namely, qr+0(q' rqr),where <8<1. Thequadratic form(3)ispositivedefinite. Hence 384 MECHANICS (20) 0<E(q r,q'r,qr,t) if(q'i, >m)isdistinct from(ft, ,gm). Proof oftheMinimum Property. Consider anarbitrary extremal (OQthrough thepointP :(qr,t<>),represented by(14): : Qr=qr(t),r=1,- ,m. LetP!:(qrl ,t\)beasecond point on<~near by.Connect P andP!byanarbitrary curve C: qr=qr(t),r=1, ,m, and letq'r(t)=dqr/dt.ThecurveCshall, however, beaweak variation, l?r(0-tfr| <>7, Let L=L(qr,q'r,t). Let(<?r, beanarbitrary point onC.Through thispoint there passes anm-parameter familyofextremals, (13) or(14). Weselect oneofthem asfollows. Letalt ,amretain the values theyhave for<~;but letft, ,ftnhavenew values, namely, those given bythesecond oftheequations (13),when qr=qr)i=t.The corresponding value ofqrwillbegiven by(7). Itisthevalue found,forthear,$rinquestion, bydif- ferentiating thefirst oftheequations (14)with respect tot.These values of r,qnqr,tsatisfy Equations (15)and (16); theprdo notenter explicitlyinthese equations, andsothefactthatthey depend on tdoesnotcomplicate theequations. Wenowapply Equation (18), setting qr=qr,giving toqr thevalue justfound, and letting q'rrefer toC,<fr=q'r.Thus (21)L=L(q r,qr,t)+%L^r(qrjqryt)(q' r-qr)+E(q r,q'r,qr,0- r The firsttwoterms ontheright of(21)canbemodified as follows.First, (22)^^=S,r(qr,f)q'r+St(qr,t). Next, from (16): (23)2s ,W" 9r+S,($-L(gr,(/=0. r HAMILTON'S PRINCIPLE. LEAST ACTION 385 Subtracting (23)from (22)wehave : ^jj^=2)S,r(qr,t)(q' r-qj+L(q r,qr>f). Finally, sincefrom (15) thefirsttwoterms ontheright of(21)have thevalue dS(q r,t)/dt, and(21)canbewritten : (24) L= -+E(q r,q'nqr,f). Wenowproceed tointegrate thisequation from ttotvOb- serve that to hasprecisely thevalue oftheintegral: ti CL<u, taken along thenatural path ofthesystem. For,along <~Equa- tion (16)saysthat r>Q~ andso ti t. /Lett=5fa r,0 to But intheendpoints, qr(t)=gr(0-Wethus arrive atthefinal result : (25)i i CLdt=/Ldt+CE(q r,$,qr,t)dt. to to o If,then,Cdiffers from cF,there willbepoints ofCatwhich E>0,andsotheintegral ofLoverC(i.e.theintegral onthe left)willbegreater than theintegral ofLover ("(i.e.the first integral ontheright) andourtheorem isproved. 386 MECHANICS 13.Jacobi's Integral aMinimum inaRestricted Region. In Jacobi's Integral: (1) thefunctions Tand'Udonotcontain texplicitly, andTishomo- geneousintheqr: (2) T= Thevaried functions, qr(u, ),arearbitrary, subject merely to thecondition that dqr ineach end-point,t t,t^Itis obvious that theintegral (1)liasthesame value astheintegral: rt (3)JTdt, subject totherestriction : (4) T=U+h. This condition shall hold forthevaried paths, too.Thus qr(u,c) isstill arbitrary; but t(u, c)isdetermined by(4).Toprove, then, that theintegral (1)isaminimum forthenatural path,it issufficient toshow that theintegral (3)has thisproperty,if (4)holds forthevaried paths. Inthepresent case, (5) L=T+U. (6) H=T-U. From(4)and(5), (7) L=2T-h. Let (nQbethepath defined in 12,and letC' : (8) r=7r(w, 0> ?=<(",), beavaried path. Consider thevaried integral. From (7) t\ *i (9) /2?dl=fldt+h(t l-*), HAMILTON'S PRINCIPLE. LEAST ACTION 387 where Theright-hand side ofEquation (9)canbecomputed asfollows. From theanalysis used in12,Equation (24),weseethat Li'du Hence Ldt=S(q r,I)"'+CEdl <b Since qr=qrforu=w,MJ,the firsttermontheright hasthe value : StorSZD-StorVo)- Hence **i (10)/2Tdf=Sfer1 ,?i)-Sfar ,o)+h(t,-t)+CEdl to to Since //isindependentoft: itfollows asinChap. XIV, 4,thatafunction Softheform : S=-ht+W(q l9-,?*, A, 2, -,a*) canbefound, where histobeidentified with avUsing this function Sin(10),wehave : (11) ArdT =I^to,1 )-Fto r)+Csdl <0 <0 Ifweallow C"tocoincide with <",then J?^0,and t\ J*2T<lt=W(q r1 )-W(q r). 388 MECHANICS Thus (11)becomes: C2Tdt =C2Tdt tot This provesthetheorem. For,ifC"isdistinct fromco'o,then J5,which isnever negative,willbepositive forsome partsofthe interval ofintegration, andhence theintegral (3),extended over C',willexceed invalue thesame integral extended over the natural path,aswastobeproved. The caseU=const, leads tothegeodesies onamanifold for which thedifferential ofarc isgivenbytheequation: ds2=5}Arsdqtdqy. r,s Thiiswehave aproof thatageodesic onamanifold obtained ar above istheshortest lineconnecting twopoints which arenottoo farapart. CHAPTER XIV CONTACT TRANSFORMATIONS 1.Purpose oftheChapter.* The finalproblem before usis theintegration ofHamilton's Canonical Equations: .. dqr8H dpr8H,A)df= Wr' dT= -W,'r=l'-'m - Themethod consists infinding alarge andimportant class of transformations ofthevariables (qr,pryt)intonew variables (q'r,p'r,t'),such thatEquations A)arecarried over intoanew system oflikeform : A/, dq'r3H' dpr 3H'tA) W=W dr=~Wr-1'-'w' or,aswesay,transformations with respect towhich Hamilton's Equations remain invariant. Themost general class ofsuch transformations weshall con- sider, aretheso-called Canonical Transformations. Aone-to-one transformation : 01,''' ,tfm,Pi,''' ,Pm, issaidtobecanonical ifthere existtwofunctions, H(<li,'' ,7m,Pi, ,Pm, andH'(q(,- ,q'm,p(, ,Pm, t') (notingeneral equal toeach other) such that *This introductory paragraph isdesigned togiveanoutline ofthetreatment contained inthefollowing chapter. Thestudent should read itcarefully, not, however, expecting tocomprehend itsfullmeaning, butrather regarding itasa guide, towhich, inhisstudy ofthedetailed developments, hewillturnback time andagain forpurposes oforientation. 389 390 MECHANICS (1)/(2V'M- H'df)=J(2Prdqr- Hdt), 1" 1' where Fisanarbitrary closed curve ofthe(2m+l)-dimensional (gv, ?>r,0-sPa(' >ail(lV'*sitsimage inthetransformed(#J,pj,<')- space, these spaces being thought ofassimply connected. Toacanonical transformation there corresponds afunction ''' i<7m,Pi,- ,Pm, such that (2)2?'M*H'dt'=^Prdqr-Hdt And conversely, when three functions H',H,^exist, forwhich thelatter relation istrue, thetransformation iscanonical. Contact Transformations. Animportant sub-set ofthese ca- nonical transformations consists inthose forwhich thelastEqua- tion I.is (3) f=t. Onequating thecoefficients ofdtonthetwosides ofEquation (2) wefind : ' Since /'= /,wemaysaythatthevariable tisnottransformed, andtreat itasaparameter. EquationsI.thustakeontheform : with<?'=q'rfal,'''iQmiPi,' ' ,Pm, t) Pr=Pr(q\y''' ,Qm,Pi,' ,Pm, t) ^(<7?> <?m,P'},'' 9Pm)_^ d(9l9'' ' ,(7m,Pi,''' ,Pm) Andnowcomes animportant modification ofEquation (2). Sincewenow areregarding the(qrjpr),andnotthe(qr,pr,t), astheindependent variables, (2)canbewritten bytheaidof (4)intheform : (5) %(p'dq'r-prdqr)=d*. r Ofcourse, d$hasdifferent meanings in(2)and(5). In(2), ^\ ^T,(6)l CONTACT TRANSFORMATIONS 391 since heretheindependent variables areqr,pr,t,whereas in(5), /*\ JT (7) * since heretheindependent variables aregr,pr;asimilar remark applying totheother differentials, dq'r.This isnotanexception, orcontradiction, inprinciple, butonlyinpractice, since the differential ofanyfunction, *(i, ,xn),depends ontheinde- pendent variables : __^<Nf and itisnotuntilwehave saidwhat these shall be i.e.defined ourfunction thatwecanspeak ofitsdifferential. Atransformation wewillhenceforth change thenotation frommton : (, fQr=?'(<?!,''' ,9n,Pi,''' ,Pn) IPr=Pifei,''' ,tfn,Pi,''' ,Pn) 3(ft',''' ,0i> Pl'''' ,Pn) ,n #ft>''' ,0n,Pi,''' ,P such that (9)Jpfdtf= jprdqr, r' r where Fisanarbitrary closed curve ofthe(QV,pr)-space, thought ofassimply connected, andF'is.thecurve intowhich itistrans- formed, shall becalled acontact transformation. There cor- responds tosuch atransformation afunction ^(q lt ,qnj Pi>' ' >Pn)forwhich (10) 2,(prdq'r-prdq r)=d*. r And conversely, atransformation (8)forwhich (10)istrue satis- fies(9)andsoisacontact transformation. Acontact transformation may, ofcourse, depend oncertain parameters, p',q'rand thusbecoming functions ofthese para- meters aswell. Thetransformation II.above isacase inpoint. Finally, thecanonical transformations form agroup;i.e.the result ofapplyingfirstoneandthen asecond such transfor- mation may itself beexpressed asacanonical transformation. 392 MECHANICS Thecontact transformations alsoform agroup. Thegroupof contact transformations II.isasubgroup ofthegroup ofcanonical transformations I. Theapproach tothecontact transformations isthrough the Integral Invariants ofPoincarg. Thecontact transformations, asdefined generally by(8)and (9),areofespecial importanceinMechanics because anysuch transformation carries anarbitrary system A)ofHamiltonian Equations over intoasecond such system, A') ;cf.infra,4. We shall treat theapplication ofthese transformations tothe integration ofHamilton's Equations atlength inChapter XV. Ifthestudent iswilling totake thisoneproperty ofcontact transformations forgranted, hecanturn atonce toChapter XV, andhewill findnoother assumptions needed inthestudyof thajtchapter. 2.Integral Invariants. Consider theaction integral: *i (1)j*L(qr,q'n{)dt, to andtheextremals, which arethepath curves, given byLagrange's Equations: /o\^**L^_n 1 (2)diWrWr~' r-V-.,n, whereListheLagrangean function, orthekinetic potential. Thegeneral solution canbewritten intheform : (3) qr=qr(t;qf, ,qJ,qf, ,tfn), r=1, ,n, where gr,qrarethe initial values ofqr,qr,i.e.their values when t=fo.Inthe(2n+l)-dimensional space ofthevariables (<7u*'* 9q*> <ii>*'' >q*9 these equations, together with then further equations: (4) <?r=<?r('; 1,,?, ?1, ,<7n)> represent acurve C, ormore properly, a2n-parameter family ofcurves C.Letaclosed curve, F : (5) ?r=?r(X), tfr=gr(X),r=l, ,n,XSXgX,, bedrawn intheplanet=.ThecurvesCwhich passthrough thepoints ofFformatubeofsolutions, whichwewilldenote byS. CONTACT TRANSFORMATIONS 393 Lettheaction integral, (1),beextended along thecurves C which form S.Itsvalue isafunction ofX : (6) where qr,qraregiven by(3) and (4)and qr,qrby (5). Differentiate /(X):FIG.147 -C"J Onintegrating byparts, observing that wehave : Henceddqr di~d\' C'^^LM =3L<^- C^^^SLM Jd</r3\d(jr3\Jdtdq r3\ t\ '(\\-CV(^Jd3L\fyr,, |ydL8q r ()~J$\Wr dtWr' 8\CU+ -f2qrd\ The integral vanishes, because qrisbyhypothesis asolution of(2). Wenowmake thetransformation, Chapter XI, 3: (7) Thus (8)dL Since Fisaclosed curve, qr(\)=^r( and (6)gives: (9) Hence X0, 394 MECHANICS andsofrom(8): *i / k P/y <l eJX=0. Let tbethought ofasconstant, but tl9which isalsoarbitrary, asvariable;denote thelatter byt.Then This equation represents thetheorem inwhich thewhole in- vestigation ofthisparagraph culminates. Insubstance itcan bestated asfollows. Wemay regard Equation (7),along with then+1further identical equations, qr=qr,t= t,asrepre- senting atransformation ofthe(qr,qr,)-space onthe(qr,prj0" space. Observe thattheJacobian - ?j Chapter XI, 3.Thus thecurves Cofthe firstspace goover intocurves C"ofthesecond space, andFgoesover intoacurve TO,Sbeing transformed intoatube S'. Letusnowmake thesecond space thespace ofthevari- ables(<?r,pr,t)ourpoint ofdeparture and,dropping theprimes, consider aclosed curve intheplanet=tQofthatspace: Consider furthermore curves Cthrough itspoints, which are obtained bytransforming thecurvesCoftheearlier space. The integrals (11)nowbecome line integrals inthepresent space. Ifwechange thenotation, setting (13) qrQ= r, Pr=Pr, then (11)assumes theform : (14)/5JPrdq r= /^0rdar, r *J r where Fisthecurve ofintersection ofthearbitrary plane/=t with thetubeSdetermined byT,anarbitrary closed curve of CONTACT TRANSFORMATIONS 395 theplanet=tQ.But this isprecisely thedefinition ofacontact transformation,tbeing thought ofasaparameter:* , . Pr(i,, ,ft, ' in,ft, ,fti, 3.Consequences oftheTheorem, a)Hamilton's Canonical Equations. Lagrange's Equations (2), 2form asystem ofn simultaneous total differential equations ofthesecond order. Bymeans ofthetransformation (7)these arecarried over into asimultaneous system of2ntotal differential equations ofthe firstorder inthe (qr,prj0-space. Letthese bewritten inthe form : (16)=Qrfop,0, Pr(?,P,0, Since theright-hand sideof(14)isindependent oft,thederiva- tiveoftheleft-hand sidewith respect totmust vanish. Hence' or*1 dC^-\ dqrj\_(\ dij2,P'^dX-u' ?1Ta\r+praTax)dx=' Integrate byparts: d\npr SinceTisaclosed curve, =0, *The geometric picture ishere slightly different from the earlier one, since thevariables (ari/3r)and (qrtpr)areinterpreted indifferent planes. Butofcourse onemay think ofacylinder onTasdirectrix, with itselements parallel tothe t-axis. Oncutting thiscylinder with theplanet=t,wehave acurveFlying in thesame plane with F.Or,tolook atthesituation from another angle,tisonly aparameter, and itisthespaces of(ar,Pr)and (qr,pr)which concern us. 396 MECHANICS andwehave : Here, dqr__dqr_n tyr __dpr_p 8t~ dt~Wr>dt~ dt" Thus Equation (17)maybewritten intheform : (Prd r-Q rdr)=0. Butrmaybeanyclosed curve oftheplanet= /,since toany suchcurve inthatplane corresponds aFintheplanet=t. Itfollows, then, thatwecandefine afunction //bymoans of theintegral: , (19) H=f-Prdqr ' where thefixed point (alt ,an,blf ,&, )oftheplanet=t isconnected with thevariable point (qly ,qniPi, ,pn, ofthissame plane byacurve lying intheplane. Because of (18) thevalue oftheintegral does notdepend onthepath, andthus //isdefined asafunction of(qr,pr)fortheparticular value of t. Letthepoint (a,&,t),fordofinitcness,lieontheextremal through thepoint (a, 0',2).ThenHbecomes afunction of(qrjpr,0- If(c/, 0',J)isreplaced byadifferent point (a", /3",tQ),the newHwill differ from theoldHbyanadditive termwhich isa function oft,butnotof(qr,pr). Moregenerally,let//bedefined bytheequation: (20) II=li+f(i), whereHisaspecific one ofthefunctions 77just defined, and f(t)isanarbitrary function of/alone. From(19)itfollows that dn CONTACT TRANSFORMATIONS 397 Thus thesystem ofequations (16)isseen tohave theform : , , dgr_SH dpr_8H (22)Hf-Wr' ~dt~~Wr'r=l,. ..,n. First fruits ofourtheorem. TheHamiltonian Function H grows naturally outofEquation (14) ;for(18)isbutanother form of(14),and (18) atonce suggests thedefinition of//by (J9)and(20). Thus ifwehadnever heard ofHthrough thetrans- formations ofChapter XI,weshould stillbeledtoitbythe theorem ofthisparagraph. TheFunction Vand ItsRelation toH.Equation (14)canbe written intheform : f Tolr- Prdctr) =0, where gr,praregiven by(15), thecurve Fbeing asbefore any closed curve intheplanet=t .Itfollows, then, thattheintegral: (<*,$) (23) extended overanarbitrary pathintheplanet=tjoining the points (a', jft'), (a, /?),isindependentofthepath and thus defines afunction of(a,0),tentering asaparameter: (J5) (24) g(Prdqr-Prdctr)=V(<*,ftt). f ('>)' Differentiate thisequation with respect to/: dqr\__dV'~ where theitalicdmeans differentiation along acurve(15). Trans- forming through integration bypartswehave : /ocx rj r, dv (25)r _ 398 MECHANICS Theintegral^onthe left isprecisely thenegative oftheintegral (19), orH(q r,pr,f).Hence (26) H=2PrQr~^, whereHisgivenby(20),and Ontheother hand, theLagrangean Function L(qr,qr,t)is connected withH(qr,pr,t)bytherelation (cf.Chapter XI, 3): (27) L+H=]Tprqr. r Hence itappears that <*>-- Just asHwasdefined onlysave astoanadditive function oft, soVcanbemodified byadding anyfunction tyandthesame istrue ofL.But itisconvenient torestrict these additive func- tions sothat (26)and(27)willhold. From theforegoing reasoning wecandraw amore general con- clusion, andthensupplementitwithaconverse. THEOREM I.Let r=l,---, n, beanarbitrary system ofsimultaneous differential equations tand let qr=<pr(t;!,--,orn,ft, ,j9) bethesolution, where ar,Prmean theinitial values ofqr,prcor- respondingtot t .LetTQbeanarbitrary closed curve lying intheplanet=tQofthe(2n+l)-dimensional space ofthevari- ables (qr,pr,t).LetSbeatube consisting ofthecurves ii)which pass through points ofF;and letTbethesection ofSbytheplane-. // Prdqr \5) CONTACT TRANSFORMATIONS 399 isanintegral invariant ofEquations i);i.e.if iv)I^Prdqr=I^LfPrdotr, r' r thenEquations i)formaHamiltonian System: . dqrdH dprdH, i-/*_=.r .= r=1 ti"'dt 8pr'dt 8qr> Tlf>n' Conversely, ifEquations i)form aHamiltonian System y),then Hi)willbeanintegral invariant, oriv)milbesatisfied. Observe, however, thatTheorem I.ismore general than its origin from theaction integral (1)andthetransformation (7) would indicate. Itapplies toanyfunctions Qr,Prforwhich Hi)isanintegral invariant;or,intheconverse, toanyfunction H,provided thatthedeterminant 02TIHu".H nn*0,Hi3= - ButasystemofEquations v)may conceivably notlead toa mechanical problem whyshould it? b)Contact Transformations. The content ofTheorem Lcan berestated interms ofcontact transformations. THEOREM II.Let qr 9r(aD''' ,<*n,ft,'*' ,ft,t) a) hr(ot lt''' ,n,ft,''' ,fti, where r=1, ,n, Pi, 0^, -,., ft,--^ftr' a?id far=gfrC^, ,an,ft, ,ft,<) I^r=Ar(a!, ,an,ft,' ,ft,O feeatransformation ofthe2n-dimensional(<xr,pr)-space onthe (qr,pr)-space; and let 400 MECHANICS bethesystem ofdifferential equations correspondingtoa);i.e. defined bya).//a)isacontact transformation;i.e.if I^Prdqr= j^Prdar, or 2J(prdqr-Prdoir)=dV(a, /3,/), r thenb)z'saHamiltonian System: C'~dt^ ~dp~r'~dt='"~ ~dqr' r==>'">n> andconversely. 4.Transformation ofHamilton's Equations byContact Trans- formations. Ifwestart outwithagiven systemofHamiltonian Equations: ,, dqr_dH dpr dH_ __1()~dt~ 8p~r'~dt" ~dq~r' r-I,---,n, andmake anarbitrary transformation : /0, f9r=/rfe,''' ,^n,Pi,' ,Pn,W 1 , , A IPr=Vr(q\y''' ,?n,Pi,''' ,Pn, thetransformed equations: (3) r=1, ,n,willnotingeneral beoftheform(1) ;i.e.they willnothave theform : where /f'=H'(q'T,p'rt)issome function ofthearguments q'T,p'rL Asufficient condition that (3)beHamiltonian,i.e.oftheform (4),isthat (2)beacontact transformation. Theproofisbased onTheoremII., 3andthefactthat the contact transformations formagroup. Let(2),then, beacontact transformation. Denote itbyT.Let(aj,#)betheinitial val- CONTACT TRANSFORMATIONS 401 uesof(?', p'r)for t=tQ.They arisefrom (ar,r)byT,formed for *=*oJ^o>letuswrite it.Thus, symbolically, (*;,#)=T(ar,r), or(r,r)=TQ-*(cl, ft). Again, wemay write symbolically: (#,PJ)=T(gr,pr). Finally, consider thesolution of(1),whereby thespace ofthe (ar,pr)iscarried over into thespace ofthe(gr,pr).This transformation istheTransformation a)ofTheorem II.,3,and sobecause of(1)isacontact transformation. Denote itbyD : D(r,r)=(qr,pr). Ontheother hand, theeffect ofthetransformation defined by thedifferential equations (3)istocarry thespace ofthe (a'Tyft) over intothespace ofthe(q'r,pr).Denote itbyA : Andnowweseethat this result thistransformation Acan beobtained asfollows :Perform firstthecontact transformation* TQIonthe(aj,$)-space, thusobtaining the(ar,/3r)-space: (ar,0J=57(;,#) Next, perform thecontact transformation Donthe (ar,fir)- space, thusobtaining the(qr,pr)-space: Finally, perform thecontact transformation Tonthelatter space, thusobtaining the(q'r,p'J fer',Pr)=T(qr,pr) Wehave inthiswayobtained Aastheresult ofthree contact transformations : A=TDT?. Hence Aisitself acontact transformation, andsothesystem (3) isHamiltonian, byTheoremII.,3. This istheresult onwhich thedevelopments ofChapter XV depend. Itmaybestated asfollows. *Theinverse ofacontact transformation isobviously itself acontact transfor- mation. 402 MECHANICS THEOREM. //asystem ofHamiltonian Equations (1)betrans- formed byacontact transformation (2),theresult isaHamiltonian system (4). Thecondition issufficient, butnotnecessary. Computation ofH'.The original system ofHamiltonian Equations (1)leads tothecontact transformationZ),forwhich therelation : (5) ^prdqr-2^dar=dV(ari r,t), r r ischaracteristic, where JT"V /nA__ dt(6) 77=2Prtfr-~ Thetransformed Hamiltonian Equations (4)lead likewise to acontact transformation D'=A,forwhich therelation (7) 2P'M-2#da'=dV>('. '> r r ischaracteristic, where jjr/ Let r r bethecharacteristic relation ofthecontact transformation T. Then (10) 2#da'"S^dc*r=dW(<*"&r>*o) r r willbethecharacteristic relation corresponding toT . Each ofthedifferentials ontheright istaken onthesupposi- tionthat tisaparameter, andsoaconstant. Moreover, (qr>pr) aregiven interms of(ar,Pr)byequations ofthetype a),3. From(5), (9),(10)weinfer that r r d[- W(ct r,ftr,<)+V(a r,0r, H CONTACT TRANSFORMATIONS 403 Hence - dt~ dt dt' providedV(ar,$,f)andW(q r,pr,0,which arcdetermined only save astoadditive functions oft,arechosen properly. From (6)and(8)wenowinfer,bymeans of(12), that (13) H'=H-+ (p'^-prqr). Each ofthefunctions H',H,dW/dt was originally defined only save astoanadditive function oft,and itisonlywhen these additive functions aresuitably restricted, that (13) holds. 6.Particular Contact Transformations. Inapplying thetheory wehave developeditwillbeconvenient todenote thetrans- formed variables byQrjPrinstead ofbyq'np'r.Thus atrans- formation : fQr=/rfoi, ,<7,Pi, ,p, IPr= <7r(<7i,- ' ,Qn,Piy' ' ,Pn, where ' ,(jft,1l)'*'?*n) ^Q '" *'' isacontact transformationif (3) 2(P rdQr-Pr.rf7r)=^^(<7r,Pr,0, r where isregarded asaparameter andthedifferentials aretaken with respect to(qr,pr)astheindependent variables. Ifsuchatransformation beappliedtotheHamiltonian system: (4\ dqr__<M ^Pr__^ff -_!... Wdt dpr'dt dqr' lj 'n' these equations goover intoanewHamiltonian system: dQ,_ff' dPr__W _j ..._W <tt~ P,' rf<~ 3Q,'' ' ' where H'(Qr,Pr) isconnected withH(qr,pr,t)byEquation (13), 4,or: (6) H'=H-+(PrQr-prqr). 404 MECHANICS The(qr,Qr,t)asIndependent Variables. Equations (1)repre- sent2nrelations between the4nvariables(gr,prjQr,Pr),and when (qr,pr)arechosen astheindependent variables, (2)and (3)hold. Itmaybepossible tochoose the2nvariables(qrjQr) astheindependent variables,talways being regarded asapara- meter in(3). Write (7) W(q r,pr,t)=W'(q r,Qr,t). Thus (3)becomes : (8) 5(PrdQr-PrdQr)=dW (qr,Qr,t). Onequating thecoefficients ofdQr,dq,in(8)wefind: Pr= (9) Equation (6)cannowbetransformed asfollows :Since dWdW dW'dQr SW'dq r8W' dt dt 'dQ rdt?8qrdt dt' wehave : Hence (6)becomes : (10) ff'-ff- OO OO TheTransformation: pr=-^-,Pr=^T'Wecanwrite Oqr Glflr down aparticular contact transformation, inwhich (qr,Qr)can betaken astheindependentvariables. Let ' ,qn,!,- ,anyt) beafunction ofthe2n+1arguments suchthat CONTACT TRANSFORMATIONS 405 Setar=Qrandmake thetransformation : /irk\ &Sr> &Si (12) Pr= Wr,Pr=~Wr, r=l,...,n. The firstnofthese equations canbesolved fortheQrinterms ofthe (qr,pr)because of(11),andthen thePraregivenbythe lastnequations. Thus atransformation (1)results, theJacobian (2)notvanishing.* Thetransformation willbeacontact transformation, for (PrdQ r~Prdqr)=~ (||dQr+Jdqr)=~d3, andwemay setW=S,since Tfandhence FT'isdetermined only save astoanadditive function of t.Equation (10)now becomes : (13) H'=H+- How suchafunction Scanbefound, which willenable usto solve Hamilton's equations explicitly, willbeshown inChapter XV. Conversely, themost general contact transformation (1)which canbewritten intheform : isgivenby(12). For,Equations (9)must betrue,and itremains only tosetSW. Itisseenatonce thattheWrof(9)must satisfy (11), since otherwise there would bearelation between thePr. *Theproof isasfollows. If Vr=fr(Xl, -,Xn), T=1, ,U, beatransformation having aninverse Xr=V?r(l/Ii' ',2/n), f=1, ,n, where fr,<f>rare allfunctions having continuous firstderivatives, then d(y\,--,yn).d(x\, ,xn)_j d(xi,--,*) d(yi,--,yn) Consequently neither Jacobian canvanish. Inthepresent case, theqr,Prcanbeexpressed interms oftheQr,Pr,since the value ofthedeterminant (11) isunchanged iftheqr,arareinterchanged. 406 MECHANICS EXERCISES OO ^Cf 1.TheTransformation: pr=T,Qr=^p- Study the oqr Ofr analogous case,inwhich (qr,Pr)canbetaken astheindepen- dent variables,tbeing, asusual, aparameter. Show that,if (#u' >Qn> <*i>'** ><*n, bechosen asbefore, and ifweset Pr=ar,then /-.^\ v*S s^ O& - (14) p,=w,Qr=W, r=l,..-,n, will giveacontact transformation. Observe that(3)canbe transformed bymeans oftheidentity d(P rQr)=PrdQr+QrdPr, sothat ittakes ontheequivalent form : rdPr+prdq r)=d(~W"+PrQr). Choose W(qr,pr,=W"(q r,I\, t),therefore,-sothat S=-W" Compute dW"/dt andshowbytheaidof(14)that (6)yields: (15)ff'=//+. State also,andprove, theconverse. 2.Computation of//'intheGeneral Case. LetTT^ ,7T2ri beanysetof2nvariables, chosen from the4nvariables far,PryQr,Pr), interms ofwhich theremaining 2nvariables canbeexpressed. Show that 06)-, where qr,Qr,andWareexpressed asfunctions of(TT*,0- /)^f OC| 3.TheTransformation:qr= -5,Pr=-^r-If(pr,Qr)can Gpr #Vr betaken astheindependent variables, and ifweset S=W+ Pr?r, CONTACT TRANSFORMATIONS 407 where qr,W,andSarenow functions of(pr,Qr,t),then the transformation takes theform : (17) and (6)yields: (18)= _^dPr'' H'=H- Conversely,ifS(q lt ,qn,alt ,an)bechosen asbefore, and ifwesetQr=QLr,then(17)willdefine acontact transformation. 6.The fi-Relations. There isonecase ofimportancestill tobeconsidered, namely, that inwhichWisafunction of (qr,Qr, t),butthe (qrjQr,t)cannot bechosen astheindepend- entvariables. Theextreme casewould bethat inwhich Qr=Wr T=n, Thogeneral case isthat inwhich <m^nindependent rela- tionsbetween the(qr,Qr,t)exist, andnomore : where therank ofthematrix : X^ Xli (2) Wn ism.ThusmoftheQ*'scanbeexpressed asfunctions ofthe remaining ju=nmQ/sandq^ ,qn,t.Asamatter of notation lettheabovemQfc'sbeQD ,Qm: /O\/") ___/'f\ f\n ff /\ __1/yyj Then thedeterminant whose matrix consists ofthefirstmcolumns of(2)willnotvanish. Among the2n (pr,Pr)itshallbepossible tochoosemvariables, TTJ, ,wmsuch that(irlt ,7rm,Qm+i, *'' >Qn, <7i, ,^n, canserve asthe2n+1independent 408 MECHANICS variables. But thefunction W(q r,pr,t),when expressedin terms ofthenew variables, doesnotdepend onirly ,irm: (4) W(q f,Pr,t)=IF*for,Or, fl. Itisnot,ofcourse, unique, because oftheQ-relations, (1). Equation (3), 5nowtakes ontheform : (5) 2)(PrdQr-prcfyr)=dW* (qr,Qr,0. Wewillrewrite itintheform : I(^-CK -?(* Itisnot,however, ingeneral true that thecoefficients ofthe differentials vanish. Bymeans ofthemequations (1)the firstmdifferentials dQu ,dQmcanbeeliminated, theresulting equation being oftheform : (7)Xm+ldQm+t++XndQn+Y.dq,+--+Yndqn=0. The differentials in(7)areindependent variables, andsowecan infer that Xm+l=0,-. ,Xn=0, Y,-0, ,Yn=0. The actual elimination canbeconveniently performed by means ofLagrange's multipliers. From Equations (1)weinfer that n n=o (8) Multiply the fc-th ofthese equations byX&andaddto(6).Then determine the X/t'ssothatthecoefficients ofdQ }, ,dQmvanish. The resulting equationisoftheform(7),andsoitscoefficients vanish automatically. Wethus arrive atthe2nequations: (9)_ '~Wr+ 'Wr r= l...m CONTACT TRANSFORMATIONS 409 The firstmofthese equations determine the\k's.Theremainder aresatisfied asshown above. The result issymmetric andholds, nomatter what setofmQk'sisdetermined by(1) ;i.e.nomatter what ?n-rowed determinant out ofthematrix (2) isdifferent from 0. Itisnoweasy todetermine Hfbymeans of(13), 4: Onreplacing Pr,prherebytheir values from(9)andobserving that dW*= dt dQrdt,dQrd{ dt> dttr^dttrdQr.^d&r^r,^r^ dt"" 2?3Qrdt^^dqrdt^ dt' wefindthefollowing result : -- -J If,inparticular, the 12'sdonotcontain texplicitly, thisequa- tionreduces to (11) H'=H CHAPTER XV SOLUTION OFHAMILTON'S EQUATIONS 1.TheProblem and ItsTreatment. Wehave considered a great variety ofproblems inmechanics, thesolution ofwhich depends, orcanbemade todepend, onHamilton's Canonical Equations:m dqr-dH dpr--mr-1 ..-n(i)~dt~~Wr'~dt~ Wr' ' ' whereHisafunction of(qr,pr,).Theobject ofthischapteris tosolve these equations explicitly intheimportant cases which arise inpractice. Themethod isthat oftransformation. Bymeans ofasuitably chosen transformation : Qr=Fr(q19 ,qn,Pi, ,P, Pr=Gr(qly'-- ,qn,Pi,'' ,Pn, Equations (1)arecarried over intoequations ofthesame type: dt~ d/V dt' butmore easily solved. Here, H'isafunction of(QT,Pr,<)>11(>t ingeneral equal toH . Thedetermination ofaconvenient transformation (2)depends onapartial differential equation ofthefirstorder, duetoJaeobi* : (A\ dV_i_(4)--+ Itisnotthetheory ofthisequation, however, butthepractice, thatconcerns us,forallweneed isasingle explicit solution, (5) V=V(q lf ,g, i, ,n,0, depending inasuitable manner onnarbitrary constants, or parameters, !,-, n.Such asolution isfound inpractice by means ofsimple devices, notably that ofseparatingthevariables. *Hamilton cameupon thisequation;butitsuseashere setforth isduetoJaeobi. 110 SOLUTION OFHAMILTON'S EQUATIONS 411 The function (5)once found, thefurther work consists merely indifferentiation and thesolution ofequations defining the qr,primplicitly. Two cases areespecially important, namely: a)Reduction totheEquilibrium Problem. Here, asolution (5)of(4)enables ussotochoose (2)that thetransformed H vanishes identically,//'=0.Equations (3)cannowbeinte- grated atsight: where ar,($rarearbitrary constants. Onsubstituting these val- uesin(2),theinverse transformation, fqr=/r(Q!, -,, PI,,P,0 yields thedesired solution : qr=/r(i,''' ,n,ft,'' ifti, (8) Pr=0r(i,''' ,<*,ft,''' ,ft, Thetransformation(2)inthis case, aswillbeshown in2,is givenbytheequations: ,. _dVp_dV _t ^ Pr~Wrr~~Wr'r-l,..-,^ whereFiswritten forthearguments qr,Qr: Thus thesolution(8)isobtained bysolving theequations: dV dV pr= Wr'&T=~^r=l,..-,n, where thepresent Vhastheform(5). b)Constant Energy,H(qr,pr)h.Thesecond case isthat inwhichHdoesnotcontain thetime explicitly: H=H(q l9 ,g, Pi, ,pn). Itishere possible tofindatransformation(2)inwhichFr,G> donotdepend ont, Qr=Fr(qlf ,gw,plf ,pn) (10) ''' ,^n,Pi,' , 412 MECHANICS such that thenewHwilldepend onlyonthePrjbutnoton Qrjt.Inparticular, #'=P^ Equations (3)nowtakeontheform : =0,r=l,...,n. Thus* Qi=t+/3},Q,=ft, 5=2, Pr=ar, r=1, ,n. Lettheinverse of(10)bewritten : (12) Then thesolution of(1)isgivenbytheformula : (13) 1 ft,ft, Thetransformation(2)inthis case, aswillbeshown in4,is givenbytheequations: HA\ dW n (14) pfSSWr9Qr= whereWisasolution oftheequation: iffm dW^'-'^ or: Tf=W(q l9 ,?n, A, 2,,) Here TFdepends onthearbitrary constant A,andalso,inasuitable manner, onn 1further constants, orparameters, a2"*' t<** These aresetequal respectively tothePr: Pi=h; P,=aa,s=2, ,n. *Thechange ofnotation whereby thea/sandthe/Vsareinterchanged ismade forthepurpose ofconforming tousage intheliterature. SOLUTION OFHAMILTON'S EQUATIONS 413 Equations (14),combined with(11),thus yield: (15),w The lastn 1ofthese equations canbesolved forq^ ,qn interms ofqlyaswillbeshown in4,thus giving theformofthe path ;andthen qcanbefound from the firstequation (15)in terms of t. Wehave characterized thiscasebythecaption: "Constant Energy," but this isnotaphysical hypothesis. Ourhypothesis is,thatHdoesnotdepend explicitly on2,and this isallweneed forthemathematical development. ThatHthen turns outto beconstant along thecurves ofthenatural path,isanimpor- tant consequence; butourtreatment doesnotdepend onthis hypothesis. Contact Transformations. Thetransformations used ina)and b),namely, (9)and (14), areexamples ofcontact transformations. Atransformation (2)with non-vanishing Jacobian wasdefined inChapter XIV, 1,tobeacontact transformationif (16) 2}(PrdQr-Prdqr)=dW (qr,pr,t), r where thedifferentials aretaken with respect tothe(qr,pr)as theindependent variables,tbeing regarded asaparameter. Such atransformation alwayscarries aHamiltonian System (1)into aHamiltonian System (3).That thetransformations (9)and (14) satisfy thecondition (16)isseen atoncebysubstituting in(16),observinginthecase of(14)that d(PrQr)=PrdQr+QrdP r. This isallthetheory thestudent needknow fromChapter XIV, toenter onthestudy ofthepresent chapter, and thisamount oftheory was alldeveloped in 1-4ofthatchapter. 2.Reduction totheEquilibrium Problem. Wehave seen in Chapter XIV, 5,thatatransformation : 414 MECHANICS where S=8(q if -,fr, -, ) isanyfunction such that andwhereQrisset=ar,willcarry theHamiltonian System (1) ofthelastparagraph over intoaHamiltonian System (3),where (2) /r-ff +f. Thetransformed function H'canbemade tovanish identically ifwecanfindasolution Vofthepartialdifferential equation: which depends onnarbitrary constants, aly ,an: V=V(q lJ ,?,!,-, an,0> and issuch that ^.......vj^ 3(ll>) Onsetting Sequal tothisfunction F,andmaking thetrans- formation(1),H'asnowdetermined vanishes identically. Thus thetransformation : (5) p,=g,Prjfcr=l,...,n, where arisreplaced byQrinF,transforms theHamiltonian System (1)totheEquilibrium Problem: t-o- T'--'-1'-'"' The solution ofthese equationsisthesystemofequations (6), 1.These arethevalues ofQr,Prtobesubstituted inthe transformation (1) ;i.e.inthepresent case, in(5): (7) p,=g,0,=~g,r-l,... f. The lastnofthese equations canbesolved fortheqr'sbecause of(4),andthen thefirstnequations givethepr. SOLUTION OFHAMILTON'S EQUATIONS 415 Thereis,ofcourse, afurther requirement inthelarge, namely, thatthear,$rcanbesodetermined astocorrespond totheinitial conditions :t=tQ,qr=qrQ ,pr=Pr.Thus theequations: Pr=V r,- ,gn,!,, n,Q,r=1,- ,n, mustadmit asolution, ar=ar,andV(ft, ,qn,al9 ,e*n, mustsatisfyalltheconditions ofcontinuity, notably (4),inthe neighborhood ofthepoint (qr,ar)=(<7r,ar). EXERCISE Pass totheEquilibrium Problem bymeans ofthetransforma- tionstudied inChapter XIV, 5,Exercise 1: dS ndS1*'~WQr=Wr>"=V--,". Here, LetV=V(QI, ,qn,alf ,an,bethesame function as that ofthotext asolution ofEquation (3). If,then, we replace arbyPrandsetS=V,thetransformed H'willvanish : //'=0,andHamilton's Equationswilltakeontheform of theEquilibrium Problem : dQr dP rn , ~W= >~di-= 'r=l,...,n. Ifwewrite their solution intheform : Qr=-Pr,Pr=Qfr,T=1, ,tt, weareledtothesame solution ofthe original Hamiltonian Equations asbefore namely, thatgivenby(7). 3.Example. Simple Harmonic Motion. Here thekinetic en- orgyTandthowork function Uareexpressible respectivelyin theform : (1) T=^q\V-- \q*,0<X. Thus (2) L=T+V-?-f (3) 416 MECHANICS (4) ff-rt-L.-Lp. +lj.. Hamilton's Equations assume theform : aldq-p dp--\na;di~m' Tt~A9' Wepropose tosolvethem bythemethod of 2.Theequation fordetermining V, 2,(3),herebecomes : Wewish tofindafunction : (6) V which satisfies this equation.* Onesuch function isenough. Letussee ifwecannot findoneintheform : (8) F=fl+W, where 12=12(f)isafunction oftalone, andW=W(q)isafunc- tion ofqalone. Ifthisbepossible, weshallhave : '+V=o. Thisequation canbewritten intheform : X d!2 Theleft-hand side of(9)depends onqalone, theright-hand side, on talone. Hence each isaconstant denote itbya;itis obvious thata^: *Letthestudent disembarasa himself ofany fearsduetohisignorance oftho theory ofpartial differential equations. Nosuch theory isneeded inthekind of application inPhysics which weareabout toconsider;itwould notevenbehelp- fulinpractice. The single function V(q, a)isobtained byasimple device fully explained inthetext. There is,ofcourse, amost intimate relation between thetheory ofHamilton's Equations aridthetheory ofthis partial differential equation, asisindicated, for example, bythe"theory ofcharacteristics";cf.Appendix C.Thepoint is,that thistheory isnotemployed insuch applications asthose illustrated here. Forthe latter purpose, asingle solution V(q\, ,qntai, , , isallthat isrequired, andsuchasolution isobtained byingenious devices ofahomely kind, assetforth inthisChapter. SOLUTION OFHAMILTON'S EQUATIONS 417 J_/dTF\2+X2==a The firstequation gives: Q=aif noconstant ofintegration being added because weneed onlya particular integral, andsochoose thesimplest. From thesecond equation, (dW\2 -j)=2ma m\q2 . Onesolution ofthisequationis : W=IV2ma-m\q*dq. Thus / (10) V=-o* + / Equations (7), 2herebecome : dq (11)__^_ ^_ fl_-^--<- m- Xg2u This lastequation gives: andthus (12) q From the firstEquation (11), (13) p=V2m^ cos\-(-0). Equations (12)and(13) constitute asolution ofHamilton's Equations, which, however,isatpresent restricted;forwehave notpaidheed toCondition (7)ontheonehand or,ontheother, 418 MECHANICS considered that thesecond equation (11)isrestricted. Here then isadifficulty.* Either wemust follow thetheory ashitherto developed, using single-valued functionsWandV;then tiscon- fined between certain fixed values. Orelsewemust introduce multiple-valued functions F,andthenwemust goback and revise andsupplement thegeneral theory. Thereis,however, athird choice away out,whereby we canremain within therestrictions ofthepresent theory. Accord- ingtothattheory thesolution givenby(12), (13)isvalid solong as Now, from thegeneral theory ofdifferential equations, Equa- tions a)admit asolution single-valued andanalytic forthewhole range ofvalues oo<2<+oo. Equations (12), (13) yield asolution forapart ofthis interval. Therefore, byanalytic continuation, thesolution (12), (13)must hold forthewhole interval. EXERCISES 1.Obtain thesolution ofEquations a)intheform : (14) p=V2ma sin^~(t bychoosing asWthefunction : W=IV2m<x m\q2dq+C(a), o andsuitably determining theconstant ofintegration C(a). 2.Solve Equations a)directly, eliminating pandthus obtain- ingtheequation *There isalsoafurther difficulty, since the firstequation (11)maynotadmit asolution (suppose p<0),butthisdifficulty canbemetbychoosing thenegative radical, V;2ma SOLUTION OFHAMILTON'S EQUATIONS 419 thegeneral solution ofwhich canbewritten intheform : -fi, ^A. nit 3.TheSimple Pendulum. Letqbetheangle ofdisplacement from thedownward vertical. Then mml2 .9 Tr jT=-g2 ,U=mglcosq; Obtain theequation formotion near thepoint ofstable equilib- rium: fdq where tisrestricted. Hence discuss thetwo cases :a)oscil- latory motion (libration) ;b)quasi-periodic motion, when the pendulum describes continually completecircles (limitation). Observe that,when tpasses beyond therestricted interval, thesign oftheradical changes, andqchanges from increasing to decreasing, orviceversa. 4.Freely Falling Body, orvertical motion under gravity. Here, qshallbemeasured downward from theinitial position. p=mq, p*- 420 MECHANICS dW Since=V2ma+2m2 gq. _dVdW P dq dq' wehavenooption astowhich radical shallbetaken. Ifthebody isprojected upward, qwillbonegative forawhile, andsowemust choose thenegative radical forthisstage ofthemotion. Atthe turning point, (7)isnot fulfilled, since d2V/dqda doesnotexist. Wehavenowanewproblem, asthebody descends. The choice ofWmust bemade onthebasis ofthepositiveradical. Nevertheless, both stages ofthemotion arecovered bythesolu- tionforthe firststage: ft12^y /> ft o\2 "%/_____ //___/9i p=mg(t /3)V2om. Why? 4.H,Independent of /.Reduction totheForm, H'=Pi. Wehave seen inChap. XIV, 5,Ex.1,that ifSbeanarbitrary function oftheqr,ofnarbitrary constants, orparameters, the ar,andoft: where *'*' '"'*S)*o, and ifwesetar=Pr,then theequations: fn\ dS ~ dS ., (2) pr=^,Qr-W, r=l,...,n, define acontact transformation whereby Hamilton's Equations (4), 5,goover into(5), 5,and (3) H'=H+ft- IfSdoesnotdepend ont,thisequation reduces tothefollowing: (4) H'=//. Suppose, furthermore, thatHisalsoindependent oft: H=H(q lf-- ,qn,plt- - ,pn). SOLUTION OFHAMILTON'S EQUATIONS 421 Then H'=H'(Q l,".,Q*,P l,---,P n). Wepropose theproblem ofdetermining SsothatHrwilldepend onlyonthePr: H'=H'(P lt,Pn\ and, infact, that //'willbeanarbitrarily preassigned function ofthePr.Begin with thecase : (5) H'(P 19- ,Pn)=Px. Tofindsuchafunction S(q lt ,qn,(xlt ,an),consider theequation: ,Rv (6) Supposeitispossible tofindasolution : W=W(q ly- ,gn,A, 2,,) depending onn 1arbitrary constants2, ,anand of course onh,which isalsoarbitrary such that* (7)(7) Itthen follows, aswewillshow later, that ,2, ,n This isthefunction which wewillchoose asS : (9)S(q l,-- ,qn,a,, ,a)=TT^, ,tfn,A, 2, , ) where at=A. Ifnowweset : (10) Pj=ai=A; P.=a.,5=2, ,n, then (6)becomes, because of(2), (9),and (10): (H) //(ft, ,<?n,Pi, ,Pn)=PI, andhence (4)gives: #'=PI, aswasdesired. *Inpractice this isdonebywriting down anexplicit function ofthenature desired, obtained bysuch artifices astheseparation ofvariables. 422 MECHANICS Thus thetransformed Hamiltonian Equations become : dQ. -dt^1' (12)0,dt'dt r=o,r=1, ,n.2, Thesolution ofthissystemisobviously: Ci=<+ft, Q>=P; s=2,-,; Pr=ar,r=1, ,n.(13) Returning, then, totheoriginal transformation(2),which now takes ontheform : {TdPr' n, (14) pr= weliave : (15) The lastn Iofthese equations canbesolved forq2, ,qn asfunctions ofq1because of(7),thus determining theform of thecurves ofthenatural path ofthesystem. And then the firstequation canbesolved forqlinterms of t.This laststate- ment isconveniently substantiated indirectly. Allnequa- tions (15)canbesolved forql9-- ,qninterms oftbecause of(8). These functions qr(t)satisfy the lastn 1equations (15),and sotheearlier solution ofthese equations forq2, ,qninterms ofqlbecome identities intwhen qrisreplaced byqr(t)given by usingallnequations. Proof ofRelation(8). Observe thatRelation(6)isanidentity inthe h,aaaswell asintheqr.Hence ondifferentiating suc- cessively with respect toh,a2, ,an,wefind : +W^+ (16) ++ SOLUTION OFHAMILTON'S EQUATIONS 423 Thedeterminant ofthese equationsistheJacobian thatappears in(8).Ifitwere 0,itwould bepossible todetermine nmulti- pliers \i, ,Xn,not all0,such thatthenequations: (17)= = aretrue,andsince (7)holdsbyhypothesis, \maybechosen at pleasure. Now multiply the fc-thequation (16)byX*andadd. The coefficient ofeachHpvanishes, andsothewhole left-hand sidereduces to0.Buttheright-hand side isXuwhich isarbi- trary. This contradiction arises from supposing that(8)isnot true,andtheproofiscomplete. TheEquation ofEnergy. When thekinetic energy Tandthe work function Uarebothindependent oft,Hisalsoindependent oft,andHrepresents thetotalenergy (sum ofthekinetic energy T andthepotential energy U).Hence* //isconstant andwe may write : h=H(q i9-- ,qn,pl9 ,p). Thus thisequation appears tobederived from thephysics of theproblem. Itis.But thisderivation isnothelpful inthe present theory. Forwearedealing with contact transforma- tions which reduce Hamilton's equations toadesired form, and Equation (6)takes itssystematic placeinthat theory. Itex- presses acondition forthefunctionWthat willmake thedesired transformation possible. Nevertheless, thephysicsofthe situ- ation throws aside lightonthesituation, which itiswelltonote. TheSymmetric Form. Wehaveset,unsymmetrically, h=Pl inEquations (10).Wemight equally well replace (10)bythe equations: (100 *(Pi, ,Pn)=A, P*= .,=2, ,n, where <i>(alt ,an)isanyfunction such thatcfa/d^ 5^0.The above reasoning, withanobvious modification indetail, shows thatthedeterminant : *ThatHishereconstant along anatural path follows fromChap. XI, 3: dHdH Tt-IT-' 424 MECHANICS d(W q,- - ,WQ) (8') ~^-^*0,3(a lf<*2, ,a) whereW=W(q ly ,q* 9h,alt ,n)isdetermined asbefore from(6),andh=$(!, 2,- , ).Thus thetransforma- tion (14)isjustified andEquations (12)become : (120dQr dt - dt1, ,n. The solution ofthese equationsisobvious, andsymmetric. First,Pr=artr=1, ,n, where theararenarbitrary constants. Next, Qr=Urt+Pr,r=1, ,H, where o>r=$r(<*i, ,an),r=1,- ,n, andthef}rarenarbitrary constants. Thuswehave, finally: (19), . ft, awholly symmetric solution ofHamilton's Equations. Ifweshould wish touseafunction $(0^, ,an),forwhich some other derivative, asd$/da z,is^0,thenweshould need asolution W(q lt ,qn,alt- ,an)such that 5.Examples. Projectile invacuo. Letaparticle ofmass mbeacted onsolely bygravity, and letitbelaunched sothat it will riseforatime. Letqltq2,</3beitsCartesian coordinates, with qlvertical andpositive downward. Then T=&2+ft2+ </s2 ), I/ SOLUTION OFHAMILTON'S EQUATIONS 425 tt=^Cpl2+p*+P^ Theequation forWbecomes : Letustrytofindthedesired function, 0(a, fa,)' bysetting W.=W,+W2+W3, whereWr=Wr(qr)isafunction ofqronly. Thus n0."j i jVd^/ Vrf? 2/ Vrf Since itisonlyaparticular functionWthat isneeded, satisfy- ingtheJacobian Relation ofInequality, 4,(8),itwill suffice toset ^J2=2m(/l-a 22-a32)- Herc, hisdetermined bytheinitial conditions from theequationH=A,and2><*3areanv ^woparameters such that initially 2m(hc*22 32 )+2m2gql>0. Wenowmaychoose : JTTf *-v/O TTZ A/O OQ where 8ispositive, negative, orzero, subject merely totherela- tionofinequality. But, inthechoice ofWlfitisthenegative root, -a,*-a,*) thatmustbechosen, since __dWl Pl J 426 MECHANICS andp1<inthestageweareconsidering. Wemaytake 0i W1=-CV2m(h- 22-a32 )+2m Cl where cxistheinitial value ofqrThus, finally, Q\ W=-A/2m(A-<*22~ 3 c, Thecondition (7), 4,issatisfied. Wearenow inaposition towritedown thesolution ofthe problem. ItisgivenbyEquations (15), 4 : t+ff=^=- mf_ ^i dh JV2m(h- 22- 32 )+2m*gq l' A=|^=2ma af8. J along with theequations: The first oftheequations ineach ofthese sets ofthree isin substance identical with theonewhich governs the vertical motion ofafalling body, 3,Exercise4,where now a=h-<*22- 32 , ]8=-ft ; andhence : - - 22- 32 ) Intheearliercase, (7=initially, andsoctmnst beset=0. The lasttwoequations inthefirst setgive: ft) SOLUTION OFHAMILTON'S EQUATIONS 427 and so,finally: s=2,3. Themethod wehaveemployed gives thesolution oftheprob- lemsolong asthebodyisrising nolonger ;forwhen itis descending, pvbecomes positive, anddWl/dq 1=dW/dqt cannot beexpressed bythenegative radical. This second stage ofthe motion,inwhich thebodyisfalling, could bedealt withbyapply- ingthemethod afresh with suitable modifications inparticular, bytaking thepositive radical fordWt/dq^ But thisstepcan beeliminated ifweobserve thattheequations weareintegrating, Hamilton's Equations, herebecome : di~mPr > dtr=1,2,3; di2,3. The solution ofthese equationsisunique, and isexpressed by functions of twhich areanalytic forallvalues ofLHence the analytic continuation oftherestricted solution found above gives thegeneral solution, andtheformulas found forqr,praretrue generally. EXERCISES 1.Central Force, twodimensions, attracting according tothe lawofnature. Letql=r,q2=<p.Then : =R 2m\dr 428 MECHANICS /dR\* n, .2wX a2 (W)=2mA+-^ Thus W= 2mh+-^dr+<*>,JT T r* where either theplus sign ortheminus sign holds throughout the first stage. Hence r dr t+=+mf~JVoz, ,2mX2mhHr 02= 2mh r r Discuss thecase that theradicand vanishes fortwo distinct positive values ofr,expressingrasaperiodic function of^>,and evaluate theintegral thatexpressest;cf. 9. 2.Thesame probleminspace. Letql=r,q%=0,qz=<p; x rcos6cos^?, 2/=rcos6sin^?,2=rsin; ~2m TT=fl++*; 2m\ JVO Complete thesolution anddiscuss thecases that theradicands have distinct roots. SOLUTION OFHAMILTON'S EQUATIONS 429 3.Discuss theproblem of4whenn=1.Show thatWis givenbysolving theequation: andintegrating: =ef"^ dh IW Thenff(q,h)dq. _8WydP Thus "-8q- 6.Comparison oftheTwoMethods. Wehave studied two methods ofsolving Hamilton's Equations, a)Reduction tothe Equilibrium Problem;b),when //doesnotdepend ont,Reduc- tiontotheForm, //'=Pv The firstmethod, being general, must apply tothesecond case. Itdoes. Letustreat thiscasebythe firstmethod, assetforth intheExercise of 2.Wewillchoose asVthefunction : (1) V=-ht+W, whereW=W(q lt ,qn,h,aa>* >n)isthefunction of4, and h,ashave been replaced byPltP8.Thetransformation of that Exercise, r)V ?)V (2) pr=~Qr r=1 n yields anH'thatvanishes identically. Thetransformed Hamil- tonian Equations thustaketheform : (Ti^ rn^ rn r 1 ...n W-dT-' dt~' r~lj 'n* 430 MECHANICS Departing from thenotation oftheExercise, write their integrals intheform : ... IQr=Pr,r=!,--,; (4) 1 IP,=h, P.=a., s=2, ,n. Thesolution oftheoriginal Hamiltonian Equationsisnowgiven bysubstituting these values in(2): =8V (5) *=2,.. ,n. But 3V_=3WW=_3W 3V= ^*dqr~ dqr'dh+dh'da,~ da,' Hence Equations (5)agree notonlyinsubstance, buteven in form, save foroneexception, with Equations (15),4.The equation arising from differentiation with respect tohinthe earlier caseread : Here itis: 7.Cyclic Coordinates. Itfrequently happens thatH,besides being independent oft,contains fewer thanng's.Begin with thecase ofoneq, (1) H=H(q l9plt---,p n). From Hamilton's Equations, (2) f-f-0. -.-., andhence (3) ?>= , 5=2, ,n. Itisnotdifficult tocomplete thesolution bymeans ofHamil- ton'sEquations andtheintegral ofenergy, (4) h=H(q l9p19,pn); SOLUTION OFHAMILTON'S EQUATIONS 431 butthis isnottheform ofsolution inwhich weareinterested. Wedesire adiscussion bythemethods of4;inparticular, by thetransformation : dW/*\ (5) where (6) W=W(qi1- ,?,h,a- ,an) isasolution oftheequation:m\ l,,...,J, ,,,..., - andPl=A,P,=aa,s=2, ,n. Tofindsuchasolution weturn totheMethod ofSeparation ofVariables, which hasrendered suchgood service inthepast. Let '(9).w=Wl+--+Wn, whereWr=Wr(<?r)isafunction ofqralone and ofthen parameters, A,a2, ,.From (5)and (3)weseethat a" 2,--- ,n, andsowetry: TT,=g.,s=2,-- ,n. LetT^!bedenoted more simply byv: (10) W,=^(fc^afc-'-.aO=t>. Then(7)becomes : (11)flr (g1,^,a2,---,n)=A. Ifweassume that fiff (12) l^ffp.feuPi."*'--.*..)*' ^Pl andsolve theequation: (13) 432 MECHANICS forp,: (14) p,=*(? A, 2,',), wehave : (15)-^=*(?i, A,a,,-,). Now choose as : i (16)v=J*(<?i> A,a2, ,a.)dqlt C where cisanumerical constant. Wearethus ledtoafunction (17) W=V+otf,++anqn ofthedesired kind, provided theJacobian relation(8)issatisfied, TheJacobian herereduces to d'2v" ' r ' dq.dh" dh' dh' where p1isdetermined by(13).Ondifferentiating (13)-wefind : M!! dp,dh andsotheJacobian doesnotvanish. Solution ofHamilton's Equations. Wecannow apply the general theory of 4.The transformation (5)ofthepresent paragraph carries Hamilton's Equations over intotheform: =0,r-l,-..,n, thesolution ofwhich is : ft=+fc, Q.=A, s=2,..-,n; Px=A, P,=aa, s=2,--- ,n. These values forQr,Praretobesubstituted in(5),andtheresult- inceauations solved forqr,pr: SOLUTION OFHAMILTON'S EQUATIONS 433 <7i dhJdhl9aw dh 8Wr, , . Pi=-=*Wi, h,a2, ,an),(18) dW p9=-7T--=a,, s=2, ,n. Theequations ofthesecond linedetermineg.asafunction offt: /'(W ~^~dq lys=2,'--,n. 8 The firstequation gives ftasafunction oft. EXERCISE Obtain thefinal result (18)directly fromHamilton's Equations. 8.Continuation. The General Case. LetHdepend on 1<v<narguments qk: (1) H=#(ft, -,??!, ,pn). Themethod oftreatment issimilar, though thesolution cannot ingeneral beobtained byquadratures. Equations (3)of7 nowbecome : (2) p,=a.,s=v+I, ,n. Byanalogy wenowseek todetermineWintheform : (3) W=v+a*+lqr+1++anqn, (4) v=*(? ,<?, h, 2, , ), (5) a^T 434 MECHANICS Equation (7), 7,forWnowbecomes : /c\ L rj (b) n= //^ft,- ,q,,TT-, ,Tjjpa^+i, , This isanequation ofthesame type as(6), 4,butwith v<n variables qr.Asinthe earlier case, only aparticular solution issought, andsuch asolution maybefound byspecial devices, notably themethod ofseparation ofvariables. Afunction vonce found, thesolution proceeds asbefore. (7) /Si=-+qi,I=v+1, ,n. From theequationsofthesecond lineg*canbefound in terms offt,k=2,- ,p.From the firstequation ftisnow found interms of t.Finally, qiisgiven bythe lastline, I=v+1, ,n. 9.Examples. TheTwo-Body Problem. Consider themotion oftwobodies (particles) that attract each other according to thelawofnature andareacted onbynoother forces. Their centre ofgravity travels inarightlinewith constant velocity, orelseremains permanently atrest.Wewillassume thelatter case. Then each ofthebodies moves asifattracted byaforce at0,thecentre ofgravity, which isinversely proportional to thesquare ofthedistance ofthebodyfrom 0. Wewill firstdiscuss themotion inaplane later, inspace. Let the particle bereferred topolar coordinates, ft=r, g2= (p.Then " 2\dt2dt2 Hence LetW=v+a2q2. SOLUTION OFHAMILTON'S EQUATIONS 435 Then visgivenbytheequation: 1f/^\2 , 221^_ 2m\\dq l/(7j2Jql' or rt 7.2mX <*22 ,2mAH 1-dr. r r2' where adefinite oneofthetwosigns holds forthe firststage of themotion. Equations (18) of 7now give thesolution ofHamilton's Equations intheform : .^ dv/* Sv . or ,=/- cdr 2mh+2m\ ft= The directness ofthe result isparticularly noteworthy. It hasnotbeen necessary tomake useofskillful devices ortoeffect complicated eliminations. From theevaluation ofthesecond integralrcanbeexpressedasatrigonometric function of<p.But thediscussion ofrinterms oftismore complicated ;cf .below thereference toCharlier. TheOrbit inSpace. Totreat themotion inthree dimensions let x=rcos cos<pty=rcos sin^>,z=rsin 0. Then Let0, Then Pi Ps 436 MECHANICS H=(v*+Si P2 2m\I tf,2 ?,2cos2?2/ql SinceH=H(qvq^,p1(pa,p3),weseethatp3=a3(const.)- Thus W=v+ci^s, where visgivenbytheequation: J_[7-^Y +A.fi?Y 4.*2 i_x=, 2mlA^/ qf\dqj q,2cos2 q.2Jql Hero,t-hcro arconlytwoindependent variables, q^=rand g.2=6.Theequation canbewritten intheform : 0. cos Onsetting v=R+ thevariables canbeseparated: -r*(~^+2mhr*+2m\r= Hence ft yft =rv^_a .,2 where thesigns aredetermined foraparticular stage ofthemotion, andc,Tarearbitrary numerical constants. Adding thefurther terma3g3,wehave: W=v+ag^,v=R+0. Wearethus ledtothesolution oftheproblem intheform given by(7), 8: SOLUTION OFHAMILTON'S EQUATIONS ,+A-.A*437 (2)&=-r dr J .9^/0T,2raX a,2 c r2\2rnh-\ \ *r r2 6 d6 -a32sec2 va,22a32sec2 The discussion ofthis solution onthehand oftheexplicit evaluation oftheintegrals andtheinverse functions thus arising presents practical difficulties. Theproblemisofsogreat impor- tance inAstronomy that ithasbeen treated atlength byCharlier, Mcchanik desHimmels, Vol.I,Chap. 4,p.167.On p.171, Equations (7)areidentical withoursolution, save astonotation. Failure oftheMethod. There arecases inwhich themethod breaks down. Consider, forexample, motion inaplane. Sup- pose thebodyisprojected from apoint A,distant afrom the centre offorce, 0,atright angles tothelineOAandwithavelocity v (}such that Itwillthen describe acircle, r=a.But theEquations (1) or(2)canobviously never yield thissolution. Why? Thefunction vwasdetermined from theequation: =2mA2raX Inthepresent case, h=- 2a'mav, andhence =0. 438 MECHANICS Thus thecondition isnotfulfilled, and so,ofcourse, there isnoreason whythemethod should apply, since thehypotheses onwhich itdepends donot hold. 10.Continuation. The Top.Wetake overfrom Chapter VI, 18,theexpressionforthekinetic energy, ByEuler's Geometrical Equations, thisbecomes : T= Let tfl^Q> <?2= <P> Then, since =dT wehave : Pi=Ad, ps=C<pcos+(Asin2+Ccos2 Thus T,expressedinterms ofthep'sand<?'s,becomes : Furthermore,* U=Mgb cos 0. Thus Hence itappears that theproblem comes under thecase of cyclic coordinates treated in 7.First, then, *Itisnecessary tochange from theearlier notation hforthedistance from thepegtothecentre ofgravity, since hplays soimportant ardle inthepresent theory. Letthedistance bedenoted by6. SOLUTION OFHAMILTON'S EQUATIONS 439 Todetermine vwehave : Irl dv* ,19.l/a2cosq 1-a3\21.,_, L 2Li55?+C^+l( singl )J+M*6COS*-* ^2 sin2 ft^2=(24A-La22-Ncosft)sin2ql- (2cosq1- 3)2 ,_ =r+V(2^ A-La22-JVcosft)sin2 g,- (2cos?,-^ t/ sinft?1> c where isanarbitrary numerical constant, notaparameter, andthesign oftheradical must bechosen with respect tothe special stage ofthemotion under consideration. Moreover, for brevity, L=4,N=2AMgb. v/ Thesolution oftheproblem, asgiven in7,nowtakes onthe form: +*-* Thus Asinft(4- C- '~e/~Hv(2Ah L22~A^cosgjsin2 gj (2cosgjas)2 Let u=cos#,. Then thisequation becomes : 9l=r^ ccose, e/+V/<W where This isthesame result obtained byelementary methods, Chap. VI, 18.Butcompare thetechnique. With only Euler's 440 MECHANICS Dynamical andGeometrical Equations towork with,* elimina- tionshadtobemade byingenious devices, whereas thepresent advanced methods free thetreatment from all artifice. The fundamental equation indesired form isevolved naturally, directly, from thegeneral theory, notuntangled from asnarl ofequations. Instead ofhaving tosolve three equations for6, <j>, \j/bymore or lessingenious methods ofelimination, thefunctions 77 ,[7,and henceHareobtained without theuseofany artifice whatever, andthemethod of7yields ql atonce asafunction oft,the further equations giving q3=<pand</3= \f/immediately. EXERCISE Study themotion ofatopwith hemispherical peg, spinning andsliding onasmooth table. Show that where F(u)=(2h-^- 11.Perturbations. Variation ofConstants. Intheproblem ofperturbations themotion which thesystem would execute if only themajor forces acted isregarded asfundamental, and then thevariation from thismotion duetothedisturbing forces, thought ofasslight,isstudied. This analysis ofthephysical problemismirrored mathematically bywriting down Hamilton's Equationsfortheactual motion : ___ -.- ~dt~Wr dt" dqr' ''' andthen setting thecharacteristic functionHoftheactual prob- lemequal totheHoftheproblem duetothemajor forces, plus aremainder, Hl: (2) H=H,+H{. *Itistruethat intheearlier treatment wehadtwointegrals ofthedifferen- tialequations ofmotion towork with attheoutset, namely;theequation ofenergy, T=U+h,andtheequation arising from thefactthat thevector moment ofmo- mentum <risalways horizontal. Buteven sothere were three equations in0,$, <f> tointegrate. SOLUTION OFHAMILTON'S EQUATIONS 441 Transformation oftheMajor Problem totheEquilibrium Problem. First, themajor problem, represented byHamilton's Equations intheform : issolved byreducing it,through acontact transformation, to theEquilibrium Problem. Thecontact transformation isgiven bytheequations: fA\ (4) Pr r where (5) F=F(<7i, ,q,P,, ,Pn, t) isobtained asfollows. Write down Jacobi's Equation, cor- respondingtoHamilton's Equations (3): Let V=Vfe, -,?,alf",an, beasolution ofthisequation such thattheJacobian O/v7~ VJ. 0(a l9- ,an) Inthisfunction, replace arbyPr.Theresulting function isthe function (5). [Inpractise, thefunction F^,- ,qnyalf , an,t)isobtained, notfromanelaborate theory ofpartialdiffer- ential equations, butbymeans ofsimple devices, adhoc.] Letthetransformation(4)bewritten intheexplicit form : ,.fQr=Fr(p lt'*' 9Pn 9qi,'** 9qn, IPr=Gr(plt'jpniQi,'",qn, or f9r=fr(P ,P,Q-^Qn, 1pr=gr(Ply' ,Pn,Q19- ,On, Tosaythat themajor problemisthereby transformed tothe Equilibrium Problem means that,when thevariables qryprthat 442 MECHANICS form thesolution ofEquations (3)aresubjected tothetransfor- mation (4),theresulting Hamiltonian Equations become : (7) f=0,^=0, r=l,..-,n. Thesolution ofthese equations canbewritten intheform : (8) Qr=0r, Pr= r, f=1, ,ft, where ar,Prareconstants. Now transform thevariables Qr,Pr that arethesolution ofEquations (7),namely, thefunctions given by(8),backbymeans ofthetransformation (4"),andwe have thesolution ofEquations (3)intheform : fQr=/r(i,''' ,n,ft, ,0n, 1Pr=0r(a lf"-,, ft,'',0, Thus thetransformations (4; )or(4"), and(9), identical exceptinnotation, represent two distinct things: a)Intheform (4") these equations represent theContact Transformation (4). b)Intheform(9)they represent theSolution oftheHamil- tonian Equations oftheMajor Problem, or(3). Transformation oftheActual Problem bytheSame Contact Trans- formation. Wenowproceed toapply thecontact transformation (4),nottothevariables (qr,pr)which satisfy Equations (3), buttothevariables(qr,pr)oftheoriginal problem, whichsatisfy Equations (1). Since this isacontact transformation, weknow thatEquations (1)willgoover intonewequations ofthesame form: dQ,_8ir dPr__9ff'j. (W)dt~dPr'dt~Wr' ' HereH'=H'(Q r,Pr,<)hasthevalue,cf.Chap. XIV, 5,Ex.1, (15): (11) H'=H+ Butfrom (6): Hence H'=H- SOLUTION OFHAMILTON'S EQUATIONS 443 Finally, from(2)itfollows that (12) Hf=fft. Thus Equations (10)take theform : dQr^dH, dP r_m, The resultmaybestated asfollows. When thevariables qrjpr which formthesolution oftheactual problem represented byEqua- tions(1)aretransformed bythecontact transformation (4)or(4'), thetransformed equations take theform (13), whereHlisthegiven, orknown, function ofEquation (2),nowexpressed through (4)or (4")interms ofQr,Pr,t. TheFinal Solution. Itisnowbutastep tothesolution of Equations (1),which represent theactual problem. Solve Equa- tions(13), thus determining Qr,Prasfunctions of t.Then transform these functions, thesolution of(13),backbymeans of(4)or(4") tothevariables qr,pr.The latter satisfy Equa- tions (1). The result canbeexpressedintheform : (14) Pr=flTr(P,, ,Pn,Qi,''' ,Qn, whereQr,Prarcdetermined byEquations (13). Variation ofConstants. Themethod above setforth hasbeen called the"variation ofconstants." This expressionisamathe- matical pun. Itisapunontheletters ar, r.These, inEqua- tions (9),areconstants theequations there representing thesolu- tion ofthemajor problem, (3).Ontheother hand, theycanbe identified with thevariables Pr,Qrof(14), these variables being determined by(13),andthenEquations (14)represent thesolu- tion oftheactual problem, (1). Wecanattain complete confusion ofideas, asisdone inthe literature, bychanging thenotation in(13)and (14)from Qr,Prto r,oLr.Thus (14)goesover intotheform of(9),and (13)isreplaced bytheequations: dar_3(-g,) df)r_B(-H,) dt~ d0r'dt~ dar'r-L>'"'n> 444 MECHANICS whereH1=H1(Qr,Prjt)isnow written asHl(ft1)a1)t))the Hamiltonian function nowbeingH1instead offf,. Thus thepunisexplained but itisapoorpunthathastobe explained. 12.Continuation. ASecond Method. Itispossible totreat theproblem ofperturbationsin still another manner. Let <f>(a lf ,an)beanygiven function whose firstpartial deriva- tives arenot all0.LettheHamiltonian Equations fortheundis- turbed motion, namely, (3),betransformed byanewcontact transformation : whereSisdefined asfollows. Consider theequation: /i/r\ / \ T (16)*(,,-..,.)= Let ,qn,!,--, an,t) beasolution such that* Now,make thecontact transformation : /1>7\ ^r>^ (17) Pr- Wr,Pr=~W, where 5=8(q ll ,q^Q,, ,Qn,0- *Inorder tofindsuchasolution, begin with theequation: , / OS dS\ ,dSh=H(qi,..., qn,~,...,--,t)+-, where hisanarbitrary constant, andseekasolution : S=S(<?,, ,qn,h, 2, ,On,Of such that tgt,''',Qn d(h,as, -, ) where as, ,Onarearbitrary. Substitute h=<p(ai, ,an) iniS.Ifd<p/dai^0,this willbethefunction desired. SOLUTION OFHAMILTON'S EQUATIONS 445 This transformation, applied toEquations (1),carries these over intoequationsofthesame type: dtdPr'dt 8Qr' ' ' ' where, byChap. XIV, 5: (19) H'=H+?j-VI But,by(16)and (17): Hence, with theaidof(2): thearguments nowbeing theQryPrintowhich gr,prhave been transformed by(17). Thus Equations (18)take theform : dQr=0/7, dP r__3/7,_3jp "r-1"'n- Solve these equations and substitute thefunctions oftthus obtained, namely, theQr,Pr,in(17). Thefunctions qr,prof t obtained from these equations arethesolution oftheactual prob- lem,orEquations (1). Carathgodory*treats Equations (20) asfollows. Hewrites X//!instead ofH1: Hethendevelops thesolution intoapower series inX: fQr=Ctr+\Cl r+X2C2r+ , <22> (Pr=r-vat+Wlr+'' , where C*r,Dkrarefunctions oft,vanishing when t=(forsim- plicity wehave set t=0).Onsubstituting these values for Qr,Prin(21)andequating coefficients oflikepowers ofX,the coefficients C*nDkrcanthenbeobtained byquadratures. *Cf.reference above, p.381.Thepage inR.-W. is211. APPENDIX A VECTOR ANALYSIS InRational Mechanics only aslight knowledge ofVector Analysisisneeded. Itisimportant that thisknowledge be based onapostulational treatment ofvectors. Thesystemof vectors isasetofelements, forming alogical class. Certain functions ofthese elements aredefined, whereby twoelements aretransformed intoathird element. These functions arecalled addition, multiplication byarealnumber (here, onlyoneelement enters astheindependent variable), theinner product (scalar multiplication), and theouter product (vector multiplication). The functions obey certain functional, orformal, laws, which happen tobeasubset oftheformal laws ofalgebra: A+B=B+A AB=BA A(BC)=(AB)C A(B+C)=AB+AC (B+C)A=BA+CA Abrief, systematic treatment such asishererequired isgiven intheAuthor's Advanced Cakulus, Chap. XIII. Forafirst approach tothesubject theHamiltonian notation ofSandVfor thescalar andvector products hasthegreat advantage ofclear- ness inemphasizing thefunctional idea theconcept:transfor- mation. Ontheotherhand thenotation pretty generally adopted atthepresent dayisthedesignation ofvectors byClarendon or boldface, thescalar product being written asaborab(read: adotb),andthevector product asaXb(read:across b). It isuseful, therefore, tohave asyllabusofdefinitions andessential formulas inthisnotation. 447 448 APPENDIX A 1.Vectors andTheir Addition. Byavector ismeant adirected linesegment, situated anywhereinspace. Vectors willusually bedenoted byboldface letters a,A,orbyparentheses; thus avector angular velocity maybewritten (w). Two vectors, AandB,aredefined asequalifthey areparallel andhave thesame sense, andmoreover areofequal length: A=B. Bytheabsolute value ofavectorAismeant itslength ;itis denoted by |A|,orbyA. Addition. Bythesum oftwovectors, AandB,ismeant their geometric sum, orthevector Cobtained bytheparallelogram law: A+B=C. Inorder that this definition may apply inallcases, itis necessary toenlarge thesystem ofvectors above defined byanulvector, represented bythe symbol0. IfBisparallel toAand ofthesame length, butoppositeinsense, then A+B=0, orB=-A.B FIG.149 Moreover, weunderstand bywA,wheremis any realnumber, avector parallel toAandmtimes aslong ;its sense being thesame asthat ofA,oropposite, according asm ispositive ornegative.Ifm=0,then raA isanulvector : OA=0.Thenotation Ammeans wA,and also aA+6B a . ,6_ r-j means .,AHrrB.a+o a+o a+o Vector addition obeys thecommutative andtheassociative law ofordinary algebra: A+B=B+A A+(B+C)=(A+B)+C Subtraction. ByABismeant that vector, X,which added toBwillgiveA : B+X=A, X=A-B. VECTOR ANALYSIS 449 Toobtain Xgeometrically, construct AandBwith thesame initial point ;thenABisthevector whose initial pointisthe terminal point ofB,andwhose terminal point istheterminal point ofA;Fig. 149. Cartesian Representation ofaVector. Letasystem ofCartesian axesbechosen, and leti,j,kbethree unit vectors lying along these axes. LetAbeanarbitrary vector, whose components along theaxes areAltA2JAz.Then evidently A=AJ+A2j+Azk. E=B,i+B,j+B,k t then A+B=(A,+BJi+(A,+B,)j+(A,+ 3)k. Also :_A^ Resultant. Ifnforces, Fx,F2, ,Fn,actatapoint, their resultant, F,isequal totheir vector sum : F=Ft+F2++Fn. Ifncouples,MuM2, ,Mn,actonabody, theresultant couple, M,isequal totheir vector sum : M=MJ+M2+-+Mn. Two ormore vectors aresaid tobecollinear ifthere isaline inspace towhich they areallparallel. Inparticular, anulvector issaid tobecollinear withanyvector. Three ormore vectors arcsaid tobecomplanarifthere isaplane inspace towhich they are allparallel. Inparticular, anulvector issaid tobeparallel toanyplane.Ifthree vectors, A,B,andC,arenon-complanar, thennooneofthem canvanish(i.e.beanulvector) andany vector, X,canbeexpressedintheform : X=ZA+mB+nC, whereZ,m,nareuniquely determined. Differentiation. Velocity. Acceleration. Osculating Plane. A variable vector canbeexpressedintheform : A= 450 APPENDIX A wherei,j,karethree fixed vectors mutually perpendicular.If /(Oi <p(t),^(0have derivatives, thevectorAwillhave ade- rivative defined as lim-rr = Itsvalue is : Moreover, Ifmisafunction ofxandAisavector depending onx>and if eachhasaderivative, thenmAwillhaveaderivative, and d(mA) dm. . ofA -~^j-=-y~A+m~T'ax ax ax IfapointPmove inanymanner inspace,itscoordinates being givenbytheequations: where/, <p,$arecontinuous functions ofthetime, having con- tinuous derivatives, and if r=xi+yj+zk, thevector velocityofPisrepresented by W/i <f>,thave continuous second derivatives, the vector acceleration ofPisgivenby Theplane determined bythevectors rand fdrawn fromP (ontheassumption that neither isanulvector)istheosculating plane. Thus thevector acceleration alwaysliesintheosculating plane. 2.TheScalar orInner Product. The scalar orinner product oftwovectors, AandB,isdefined astheproduct oftheir absolute values bythecosine oftheangle between them. Itisdenoted byA-B orABand isread:"AdotB." Thus A-B=AB= |A | |B |cos c. VECTOR ANALYSIS 451 Ifoneofthefactors isanulvector, thescalar productisdefined asO. Thecommutative andthedistributive lawshold : AB=BA A(B+C)=AB+AC. The associative lawhasnomeaning. The scalar product vanishes when either factor isanulvector; otherwise when andonlywhen thevectors areperpendicular to each other. Furthermore : often called thenorm ofthevector. :2 i ;2 i k2_i ii, j i,Ki, jk=0, ki=0, ij=0. Cartesian Form oftheScalar Product : =AB Differentiation: Ifaisaunit vector,i.e. if |a |=1,then a2=1, and aa'=0. 3.TheVector orOuter Product. Lettwovectors, AandB, bedrawn from thesame initial point. Then they determine aplane, M,andaparallelograminthat plane. The vector orouter productisdefined asavector perpendicular toMand oflength equal tothearea oftheparallelogram. Itssense isarbitrary. Itisdefined with ref-AXB erence tothe particular systemof Cartesian axes tobeused later. Itis denoted by AXB and isread :"AcrossB." FIG.150 452 APPENDIX A Ifoneofthese vectors is0,orifthevectors arecollinear, neither being 0,thevector productisdefined as0,andthese arethe only cases inwhich itis0.Otherwise, letebetheangle between thevectors. Then |AXB |= |A | |B |sin c. Thecommutative lawdoesnothold ingeneral, for AXB=-BXA. The associative lawdoesnothold;e.g. (iXj)Xj5^iX(jXj). Butthedistributive law istrue : andC)=AXB asc^nbeproved geometrically, orstillmore simply, analytically, bymeans oftheCartesian form;cf .infra. Itisconvenient tochoose thesense ofthevector product so that iXj=k, jXk= i,kXi= j. Inanycase AXA=0, and so,inparticular, iXi=0, jXj=0, kXk=0. Cartesian Form oftheVector Product : AXB= iJ **l-^- Differentiation: - dx dx- dx 4.General Properties. LetA,B,Cbethree non-complanar vectors drawn from thesame point. Thevolume oftheparal- lelepiped determined bythese vectors isnumerically A(BXC). VECTOR ANALYSIS 453 Anecessary and sufficient condition that three vectors A,B, Cbecomplanaris : A-(BXC)=0. Linear Velocity inTerms ofAngular Velocity. Letspace be rotating about anaxis/with vector angular velocity (w).Then thevelocity vofanarbitrary pointPwillbe : v=()Xr, where risthevector drawn fromanypointOoftheaxis to thepoint P. Iftheaxispasses through theorigin, thenFia.151 v=J xy and vx=zwy ywz Vy=Xtl)g Z<l)X Ifitpasses through thepoint (a,6,c),then vx=(z- c)coy-(y-6)wz vz(y 6)o)x(x a)uy Inthogeneral case ofmotion ofarigidbody (i.e.motion of rigid space), lot0' :(z ,y ,z)beapoint fixed inthebody, and let(f, ??,f)bethecoordinates ofanypointPfixed inthebody, the origin being at0';butotherwise the(, rj,f)-axes may Mpmove inanymanner. Then FIG.152O'where P7 COf CO,C0 *irV=V+V', v= 454 APPENDIX A Localized Vectors. Itissometimes convenient toprescribe the initial point ofavector, orthelineinwhich thevector shalllie, asinthecase ofaforce acting onaparticle, oraforce acting on arigid body. Itiswith reference tosuch vectors thatthefollow- ingdefinitions areframed. Bythemoment ofavectorFwith respect toapointismeant thevectorM=rXF, where risthevector drawn from toanypoint ofthelinein whichFlies. Inpractice, Fmaybeaforce acting onarigidbody, orFmaybethevector momentum, mv,ofaparticle. Themoment ofacouple canbeexpressed as TIXF!+r,XF2, where FDF2aretheforces ofthecouple and rt,r2arevectors drawn fromanypointofspace toanypointsPltP2ofthelines ofaction ofFDF2,respectively. Bythemoment ofavectorFabout adirected lineLismeant the vectorM=Ma,M=a-(rXF), where aisaunitvector having thedirection andsense ofL,and risthevector drawn fromanypoint ofLtoanypoint oftheline inwhichFlies.Thus ifFisaforce acting onarigid body,let itspoint ofapplication betransferred tothepointPnearest to L,and let bethepointofLnearest toF;i.e.OP isthecommon perpendicularofLandthe line ofaction ofF.Decompose F atPintoaforce parallel toLandoneperpendicular toL.The vector moment ofthelatter component atPwith respect to isMa. 5.Rotation oftheAxes. Direction Cosines. Atransforma- tionfrom onesetofCartesian axes toasecond having thesame origin (both systems being right-handed, orboth left-handed) ischaracterized bythescheme ofdirection cosines : X y zn^n%7i3fcHIw2 VECTOR ANALYSIS 455 Between thenine direction cosines there exist thefollowing relations : mS+m22+m32=1 n*+n22+n32=1IS+mS+ 1 n +tn2w3+n2na= +m^m l+n3nx= -fw,m 2+nxn2=== n^j+n2Z2+n3i!3= JiW,+ Z2m2+ J3ms= m3n2ml=n2Z3n3Z2 n2=Z3mIZ^j Z3= 1. APPENDIX B (dij\^ ~jj)=/() Differential equationsoftheform : where I. f(u)=(t*-a)(6 -*) or II.' /(w)=(w-a)(6-tO and^(w)iscontinuous andpositiveintheinterval a^u^bj playanimportantroleinMechanics. Letusstudy their integrals. CASE I.Aparticular integral of(1)isfound byextracting thesquare root : du andseparating thevariables : du dt= <2>'-Ajsra^u^b. Geometrically, thefunction ontheright of(2)canbeinter- preted astheareaunder thecurve, a b(3)\/w uThegraph ofthefunction FIG.153 (4)y=V(u-a)(6-u 456 ADIFFERENTIAL EQUATION 457 isrepresented byFig. 153.The reciprocal ofanordinate ofthis curve gives thecorresponding ordinate ofthegraph ofthefunc- tion (3),Fig.154 : /r\ .. _ m V(u-a)(b -u)t(u) Theareaunder thecurve(5),shaded in thefigure, represents theintegral (2),or:v (6)trdu_ ~~JV^-~Mb~^ Thus thisarea expressestandbrings out the fact that tincreases asuincreases. Conversely, uincreases as tincreases. LetAbedefined bytheequation:a u b FIG.154 (7)rdu^JV(1T^~ci)(6~- Then thegraph ofu,regarded asafunc- tion ofty (8) u= FIG.155isasshown inFig. 155. Itsslopeisat eachextremity andpositiveinbetween. The definiteintegral, (2)or(6),hasnowserved itspurpose. Ithasyieldedforarestricted interval, ^tgA, aparticular solution of(1). Continuation byReflection. Reflect thegraph ofthefunc- tion(8),Fig. 155,intheaxis ofordinates, and letthecurve thus obtained define acontinuation ofthefunction ^(0throughout theinterval A^trg0.Analytically thereflection isrepre- sented bythetransformation : Thus =*(- 0,-A 458 APPENDIX B Theextended function : isseentosatisfy thedifferential equation fdu\(w)= Hence thefunction<p(t)thus defined intheinterval(A,A),or u=<*(0,-A^t^A, isasolution of(1). Tocomplete thedefinition of<p(f) for allvalues oft,i.e. oo<t<oo,wecould repeat theprocess ofreflection, using next thelines /=Aand t=A; anc *soon '^ut***ss^mP^erto !/i\introduce theidea ofperiodicity. \v Periodicity. Letthefunction nowbeextended toallvalues of tbytherequirement ofperi--2A-A FIG.156odicity. (9) <f>(t+2A)=<p(f),-oo<t<oo. Thenwehave onesolution ofEquation (1). TheGeneral Solution. Thegeneral solution ofEquation (1)in thepresent casecannowbewritten intheform : (10) u=<p(t+7), where 7isanarbitrary constant. Observe that (U) *>(-=<f>(i). Hence /io\ '/ t\ ft\ Toanarbitrary value uofusuch that a<w<6there correspond twoandonlytwovalues oftintheinterval(A,A), forwhich (13) UQ=^>(0, namely Ifw= ,there isonly one value, namely,t=0;and if w=&,then tAyA.Butonly oneshould becounted, ADIFFERENTIAL EQUATION 459 since thefundamental interval ofperiodicity should betaken as anopen interval, c<tgc+2A or cgt<c+2A, where cisarbitrary. Moreover, du/dt hasopposite signs in J and t'Q,because of(12). Wecannowprove that there isasolution ofthegiven differ- ential equation, which corresponds toarbitraryinitial condi- tions :u=ul9t=t19provided merely that agH!^6. Suppose that itisknown from thephysics oftheproblem that du/dtisnegative initially. Now, setUQ=^anddetermine tQasabove sothat w=^(a *'(*o)<0. Finally, define 7bytheequation: *i+7=tQ, 7i=<o- 'i- Thus7=7iisuniquely determined andthefunction (14) u*(+7i) isthesolution wesetouttoobtain. But isthissolution unique, orarethere stillother solutions which satisfy thesame initial conditions? Ifa<u^<6,the answer isaffirmative forvalues oftnear^;butforremote values, thequestion ofsingular solutions arises, towhich wenow turn. Singular Solutions. The given differential equation admits, furthermore, singular solutions. Thefunctions u=a, u=6 areobviously solutions ofthedifferential equation: (15) eachbeing considered inanyinterval fort,finite orinfinite. Such asolution, moreover, may becombined with asolution(10) at any point. The solution nowmay follow (10) indefinitely; or itmayswitch offonasingular solution again. These solutions donot, however, have any validity inthe problems ofmechanics, forwhich theabove study hasbeenmade. Themechanical problems depend each timeondifferential equa- 460 APPENDIX B tions ofthesecond order, andthese haveunique solutions, depend- ingonthe initial orboundary conditions. Equation (14) repre- sents anintegral ofthese equations. But theconverse isnot true, namely, thatevery integral of(15)isanintegral ofthe second order equations why should itbe?We see, then, thatwemaybeondangerous ground whenwereplace thelatter equations, inpart,bytheintegral ofenergy, forexample ;since the modified system mayhave solutions other than that ofthegiven mechanical problem. Cf.theAuthor's Advanced Calculus, p.349. Does thisremark not callinto question thevalidity ofthe treatment inChap. XV, since theequation: isessentially theintegral ofenergy? Not ifweapply thatmethod assetforth inthetext. Forinasuitably restricted region there isonlyonesolution yielded bythose methods, andwewere careful topoint outthat itistheanalytical continuation ofthissolution that yields thesolution ofthemechanical problem beyond this region. Thus thesingular solutions areautomatically eliminated. CASE II.This case : (16)(J~J=(u-a)(6-u)*t(u), ismore easily dealt with.Aparticular integral of(16)isgiven bytheformula (17)t= f-(b_dM=== , a^u<b a Theinverse function, (18) u=*>(0, ^t<oo, represents anintegral of(16)intheinterval indicated. And now thissolution canbecompleted bythedefinition : (19) <p(- t)=<p(t). Thuswehave onesolution : u=b (20) u=,(<), oo<t<oo. Itisnowshown asbefore that thegeneral solution is ADIFFERENTIAL EQUATION 461 (21) u=?(+7). Afurther case,namely: (22) canbetreated inasimilar manner; or,more simply, bethrown backonthecase justconsidered byalinear transformation. Finally, thecase (notmentioned above): (23) breaks upintothetwodistinct equations: a)-=+(u-a)(6-t)vV(tt); b) ft=-(u-a)(b-u)Vt(d. Each ofthese issolved atoncebyaquadrature. FURTHER STUDY OFCASE I.There isanother treatment of Case Iwhich brings outtheimportant factthatthefunction<p(t) isessentially asineorcosine function : (24) u=Ccos+C', where0,inthesimplest case,isproportional tothetime : *= j<, andinthegeneral case isoftheform : where h(/)isperiodic with theperiod 2A : h(t+2A)=h(t). Thismethod, moreover, may simplify thecomputation incase itisdesired totabulate thefunction(p(t). Thegiven differential equation: (25) 462 APPENDIX B canbereduced byalinear transformation : ,2u-a-6U=b-a' totheform, afterdropping theaccent : (26) Make thesubstitution : (27) u=cos6, <<IT. Equation (26)becomes, onsuppressing thefactor*sin2 : (28) Thisequationisequivalent tothetwoequations: (29) (30)= The solution of(30)isobtained from thesolution of(29)by changing thesign oft. Aparticular solution of(29)isgivenbythequadrature: (3D Write (32) where (33)9 "/:de ^(cosfl)'-oo<e<oo. f.d9-24. _JV^(cose) Theng(6)isperiodic with theperiod 2v. For, (34) *Insodoingwesuppress thesingular solutions of(26). ADIFFERENTIAL EQUATION 463 Butthevalue oftheintegral, because oftheperiodicity ofthe integrand,is2Aforallvalues of0.Hence (35) g(e+2r) Equation (31), oritsequivalent, (36)t=^+ defines 6asasingle- valued function oft,since theintegral (31) represents amonotonic function of0.Let bewritten inthe form: (37) 0=J+/KO- Then h(t)hastheperiod 2A : (38) h(t+2A)=h(t). For, lethaveanarbitrary value in(36)and letthecorrespond- ingvalue oftbe t : Let=+27r,and let t'bethenewvalue oft: ,,*('.+20+,(,.+ar). Byvirtue of(35), or t'=<+24. From (37)wenow infer : Hence andtheproofiscomplete. Ifwemultiply (36)byvand(37)byAandadd,wefind : (39) 0()+AA(0 =0. 464 APPENDIX B Wearenowready toexpress uinterms of/.InEquation (27) Bwasrestricted. Now, setgenerally: (40) This function isseenbydirect substitution tobeasolution of(26). Ifwedenote itby<p(f) ythegeneral solution of(26)willbe : (41) u=*>(*+7). Theother equation, (30), leads tothesame result. Ifitisaquestion actually ofcomputing h(t),then theintegral (31)canbetabulated forvalues offrom toTT,thereckoning being performed bytheordinary methods forevaluating definite integrals Simpson's Rule, etc. Integral ofaPeriodic Function. Letf(x)beacontinuous peri- odicfunction : f(x+A)=/(*),-oo<x<oo, whereAisaprimitive period, corresponding to2Aabove. Let A c= Then J+AJf(x)dx. L f(x)dx=C, where xisarbitrary. For, X+A //(*)dx=f(x+A)- /(*)=0. Let X/Vf(x)dx- jx. c Then<p(x)isperiodic: <p(x+A)=<p(x). For,x+A v(x+A)-*(x)= ff(x)dx-( x+A)+ ADIFFERENTIAL EQUATION 465 Hence where X=C/A. The resultmaybestated asfollows. THEOREM. The integral ofaperiodic functionisthesum ofa periodic function andalinear function: where and A C=ff(x)dx,X= |- Instead ofthelinear function \xwemay write Xz+7 or X(z-xn), thefunction <p(x) being changed byanadditive constant. In particular, X=ifandonlyif A ff(x)dx=0. o APPENDIX C CHARACTERISTICS OFJACOBI'S EQUATION Although Jacobi's partial differential equation ofthefirstorder : ANA) hasplayed animportant roleinthesolution ofHamilton's Equa- tions : rnd(Jr_dH dpr_m *>~dt~Wr ~dt~~Wr'r-l,..-,m, where //=H(q ly- ,qm,ply ,pm,t),wehave notfound it necessary torefer tothetheory ofcharacteristics, partly because wehave sought certain explicit solutions bymeans ofingenious devices (separation ofvariables, forexample) ;partly because, whenwehave needed anexistence theorem,itwassupplied at oncebyreference totheCauchy Problem. Nevertheless itis ofinterest forcompleteness toconnect theequation with its characteristics. 1.The Analytic Theorem. Consider thegeneral partialdif- ferential equation ofthe firstorder : T .. . Ll' 'n> > 'dx LetF(x ly-- ,xn,z,2/i, ,2/n),together with itspartial deriva- tives ofthefirsttwoorders, becontinuous forthose values ofthe arguments forwhich (xl9 ,xn,z)isaninterior point ofan (n+l)-dimensional region*Rofthespace ofthevariables (x19 ,xn,z),andtheykarewholly unrestricted. Usethe notation : /i\ y__dF7__dFv3F (1) Xk~^'Z-~te>Yk~^ Atagiven pointA :(a,, ,an,c,b19 ,bn)=(a,c,b)ofR lettheYknot allvanish;inparticular, letYn^0. *Rshallnotinclude anyofitsboundary points. 466 CHARACTERISTICS OFJACOBFS EQUATION 467 Thecharacteristic strips aredefined bythesystem of2nordinary differential equations: IIdxk=dz="""dykk=lYkSykYkXk+ykZ*' ' The solution ofII.shall gothrough thepoint (x,2,yQ ),which shall lieintheneighborhood ofAandmoreover onthemanifold F=0,or F(x xzv '2/^^=0 Although there are2n+1initial values the (2,2,y) there isonlya2n-parameter family ofsolutions ofII.,forwe may without loss ofgenerality setxn=anonce for all.The solution ofII.cannowbewritten intheform : T.= f-(r -r.r 7 11 11^^t ^i\^n ,-^i , ,-^n 1)*9Ml 9 9if*/9 i=l,---,n-l; (2) Along anycurve(2)thefunction F(x lt ,xn,z,y^ ,yn) isconstant, since dF=%Xkdxk+Zcfe+5)y*d*- Onsubjecting dxk, dz,dyktotheconditions imposed byII. it appears thatdF=0.Hence (3) F(x l9 ,xn,z,y l,.-,y n)=C isanintegral ofthesystem ofdifferential equationsII. Characteristic strips arecurves (2)forwhich (4) F(x,-- ,4-i, On,2,2/i, , )=0, i.e.C= in(3). This equation canbesolved forynQsince Thus there isa(2n l)-parameter family ofcharacteristic strips. Consider now the(n+l)-dimensional space ofthevariables (&!,, xnyz),inwhich asolution : (5) z=*(*!,--,*), 468 APPENDIX C ofthepartial differential equationI.will lie.Inthe(hyper-) plane xn=anofthisspaceletamanifold bedefined bythe equation: (6)2=Cdfo ,'' ,Xn-l), where o)(x lt ,xn-\),together with itsfirst partial derivatives, iscontinuous intheneighborhood ofthepoint (a,, ,an~0, and Furthermore,let and letynbegivenby(4)or(4'). If,now,weregard the asn 1independent parameters and forsymmetryinnotation set Xn=Un, the firstnequations (2),combined with this lastequation, will represent a(hyper-) surface parametrically, theequation ofwhich canbethrown intotheform(5)byeliminating the(ulf ,wn), and thisfunction (5)isasolution ofthegiven partial differential equationI.Moreover itisthemost general solution;i.e.any solution (5),such thatVsatisfies theabove requirements of continuity, canbeobtained inthismanner.* This isthegeneral theorem ofthesolution ofI.bymeans of characteristics. Weproceed toapply the result toJacobi's Equation A). 2.Jacobi's Equation. Let (8) (Xr=q" y'=Pr> r==1, ,m=n-1; 1z=F, xn=*. Asregards yn,weseefrom I.andA)that itisgivenbytheequa- tion: (9) FssH(x lt ,xm,*,Vi,- - ,2/m,xn)+yn=0. EquationsII.nowtaketheform : *For theproof cf.Goursat-Hedrick, Mathematical Analysis, ortheAdvanced Calculus, Chap. XIV, p.366. CHARACTERISTICS OFJACOBFS EQUATION 469 (10)m im=-dpr dyn d# dff dqr dt The initial values are : Xr Qr3 (H)/7 ?/=7i r=s1 . *n</r > JfrPr , r1, ,771, ^o> 2/w="""> ^0* From (10)follow firstHamilton's Equations: (12)dqr dt dpr'dt dqr' Furthermore, bytheaidof(9), (.dtf^m.\/ fit QJ Jat ot andfinally, sincefrom (12)m. (14)dV dt=ZPrQr- Observe inpassing that theright-hand side of(14)istheLa- grangean Function, L : H+L=5)prqr, r andso (15)dV-L~-L - Butwehave anticipated theresults ofthegeneral theory and although obtaining thefacts ofthecase inEquations (12) ;(13), and(14),wehave notbrought outthedirect testimony ofthe general theory inthepresent case. Letusturn back, then, to Equations (2)andCondition (4)or(4'). Itappears that the solution ofEquations (10)takes theform : (16)n__ ///. Qr Jr(J> , 7) / (tPr~Jn+r \f,-<)Pi nV >Q.m ,'0> ,<7m ,F0) Pm , 470 APPENDIX C where r=1, ,m=n 1and (17) y=-H(<?,, ,qn,Pl ,-,pm,t) Inthecase before usthefunctions frt/w+r,r=1, ,m, arising astheydofrom thesolution ofEquations (12),donot depend onF,andynQisgivenby(17). Thuswehave (18)V'' andalsothefurther integral of(10), givenby(9): (19) 2/n=-//(ft,- ,g,p,, ,?m,0- ButVin(16)doesdepend onV .Itisgivenby(14): (30) or t V= to Somuch, then, forthediscussion ofthesolution ofEquations II.,i.e.(10). Asregards nowthesolution ofEquation I.,i.e.A): (21) + wechoose wsubject totheconditions under (6): (22) F=cofoo, ,qm) andset n= r's 3.Application. Wehave seen inChap. XV, 2,thatHamil- ton'sEquations canbesolved byacontact transformation : /f\n\ d$ n d$-i (23) Pr=w,Pr=~W,r=l,--,m, which transforms thegiven dynamical problem intotheEquilib- riumProblem, thesolution ofwhich is Qr=ar,Pr=r,r=1, ,m, CHARACTERISTICS OFJACOBFS EQUATION 471 where ar,@rarearbitrary constants, wholly unrestricted sofar asthetransformed Hamiltonian Equations: dQr_Qd^-n r-1 m"rfT"'~3T~'r-l,---,m, areconcerned. Thedemands that thefunction Sfa,- ,qm,Qif ,Qm, fulfil arethefollowing. First,itmust bepossible tosolve the equations: (24) br=Sqr(alt ,Om,alf ,ctm,t) forthear: (25) oj=aj ,--- ,am=am, where ar=#r,6r=prareanarbitrary setofinitial values of</r,pr. Furthermore, S(q l9- ,</m, 1? ,am,shallbecontinuous, together with itsderivatives : dS dS d*S dqr' dctr dqrdoL8' intheneighborhood ofthepoint (ar,ar,tQ),and !,,, , ,m Finally, thefunction V=S(q r,<xr,t)shall satisfy Jacobi's Equa- tionA). Theproof oftheexistence ofsuch afunction Sisgivenbythe theorem of2bysetting (27) cofo ,- ,qm)=a^i++<*mqm. Fornowthecorresponding solution ofJacobi's Equation A): (28) V=S(q ly--,qm,an ,aw,0, hastheproperty that Moreover, theJacobian determinant (26)isseen tohave thevalue 1,andwearethrough. Wehave obtained thisexistence theorem forthefunction bymeans ofthetheorem of2,theproof ofwhich isbased on 472 APPENDIX C characteristics. But itmight equally wellhave been derived directly from theexistence theorem which isusually referred to asCauchy's Problem, 5below, provided wearewilling toassume thatH(q r,pr,isanalytic inthepoint (gv,pr,tQ). 4.Jacobi's Equation:H,Independent of t.Consider the casethatHdoesnotdepend on t: (1) H=H(q l9--,<?m,plf-,pm). Jacobi's Equation nowtakes theform : A/, * A')_+ Weseekthespecial solution (2) V=S(q l9-,?, -,ob,0 defnanded in 3. Itispossible toobtain Sasfollows. Asolution ofA')canbe found bysetting (3) V=-ht+W, wherew=w(qi,..-,?) doesnotdepend on t.ThenWwillsatisfy thepartial differential equation: r,x v( dW 8W\ , C)ff^,...,^-.,... ,__)-*. The derivatives ofH(q l}- ,qm,pl9 ,pm)with respect to theprarenot all0.*Let * Then theequation (5) fffe,''' ,?m,Pi,'' ,Pm)=ft canbesolved forpl: (6) Pi=x(7i,",?, A,P2i'Pm), andC)isequivalent totheequation: ^ dW .dW dWc)= *Either because ofthehypotheses ofChap. XI, 3orbecauseHisapositive definite quadratic function ofthep/s. CHARACTERISTICS OFJACOBFS EQUATION 473 Let (7) W=W(qi,*--,q m,h,a 2,-,) bethat solution ofC)which reduces to (8)W=a>(<7 2, ,fr) when ql=q^. If,now,weseth= !,thedesired function S isgivenbytheequation: (9) S(q l9 ,qmy!,--, a,= -o^+TF(ft, ,qm,al9 ,am). For, dW Hence pr=ar,r=2, ,m, areasystem ofequations which canbesolved forthear,r=2, ,m,andc^isgivenby(5). Itremains toexamine theJacobian, Since (J1S\ ={0, \dardq a/ I1, wehave only toshow that0, r?s r,s=2, ,m,r=s Now, (12)xi isgivenbytheequation (5): Hence with theaidof(4) andtheproofiscomplete. 474 APPENDIX C Summary ofResults. Tosum up,then :thesolution of Hamilton's Equations B)isgivenbytheequations: 98 dW8S (14) or (15) Theararedetermined interms ofthe(q,p)bytheequations: r=2,--- ,m; (16)=Pr, The#rarenowgivenby(15)onsetting qr=qrQandsubstituting forarthevalue givenby(16). TheFunction W.The total differential equations which deter- mine thecharacteristics ofC)are : (17) Since wehave (18) or (19)dqrdW -dp r ^L2 SprPr8pr dqr m_. dPr~9" dW Wr-1, m. Ifffisahomogeneous quadratic function ofpM ,pm,then (20) W=2Hdt+WQ. CHARACTERISTICS OFJACOBFS EQUATION 475 6.TheCauchy Problem. LetF(xi, ,xn,z,y\, ,yn) beanalyticinthepoint (a,c,b)=(a1? ,an,c,bl9 ,&n) and letdF/dx^^ there. Consider the partial differential equation: fa 2Z\ n,...,x n,2,,..-,)=o. Let beanalyticinthepoint (a2, ,an)and let ^( 2,,an)=c, ^*(oa, ,o)=6*,A;=2, ,M. Then there exists oneandonlyonefunction, Z=^(Xj,- - ,Xn), which isanalytic inthepoint (a^ ,an),has ^(i, ,On)=c, ^/(ai>''' >n)==6/, y=1, ,n, and satisfies thegiven differential equationintheneighborhood ofthepoint (o1; ,an). This istheexistence theorem known astheCauchy Problem. Cf.Goursat-Hedrick, Mathematical Analysis, Vol. II, 446. APPENDIX D THEGENERAL PROBLEM OFRATIONAL MECHANICS I PATHS Consider* asystem ofnparticles mt:(x,y^Zi)acted onby forces (Xi, Yi,Zi). Their motion isgoverned byNewton's Law: A) niiXi Xi niiiji=Yi mi'Zi=Z,- Here areQndependent variables, thex,,yitz,-,X^Yi}Zt,con- nected by3nequations. Theproblem ofmotion istofind3n supplementary conditions whereby these 6nvariables willbe determined asfunctions ofthetime, t,andsuitable initial condi- tions, and tosolve forthese functions. Eachmember ofthe family which forms thesolution, namely thecurve : Xi Xi\l)) y\ yi\'/j Zi Zi\tj Xi=Xi(t), Yi=Yi(t), Zi=Zi(t) determines acurve : Xi=xt(t), yi= 2/t(0, *<=*;(0 (2) _.__.__, Xi~ ~dT'Vi~ "df' Zi" ~dt inthe(6n+l)-dimensional space ofthe(a;,-, 2/t-,2,ft,?/, ,0> andsuch acurve iscalled apath. Obviously thepaths (2) stand inaone-to-one relation tothecurves(1). Theproblem ofmotion assoformulated transcends thedomain ofRational Mechanics. Inorder torestrict ourattention tothe latterfield,wenow laydown thefurther postulate which, be itnoted, isnotsatisfied bycertain systems which occur innature, viz., certain systems inwhich electro-magnetic phenomena are present. *Thefollowing treatment istheresult ofajoint study oftheproblem by Professor Bernard Osgood Koopman andmyself. 4.76 GENERAL PROBLEM OFRATIONAL MECHANICS 477 POSTULATE I.DYNAMICAL DETERMINATENESS. Inagiven dynamical system, when6n+1constants (x^, yi() ,zt-,Xi, 7/,, z,-,J) arearbitrarily assigned,notmore than onepath (2)exists which passes throughthispoint: THEDOMAIN ZXThose points (#;,yi}zi}x,#;, ,-, ofthe (6n+1)-dimensional space, through which paths pass, con- stitute thedomain D.Thisdomain may consist oftheentire space, orofaregion ofit;butingeneral neither ofthose things willbethecase. Itisapoint set,concerning theconstitution of which weneedmake nohypothesisatthepresent moment. It willberestricted bylater postulates. THEOREM I.The variables Xi}Ft,Ziareuniquely determined inthepoints ofD: Xi=Xifa,y,,Zj,xhyitZj,t) (4) Yi=Yi(x/, y,-,Zj,Xj,fa,Zj,t) Zi=Zi(Xj,yhZ3;Xj, 7/y,Zj,/) where(xj,yhzitXj, y,-,z},t)isanypoint ofD. For,through each point ofDpasses apath, unique invirtue ofPostulate I.Along agiven path Xi,Yi,Ziareuniquely deter- mined asfunctions of tbyA).Hence Xi,Yi,Ziareuniquely determined atthepoint ofDinquestion, butnotingeneral in points notlyingonZ). THEDOMAIN R.Inthe(3n+l)-dimcnsional space ofthe variables (x^ yi,z^f)those points which participateinpaths form apoint setR,whichmaybedescribed astheorthogonal projection onthisspace ofthedomain D.Inparticular Rmay consist ofthewhole space, orofa(3n+l)-dimensional region init.Butingeneral neither ofthese things willbethecase. LetPbeapointofR.ToPthere corresponds atleastone path given by(2).Thepoints (xiyy^Zi,t)represented bythe first lineof(2),namely: (5) Xi=Xi(0, Vi=Vi(0, Zi=^(0, allbelong toR.Hence thecurve (5)lieswhollyinR. 478 APPENDIX D Consider anarbitrarylinethrough P,butnotperpendicular totheaxis of t.Let itsdirection components bea,-, ft, 7,-,K, where K^0.Theremay beapath corresponding toP,such that atthispoint Xi:yi:Zi=<*<:ft:7*. When this isnotthecase, not alllines through Pcorrespond topaths, and socertain relations between thedirection com- ponents (ca,pi,7,,K)must exist. Thusweareledtoasecond postulate. POSTULATE II.Thedirection componentsatpoints ofR,towhich paths correspond, aregiven bytheequations: (6) 2}(A3iai+B8ift+Cai7t)+D8K=0, s=1, ,cr, ii whereA8i,B8i,C,,D8arefunctions* of(Xi,yl,zt,t)such that the rank ofthematrix : All'''AlnBn'Bin Cll'Cln (7) is<r. Since along acurve(5) i~~ ft~~ 7i;~~ K atthepoint P,itfollows that n B) ^(AtiXi+B8iyi+C,Zi)+D8=0,s=1,- ,<r. These equations form anecessary and sufficient condition for (, yi,Zi)if(Xi,yiyziyxiyyitzitt)istobeapoint ofD. Itmayhappen that thesystem ofEquations B)(asystem of Pfaffians) admits certain integrals: where therank ofthematrix : *Throughout thewhole treatment, thecontinuity ofthefunctions which enter, andtheexistence andcontinuity ofsuch derivatives asitmaybeconvenient touse, areassumed. GENERAL PROBLEM OFRATIONAL MECHANICS 479 (8) is I.Since thesystem B)may obviously bereplaced byany non-specialized linear combination ofthese equations, itisclear thatEquations B)maybesochosen thatthelast Iofthem are : (9)dt=0, Theconstants Ckcome tousasconstants ofintegration inthe system ofintegrals C)ofthePfaffians B).They contribute toward determining theparticular dynamical system weare defining, different choices oftheCkleading toseparate dynamical systems. They arenottobeconfused with constants ofintegra- tion that aredetermined bythe initial conditions within apar- ticular dynamical system. Holonomic andNon-Holonomic Systems. If,inparticular, I= or,Equations B)canbereplaced byEquations C)andthus become completely integrable. Thedynamical system weare inprocess ofdefining isthen said tobeholonomic. But ifthere remain a I=/*>Equations B),which then arenon- integrable, thesystemissaid tobenon-holonomic. Equations B) shallnowbereplaced bythe firstpofthem, andEquations C): B') C)(Aaii+Baiyi+CaiZi)+Da=0,a=1, ,/*; = /x+ I. II THEFORCES. D'ALEMBERT'S PRINCIPLE TheforceXt,Y^Ziwhich actsonmismadeupingeneralof aforce-X'iyF',Z\which isknown interms ofzt-,y^zt-,xiyfa,zitt, and offurther forces XJ/, F{,-,Z/,wherej=1,2, ,p,the componentsofthese latter forces being wholly orinpart un- 480 APPENDIX D known. Denote theunknown components byS19 ,SK. Then ourpostulates must provideforenough known equations between theS'sandtheXi,y^zt-,Xi, ?/,-, 2,-,ttomake possible the elimination ofthe$'sbetween these equations andEquations A), with theresult that theequations thus obtained, combined with Equations B')andC),willjust suffice todetermine xify*,z>asfunc- tions oftandtheinitial conditions. Weproceed tothedetails. D'ALEMBERT'S PRINCIPLE Inpractice theequations which theSi, ,SKsatisfy are usually linear. Ourproblem shall berestricted tosystems which obey thefollowing postulate. POSTULATE III. Theforce Xi,YitZ{isthesumoftwoforces: (10)'Xi=xi+A7, Yi=y;+17; Zi=z;+z;, where X\, FJ,Z\areknown interms ofthecoordinatesa:,-,y^ z,-, Xi, ilijZi,tfanarbitrary point ofD,amiwhere (11) ijxrfc +rrih.+z; $-,=(> i=l forallfi, r/i, f,-such that (12) 2JAbb+Shu+C'pift=0, j8=1,-- ,i/. i=l Here, A'ai,B'ai,C^iareknown functions oftheabove#,-, 7/t-,z^ Xi9tit* *i) t>andtherank ofthematrix : (13) v.Conversely, when Equations (12)aresatisfied. Equation (11) Turning now toEquations A),wehavewhat isknown asthe General Equation ofMechanics : (14) =0, GENERAL PROBLEM OFRATIONAL MECHANICS 481 where,-,T/,-,fiare3narbitrary quantities. Under thesanction ofPostulate III.thisequation canbereplaced bythefollowing: (15)2)(mt i~XI) fc+(m<fr-Ffl*+(w2-Z')f-0, <-l where,-,?7t,f,-areany3nquantities whichsatisfy thecondition (12). Multiply the0-thequation (12)byX^andsubtract the re- sulting equation from (15): (16)2)(mtfi-X,'-5)A'fiiljh +(mtfr-Y'<-%B'^n <=1 0-1 /3-1 +(mizi-Z'i-^C'ei\p)t<=0. 0=1 Suppose fordefiniteness that thedeterminant whose matrix consists ofthe first vcolumns ofthematrix (13)is^0.Then theX'scanbesodetermined that thecoefficients ofthe first v ofthequantities 1, ,n,i?i,- ,*7n,fi, ,fnin(16) will vanish. Substitute these values ofX1? ,\vintheremaining coefficients of(16). Thus anew linear equation inthe,-,17,-,f$- arises,inwhich only thelast3n vofthese quantities appear. Butthelatter arearbitrary. Hence each coefficient must vanish. The3n vequations thusobtained express theresult ofelim- inating theunknown Sl ,SK,i.e.theX*,Yf,Zf,from the problem. They contain only a?,yiyziyXi,#,z^xityiyzi}t,and canbewritten intheform : E)%(EyiXi+Fyiy<+Gyi 2,)+#*=0,7=1,. ,3n- v, <-i where the coefficients EyilFyi,Gyi,Hyareknown functions of Zt, IJi,Zi,Xi,y*,Zi,tateach pointofD. Equations E)andB)form anecessary condition forthefunc- tionsXi(f), yi(t), Zi(t)which define apath (2).Hence ifP : (xfjyfjZiQ ,XiQ ,yfjzP,tQ)isanarbitrary point ofZ),Equations E) andB)admit asolution having asitsinitial values thecoordinates ofJP .Furthermore, byvirtue ofPostulateI.,this solution is unique. Wehavenowarrived atacomplete analytical formula- tionoftheproblem,forwecanretrace oursteps. Let 482 APPENDIX D beacurve lyingonC)andsatisfying B').Then a)gives risetoa curveTwhich liesonD.Consequentlyallthecoefficients inB'), (12),andE)aredetermined inthepoints ofa).Leta)also satisfy E). Since E)holds,itfollows thatXD ,X,canbedetermined so astomake each parenthesisin(16) vanish. Next, determine X*9Yi fZ*from these X'sbytheequations: X*=2*Afii X/3, Y*= jBfii Xj3, Z*= jCfti X/3. 0=1 /3=i 0=1 These quantities satisfy (11)and (12).Onsubstituting them in (10), values ofXt,F,Z{areobtained forwhich A)istrue,because each parenthesis in(16)vanishes, andsoTisapath. Butthere isonlyonepaththrough anarbitrary pointPofD.Hence a)is unique. Retrospect. These Postulates complete theformulation ofthe class ofproblems inRational Mechanics which wesetoutto isolate. The rolewhich d'Alembert's Principle*playsistwo- fold. First, itrequires that therelations between theunknown S\,'m 9&*shallbelinear. Secondly,itperforms theelimination byatechnique such that themultipliers {,-, T/;,fcanalways be interpreted asvirtual displacementsofthesystemofparticles m : (xt,yi,Zi)bysetting (17) bXi=fc byt= ??i, bZi=ft. Remark. Ingeneral there isnorelation between thecoefficients A9i,B9i,C8iofEquations B)andtheAp itB^i, C'^ofEqua- tions (12). Hence thevirtual displacements to,-, 6r/, faiof(17) willnotcoincide save astoinfinitesimals ofhigher order withany possible displacement A# t,Ay iyAs*duetoanactual motion ofthe system intime A/. Inasub-class ofcases ithappens, however, thatthe4, J3, t,Ci inB)andtheApt,B'ftiyC'piin(12)arerespectively equal toeach *Historically d'Alembcrt's Principle took itsstart intheassumption ofacon- dition, necessary and sufficient, thatasystem offorces, acting onasystem of particles, beinequilibrium, namely, that thevirtual work corresponding toa virtual velocity benil.When asystem offorces notinequilibrium actsonasystem ofparticles, theformer canbereplaced byasystem offorces inequilibrium through theintroduction of"counter effective forces" or"forces ofinertia" (sic),andthusd'Alembert arrived attheGeneral Equation ofDynamics. GENERAL PROBLEM OFRATIONAL MECHANICS 483 other. Buteven so,iftheDaarenot all0,thevirtual displace- ment willnottally save astoinfinitesimals ofhigher order with anypossible actual displacement. Finallyitcanhappen that, inaddition, theD,are all0. Then thevirtual displacement corresponds toapossible displace- ment. But this isavery special, though highly important, case. Let (18)Ill LAGRANGE'S EQUATIONS ',0m, -,ffm, -,ff, where therank ofthematrix : (19)dq, isw,andwhere, moreover, theregion ofthe(x,y^z,0-space which corresponds tothepoints (qly ,qm,f)inwhich /, ??,,^i aredefined, atleast includes thepoints ofR. LetTdenote thekinetic energy: Then, foranarbitrary choice oftheqr,since Tgoesover intoafunction ofgr,qrtt: T=T(q r,qr,t). Conversely,if,^, ,areanysetofnumbers forwhich these equations aretrue, theqrareuniquely determined. 484 APPENDIX D Consider apath (2). Since thepoints #=Xi(t), yi=yi(t), Zi=Zi(0alllieinR,acurve ofthe(qr,0-spaceisthus defined : (20) qr=qr(t), r=1,...,m. Forthepath inquestion wehave :__ dtdqr Sqr~Q" ' where 8xi4-Y8yi4-78z<+i+ Equations (21)arealways trueunder theforegoing restrictions. Theywillbesufficient todetermine themotion ifthesystemis holonomic and if (22) r=1, ,m.Forthen 'dXi-dyi-'dzr" andthusQrisknownfirst, interms ofx<,y^zltxiyyiyziyt,and sofinallyinterms ofqr,qr,t.That Equations (21)canbesolved forql9 ,qmfollows from thefactthatTisapositive definite quadratic form intheqly ,qm. Thismeans interms oftheforegoing treatment thatasuitable choice ofthemultipliers ,,7?t,ftinEquation (12)is : \ vhere the8qrarearbitrary. Equations C),ifpresent, are all satisfied identically when the Xi,y^Ziareexpressedinterms of tteqrand tbyEquation (18)." Equations B')arenotpresent inthe problem. Thesystem is,tobesure, holonomic, but itisnotthe onlycase inwhich this isso. TheGeneral Case.Weassumed inPostulate III.thatX't,F{,Z( are fleefrom the /S's,and that theS'scoincide with the X*,Yf)Z*.Wenowdivide theS'sintotwocategories: t)asub-set, denoted anew byXf,F*,Z*,which fulfil the former requirements (11), (12), (13); GENERAL PROBLEM OFRATIONAL MECHANICS 485 ii)asecond sub-set, R19 ,RT>onwhich theXI,Y( 9Z\shall nowdepend linearly. Thus Equation (21) holds, whereQrisgivenby(23). Let (25) Qr=Qr+Q?,r=1,- -,m, where Q'risknown interms ofsuch values ofqr,qrjtascorrespond topointsofD. Intheparticular case before us,namely, Equations (18),it canhappen thattheequation: (26) Qi* *-!++Qi*= istrue for allvalues ofthemultipliersirrforwhich thefollowing equations hold : (27) Oft!Tt+---+a'tmKm=0,=1, ,vl9 wheren'prdepends onvalues ofqr,qr,t,which correspond topoints ofD,andtherank ofthematrix : / (28) isvl' 9andconversely, when Equations (27)hold, then (26) istrue. Equations B'),C)goover inthepresent caseinto : Bq) al*ll 4""""~T~ttam^wt ~f"&a==0,Oi==1, ',/Zj 9 Cq) <&k(fir)==T*k) k=1,** ',lu where^^n;^gZ,andwhere therank ofthematrix : (29) is/zt;therank ofthematrix : (30) Fg~ being ^. 486 APPENDIX D Finallyitcanhappen thattheRv ,Rrcanbeeliminated between these equations, thus leaving asystemofequations be- tween theqrjqr,qr-Suchasystem yields aunique solution, cor- respondingtoeachpathofthedynamical system withwhich we setout.Thus thedynamical problemiscompletely formulated by means ofLagrange's Equations. Alloftheforegoing assumptions areintentative form"It mayhappen"Attheoneextreme, thechoice ofthefunctions fit9*ticanalways bemade sothat allthese things dohappen ; fortheqrcan, inparticular, beidentified with thexi}yi}z<: Attheother extreme, mmaybechosen sosmall thatEquations (21), though true, willcontain unknown functions which cannot beeliminated namely, theRlt ,RT.Thismeans that, for such' achoice ofthefunctions (18), the t-,rjf,ft-asgiven by (24)aretoorestricted. The &, rjiff.ofEquations (11)arequan- tities which must beable totakeonevery setofvalues which satisfy (12). The.-,77,-, {*which here figure, given by(24), are notfreeunder thecondition (24),butarcunwarrantably restricted by(24). Inagiven problem thedesideratum usually is,tochoosemas small aspossible, subject totherequirement thatthesame degree ofelimination ofthe'S's through (24) shallhave been attained, asifEquations (11)and(12)hadbeen used. IV NOTES Consider thedynamical system that consists ofabead sliding onafixed circular wireandacted onbynoother forces than the reaction ofthewire. Equations A)take theform : mx=X,my=Y,mz=Z. Letthewirebeacirclewhose axis istheaxisof2.ThenEqua- tionsB)become : xx+yy= B)o Thissystem ofPfaffians iscompletely integrable: fx2+ 2/2=a2 I z=c GENERAL PROBLEM OFRATIONAL MECHANICS 487 Different values oftheconstants ofintegration, aandc,give different systems ofpaths, (2) ;butapath ofonesuchsystem hasnopoint incommon withapath ofasecond system. Proceeding totheforces weseethatZ=0,since 3=0,and sowehave atwo-dimensional problem. TheSmooth Wire. Assume firstthatthewire issmooth. Then thereaction isalong theinner normal. X=X*, Y=Y*, andX*+F%= provided ^+yt]=0. Turning toLagrange's Equations wesetm=1andtake x=acos#, y=asinq. Then =X*(-asin?)+7*(acos?) =;T(-2/) +F**=0. Hence, finally: and itremains merely tointegrate this differential equation. TheRough Wire. Suppose, however, thewire isrough. Let q>0.Then X*=Rcosq+pRsinq Y*=RsinqpRcosq. Lagrange's Equation: Jt~dij~ ~dq*Q' isstilltrue. But 488 APPENDIX D (aresult atonce obvious) andLagrange's Equation becomes : Wehave notenough equations tosolve theproblem. This is thecase inwhich Lagrange's Equations aresaid to"fail" orbe "inapplicable." The failurelies,notinLagrange's Equations, butinamisuse ofthem.Weshould takem=2.Letusfirst treat theproblem, however, bythemethods ofPartsI.,II.,before Lagrange's Equations were introduced inPart III. Here, then, r72rrm=X*=-RcosB+Rsin at* m-j%=F*=-juftcos0-Rsin0. ut Equation (11)nowtakes theform : x**+7*77=0, or (Rcos+p,Rsin6)+(-nRcos6-Rsin0) rj=0, or,finally, (x+y)+(MZ-y)v=0, andthis istheform ofEquation (12). Hence wemaytake =IJLX+y, rj=-x+ny. Onsubstituting these values intheGeneral Equation of Dynamics wehave : / \d2x ,t.^d?y ~ (i*x+V)-fc+(-x+/iy)~=0. Thisequation andEquation C),namely: x*+y*=a2 , provide uswithtwoequations fordetermining xandyasfunctions ofI,andthus theproblemisreduced toapurely mathematical problem indifferential equations. Observe, however, that the virtual displacement used inthissolution : GENERAL PROBLEM OFRATIONAL MECHANICS 489 isnotonewhich iscompatible with theconstraints,i.e.thecircular wire even save astoinfinitesimals ofhigher order than e.It corresponds toadisplacement along alineatright angles tothe resultant ofRandpR. Turning now toLagrange's Equations letuschoose qlandqz asthepolar coordinates ofthemass m.Then Lagrange's Equa- tions (21)become : (31) 7<S Now, Equations Cq)herebecome : Cfl)r=a. OntheotherhandEquation (26): herebecomes : 0, andthusEquation (27)takes theform : *"l+M^2=0- If,then,weset : TTi= /i, ^2 Equations (31)andCq)yield: dO* ,d*0-.ma_ 7ri+ma_ or #8 <W and itremains merely tointegrate thisequation. Asafurther illustration oftheuseandabuse ofLagrange's Equations maybementioned theLadder Problems ofpages 322 and323. INDEX Absolute unitofforce, 52 ofmass, 79 Absolute value, 24 Acceleration, 50,52,287 d'entrainement, 288 ofgravity, 56 Vector, 90 Addition ofvectors, 4 d'Alembert's Principle, 345,480 Angle offriction, 10 Angular velocity, Vector, 170,285 Appell, 225, 244,246,307,337 Areas, Lawof,108 Atwood's machine, 134 Axes, Principal, ofacentral quadric, 194,196 Rotation ofthe,454 B Bending, K,226 Centre of,234 Billiard ball,with slipping, 143,237, 314 without slipping, 145,240,314 Blackburn's pendulum, 184 Bocher, 334 Bolza, 372,375 Brah6, Tycho, 115Centripetal force, 102 Centrodes, 159 Space andBody, 174 Change ofunits, 76 Characteristics ofJacobi's Equation, 466 Charlier, 437 Check ofdimensions, 79 Coefficient offriction, 10 ofrestitution, 271 Componentofforce, 2 ofvelocity, 87 Compound pendulum, 130 Cone, Body, Space, 213 Conservation ofenergy, 256 Conservative field offorce, 255,258 Constrained motion, 95 Constraint, Forces of,315,325 Contact transformations, 390,399 Particular, 403 Coordinates, Cyclic, 430 Generalized orintrinsic, 297 Normal, 335 Coriolis, 288 Couples, 25,29,34,37 Composition of,31 Nil,31 Resultant ofn,31 Vector representation of,38 Cyclic coordinates, 430 Canonical equations, 338,395 transformations, 389 Carathe'odory, 381,445 Cart wheels, 241,314 Cauchy problem, 475 Central force, 108,379,427,434 Centre ofbending, 234 Centre ofgravity, 26,27,42 Motion ofthe,120 Centre ofmass, Motion ofthe, 123 Centrifugal force, 101 field offorce, 106,291 oilcup,1055,Definition of,356 Critique of,379 Dancing teacup,165 Decomposition offorce, 2 Dimensions, Check of,79 Direction cosines ofthemoving axes, 216,454 Dyne, 56 120,E Elastic strings, 58 Elasticity, Perfect, 272 Electromagnetic field,254 Ellipsoid ofinertia, 192 491 492 INDEX Energy, Kinetic, 75,260 Conservation of,256 Potential, 255 Work and, forarigidbody, 266 Equation, Solution ofatrigonomet- ric,12 Fundamental, 367 ofmoments, cf .Moments Equilibrium ofcouples, 31 ofadynamical system, 330 offorces inaplane, 32 offorces inspace, 36,41 ofnforces, 9 ofthree forces, 5 ofarigid body, 26 Problem, 413 Escalator, 265 Euler's Angles, 214,215 Dynamical Equations, 210, 325, 352 Equations, 359 Geometrical Equations, 214 Field offorce, 253 Centrifugal, 291 Gravitational, 254 Electromagnetic, 254 Force,1 Absolute unit of,52,55 Central, 108,379,427,434 Centrifugal, 101,291 Centrifugal field of,291 Centripetal, 102 Component of,2 ofconstraint, 315,325 Equilibrium ofthree, 5,43;cf. Equilibrium Fieldof,253 function, 253 Moment ofa,28 Parallel, inaplane, 21;inspace, 36 Parallelogram of,2 Polygon of,7 Triangle of,4 Foucault Pendulum, 292 Friction, 9 Angle of,10 Coefficient of,10 Problems in,19 Function, Lagrangean, 338 Hamiltonian, 342 Fundamental equation, 367Generalized coordinates, 297 Geodesies, 308 Goursat, 468 Gravitation, Motion under theattrac- tion of,69 Law ofuniversal, 116 Gravitational constant, 116 Gravity, Accelerationof,56 Gyration, Radius of,129 Gyroscope, 217 Intrinsic treatment ofthe,225 H Hadamard, 224 Hamilton's Canonical Equations, 338, 395 Proofof,342 Solutionof,410,432 Reduction of,totheEquilibrium Problem, 411,413 forconstant energy, 411,420 Hamiltonian Function, 342 Hamilton's Principle, 371 Integral, 356 Integral aminimum, 381 Harmonic Motion, Simple, 64,415 Haskins, 228 Hedrick, 468 Helical motion, 168 Hertz, 244 Holder, 370 Holonomic, 313,479 Hooke's Law, 59,74 Huntington, 208 Huygens, 133 Impactofparticles, 270 Oblique, 274 ofrigid bodies, 277 Impulse, 271 Inertia, 118 Ellipsoid of,192 Moment of,128, 137,191 Product of,191 Instantaneous centre, 154, 157,160 axis, 168,173 Integral invariants, 392 Integral ofkinetic energy, 362 ofaperiodic function, 464 ofrational mechanics, 360 INDEX 493 Internal work, 258 Intrinsic treatment ofthegyroscope, 225 coordinates, functions, 297 equationsofthegyroscope, 236 Invariable lineandplane, 201 Inverse problem, 114 Isolate theSystem, 102 Jacobi's Equation, 410, 468,472 Characteristics of,466 Integral aminimum, 386 Principle ofLeast Action, 377 K,bending, 226 Rater's pendulum, 133 Kemble, 234,263 Kepler's laws, 115 Kinetic energy, 75 ofarigid system, 166,260 Integral of,362 Klein-Sommerfeld, 236,246 Koopman, 123,476 Kreisel, 236,246 Ladder, 147,322, 323,353 Lagrange's Equations, 299, 304, 312, 348,350,482 multipliers,194,316,375 Principle ofLeast Action, 374,377 Solution of,Equations, 326 Lagrangean function, 338 Lagrangean integral, 372 Lagrangean integral aminimum, 381 Lagrangean system, 338 Law ofareas, 108 ofnature, 109,116 ofuniversal gravitation, 116 ofworkandenergy, 258 Least Action, 374,377 Leval, Turbine of,247 Lissajou's curves, 182,190 M Mass, Absolute unit of,79 Moments about centreof,139,205 Motion ofthecentreof,120,201 Notion of,118Material point, 50 Maxwell, 119 Moment ofaforce, 28 ofacouple, 29 ofavector, 37 ofalocalized vector, 197 ofmomentum, 197,205 ofinertia, 128, 137,191 Theorem ofMoments, 127,200 Moments about thecentre ofmass, 139,205 Moments about theinstantaneous centre, 207 Moments about anarbitrary point, 205,208 ofavector about aline,40 Momentum, 50,201,350 Momentof,197,200,350 Motion under theattraction ofgravi- tation, 69 Newton's Laws of,50 Simple Harmonic, 64,415 Constrained, 95 Simple Pendulum, 97 Spherical Pendulum, 306 inaresisting medium, 81 inaplane andinspace, 86 ofaprojectile, 93,424 onasmooth curve, 99 onaspace curve, 100 ofthecentre ofgravity, 120 ofspace, General caseof,175 about afixed point, 212 Moving axes, 172,216 curve, 299 surface, 303 N Newton's Laws ofMotion, 50 Second Law, 92,290 Non-holonomic, 244,313,479 Normal, 9 Principal, 90 Normal coordinates, 335 Nulvector (ornilvector), 5,447 couple, 31 Numerical value, 24 Operator, Symbolic vector, 254 Orbit ofaplanet, 111, 113,435 Oscillations, Small, 333 Osculating plane, 90,92 494 INDEX Parabolic motion, 93 Parallel forces inaplane, 21,23 inspace, 36 Parallelogram offorces, 2 Particle, 50 Pendulum, Blackburn's, 184 Compound, 130 Foucault, 292 Rater's, 133 Simple, 97,419 Spherical, 306 Torsion, 139 Periodic time, 111 Perturbations, 440 Poincar6, 392 Poinsot, 213 Potential, 253 energy, 255 Poundal, 56 Principal axes ofacentral quadric, 194 Principle ofthemotion ofthecentre ofmass, 123 ofmoments, 139 ofmoments with respect tothe centre ofmass, 205 d'Alembert's, 345,480 Hamilton's, 371 ofLeast Action, 374 Variational, 370 Product ofinertia, 191 Projectile, Motion ofa,93,424 Quadric, Central, 194 Radius ofgyration, 129 Eankine, 10 Rectilinear motion, 49 Relativevelocities, 177 Resistance, Graph ofthe,84 Resisting medium, Motion ina,81 Resultant, 2 ofparallel forces inaplane, 21,23 ofncouples, 31,36 ofnforces inaplane, 32 ofnforces inspace, 36,38 axis,39 oftwovelocities, 87 Riemann, 381,445Rotation oftheaxes,454 about afixedaxis, 127,136 ofaplane lamina, 139 ofarigidbody, Chap. VI Routh, 134,202,225,246,337 Ruled surfaces, 176 S <r,197 Evaluationof,forarigid system. 208 Transformationof,202 Sabine, 190 Sand tunnel, 185 Ship's stabilizer, 236,247 Simple Harmonic Motion, 64,415 Simple pendulum, 97,419 Smalloscillations, 333 Smooth curve, 99 Solution oftrigonometric equation, 12 Hamilton's Equations, 410,432 Sommerfeld, 236,246 Space curve, Motion ona,100 Spherical pendulum, 306 Stabilizer, Ship's, 236,247 Stationary, 359 Strings, Elastic, 58 Symbolic vector operator, 254 Tautochrone, 98 Tennisball,282 Top, 220,438 Torque, 29 Torsion pendulum, 139 Transformation of<r,202 Contact, 390, 399,413 Canonical, 389 ofHamilton's Equations bycon- tacttransformations, 400,413 Translation, 159 Transmissibility offorce, 22 Triangle offorces, 4 Trigonometric equation, 12 theorem, 44 Twobody problem, 114,879,427,434 TychoBrah6, 115 Tyndall, 166 Units, Absolute, 52,55,79 Change of,76