Synge Griffith Principles of Mechanics 2nd 1949 no TOC

w a:OU158934>m -7] OSMANIA UNIVERSITY LIBRARY CallNo.^3 / OsmaniaUniversity Library CallNo. Accession No. Author Title Thisbookshould bereturned onorbefore- thedate last marked below. / PRINCIPLES OFMECHANICS PRINCIPLES of MECHANICS BY JOHN L.SYNGE Professor ofMathematics Carnegie Institute ofTechnology AND BYRON A.GRIFFITH Assistant Professor ofMathematics University ofToronto SECOND EDITION NEWYORK TORONTO LONDON McGRAW-HILL BOOK COMPANY, INC. 1949 PRINCIPLES OFMECHANICS Copyright, 1942, 1040,bytheMcGraw-Hill BookCompany, Inc.Printed in theUnited States ofAmerica. Allrights reserved. Thisbook, orparts thereof, may notbereproduced inanyform without permission ofthepublishers. PRINTED BYTHEMAPLE PRESS COMPANY, YORK, PA. Lesavant doitordonner; onfaitlascience avec desfaitscommc unomaison avcc despierrcs, inaisuneaccumulation defaits n'est pasplus une science qu'un tasdepicrresn'est une maison. HKNRI POINCARE: PREFACE TOTHESECOND EDITION This edition differs innoessential wayfrom the first. The principal revision occurs inChap XIII, where theaccount of themotion ofaparticle inanelectromagnetic fieldhasbeen completely rewritten. Thetreatment ofprincipal axes ofinertia inChapXIhasbeen amplified, andsome revisions have been made inthetreatments ofFoucault's pendulum, thespinning projectile, andthegyrocompass. Theemphasis onunits and dimensions hasbeen increased bytheinclusion intheearlier part ofthebook ofafewshort paragraphs, with references tothe Appendix, where these matters arediscussed indetail. Afew additional exercises have been inserted, andnumerous minor corrections havebeenmade. Wewish tothank allthose readers whohave contributed totheimprovement ofthissecond edition bytheir suggestions, arid, inparticular, Professors L.Infeld, A.E.Sehild, andA.Weinstein. JOHN L.SYNGE BYRON A.GRIFFITH PITTSBURGH, PA. TORONTO, ONT. July, 1948 PREFACE TOTHEFIRST EDITION Inasense this isabook forthebeginner inmechanics, butin another sense itisnot.From thetimewemake our firstmove- ments, crude ideasonforce, mass, andmotion takeshape inour minds. Thisbody ofideas might bereduced tosome order at high school (ascrude ideas ofgeometry arereduced toorder), butthat isnottheeducational practice inNorth America. There israther anaccumulation ofmiscellaneous facts bearing onmechanics, some mathematical andsome experimental, until astate isreached where thestudent isindanger ofbeing repelled bythesubject, asachaotic jumble which isneither mathematics norphysics. Thisbook isintended primarily forstudents atthis stage. Theauthors' ambition istoreveal mechanics asanorderlyself- contained subject. Itmaynotbequite sologically clear aspure mathematics, but itstands outasamodel ofclarity amongall thetheories ofdeductive science. The artofteaching consists largelyinisolating difficulties andovercoming them onebyone,without losing sight ofthe main problem while attending tothe details. Inmechanics, themainproblemistheproblemofequilibrium ormotion under given forces thedetails aresuch things asthevector notation, thekinematics ofarigid body, orthetheory ofmoments of inertia. Ifwerush straight atthemain problem, webecome entangled inthedetails andhave toretrace oursteps inorder to dealwiththem.If,ontheother hand,wedecide tosettle all details first,weareapttofindthem uninteresting because we donotseetheir connection withthemain problem. Acompro- mise isnecessary, and inthisbook thecompromise consists ofthedivision intoPlane Mechanics (Part I)andMechanics in Space (Part II). These titles must, however, beregarded only asrough indications ofthecontents. Part Iincludes some of theeasier portions ofthree-dimensional theory, while Part II contains anintroduction tothespecial theoryofrelativity, with mechanics inonlyonespatial dimension ! X PREFACE TOTHEFIRST EDITION Thereis,ofcourse, nothing novel inregarding plane mechanics asthepreliminary field; but itisrather unusual todivide the subject inthisway inasingle volume, oreven inasequence of volumes. Ithasmade thetask ofwriting moredifficult, but theauthors have felt itworth while. Many ofthemost interest- ingresults instatics anddynamics belong totheplane theory, and itisunfair todeny thereader access tothem until hehas mastered themore elaborate technique required forthree dimensions. Part Iiscomplete initselfandmight beused asatextbook inplane statics anddynamics, withsome excursions into three- dimensional theory. Vector notation isintroduced, butused sparingly. Thereader should haveafairknowledge ofcalculus, elementary differential equations, andsome analytical geometry. Practical experience inphysicsisnotessential butvery desirable; mechanics isatrootaphysical subject andshould notbetreated merely asanexcuse fortheexercise ofmathematical techniques. InPart IIthelanguageofvectors isusedextensively. A knowledge ofthree-dimensional analytical geometry isrequired andgreater power intheuseofmathematical processes. This part iscomplete initself, except foroccasional references to Part I.The selection ofparticular applications follows con- ventional lines, except foronenovel feature asection onelectron optics. Chapters onLagrange's equations andonthespecial theory ofrelativity areincluded. Thebook hasdeveloped from lectures delivered byboth authors toHonor Students intheirsecond andthird years atthe University ofToronto. These lectures cover about 110periods of50minutes, and ithasbeenfound thatthework canbedone fairly adequately inthat time. But thisdoesnotallow suffi- ciently fortheworking ofproblems withtheclasses; itisfeltthat 150periods might wellbespent onthecontents ofthebook, were itnotforotherdemands onthestudents' time. Each chapter isfollowed byasummary. Thesummaries to thechapters dealing with ijiethods arenaturally themore funda- mental there islittle hope ofbeing able toattack problems unless one isthoroughly familiar with thegeneral principles outlined there. Ontheother hand, thesummaries tothechap- tersdealing with applications areintended toprovide only a synopsis ofwhat hasbeen done. PREFACE TOTHEFIRST EDITION xi Many oftheexercises aretaken withpermission fromexamina- tionpapers setintheUniversity ofToronto andprinted by theUniversity Press. Ineach setofexercises, the firstfew problems aresosimple that failure tosolvethem willreveal a lack ofunderstanding ofbasic methods, rather thanadeficiency inskillandingenuity. Theequations arenumbered insuch away that,when read asdecimals, they stand intheir proper order. The integer represents thechapter, the firstdecimal place represents the section, andthelasttwodecimal places thepositionoftheequa- tion inthesection. Debts toother textbooks aretoonumerous toacknowledge. Butwewould liketopay tribute totwobooks andrecom- mend them tothereader who wishes topursue thesubject further. They areE.T.Whittaker's Analytical Dynamics (Cambridge University Press) and P.AppelPs Mgcanique rationnelle (Gauthier-Villars).These books have suggested the possibility ofreconciling inatextbook onmechanics twoopposing goals thereduction ofthesubject toacompact and classified form and itsexposition with sufficient fullness tomake the arguments easy tofollow. Wegratefully acknowledge assistance andadvice received from ourcolleagues, Professor H.S.M.Coxeter, Professor A.F. Stevenson, Dr.A.Weinstein, andMr.A.W.Walker. Weare under aparticular debttoProfessor L.Infeld, whoreadmost ofthe manuscript andhasbeen unsparing infrank criticism andsug- gestions;ifwehavesucceeded inavoiding dullness andobscurity, itisdueinnosmall measure tohim. J.L.SYNGE B.A.GRIFFITH TORONTO, ONTARIO MEDICINE HAT, AIJBEUTA December, 1941 PART I PLANE MECHANICS CHAPTER I FOUNDATIONS OFMECHANICS 1.1.SOME PHILOSOPHICAL IDEAS Whydowestudy mechanics? There areatleast three reasons. First,weliveinanageofmachinery, which cannot bedesigned without aknowledgeofmechanics; infact,itisthemost funda- mental subject inengineering. Secondly, mechanics plays a basic part inphysics andastronomy, contributing toourknowl- edge oftheworking ofnature. Thirdly, themathematician is interested inmechanics, both inthelogic ofitsfoundations and inthemethods employed; aconsiderable portion ofmathematics wasdevelopedfortheexpress purpose ofsolving mechanical problems. Thesubject ofmechanics isnotamere collection offacts. From certain simple hypotheses anelaborate theoryisbuilt up. Anyone whohasstudied thesubject should beable toanswer questionsofinterest toengineers andphysicists; that istosay, heshould beable toapplyhisknowledge. Butheshould also have afairidea ofthelogical structure. Asuccessful textbook hastosteeramiddle course between undue concentration onthe mere working outofproblems ontheonehand, andanover- elaborate developmentoflogical structure ontheother. Thetwoways ofthinking. What thestudent ofmechanics requires more than anything else isthedevelopmentofacertain pointofviewwhich isdifficult todescribe inafewwords. Since thereader isexpected tohave afairknowledgeofgeometry,itwillbehelpful toconsider the ways inwjiichwethink about that subject. Every student ofgeometrylearns tothink intwoways. First, there isthephysical way,inwhich apointisasmall dot onasheet ofpaper, astraightlineamarkmade bydrawing asharp pencil along astraight edge, acircle amarkmadebya pair ofcompasses, andsoon.Secondly, there istheideal or 3 4 PLANE MECHANICS [Sac. 1.1 mathematical way,inwhich apointisnolonger adotonpaper, butanideal thing which thedotserves only tosuggest. Anyone whousesgeometry hasboththeseways ofthinking athisdisposal, switching fromonetotheother without confusion. Theengineer andthephysicist generally think inthephysical way,butwhen there isatheorem tobeproved theysubconsciously switch tothe mathematical way. Ontheotherhand themathematician will think primarily inthemathematical way,buthewillchange to thephysical waywhen hewants toaidhisthought with a diagram. This duality inpoint ofview isconfusing tothebeginner ingeometry. But itisfortunate thathehastofacethisdifficulty atanearly stage inhiscareer, because itprepares him fora similar dualityinmechanics, about which hehasalso tolearn tothink intwodifferent ways. First, there isthephysical way.Wethink ofactual physical things, natural orman-made. Weseek tounderstand thelaws governing their behavior andtopredict howthey willbehave under given circumstances tobeable totrace thepaths of comets inadvance, ordesign machinery andbridges with con- fidence astotheirbehavior when constructed. Ontheother hand, there isthemathematical way. Often without realizingitconsciously, thephysicist, astronomer, or engineer slips overfrom thephysical way ofthinking tothe mathematical. Thus theastronomer may treat theearth asa perfect sphere anabstract mathematical concept which does not exist innature ortheengineer may discuss awheel as ifitwereaperfect circle. The transition from thephysical tothemathematical and back againisasource ofmore confusion thanmaybesuspected, but itisunavoidable. There isnodoubt that thephysical way ofthoughtisthemore natural; butaslong asitistheonly way, progressisslow. Physical things arevery complicated andhard tothink about. Slowly wecome todistinguish between properties which areessential andproperties which areincidental. Welearn tosimplify problems byforgetting theincidental properties andconcentrating onthose which areessential. Toillustrate, suppose weareinterested intheperiodic time ofabarsuspended from oneend, oscillating asapendulum. Which properties ofthebar isitessential forustobear inmind, SEC. 1.1] FOUNDATIONS OFMECHANICS*5 andwhichmayweneglect asincidental? Canwepredict the periodic time ofoscillation without knowing thematerial of which thebar isconstructed? Does theform ofthecross section ofthebarmatter? Does itmake anydifference whether the bar issupported onaknife-edge orbybearings? Thecautious well-informed physicist would saythat allthese things mattered andmany others. Onematerial yields more than another, the form ofcross section influences thedistribution ofmaterial, and achange inthemode ofsuspension may alter theaxisabout which thependulum oscillates. But ifwewere ascautious as thisweshould havenoscience ofmechanics. Tostartonthe problem, atany rate,wemust simplifyitruthlessly. Sowe think ofthebarasarigidmathematical straight lineandthe support asafixedmathematical point. Nowwehaveaproblem which isreasonably simple tohandle mathematically. Strictly speaking, noproperties areincidental. Even thecolor ofthe baraffects thepressure oflightonit;asubway train stopping fivehundred milesawaymaycause avibration inthesupport and affect themotion ofthebar.Common sense, which istheaccu- mulated experience ofcenturies, gives ussome guide astothe factors which wemay neglect. Mathematical models. Gradually stripping physical thingsofattributes which are unimportantforthequestioninhand, wearrive atamathe- matical wayofthinking about nature. The particular mathe- matical model* tobeusedonagiven occasion depends onthat occasion. Consider theearth, forexample. Thesimplest model oftheearth isaparticle, amathematical point withmass. This model suffices toobtain theearth 'sorbit round thesun,but obviously willnotdoforthediscussion oftides orlunar eclipses. Forthesephenomena wemaythink oftheearth asa rigid sphere, but thismodel willnotserve forthediscussion oftheprecessionoftheequinoxes (forwhich werequire anellip- soidal rigidbody) orforthediscussion ofearthquakes (forwhich werequire anelastic sphere). Thus there aremany mathe- *Thereader willofcourse understand thatwhenwespeak ofa"model" wedonotmean anactual physical reproduction onasmall scale.Weuse theword forwant ofabetter todescribe oursimplified mental picture ofaphysical object. 6 PLANE MECHANICS [SEC. 1.1 matical models fortheearth, andtheonewhich wechoose depends onthequestion wearediscussing atthemoment. Infact, mechanics andindeed alltheoretical science isa game ofmathematical make-believe. Wesay: //theearth wereahomogeneous rigid ellipsoid acted onbysuchandsuch forces, howwould itbehave? Working outtheanswer tothis mathematical question, wecompare ourresults with observa- tion. Ifthere isagreement, wesaythatwehave chosen agood model;ifdisagreement, then themodel orthelawsassumed are bad. Letusnowsumupthegeneral procedure intheoretical mechanics inthefollowing fivesteps. (1)Aphysical systemisanobject ofcuriosity; wewish to predictitsbehavior under various circumstances. (Thesystem inquestion might beapendulum, orapair ofstars attracting oneanother.) (2)Anideal ormathematical model ofthephysical systemis constructed mentally. (Thependulumisregarded asarigid straight line,andthestars areregarded astwo particles.) (3)Mathematical reasoningisapplied tothemathematical model. (Thismeans that differential orfinite equations areset upand solved. Formulas aredeveloped togiveanswers to interesting questions, such asthose concerning theperiodic time ofthependulum ortheorbit ofonestar relative tothe other.) (4)Themathematical results areinterpreted physically in terms ofthephysical problem. (5)The results arecompared with theresults ofobservation, ifpossible. Certain remarks should bemade about these fivesteps. First, (1)implies aphysical curiosity. Inspite ofthefactthattheo- retical mechanics isapartofmathematics, weshould notforget that itsroots lieinphysics andtheactual world around us. Secondly, ashasbeenremarked above, theconstruction of amathematical model (2)atonce simple andadequateisbyno means easy inallcases. However, mechanics isanoldsubject, andthere ismuch accumulated experience tofallback on.The concepts ofparticles, rigid bodies, forces, etc. (allmathematical idealizations), have bendesigned forthispurpose. These will bediscussed^v SEC. 1.1] FOUNDATIONS OFMECHANICS 7 Step (3)belongs largely topure mathematics, requiring no particular knowledge of,orinterestin,thephysical problem. Nevertheless,itisoften ofthegreatest assistance tothemathe- matician tobearthephysical problem continually inmind; in thisway,methods ofattack maybesuggested tohim. Thefourth step ingeneral presents nodifficulty, provided thatweareclear astothethings innature which correspond to thethings inourmathematical model. Thetechnical details ofthefifth stepbelong toexperimental physics orobservational astronomy, andwiththemweshall notbeconcerned. Butweareinterested inthefactthat the conclusions drawn from amathematical theory are,orarenot, physically true, within thelimits ofaccuracyofobservation. Itisnecessary todistinguish between mathematical truth andphysical truth. Indeveloping thetheory ofmechanics, we shall trytomake themathematical arguments fairly complete, so thatwecanhave confidence thattheconclusions follow logically from thehypotheses, i.e.,thatthey aremathematically true. Weshould notundertake thiswork, however,ifwehadnot confidence thatourconclusions arealsophysically true, inthe sense thattheyagree with observation. Avastaccumulation of physicalresults confirms ourconfidence. Nevertheless,itwould betoomuch toclaim that allourconclusions arephysically valid. Attempts toconstruct asuccessful model ofanatom onthe basis ofNewtonian mechanics have failed. This failure led totheinvention ofquantum mechanics. Wemaysayingen- eralthatNewtonian models ofsmall-scale phenomena have notbeen successful, whereas attheother end ofthescalewe find difficulty also inthelarge-scale phenomena ofastronomy. Inspiteofthemany triumphs ofNewtonian mechanics in dynamical astronomy, there remain afewphenomena which areinapparent disagreement withit;thebest-known concerns theorbit oftheplanet Mercury. Thisdifficulty wasovercome when Einstein created thegeneral theory ofrelativity. Toexplore withanydegree ofcompleteness thetheories referred toabove would demand acourse ofstudy farwider than thatcovered inthisbook. Thereader may feeldisappointed thatatthisstage hecannot reach theforefront ofourmechanical knowledge. Toencourage him,however,itmaybepointed out that aslong asthephysical problems concern only apparatus 8 PLANE MECHANICS [Sac. 1.2 ofanintermediate scale, i.e.,neither atomic ontheonehand nor astronomical ontheother, onemayhave complete confidence thatnoexperimental technique canreveal anydiscrepancy between observation andtheconclusions drawn fromNew- tonian mechanics. This confidence mayevenbeextended to astronomy, because there therelativistic effects areextremely minute; thevastbodyofcalculations ofdynamical astronomy are stillsafely based onNewtonian mechanics. Relativity andquantum mechanics notonly enable usto obtain results which arephysically true they alsothrow light onsuch basic philosophical ideas assimultaneity andcausality. Chapter XVI contains anintroduction tothespecial theory of relativity. The general theoryofrelativity andquantum mechanics both lieoutside thescope ofthisbook. 1.2.THEINGREDIENTS OFMECHANICS Inanysubject there arewords which occur again andagain, likethewords "point/7 "line," and"circle" inelementary geom- etry. Aswell asthese, technical words, there occur ordinary words withthemeanings ofwhichwearesupposed tobefamiliar. When westart anew subject, wearenotexpected toknow what thetechnical words mean. They areintroduced withsome formality, being infactgiven definitions. Adefinition isitself onlyasetofwords andmaynotmeanmuch; thegeneral idea is toexplain anewthing interms ofthings already familiar. Wearenowtotrytocreate mathematical models ofphysical things. Westartwithafairgeneral unprecise knowledge ofthe world around us;theplaces inourminds reserved forthemathe- matical models aresupposed tobeabsolutely blank. Ifwe opened these places fortheactual world torushin,weshould be overwhelmed with confusion. Weguard thedoorandadmit onlyafewingredients ofsimple mathematical character. Particles. The firstthingweadmit isaparticle. Wehave seen tiny scraps ofmatter and itisnotdifficult forus,withourtraining ingeometry, tothink ofascrap ofmatter withnosizeatall, butwith adefinite position; that isaparticle. When wehave todealwithaphysical problem inwhich abodyisvery small incomparison with distances orlengths involved (forexample, SBC. 1.2] FOUNDATIONS OFMECHANICS 9 theearth incomparison with itsdistance from thesun,orthebob ofapendulumincomparison withthestring), wemayrepresent thatbody inourmathematical model byaparticle. Mass. Primitive tradewasamatter ofbarter; later,money wasintro- duced asastandard scale forcomparisonofvalues, andequiva- lence invalue isnow expressed byequalityofprice. This exemplifies aprocess ofdeep importance inscience, namely, aconcentration onsome characteristic (value) ofathing and itsexpression bymeans ofanumber (price). Abarrel ofapples isvery different from apairofshoes, buttheymaybeequivalent ifvalue istheonly characteristic inwhich weareinterested. That thepriceisthesame expresses complete equivalenceas farasourpurseisconcerned. Consider now agreat varietyofbodies piecesofstone, iron, gold, wood,etc.andmechanical experiments' performed onthem. Asexamples, wemention twoexperiments: (i)Thebodyisplacedinthepanofaspring balance andthe reading noted. (ii)Thebodyisfiredfrom agunbymeans ofadefinite explo- sivecharge andpassesintoablock ofwood, theresulting dis- placement ofwhich isnoted. IfAandBaretwopieces ofiron, asnearly identical inshape and sizeasitispossibletomake them, theywillofcourse give thesame results when used inanyexperiment, performedfirst usingAandthenrepeated usingBinstead. But itisaremark- able fact, resting onlong experience, thattwobodiesAandB maydiffer inmaterial, size, shape, etc.,andyetgivethesame result inagreat variety ofmechanical experiments. Wethen saythatthey aremechanically equivalent. Apiece ofwoodanda pieceofgoldmaybemechanically equivalent, justasabarrel of apples andapairofshoesmaybeequivalent invalue. Asjwe assign apricetoeach article oftrade, sowemay assign anumEer toeach pieceofmatter, equality ofthese numbers implying mechanical equivalence. Thisnumber iscalled mass and isusually denoted bym.Following theanalogyofmoney, based onastandard substance (gold),itiseasy toseehowa scale ofmass istobeconstructed. Westart withanumber of identical piecesofsome standard material such asplatinum, 10 PLANE MECHANICS [SEC. 1.2 andweassign tothemass ofeach thevalue unity (m=1). When nofthese pieces arelumped together, weassign tothe mass ofthelump thevaluen(m=n).Bycutting thepieces, wecanconstruct bodies with fractional masses andsoobtain asetofstandard bodies ofallpossible masses. Then, toassign amass toabodyA(not ofthestandard material), wesubjectit toexperiments and findthatstandard bodyBtowhich itis mechanically equivalent. Wethen saythat themass ofA isthesame asthemass ofB. Thecomparison ofmasses isusually made byweighing as intheexperiment (i)mentioned above, except that forreasons ofaccuracy thespring balance isreplaced byalaboratory balance. Thus, inpractice, twobodies aresaid tohave thesame mass when theyhave thesame weight. Theabove considerations dealwith physical bodies. Inthe mathematical model inwhich these bodies arerepresented by particles, wearetoregard each particle ashaving attached toit apositive number m,itsmass, which doesnotchange during the history oftheparticle. Indealing with asystemofparticles, wedefine themass of thesystem tobethesum ofthemasses oftheparticles which composeit. Rigid bodies. Wehavenowadmitted asamathematical model theparticle withmass. Thenextthing toconsider istherigid body. Itisamatter ofcommon experience thatbodiesmaybesoft likerubber orhard like steel. Even thehardest body, however, changesitssizeandshape bymeasurable amounts under the action ofsufficiently greatforces. Butjust asweidealized the small body ofourexperience intotheparticle with position butnosize, soweidealize thehardbody ofourexperienceinto therigidbody, which never undergoes anychange jrfsize or_shape. The rigidbodyisnowadmitted asamathematical mocfeL Wepause foramoment toexamine critically something written justabove. Wespoke ofabody changingitssizeandshape. What does thisreally mean? Suppose, forexample, wehave a barofsteelwithtwomarks onit.Alongside thebarwelaya graduated measuring scaleandnote thereadings onthescale opposite thetwomarks onthebar.Thenwepulltheends of thebarandnote thereadings again. The difference between SBC. 1.2] FOUNDATIONS OFMECHANICS 11 them isgreater than itwasbefore; hence wesaythatthebar hasincreased inlength. However, anargumentative person might assert that thiswas anincorrect statement;hemight holdthatthelength ofthebar wasthesame asbefore butthatthemeasuring scalehadshrunk. Wecannot saythatheiswrong intaking thispoint ofview until weclarify ourideas astothemeaning oftheword "length." The idea oflengthisonethat involves thecomparison of twobodies. Wedecide once forallonaunitoflength bymaking twomarks onapiece ofmetal andstating conditions withregard totemperature andpressure under which measurements are tobemade with thispiece ofmetal. Wewereperhaps alittle hasty inadmitting arigidbody asamathematical model, because there isnosense intalking about asingle rigidbody;wemust havesomemeans ofmeasuringitandtesting that itisrigid. Sowhenweadmit therigidbody,weshall atthesametimeadmit ameasuring scale. When wesaythatabodyisrigid,wemean thatmeasurements ofdistances between marks onitalways have thesame values, themeasurements beingmade with the measuring scale. Events. Theword event isfamiliar inordinary speech. Itusually denotes something alittle outoftheordinary, something that occurs inafairly limited region ofspace and isoffairly short duration. Thus afootball game orthearrival ofatrainmight be described asanevent. Theword hasnowacquired anidealized scientific meaning, theidealization involved being rather similar tothatbywhich wecreated theconcept ofparticle. Instead of occupying afairly limited region inspace, anevent (inourmathe- matical model) occurs atamathematical point; andinstead of beingoffairly short duration,itoccurs instantaneously. Wedo notcarry over intoourmathematical model theslightly dramatic meaning attached totheword inordinarylife. Anything that happens maybecalled anevent. Even thecontinued existence ofaparticle forms aseries ofevents. Frames ofreference. Indescribing anevent inordinary life, itisusual tospecify theplace andtime. Thus itisrecorded ofthesinking ofa ship that itoccurred atacertain latitude andlongitude, and 12 PLANE MECHANICS [SBC. 1.2 atacertain Greenwich mean time. Latitude andlongitude define position ontheearth's surface; wearehereusing theearth asaframe ofreference. This isthemost familiar frame ofrefer- ence, butothers maybeused. Astronomers prefer aframe of reference inwhich thesun isfixedandwhich doesnotshare in theearth's motion ofrotation. Also, theinterior ofatrain, streetcar, elevator, orairplane may beused. The essential thing about aframe ofreference isthat itshould befairly rigid. Inourmathematical model, weemploy arigidbody asframe ofreference. Aswemayintroduce anynumber ofrigid bodies moving relative tooneanother, wehave thus atourdisposal anynumber offrames ofreference. Selecting one ofthese andtaking rectangular axes ofcoordinates init,weassign to anyevent asetofthreenumbersx,y,z,thecoordinates inthe frame ofreference ofthepoint where theevent occurs. Time. Anevent hasnotonly position;italsohasatime ofoccurrence. Thiswehavenow toconsider. The possibilityofrepeating anexperiment forms thebasis ofexperimental science. Itisassumed that,ifanexperimentis repeated under thesame conditions, thesame results willbe obtained. Consider, forexample, atank ofwater drained through aholeinthebottom, andthen refilled anddrained again. Strictly speaking,itisimpossible toreproduce conditions exactly, andwehave tousejudgment todecide whether thenew conditions aresufficiently near theold.Butinanideal sense wemay think ofanexperiment repeated overandoveragain under exactly thesame conditions. Todefine time,wethink ofsome experiment which canbo repeated overandover again, anewexperiment starting just when thepreceding oneends. Denoting timeby t,weassign thevalue t=tothebeginning ofthe firstexperiment,t 1 tothebeginning ofthesecond experiment,t=2tothebeginning ofthethird experiment, andsoon.Therepeated experiment thusforms aclock forthemeasurement oftime;weshall callthe unit oftime given bysome such ideal experiment aNewtonian unit. This istheprocedure actually adoptedinpractice. In awatch, theexperimentisanoscillation ofthebalance wheel; in apendulum clock,itisanoscillation ofthependulum. In SBC. 1.2] FOUNDATIONS OFMECHANICS 15 Ashasbeen pointed outalready, wearenottoexpect a mathematical model tohave allthecomplexityofnature. The model which weshall useresembles insomeways themodern physicist's concept ofasolid body, but itisgreatly simplified. Itwasinvented long before thedevelopment ofmodern atomic physics, andwas originally supposed tobeamore complete representationofnature thanwenowknow ittobe.Neverthe- less,itenables ustopredict toahigh degree ofaccuracy an immense number ofphenomena; itisinfactthebasis ofagreat deal ofNewtonian mechanics. Thismathematical model ofasolidbodyisdiscontinuous acollection ofavastnumber ofparticles. Inarigidbody the distances between theparticles remain invariable, butinanelastic body these distances may change. Since thismodel involves avery largenumber ofthings, statistical methods maybeused; instead offollowing individual particles, wemay direct our attention totheiraverage behavior. Infact,wementally replace thediscontinuous body, consisting ofagreatnumber ofpar- ticles, byacontinuous distribution ofmatter. This simplifies thedetermination ofmass centers andmoments ofinertia, because themethods ofintegral calculus canthenbeused. Toavoid lengthy andperhaps uninteresting arguments, we shall leave certain gaps inthelogical development ofoursubject. Weshall notgivearguments ofastatistical nature inorder to passfrom aresult established foradiscontinuous system tothe corresponding result foracontinuous one. Itisusually easier toestablish general theorems fordiscontinuous systems andto solve special problems forcontinuous systems. Force. Letusnowintroduce intoourmathematical model theconcept offorce, idealizing asusual from oursomewhat vague physical concepts. Ourprimitive concept offorce arises outofoursensa- tionofmuscular exertion. Wepushandpullobjects, sometimes withsmall exertion, sometimes with great effort. Butthesame effects asthose produced bymuscular effortmaybeproduced in other ways. Inthismachine age, direct muscular effort isused toagreat extent only tocontrol much greater forces duetothe pressure ofsteam, theweight ofwater, theexplosive pressure of gasoline, orforces ofelectromagnetic type. One alsoadmits the 16 PLANE MECHANICS IBC. i.z existence ofhuge forces beyond human control, such asthegravi- tational attraction exerted bythesunontheearth. Onthebasis ofourexperience with simple muscular forces, wethink oftheidealized force ofourmathematical model as something which has (i)apoint ofapplication, (ii)adirection, (iii)amagnitude. Forthedevelopmentofgeneral results intheoretical* mechanics, itwould besufficient torepresent themagnitude ofaforcebya letter, standing forsome unspecified numerical value. Butwhen wewish tomake predictions regarding aphysical system subject toforces, werequire adefinite procedure bymeans ofwhich we may assign numerical values totheir magnitudes. Wemust, infact, define aunit offorce. There hasbeensome controversy about thisquestion. Though allareagreed astotheform oftheory whichweshould ultimately obtain, there hasbeen disagreement regarding theproper order ofintroduction ofthevarious partsofthetheory. Thuswe might assume astatement Aasanaxiom anddeduce astatement Bfromit,oralternately wemight assume Banddeduce A. The order ofpresentation chosen inthisbook seems tothe authors themost natural; but itishoped that thereader will explore forhimself thepossibility ofadifferent approach.* Wedefine theunit offorce interms ofastretchedspring; itisthat force which produces some standard extension insome standardspring. Later weshall linkuptheunit offorce with theunits ofmass, length, andtime; butforthepresent theunit istoberegarded asarbitrary. Tomeasure aforce,weexamine theextension which itproduces inabattery ofstandard springs sidebyside,allidentical with oneanother. Ifthestandard extension isproduced innsprings, then theforce isofmagnitude n. Ifthemagnitude oftheforce inquestionisnotaninteger, wereproduce itmtimes soastoget *Seetreatments inE.T.Whittaker, Analytical Dynamics (Cambridge University Press, 1927), p.29;H.Lamb, Statics (Cambridge University Press, 1928), p.12,andDynamics (Cambridge University Press, 1929), p.17; J.S.AmesandF.D.Murnaghan, Theoretical Mechanics (Ginn andCom- pany, Boston, 1929), p.104. Foracritical and historical account ofthe development ofmechanics, seeE.Mach, TheScience ofMechanics (Open Court Publishing Company, Chicago, 1919). SEC. 1.3] FOUNDATIONS OFMECHANICS 17 thestandard extension insomenumber nofstandard springs; then themagnitude oftheforce isn/m. Having thus given ameans ofmeasuring force,wemay con- struct asimplified apparatus. Taking any spring, fixed at oneend,wemark itsextensions under theaction ofmeasured forces. Inthisway,wecalibrate thespring; thecalibrated spring maybeused directly forthemeasurement ofaforce. Wehave preferred tousethetension inaspring rather than theweight ofabody asthefoundation ofourdefinition offorce. Itiscustomary formany practical purposes tospeak ofaforce ofsomany pounds weight (Ib.wt.);byaforce of10Ib.wt., wemean aforce equal totheweightofamass of10Ib.Although this* practiceisconvenient andadequate formany purposes1 ; itisopen^to objection onthefollowing ground: Iftheweight ofabodyismeasured bymeans ofacalibrated spring attwo different latitudes, theresults arenotthesame (see Sec. 5.3). Adefinition offorce based ontheextension ofaspring gives a consistent theory without contradictions, whereas adefinition based onweight would involve usinexplanations astowhya spring, showing thesame definite extension inToronto andin Panama, should exert different forces inthetwoplaces. Indescribing particles, rigid bodies, andforces wehave intro- duced thebasic ingredients ofmechanics. Asweproceed, other ingredients willappear, but itisremarkable howmuch ofthe subject turns onthesimple concepts justmentioned. 1.3.INTRODUCTION TOVECTORS. VELOCITY ANDACCELERATION Definition ofavector. Inorder toreproduce agameofchess, wemust beable to describe themoves. Thereis,ofcourse, anaccepted wayof doing this, butweshall describe another. Let letters A,B, C, (supplemented with other symbols tomake up64) beassigned tothesquares oftheboard, oneletter toeach square. Then symbols such asAB,CF,UBwillrepresent definite moves, > thesymbol AB, forexample, meaning thatapieceismoved from thesquareAtothesquare B. More generally,ifwecarry aparticle from aposition Atoa position Binspace, theoperation which weperform maybe 18 PLANE MECHANICS [Sac.13 > represented symbolically byAB.Thedirected segment drawn fromAtoJ3,orthecarrying operation orindeed anyphysical quantity which canberepresented bythedirected segmentis called avector,* andthesymbol ABisused foranyofthem. AvectorABhasthefollowing characteristics: (i)anorigin orpoint ofapplication (A); (ii)adirection (defined analytically bythethree direction cosines ofABwith respect torectangular axes); (iii)amagnitude (thelength AB). Anumber offundamental physical quantities have these char- acteristics forexample, aforce orthevelocity ofaparticle; each ofthese quantities mayberepresented byadirected seg- ment and istherefore avector. They aretobedistinguished from quantities such asmass orkinetic energy, which donot involve theidea ofdirection andaredescribed eachbyasingle number. Quantities ofthislatter type arecalled scalars. Itisconvenient toemploy theword "vector" inaslightly wider sense than thatgiven above andtodefine thefollowing: (i)freevector; (ii)sliding vector; (iii)bound vector. A.freevector isanyoneofasystem ofdirected segments having acommon direction andmagnitude butdifferent origins. A physical quantity equally wellrepresented byanyoneofsuch directed segmentsisalsocalled afreevector. Such, forexample, isthedisplacement, without rotation,ofarigid body, which isequally wellrepresented byanyoneofthedirected segments giving thedisplacements ofitsvarious points. Asliding vector isanyoneofasystem ofdirected segments obtained bysliding adirected segment alongitsline.Aphysical quantity equally wellrepresented byanyoneofsuch directed segmentsisalso called asliding vector. Such, forexample, isaforce acting onarigidbody, which (bytheprinciple ofthe transmissibility offorce proved onpage 64)may equally well beapplied atanypoint onitslineofaction. Abound vector isaunique directed segment, oraphysical quantity sorepresented. Such, forexample,isaforce acting on *Theword "vector" isderived fromtheLatin veho, Icarry. SEC. 1.3] FOUNDATIONS OFMECHANICS 19 anelastic body; wecannot ingeneral alter this force, byany displacementofthedirected segment representing it,without changingitseffect. Bytheword" vector," without anadjective, weshall generally understand "free vector"; butwherewehave tospeakofbound orsliding vectors,itwillbeunnecessary tousethequalify- ingadjective when itmay beunderstood from thecontext. Throughout therestofthissection, thevectors aretoberegarded asfree. Notation. Avector isindicated inprint byaboldface letter (P): inmanuscript work thesymbol maybeunderlined (P)oran arrowmaybewritten ontop(P). Itsmagnitudeisdenoted by thesame letter inordinary type, orbyanunmarked symbol inmanuscript work (P).Avector ofunitmagnitudeiscalled aunit vector. Torefer toabound orsliding vector, wemay write"Pacting atthepointA11or"Pacting ontheline I/," ifthere isanydoubt astotheorigin orline. Two vectors areequal tooneanother when theymaybe represented byequal parallel directed segments with thesame sense.Weusetheusual sign ofequality andwrite P=Q. Thesign ofequalitycarries theusual algebraic property: vectors equal tothesame vector areequal tooneanother. Multiplication ofavector byascalar. LetPbeavector andmascalar. Wedefine theproductof raandP(written raPorPw) asfollows: Ifmispositive, then rriPhasthesame direction asPandamagnitude wP;ifmis negative, thenmPhasadirection opposite tothat ofPanda magnitude raP. Wewrite(-l)P=P;thusPisthevectorPreversed. Addition ofvectors^ Thesumoftwovectors PandQiswritten P+Q;itisdefined asthevector represented bythediagonal ADofaparallelogram ofwhich twoadjacent sidesAB,ACrepresent PandQ,respec- 20 PLANE MECHANICS [SBC. 1.3 tively (Fig. 1).Obviously, analternative wayofconstructing P+Qisthefollowing (Fig. 2):Draw asegment ABtorepresent P,andfrom itsextremity drawBDtorepresent Q;thenAD represents P+Q. This isamathematical definition ofP+Q.It'does not contain theimplication thatP+Qisthephysical resultant of PandQ,although inalmost allcasesweshall findthatP+Q isactually thephysical resultant. Finite rotations aretheout- standing exceptions; afinite rotation isavector, buttheresultant oftwo finite rotations isnotthesum ofthevectors (cf.Sec. 10.5). FIG. 1.Addition ofvectors by parallelogram.FIG. 2.Addition ofvec- torsbytriangle. When twovectors PandQhave thesame direction oropposite directions, theparallelogram constructed togive theirsum collapses intoastraight line. But that does notprevent us fromapplying theabovedefinition, regarded asalimiting process. Itiseasily seen that,ifPandQhave thesame direction, then P+Qhasalsothat direction andamagnitude P+Q;ifthey have opposite directions andPisthegreater, thenP+Qhas thedirection ofPandamagnitude P Q.Comparing this with thedefinition oftheproduct ofavector byascalar, we findinparticular that P+P=2P, andweverify generally thatthemultiplication ofavector bya scalar isdistributive both with respect tothe*scalar andtothe vector; thismeans thatwehave (1.301) (1.302)(m+ri)P=raP+wP, m(P+Q)=mP+mQ. SEC. 1.3] FOUNDATIONS OFMECHANICS 21 Itisanimmediate consequence ofthedefinition that the addition ofvectors iscommutative, thatis, P+Q=Q+P Thesubtraction ofvectors isimmediately effected bywriting P-Q=P+(-Q) andapplying therule foraddition. The difference between P andQiseasily constructed asfollows : p DrawAB,ACtorepresent P,Q,re- spectively (Fig. 3);thenCBrepre-Q sentsP-Q. C Applying therule forsubtractionFlQ -3.-subtaotion ofvectors, tothecasePP,weobtain avector ofzeromagnitude, which we denote by0,sothat P-P=0. Wecall thezero vector; allvectors ofzeromagnitude areregarded asequaltooneanother. Anyunfamiliar symbol containing vectors mustbeapproached with caution. .Itmaymean nothing atall(forexample, wenever attempt todefine thesum ofascalar andA avector,m+P) ;ontheother hand,it maybegiven ameaning. Onthebasis ofprevious definitions, P+Q+Rhas nomeaning, because wehave defined the sum oftwovectors, not three. But (P+Q)+R hasameaning,ifweregard theparen- theses ascarrying theinstruction toadd PandQ,andthenaddRtothatsum;^ FIG. 4.The associativeP~t~(Q "4"R) property ofvector addition. hasameaning,also. Itistheneasy toseethat (P+Q)+R=P+(Q+R), > > bymeans oftheconstruction shown inFig. 4,whereAB,BC,CD represent P,Q,R,respectively, andADeither oftheabove sums. 22 PLANE MECHANICS [SBC. 1.3 Thus aunique meaning canbeattached toP+Q+R.Wesay thatvector addition isassociative. Exercise. If P4-2Q=R, P-3Q-2R, show thatPhasthesame direction asR,andQtheopposite direction. Componentsofavector. LetLbeastraight lineandPavector represented inFig.5 byAB. Ifwedraw through AandBplanes perpendicular toL, these planes cutoffonLadirected segment CZ),theorthogonal projection ofAB]CD isacommon perpendicular tothepair of planes.Ifwetakeadifferent directed segment ABftorepresent > * P,wegetaprojection C'D'onL.Now ifwegiveABandthe B' B ^P C D C7T)' FIG. 6.Thecomponent ofavector onalino. pairofplanes associated with itadisplacement (without rotation) which carriesAtoA',thenBwillgotoB'andthepair ofplanes associated withABwillcoincide with thepair ofplanes asso- ciated with A'B'.CD willbecome acommon perpendicular tothelatter pair ofplanes, andsoCDandC'D' areequal in magnitude anddirection. Thus thevector obtained byproject- ingonLadirected segment representing Pisindependent ofthe particular segment chosen. Writing Qforthevector repre- sented byCDorC'D',wesaythatQisthevector component ofP on!/. Ofthetwosenses onthelineL,letuspickoutoneand call itthepositive sense, theother being negative; thelineListhen said tobedirected. Let ibeavector ofunitmagnitude, lying onLandpointinginthepositivesense. Then itispossible to SEC. 1.3] FOUNDATIONS OFMECHANICS express theprojection Qintheform23 where cisascalar, positiveifQhasthesame sense asiand negativeifQhastheopposite sense toi.Thescalar ciscalled thescalar component ofPonthedirected lineL. SinceQisrepresented byanycommon perpendicular (inthe proper sense) totheprojecting planes,itiseasy toseethatthe scalar component ofPonLisPcos0,where istheangle between Pandthepositive sense ofL. Inspeaking ofacomponent (without qualification),itwillbe clearfrom thecontext whether thevector orscalar component istobeunderstood. Unitcoordinate vectors. Inaframe ofreference S,letOxyzborectangular Cartesian axes. Leti,j,kbeunitvectors lyingonOx,Oy,Oz,respectively, each inthepositive sense. The soti,j,kiscalled aunit ortho- gonaltriad. LetPbeanyvector, andPI,P2,P3itscomponents onOx,Oy,Oz,respectively. These components, obtained byprojec- tion, areindependentofthepar- ticular directed segment chosen to represent P;wemay therefore take therepresentative segment with its origin at (Fig. 0).From therule ofvector addition,itisclear that (1.303) P=Pii+P2j+P3k 'P.I FIG. C.-Resolution along unit coordinate vectors.This reduction ofavector tothe sum ofthree vector components along aunitorthogonal triad isof great service inmechanics, for it represents thelinkbetween thevector methods andthemore usualmethods ofanalysis. Vector notation isonlyashorthand fortheexpression offairly general statements. Intheend,we mustwork inordinary numbers, andtheabove formula isthe bridge bywhich wepassfrom vectors toordinary numbers. WenotethatthemagnitudeofavectorPisexpressed interms ofitsscalar components by 24 PLANE MECHANICS [SEC. 1.3 (1.304) P=VPl+P\+PI Thecomponents aregiveninterms ofPandX,/z, *>,thedirection cosines ofP,by (1.305) P!=AP,P2=/*P,P3=*P. Thefollowing facts areimportant butmaybelefttothereader toverify, using Fig.2inthecase ofthesecond: (i)Thecomponents ofmParemPi,raP 2,mP 3. (ii)Thecomponents ofP+QarePi+Qi,P2+Q2,PS+Q3. IfPI,P2,Paarethecomponents ofavectorPonthecoordinate axesandLisadirected linewith direction cosines A, /u,v,then theangle between PandLisgivenby Hence thescalar componentofPonLis (1.306) Pcos-APi+/*P2+pPs. Exercise. Thecomponents ofavector onaxesOxyinaplane areX,Y. What arethecomponents X',YronaxesOx'ij', obtained byrotating Oxy through anangle 0? Position vector. LetAbeaparticle, moving relative toaframe ofreference Sin which Oxyz arerectangular Cartesian axes. ThevectorOA is called theposition vector ofArelative to0. Ifwedenote itbyr andthecoordinates ofAbyx,y,z,wehave (1.307)r=xi+yj+zk, wherei,j,kistheunitorthogonal triad along theaxes. Astheparticle moves, thevector rchanges. Infact,rmaybe regarded asavector functionofthetimet\wemay expressthis bywriting r=r). Differentiation ofavector. Theabove considerations lead naturally toamore general concept, namely, avectorPwhich isafunction ofascalar u,a relation expressed bywriting P=P(u). SEC. 1.3J FOUNDATIONS OFMECHANICS 25 Weneednolonger think ofPasaposition vector orofuasthe time. Wearefamiliar withtheideaofdifferentiating ascalar function ofascalar: canweenlarge thefamiliar method toobtain aprocess fordifferentiating avector function ofascalar? Wemay follow thefamiliar plan almost word forword.Weconsider twovalues oftheparameter, uand u+Aw,andthecorrespondingin- _^^^ crement inP,P* *^/AP FIG. 7.Differentiation ofa AP=P(u+Aw)~P(u).vector. Wemultiply byI/Aw, toform thequotient AP/Aw (Fig. 7)and letAwtend tozero. Thuswegetalimiting vector (1.308) ^=lim~tV 'du Att-K)AW whichwecallthederivative ofPwith respect tow. Tofindthecomponentsofthederivative weintroduce unit coordinate vectors, sothat (1.309) P=P1i+P 2j+P 3k; Pi,P2,PSarescalar functions ofw.Increasing wtow+Aw,we have P+AP=(Pi+APj)i+(P2+AP 2)j+(P3+AP,)k. Subtraction gives AP=APii+AP 2j+AP 3k. Dividing byAw,weget ^^^;+^J?i4-^L3kAw Aw"*"AwJ^Aw' and so,letting Awtend tozero, thederivatives ontheright being ofcourse derivatives ofscalars theusual derivatives ofthedifferential calculus. Thus the 26 PLANE MECHANICS [SEC. 1.3 components ofdP/du are dl\ f dP*^dP* du' du1du Inwords, thecomponents ofthederivative areequal tothe derivatives ofthecomponents. Thefollowing results areeasy toprove, either directly from thedefinition ofthederivative orfrom (1.310): -7-p=-T- P~T>durdu du where P,Qarevector functions ofu,andpisascalar function ofu. Itwillbeobserved that (1.310) canbeobtained directly from (1.309) bydifferentiation, using therules (1.311); inthisprocess thevectorsi,j,karetreated asconstants. Toavoid possible confusion, letusaskthoquestion: What do wemeanbysaying thatavectorQisconstant? This ismeaning- lesswithout astatement (explicit orunderstood) regarding the frame ofreference employed. Inaframe ofreference S,avector Qisconstant ifitmayberepresented permanently byadirected segment joining twopoints fixed inS.Butviewed fromanother frame ofreference thissame vectorQmay notbeconstant. Thus,in(1.309), i,j,kareconstants inSbecause they areunit vectors along theaxes; theymaynotbeconstants inanother frame ofreference moving relative toS. Ifthevector P(u)isofconstant magnitude, then thetriangle shown inFig.7isisosceles. AsAMtends tozero, thevector AP/Aw tends toperpendicularity with P,or,inother words, for avector P(u) ofconstant magnitude, dP/du isperpendicular to P(i). Velocity andacceleration. LetSbeaframe ofreference andOapoint fixed inS.LetA beamoving particle,itsposition vector relative toObeingr. Wedefine thevelocity ofArelative toStobethevector (1.312) q=* Using dotstoindicate differentiation withrespect totime,wehave (1.313) q=f=*i+ j+*k, SBC. 1.3] FOUNDATIONS OFMECHANICS 27 where x,y,zarethecoordinates ofArelative torectangular axes Oxyz coincident withtheunitorthogonal triadi,j,k.Thus the componentsofvelocity are(x,y,z).Themagnitude qofthe velocity iscalled speed.*Wedefine theacceleration ofArelative toStobethevector (1.314) f-, or (1.315)f-q=*i+7/j+zk. Thecomponents ofacceleration are(z,y,z). Ifthevelocityisresolved intocomponents, q=u\+vj+wk, then (1.316) f=u\+vj+wk. Wecould,ofcourse, continue thisprocess, defining asuper- acceleration df/dt. However, acceleration istheimportant vector inNewtonian mechanics, andsowestopourdefinitions here. Asasimpleillustration ofthese ideas, consider acartraveling along astraight road. Itfollows from (1.312) andthedefinition ofthederivative ofavector thatthevelocity qliesalong theroad andpointsinthodirection inwhich thecar isgoing. Similarly, itfollows from (1.311) thattheacceleration fliesalong theroad, butnowthere isanimportantdifference. Thevector ofaccelera- tiondoesnotnecessarily pointinthedirection ofmotion; this isthecaseonlyifthespeedisincreasing.Ifthespeedisdecreas- ingunder applicationofthebrakes, thevector fpoints backward. Itisnothard toshow thatwhen thecarrounds acurve the velocity continues topoint along theroad, buttheacceleration pointsofftheroadtoward theinside ofthecurve. Theformulas (1.313) and (1.315) areparticularly useful for direct calculation when themotion isdescribed bygiving x\y,z asfunctions oft.Suppose,forexample, that *Thus velocity isavector andspeedisascalar. However, when no confusion islikely toarise, theword "velocity" isoften used inthescalar sense todenote themagnitudeofthevelocity vector; forexample, tho expression "velocity oflight"isused instead ofthecorrect expression "speed oflight." 28 PLANE MECHANICS [SEC. 1.3 x=acosa>t, yasino>2, z=0, where aand o>areconstants. Then r=acosut-i+asin a?j, q=acosin atf i+acocos coj, f=aw2cosut iaw2sin a>j. Itiseasy toseethat this ismotion inacircle, thevelocity pointing along thetangent andtheacceleration inalong theradius. Units ofvelocity andacceleration. Inthec.g.s.system ofunits, thecomponents ofraremeasured incentimeters. Anycomponent ofthevelocity qisobtained by dividing anumber ofcentimeters byanumber ofseconds, and theresult isexpressed assomany centimeters persecond or, briefly, cm. sec.""1Theunitofvelocity isonecentimeter persecond. Anycomponentoftheacceleration fisobtained bydividing a velocity component bytime or,more precisely, anumber of centimeters persecond byanumber ofseconds, andtheresult is expressed assomany centimeters persecond persecond or, briefly, cm.sec.~2Theunitofacceleration isonecentimeter per second persecond. Iff.p.s. units areused, theabove italicized statements are modified bychanging theword "centimeter" to"foot." Although the c.g.s. and f.p.s. units arecommonly used in scientific work, there isnonecessity tolimit ourselves tothem. Theunits oflength andtimemaybechosen arbitrarily. Indeed, velocities arefrequently expressed inmiles perhour (m.p.h.). Since velocityisobtained bydividing length bytime,wesay ithasthe"dimensions" [LT~1 ]',similarly, acceleration hasthe dimensions [LT~2 ].Thisnotation isdiscussed intheAppendix. Gradient vector. When toeach point ofspace, ortoeach point ofaplane, a scalar isassigned, wesaythatwearcdealing withascalarfield. Thus thedistribution ofpressure (ortemperature) intheatmos- phere atacertain time gives ascalar field. Orconsider amapon which heights aremarked; theheight gives ascalar fieldover themap. When toeach point ofspace ortoeach point ofaplane a vector isassigned, wehaveavectorfield. Thewind velocity inthe atmosphere gives avector field. Weshallnowshow that ascalar field defines anassociated SEC. 1.3] FOUNDATIONS OFMECHANICS 29 vector field inaverysimple way. Forgenerality, weshallmake theargument three-dimensional, but thereader will find it interesting toconsider byway ofillustration themapmarked with heights. LetOxyzberectangular Cartesian axesandV(x, y,z)ascalar field. (For themap,wesuppress z]V(x,y)istheheight oftheland atthepoint x,y.)The surfaces V=constant arecalled level sur- faces. (Onthemap thelevel sur- faces become thecontour lines.) LetAbeanypoint and8thelevel surface passing through A(Fig. 8). Letusdraw thenormal to8onthe sideonwhich Vincreases, and t t., j.. FIG. 8.Thegradient vector. proceed anarbitrary distance n along thisnormal. ThenVisafunction ofn,and atAysinceVisincreasing. >Wenowintroduce avector AB, called thegradient ofVatA, orbriefly grad V. (Itisalsodenoted byW.) The defining propertiesofgradVareasfollows: (i)Itsdirection isalong thenormal toSatA,inthesense of Vincreasing. (ii)ItsmagnitudeisdV/dn, calculated atA. Inthecase ofthemap, gradVisperpendicular tothecontour lineandpointsintheuphill sense;itsmagnitude istherate of increase ofheight. Wenowproceed tofindthecomponents ofgradVontheaxes ofcoordinates. Leta,ft7bethedirection cosines ofthenormal toSinthesense ofVincreasing. Thenanyinfinitesimal dis- placement lyingin$,i.e.,making JTrdV ,.dV,,dV .A^-te^a^****-' alsomakes adx+ftdy+ydz=0.Hence, (U17) _,-*-. where 4>issome factor ofproportionality. Asweproceed along 30 PLANE MECHANICS [Sac. 1.3 thenormal toSatA,wehave, sinceVisafunction ofx}y,z, which inturn arefunctions ofn, ^ dn dxdn dydn dzdn dVtdV-.dV "tea+a^+aT- Substitution from (1.317) gives /iv (1.318)g=<!>(<**+(!*+r2 )=*. When wesubstitute thisvalue of<#>in(1.317), weget ,,o,mW dv dv' adv dv dv (1.319)^=5- a7-*5T ar-*ST' andsoby(1.305)thecomponents ofgradVontheaxesare dV dV dF dx1 dy'dz' Ifi,j,kistheunitorthogonal triad along theaxes, then (1.320) By(1.306) thecomponentofgradVonadirected lineLis where X,M>?arethedirection cosines ofL.Since dx dy dz \=-T-> /*=j->v^~r>ds ds ds where dsisanelement ofL,thiscomponentis +<>Y.dy,dVdz=dV f dxds dyds dzds ds Thecomponent ofgradVinanydirection istherateofchange ofV inthatdirection. Wehave seenhow toobtain avector field (grad V)from a scalar field (7). Itisnotingeneral possible toexpress an arbitrary vector fieldas-thegradient ofascalar field,butwefind inmathematical physics many vector fields thatcanbeso expressed. InSec. 2.4weshall discuss fields offorce; inmost cases ofphysicalinterest afield offorce isthegradientofascalar SBC. 1.4] FOUNDATIONS OFMECHANICS 31 field ofpotential energy (with achange ofsign). Since the description ofavector fieldrequires three functions andascalar fieldonlyonefunction, aconsiderable simplification results from theuseofthescalar field. Exercise. IfinaplaneVx2+j/2 ,findthecomponentofgradVat thepoint (1,0)inadirection making anangle of45withthes-axis. 1.4FUNDAMENTAL LAWS OFNEWTONIAN MECHANICS Letussuppose thatweareconducting experimentsinwhich small bodies ofmeasured masses areacted onbymeasured forces. Wechoose someframe ofreference andobserve themotions ofthe bodies relative toit.Weshall usethefollowing notation: P=force, m=mass, f=acceleration. Theunits offorce, mass, length, andtime arechosen arbitrarily. Weaskthisquestion: Asaphysical fact,isthereanysimple relation connecting P,m,and fforthemotion ofabody? The answer tothisquestion is,ingeneral: There isnosimple relation. Wehavethoughtoftheexperimentsasconducted inanyframe ofreference itmight bethecabin ofanairplane looping the loop, oritmight beanordinary laboratory. Weaskasecond question:Isitpossibletochoose aframe ofreference sothat there isasimplerelation connecting P,m,and f?Theanswer is:Yes. Oneframe ofreference yielding asimple relation among P,w,and fistheastronomical frame ofreference,inwhich the sun* isfixedandwhich iswithout rotation relative tothefixed stars asawhole. Thesimple relation is P=kmf, where A-isauniversal positive constant, thevalue ofwhich depends onlyontheunits employed, f *More accurately, themass center ofthesolar system (see Sec. 3.1); actually thispointisnotfarfrom thecenter ofthesun. fThis relation isinexcellent agreement with astronomical observations, butthere areexceptions;theorbit oftheplanet Mercury reveals aminute discrepancy. Although fewmodern astronomers acceptthis relation as absolute physical truth,itsvalidityissohighthat itistaken asthebasis of celestial mechanics. Totakeadeeper pointofview,wemust recast our whole mode, ofthought andfollow thegeneral theory ofrelativity. 32 PLANE MECHANICS [SBC. 1.4 Thethree laws. Weshallnow state thefundamental lawsonwhich Newtonian mechanics isbased. These arethelawsaccording towhich our mathematical model ofnature works. Thelaws asstated here areequivalent tothose usedbyNewton, butthey areexpressed inadifferent form.* LAWOFMOTION. Relative toabasicframe ofreference aparticle ofmassm,subjecttoaforce P,moves inaccordance with theequation (1.401) P=knd, where fistheacceleration oftheparticle andkauniversal positive constant, thevalue ofwhich depends onlyonthechoice ofunits of force, mass, length, andtime. Anyframe ofreference relative towhich (1.401) holds is called Newtonian. IfP=0,then f=by(1.401). Since f=dq/dt asin(1.314), itfollows thatqisaconstant vector. Inwords, aparticle under theinfluence ofnoforcetravels withconstant velocity; i.e.,ittravels inastraight linewith constant speed. Thisimportant special casewasstated separately byNewton ashisfirstlawofmotion, hissecond lawdealing with thecasewherePisnotzero. Thus, hisfirsttwolaws areincluded inourlawofmotion asstated above. LAW OFACTION ANDREACTION. When two particles exert forces ononeanother, these forces areequal inmagnitude and opposite insenseandactalong thelinejoining theparticles. This isoftensummed upbysaying: Action andreaction are equal andopposite. LAWOFTHEPARALLELOGRAM OFFORCES. When twoforcesP andQactonaparticle, theyaretogether equivalenttoasingle force P+Q,thevectorsumbeing defined bytheparallelogram construc- tionasinSec. 1.3. Itisusual tocallP+Qtheresultant ofPandQ;PandQare called thevector components ofP+Q.t Insetting upasystem oflaws oraxioms,itisgenerally con- *Fortheoriginal form ofNewton's laws, seeSirIsaac Newton's Mathe- matical Principles, translation revised byF.Cajori (Cambridge University Press, 1934), pp.13,644. fInvSec. 1.3weusedtheexpressionvector component only inthecasewhere thecomponents were perpendicular tooneanother. Itisconvenient to use italso inthepresent more general sense. SEC. 1.4] FOUNDATIONS OFMECHANICS 33 sidered desirable tomake them independent, inthesense that nooneofthem canbededuced from theothers. Many attempts havebeenmade todeduce theparallelogram offorces, but all these deductions require thestatement ofother laws, which are individually simpler than theparallelogram lawbutrather long tostate; so,forbrevity, weaccept theparallelogram lawdirectly.* Thethree lawsstated above form thelogical basis ofmechanics. Their fullmeaning canbeunderstood onlybyapplying them, and weshall notdelay thedevelopment ofthesubject byfurther general discussion. We should, however, point out their significancefortheprediction oftheresults ofordinary laboratory experiments. Letussuppose thatwearediscussing themotion ofabilliard ballwhich rollsdown aninclined planeinalaboratory. Our problemistopredict thebehavior oftheballbymathematical reasoning. Ontheonehand, wehave theactual physical apparatus, ontheother ourmathematical model, inwhich the ball, theplane, and theforces acting arereplaced bytheir mathematical idealizations. There isaprecise correspondence between thephysical things andtheingredients ofourmathe- matical model. Buthere there enters animportant question: What physical frame ofreference corresponds tothebasic Newtonian frame mentioned inthelaw ofmotion theframe relative towhich (1.401) holds? Wehave already indicated theanswer: The physical frame inquestionistheastronomical frame ofreference. Butwedonotwant toknow thebehavior ofthebilliard ball relative totheastronomical frame ofreference; wewant toknow itsbehavior relative tothewallsand floor ofthelaboratory, i.e., relative totheearth's surface. Isitlegitimate toregard the earth's surface asaphysical frame corresponding totheNew- tonian frame of(1.401)? Strictly speaking,itisnot. Very refined experiments would enable ustodetect differences between thephysical behavior ofthebilliard ballandthetheoretical predictions based onthat correspondence. These differences areduetotherotation oftheearth, fButthey areveryminute *Foraninteresting "proof"oftheparallelogram offorces, seeW.R. Hamilton, Mathematical Papers (Cambridge University Press, 1940), Vol. II,p.284. fInSec. 13.5weshall discuss some ofthedynamical consequences of theearth's rotation. 34 PLANE MECHANICS [SEC. 1.4 inthevastmajorityofexperiments made inalaboratory andin allproblems connected with engineering structures; excellent predictions maybemade from ourthree lawsbytaking the earth's surface tobethephysical frame ofreference corresponding totheNewtonian frame. Unitsanddimensions. Intheequation (1.401), aconstant kappears, and iftheequa- tion isleftinthisform, thisconstant kwilloccur throughout our dynamical equations. Toavoid this,wemake k=1bya special choice oftheunit offorce; theunitsochosen iscalled the dynamicalunit.When itisused, (1.401) reads (1.402) P=mi. Clearly thedynamical unit offorce prodiices unit acceleration inunitmass, since ifP=1andm=1,then/=1.Inthe c.g.s.and f.p.s. systems, thedynamical units offorce arecalled thedyneandthepoundal, respectively. Aforce ofonedyne produces anacceleration ofonecentimeter persecond per second inamass ofonegram, andaforce ofonepoundal produces anacceleration ofonefootpersecond persecond inamass of onepound. Since, in(1.402), force istheproductofamassbyanaccelera- tion, thedynemaybedescribed asonegram centimeter per second persecond, orbriefly, 1dyne=1gm.cm.sec."2 Similarly, 1poundal=1Ib.ft.see."2 Force measured indynamical units hasthedimensions [MLT~2 ]. Indynamical problems, weshall always assume that the dynamical unit offorce isused, sothat thelawofmotion is (1.402). Instatical problems, ontheother hand,weshall leave theunit offorce arbitrary, since there isnoadditional simplicity tobegained byrestrictingit. Atthispoint thereader isadvised tostudy theAppendixat theendofthebook inorder thathemayunderstand therefer- ences inthetexttounitsanddimensions. Inparticular, atten- tion isdirected totheuseofthetheoryofdimensions asaneasy andrapid check against slips incalculation acheck which isof great value both inelementary andadvanced work. Inany equation inmechanics, allterms must have thesame dimensions. SEC.1.5] FOUNDATIONS OFMECHANICS 35 Forexample,ifwehad carelessly derived theformula(cf. page 28) f=acocos co i aa>3sin&tj, aglance would show thatsomething iswrong with thisformula. The acceleration fhasdimensions [LT~2 ],awhasdimensions [LT~l ]yaw3hasdimensions [L77~3 ],while thetrigonometrical functions andthevectors i;jaredimensionless. 1.6.SUMMARY OFTHEFOUNDATIONS OFMECHANICS Thepurposes ofthis firstchapter havebeen(i)todigdown tothefundamental physical ideasand(ii)tolaythefoundations ofalogical structure. Before proceeding tothenext chapter, we nowextract andemphasize those concepts andlawswhich willbe required later. I.Theingredients ofmechanics. (a)Aparticle hasposition andmass(ra). (6)Arigid bodyisasystem ofparticles, thedistances between which remain unchanged. Itmay alsoberegarded asacon- tinuous distribution ofmatter. (c)Aframe ofreferenceisarigidbody inwhich axes ofcoordi- nates aretaken. (d)Aforce haspoint ofapplication, direction, andmagnitude. (e)Theunits ofmass,. length, andtime arearbitrary. Soalso istheunit offorce, but itisconvenient indynamics toconnect theunit offorce with theunits ofmass, length, andtime. II.Vectors. (a)Addition ofvectors iscarried outbymeans ofaparallelo- gram. (6)Theusual simple algebraic rules apply tovectors, butwe donotyetdefine themultiplicationofvectors byoneanother. (c)Avector function ofascalarmaybedifferentiated; the derivative isanother vector function. P=Pa+P2j+ then Pi,P2,PSarethescalar componentsofthevector ontheunit orthogonal triad(i,j,k). (e)ThecomponentsofgradVare dV dV }dV m dx* dy' dz' 36 PLANE MECHANICS [Kx. I (/)ThecomponentofgradVinanydirection isdV/ds. III.Velocity andacceleration ofaparticle. (a)Position vector: r=xi+yj+zk. (b)Velocity vector: q=f=xi+yj+zk. (c)Acceleration vector: f=q=xi+yj+zk. IV.Basic laws ofmechanics. (a)Law ofmotion: P=raf, (P=force). (b)Law ofaction andreaction: Action andreaction areequal andopposite. (c)Law oftheparallelogramofforces :P+Qistheresultant ofPandQ. EXERCISES I 1.Iftwoforces ofmagnitude PandQactataninclination toone another, prove thatthemagnitude oftheresultant Risgivenby 7t!2=P2+Q2+2PQcos B. 2.Aparticleisacted onbyforces ofmagnitudes PandQ,their lines of action making anangle withoneanother. They aretobebalanced by twoequal forces, acting atright angles tooneanother. Find thecommon magnitude ofthese forces. 3.Show that ifthemagnitudes ofanumber ofcoplanar vectors are multiplied byacommon factor, thedirections ofthevectors being unchanged, themagnitude ofthesum ismultiplied bythesame factor and itsdirection isunchanged. Show alsothat ifallthevectors arerotated through acommon angleintheir plane, without change ofmagnitude, their sum isrotated through thesame angle without change ofmagnitude. 4.Forces ofmagnitudes 3,4,and5Ib.wt.actatapoint indirections parallel tothesides ofanequilateral triangle taken inorder. Find their resultant. 5.Two forces acting inopposite directions onaparticle have aresultant of34Ib.wt.;ifthey acted atright angles tooneanother, their resultant would have amagnitude of50Ib.wt.Find themagnitudes oftheforces. 6.Anairplane dives at400miles perhour, losing height attherateof 220 ft.persec.What isthehorizontal component ofitsvelocity inmiles perhour? 7.Coplanar forces ofmagnitudes P,2P,4Pactonaparticle. How should theybedirected tomake theresultant(i)amaximum, (ii)amin- imum? 8.Ifanynumber ofcoplanar vectors allofthesame magnitude are drawn from apoint, arranged symmetrically sothat theangles between adjacent vectors areallequal, prove that theirsum iszero. 9.IfV=x*+y*+z2-fxy+x,atwhat points inspace isthevector gradVparallel tothez-axis? 10.Ascalar field isgiven overaplaneby i,-*'+y Ex. II FOUNDATIONS OFMECHANICS 37 What arethelevel curves? Show that, atthepoint with polar coordi- nates(r,0),gradVisinclined tothere-axis atanangle 20and itsmagnitude is$sec20. 11.Atrain, starting attime t=0,hasmoved intime tadistance x=at(l-e-6 '), where aand6arepositive constants. Find itsvelocity andacceleration; what dothesebecome afteralongtimehaselapsed? 12.Aparticle travels along astraight linewith constant acceleration /. Prove .s=ut+%ft*, v=u+ft,v*=uz+2/s, where sisthedistance covered from theinstant t=0,utheinitial velocity, and vthefinal velocity. 13.Auniformly accelerated automobile passes twotelephone poles with velocities 10m.p.h. and20m.p.h., respectively. Calculate itsvelocity when itishalfway between thepoles. 14.What curve isdescribed byaparticle moving inaccordance with the equation r=acos ct i+bsinct- j, wherea,6,careconstants aridi,jfixed unitvectors perpendicular toone another? Show that theacceleration isdirected toward theorigin. 15.Anelevator weighing onetonstarts upward with constant accelera- tionandattains avelocity of15ft.sec."1inadistance of10ft.Find in tonsweight thetension inthesupporting cable during theaccelerated motion. 16.Acarweighing 2tonscomes torestwithuniform deceleration from a speed of30m.ph.in100 ft.Find theforce exerted bythecar011theroad, showing itsdirection inadiagram. 17.Aparticle ofmassmmoves ontheaxisOxaccording totheequation x=*asinpt, where aandpareconstants. Express theforce acting onitasafunction ofx. 18.Atugtowsabarge A,which inturntows another barge B.They start tomove withanacceleration /.Find thetensions inthetowing cables,interms of/andm,mf(themasses ofthebarges). 19.Indicate thefallacy inthefollowing argument: Alocomotive pulls atrain. But toevery action there isanequal andopposite reaction. Therefore thetrain pulls thelocomotive backward withaforce equal tothe pullofthelocomotive, andsothere canbenomotion. 20.Ifthefundamental lawofmechanics foraparticle moving ona straightlinewere d( instead of(1.402),mand cbeing constants, findthedistance traveled from restintime tunder theaction ofaconstant force P.(This istherelativistic equation ofmotion; seeChap. XVI.) CHAPTER II METHODS OFPLANE STATICS 2.1.INTRODUCTORY NOTE Inorder todealsystematically with thesubject ofmechanics, wehave tobreak itupinto parts. The first division isinto statics anddynamics: statics dealswiththeequilibrium ofsystems atrest,anddynamics withthemotion ofsystems. Aswehave already seen, restandmotion areterms which havemeanings onlywhen aframe ofreference hasbeen specified. Thus the question atonce arises: When wesaythat statics deals with systems atrest,what physical frame ofreference havewein mind? Themathematical theory ofstatics isbased onthe fundamental laws ofSec. 1.4.Thusweshould beconfident ofa close agreement between theory andobservation ifwewere to develop statics relative totheastronomical frame. Butthat would notbephysically interesting, because there areactually nosystems atrest inthat frame; theearth's surface isthe physically interesting frame ofreference forstatics. Although (1.401)isnotsatisfied with great precision relative totheearth's surface,itispossible byamodification oftheforces(i.e.,by inclusion ofcentrifugal force) toobtain extremely satisfactory results instatics relative totheearth's surface. Thus inthe mathematical theory ofstatics theframe ofreference willbesuch thatthefundamental laws hold,andinthephysical interpreta- tion oftheresults theframe ofreference willbetheearth's surface. But there isanother division ofthesubject ofmechanics, namely, adivision intoplane mechanics andmechanics inspace. This division isartificial from aphysical point ofview but is convenient inlearning thesubject, because themathematics oftheplane theoryissimpler thanthemathematics ofthespace theory. This isduetothefactthatcertain quantities (moments offorces, angular velocity, andangular momentum) appear as scalars intheplane theory butasvectors inthespace theory. 38 SBC. 2.2) METHODS OFPLANE STATICS 39 Accordingly, toavoid undue mathematical complicationsinthe development, weshall deal firstwithplane mechanics, butwhere thespace theory presents nodifficulty weshall developit simultaneously. Tobeprecise, thesubject ofplane mechanics deals with (i)The statics anddynamics ofasystemofparticles lyingina fixed plane. (ii)The statics anddynamics ofrigid bodies which canmove only parallel toafixed plane, thedisplacement orvelocityof every particle being parallel tothefixed plane. Asanexampleof(i),wemaymention theproblemofthe motion oftheearth relative tothesun,both being treated as particles, and asanexample of(ii)themotion ofacylinder rolling down aninclined plane. Theplaneinwhich thesystem lies,ortowhich themotion is parallel,willbecalled thefundamental plane. We shall find inboth statics anddynamics that thebasic laws lead tocertain general principlesormethods. When any specific problem presents itself,wedonotattack itdirectly from first principlesasarule;wecanavoid waste ofenergy by applyingtotheproblem one ofthegeneral methods. The present chapterisdevoted togeneral methods inplane statics, with inclusion ofthespace theory where itpresents nodifficulty. 2.2.EQUILIBRIUM OFAPARTICLE According to(1.401) aparticlewillhaveanacceleration unless theforce acting onitvanishes. Thus ifPistheforce acting onaparticle, thenecessary and sufficient condition forequilib- rium is (2.201)P=0. Ifseveral forces P,Q,R actonaparticle, thenecessary and sufficient condition forequilibriumisthevanishing ofthe resultant force, i.e., (2.202) P+Q+R+''=0. Thisvector condition may alsobeexpressedinscalar formby means ofcomponents.Letaparticle beacted onbyforces whose components onanorthogonal triad areindicated as follows: 40 PLANE MECHANICS [Sac. 2.2 P(P 1,P2,P,), Q,Oi), #2,Ra), Then theconditions forequilibrium are (Pi+0i+fii+ -o, (2.203))P2+Q2+#2+ =0, (P,+Q3+Rs+'=0. Alltheabove remarks holdwhether theforces acting onthe particlelieinaplane ornot.Theparticular feature oftheplane case isthatonlytwocomponents aretobeconsidered instead of three. Nodifficultywillbefound inusing (2.202) toprove thefollow- ingtheorems which areoften useful : (i)THETRIANGLE OFFORCES. Ifaparticleisinequilibrium under theaction ofthree forces, these forcesmayberepresented inmagnitude anddirection bythethree sides ofatriangle, taken inorder(Fig. 9). R Forces onpatficle"VQ Triangle offorces Fio. 9.Thetriangle offorces. (ii)THEPOLYGON OFFORCES. Ifaparticleisinequilibrium under theaction ofseveral forces, these forcesmayberepre- sented bythesides ofaclosed polygon, taken inorder. (iii)LAMY'S THEOREM. Ifaparticleisinequilibrium under theaction ofthree forces P,Q,R,then P__Q __R (2.204)sina sinft sin7 where aistheangle between QandR, fttheangle between Rand P,and7theangle between PandQ. Exercise. Aparticleisinequilibrium under three forces. Two ofthe forces actatright angles tooneanother, onebeing double theother. The SEC. 2.3] METHODS OFPLANE STATICS 41 third force hasamagnitude 10Ib.wt.Find themagnitudesoftheother two. 2.3.EQUILIBRIUM OFASYSTEM OFPARTICLES Systems ofparticles. Letusnow consider asystem ofparticles. Wemust inall casescome toaclearunderstanding astowhat isincluded inthe system under consideration. Letussuppose thatwearedealing with abook which restsonatable, thetable standing onthe floor. Thebookandthetable areregarded (inourmathematical model) ascomposed ofavery greatnumber ofparticles. In talking about thisarrangement ofmatter, wearenotcompelled tothink ofthe"system" under consideration ascomposed ofthebookandthetable. Ifwelike,wemaythink ofthetable alone asasystem, orthebook alone asasystem, orthehundredth pageofthebook asasystem. Itisimportant torealize thatthesystem under consideration issomething wepick outfrom thegiven arrangement inan arbitrary manner. Itisnecessary tounderstand this inorder toappreciatethedistinction between external andinternal forces. External andinternal forces. Thebook presses down onthetable andthetable presses uponthebook withanequal andopposite force (law ofaction andreaction).Ifthesystemisbook+table, these forces are both exerted byparticlesofthesystem. But ifthesystem consists ofthebook only,this isnolonger thecase; theforce exerted bythetable onthebook isdue,nottoparticles ofthe system, buttoparticles lying outside thesystem. Wemake thefollowing generaldefinition: Aforce acting onaparticleofagiven systemisaninternal forcewhen itisexerted byanother particleofthatsystem; otherwise itisanexternal force. Inaccordance with thelawofaction and reaction, internal forces occur inequal andopposite pairs, each pairrepresenting themutual interactions ofapairofparticles ofthesystem. Figure 10shows three particles A,B,C,thebroken lineindi- cating theboundary ofthesystem. The forces areclassified asfollows: 42 PLANE MECHANICS ISnc. 2.3 External: PatA,QatB,RatC. Internal: UatB,-UatC,Vat(7,-VatA,Wat A,-Wat B. Figure 11shows these particles situated precisely asinFig.10 andsubject tothesame .forces. Theonly difference between the twodiagrams liesintheposition ofthebrokenline, indicating the-boundary ofthesystem under consideration. InFig.11the system contains only theparticles AandB.NowPat"A,-VatA,QatB,UatBareexternal; andWatA,WatBare internal. Ifwereduce thesystem under consideration toone particle only, there arenolonger anyinternal forces. FIG. 10.External and internal forces. The"system" isenclosed bythebroken line.FIG. 11.Thesame particles and forces asinFig. 10,butadifferent "system." Theonly essential difference between thetreatment ofthis question ofinternal andexternal forces inaplane and itstreat- ment inspaceisthat intheformer casewemay delimit the system byaclosed curve, whereas inthelatter casewerequire a closed surface. Necessary conditions forequilibrium (forces). Weshallnowobtain necessary conditions fortheequilibrium 6fanysystem ofparticles. Themeaning oftheword"neces- sary" should beemphasized: these conditions must besatisfied if thesystemisinequilibrium, butthesatisfaction oftheconditions doesnotimply thatthere isequilibrium. Consider any particle ofthesystem, assumed inequilibrium. Thatparticleisitself inequilibrium, andhence thevector sum of allforces acting onitiszero. Similarly forallparticles. Hence, Thevectorsumofallforces acting onallparticlesiszero. Buttheforces aresome external, some internal. Wemay state SEC. 2.3] METHODS OFPLANE STATICS 43 Thevectorsumofallexternal forces and allinternal forces iszero. From theequality ofaction andreaction, Thevectorsum ofallinternal forces iszero. Comparing thiswith thepreceding statement, wehave (2.301) Thevectorsumofallexternal forcesiszero. Herewehave thefirst ofthenecessary conditions fortheequilib- rium ofasystem ofparticles. Exercise. Consider aglass ofwater standing onatable, taking asthe system (i)thewateronly, (ii)thewater andthe glass. What arethe external forces ineachcase,andwhat does (2.301)tellusabout them? Themoment ofavector about aline. Consider alineLandabound vector P,perpendicular toL butnotintersectingit.Letabethelength ofthecommon L L P A FIG. 12.AlineLand a vector Pperpendicular to it. FIG. 13. (a)Af Themoments inthetwocases have opposite signs. perpendicular toLandtheline ofaction ofP.Themoment of PaboutLisdefined tobe (2.302) M=aP. Theambiguous signisintroduced inorder thatwemay distin- guish between thetwocasesshown inFig. 12,inwhich thevector Pindicates rotations inopposite senses about L.Having decided tousethe+sign forvectors indicating rotations inonesense, weusethe signforvectors indicating rotations intheopposite sense. The significanceofthesignswillbebetter understood when Sec. 9.3hasbeen read, butforthepresent thefollowing descriptionoftheconvention willserve. Letussuppose thelineLdraAvn perpendicular totheplane ofthe paper, intersectingitatthepointA(Figs. 13aand b) ;themoment is+aPwhenPindicates acounterclockwise rotation around A intheplane, and aPwhen aclockwise rotation isindicated. Consider nowalineLandabound vectorPwhich isnotper- pendicular toL.Themoment ofPaboutLisdefined tobethe 44 PLANE MECHANICS [SEC. 2.3 moment aboutLoftheprojectionofPonaplane perpendicular toL,thelattermoment having already been defined above. Thefollowing facts arenowobvious : (i)Themoment ofPaboutLisunaltered ifPismade toslide alongitsline ofaction without change ofmagnitude orsense. (Itisseen atonce that thisonlyslides theprojection along its lineofaction, without change ofmagnitude orsense.) (ii)Thesum ofthemoments about any lineLoftwovectors P,Psituated onthesame line iszero. (Their moments are equalinmagnitude butoppositeinsign.) (iii)Themoment aboutLofavector Pisunaltered ifPis moved without change ofmagnitude ordirection inadirection parallel toL.(This doesnotchangeitsprojection onaplane perpendicular toL.) When wedealwith themoments ofcoplanar vectors about a lineperpendicular totheir plane, weoften refer tothesemoments asmoments about apoint, namely, thepoint where thelinecuts the plane. Thetheorem ofVarignon. Consider alineLandabound vectorRatapointB(Fig. 14). LetAbethefoot oftheperpendic- ulardropped fromBonL.LetQ betheprojectionofRontheplane through Bperpendicular toL.Let Nbothelinethrough Bperpendic- ular toABandtoL,and letPbe theprojection ofQonN. Itis clear thatPisalsotheprojectionof RonN.Wewish toprove that themoment ofRaboutLis equaltothemoment ofPabout L. Figure 15shows theplane containing P,Q,andAB. LetAC bedrawn perpendicular totheline ofaction ofQ,and letthe angleCABbedenoted byB.Then themomentMofRaboutL equals themoment ofQaboutL(bydefinition), sothat*M=Q-AC=QABcos=ABQcos6=AB-P, which isthemoment ofPabout L.Hence, themoment ofavector *Forsimplicity, wehave taken thecasewhereMispositive; theother case isdealt with similarly.Q FIG. 14.By definition, Q andRhave thesamemoment about L\itistobeproved thatP alsohasthesamemoment. SEC. 2.3] METHODS OFPLANE STATICS 45 atBabout alineLisequaltothemoment aboutLoftheprojection ofthevector onthelinethrough Bperpendicular totheplane con- tainingBandL. f Since theprojection onany lineofthesum ofanynumber of vectors isequal tothesum oftheir projections onthat line,the theorem ofVarignon follows immediately: Thesumofthemoments about alineLofvectors P,Q,R, , withcommonorigin B,isequaltothemoment aboutLofthesingle vectorP+Q+R-f-with origin B. FIG. 15.Thoplane contain- ingPandQ.O Fia. 16.Analytical method of finding themoment ofavector. Instatics theonlyvectors whose moments wehave occasion to consider areforces. But itshould benoted that theabove definitions andtheorems hold foranyvectors. Weshallhave occasion tousethem inChap.Vindiscussing angular momentum. Thefollowing analytical result isimportant;itisobvious from Fig. 16,which shows theprojection ontheplane Oxy. Ifa vector withcomponents (X,Y,Z)actsatthepoint (x,y,z),then its moment about theperpendiculartotheplaneOxy at is (2.303) M=xY-yX. Themoment about theperpendicular totheplane Oxy atthe point (a,6)is (2.304) M-(x-a)Y-(y-b)X. These formulas take care ofthesign ofMautomatically. Exercise. Aforce, offixedmagnitude Randvariable inclination Btothe z-axis, acts intheplane Oxyatthefixed point (a,6).Find itsmoment about theorigin asafunction of0,andobtain thevalues offorwhich this moment (i)isamaximum, (ii)isaminimum, (iii)vanishes. Necessary conditions forequilibrium (moments). Letusnowreturn totheconsideration ofasystem ofparticles inequilibrium. LetLbeanyline. Consider any particle. Since itisinequilibrium,theresultant ofalltheforces acting on 46 PLANE MECHANICS [SEC. 2.3 itiszero; hence, byVarignon's theorem thesum ofthemoments aboutLofallforces acting ontheparticleiszero. Similarly forallparticles. Hence, Thesum ofthemoments aboutLofallexternal forces and all internal forces iszero. Butthesum ofthemoments ofapair ofequal andopposite internal forces iszero. Hence, Thesumofthemoments aboutLofallinternal forces iszero. Comparing thiswith thepreceding statement, wehave (2.305) Thesum ofthemoments aboutLofallexternal forcesis zero. From(2.301) and (2.305) wemaynow state necessary condi- tions fortheequilibrium ofanysystem ofparticles. //asystem ofparticles isinequilibrium, then (i)ThevectorsumofallEXTERNAL forcesiszero. (ii)Thesum ofthemoments ofallEXTERNAL forces about anyline iszero. This result isthekeytothesolution ofstatical problems, and itshould beremembered. The particular form oftheabove statement applicable to plane statics isasfollows: //asystem ofparticlesisinequilibrium, then (2.306) F=0,N-0, whereFisthevectorsum oftheprojections ofallEXTERNAL forces onthefundamental plane, andNthesum ofthemoments ofallEXTERNAL forces about any lineperpendiculartothe fundamental plane. Itisconvenient tohaveanexplicit form of(2.306). Letaxes bechosen sothatthefundamental planeisz=0.Weshalltake moments about the z-axis. Suppose now that thesystemis acted onbyexternal forces withcomponents (Xi, YI,Zi), (X*,72,Z2), (Xn,Yn,Zn)atpoints (xi, y\,Zi), (x*,yz,z2), (x, 2/n,Zn).Byprojection onthefundamental plane,it follows thatFandNareunaltered ifwereplace thissystem.by forces inthefundamental plane with components (Xi, FI), (X2,Y2),--(Xn,Yn)atpoints (xi, ?/i), (#2, 2/2), (xn,2/). Hence, by(2.303) weseethat if(X,Y)arethecomponents ofF, then (2.307)X-2Xt,Y=2Kt,N=5(x%Yi SBC. 2.3) METHODS OFPLANE STATICS 47 andsonecessary conditions forequilibrium are (2.308) Xi=0,VYt=0,V(.rtr<-yXj-0. <-i 1=1 t~i Itmight bethought that,bytaking moments about other lines perpendicular tothefundamental plane, new conditions might beobtained; butthis isnotso. For,by(2.304),ifmoments aretaken about aperpendicular totheplane at(a,6),then the totalmoment is -(y.-6)AM=V(x.Y< and thisvanishes automaticallyif(2.308) aresatisfied. Thus conditions ofthetype (2.306) or(2.308) areactually three in number, andnomore. Thefollowing important results areeasy toestablish from the principles laiddown above: (i)Ifasystemisinequilibrium under theaction ofonlytwo external forces, then these forces have acommon lineofaction, equal magnitudes, andopposite senses. (ii)Ifasystemisinequilibrium under theaction ofonlythroe external forces, these forces lieinapiano andthoir lines ofaction areeither concurrent orparallel. Ifaforce ismeasured indynamical units,itsmoment hasdimensions [ML2?1"2 ](seeAppendix). Inthec.gs.system, theunitmoment is 1 dyneem.or1gm.cm.2sec."2 ;inthef.p.s. system,itis1ft.poundal or1Ih. ft.2see."2Instatics wefrequently useaunit offorcowhich isnotadynami- calunit, such ustheIb.wt.orthetonwt.(contracted toread Ib.andton). Thecorresponding moments aremeasured inft.Ib.orft.ton. Equipollent systems offorces. Two systems offorces aresaid tobeequipollent* when the following conditions aresatisfied: *Theword equivalentisoften used. Butthere isadanger ofconfusion inusing acommon word inatechnical sense. Wemight think thatthe effects oftwosuch force systems were thesame; this isnotalways thecase. Ifwepullastring with equal andoppositeforces atitsends,weproduce a very different effect from thatcaused bypushingitwiththese forces reversed indirection; yetthetwoforce systems areequipollent. 48 PLANE MECHANICS [Sic. 2.3 (i)Thevector sumofalltheforces ofonesystemisequal tothe vector sum ofalltheforces oftheother system. (ii)Thesum ofthemoments ofalltheforces ofonesystem about anarbitrary line isequal tothesum ofthemoments of alltheforces oftheother system about that line. Forthediscussion ofplane mechanics, werequire only a restricted type ofequipollence, which weshall callplane equi- pollence. Twosystems offorces aresaidtobeplane-equipollent forthefundamental planeif (i)The vector sum oftheprojections onthefundamental plane ofalltheforces ofonesystemisequal tothevector sum oftheprojections onthatplane ofalltheforces oftheother system. (ii)Thesum ofthemoments ofalltheforces ofonesystem about anarbitrarylineperpendicular tothefundamental plane isequal tothesum ofthemoments ofalltheforces oftheother system about thesame line. Itisevident that ifF,F'arethevector sums oftheprojections oftheforces ofthetwosystems, andN,N'themoments about someline,thentheconditions forplane equipollence are (2.309) F=F',N=N'. IfF=0,N=0,wesaythatthesystemisplane-equipollentto zero. Thus,ifasystem ofparticlesisinequilibrium, theforces acting onitareplane-equipollent tozero. Theidea ofequipollenceisextremely useful instatics. The solutions ofproblemsinplanestatics turnontheconditions (2.306), and difficulties may arise inthecalculation ofFandN. These difficulties arereduced bysplitting upthecalculation into parts, each ofwhich issimple. This reduction depends onthe following fact(obvious from thedefinition ofequipollence):Forthe calculation ofthevectorsumoftheprojections oftheforces ofasys- temonthefundamental plane andthesumoftheirmoments about a lineperpendiculartothatplane, anysystem offorcesmay bereplaced byasystem plane-equipollenttoit. Inwhat follows below, weshallspeak only offorces inthe fundamental plane. Thismakes forsimplicity ofexpression without anyreal lossofgenerality, because inproblemsofplane mechanics weareactually interested inprojections onthefunda- mental plane andmoments about lines perpendicular toit; SBC. 2.3] METHODS OFPLANE STATICS 49 forthecalculation ofthese,wemay replace agiven forcesystem byaplane-equipollent system inthefundamental plane. Exercise. Find asystem oftwoforces equipollent toasystem ofthree forces represented bythesides ofanequilateral triangle taken inorder. Couples. Acouploisdefined asapair offorces acting onparallel lines, equalinmagnitude andopposite insense. Infact, acouple consists ofapair offorces P, P.Alinedrawn perpendicular tothetwo lines ofaction andterminated bythem iscalled the arm ofthecouple. A A 17cr. 176. FIG. 17. (a)Couple with positive moment, (b)Couple withnegative moment. Thus thevector sum oftheforces constituting acoupleiszero. Consider nowthesum ofthemoments oftheforces forming a couple about anylineL,perpendicular totheplaneofthecouple andcuttingitatA(Figs. I7aand17fr). IfAB,ACarcthe perpendiculars dropped fromAonthelines ofaction, thesum ofthemoments is M=P-AC-PAB=PBC=Pa (Fig. 17a),M=P-AB-PAC=-PBC=-Pa (Fig. 176), where aisthearmofthecouple. Theruleforsigniseasily seen tobeasfollows: (2.310) M=Pa, where the+or signistobetaken according astheforces indicate apositive (counterclockwise) rotation oranegative (clockwise) rotation intheplaneofthecouple about anypoint taken between their lines ofaction. Weobserve thatthesum ofthemoments oftheforces forming acouple about alineperpendiculartoitsplaneisthesame forall 50 PLANE MECHANICS [SBC. 2.3 such lines. Hence, wemayspeak inanabsolute sense ofthe moment ofacouple. Since thevector sum offorces inacoupleis zero,itfollows that twocouples inthefundamental plane are plane-equipollent iftheyhave thesamemoment. Itisevident thattwocouples ofmoments M,Mrinthefunda- mental plane aretogether plane-equipollent toasingle couple of momentM+Mrinthat plane. Reduction ofageneral plane force system. Consider asystem offorces inthefundamental plane. Let Fbetheir vector sumandNthesum oftheirmoments about some point intheplane.*Consider ontheotherhand asingle forceFatandacouple ofmomentNinthefundamental plane. Obviously, thissimple systemisplane-equipollent tothegiven system. Hence, wemaystate thefollowing general result: Asystem offorces inthefunda- mental planeisplane-equipollent toasingle force applied atanarbi- trary point intheplane, together withacouple. FIG. 18.Reduction ofaforceanda Itiscustomary alsotoexpresscouple toasingle force. ., . , . ,,.fthisbysaying that asystem of forces inthefundamental planemaybereduced toaforceanda couple. Expressed analytically, asystem offorces (Xi, Fi),(Xz,F2), -(Xn,Fn)atpoints (xi,yi), (x2,2/2), (x,y)may bo reduced toasingle force attheorigin withcomponents X,F, together withacoupleN9where (2.311)X=Xi9Y=Ft,N Weshallnowshow thatastillgreater reduction ispossible. Let beanarbitrary point. Theforce system, aswealready know,maybereduced toaforceFat0,together withacouple ofmoment N.InFig. 18,theforceFat isshown asOB.We nowconsider thefollowing twocases: *Thatis,themoments about alineperpendicular tothefundamental plane, cuttingitatO(see p.44). SBC. 2.3] METHODS OFPLANE STATICS 51 CASE(i):F7*0.LetussupposeNpositive forsimplicity, thecasewhereNisnegative being similarly dealt with.We drawOAperpendicular toOBasinFig.18andmeasure offOA equal toN/F. Then thecoupleisequipollent tothepairofforces OC(or-Fat0) ;AD(orFatA). Thus thegiven system isreduced tothethree forces Fat0,-Fat0,FatA. These areequipollent toasingle forceFatA. CASE(ii):F=0.Here thesystemisreduced toacouple. Hence wemay state thefollowing general result : Anyforce system inthefundamental planemay bereduced either toasingle forceortoacouple. Itmaybenoted thatreduction toasingle force willoccurmuch more frequently than reduction toacouple, because reduction toacouple occurs onlywhen aspecial condition issatisfied, viz.,F=0,or,inother words, when thevector sum ofthe forces inthesystemiszero. Letusnowcarry outthisreduction toasingle force ortoa couple analytically, thegiven plane force system being specified asconsistingofforces withcomponents (Xi, FI),(X^ Y%), (Xn,Yn)atpoints withcoordinates (xi,y\\ (#2, 2/2), (#n, 2/n). Letussuppose that thissystem may bereduced toaforce withcomponents (X,F)at(xty).The conditions ofplane equipollence are (2.312) X=2*X^F=2)Ft, n xY z/X=V(x*^" The firsttwoequations determine thecomponents ofthesingle force. The lastequation gives onerelation between thecoordi- nates ofthepointofapplication; this relation, being linear, defines astraightline. Itisseen atonce that this linepoints inthedirection oftheforce withcomponents (X,F);itis,in fact, the line ofaction ofthat force. The lastequation of (2.312) leaves thepointofapplication indeterminate toacertain extent itmaytakeanyposition onacertain line. Oneparticu- 52 PLANE MECHANICS [Sue. 2.3 larpointonthisline isgivenbythesymmetrical formulas (2.313) x=-=--,y Ifitshould happen that (2.314) j\X>=0,V7,=0,2feF<-yA)^0, t=i i=I~ itisevident thatwecannot findZ,F;a;,ytosatisfy (2.312). Then thesystem cannot bereduced toasingle force. Tofind thecouple towhich itcanbereduced, wecompareitwith the couple consisting offorces(0, 7), (0,F)atthepoints (0,0), (0,x\respectively. Theconditions ofequipollence are +=0,-7+F=0,xY==V i=l These equations aresatisfied provided thatthemoment ofthe coupleis (2.315) M=(x,y.- 2/tXt). Tosumup:Ingeneral aplane system offorces isplane-equi- pollenttoasingle force whose components and lineofaction are given by(2.312). // (2.316) iX=0, thesystem isplane-equipollenttoacouple withmoment given by (2.315). //(2.316) aresatisfied andalso (2.317) 2}(xtyf-ytZt)-0, then thegiven systemisplane-equipollenttozero. Exercise. Forces ofmagnitudes 2and3actparallel tothex-axis atpoints (1,3)and (2,4),respectively. Reduce them(i)toaforce attheorigin and acouple, (ii)toasingle force. SEC.2.4] METHODS OFPLANE STATICS 53 2.4.WORK ANDPOTENTIAL ENERGY Definition ofwork. Consider aparticle Aonwhich aforcePacts. Lettheparticle begiven aninfinitesimal displacement ofmagnitude 5s,repre- sented byAB(Fig. 19).TheworkdonebyPinthisdisplace- ment isdefined tobetheproduct of5sandthecomponentof Pinthedirection ofthedisplacement;infact, thework5W is (2.401) 8W=Pcos6-5s, where 6istheangle between Pandthedisplacement.*Itwill bepositive ornegative according as Qisacute orobtuse.A Since thesum ofcomponents inany direction isequal tothecomponent of ,,.,, ,,. ,. ., Fia. 19.The forcePdoesthesum inthat direction,itfollowsworkwhcntheparticle onwhich thatthetotalworkdonebyanynum-itactsreceives thedisplacement berofforces P,Q,R, ,acting on aparticle,isequal tothework donebytheir resultant P+Q +R+.-.. Itisevident that8W isalsoequal totheproduct ofPandthe component ofthedisplacementinthedirection ofP.Hence thework donebyPinasuccession ofsmall displacementsis equal totheworkdonebyPintheresultant displacement. LetX,Y,ZbethecomponentsofPinthedirections ofthe axes ofcoordinates, and letthecoordinates ofA,Bbe(x,y,z), (x+dx,y+dy,z+5z),respectively. Then, since thedirection cosines ofPareX/P, Y/P,Z/Pandthose ofthedisplacement ABare5z/5s, 5?//5s, 5s/5s,wehave Xdx ,F5?/.Zdz (2.402) cosQ=-D-r-+-Q-r+DT-> tos ios i5s andsotheworkdonebyPinthedisplacement AB is (2.403) 8W=Xdx+YSy+Z5z. *Theuseofthesymbol5instead ofthemore usual d(fordifferential) is traditional, andnotofmuch importance asfarasstatics isconcerned. But indynamicsitisnecessary todistinguish between apurely hypothetical (orvirtual) displacement dxandthedisplacement dxactually occurring in time dt(ie.,dx xdt). 54 PLANE MECHANICS [Ssc. 2.4 Wenote thatnowork isdonewhen thedisplacementis perpendicular totheforce. Ifforce ismeasured indynamical units, work hasdimensions [AfL2jT~2 ] (seeAppendix). Inthec.g.s. system, theunit ofwork istheerg,which is1 dynecm.or1gm.cm.2sec."2 ;inthefp.s.system,itis1ft.poundal or1Ib. ft.2sec*"2Instatics, the ft.Ib.and ft.tonareused. Rate ofworkingiscalled power. Indynamical units, power hasdimen- sions[ML2r~3 ].Aunitcommonly employedisthehorsepower (550ft.Ib. wt.sec.""1=1hp.). Forces which donowork. Consider aparticleincontact with thesurface ofafixed rigid body, andsuppose thataforce actsontheparticle tending to drive itinto thebody.Ifnoother force acted, theparticle would have topenetrate thebody, inaccordance with (1.401), Since, however, weregard such penetration asimpossible, wemust assume theexistence ofanother force, thereaction of thesurface, which prevents thepenetration from taking place. The particle inquestion maybeanisolated particle, oritmay beoneoftheparticles ofarigidbody. Inamechanical problem thereactions between particles and surfaces, orbetween pairs ofsurfaces, arenotingeneral tobe regarded asknown forces. They arecalled intoplay solely to prevent violation ofthecondition ofnon-penetration. Thewords "smooth" and"rough" arefamiliar inordinary life;wespeakofpolished steel, glass, ice, etc., assmooth, and sandpaper, cloth, etc.,asrough. Wemake suchaclassification primarily onthebasis ofoursense oftouch. Amore scientific classification isobtained byexamining thedirections ofthereac- tionsbetween bodies. Itisfound thatwithsmooth bodies the reactions alwayslieveryclose tothecommon normal ofthe surfaces incontact. Asanidealization ofsuch bodies, weadmit intoourmathematical model theconcept ofasmooth surface with thefollowing property: Thereaction atasmooth surface isnormal tothesurface, and isofsuchamagnitude asjusttoprevent penetration oroverlapping inspacefrom taking place. Thereaction atarough surface hasmore complicated proper- tieswhich willbediscussed inSec. 3.2. Since thereaction atasmooth surface isnormal tothesurface, thefollowing result isevident from (2.401): SBC.2.4] METHODS OFPLANE STATICS 55 Nowork isdonebythereaction atasmooth fixed surface inan infinitesimal displacement which preserves thecontact. Inanycontact there areactually twoequal andopposite forces involved, oneacting oneach body. Inwhat hasbeen saidabove, onebodywassupposed fixed, sothattheforce acting onitdidnowork. Suppose now thatboth bodies, having smooth contact, aredisplaced infinitesimally. The displace- ments oftheparticles ofthetwobodies incontact withone another maynowhavecomponents along thecommon normal, andsowork isdonebyeach ofthetwoforces ofreaction. But itis*not difficult toseethat the sum ofthese twoworks iszero. Letusnow consider arolling contact, confining ourattention here, forsimplicity, totherolling ofarigid circle onafixed linein thefundamental plane (Fig. 20). __....._. Wesaythat thecircleCrolls A'B onthelineLifitpasses con-F''20-AcircleroUmg on tinuously through asequence ofpositions such that(i)Lis always tangential toC; (ii)ifanytwopoints A,BofCmake contact withpoints A',B'ofI/,then arcAB=A'B'. (Consider amotor tireandthepatternitleaves ontheroad.) IfCadvances adistance Aswhile itturns through anangle A0,itisclear that (2.404) Az=aA0, where aistheradius ofC.During thisadvance thepoint ofC initially incontact withLreceives thefollowing displacements: Horizontal: Ax asinA0, Vertical: aacosA0. Ifthedisplacement As isinfinitesimal, then these displace- ments areinfinitesimals oftheorders (Ax)3 ,(Ax)2 ,respectively. This iseasily seenonusing thewell-known series forsineand cosine. Suppose nowthat there isareaction Rexerted byLonC. Weshall notassume thatRisperpendicular toL.Thework WdonebyRinasuccession ofinfinitesimal rolling operations 56 PLANE MECHANICS [SBC. 2.4 willbeafunction ofxythefinaldisplacement. Itmaybewritten Butsince thedisplacementofthepoint ofcontact isaninfinitesi- malofhigher order than theincrement inx,wehavedW/dx= andhenceW=0.Thus, nowork isdone bythereaction ata rolling contact. Itistruethatourproof deals onlywith thecase ofacircle rolling onaline; thegeneral case ofamoving curve rolling onafixed curvemaybediscussed byaslightly more complicated argument leading tothesame result. Thecondition ofrollingistheequality ofarcsonthetwocurves between points ofcontact; theessential point intheproofisthefactthatthe displacement ofthepoint ofthemoving curve instantaneously incontact isaninfinitesimal ofhigher order thantheinfinitesimal angle through which themoving curve turns. When rolling takes place between twomoving surfaces, the sumoftheworks donebytheequalandopposite reactions iszero. This follows from thefactthat thedisplacements ofthetwo particles incontact (onebelong- ingtoeachbody) areequal, to thefirstorder ofsmall quantities. The factthatnowork isdone atarolling contact isofenormous importance inmodern transport, -Internal reactions inawhich moveg Qnwhcels inCOn- rigid body. trast tothedragged vehicles of primitive civilizations. Itexplains whyboats arelaunched or hauled outofthewater withmuch greater easewhen placed on rollers, andwhymachinery operates more easily onballorroller bearings thanonplain bearings. Letusnow consider thework donebyapair ofequal and opposite reactions, exerted ononeanother bytwo particles of arigid body, when thebody receives aninfinitesimal displace- ment. LetA,Bbethepositions ofthetwo particles before displacement, and A',Bftheir positions after displacement (Fig. 21).The linesAB,A!B'make aninfinitesimal angle with oneanother, and (2.405) AB-A'B', SEC. 2.4] METHODS OFPLANE STATICS 57 since thebody isrigid. IfAQ,BQaretheprojections ofA',B' respectively onthelineAB,wehave obviously (2.406) AA+AB=AB+BB Q. Since theinclination ofA'B' toAB isinfinitesimal, AoB=A'B' tothefirstorder ofinfinitesimals; hence,AQBQ=ABby(2.405) andso(2.406) gives (2.407) AA=BB Q. Iftheforces onA,Barcrespectively P, P,theamounts of workdonebythem are P-AA,-PBB, andthesum ofthese iszero. Hence, nowork isdonebyapairof equalandopposite reactions, exerted ononeanother bytwoparticles ofarigid body. Tosumup,wehave seenthatnowork isdoneby (i)thereaction onamovable body insmooth contact witha fixedbody; (ii)thepair ofreactions atasmooth contact; (iii)thereaction onabody rolling onafixedbody; (iv)thepair ofreactions atarolling contact; (v)thepairofreactions between twoparticlesofarigidbody. Inthecases considered above, theparticles forming thesystems arenotwhollyfree. Thesystems are, infact, subject tocon- straints, andthereactions arebrought intoplay toprevent the violation ofthese constraints. Since these reactions ofconstraint donowork inpermissible displacements, wespeak ofthese constraints asworkless. The principleofvirtual work. Letusstartbyconsidering thesimple system shown inFig. 22. AB isarigid bar;asmall hole isdrilled initatC,andasmooth pinpasses through thehole, attaching thebartosome fixed support (notshown). Thus thebarcanturnaboutCinthe plane ofthepaper. ForcesPandQareapplied atAandB, respectively,indirections perpendiculartoAB. There aretwowaysofregardingthissystem: (i)Itisjustarigidbody, which canturnroundCandwhich issubjected totheforcesPandQ. 58 PLANE MECHANICS [SEC. 2.4 (ii)Itisacollection ofavastnumber ofparticles, subjected notonly totheforcesPandQ,butalsotoavastnumber of reactions between theparticles andareaction atC,alladjusted sothatthedistances between theparticles remain constant and theparticles nearCdonotmoveawayfrom thepin. Forpresent purposes weregard (ii)asthemore useful view. Indeveloping theprinciple ofvirtual work, wehave todeal with displacements, forces, andwork. Theonly displacement consistent with theconstraints isa rotation around C.But ifwetakethepoint ofview(ii)above, wemay think ofother displace- Bments inwhich theconstraints areviolated forexample, only oneparticle ofthebarmight be moved from itsposition. Such adisplacementismerely a mathematical device. Ineither Fm.22.-A rigidbarpinned atC.CMe (whether the constraints aresatisfied ornot)thedisplace- ment iscalled "virtual/7thisword implying thatthedisplace- ment isahypothetical oneandnotadisplacement actually experienced. Wenote then that virtual displacements areof twotypes: (a)virtual displacements satisfying theconstraints, (6)virtual displacements violating theconstraints. Asforforces, wehavePandQandthereactions ofconstraint. Weneed aword todistinguish PandQfrom thelatter; we cannot usetheword "external" because thereaction exerted by thepinatCisexternal. Soweshall callPandQapplied forces, with thisgeneral definition foranysystem with workless con- straints :Forces other than reactions ofconstraint arecalled applied forces. Thus theforces acting onthesystem areoftwotypes: (a)applied forces, (&)reactions ofconstraint. Asforthework done inavirtual displacement (called virtual work), wearetoobserve thatnowork isdonebythereactions ofconstraint provided that theconstraints areoftheworkless typeandaresatisfied bythevirtual displacement. We shallnowproceed tostate andprove theprinciple of virtual work. Theargument willbequite general, covering SBC. 2.4] METHODS OFPLANE STATICS 59 thecaseofanysystem inwhich theconstraints areoftheworkless type. PRINCIPLE OFVIRTUAL WORK. Asystem with workless con- straints isinequilibrium under applied forces if,andonly if,zero virtual work isdonebytheapplied forces inanarbitrary infinitesi- maldisplacement satisfying theconstraints. Itwillbenoticed that thisstatement contains both asuffi- cient(if)condition forequilibrium andanecessary (only if) condition. There aretwo.theorems here, requiring separate proofs. Letusfirstprove thenecessity ofthecondition; thatis,we aregiven thatthesystemisinequilibrium, andwehave toprove that zero virtual work isdonebytheapplied forces inany infinitesimal displacement satisfying theconstraints. Consider anyparticle ofthesystem;itisinequilibrium, andsotheresult- antofallforces acting onitiszero. Thus, zerovirtual work is donebytheforces acting onthat particle inanydisplacement of it.This holds for allparticles; andso,inanydisplacementof thesystem, zero virtual work isdonebyallforces acting. But thereactions ofconstraint donowork inanydisplacement which satisfies theconstraints. Hence theapplied forces donowork in such adisplacement, andsothenecessity ofthecondition is proved. Letusnowproveitssufficiency; thatis,wearegiven that zerovirtual work isdonebytheapplied forces inany infinitesi- maldisplacement satisfying theconstraints, andwehave to prove that thesystemisinequilibrium. Supposeitisnotin equilibrium; then itstarts tomove. Itisclearfrom (1.401) thateach particle starts tomove inthedirection oftheresultant force acting on it.Referring to(2.401), weseethatapositive amount ofwork isdone intheinitial displacement, since = andcos=1.This istrue forevery particle; andso,inthe initial displacementofthesystem, positive (not zero) virtual work isdonebyalltheforces. Butsuchaninitial displacement must ofcourse satisfy theconstraints, andsothereactions of constraint donowork. Thus, onthebasis ofourassumption that thesystemisnot inequilibrium, weseethat itmust undergo adisplacement which satisfies theconstraints and in which theapplied forces dopositive work. But thiscontradicts thegiven information, according towhich zerowork isdone. 60 PLANE MECHANICS [SEC. 2.4 Since ourassumption leads toacontradiction,itmust befalse, andsothesystem doesremain inequilibrium. The sufficiency ofthecondition isproved. Returning tothesystem shown inFig. 22,letusgivethebar arotation about Cinacounterclockwise sense through an infinitesimal angle60.Thework donebyPisPa50andthe workdonebyQisQb50,andsothetotalwork is 6W=(Pa-Qb)80. Ifthesystemisinequilibrium, then8W=andhence P^b Qa Ontheother hand,ifP/Q=6/a,then8W=forthisdisplace- ment. But this isthemost general displacement satisfying the constraints; hence thebarmust beinequilibriumifthecondition P/Q=b/aissatisfied. Although thechief merit oftheprinciple ofvirtual work lies inthefactthat itdoesnotinvolve thereactions ofconstraint, nevertheless itcanbeused tofindthese reactions should theybe required. Suppose, forexample, wewish toknow thereaction atCinthesystem considered. Ifthisreaction isR,itIsobvious thattheequilibrium willnotbedisturbed ifweremove thepin atCandapply atCaforce R.Since there isnownoconstraint atC,other virtual displacements arepermissible. Wemay slidethebaralongitslength. Inthisdisplacement, PandQdo nowork; henceRdoesnowork, andconsequently Racts at right angles toAB. Ifwenowpush thebarthrough asmall distance 8x,perpendicular toABintheupward direction, the workdonebyPandQis(P+Q)dx.Hence thework done byRis(P+Q)8x;therefore, Rhasamagnitude P+Qand acts inthedirection opposite tothecommon direction ofP andQ. Wehave developed theprinciple ofvirtual work forthesim- plestandmost interesting systems, namely, those forwhich the constraints areworkless. But itiseasily extended tocovermore general casesbymeans ofthedevice employed above, namely, thereplacement ofaconstraint byanunknown "applied" force. Exercise. Alever, intheform oftheletter L,ispivoted attheangle. Itisinequilibrium under forces applied attheends ofthearms, andper- SEC. 2.4] METHODS OFPLANE STATICS 61 pendicular tothem. Usethemethod ofvirtual work tofindtheforce at theendofonearmandthereaction atthepivot, theother forceandthe lengths ofthearms being given. Infinitesimal displacements ofarigidbody paralleltoafixed plane. Letusconsider arigidbody which ispermitted tomove only parallel toafixedfundamental plane. Thesection ofthebody bythisplaneisitself atwo-dimensional rigidbody; wecall it therepresentative lamina. Adescription ofthemotion ofthis laminaspecifies themotion ofthebody, andvice versa.We may therefore discuss infinitesimal displacements ofthelamina B A FIG. 23.-Genoial dis- placement ofalamina in itsplane.FIG. 24. Inaninfinitesimal displacement thequantities a, b,and Breceive infinitesimal increments, while rand re- main constant. inthefundamental plane instead ofdisplacements oftherigid body paralleltothatplane; theycome tothesamething. InFig. 23,Listhelamina before andL'thelamina afteran arbitrary displacement. LetA,Bbeanytwopoints ofthe lamina before displacement andA',B'their positions after dis- placement. Obviously thedisplacement fromLtoL'maybe achieved intwosteps: (i)atranslation,inwhich each pointofLreceives adisplace- > ment equal andparallel toAA'] (ii)arotation about A'through anangle equal totheangle between ABandA'B'. Thepoint A,used indescribing thedisplacement,iscalled a base point. 62 PLANE MECHANICS [SEC. 2.4 Letusnow consider aninfinitesimal displacement. Forfixed axesOxy inthefundamental plane (Fig. 24),let(a,6)bethe coordinates ofA,and 6theinclination ofABtoOx.Then theincrements 5a,56describe thetranslational displacement, andtheincrement 86describes therotation. LetPbeany particleofthelamina; letusputr=AP, <t>=BAP. Since thelamina isrigid,rand<j>remain constant asthelamina moves. Now if(x,y)arethecoordinates ofP,wehave (2.408) x=a+rcos(0+ ), y=b+rsin(0+4). Hence thedisplacementofPintheinfinitesimal displacement 5a,56,50ofthelamina is dx=5a-rsin(0+<)50, 5t/=56+rcos(0+<)50. Substituting forrsin(0+<),rcos(0+<)from (2.408), we obtain (2.409) 8x=5a-(y-6)50, 5t/=56+(x-a)50. This gives theinfinitesimal displacement ofanypoint inthe lamina (orintherigidbody ofwhich thelamina isasection) in terms ofthetranslation (5a, 56)ofthebasepoint (a,6),andthe rotation 50.Bygiving arbitrary infinitesimal values to5a, 56,50,wegetthemost general infinitesimal displacement ofa lamina inaplane orofarigidbody parallel toaplane. Exercise. Findthedisplacement oftheparticle atthehighest point ofa rolling wheel, when thewheel advances asmall distance 3s. Sufficient conditions fortheequilibrium ofarigidbodymovable parallel toafixed plane. Wenow consider theequilibrium ofarigidbody which can move only parallel toafixed fundamental plane. There are noconstraints limiting themotion ofthebody parallel tothe plane, butthere areconstraints preventing other motions. These aresupposed tobeoftheworkless type. Inaddition to thereactions ofthese constraints, there actapplied forces, not necessarily parallel tothefundamental plane. Let(x\, 2/1), (^2, 2/2),* **(xn,2/n)betheprojections onthefundamental plane (z=0)ofthepoints atwhich these forces areapplied, and let(Xi, FI),(Z2,72), (Xn,Yn)bethecomponents of these forces inthedirections oftheaxesOxy. Inaninfinitesimal virtual displacement 5a,56,50,asdescribed above, theworkdone is SBC. 2.4] METHODS OFPLANE STATICS 63 (2.410) SW-VXt[Sa-fa-b)38] =X8a+Ydb+N50, where X,Yarethecomponents ofthevector sumoftheapplied forces andNistheirmoment about thepoint (a,6). Thus,if (2.411) X=0,F=0,N=0, wehaveBW=forthemost general infinitesimal displacement consistent with the constraints. Thus, bytheprinciple of virtual work, thebodyisinequilibriumif(2.411) aresatisfied. These aretherefore sufficient conditions fortheequilibrium of thebody. Letusrestate thisimportant result, asfollows: //arigid bodyisconstrained tomove paralleltoafixedfunda- mental plane, then thebodyisinequilibrium under theaction ofanysystem ofexternal forces plane-equipollenttozero; i.e., there isequilibrium providedthat thevectorsum oftheprojections of these forces onthefundamental plane vanishes, andthemoment of these forces about some onelineperpendiculartothefundamental plane vanishes also. Wenote that (2.411) arethesame as(2.306) or(2.308), written inaslightlydifferent form. InSec.2.3these conditions wereshown tobenecessary fortheequilibrium ofanysystem; nowwefindthem tobesufficientfortheequilibrium ofarigid bodymovable parallel totheplane=0. LetSand S'betwoplane-equipollent force-systems acting onarigidbody; thenX,F,Nhave thesame values forSand S'. Itfollows from (2.410) that,ifdW isthevirtual workdoneby SandBWthatdonebyS'inthesame displacement ofthebody parallel tothefundamental plane, then SW=8W. Itisevident thattwo plane-equipollent force systems are equivalent inallstatical problems concerning theequilibrium 64 PLANE MECHANICS [SEC. 2.4 ofarigidbody movable parallel tothefundamental plane, in thesense thatonesystem maybereplaced byaplane-equipollent system without disturbing equilibrium. Inparticular, twoforces areequivalentifthey areequal inmagnitude, with thesame sense andwithacommon line ofaction. Thus, wemay slide aforce alongitslineofaction without changingitseffect. This isknown astheprinciple oftransmissibility offorce forarigid body, and itenables ustoregard aforce acting onarigidbody asasliding vector. Inthesame way,twocouples inthesame plane areequivalent iftheyhave thesamemoment N.SinceX=Y= fora couple,itfollows from (2.410) that thework donebyacouple of momentNinaninfinitesimal rotation 3disN 0. There isnounique way inwhich theprinciples ofmechanics must bedeveloped. Two rivalmethods exist, themethod of forces asused inSec. 2.3andthemethod ofvirtual work asused here. Indeveloping thetheory uptothisstage,wehave used onemethod toestablish some points andtheother method to establish other points. Thereadermay prefer towork outsome other logical development ofthesubject, andindeed ismost likely toappreciate thecritical points inthechain ofreasoning bysodoing. Exercise. Acard liesonatable. Along theedges, there areapplied forces representedinmagnitude anddirection bytheedges, taken inorder. Thecard isgiven anysmalldisplacement. Show thatthework done is represented bytwice theareaofthecard, multiplied bytheangle ofrotation. Potential energy. Consider anysystem ofparticles. Weshall refer toasetof positions ofalltheparticles asaconfiguration ofthesystem. LetAObesome configuration selected asastandardconfiguration, and letAbeanyother configuration. Letustakethesystem fromAtoAo,anddenote byWthework donebyallforces acting onthesystem during thisprocess. Ifthesystem consists ofasingle particle intheplane Oxy, wemight taketheorigin asstandard configuration A .Then ifX,Yarethecomponents offorce acting onit,thework done inbringing itfrom theposition Ais (2.412) W theintegral being taken along thecurve bywhich theparticle is brought totheorigin. Forexample,iftheforcedepends onthe SEC. 2.4] METHODS OFPLANE STATICS 65 position ofthe particle according totheequations X=2x,Y=6yand ifthecoordinates ofAarea,6,then (2.413a)W=JT*(2xdx+Qydy)- [s+3*/2 ];' ='_as-352. This value isindependent oftheparticular path along which theparticle isbrought to0. Totakeanother example,ifX=6y,Y=2x,wehave (2.4136) W=f'(Gyda:+2xdy).Jo>,o Thevalue ofthis integralisnotindependentofthepath of integration, asiseasily seenbytaking thetwopaths along the sides oftherectangle x0,y=0,x=a,y=b.Thus itis only insome cases thatWisindependent ofthepath. Passing from thecase ofasingle particle toageneral system, wemake thefollowing definition: When theforces acting onasystem aresuch thatthework donebythem,inthepassage ofthesystem from aconfigura- tionAtothestandard configiiration AQ,isindependent of theway inwhich thispassage iscarried out,then thesystem issaid tobeconservative. Thework donebytheforces inthe passage fromAtoAoiscalled thepotential energy* ofthesystem attheconfiguration A. Since thestandard configuration maybechosenarbitrarily, thepotential energy ofaconservative system isindeterminate towithin anadditive constant, thevalue ofwhich depends on thestandard configuration chosen. Potential energywillbedenoted byV.Thus, inthecase of(2.413a), thesystem (asingle particle) isconservative and thepotential energy Vattheposition x,yis (2.414) V=-z2-Si/2 . Generally speaking, most ofthesystems considered inme- chanics are conservative. The outstanding exceptions are systems inwhich frictional resistances areinvolved. Cases likethat of(2.4136) occur rarely. Letthere beaconservative system, AQbeing thestandard configuration. Weshall usethefollowing notation: *Since potential energyisdefined asworkdone,"ithasthesame dimen- sions[ML*T~*] and ismeasured inthesame units aswork (cf.p.54), 66 PLANE MECHANICS (SEC. 2.4 V(A)=potential energy atconfiguration A, W(A, B)=workdonebyforces inpassage fromAtoB. Bythedefinition of7,wehave (2.415) V(A)=W(A, A*). Now givethesystem aninfinitesimal displacement fromAtoB] let8Vbetheincrement inpotential energy and8Wthework done. Wehave 8V=V(B)-V(A) Butsincework done isindependentofpathandadditive, we have W(B, A*)=W(B,A)+W(A, Ao), and W(B,A)=-W(A,B)=-ST7. Hence wehave (2.416) dV=-dW. Inwords, Jtaincrement inpotential energy equals thework done, with itssignchanged. Consider aparticle which canmove inaplane subject to theaction ofaforcewhich depends onlyontheposition ofthe particle.IfX,Yarethecomponents offorce andx,ythe coordinates ofthe particle, thenX,Yarefunctions ofx,y. This iscalled afield offorce. Itissaid tobeaconservative field iftheparticle under itsinfluence isaconservative system. In that case, denoting thepotential energy by7,wehaveby (2.403) and(2.416) theequation (2.417) X5x+YSy--57, where 6x,dyarethecomponents ofanarbitrary infinitesimal displacement given totheparticle. Itfollows that (2.418) X=-g,Y=-f- Thepreceding resultmaybeextended without any difficulty toaconservative field inspace; wehavethen 9V dV dV SBC. 2.4] METHODS OFPLANE STATICS 67 Inthelanguage ofSec. 1.3:Inaconservative field,theforceisthe gradient ofpotential energy, withsign reversed. Auniformfield offorce isoneinwhich X,Y,Zareconstants. Such afield isconservative, with potential energy (2.420) V=-(Xx+Yy +Zz). Returning toageneral conservative system,letusnote a useful consequence of(2.416) inconnection with theprincipleof virtual work. //aconservative system isinequilibrium,thechange inpotential energy inanyinfinitesimal displacement iszero. This isalsoexpressed bysay- ingthat thepotential energy hasa stationary value. Example. Asanillustrative example ofthe principleofvirtual work, consider thesys- temshown inFig.25.Twoheavy particles of weights w,w'areconnected byalight inex- tensible string andhang over afixedsmooth circular cylinder ofradius a,theaxisofwhich ishorizontal. Wewish tofindtheposition of equilibrium. Herewehaveasystem oftwoparticles. Theforces acting onthem are (i)gravity, (ii)forces duetothetension inthestring, (iii)reactions exerted bythecylinder. Ifwegiveavirtual displacement satisfying theconstraints, onlygravity does work. Thesystem isconservative, andthepotential energy is,by(2.420), with suitable choice ofthestandard configuration,FIQ. 25.Twoheavy par- ticlea balanced onasmooth cylinder. V=wacos6 cos0', where 0,0'aretheinclinations tothevertical oftheradiidrawn tothe particles. Inaninfinitesimal displacement, dV wasin 60 w'asin0'60'. But+0'isconstant, since thestringisinextensible. Thus 60'**80, and 57=sa50(w' sin 0'wsin 0). Hence, when thesystem isinequilibrium, thefollowing condition mustbe satisfied : sin_w' sin 8'~~w 68 PLANE MECHANICS [Sue. 2.5 2.5.STATICALLY INDETERMINATE PROBLEMS Consider arigidbody which canmove parallel toafixed fundamental plane, inequilibrium under external forces inthat plane. LetXandYbethetotalcomponentsoftheexternal forces onaxes ofcoordinates inthefundamental plane andN their totalmoment about theorigin. Then, ashi(2.411), wehave thescalar equations ofequilibrium (2.501) X=0,Y=0,N=0. Itisimportant tonotethatthese equations arethree innumber. Thismeans that, inanyproblem concerning theplane statics ofarigid body, wecanfind threeunknowns andnomore. If there aremore than three unknowns, theproblem isstatically AXxP BX2 FIQ. 20.-Astatically indeterminate problem. indeterminate,which means that itcannot besolved bymeans of theconditions ofequilibrium alone.Weshall illustrate with a simple example. Arigid barAB(Fig. 26) isfixed atitsends; atitsmiddle point there isapplied aforce withcomponents P,Qalong and per- pendicular tothebar. Find thereactions onthebaratA andB. Letthecomponents ofthereactions beXi,Y\atAand X2,Yzat#,and letAB=2a.Taking components along and perpendicular tothebarandmoments about A,by(2.501) wehave (X=Xi+P+X=0, (2.502);Y-Yi+Q+Y2=0, (N=aQ+2aY 2=0. Hence, (2.503) 72=-Q, 7i=-Q, Xl+X2=-P. Wehave only three equations forfourunknowns; theproblem isstatically indeterminate, andnothing more canbefound out about thereactions from theconditions ofequilibrium alone. SEC. 2.6] METHODS OFPLANE STATICS 69 Ifsuchaproblem were tooccur inphysical reality, thefour unknown quantities would have values which might bemeasured. Apparently ourmathematical methods have failed us;theyhave notprovided uswiththeanswer toaquestion ofphysical inter- est.The faultactuallyliesintheselection ofamathematical model. Rigid bodies donotexist innature, and thisproblem isonewhere theuseofarigidbody asamathematical model is notjustified. Weshould takeanelastic barasamodel instead (cf.Sec. 3.3). Asimple modification intheconstraints may render an indeterminate problem determinate. Consider thesame prob- lem, altered bythecondition that theendB,instead ofbeing A X! FIG. 27.Astatically determinate problem. fixed, slides onasmooth plane inclined atanangle of45to AB (Fig. 27). Again, wehave (2.503), butalsoanadditional equation (2.504) X2+F2-0, arising from thecondition thatthereaction atBisperpendicular totheplane. Now theproblemisstatically determinate, and wehave nwn(2.505) Hadwetaken oneoftheaxes paralleltotheplaneofconstraint atB,weshould have obtained aslightly simpler treatment, because only threeunknowns would have appeared. 2.6.SUMMARY OFMETHODS OFPLANE STATICS I.Conditions ofequilibrium. (a)Forsingle particle (necessary and sufficient): (2.601) Vector sum ofallforces vanishes, 70 PLANE MECHANICS [SEC. 2.6 or (2.602) Total componentsintwo perpendicular directions vanish. (6)Foranysystem (necessary) orforrigidbody (necessary and sufficient): (2.603) F-0,N=0; orX=0,Y=0,N=0. (c)Foranysystem with workless constraints (necessary and sufficient): (2.604) dW=(work donebyapplied forces). II.Moment ofavector inaplane about apointinthatplane. (2.605) M=aP (+forcounterclockwise) ; (2.606) M=xY-yX (about origin). III.Plane equipollence. (a)Conditions forplane equipollence: (2.607) F=F,N=N'. (6)General system offorces canbereduced toasingle force atanassigned point, together withacouple. (c)General system offorces canbereduced toasingle force ortoasingle couple (latter caseexceptional). IV.Work andpotential energy. (a)Definition ofwork: (2.608) dW=Pcos-6s=X8x+YBy. (&)Reactions donowork atsmooth contacts, rolling contacts, andinside rigidbody. (c)Fortheinfinitesimal displacement ofarigidbody, (2.609) 8W=Xda+Yfib+N86. (d)Potential energy: (2.610) dV=~dW,~ (2.611) *--?,V=~~v 'dx dy (Forceisgradient ofpotential energy with signreversed.) Ex. II] METHODS OFPLANE STATICS 71 EXERCISES H Themethod ofvirtual workmaybeused inanyofthese problems;itwill befound particularly useful inthecaseofthosemarked withanasterisk. 1.Aparticleisinequilibrium under theaction ofsixforces. Three of these forces arereversed, andtheparticle remains inequilibrium. Prove that itwill stillremain inequilibrium ifthese three forces areremoved altogether. 2.Aladder ofweightWrests atanangle tothehorizontal, with itsends resting onasmooth floorandagainst asmooth vertical wall. The lower end isjoined byarope tothejunction ofthewallandthe floor. Find, interms ofWand a,thetension oftheropeandthereactions atthe wallandtheground. (Assume thattheweight oftheladder actsatits middle point.) 3.ThecornerAofasquare plateABCD isheld fixedbymeans ofa smooth hinge which permits theplate toturn freely initsownplane. Four forces, each ofmagnitude P,actalong thefour sides inorder. Find the single additional force which, applied atthecenter oftheplate parallel to thesideAB, willkeeptheplate inequilibrium. What isthecorresponding reaction atthehinge? 4.Adoor ofweight W,height 2a,andwidth 26ishinged atthetopand bottom. Ifthereaction attheupper hinge hasnovertical component, findthecomponentsofreaction atboth hinges. (Assume thattheweight ofthedoor actsatitscenter.) 6.Aparticle ofweightWissuspended from afixed pointbyalight string. Ahorizontal forceHisapplied toit, andtheparticle takesupaposition ofequilibrium with the string inclined tothevertical. Ifthestring breaks when the tension initroaches avalue To,findthesmallest value of// necessary tobreak thestring. 6.Aheavy beamAB,8ft.long, rests horizontally ontwo supports, oneatAandtheother 3ft.from B. Ifthegreatest weight thatcanbehungfromBwithout upsetting thebeam is 20lb.,findtheweight ofthebeam. (Assume thattheweight ofthebeam actsatitsmiddle point.) 7.Show that ifalight cable passes round apulley mounted onsmooth bearings, thetensions intheportions on either sideofthepulley areequal. Hence findthetension T inthecable forthepulley system shown, supporting aweight W,thepulleys andcable being supposed lightandthedistance between theupper andlower pulleyssogreat thatthecables mayberegarded asvertical. 8.Aforce ofmagnitude P,acting upandalong asmooth inclined plane, cansupport aweight W;when acting horizon-|W tally, itcansupport aweight w.Findarelation among P,Wt andw,notinvolving theinclination oftheplane. 9.Fourlamps eachweighing 4lb.aresuspended across aroadbetween posts 40ft.apart bylight cords attached atpoints B,C,D,Eofacord 72 PLANE MECHANICS [Ex. II ABCDBF, whose endsAandFarcfixed atthesame level totheposts. The cords supporting thelamps divide thehorizontal distance between theposts intoequal parts. CandDare12ft.below AF. Findthetension inCD. *10.Alight rigid rodoflength 26,terminated byheavy particlesof weights w,W,isplacedinside asmooth hemispherical bowl ofradius a, which isfixed with itsrimhorizontal. Iftheparticle ofweight wrests justbelow therimofthebowl, prove that wa*-W(2b*_a2). 11.Asystem offorces acting onarigidbody consists ofnforces acting along thensides ofaclosed polygon taken inorder. Ifthemagnitudes of theforces areproportionaltothelengths ofthesides along which they act, show thatthesystem reduces toacouple whosemoment isproportional to theareaenclosed bythepolygon, aproper convention asregards thesign ofthisareabeing made. Give asimple example ofsuchasystem offorces which would keeparigidbody inequilibrium. 12.Explain whyinamotion picture thespokes ofarotating wheel some- times appear tobemoving thewrong way. 13.AforcePofconstant magnitude andfixed direction isapplied toone endofanarm oflength a,which canturnabout theother endinaplane containing thedirection ofP.Find thetotalworkdonebytheforce asit pullsthearminto itsowndirection from aposition perpendicular toit. 14.Show thatafield offorce withcomponents (X,Y)isconservative if, andonly if, . dy dx' 16.Find thepotential energy ofaparticle attracted toward afixed pointbyaforce ofmagnitude k*/rn ,rbeing thedistance from thefixed pointandfc,nanyconstants, *16.Alight lever, intheform ofaletterLwitharmsaand6,ispivoted attheangle sothat itcanturn freely inavertical plane. Weights W,w aresuspended from theends. Show that there arejusttwopositionsof equilibrium. *17.Toanumber offixed points Ai,A-2, ,An,situated atequal intervals aonastraight lineinclined atanangle tothehorizontal, there areattached rods allofthesame length aandweight w.Theother ends ofthese rods,B\ t#2, ,B*.,areconnected byrods ofthesame length a andweight w.Thesystem hangs inavertical plane, forming asetof squares, A\andB2being connected byalight rigid rod. Findthereaction inthisrod,assuming that allthejoints aresmooth andthattheweight of eachrodactsasitsmiddle point. *18.Aframework ABCD consists offour equal, light rodssmoothly jointed together toform asquare;itissuspended from apegatA,anda weightWisattached toC,theframework being keptinshapebyalightrod connecting BandD.Determine thethrust inthisrod. Ex.II] METHODS OFPLANE STATICS 73 19.Anumber ofcoplanar forces actonarigidbody. Alltheforces are turned intheir plane through thesame angle about their pointsofapplica- tion, without change ofmagnitude. Show that their resultant turns through theangle about afixed point inthebody. (This pointiscalled theastatic center.} 20.Four forces ofmagnitudes 1,3,4,6actinorder along thesides ofa squareABCD ofsidea,theforce ofmagnitude 1acting alongAB. Choosing asaxes intheplane thelinesABandAD, findtheequation ofthelineof action oftheresultant force. Find alsotheposition oftheastatic center (seeExercise 19),ifoneforce onlyisconsidered asacting through each corner ofthesquare andtheforce ofmagnitude1actsatA. CHAPTER III APPLICATIONS INPLANE STATICS Inthischapter weshallbeconcerned chiefly withsystems lying inaplane. However, mass centers andcenters ofgravity are here discussed forsystems inspace; thepresence ofathird coordinate causes norealcomplication. 8.1.MASS CENTERS ANDCENTERS OPGRAVITY Definition ofmass center. Consider asystem ofnparticles ofmasses mi,ra2,wn, situated atpoints Pi,P2,P. Iftheposition vectors of these points relative tosome assigned point areri,r2,rn, wedefine thelinear moment ofthesystem with respect tothat point tobethevector wtrt. Themass center ofthesystemisdefined tobethat point with respect towhich thelinear moment vanishes. Toshow that thisdefinition issignificant, wehave toprove twothings: (i)a mass center exists; (ii)there isonlyonemass center. Toestablish theexistence ofamass center, wetakeany point 0;lettheposition vectors ofPi,P2,Prelative to beTI,r2, rn.LetCbethepoint such that (3.101) OC= Then theposition vector ofthepoint P<relative toCis it-OC, 74 SEC. 3.1] APPLICATIONS INPLANE STATICS 75 andsothelinearmoment ofthesystem with respect toCis _^ n v mt(rt-OC)=mlr,-OC But thisvanishes by(3.101), andtherefore Cisamass center. Toestablish theuniqueness ofthemass center, weassume thatthere aretwomass centersC,C',relative towhich theposi- tionvectors oftheparticles areTI,r2,rnandr{,r, r^ } respectively. Then, (3.102) But r.=rj+CC'; combined with (3.102), thisleads toCC'=0,sothatCandC' coincide. Equation (3.101) gives theposition vector ofthemass center relative toanarbitrary origin 0;thispositionvector isthequotient ofthelinear moment bythetotalmass. Itfollows thatthelinear moment ofasystemisthesame asthat ofaparticle, having a mass equal tothetotalmass ofthesystem, situated atitsmass center. Ifthesystem consists ofonlytwo particles, with masses mi,w2,thedefinition shows thatthemass center liesontheline joining them anddivides itintheratiom^:m\. Forthecalculation ofmass centers,itisconvenient tohave (3.101) inscalar form; referred toany axes, thecoordinates of themass center are (3.103) x Thenumerators are, ofcourse, thecomponents ofthelinear 1moment with respect totheorigin. Itisimportant tonotethatwhenwemove asystem ofparticles rigidly (i.e., without changing mutual distances), themass renter iscarried along asifrigidly attached tothesystem. 76 PLANE MECHANICS [SBC. 3.1 This follows from (3.103). For letOxyz beanysetofaxesand O'x'y'z' anew setofaxes, such that thenew position ofthe system relative toO'x'y'z' isthesame astheoldposition relative toOxyz; thismeans thatx\xl,y(yt,z(=zt.Then the coordinates ofthenewmass center relative toO'x'y'z' willbe thesame three numbers asthecoordinates oftheoldmass center relative toOxyz, andhence thenewmass center occupies thesame position relative tothesystem astheoldonedid.* Letusnowconsider acontinuous distribution ofmatter instead ofasystem ofparticles. Viewing thecontinuous distribution as thelimit ofthediscontinuous system, wearoledtoassociate a definite mass withanyvolume inthecontinuous distribution. Densityisdefined asmass perunitvolume; bythiswemean that thedensity pis (3.104) p=lim~?> whoreAw isthemass inthevolume Avandthesign"lim" means "limit asAvcontracts toapoint.7'Inaninfinitesimal volume dvthemass is (3.105) dm=pdv. Inhomogeneous bodies (withwhichweshallbechiefly concerned), pisaconstant. Ifpvaries from point topoint inabody, the bodyissaidtobeheterogeneous. The definition given above forthelinearmoment ofadis- continuous system suggests that thelinear moment ofacon- tinuous system should bedefined as JJJrpdxdydz, where ristheposition vector ofageneral point ofthesystem and pthedensity atthat point. Thisvector hascomponents fffxp dxdydz, J7/2/P dxdydz, fffzPdxdydz. Theprevious definition ofmass center leads ustothestatement thatthemass center isthatpoint forwhich, taken asorigin, we have (3.106) ffjxp dxdydz=0, JJJ?/p dxdydz=0, J7/zp dxdydz=0. SEC. 3.1]fAPPLICATIONS INPLANE STATICS 77 Tofindthemass center wemay use (3.103), changed into continuous form. Thus, forany axes, themass center haa coordinates r- dxdydz _JffypdxdydzXMJpdxdydz'J" J/Jpdzdydz' -_"~ JJJPda;^2/dz Consideration ofasystem ofparticles lying inorvery close toaplane orsurface leads totheidealized concept ofacon- tinuous distribution ofmatter onaplane orsurface; weintro- duce aquantity acalled surface density, suchthatthemass ofan infinitesimal areadSofthesurface is<rdS.Themass center ofa surface distribution hascoordinates (3.108)v ' Similarly, weconsider acontinuous distribution ofmatter along alineorcurve; weintroduce aquantity Xcalled theline density, such thatthemass ofanelement dsisXds.Themass center ofacurvilinear distribution hascoordinates . f?/Xds _ fzX r/s- - In(3.107), (3.108), and (3.109) thedenominator ineach caso represents thetotalmass ofthesystem. Inthecase ofuniform distributions ofmass(i.e., distributions ofconstant volume density, surface density, orlinedensity, asthe casemay be),thedensity factor comes outside thesigns ofinte- gration andsodisappears bycancellation from (3.107), (3.108), and (3.109). Methods ofsymmetry anddecomposition. Incases ofsymmetry,itispossible tolocate themass center (or,atanyrate,limit itsposition) without anycalculation. Asystemissaid tohave central symmetry with respect toa pointifthesystemisleftunchanged byreflection inthe point 0.(Byreflection wemean thataparticle orelement of massmatAisreplaced byaparticle orelement ofmassmat B9whereOB=OA.) Forsuchasystem,itisimmediately seen that themass center coincides with thecenter ofsym- 78 PLANE MECHANICS [SEC. 3.1 metry, because thelinearmoment about thatpoint consists of contributions which cancel inpairs. Asystem hasaplane ofsymmetryifthesystemisleftunchanged byreflection inaplane. Itiseasily seenthat insuch cases the mass center liesintheplane ofsymmetry. Asystem hasanaxisofsymmetryifthesystemisleftunchanged byarotation ofarbitrary magnitude about the axis. Itis not difficult toshow that themass center liesontheaxis of symmetry. Thus, forexample,itisevident that (i)Themass center ofasolid sphereliesatitsgeometrical center, when thesphereishomogeneous orwhen thedensity depends onlyonthedistance from thecenter. (ii)Themass center ofasolid homogeneous hemisphereliesonthe radius which isperpendicular toits plane face. (iii)Themass center ofaplate intheform ofanequilateraltri- angle (ofuniform density andthick- ness)liesatthecentroid. Sometimes wemeet distributions ofmatter whichmaybedecomposed intosimple parts, themass centers ofwhich canbefound. Suchasys- tem isshown inFig. 28,thelines ofdecomposition being dotted. Weshallnow establish thefol- lowing principle ofdecomposition:Ifasystemisdecomposed into parts withmasses M\,M*,Mnandmass centers attheFIG. 28.AletterFiscutout ofmetal sheeting. The position ofthemass center isrequired. points PI,P2,P,then themass center ofthecomplete systemisatthemass center ofthesystem ofnparticles ofmasses Mi,Mz,-Mn,situated atthepoints PI,P2,Pw. Weshall provethis principleforasystem ofparticles, the proof foracontinuous system being similar. Further, for simplicity weshallsuppose thatthesystemisdecomposed into three parts, since theproof fornpartsissimilar. Theproof restsonthefactthat linearmoments areadditive; this isobvious from thedefinition oflinear moment. Thus thelinearmoment SBC. 3.1] APPLICATIONS INPLANE STATICS 79 ofthecomplete system isthesum ofthelinear moments ofthe three parts. Butby(3.101) thelinearmoment ofeach partis thesame asthelinearmoment ofaparticle situated atitsmass center, having amass equal tothemass ofthepartinquestion. Hence thelinear moment ofthecomplete systemisequal to thesum ofthelinear moments ofthethree particles, andboth vanish when they arecalculated relative tothemass center of thecomplete system. This pointistherefore themass center of thethree particles. Inthecase oftheplateshown inFig. 28,thereader should verify bythismethod that themass center liesatxf-, y=^. The principle ofdecomposition may alsobeexpressed as follows :Forthecalculation ofmass centers, anypartofasystem maybereplaced byarepresentative particle, situated atthe mass center ofthepartandhaving amass equal tothemass of thepart. Inapplying themethod ofdecomposition,itisoften convenient todecompose thesystem into infinitesimal portions. Generally theuseofsuchadecomposition willrequire aprocess ofintegra- tion,butsometimes thiscanbeavoided. Thus,ifatriangular plateisdecomposed intothin strips, therepresentative particles lieonthemedian ofthetriangle which bisects thesestrips. Hence themass center liesoneach ofthemedians; themass center ofatriangleistherefore attheccntroid. Byanextension ofthesame method,itiseasily seenthatthe mass center ofasolid tetrahedron liesatthepoint ofinter- section ofthelines joining thevertices totheccntroids ofthe opposite faces. Ifwewish tofindthemass center ofabody with ahole init, wecanregard thebody asasuperpositionofthecomplete body withnoholeandafictitious body ofnegative density (equal in absolute value tothedensityofthebody) occupying theposition ofthehole. Thus,ifacircular hole ofradius 1in.ispunched fromacircular disk ofradius 4in.,theedge oftheholepassing through thecenter ofthedisk, themass center isthat ofapair ofparticles withmasses intheratio 16 :1situated atthecenters ofthecircles. Hence themass center liesatadistance of-^rin- from thecenter ofthelargercircle. 80 PLANE MECHANICS [Sao. 3.1 Theorems ofPappus. Ourknowledge ofcertain surface areasandvolumes enables us tocalculate somemass centers quickly bymeans ofthetheorems ofPappus, which state I.Letthere beauniform distribution ofmass along aplane curve C,which doesnotcross astraight lineLinthesame plane. Letpbethedistance ofthemass center from L,Ithelength ofC,andSthesurface areagenerated byrotating Cabout L, toform asurface ofrevolution. Then (3.110) 2wpl-8. II.Letthere beauniform distribution ofmassonaregionR ofaplane. LetLbealineintheplane, notcrossing R.Letp bethedistance ofthemass center from L,Athearea of72,and Vthevolume generated byrotating Rabout L,toformasolid ofrevolution. Then (3.111) 2*pA=V. Toprove these theorems, wetake axesOxy,Oxlying along Lineach case. Then,inthecase ofI,by(3.109) wehave p=$yds/l theintegral being taken along C.But S=fay ds, andhence (3.110) follows. InthecaseofII,by(3.108) wehave p=Jydxdy/A, theintegral being taken overR.But V=faydxdy, andhence (3.111) follows. Thus thetheorems ofPappus are established. Asanexample oftheuseofthe firsttheorem, consider awire bent intotheform ofasemicircle ofradius a.Wetake forLthe diameter joining theends. Then (3.112)I=Tra, S=47ra2 ,p=~= - Airl TT Asanexample oftheuseofthesecond theorem, consider aflat semicircular plate.Wetake forLtheterminating diameter. Then (3.113) A=^a*, V=fra', p=JL=g- SEC. 3.1] APPLICATIONS INPLANE STATICS 81 Mass centers found byintegration. Though much labormaybesaved byusing themethods of symmetry anddecomposition orthetheorems ofPappus,itis evident from(3.107), (3.108), and (3.109) thatwhen these methods failwecan fallbackondirect integration. Usually a judicious mixture oftheseveral methods willyield theresult most rapidly. Asillustrations, weshall calculate themass centers ofawirebent toform aquadrant ofacircle, asolid hemisphere, andathinhemispherical shell. Interms ofpolar coordinatesr,initsplane, theequation ofa quadrant ofacirclemaybewritten r=a,with running from to-JTT.Thelength ofanelement isrdB]aridwithx=rcos0, yrsin0,theCartesian coordinates ofthemass center are, by(3.109), 2ax= jnacos aav//naav (3.114) Themass center liesontheradius bisecting thearcatadistance 2-\/2*a/irfrom thecenter. Thereader maycompare (3.114) with (3.112) andconsider how (3.114) might havebeendeduced from (3.112) without calculation. Wemaydecompose asolidhemisphere intothin circular platen parallel totheplaneface. Thedistance ofthemass center from theplane face isthus 'irr2dz/fa irr*dz,IJz=* where ristheradius ofthecircular section atadistance zfrom theplaneface. But r2a2z2 ,where aistheradius ofthe spherical surface. Hence, (3.115)I=|o. Wemay decompose athin hemispherical shell into thin circular bands bymeans ofplanes drawn parallel totheopen face. If istheangle between anyradius andtheradius perpendicular totheopen face, thearea oftheband between and+d6is 2?ra2sin J0,where aistheradius ofthe shell. Hence the height ofthemass center above theopen face is 82 PLANE MECHANICS [SEC. 3.1 f**acos6-2?ra2sin6dB (3.116) z=*^- =ia. /27ra2sin0d0 Historically,thisresult isfamous;itwasobtained byArchi- medes through comparison oftheshell with acylinder ofthe same radius, andlength equal totheradius, containing the hemisphere andtouchingitalong theedge ofitsopen face. Itiseasy toshow thattwoadjacent planes parallel totheopen faceintercept thesame areasonthehemisphere andthecylinder. These twoareas contribute thesame linear moment, andsothe mass centers ofthehemisphere andthecylinder coincide; from thisfacttheresult follows. Gravitation. Abodyfalls totheground unless itisheldupbysuitable forces. This isduetogravitational attraction between thebody andtheearth. Every body attracts every other body, andwe accept asoneofourhypotheses thefollowing law: NEWTON'S LAWOFGRAVITATION. //twoparticles ofmasses mi,m%areatadistance rapart, each attracts theother witha gravitational force ofmagnitude whereGisauniversal constant, called theconstant ofgravitation. Theforcesactalong thelinejoining theparticles, inaccordance with thelawofaction andreaction stated inSec. 1.4. Ifwethink oftheparticle ofmassmiasfixedandthat ofmass m2asfreetotakeupvarious positions, werecognize that the massmiproduces afield offorce. Itisusual totakem2=1for simplicityindiscussing thisfield; then themagnitude oftheforce ofattraction isGmi/r2 . Ifwetakecoordinates with origin atmi,thedirection cosines ofthelinedrawn from toanypointAwith coordinatesx,y,z arex/r, y/r, z/r.Hence thecomponentsofforceonunitmass atAare /'QIITN vivi (3.117)X=-- -,Y=--3-, Z= theminus sign occurring since theforce isdirected fromA toward 0.Now SEC. 3.1] APPLICATIONS INPLANE STATICS 83 r2=z2+2/2+22r|^=s,-=-,d'dz r therefore(3.117) maybewritten (3.118) X=)Y--?Zd l where (3.119) 7=- FIG. 29.Aspherical shell di- vided into thin rings forthe cal- culation ofthepotential atA.This isthepotential energy (cf.2.419) ofaparticle ofunitmass inthegravitational field ofaparticle ofmass mi,or,briefly, the potential ofthe field. Thus the force ofattraction isthegradient of thepotential, with sign reverced. When anumber ofattracting particles arepresent, theresultant force ofattraction isthevector sum oftheindividual forces ofat- traction. This resultant force is equal tothenegative ofthegradi- entofthetotal potential, i.e.,the sum ofthepotentials duetotheseveral particles. Incalculat- ingtheforce ofattraction duetoasystem ofparticles (ora continuous distribution ofmatter),itisoften convenient tofind thepotentialfirst. Letusconsider athin spherical shell ofmatter ofradius a (Fig. 29).Wewish tofindthopotential atanexternal point A, atadistance rfrom thecenter 0. Letusdraw cones with forvertex, OAforaxis,andsemi- vertical angles 0,9+dO.These cutofffrom theshellaring of area 2ira2sin6dd.Theelements ofthisringare allatthesame distance (R)from A,andsothepotential duetotheringis -2irG<ra2sinOd6/R, where aisthemass perunitarea oftheshell. Expressing Rhi terms ofa,r,6andintegrating over theshell,wefind forthe potential raianv-r 2wG<ra*s{nede (3.120)V- 84 PLANE MECHANICS [Sac. 3.1 thepositive values ofthesquare roots being understood. Since r>a,thelastsquare root isra,andso (3.121) . V=- / i whereMisthetotalmass oftheshell. Thuswehave thefollowing result: Thepotential (and hence theforce ofattraction) ofathinspherical shell atanyexternal point isthesame asifthewhole mass oftheshellwere concentrated atits center. IfthepointAliesinside the vshcllinstead ofoutside, weproceed asbefore down to(3.120). Butnow a>r,andsothelast square root isa r.Hence V=-4wO<ro, aconstant. Thus, inside athin spherical shell thepotentialis constant, andtheforce ofattraction iszero. Wecannow discuss thegravitational field oftheearth, sup- posingittobecomposed ofthinspherical shells, each ofconstant density. Each shell attracts asifitsmass were concentrated atthecenter oftheearth. Hence wehave thefollowing result: Atapoint A,outside theearth,theforce ofattraction isdirected toward thecenter oftheearthand isofmagnitude (3-122) % whereMisthemass oftheearthand rthedistance ofAfrom the center oftheearth. Inparticular,ifristheradius oftheearth, (3.122) gives the force ofattraction attheearth's surface. Theconstant Gisvery small (6.67X10~8 c.g.s. unit), andsogravitational forces are insignificant unless themasses involved aregreat. Forthisreason weusually neglect themutual attractions ofbodies ontheearth's surface incomparison withtheearth's attraction. Centers ofgravity. Weconsider nowabody near theearth's surface, thebody being small incomparison with theearth's radius. (Wehave in mind apiece oflaboratory apparatus orevenalarge engineering structure, butnotanything which would beofappreciable size onamap oftheworld.) Throughout thisbody thedirection and SEC. 3.1] APPLICATIONS INPLANE STATICS 85 magnitude oftheearth's attraction arenearly constant. This leads ustotheconstruction ofthefollowing model forthediscus- sion ofgravity neartheearth's surface: Theearth's surface (orthe ground)isrepresented byaplane (thehorizontal plane). The earth's attraction onaparticle ofmassmisofmagnitude mg,where g isaconstant; itisdirectedvertically downward(i.e.,perpendicular toandtoward theground). Thevalue ofgisapproximately 32ft. sec.~2 ,or980cm. scc.~~2 Weshallnowshow thatthere isjustonepoint C,thecenter of gravity ofabody, which satisfies thefollowing conditions: (i)Thepotential energy ofthebodyisequaltothatofasingle particle withmass equaltothetotalmass ofthebody, situated atC. (ii)Thewhole system offorces due togravityisplane-equipollent (with respecttoanyvertical plane)toasingle vertical force through C. Letustake axes Oxyz, OxandOzbeing horizontal andOy vertical. Letuschoose asstandard position foreach ofthe particles forming thebody. Then aparticle ofmass ratatthe point (xj, yi,Zi)hasby(2.120) potential energy n^gy*, andso thewhole potential energyis(fornparticles) (3.123) V=g Letthecoordinates ofCbex,y,z;condition(i)isequivalent to (3.124) V=Algy, whereMisthetotalmass ofthebody. Henco, comparing the twoexpressionsforV,wehave n ,1 Thecondition ofplane equipollencc with respect totheplane =demands that these beingmoments about Oz.Hence M 86 PLANE MECHANICS [SEC. 3.2 Similarly, Thus thecenter ofgravity Cexists, with coordinates 7 7t It, 2}mtX *2)m#* 2)m^ (3.125)35=r^>y= Tjr; %~ jTr* Wenote,onreferring to(3.103), thatthecenter ofgravityisin factthesame point asthemass center. The force Mg, directed downward through thecenter of gravity,iscalled theweightofthebody. Anaccurate treatment ofstatics ontheearth's surface iscom- plicated bytheearth's rotation about itsaxisand itsmotion round thesun. However, theeffects duetothese causes are very small, andwemay neglect them without making serious physical errors. Infact,wegetsatisfactory results bytreating theearth asaNewtonian frame ofreference. Likewise, another simplification introduced above (theassumption thattheearth is flat,with auniform gravitational field) doesnotcause serious physical errors. So,ifwedonotwish toobtain results of extremely highphysical accuracy, wemayusethemodel described above; thisis,infact, theprocedure throughout therestofthe chapter. The effects oftherotation oftheearth areconsidered inSec. 5.3andalsoinSec. 13.5. Itwillbeshown that, asfarasstatics isconcerned, thisintroduces norealcomplication;itmerely modifies thevalue ofg. 3.2.FRICTION InSec.2.4weintroduced theconcept ofasmoothsurface; the essential propertyisthat, atasmooth contact, thereaction is normal tothesurface. Weshallnow discuss thereaction ata rough contact andstate thelaws offriction. SEC. 3.2] APPLICATIONS INPLANE STATICS 87 Fio. 30.Thereaction Rata rough contact resolved intothe normal reaction (N)and the force offriction (F).Laws ofstatic andkinetic friction. LetAandB(Fig. 30)betwobodies incontact. LetRbethe reaction exerted byBonA.Rcanberesolved inaunique manner intotheforcesNandF,N lying along thenormal atthepoint ofcontact andFlying intheplane of contact. Niscalled thenormal re- action andFtheforce offriction. (Atasmooth contact, F=0.) Onthebasis ofexperiment, certain laws offriction areaccepted. These aremathematical idealizations from theexperimental results, andahigh degree ofaccuracy inpredictions based onthese laws isnottobeexpected. LAWOFSTATIC FRICTION. When twosurfaces areincontact and noslipping takes place, theratioF/Ncannot exceed anumber/u,the coefficient ofstaticfriction, which depends only onthenature ofthesurfaces. Instatical problems thetwobodies willbeatrest,butthe above statement issufficiently general tocover thecasewhere one body rollsonanother. Theacute angle Xdefined by (3.201) tanX=M iscalled theangle offriction.Itisseen atonce thatthelawof static friction (3.202) F~M implies (3.203)B<>X, where 6istheinclination ofthereaction Rtothenormal. Thus thedirection ofRmust lieinside thecone ofstaticfriction, formed bydrawing alllines inclined tothenormal atanangleX. When onebodyslidesonanother, thebehavior ofthereaction iscontrolled bythelawofkinetic friction. Weshall state this lawforthecasewhere onebody isatrest. LAW OFKINETIC FRICTION. When one surface slides on another which isatrest,theforce offriction Fontheformer acts 88 PLANE MECHANICS [Ssc. 3.2 inthedirection opposedtothedirection ofmotion oftheparticle atthepoint ofcontact, and (3.204)- ', whereit!isthecoefficient ofkineticfriction, which depends onlyon thenature ofthesurfaces.* Ifboth surfaces aremoving, thelawhasthesameform except thatthedirection oftheforce offriction isopposed tothedirec- tionofrelative motion. Theangle ofkinetic friction X'isdefined by (3.205) tanX'=/. Asanexperimental result, //islessthan/z; jj,isalways lessthan unity, t Problems instatic friction often present considerable difficulty because thefundamental relation (3.202)isaninequality and,in mathematics, inequalities areusually more difficult tohandle than equations. This difficulty may, however, beovercome by treating cases oflimiting friction,forwhich (3.206) ~=/i. When thisrelation holds, thesystem isonthepoint ofslipping. Some problems onfriction. Example1.Alight ladder issupported onarough floorandleans against asmooth wall.Howfaruptheladder canamanclimb without slipping taking place? InFig. 31,AB istheladder andCistheman (replaced byaparticle). Only three forces actontheladder:(i)theweight oftheman(W}\ (ii)the reaction atthewall, thisreaction being horizontal onaccount ofthesmooth- ness ofthewall; (iii)thereaction oftheground. The lines ofaction ofthe firsttwomeet atD.Hence thelineofaction of(iii)must passthrough D, andhence theangleDBEtwhereBE isvertical, must notexceed theangle offriction X.Thus thehighest position thatthemancanreachmaybe found asfollows: Draw alinethrough B,making anangle XwithBE;letit cutthehorizontal through AatD;through D,draw avertical; thepointC *Thisquantity willbedenoted by juwhen there canbenoconfusion with thecoefficient ofstatic friction. tForfurther details regarding friction, seeP.P.Ewald, Th.Poschl, and L.Prandtl, ThePhysics ofSolids andFluids (Blackie &Son, Ltd., Glasgow, 1930), p.67. SEC. 3.2] APPLICATIONS INPLANE STATICS 89 where this linecuts theladder istherequired highest position. This method iscalleddescriptive orgraphical, because theresultmaybeobtained bydrawing toscale. F FIG. 31.--The ladder problem for asmooth wal) (desciiptivo method).w Fio. 32.-Theladder problem forasmooth wall (analytical method). Letusnow discuss thosameproblem analytically. Figure 32shows the forces acting ontheladder. Lot betheinclination oftheladder tothe vertical. The total vortical component must vanish; thus tf-W=0. The total horizontal component must vanish; thus N'-F=0. Thetotalmoment aboutBmust vanish; thus WBCsinaN'-ABcos 0. Hence Thus, by(3.202),F=N'-WBC 'ABtana, N F Nw,BC "AB BC TBtana. .tana<> Thehighest pointCattainable isgivenby (3.207) BCABn cota. The analytical method appears more complicated than thedescriptive, but ithastheadvantage ofbeing more systematic. Moreover, since the 90 PLANE MECHANICS [3EC. 3.2 three conditions ofequilibrium giveallpossible information, thesolution of theproblem isreduced toalgebra assoon asthey arewritten down. Itmight bethought that indrawing thearrow fortheforce offriction to the leftinFig. 32,wewere antici- pating theresult. This isnotac- tually the case. When wedraw anarrow inconnection withacom- ponent ofaforce, wearesimply indicating thesense inwhich this component isconsidered positive. Hadwedrawn thearrow tothe right inFig.32,weshould have ob- tained equations asabove, butwith thesign ofFreversed. The final physical result would havebeenthe However, since positive quanti- tiesareeasier tothink ofthannega- tive quantities, itisadvisable whenever possible todraw the arrows inthesenses inwhich the forces really act. Thus, inthecase ofN,wedraw thearrow upward. Asforfriction, itisgenerally foundB W1 FIG. 33.The ladder problem for rough wall (descriptive method). thattheforce offriction acts inthedirection opposed tothemotion which would takeplaceinitsabsence. That iswhythearrow forFinFig.32was drawn totheleft. Example 2.Thepreceding problem modified bysupposing both walland floor toberough, with thesame coefficient offriction /*. Consider thecones offriction atA andB.They willcuttheplane ofthe paper infour lines asshown inFig.33, these four lines giving thequadrilat- eralFGHJ. Draw thevertical through C,theposition oftheman, and letthisvertical cutthesides ofthe quadrilateral atK,L.LetMbeany point onthesegment KL.Now the weightWmayberesolved intoforces alongMA,MB,andhenceWcanbe balanced byforces alongAM,BM.B FIG. 34.Theladder problem fora rough wall (analytical method). Since these lines lieinside thecones offriction, thelawoffriction issatisfied. Wehave hereacaseofstatical indeterminacy (cf.Sec. 2.5):provided that thevertical through Ccutsthequadrilateral FGHJ, theladder willbein equilibrium, butwecannot tellprecisely what thereactions ofthewalland floor willbe. SEC. 3.2] APPLICATIONS INPLANE STATICS 91 Now letusask:How farcanthemangouptheladder before slipping takes place? Obviously, hecanclimb untilthevertical throughhisposition passes through thepoint J.When hepasses that position,itwillnolonger bepossible tofindreactionssatisfying theconditions ofequilibrium andthe lawoffriction. Thequestion may alsobetreatedanalytically. Consider themanslowly climbing theladder. Iftheladderslips atall,justatthepoint ofslipping thereactions atbothcontacts must correspond tolimiting friction. Thus, atthepoint ofslipping, theforcesystem isasshown inFig.34,withF=pN. F' pN'. Taking vertical andhorizontal components andmoments about B,wehave thethree equations WBCsina-+N-W-0, N'-N=0, -ABsino-AT'ABcosa-0. These three equations determine N,Nr ,BC:wefind W,W BC.._ 1+//(3.208)N N'AB). w Fio. 36.Aheavy block pushed byahorizontal force P. Example 3.Ablock restsonarough horizontal floorand ispushed bya gradually increasing horizontal force. Will theblock slidetorwill ittopple overanedge? Letthethickness oftheblock be2a,itsweight W,andthecoefficient of static friction /*.Letthehorizontal forcePbeapplied ataheight habove the floor. The firstquestionis:GivenWandPasshown inFig. 35,can there beasystem ofreactions exerted bytheground, satisfying simultane- ously thelawoffriction andtheconditions ofequilibrium fortheblock? Anysuchsystem ofreactions willbeplane-equipollent toforces X,YatAas shown, together withacouple N. Ifequilibrium exists, itisclear thatthe following conditions aredemanded bythelawoffriction andthefactthat thefloorcannot pulltheblockdownward : (3.209) Y>.0,N2>0. 92 PLANE MECHANICS [SEC. 3.3 Taking horizontal and vertical components andmoments about A,we have X=P,F-W,N-aTF-AP, andso(3.209) give (3.210) P<ZnW, P^^ Starting with asmall value ofP,these inequalities areboth satisfied; butasPisincreased, oneorother willbeviolated, andthenequilibrium will cease. If (3.211) ft<*> the firstinequality of(3.210) willbebroken first. Attheinstant when P=nW,wehave X-pY, N>0. This isastate oflimiting friction; and so,if(3.211) holds, equilibrium ofthe block willbebroken bysliding along theplane. Ontheother hand,if (3.212) M> , then thesecond inequality of(3.210) willbeviolated first. Attheinstant whenP=aW/h, wehave X<pY, Y>0,N 0. The friction isnotlimiting, andsoslipping cannot take place. Butany further increase inPwillcause violation ofthelastinequality of(3.209). Hence weconclude that,if(3.212) holds, equilibrium willbebroken by theblock turning overtheedgeA. The result isinagreement withcommon experience: thesmaller we make h,themore likely issliding tooccur. 3.3.THINBEAMS Tension, shearing force, andbending moment. Letusconsider astraight beam ofuniform section (Fig. 36) andaplanePparallel toitslength. Pmayberegarded asthe S FIG. 36.Reactions across asection ofabeam. plane ofthepaper. External forces, parallel toP,actonthe beam. (These forces arenotshown. Theymay consist of theweight ofthebeam orloads placed onit.)Letacross section SEC. 3.3] APPLICATIONS INPLANE STATICS 93 bedrawn through apoint O,perpendicular tothelength ofthe beam. Letustake asour"system" theportion ofthebeam extending from theendAuptothissection. Theexternal forces acting onthissystem willconsist of (i)theexternal forces already mentioned, acting onthispor- tion ofthebeam, (ii)thereactions exerted across thesection bytheparticles intheportion ofthebeam extending from thesection totheend A (J5B FIG. 37.Athinbeam?" B.These reactions areinternal forces asfarasthewhole beam isconcerned, butthejr areexternal forces forthesystem at present under consideration. LetustakePasthefundamental plane. Thereactions across thesection areplane-equipollent toaforce acting at0,together withacouple M.The forcemayberesolved intocomponents T,Salong thebeam andper- pendicular toitslength, re- spectively. Wedefine the A following terms: T=tension, S=shearing force,M=bending moment. Weshall confine ouratten- (6) tion tothinbeams. Thethin Fio. 38. (a)Reactions exerted on beam isamathematical ideal-$%$(&)RcaPtlons oxorted on ization, inwhich thecross sec- tion isreduced toapoint andthebeam toastraight line. Figure 37shows athinbeamAB;Cisanypoint ofit.To draw thereactions onACacross thesection atCwithout con- fusion, wedelete thelineCBasinFig.38o. Figure 386shows thereactions onCB;these have thesame magnitudes as,but opposite senses to,those shown inFig.38a,onaccount ofthe lawofaction andreaction. Letustakeanorigin onthebeam, therr-axis along thebeam andthey-axis perpendiculartoit.Consider asmall length of thebeam extending from xtox+dx(Fig. 39). LetT,S,M bethevalues oftension, shearing force, andbending moment at x,andT+dT,S+dS,M+dMthevalues atx+dx.To allow forgravity orother continuous external loading, weshall 94 PLANE MECHANICS [SEC. 3.3 addaforce withcomponents-X"dx,Ydx(notshown inFig.39) acting atthemiddle point oftheportion x,x+dx.Bytaking components andmoments about thepoint xand neglecting y S-hdS FIG. 39.Reactions ontheends ofasmall element ofabeam. infinitesimals ofthesecondorder, wehave, asconditions of equilibrium forthesmall length ofthebeam, dT+Xdx=0, dS+Ydx0,dM+Sdx=0. Thus (3.301) --Z, --Y, --*v ' ' 'dx dx dx BThese arethegeneral differential equations fortheequilibrium of thinbeams. Butinstatically determinate caseswecanobtain allrequired information regard- inginternal reactions without us- ingthese equations, orrather byusing them inintegrated formw a.40.Alightbeam loaded atits middle point. Statically determinate problems. Weshall illustrate themethod bythesolution ofaproblem. AlightbeamABoflength 2aishinged atAandsupported ona smooth horizontal plane atB(Fig. 40).AloadWisplacedat themiddle point C.Find thebending moment andshearing force along thebeam. First, byapplication oftheconditions ofequilibrium (2.306) tothewhole beam,wefindthereactions onthebeam atAandB. SEC. 3.3] APPLICATIONS INPLANE STATICS 95 These areeach ofmagnitude %W ,directed upward. Letustake ourorigin atAandthez-axis along thebeam. Consider the rtsf |<-X >l>ipA D Fio. 41. External forces onaportionof thebeamshown inFig. 40(AD<AC).D W Fio. 42. External forces ona portion ofthebeam shown in Fig.40(AD>AC).>T (3.302)portion ofthebeamADtwhereDliesinA(7;letAD=x(Fig. 41).From theequilibrium ofAD,wehave (T=0,8=-iTF, \M=-xS=&W, (x<a). These givetheshearing forceandbending moment foranypointmAC; there isnotension. SinceSisnegative, theshearing force actually acts inthedownward direction. Now takeDinCB(Fig. 42). Instead of(3.302), wehave -xS=(a- Theshearing force isnow positive. (3.301)issatisfied by(3.302) and(3.303). Thegraphs ofSandMalong thebeam areshown inFig. 43.AWenote that thelast of M The Euler-Bernoulli theory of thin elastic beams. 8 FIQ. 43.Graphs ofshearing force (S)andbending moment (M)along thebeam shown inFig. 40.Ifastraight beam rests on three supports, theproblem of finding thereactions duetothesupportsisstatically indeter- minate(cf.Sec. 2.5),andwecannot findtheshearing forceand bending moment byelementary statical principles. But this indeterminacy disappears whenwetake intoconsideration the elasticity ofthebeam. Although straight initially, anelastic beam willstretch andbend under theinfluence offorces. We 96 PLANE MECHANICS [SEC. 3.3 suppose thestretching andbending tobevery small andaccept thelawofHooke forstretching andthelawofEulcr andBer- noulli forbending.* HOOKK'S LAW. When abeam isslightly stretched, (3.304) T=k'e, where cistheextension (increase inlength perunitlength) and kraconstant forthebeam. (Actually kr=EA,whereEis Young's modulus forthematerial andAthearea ofthecross section.) THEEULER-BERNOTJLLI LAW. When abeam isslightly bent, thebending moment isconnected with thecurvature bythe relation (3.305) M=-> P where pistheradius ofcurvature andkaconstant forthebeam. (Actually k=El,whereEisYoung's modulus and7the "moment ofinertia" ofthecross section about anaxisthrough itsmean center perpendicular totheplaneofthecouple M.) When thebeam isapproximately straight andtheaxes asin Fig. 39, p= ~dx*approximately, and(3.305) maybewritten (3.306) M=kg- Letusrefer toFig.39andtotheequations (3.301). Weshall suppose thatthebeam issubject toaforcewperunitlength inthenegative sense ofthe 2/-axis, dueeither toitsownweight ortoaloadplaced onit.ThenX=0,Y=w,and (3.301) read (3.307)f=0,f= ,d-f=-S.^ 'dx'dx'dx Weseethatthetension Tisconstant. Elimination ofMandS from (3.306) and(3.307) gives (3.308) *g-u,. *ThelawofEuler andBernoulli follows from that ofHooke; theproof belongs tothetheory ofelasticity. SEC. 3.3J APPLICATIONS INPLANE STATICS 97 This isthefundamental differential equation inthetheory of thin elastic beams. Ifitissolved, thebending moment and shearing force aregiven by d*ys=_dM dxv~ dx(3.309) M S+AS AM+AMItmust berealized that thedifferential equation (3.308) holds onlybetween isolated loads orsupports. Todealwith these a special treatment isnecessary. Figure 44shows anclement Ax ofathinbeam withanisolated loadWsuspended from its middle point P.(The case ofasupport iscovered by makingWnegative.) The element isinequili- brium under four forces and two couples: thecontinuous loadonAx(notshown), the isolated loadW,theshearing forceS+AS,andthebend- ingmomentM+AMonthe right, andtheshearing force Sandthebending momentM ontheleft, positive senses being asindicated. IfAxtends tozero, thecontinuous load tends tozeroandso does themoment ofthisloadabout P.Hence theconditions ofequilibrium give, inthelimit,AS=W,and(taking moments about P)AM=0. Thismeans thatthebending momentMiscontinuous across anisolated load orsupport, buttheshearing forceSchanges abruptly. Interms ofyand itsderivatives (since thebeam is notbroken attheisolated load orsupport) wehave continuity iny,dy/dxy d^y/dx*, butdiscontinuity ind*y/dx*. Example. Auniform heavy beamOPoflength 2aandweightWishinged atOand restsontwosmooth supports, oneatPandtheother atitsmiddle point Q.Find thereactions onthesupports, if0,P,Qareallatthesame height. Weshalltaketheorigin ofcoordinates atO,there-axis horizontal, andthe y-axis directed vertically upward. Integration of(3.308) alongOQgivesFIG. 44.--Element ofbeam containing ibolatcd load. (3.310) ky--faux* -fAx*+Bxt (OQ) 98 PLANE MECHANICS [Sac. 3.4 where A,Bareconstants ofintegration; twoother constants ofintegration havebeenputequal tozeroonaccount ofthevanishing ofyandd*y/dx* atO (There canbenobending moment atahinge orfreeend )Similarly, we have alongQP (3.311) ky=-ftw(x-2a)<+A'(x-2a)+B'(x-2a), (QP} where A',B'areconstants ofintegration. Inthesetwoequations, wehave fourunknown constants; they aretobefound from theconditions that y=atQ,while dy/dx andd*y/dx* arecontinuous there. Thus,wehave thefourequations Aa3+Ba fWa*=0, A'a8+B'a+fawa*=0, 3Aa2+B-\wa*-3A'a2+B' 6Aa-Jfl=6A'a Wefind A.-A'=^wa, #=B'= andsubstitution in(3.310) and (3.311) gives theequations ofthetwopor- tions ofthebeam ~ar)44. InOQthebending moment is Itsmaximum value occurs atx=|<z.Theportion OQisasystem inequi- librium;hence, taking moments about Q,wehave forthereaction JRoatO Rod-MQ+$waz-|u>a2-&Wa. When onereaction hasbeen found, theothers follow from theusual statical methods. Hence (3.313) Ro=&W, RQ-i$JK, RP= 3.4.FLEXIBLE CABLES Aflexible cable differs from astiffrodintheeasewithwhich it canbebent intoacurve. Thebending moment perunitcurva- ture ismuch lessforthecable. Inmechanics, weidealize this property andunderstand byaflexible cable amaterial curve such that there canbenobending moment across any section. By considering theequilibrium ofasmall portion ofthecable,itis easily seenthattheshearing forcemust alsovanish. Hence the only surviving component ofthereaction across asection ofa flexible cable isatension 27 ,which actsalong thetangent tothe curve inwhich thecable lies. Weusetheword "cable" exclusively, but itistobeunderstood that thepractical applications cover chains, ropes, strings, and threads. The theoretical predictions willagree wellwith the SEC. 3.4] APPLICATIONS INPLANE STATICS 99 Wdftresults ofexperiments conducted oncables inwhich thebending moments aresmall. General formulas forallflexible cables hanging freely. Letusconsider aflexible cable hanging under theinfluence ofitsown weight, andperhaps additional continuous vertical loads attached to it.Forthe present, weshall notmake any special assumptions regarding the nature ofthecable ortheload. Wepass over the trivial case inwhich thecable hangs from oneendinavertical line.When suspended from two points,it hangs inavertical plane. Let Oxybeaxes inthisplane, Oxbeing horizontal andOydirected verti- callyupward (Fig. 45). LetA beapoint onthecable withcoordinates(.r,y),andBanadjacent point with coordinates (x+dx,y+dy). Letdsbetheinfini- tesimal length ofAB,and letwdsbethetotal loadonAB, including theweight ofthecable. TheportionAB isasystem inequilibrium under theaction ofthetensions atitsendsand theload. Let Bbetheinclination ofthetangent atAtothe horizontal. Then dx/ds=cos0,dy/ds=sin0;andso,taking horizontal andvertical components, wehave 0. Bythe first ofthese equations,thehorizontal component ofthe tension isconstant. Thesecond equation maybewrittenFIG. 45. Forces acting onanele- ment ofahanging cable. (3.401) If//istheconstant horizontal component oftension, wehave (3.402) rg-H; substitution in(3.401) gives <n> sSD-r 100 PLANE MECHANICS [SEC. 3.4 This isadifferential equation satisfied bythecurve inwhich the cable hangs. When thisequation hasbeen solved, thetension maybefound from (3.402). Thesuspension bridge. Letusnowsuppose thataweightless cable supports aload uniformly distributed onahori- zontal line;fortheloadonahori- zontal length dx,wewritewdor. This approximates tothecondi- ;tion ofacable ofasuspension bridge (Fig. 46),theload consist- ingoftheroadway AB, ofweight WQperunit length. With thenotation used above, wehavewds=WQdx,andso dxA B FIG. 46.Suspension bridge. thus (3.403) reads ds\dx 775? dx*H Iftheorigin ischosen atthelowest pointofthecable, sothat y=dy/dx=when x=0,weobtain astheequation ofthe cable (3.404) y=i^ This isaparabola. Thetension inthecable isgivenby(3.402). Since (3.405) wehave (3.406) T=H Thecommon catenary. Weshallnowconsider auniform cablehanging freely under its ownweight, wperunit length. Thefundamental equationis SBC. 3.4] APPLICATIONS INPLANE STATICS 101 (3.403),inwhichwisnowaconstant. Wewrite itintheform d*y_wds dx*~Hfa' or,by(3.405), (3.407)- Introducing avariable zdefined by (3.408)sinh z= ~|, wereduce (3.407) to dxw H' andso 47.Tho common catenary.whereAisaconstant ofintegration. Choosing theoriginOatthelowest point ofthecable (Fig. 47),wehave y=dy/dx= fora;=0, andhence z=forx=0.ThusA=0,and(3.408) reads (3.409) Hence (3.410)dy.twx=smh H(y=IJw\,wx .cosh-jf1 when account istaken oftheconditions at0.Thiscurve is called thecommon catenary; thelowest pointiscalled its vertex. Itiscustomary todefine theparametercofthecatenary by (3.411)c- ; then (3.410) reads (3.412) y=c(cosh^-1 Tofindthetension from (3.402), wenote thatfrom (3.405) and (3.409) (3.413) 102 PLANE MECHANICS [SBC. 3.4 andso (3.414) T=H~=Hcosh-=H+wy.ax c Sofarwehave concentrated ourattention ontwo things, thecurve inwhich thecable hangs andthetension atany pointinit.These have been found in(3.412) and (3.414). But other problems suggest themselves, andweneed other formulas tosolve them. Suchproblems mayinvolve thelength ofthecableandtheinclination ofitstangent tothehorizontal. Letusdenote thelength bys(measured from thevertex toa general point) andtheinclination by0}there arethen five variables involved inthetheory ofthecable, x,y,T,5,6. Anyoneofthese variables isexpressible interms ofanyother, and itisaninteresting exercise toprepare atable offiverowsand columns showingallthetwenty expressions. Weshallnotehere onlytheexpressions giving sinterms ofx,y,and0,asfollows: (3.415) (3.416)s=csinh -,cs2=y2+2yc t ctan 8. These equations areeasy toobtain from (3.413), combined with (3.412) ;toget(3.416), weusethefactthatdy/dx=tan 0.The equation (3.416)istheintrinsic equa- tionofthecatenary. Examples. Problems connected with freely hanging cables usually involve theso- lution ofatranscendental equation. As illustrations, twoproblemswillbeconsidered. These problems may bestated briefly as follows : (i)Given thespanandlength,tofindthe maximum tension. (ii)Given thelength andsag, tofindthespan.FIG. 48.Ahanging Acable, ofweight wperunitlength andlength 2Z,hangs fromtwopoints AandB,atthesame height andatadistance 2aapart (Fig. 48).Wewish tofindthemaximum tension inthecable. Itisclearfrom (3.414) thatthemaximum tension occurs atAandB,and thevalue is (3.417) nM=Hcosh-=wecosh c c SEC. 3.4] APPLICATIONS INPLANE STATICS 103 Here, asinmost problems onthecatenary, thesolution depends onfinding theparameterc.Applying the firstof(3.415) atthepoint B,wehave (3.418)Icsinhi This isanequation todetermine cinterms ofaandI;itmaybewritten (3.419)B-^/)-L Iftables of(sinhX)/X areavailable, thenumerical value ofa/cmaybe obtained atonce.* The solution oftheproblemisgivenby(3.417) on inserting thevalue forc,found from (3.419). Iftheratio I/aisnearly unity, i.e.,ifthecable isonlyalittle longer than thespan, thesolution of(3.419) fora/cissmall, because ,sinhX ..hmy=1. Infact,theparameter cislarge. Thenwecanobtain anapproximate solu- tion of(3.419) without recourse tonumerical tables. Retaining only the firsttwoterms oftheexpansionforsinh a/c,wehave (3.420) Since cislarge,Tmt*asgivenby(3.417) islarge;itisapproximately equal toH,where (3.421)//=we-wa^^~ ay Thesecond problemariseswhen thedistance between twopointsAandB atthesame heightismeasured byameasuring tapewhich sagsunder its own weight. With thenotation ofFig. 48,wearcgiven h,I;wewish to find a. Applying thesecond of(3.415) atthepoint B,wehave (3.422)c=~- - Theanswer totheproblemisgiven by(3.418). This isaquadratic equa- tion forea/c ,andthepositive root gives t *J.W.Campbell, Numerical Tables ofHyperbolic andOther Functions (Houghton Mifflin Company, Boston, 1929), p.30.These tables were prepared with thesolution ofcatenary problems inmind. tThroughoutthisbook "log" moans thenatural logarithm, thatis,log.. 104 PLANE MECHANICS [Sac. 3.4 Iftheratio h/lissmall, wehaveapproximately (3.424)lc andhence (3.425) withanerror oftheorder of Cables incontact withsmooth curves. Sofarthecables considered havebeen unconstrained. Letus nowconsider thecase ofacable lying against asmooth surface, or, aswemaysayintwo-dimensional language, against asmooth curve. Gravity willbeneglected. Figure 49shows asmall portionABofacable lying inequilib- rium incontact with asmooth curve. Let beanyassigned point onthecableand sthelength ofthecable between andA. LetthelengthABbeds,and let theinclinations tosome fixed di- rection ofthetangents tothecable atAandBbeand+d9.The element ABisinequilibrium under three forces, namely, thetension T atA,thetension T+dTatB,and anormal reaction duetothecurve. This lastmaybewrittenNdsand maybesupposed toactalong the normal atA.Resolving forces along thetangent andnormal at A,weobtain from theconditions ofequilibriumNds FIG. 49. Forces onanelement oflight cable incontact with a smooth curve. (3.426) dT=0,Nds=TdO. Hence, thetension isconstant along alight cable incontact witha smooth curve. Also, since ds/dO=p,theradius ofcurvature, wehave (3.427)TN=- Anexampleofthesignificance ofthis lastformula occurs in tyingupaparcel:thesharper theedge oftheparcel, thesmaller p andhence thegreater thetendency ofthestring tobiteintothe parcel. SBC. 3.4] APPLICATIONS INPLANE STATICS 105 Cables incontact withrough curves. Letusnowsuppose thatthecurve shown inFig.49isrough andthatthecable isjustonthepoint ofslippinginthedirection AB. Inaddition totheforces already considered, there isnow aforce offriction Fdsontheelement, acting along thetangent atAandopposing motion. Theconditions ofequilibriumare now (3.428) dT=Fds,Nds=TdO. ButF=pN,where/xisthecoefficient offriction. Hence (3.429) g.^, tf-rg andso (3.430) =,T. Integration gives (3.431) T=To&', whereTisaconstant ofintegration. Therapid increase oftheexponential with increasingisof great practical importance. Asanumerical example, consider a ropewrapped twice around apost, forwhich thecoefficient of friction is .Then T=TQC**=Toe2 *, where T,Tarethetensions intheropewhere itmeets and leaves thepost, slipping being about tooccur inthedirection ofT.Wehave ?J>=c-2ir=0.0019. IfT=2000 lb.,To=3.8 Ib.Thus, aload ofonetoncanbe sustained byapplicationofaforce oflessthan 4lb.;and, of course, amuch greater loadmight besustained iftheropewere wrapped more often round thepost. This principleisused in holding shipsbyropes passed round mooring postsandinhoists inwhich aropeispassed round arevolving drum, theendbeing held inthehand. 106 PLANE MECHANICS [SEC. 3.5 3.6.FRAMES Just-rigid frames. Figure 50shows asimple frame ortruss, asused inbridges. Itconsists ofsteel girders riveted together atthejoints. For mathematical discussion wesimplify thesystem asfollows: (i)thegirdersaretreated aslight rigid bars, (ii)thejoints are C D FIG. 50.Ajust-rigid frame with loads applied atEandF. supposed tobesmoothly working hinges, eachbarbeing capable ofrotation about thejoints onitwithout anyresisting couple. Weshall discuss onlyframes with joints lying inaplane, and weshallnotconsider displacements outofthat plane. InFig. 50,wesuppose thejointAfixedandthejointBcon- strained toslideonahorizontal line. Inspection shows thatthe whole frame isfixedbythese conditions. Infact,theframe isa rigidbodyand isfixedwhen oneofitspointsisfixedandanother ofitspoints constrained tomove onaline. Ifone bar, for example CD,wereremoved, theframe would cease tobearigid body. Hence itiscalled just-rigid. Thefollowingisthegeneral definition :Aframe isjust-rigid when theremoval ofanyoneofits bars destroysitsrigidity, Ifanadditional bar isinserted inajust-rigid frame,itbecomes over-rigid. Weshall dealonlywith just-rigid frames. We shallnowshow thatajust-rigid frame withjjoints has 2j 3bars. Taking anyaxes intheplane oftheframe, we denote thecoordinates ofthejoints by (xi, 2/1), (x2,2/2), (xj, 7/7);there are2jcoordinates altogether. Ifthe firsttwo joints areconnected byabaroflengthlytheir coordinates must satisfy (xi-*2)2+(2/1- 2/2)2l\ Thus ifthere are6bars, the2jcoordinates aresubjected to6 relations ofthistype.Ifwefixonejointandconstrain another SEC. 3.5] APPLICATIONS INPLANE STATICS 107 joint tomove onaline,weimpose 3more conditions. Ifthe frame isjust-rigid, these 6+3conditions suffice tofixthewhole frame, i.e.,todetermine the2jcoordinates ofthejoints. Hence b+3=2jjwhich gives thestated result: (3.501) 6=2j-3. If6<2j-3,theframe isnotrigid.r FlG> 51.__Frame Thesmallest number ofjoints possible inawith three joints just-rigid frame isj=3.Then thenumberandthree bar9 ' ofbars is2j 3=3.Inthiscase,wehave atriangular frame (Fig. 51). Now take j=4;then thenumber ofbars is2j 3=5. Examples areshown inFig. 52.(When twobars cross ina FIG. 52.Frames \uth four joints andfivebars. diagram, without indication ofajoint, they aresupposed capable offreemotion pastoneanother.) Ifj=5,thenumber ofbars is2j 3=7.Examples areshown inFig. 53. FIG. 53.Frames with fivejoints andseven bais. Asimple waytobuildupajust-rigid frame istostart witha triangle andaddtwobars atatime. Since,ineach operation, weaddonejointandtwobars,afterpoperations wehave 3+p joints and3+2pbars; theidentity 3+2p 2(3+p)-3 shows that thecondition forajust-rigid frame issatisfied. However,alljust-rigidframes cannot beconstructed inthisway. 108 PLANE MECHANICS [Sac. 3.5 Stresses inbars. Suppose thatajust-rigid frame isfixedbyexternal constraints sothat itcannot move. (Thenormal planistofixonejointand constrain another joint tomove onaline, asinFig. 50.)Now letexternal forces, orloads, beapplied tosome orallofthejoints. Each bar isinequilibrium under two forces, thereactions atits ends. These twoforces must beequal inmagnitude andactin oppositesenses along thebar. Iftheforces actaway from one another (sothatthebartends tobetorn intwo), thebar issaid to beintension;iftheforces acttoward oneanother (sothatthebar tends tobuckle), thebar issaid tobeinthrust. Theword stress Tension Thiust Fio. 54.Reactions exerted byabaronthojoints atitsends. isused tocover both cases.Aplus signisassociated with tension andaminus signwith thrust. Thus,ifwesaythatthe stress inabar is+3tons,wemean thatthere isatension of3tons init;ifthestress is5tons,wemean thatthere isathrust of 5tons. InFig.54thearrows indicate forces exerted onthejoints bythebars. The forces exerted onthebarsbythejointsart inthereverse directions. Foraframe inequilibrium, twoproblemsarise: (i)todetermine theexternal reactions atthesupported joints; (ii)todetermine thestresses inthebars. The firstproblemiselementary.Itisaquestionoftheequi- librium ofasystem,asdiscussed inSees. 2.3and2.4andsum- marized inSec. 2.C. Itiswith thesecond problem thatwearc concerned. Method ofjoints. Thefollowing argumentisgeneral, butthereader may con- sider theframe shown inFig.50asanexample, theloads being indicated byarrows atthejointsEandF.Theloads arcgiven, andthestresses aretobefound. Each jointmaybeconsidered asaparticleinequilibrium, under thoaction ofaload (ifany) andthereactions ofthebarsmeetingthere. (Since thejointis thesystem considered, thismethod iscalled themethod ofjoints.) Astheforces lieinaplane, there arctwoequations ofequilibrium foreach joint, andthus 2jequationsinallifthonumber ofjoints isj.These equationsinvolve 3unknown components ofexternal SEC. 3.5] APPLICATIONS INPLANE STATICS 109 reactions atthesupports andanumber ofunknown stresses equal tothenumber ofbars, i.e., 2j3.Thus the total number ofunknowns is2j,andwehave 2jlinear equationsto findthem. Thus, inajust-rigid frame theproblem offindingthe external reactions atthesupports and thestresses inthebars isa determinate problem, involving thesolution ofanumber ofsimul- taneous linear equations equaltotwice thenumber ofjoints. Iftheframe wereover-rigid, thenumber ofunknowns would exceed thenumber ofequations, andtheproblem would be indeterminate. Weshould have toconsider theelastic properties ofthebars. FIG. 55a.Aframe with 14joints, supporting aload atM. Theproblem ofthejust-rigid frame having been thusreduced tothesolution ofsimultaneous linear equations,itmight be thought that nothing remained tobesaid. However, the system ofequations obtained inthemanner described above may bevery involved, andmuch labormay beavoided by modifying themethod. This isparticularly true ifweonly require the stresses incertain bars. Butthereader should realize that these areonly laborsaving devices. Ifhecannot discover theparticular device suited toacertain problem, he canalways fallbackonthedirect laborious method. Before turning tothespecial devices, letusseehowthemethod ofjointsmay beapplied without undue complication tothe frame shown inFig.55a. Thisframe has14joints, andhence a direct attack involves 28simultaneous equations. Theload atMisW.Wefindatonce(bytaking components andmoments) thatthereactions atHandNareboth vertical and ofmagnitudes RH=^W,RN=W> LetSAB,Sac bethe stresses inthebars. From theequilibrium ofthejoint AT,we have -fiir=-0. 110 PLANE MECHANICS [SEC. 3.5 Passing toG,wehave SQMsina=SGN SFG=SGMCOSa-%W cota, where aistheinclination oftheoblique bars tothehorizontal. Proceedinginthisway, stepbystep,wecanfind allthestresses. Incidentally, weshall getacheck onourworkwhenwereach the last joint.Itwillbenoted that, tostart themethod, wemust begin withajointwhere onlytwobarsmeet. Method ofsections. When werequire thestresses inonlysome ofthebars, the method ofjointsmay prove unnecessarily laborious. Letus recall thefact,emphasizedinSec. 2.3,thatwemaychoose any part ofthegiven system asthesystem towhich theconditions of FIG. 55fc.Method ofsections: the"system" isenclosed bythebroken line. equilibriumareapplied. Uptillnowwehavebeen thinking of asingle bar,thewhole frame, orasingle jointasthesystem. Butherewetakeadifferent approach, following themethod of sections. Figure 556shows thesameframe asthat ofFig.55a,withthe same load.Wewish tofindthestresses inKL,KE,DE.We consider asasystemthepart oftheframe enclosed within a curved linecutting thebarsKL,KE,DE,butnoothers. This systemisacted onbythefollowing external forces: theloadWatM; thereaction RNatN; thestresses inKL,KE,DE. Taking moments about E,wehave SKIEL+W-EF=RK'EG. SBC. 3.5J APPLICATIONS INPLANE STATICS 111 ButRNmaybefound byconsideration oftheequilibrium ofthe whole frame; hence SKL=\WCOta. From consideration ofthetotal vertical component offorce,we have SKE=?Wcosec a; and,from thetotal horizontal component, SDE SKL SKECOS a.=%-WCOta. Wenotethatthemethod would nothaveworked hadthethree barscutbythesection metinapoint. Method ofvirtual work. Themethod ofvirtual workmaybeapplied totheproblem just treated. TofindSKL,wesuppose thebarKLremoved and forces applied tothejointsKandLequal tothe(unknown) stress inKL.Theframe isnolonger rigid, but itisinequilibrium. Hence thevirtual work done inaninfinitesimal displacement is zero. For infinitesimal displacement,letustake arotation aboutEoftheright-hand portion oftheframe. Theonly forces todowork aretheloadW,thereaction RN,andtheforce atL replacing thestress SKL. Equating thework donebythem to zero,weobtain theexpression forSKLgiven above. Togetthestress inEK,wereplace thebarKLandremove EK,atthesame timeapplying tothejointsEandKforces equal tothe(unknown)stress inEK.Nowwegiveavirtual displace- ment, holding theleft-hand portionfixed. Theright-hand side rises slightly with parallel displacementofitsbars, thebarDE hinging atDandKLhinging atK.Theonlyworking forces aretheloadW,thereaction RN,andtheforce atEreplacing the stress SKE. Equating tozerothework done,wefind forSKs thevalue given above. SDS isfound similarly without difficulty. Complex frames. Frames constructed byadding successive pairs ofbars toa basic triangular frame arecalled simple frames. Those sofar discussed have been ofthis type. But there arealso just- rigidframes which cannot bebuiltupinthisway; such frames arecalled complex. Anexampleisshown inFig. 56,inwhich 112 PLANE MECHANICS [SBC. 3.5 thebars aresupposed tocross without touching. The stresses may befound bysolving the12equationsofequilibrium of thejoints, butthatmethod iscom- plicated. Wecannot usethestep- by-step method ofjoints, because there isnojoint atwhich onlytwo bars meet.Wemodify themethod byassigning anunknown value to oneofthestresses; wefindtheother stresses interms ofthisoneun- known bytheconditions ofequilib- rium ofthejoints and finally, on closing the calculation, determine theunknown stress andhence allthe stresses. Letuswork thisoutinthecase shown inFig. 56,inwhich thebars FE,ED,AD,FCareinclined tothe horizontal at45,andAB,BC in- clined tothehorizontal at30. Write SEB=S.ThenFIG. 66.Acomplex frame. atE} 8*0=Sn= atD, SAD=S*D=$/\/2, (=SFC,bysymmetry), atZ>, SCD=(S,D-&UOA/2 =-S, atC, SCD+SFC/VZ+&BC/2+Re=0, atC, &c/\/2+/SW-v/3/2=0. Elimination ofSBCfrom thelasttwoequations gives ScoV*+&c(V3-1)/V2+RcV3=0. SinceRc=W/2,SCD=-S,S,c=S/A/2, weobtain S=iTF(3-V3); allthestresses arenow easily written down. Concluding remarks. Themethods described above areadequate insimple cases, andthereduction oftheproblem ofdetermining thestresses tothesolution ofasetof2jsimultaneous linear equationsis complete andsatisfactory mathematically, although often com- plicated. When wehave written down theequations, weknow thatwehave given complete mathematical expression toallthe SEC. 3.6] APPLICATIONS INPLANE STATICS 113 conditions ofequilibrium andthatthestresses canbefound from theequations provided theyareconsistent. Itmayhappen thattheequations ofequilibrium areincon- sistent; thisoccurs inthecase ofcritical forms,ofwhich an exampleisshown inFig. 57.This frame isjust-rigid ;butsince thebars AB,EC lieinastraight line,no stresses inthem cangive equilibri- umofthejoint B,when aloadWis applied there. Such aframe would beanunsound engineering struc- turc. Actually thejointfl would beFlQ .57._Acnticid form . slightly depressed (owing tostretch- ingofthebars), andtherewould beverygreat tensions inthebars AB,BC. Onaccount ofitsimportance inengineering, thetheory of frames hasbeen elaborately developed. Foramore complete account, with special reference toengineering problems, the reader isreferred toS.Timoshenko andD.H.Young, Engineer- ingMechanics (McGraw-Hill BookCompany, Inc.,NewYork, 1940). Most statical problems admit twomethods ofattack. Onthe onehand,wemayreduce theproblem tothesolution ofequa- tions; this istheanalytical method. Ontheother hand,wemay represent forces bysegments, andcompound andresolve them byactual drawing; this isthegraphical method. Eachmethod has itsadvantages, butthroughout thisbookwehave preferred tousetheanalytical method, because itiseasier toexplain and hasawider range ofapplication. Forthegraphical method in statics and itsapplicationtoframes, thereader may consult forexample H.Lamb, Statics (Cambridge University Press, 1928). 3.6.SUMMARY OFAPPLICATIONS INPLANE STATICS I.Mass centers andcenters ofgravity. (a)Formulas forthemass center: (3.601)r=~-(system ofparticles); 114 PLANE MECHANICS [Sue. 3.6 (3.602)f-Vpf*fyf8 (continuous system).JJjpdxdydz (6)Devices forfinding mass centers: (i)symmetry, (ii)decomposition, (iii)theorems ofPappus. (c)Center ofgravity coincides withmass center. Potential energy=Mgy. With respect toanyvertical plane, theweights ofalltheparticles ofasystem areplane-equipollent toasingle force (total weight) acting through thecenter ofgravity. II.Friction. Static friction: F/N^/* or 6<X;(tanX=ju). Kinetic friction: F/N=n' or =X'; (tanX'=/*') III.Thinbeams. (a)S -^T T=tension, S=shearing force,M bending moment. (6)Basic assumptions: (i)Hooke's law:T=k'e, (e=extension,&'=EA). (ii) Euler-Bernoulli law:M=k/p, (p=radius ofcurvature, k=El). (c)Differential equationofathinheavy beam: (3.603)kj=-w. (3.604) If-jg, S=-f- (d)Continuity conditions:T/,dy/dx, d2y/dx* arecontinuous. IV.Flexible cables, (a)General formulas: (3.605) Tdfa-H (aconstant);(g). J. (6)Cable ofsuspension bridge hangs inaparabola. Ex.Ill] APPLICATIONS INPLANE STATICS 115 (c)Commoncatenary: (3.606) y-c (cosh5-A c^, (3.607) s=csinh -> (3.608) s2+c2=(*/+c)2 , (3.609) r=77+wy. (d)Light cable incontact with asmooth curve: (3.610) T=constant, N=- P (e)Light cable incontact witharough curve: (3.611) T=T&' (forcableonpoint ofslipping). V.Frames. (a)Just-rigid frame: (3.612) b=2j-3 (b)Method ofjoints. Begin withajointwhere onlytwobars meet. (c)Method ofsections. Section mustnotcutmorethan three bars,andthese three barsmust notmeet atapoint. (d)Method ofvirtual work. Remove abar. (e)Complex frames: (6)and (c)notapplicable directly. Assume onestress<S,anduse (6). EXERCISES III 1.Findthemass center ofacubical boxwithnolid,thesidesandbottom beingmade ofthesame thinmaterial. 2.Aladder leans against asmooth wall,thelower endresting onarough floor forwhich thecoefficient offriction is\.Find theinclination ofthe ladder tothevertical,ifitisjustonthepoint ofslipping. 3.Asquare frame isbraced bytwodiagonal bars. Oneofthese con- tains aturnbuckle, which istightened until there isatension Tinthebar. Find thestresses intheother bars. 4.Aman ofweightWwalks slowly along alight plank oflength a, supported atitsends. Find thebending moment intheplank directly beneath hisfeetasafunction ofhisdistance from oneendoftheplank. Find alsotheshearing forcesjustinfront ofhimandjustbehind him.Draw diagrams toshow thesenses ofthebending moment andshearing forces. 5.Findthemass center ofawirebent intotheform ofanisosceles right- angled triangle. 6.Arod4ft.long restsonaroughfloor against thesmooth edge ofa table ofheight 3ft. Iftherod isonthepoint ofslipping when inclined atan angle of60tothehorizontal, findthecoefficient offriction. 116 PLANE MECHANICS [Ex. Ill 7.Abody ofweight wrestsonarough inclined plane ofinclination i, thecoefficient offriction(/*)being greater thantan *.Findtheworkdone inslowly dragging thebody adistance auptheplane andthendraggingit back tothestarting point, theapplied force being ineach case parallel to theplane. 8.Aheavy cable rests incontact withasmooth curve inavertical plane. Show thatthedifference inthetension attwopoints ofthecable ispropor- tional tothedifference inlevel atthese points. 9.Two light rings canslideonarough horizontal rod.The rings are connected byalight inextensible string oflength a,tothemid-point of which isattached aweight W.Show thatthegreatest distance between therings, consistent withtheequilibriumofthesystem,is +M2 , where pisthecoefficient offriction between either ringandtherod. 10.AheavybeamABCD, ofweight 2Ib.per ft.,issupported horizontally byknife-edges atBandD.Thebeam issubjected toanadditional vertical load of20Ib.atC. IfAB=BC=CD=4ft.,determine theshearing forceandbending moment forallpoints ofthebeam. 11.Aportion ofacircular disk ofradius riscutoffbyastraight cutof length2c.Findtheposition ofthemass center ofthelarger portion. Ifr 1ft.,c=6in.,calculate thedistance ofthemass center from tho center ofthe circle. 12.Alight cable connects twoweights W,w(W>w)andpasses overa rough circular cylinder whose axis ishorizontal. Wrestsontheground, andwissuspendedinthe air.Find theleast value ofthecoefficient of friction between thecylinder andthecable inorder thatWmayberaised from theground byslowly rotating thecylinder, and findexpressions for thework done inturning thecylinder through onerevolution(i)ifWis raised, (ii)ifWisnotraised. Find alsothework done inturning thecylinder through onerevolution intheoppositesense. 13.Fortheframe shown inFig.55a,takea=45,andfindthestresses inallthebars. Make asketch oftheframe, marking withadouble line eachbarinwhich there isathrust. 14.Auniform semicircular wirehangs onarough peg,thelinejoiningits extremities making anangle of45withthehorizontal. Ifitisjustonthe point ofslipping, findthecoefficient offriction between thewireandthepeg. 15.Anelastic beam restsonthree props, twobeing situated attheends ofthebeam andatthesame height, andthethird atthemiddle point oftho beam. Find theheight ofthiscentral propifthepressures onallthree props areequal. 16.Acable 200 ft.longhangs between twopoints atthesame height. Thesagis20ft.,andthetension ateither point ofsuspensionis120 Ib.wt. Find thetotal weight ofthecable. 17.Auniform cable hangs across twosmooth pegs atthesame height, theends hanging down vertically. Ifthefreeends areeach 12ft.long Ex.Ill] APPLICATIONS INPLANE STATICS 117 andthetangent tothecatenary ateachpegmakes anangle of60withthe horizontal, findthetotal length ofthecable. 18.Consider aframe asinFig.50,thetriangles being equilateral. Itisto carry aload2W, either asasingle load atEorequally divided between E andF.Inwhich case isthere greater danger ofcollapse,itbeing assumed that collapseisduetoathrust inabarexceeding some definite value, the same forallbars? 19.Four rodseach oflength aandweightwaresmoothly jointed together toform arhombus ABCD, which iskeptinshapebyalightrodBD.The angleBAD is60,andtherhombus issuspended inavertical plane fromA. Find thetension orthrust inBDandthemagnitude anddirection ofthe force exerted bythejointContherodCD. 20.Two equal spheres, each ofweight W,restonahorizontal plane in contact with oneanother. Allthree contacts areequally rough, with coefficient offriction p.The spheres arepressed together byforces of magnitudes P,Q(P>Q)acting inward along thelineofcenters. Show thnt there willbeequilibrium if,andonly if, P-Q<,n(P+Q),P-Q<A1W-(P-Q)]. IfPandQareincreased, their ratioremaining fixed,how willequilibrium bebroken? 21.Findthestresses intheframe shown inFig.56ifthejointFisfixed, instead ofA. 22.Ahanging cable consists oftwoportions forwhich theweights perunit length arew\andWz.Show thatthere isadiscontinuity ofcurvature where thetwoportions areconnected, theradii ofcurvature(pi,p2)onthetwosides ofthejoinsatisfying theequation p\w\=pzWz. 23.AbeamAB, oflengthIandweight W,rests inahorizontal position withAclamped andaloadWissuspended from B. Iftheweight perunit length ofthebeam varies asthesquareofthedistance from B,show thatat distance xfromAtheshearing forceSandthebending momentMaregiven by WM-W'(l-x)+~(I-*)*. 24.Prove thatatapointinside auniform solidsphere theforce ofattrac- tionvaries directly asthedistance from thecenter. 25.Find thepotential ofacircular disk atapoint onitsaxis. Usethe result tocalculate thepotentialofasolid sphere atanexternal point. CHAPTER IV PLANE KINEMATICS 4.1.KINEMATICS OFAPARTICLE Having completed ourstudy ofplane statics, wenowprepare forthestudyofdynamics bydeveloping some results inkine- matics; thissubject deals with themotions ofparticles andrigid bodies without anyconsideration oftheforces required toproduce these motions. Inthepresent chapter wediscuss kinematics inaplane. Tangential andnormal components ofvelocity andacceleration. Consider aparticle Pmoving inaplane, inwhich Oxyare fixed axes. The position vector oftheparticle (cf.Sec. 1.3) isr=OP,andthevelocityisq=dr/dt. Ifthepath ofthe particleisthecurve(7,then drisan infinitesimal displacement along C,so that dr=ids, where dsisanclement oflength on Cand iistheunitvector tangent to C.Thus O *or,inwords, thevelocity ofamoving FIG. 58a.--Resoiution alongpartidehasadirection tangenttothe tangent andnormal.r pathandamagnitude ds/dt. Let jbetheunitnormal vector toC(Fig. 58a),and let <be theinclination ofitotheoxixis. Aswemove along C,iandj arefunctions of0.Figure 586shows thevectors iand i+Ai (corresponding to <and+A<,respectively) transferred toa common origin. Since these areboth unit vectors, thetriangle formed bythethree vectors i,i+Ai,Aiisisosceles. Themagni- 118 SBC. 4.1] PLANE KINEMATICS 119 tude ofAiis2sin|A0,andsothelimit ofthemagnitude of Ai/A0isunity. Thus di/d<f> isaunit vector, pointing inthe limiting direction defined byAiasA<tends tozero. This direc- tion isclearly that ofj,andsodi/d<f>=j.When asimilar argumentisused toevaluated]/d<f>, wereadily seethat dj/d<t>is perpendiculartoj,butsince thelimiting direction ofAjisthat FIG. 586. Change inunittangent vector. ofi,wegetdj/d(j>=i.Combining these results wehave /Air\n\ dl . rfj (4.102) ^= j,-I=-L Tofindtheaccelerationf,wedifferentiate (4.101); thisgives (4103) |=*S.!*!+.*.* (1.W6) I dfI^-r-^d Hence, by(4.102) andthefactthat theradius ofcurvature of Cisp=ds/d<t>, wehave (4.104)f=iS+i^; or,inwords,theacceleration ofamoving particle hasacomponent dq/dt along thetangent andacomponent q*/palong thenormal to thepath. Thetangential component may alsobeexpressed in theform qdq/ds. Itiseasily seen that thenormal componentofacceleration always points totheconcave sideofthepath. Asanexample, consider aparticle traveling inacircle ofradius rwitha speed qwhich is(a)constant, (6)proportionaltot.Incase (a),theaccelera- tionvector isdirected inward along theradius andhasamagnitude ga/r; incase(6),theacceleration vector hasaconstant component along the tangent andacomponent along theradius which varies as '. 120 PLANE MECHANICS [SEC. 4.1 Radial andtransverse components. Consider aparticle Pmoving inaplane,itsposition being described bypolar coordinatesr,6(Fig. 59). Let ibetheunit vector alongOPand jtheunitvector perpendicular toi,drawn inthesense shown. Wenote that x FIG. 59. Resolution along and perpendicular toradius vector. (4.107) Thus(4-105)de Wehavethen r=ri,andthevelocity is (4.106) q=f=ri+rtj, thedotindicating d/dt. Thus(f,r6) arethecomponents ofvelocity along andperpendiculartotheradius vector. For the acceleration wefind, on differentiating (4.106) and using (4.105), arethecomponents ofacceleration along andperpendicular tothe radius vector. Itisusual tocallthecomponentsinthedirections iandj theradial andtransverse components, respectively. Thehodograph. Itiseasy toform anintuitive pictureofthevelocity ofa particle; wehave merely tovisualize thesmall displacement itreceives inasmall timeandthenimagine thatsmall displace- ment greatly magnified without change ofdirection. But itis much more difficult toformanintuitive picture oftheaccelera- tion. Thehodographisadevice tofacilitate this. Figure 60a shows thepathCofaparticle, with itsvelocity qandacceleration fattheposition P.Imagine nowafictitious particle P'moving intheplane (Fig. 606)with amotion correlated tothemotion SEC. 4.2] PLANE KINEMATICS 121 ofPbythefollowing rule: theposition vector ofP',relative to some chosen origin 0',isequal tothevelocity ofP.Thepath described byP'iscalled thehodograph ofthemotion ofP. 0' (a) (6) FIG. 00.(a)Motion, (b)Hodograph. Denoting byr',q'theposition vector andvelocity ofP',we have (4.108) r'=q. Hence, ondifferentiation, (4.109) q'=f; inwords,thevelocity inthehodographinequaltotheacceleration intheactual motion. Exercise. Verify thefollowing statements: (i)Ifaparticle hasanacceleration which isconstant inmagnitude and direction, thehodographisastraight linedescribed withconstant speed. (ii)Ifaparticle moves inacircle withconstant speed, thehodograph isa circle described with constant speed. 4.2.MOTION OFARIGID BODY PARALLEL TOAFIXED PLANE Description ofthemotion. Asalready remarked inSec. 2.4,themotion ofarigidbody parallel toafixed planeiscompletely described bythemotion oftherepresentative lamina, i.e.,thesection ofthebodyby theplane. Wemay therefore confine ourattention tothe representative lamina. Wealsodiscussed inSec.2.4thegeneral infinitesimal displacementofalamina initsplane. This dis- placement wasdescribed byselecting abasepointAinthelamina 122 PLANE MECHANICS [Sac. 4.2 andgiving (i)theinfinitesijnal displacement ofAand(ii)the infinitesimal angle through which thelamina isturned. Acontinuous motion ofalamina maybeconsidered asa sequence ofinfinitesimal displacements received ininfinitesimal intervals oftime.Weselect some particle Aofthelamina as base point. Atanytimet,Ahasavelocity, say q^.Attime t theangle between alinefixed inthelamina andalinefixed inthe planeisincreasing atsome ratewhich weshall denote by: wiscalled theangular velocity ofthe lamina.* Inasmall time interval dttheparticle Areceives asmall displacement q^dt,andinthesame interval thelamina isturned through asmall angle wdt.Hence the specificationofqAand coasfunctions of tdescribes the succession of infinitesimal displacements which thebody undergoes. Tosumup:xThemotion ofalamina inaplane is Fio.61. Themotion ofalamina 7 ., ,7 ,.^ ,..* described by<uand *>.described by(i)selecting abasepoint Ainthelamina,(ii)specifying the velocity q^ofAasafunction ofthetimet,and(iii)specifying the angular velocitycoofthelamina asafunction oft. Thisdescription, fortheinstantt,isshown diagrammaticallyin Fig. 61,thecurved arrow being used toindicate angular velocity. Theinstantaneous velocity ofanypointPofthelamina can befound from(2.409), which gives theinfinitesimal displacement ofapoint. Let(a,6)bethecoordinates ofAand(x,y)those ofP,bothmeasured onfixed axesOxy. Then theinfinitesimal displacement ofPinthetime interval dthascomponents dx=u*di~(y~6)wdt>-a)co dt, where UA,VAarethecomponents ofq^.Thus thevelocity ofP hascomponents u-tu- fc,- (4.202) (v=VA+(x a)co. *Since theangle between two lines fixed inarigidbody isconstant, itis easily seenthatthevalue ofwisthesamenomatter what lines arechosen in thelamina andintheplane. SBC. 4.2) PLANE KINEMATICS 123 Instantaneous center. Atanyinstant, there isjustonepoint ofamoving lamina which hasnovelocity. Itscoordinates arefound from (4.202), on putting u=v=0;they are (4.203) x=a y=b+UA/W- This pointiscalled theinstantaneous center. Under oneexcep- tional condition noinstantaneous centerexists, namely, when co=0.Wemay thensaythat theinstantaneous center isat infinity. C FIG. 62.Determination ofthe instantaneous center from the velocities oftwopoints.FIG. 63.Thebody centrode B rollsonthespace centrode S. Once theinstantaneous centerCisknown,itisveryeasy to visualize whathappens tothelamina inasmall interval oftime dt:thelamina rotates aboutCthrough asmall anglecodt.This factenables ustofindCwhen thedirections ofthevelocities of twopointsAandBofthelamina areknown. For, since the lamina isturning aboutCattheinstant, thevelocity ofany pointPisperpendicular toCP. Hence, Cislocated atthe intersection ofthelinesdrawn through AandBperpendicular tothevelocities ofthose points (Fig. 62). From thedefinition oftherollingofonecurve onanother, given inSec. 2.4,itisnow clear thatamoving curve rollsona fixed curvewhen thecurves touch and theinstantaneous center of themoving curve isatthepoint ofcontact. Infact, thisstatement maybetaken asadefinition ofrolling, instead ofthatgiven in Sec. 2.4. Ifboth curves areinmotion, wedefine rolling bythe conditions that thecurves touch andthat theinstantaneous 124 PLANE MECHANICS [SBC. 4.2 velocities ofthetwo particles atthepoint ofcontact (oneon each curve) areequal tooneanother. Asalamina moves, theinstantaneous centerCmoves inthe fixed plane; thecurve described byitiscalled thespace centrode (S).ButCalsomoves inthelamina; thecurve described byC inthelamina iscalled thebody centrode (B).Atanyinstant, 8 andBhave thepointCincommon, andBisturning about(7, sinceBiscarried along with thelamina. Thesituation atime t isshown inFig. 63.Alittle later, attime t+dt,apointDofB hasmoved intocoincidence withapointEofStoform thenew instantaneous center, and this isdonebyturning BaboutC through thesmall anglewdt.Hence,itisevident thatBcannot cutSatafinite angle; therefore Btouches Sat(7,andsinceCis theinstantaneous center ofB,wehave thefollowing result: The body centrode rollsonthespacecentrode. Exercise. Verify thefollowing statements: (i)When awheel rollsonatrack, thespace centrode isthetrack itself andthebody centrode thecircumference ofthewheel. (ii)When arodoflength 2aslides with itsextremities ontwolineswhich intersect atright angles, thebody centrode isacircle ofradius aandthe space centrode acircle ofradius 2a. Example. Asanexample ofouranalysis ofthemotion ofarigidbody, letusconsider twowheels WiandWzofradii aiandaz,respectively, lying inaplane. Their centers areconnected byarodRoflength ai+ 2,and thewheels engage without slipping.Ifwefixthecenter ofWi,thenW\and Rcanturnindependently about thiscenter, andWzwillrollonW\.Each of thethree bodies W\,Wz,Rhasanangular velocity, say coi, 2,ft.These three angular velocities arenotindependent; letusfindtherelation con- necting them. The particles ofW\andWzattheir point ofcontact have thesame veloc- ity.Wecanfindtwodifferent expressions forthiscommon velocity; equat- ingthem,weobtain therequired relation. SinceWiturns about itscenter withangular velocity coi,thevelocity ofitsparticle atthepoint ofcontact is tangential andofmagnitude anu\. ThewheelWzhasamotion whichmay bedescribed bymeans ofabase point taken atitscenter. Thevelocity ofthisbase pointisperpendicular toRand ofmagnitude (cti-faa)S2. Hence thevelocity oftheparticle ofW*atthepoint ofcontact withWiis tangential andofmagnitude (ai+az)fta2co2. Therefore wehave, astherequired relation insymmetric form, Ex.IV] PLANE KINEMATICS 125 Asanalternative method offinding w2when wiand ftaregiven, thefollow- inggeneral method offinding angular velocity maybeused. Take two particlesofthebody, sayAandB.Resolve their velocities perpendicular toAB.Thedifference ofthesecomponent velocities, divided byAB, isthe required angular velocity. Theproof ofthis isleftasanexercise. 4.3.SUMMARY OFPLANE KINEMATICS I.Kinematics ofaparticle. (a)Componentsofvelocity andacceleration: (6)Position inhodographisvelocityinmotion; velocityin hodographisacceleration inmotion. II.Kinematics ofarigid body. (a)Theangular velocitycoofalamina istherate ofchange oftheangle between alinefixed inthelamina andalinefixed intheplaneofreference. (6)ForbasepointA(a,6)thevelocity at(x,y)hascomponents (4.301) u=UA-(y-6),v=VA+(x-a)co. (c)Thespacecentrode (S)isthelocus oftheinstantaneous center intheplaneofreference. Thebody centrode (B)isthe locus oftheinstantaneous center inthebody. BrollsonS. EXERCISES IV 1.Aparticle moves inaplane with constant speed. Prove that its acceleration isperpendiculartoitsvelocity. 2.Aparticle moves inanelliptical pathwithconstant speed. Atwhat pointsisthemagnitudeoftheacceleration (i)amaximum, (ii)aminimum? 3.Aparticle moves along acurve y=asinpx,where aandparecon- stants. Thecomponentofvelocity inthe^-direction isaconstant (w). Find theacceleration, anddescribe thehodograph. 4.AB,BCaretworods,each2ft.long,hinged atB.AandCaremade toslide inastraight grooveinopposite directions, eachwithaspeed of8ft. 126 PLANE MECHANICS [Ex.IV persec. Find thevelocity andacceleration ofBattheinstant when the rodsareperpendicular tooneanother. 6.Starting from x*rcos0, yrsin0, calculate $and #.Hence, byresolving theacceleration vector along and perpendiculartotheradius vector, establish theformula (4.107). 6.Awheel ofradius arollswithout slipping along astraight road. If thecenter ofthewheel hasauniform velocity v,findatany instaiit the velocity andacceleration ofthetwopoints oftherimwhich areataheight h above theroad. Examine inparticular thecases h=0,h=2o. 7.Isitpossibleforaparticle tomove inacircleandhave ahodograph which isastraight line? Give reasons foryour answer. 8.Awheel ofradius arollsalong astraight track, thecenter having a constant acceleration/.Show thatthereis,atanyinstant, justonepoint ofthewheel withnoacceleration;find itsposition relative tothecenter of thewheel. 9.Arectangular plateABCDmoves initsplane withconstant angular velocity.Atagiven instant thepointAhasavelocity ofmagnitude V along thediagonal AC. Find thevelocity ofBatthisinstant interms of V,u>,andthedimensions oftherectangle. 10.Auniform circular hoop ofradius arollsontheouter rimofafixed wheel ofradius6,thehoopandthewheel being coplanar.Iftheangular velocity uofthehoop isconstant,find (i)thevelocity andacceleration ofthecenter ofthehoop; (ii)theacceleration ofthat point ofthehoop which isatthegreatest distance from thecenter ofthewheel. 11.Amotorboat experiences aresistance proportional tothesquare ofthe speed. Theengine isswitched offwhen thespeedis50ft.persec.When theboathasmoved through adistance of60ft.,itsspeed hasbeenreduced to20ft.persec. Find(tothenearest foot) thetotal distance traversed when thespeed hasbeenreduced to10ft.persec. 12.ApointAhasauniform circular motion about afixed point with angular velocity m.ApointBhasauniform circular motion aboutAwith angular velocity n.What relation connects mandniftheacceleration of Bisalways directed toward 01 13.Acircular ring ofradius 6turns initsplane about itscenter with constant angular velocityft.Asecond circular ringofradius a(<&) rolls inthesame plane ontheinner sideofthe first ring. Theangular velocity ofthecenter ofthesmaller ringabout thecenter ofthelarger ring is,a constant with thesame signas0.Find thespace andbody centrodes for thesmaller ring. 14.Aparticle Pmoving inaplane hasanacceleration directed toward a fixed point intheplane andvarying asI/OP2 .Show thatthecurvature ofthehodograph isconstant andhence thatthehodograph isacircle. CHAPTER V METHODS OFPLANE DYNAMICS 6.1.MOTION OFAPARTICLE Equations ofmotion. Inaccordance with (1.402) aparticle, under theinfluence ofaforce P,moves soastosatisfy theequation (5.101) mi=P, wheremisthemass oftheparticle and fitsacceleration rela- tivetoaNewtonian frame ofreference. IfOxyz arerectangu- laraxes inthisframe, then (5.101) gives, onresolution into components, (5.102) mx=X,my=Y, mz=Z, where X,Y,Zarethecomponents ofPalong theaxes. Theabove statements hold forapar- ticlemoving inspace; letusnowconfine ourattention toaparticle moving ina plane, theforcePbeing supposed toact intheplane ofmotion. Wemay resolve Pintocomponents along thetangent andnormal tothepath oftheparticle (Fig. 64). Ifthese com- ponents areP,Pn,respectively,the vector equationofmotion (5.101) gives, onresolution along thetangent and normal Icf.(4.104)], (5.103) "S?-P*?-'-*- Aninteresting deduction maybenoted. Iftheforce acting onaparticleisalways perpendiculartoitsvelocity (sothat Pt=0),then thespeed oftheparticleisconstant.If,further, theforce isofconstant magnitude, thenPnisconstant; then 127O x Fio. 64.Resolution of force along thotangent andnormal tothepath. 128 PLANE MECHANICS [SBC. 5.1 pisconstant, sothattheparticle describes acircle. This occurs when anelectrically charged particle moves inauniform mag- netic fieldwith lines offorce perpendic- ulartotheplane ofmotion. Letusnow consider aparticle mov- inginaplane under theinfluence ofa force always directed away from, or toward, theorigin (Fig. 65). Such a force iscalled acentral force. LetR O x bethecomponent offorce inthedirec- Fio.05.-A central force.away from^^^^^^R positive when theforce isrepulsive andnegative when itis attractive. Then, by(5.101), onresolution along andperpen- dicular totheradius vector[cf.(4.107)], (5.104) m(r-r02 )=R,^(r2d)=0. Equations (5.102), (5.103), and (5.104) are allvery useful forms oftheequations ofmotion ofaparticle. Exercise. Referring to(3.122), writedown theequations ofmotion of aparticleintheearth's gravitational field, using (i)polar coordinates and (ii)rectangular Cartesians. Principle ofangular momentum. Themomentum ofaparticle ofmass w,moving with velocity q,isdefined asthevector mq. (Thisissometimes called linear momentum, todistinguish itfrom angular momentum, defined below.) Thecomponents ofmomentum formotion inaplane are mx, my, where Oxyarerectangular axes intheplane. Sincemomentum isavector,ithasamoment about anypointAintheplane; this moment iscalled moment ofmomentum orangular momentum about A.InChap. II,wediscussed moments ofvectors; we sawthatthemoment ofavector isthesum ofthemoments of itscomponents, andsoby(2.303) theangular momentum ofa moving particle about theoriginis (5.105) h=m(xy yx). If,instead ofresolving themomentum vector along theaxes,we SEC. 5.1] METHODS OFPLAXE DYNAMICS 129 resolve italong andperpendicular totheradius vector drawn from theorigin, weobtain components [cf.(4.10(5)] mr, mrd. Theformer component hasnomoment about theorigin; hence (5.106) h Consider nowaparticle moving inaplane under theaction ofaforce withcomponents X,Y.Therate ofchange ofangular momentum about theoriginis h=m(xy yx)=xYyX=JV, whereNisthemoment oftheforce about theorigin. Hence we have theprinciple ofangular momentum: Foraparticle moving in aplanetherateofchange ofangular momentum about anyfixed point intheplaneisequaltothemoment oftheforce about that point. Wenote that, inthecase ofacentral force, thesecond equation of(5.104)isequivalent tothestatement thattheangular momen- tumabout theoriginisconstant. Linear momentum, being theproduct ofmassandvelocity, hasthedimen- sions[MLT~1 ].Inthecgs.system,itismeasured ingin.cm.sec."1 ;inthe fp.s.systeminIb.ft.sec."1Onaccount oftheequivalenceofdimensions (seeAppendix),these unitsmay alsobecalled dyne sec.andpoundal sec., respectively. Angular momentum hasthedimensions [MLZT~1 ]and is measured ingm.cm.2sec.~lorIb.ft.2sec."1 Principleofenergy. Thekinetic energy* ofaparticleofmassmmoving with velocity qisdefined tobe%mq2 .Itwillbedenoted byT.Thus,fora particle moving inspace, (5.107) T=frnq2=?m(x2+y*+32 ). Therate ofchangeofkinetic energyis T=m(xx+yy+zz)=Xx -f-Yy+Zz, where X,Y,Zarethecomponentsoftheforce acting onthe particle. Theincrease inkinetic energy intheinterval(t ,ti) *Dimensions (ML*T~*], asforwork orpotential energy andmeasured in thesame units (cf.p.54). 130 PLANE MECHANICS [SBC. 5.1 istherefore rx-To-(Xx+Yy+Zz)dt. LetWdenote thework donebytheforce during thistime interval. By(2.403) theworkdone inaninfinitesimal displace- ment is dWXdx+Ydy+Zdz=(Xx+Yy+Zz) dt, andso W= J["(Xx+Yy+Zz) dt. Hence (5.108) Ti-To=W. This establishes theprinciple ofenergy: Theincrease inkinetic energyisequaltotheworkdonebytheforce. Differentiating (5.108) withrespect tot\andthendropping the subscript 1,wehave (5.109) T=W; inwords,therateofincrease ofkinetic energy equals therateof working oftheforce. Iftheparticle moves inaconservative field offorce with potential energy V,then, asin(2.419), v dV vdV dVZs3-arF~~Vz=-Tz' Then W=f1 (Xx+Yy+Zz)dt=-Vdt=-Ft+7, where V\yVQarethepotential energiesattimest\,t,respectively. Comparison with (5.108) gives T,-To--.7i> 7o, 21 !+7i-To+V . Hence, ingeneral, (5.110) T+V-E, whereEisaconstant, called the total energy. Thus thesum of thekinetic andpotential energiesisconstant. This iscalled the principle oftheconservation ofenergy. The principle expressed mathematically by(5.110) isoneof SEC. 5.2] METHODS OFPLANE DYNAMICS 131 thefundamental formulas ofmechanics, and itisofgreat usein thesolution ofproblems. Itrepresents onerelation among the three coordinates andthethree components ofvelocity, Vbeing supposedly known asafunction ofthecoordinates. Inthe absence ofaconservative field,wenolonger have (5.110), only (5.108). This ismuch lessuseful becauseWisnotafunction of thecoordinates. Itisanintegral thevalue ofwhich depends onthepath oftheparticle andthus isunknown, since thepath of theparticleisprecisely whatwehave tofind inthemajority of problems onthedynamics ofaparticle. Exercise. Aparticle slidesdown asmooth inclined plane. Use(5.110) tofind itsspeed interms ofthedistance traveled from rest. 6.2.MOTION OFASYSTEM Anyone familiar with theusual type ofproblems posed as exercises inmechanics must havebeen struck bytheir artificial character. Themechanical systems considered areoften too simple tobeofmuch practical interest. Attention isconcen- trated onsuch simple systems asrigid bodies swinging about fixed axes orwheels rolling along lines, instead ofoncomplicated realities like trains, automobiles, orairplanes. This isbecause ithasbecome traditional inthestudy ofmechanics todirect attention toproblems which aresoluble, inthesense that the behavior ofthesystem canbedescribed bysimple formulas. This isanunfortunate practice, because itfailstoemphasize one ofthegreatest achievements oftheapplied mathematician, namely, hiscapacity tomake general statements about com- plicated systems without paying much attention tothedetails ofthesystems. Toextract thefullest interest from thepresent section, the reader should bear inmind thestriking generality ofthestate- ments. Since, however,itistiring andconfusing tothink too much interms ofgeneralities, heshould bear inmind afew concrete examples andthink ofthem inconnection with the various principles about tobediscussed. Thefollowing systems aresuggested assuitable examples: (i)astick sliding onafrozen pond; (ii)acomplete automobile; (iii)awheel ofanautomobile; 132 PLANE MECHANICS [SEC. 5.2 (iv)anairplane; (v)amanonatrapeze; (vi)thesolar system. Asattention isatpresent directed toward plane dynamics,it isadvisable tothink primarily oftwo-dimensional motions of theabove systems; e.g.,theautomobile andtheairplane are traveling straight ahead. Forthe first fivesystems, wemay accept theearth's surface asaNewtonian frame. Asforthe solar system, wemaymerely assume that there issomeNew- tonian frame andtrytoidentifyitbyexamining theconsequences ofthelaws ofmotion. Thesystem under consideration isregarded ascomposed of particles. Theforces acting ontheparticles areinpart internal and inpart external, theinternal forces satisfying thelaw of action andreaction (Sec. 1.4).Arigidbodyisaparticular type ofsystem, inwhich theinternal forces aresuch astoprevent the alteration ofthedistances between theparticles. The internal forces inasystem are,asarule, complicated; thepurpose of thegeneral principles which weareabout toestablish istomake important statements about themotion ofthesystem which involve, notthese complicated forces, butonly theexternal forces which asarulearecomparatively simple. Forexample, inthecase oftheautomobile, theonly external forces are(i) gravity, (ii)reactions atthecontacts ofthetireswiththeground, and(iii)resistance ofthe air. Principle oflinearmomentum; motion ofthemass center. The linearmomentum ofasystemisdefined asthesum ofthe linear momenta oftheseveral particles ofthesystem. Thus, ifthemasses oftheparticles aremi,w2,m,and* their velocities qi,q2, q,thelinearmomentum ofthesystemis thevector n (5.201) M=2)m'<l" ti We shallnowprove theprinciple oflinear momentum: The rateofchange oflinearmomentum ofasystem isequal tothevector sumoftheexternal forces. From (5.201), wehave (5.202) M=mA, SEC. 5.2] METHODS OFPLANE DYNAMICS 133 where ftistheacceleration oftheithparticle. Thus, (5.203) &=(P,+PO,=1 wherePistheexternal force ontheithparticle andPjthe internal forceonit.But,from theequality ofaction andreac- tion,weseethat (5.204) p;=o, 1=1 because thissummation consists ofvectors which areformed from pairs offorces thatareequalandopposite. Hence, (5.203) gives (5.205) M=J)P,, t=1 which proves theprinciple oflinearmomentum. The lastequation maybewritten (5.206) M=F, whereFisthevector sum oftheexternal forces. Weshallnowprove thelawofmotion ofthemass center: The mass center ofasystem moves likeaparticle, having amass equal tothetotalmass ofthesystem, actedonbyaforce equaltothevector sum oftheexternal forces acting onthesystem. From (3.101)itfollows that thevelocity ofthemass center ofasystem ofparticlesis n (5.207) q-XMUM t=i wheremisthetotalmass ofthesystem. Thus iffistheaccelera- tion ofthemass center, wehave (5.208) ml=w4 and so,by(5.206), (5.209)mi=F. This istheequationofmotion ofaparticleofmassmacted onby aforce F,andsothelaw isestablished. 134 PLANE MECHANICS [Sac. 5.2 Wenote that,by(5.207), n (5.210) mq-]mtqt-, sothat thelinearmomentum ofthe fictitious particle moving with themass center isequal tothelinearmomentum ofthe system. The conclusions tobedrawn from thepreceding principles aresimple andinteresting when thevector sum oftheexternal forces iszero. Then thelinearmomentum ofthesystem remains constant, and itsmass center travels inastraight linewith constant speed. This istrueinparticularforastick sliding ona frozen pond. Asforthesolar system, weseethatanyNew- tonian frame ofreference must besuch thatthemass center of thesolarsystem hasaconstant velocity relative toit. Principle ofangular momentum; motion relative tomass center. Theangular momentum ofasystem about aline (orabout a point initsplaneifthesystemisconfined toaplane)isdefined asthesum oftheangular momenta oftheparticles composingit. Thus iftheparticleofmass m*hascoordinates #t,y^z%and velocity components &,#, z,,theangular momentum ofthe system about Ozis (5.211) h ]~i Therate ofchange ofangular momentum about Ozisthen n (5.212) =! But m&i X.+X'i, mjjv=Yt+Y'it where Xi,Yiarethecomponentsofexternal force acting onthe particle andXf i}FJthecomponentsofinternal force. Thus, (5.213) h-2&Y<~y^+2 -ffi ffi N+N', where -ATisthetotalmoment about Ozofalltheexternal forces acting onthesystem andN'thetotalmoment ofalltheinternal SBC. 5.21 METHODS OFPLANE DYNAMICS 135 forces. But since theinternal forces occur inbalanced pairs, their totalmoment iszero; hence (5.214) h-N. This equation expresses theprinciple ofangular momentum: The rateofchange oftheangular momentum ofasystem about afixedline isequaltothetotalmoment oftheexternal forces about that line. Ifwethink, forexample, ofamanonatrapeze, theonly external forces are(i)gravity and(ii)areaction atthepoint of suspension. Butthelatter hasnomoment about thepoint of suspension. Hence therate ofchange ofangular momentum about thepoint ofsuspensionisequal tothemoment ofthe gravitational forces about that point. Letx,ytzbethecoordinates ofthemass center ofasystem and let#(,y{,z(bethecoordinates oftheithparticle relative to themasscenter, sothat (5.215) x,=*+*J, y<=y+y(. Thecomponents ofvelocity oftheparticle relative tothemass center arexf ity(,z(,andsotheangular momentum relative toa linethrough themass center parallel toOzis n (5.216) h=%mv(x(y(-y(x(\ where weuse,incomputing angular momentum, thevelocities relative tothemass center. Weshall refer tothisbriefly asthe angular momentum relative tothemass center. From (5.216), weobtain (5.217) =i andhence by(5.215),differentiated twice, (5.218) h= 136 PLANE MECHANICS [SEC. 52 asbefore,Xt,Ftarecomponents ofexternal forceandX{,Y( components ofinternal force. But,from thedefining property ofthemass center, wehave m%x(- sothat the firsttwoterms ontheright-hand side ofourlast equationvanish. Thefourth term vanishes through thebalanc- ingoftheinternal forces inpairs, andsoWehave (5.219) h=N, whereNisthetotalmoment oftheexternal forces about the mass center. This istheprinciple ofangular momentum relative tothemass center: Therateofchange ofangular momentum relative tothemass center isequaltothemoment oftheexternal forces about themass center. Exercise. Check toseethat thetwo sides of(5.219) have thesame dimensions. Itwillbenoticed thatwehave aprinciple ofangular momen- tum relative toafixed axisandaprinciple ofangular momentum relative tothemass center. Theprinciple doesnothold foran arbitrarily moving axiswith fi,xcd direction. Asanillustration oftheprinciple ofangular momentum relative tothe mass center, consider thefront wheel ofanautomobile. Theexternal forces onitare(i)gravity, (ii)thereaction oftheaxle,and(iii)thereaction of theground. The force ofgravity andthereaction oftheaxlehaveno moment about thecentral lineoftheaxle,which passes through themass center. Hence therate ofchange ofangular momentum relative tothe center ofthewheel equals themoment ofthereaction oftheground about thecenter. Inparticular,ifthecar istraveling atconstant speed, the angular momentum isconstant, andsothereaction oftheground must actvertically upthrough thecenter ofthewheel; noforce offriction iscalled intoplay. Further,ifthewheelbumpsofftheground, itsangular momen- tum willremain constant aslong asitisinthe air.These statements are made ontheassumption that thebearings aresmooth; thereader can supply thequalitative description ofthemodifications which arisewhen there isfriction inthebearings. Theprinciple ofenergy. The kinetic energy ofasystemisdefined asthesum ofthe kinetic energies ofitsconstituent particles; theformal expression SEC. 5.2] METHODS OFPLANE DYNAMICS 137 is (5.220) T=1|)mt(xt2++*J). Then, (5.221) T=V^(iA+frfr+A) 1=1 where Xi,F-,Ziarethecomponents ofthetotal force, external andinternal, acting ontheithparticle. Thus,ifWisthework donebytheforces from time tototime t}wehave (5.222) T=W. This isformally thesame as(5.109), butherewearecon- sidering asystem instead ofasingle particle. Forasystem, theprinciple ofenergy takes thefollowing form: The rateof change ofkinetic energy ofasystemisequaltotherateofworking ofalltheforceSjexternal andinternal. There isasharp difference between theprinciples oflinear andangular momentum ontheonehandandtheprinciple of energy ontheother. Intheprinciples ofmomentum theinternal forces areeliminated; intheprinciple ofenergy they arenot eliminated, exceptinthespecial casewhere theydonoworkand socontribute nothing toW.Inouridealized mathematical models, consisting ofrigid bodies withsmooth contacts, nowork isdonebytheinternal forces, andsothey disappear from the principleofenergy. Incases ofcollision, however, workmay bedonebytheinternal forces (seeChap. VIII) ;that isbecause, insuch cases,itisimpossible toregard thebodies asabsolutely rigid. When thesystemisconservative, with potential energy V, wehaveW=-Vby(2.416); then (5.222) leads totheprinciple oftheconservation ofenergy (5.223) T+V=E, whereEistheconstant total energy. 138 PLANE MECHANICS [Ssc. 5.2 D'Alembert's principle. The principle about tobediscussed adds nothing essential totheprinciples already given, but itisinteresting asanalter- native expression. Wehaveregarded "force"asaprimitive concept inmechanics, andweshallnotabandon thatpointofview. Onemust guard against logical confusion inaccepting thefollowing definitions, inwhich weusetheconventional terms. Consider aparticle of massmthaving atacertain instant anacceleration f.The vector mfiscalled the"effective force"acting ontheparticle, and thatvector reversed, i.e., mf,the"reversed effective force." Now consider asystem ($)ofnparticles inmotion, thereversed effective force ontheithparticle being mdi. Alongside the mental picture ofthissystem, think ofanother ($') inwhich theparticles areatrest atthesame positions astheyhave instantaneously inSandareacted onbythesame forces, external and internal, asinS;inaddition letthere actinthestatical system S'asetofrealforces identical withthereversed effective forces ofS.Now,bytheequations ofmotion oftheparticles inSjwehave Pi-m&=0, (i=1,2,-- .n), wherePistherealforceontheithparticle inS',hence itfollows that S'isinstatical equilibrium since thetotal force oneach particleiszero. Thuswehave D'Alembert's principle: The reversedeffective forces and therealforces together give statical equilibrium. Todealwith problems inplane dynamics, weintroduce a fundamental plane towhich themotion isparallel. Since the internal forces areplane-equipollent tozero,itfollows that theexternalforces, together with thereversed effective forces, form a system plane-equipollenttozero.Werecall that thismeans that thevector sumandthemoment vanish. Thestatement ofD'Alembert's principle may givetheimpres- sionthat itreduces dynamics tostatics. This ispartly true in thesense that thestatement involves only theconditions of statical equilibrium. However,itmustberemembered thatthe reversed effective forces involve derivatives ofcoordinates and that therefore conditions ofstatical equilibrium involving these forces areactually differential equations ofmotion. Todetermine SEC. 5.3] METHODS OFPLANE DYNAMICS 139 motion under given forces, these differential equations must be solved adynamical, rather than astatical, problem. Onthe other hand,ifthemotion isknown, D'Alembert's principle enables ustousethemethods ofstatics todetermine theforces acting onthesystem. Exercise. Three equal particles arejoined bylight rods toformanequi- lateral triangle. Ifthetriangle rotates initsplane about itscentroid with constant angular velocity,findthetensions intherods. 5.3,MOVING FRAMES OFREFERENCE Indeveloping dynamics uptothis point, wehaveassumed theexistence ofaframe ofreference relative towhich bodies move inaccordance with theNewtonian laws. Togetaccurate agreement between theoretical prediction andobservation, we take forframe ofreference oneinwhich themass center ofthe solarsystemisfixedandwhich hasnorotation relative tothestars asawhole. Foraslightly lessaccurate agreement inthecase ofexperiments ontheearth, wemay take theearth itself as frame ofreference. Wenow raise thequestion: Knowing that abody behaves relative toaNewtonian frame ofreference inaccordance with thelawsandprinciples discussed earlier, howdoesabody appear tobehave when viewed fromaframe ofreference moving relative tothe Newtonian frame? Oy' s s' xfFrames ofreference with uniform translational velocity. LetSbeaNewtonian frame of reference andS'aframe ofreference which has, relative toS,auniform (i.e.,unaccelerated) translational mo- tion. (IfSistheearth's surface,' might beatrain running smoothly on straight tracks atconstant speed.) Weshall consider onlytwo dimensions, buttheargument canbeextended tospace immediately. InSwetake axesOxyandinS'wetake parallel axes O'x'y' (Fig. 66). Let,rjbethecoordinates of0'relative to0.ThenFIG. 66.Frames ofrefer- ence inrelative motion with- outrotation. (5.301) wo, t>o, 140 PLANE MECHANICS [Sjuc. 5.3 where u,varetheconstant components ofthevelocity ofS' relative toS.LetAbetheposition ofanymoving particle;it hascoordinates (x,y)relative toOxyandcoordinates(#', y') relative toO'x'y'. These coordinates areconnected bythe relations x=xr+,y=y'+ 97, and,ondifferentiation, (5.302) x-x'+,y=yr+ 17. Ifwedenote byqthevelocity ofArelative toS,byq'thevelocity ofArelative toS',andbyqthevelocityofS'relative toAS', (5.302) maybeexpressed intheform (5.303) q=q'+q<>. This iscalled thelawofcomposition of velocities and isexhibited graphicallyin FIG. 67.Composition Fig. 67. ofvelocities. Exercise. Aman stands onthedeck ofa steamer, traveling eastat15miles perhour. Tohimthewind appears to blowfrom thesouth withaspeed of10miles perhour. What isthetrue speed anddirection ofthewind? Sincej,ijareconstants,differentiation of(5.302) gives (5.304) x=*', y=y'; thus theacceleration relative toS'isequal totheacceleration relative toS,andthismaybeexpressed invector form asf'=f. Thus thelawofmotion (5.305) mi=P may alsobewritten (5.306) mi'=P, andsoNewton1slawoflmotion holds inS'aswellasinS. From thiswedraw animportant conclusion. Given one Newtonian frame ofreference S,wecanfindaninfinity ofother Newtonian frames ofreference, namely,allthose frames of reference which have auniform motion oftranslation relative tOAS. TnSec. 5.2wesawthat ifaNewtonian frame exists and of SEC. 5.3] METHODS OFPLANE DYNAMICS 141 course wesuppose that itdoes, since otherwise* there would be noNewtonian mechanics then themass center ofthesolar system musthave aconstant velocityrelative toit.From what hasbeenshown above,itfollows thatwemaychange toanother Newtonian frame inwhich themass center ofthesolarsystemis atrest. Thisis,infact, theastronomical frame, towhich we have referred before. Iftheearth isregarded asasatisfactory Newtonian frame ofreference, thenwemust regard asequally satisfactory the interior ofanyvehicle which moves overtheearth withconstant velocity.This isinaccordance withcommon experience: wearenotconscious ofthesmooth uniform motion ofatrain whenwearetraveling init;webecome conscious ofthemotion onlywhen thetrain lurches orbrakes orrounds acorner. Frames ofreference with translational acceleration. Letusnowsuppose that ASisaNewtonian frame andthat *S" hasrelative toitatranslational motion with constant accelera- tion. Then (cf.Fig. 66),wehave (5.307)= ,rf=ft, whereo,ftareconstants. The relations (5.302) hold inthis case also,anddifferentiation gives (5.308)-jc=x'+ao, y=y'+ft. Let f,fdenote, respectively,theaccelerations ofArelative to SandS'9and letfdenote theacceleration ofA"relative to8; then (5.308) maybewritten (5.309)f=f+fo. This iscalled thelawofcomposition ofaccelerations. Theequationofmotion (5.305) nowleads to (5.310)tnf-P-mf .' Thus theNewtonian lawofmotion does nothold relative toS'. ButwecansaythattheNewtonian lawholds provided thatwe addtothetrueforcePafictitiousforce wf . Asanillustration,consider anelevator descending withconstant accelera- tion/o.Relative totheelevator, everything takes place asiftheelevator 142 PLANE MECHANICS SBC. 5.3 were atrestandevery particle experienced anupward forcemfQ,wherem isthemass oftheparticle,inaddition tothedownward force mg.These fictitious forces alter thereactions among theparticles constituting the human body, andsoweareconscious ofanacceleration, eventhough we cannot lookoutside ourframe ofreference. Frames ofreference rotating withconstant angular velocity. Letusnowsuppose thatSisaNewtonian frame ofreference andS'aframe ofreference rotating about apoint ofSwith constant angular velocity.Leti,jbeperpendicular unit vectors, fixed inS'(Fig. 68). LetAbeamoving particle. (We maythink ofAasaflywalking onarotating sheet ofcardboard.) Taking axesOxyinS',inthedi- rections ofiandj,theposition vector ofAis r-xi+2/j. Fio. 68.Rotating frame of dt dt' reference. andsodifferentiation of(5.311) gives, forthevelocity ofA(relative toS), (5.313) q=f=(x- ort/)i+(y+<*c)j. Another differentiation gives, fortheacceleration ofA(relative toS), (5.314) f=q=(x-2o>-rfx)\+(y+2ax-co2 i/)j. Thus,ifX,Yarethecomponents oftrue force inthedirections ofi,j,respectively, wehave theequations ofmotion (5.315) m(x-2uy-rfx) X, m(y+2<*x-tfy)=Y. Thesemay alsobewritten (5.316) mx-X+X'+X", my-Y+Y'+7", where X'=2ma$ Y'=-- SEC. 5.3] METHODS OFPLANE DYNAMICS 143 Thuswemaysaythat theparticle moves relative totherotating frame ofreference inaccordance withNewton's lawofmotiont provided thatweadd tothetrueforce thetwofictitious forces (X1 ,Y') and(X", 7"). The fictitious force (X', Y')iscalled theCoriolis force.Its magnitudeisproportional totheangular velocity ofS'andto thespeed qroftheparticle relative toS';itsdirection isper- pendicular tothevelocity q'relative toS',and isobtained from thedirection ofq'byrotation through aright angle inasense opposite tothesense oftheangular velocity (Fig. 69).The fictitious force (X" yY")iscalled the centrifugal force. Its magnitudeisproportional tothesquareoftheangular velocity ^S^O)mw2r (Centrifugal force) 2mwq' (Coriolis force) FIG. 69. Centrifugal forceand Coriolis force inarotating frame ofreference. ofS'andtothedistance oftheparticle from thecenter ofrota- tion;itisdirected radially outward from thecenter ofrotation. Theframe ofreference whichweemploy inordinary life isthe earth. Itrotates relative totheastronomical frame withan angular velocity of2irradians persidereal day; since onesidereal daycontains 86,164.09 seconds (cf.page 14),theangular velocity oftheearth is7.29X10~6radians persecond. This isavery small angular velocity, andhence theCoriolis force andthe centrifugal force arising from theearth's rotation arenotnotice- able inourdaily lives. They areimportant geographically, however; thecentrifugal force isresponsible fortheequatorial bulge ontheearth, andtheCoriolis force isresponsible forthe trade winds. When frames ofreference turning rapidly relative totheearth areemployed, these fictitious forcesmayassume serious pro- portions. Thus,inanairplane turning inaerial combat or coming outofadive, centrifugal forcemaybemuch greater than theforce ofgravity. 144 PLANE MECHANICS [SEC. 5.3 Statical effects oftheearth's rotation. InSec. 3.1wegaveanintroductory discussion oftheforce ofgravity andtheweight ofabody .near theearth's surface, leaving theearth's rotation outofaccount, i.e.,treating the earth asaNewtonian frame. Wenowseethat itmayindeed be sotreated provided thattheproper fictitious forces areadded. Ifwedealonlywith statical problems, i.e.,those inwhich the systemisatrestrelative totheearth, there isnoCoriolis force, andsotheonlyfictitious force iscentrifugal. The"weight" ofaparticleistheresultant oftheforce ofgravity andthe FIG. 70.Plumb lineontherotating earth. centrifugal force, instead ofbeing merely theforce ofgravity alone. Thus theweight ofaparticleisproportional toitsmass, andthetheory ofSec.3.1isvalid provided thatweunderstand by mgtheweight asjustdefined. Weshallnowbring ourtheorystillcloser toreality bytaking amore accurate model oftheearth. Asafirstcrude approxima- tion, theearthmayberegarded asasphere ofradius R,where R=3960 miles. More accurately,itisanoblate spheroid with anequatorial radius of3963 miles andapolar radius of3950 miles. This isthemodel which weshall accept forthepresent discussion, andweshallassume thatthemodel rotates about its polar axiswith constant angular velocityft. InFig. 70,SN istheearth's axis,Aanypoint onitssurface, andABtheperpendicular dropped onSN.Thegravitational attraction oftheearth onaparticle atAactsalong some line SBC. 5.3J METHODS OFPLANE DYNAMICS 145 ACwhich intersects SN\itsmagnitudeisproportional tothe massmoftheparticle, andweshall denote itbymg'. The centrifugal force isdirected along BA,and itsmagnitudeis rapQ2 ,where p=BA. Inorder thattheparticle mayremain inequilibrium relative totheearth, athird forcemust beapplied tobalance thegravita- tional forceandthecentrifugal force. This forcemust liein theplaneABC, and itsmagnitude must beproportional tom. Wedenote itbymg]itmaybesupplied bythetension inastring orplumb line, orbythereaction ofasmooth plane. Wedefine theverticalAVatAasthedirection ofthisforceandthehori- zontal planeHAH' astheplane perpendicular toit. Wenowask :Aswerange overtheearth's surface, what isthe relation between gandg',andwhat istheinclination ofthe vertical tothedirection ofthegravitational force? Theastronomical latitude Xisdefined astheelevation ofthe astronomical poleabove thehorizontal plane, i.e.,theangle between SNandH'AH, or(equivalently) theangle between BA andAV. Letusdenote by theangle between CAandAV. Resolution offorces along andperpendicular toACgives as conditions ofequilibrium. OS(X- 0), Now theratiopW/gissmall; hence 6issmall, andcos6differs from unity byasmall quantity ofthesecond order. Tothe firstorder ofsmall quantities, wehave (5.319) gf=g+ptt2cosX,6=^sinX. a Theterms involving12aresmall, andtoourorder ofapproxima- tionweareentitled toreplace pandXbyapproximate values. Now Xisapproximately equal totheangleBAC, andso p=CAcosX, approximately. According toawell-known lawofhydrostatics, thesurface oftheocean must betangent tothehorizontal plane ; infact,AV isnormal tothesurface ofourmodel oftheearth. Thus, since 6issmall,CA isvery nearly normal tothissurface, andsoCA=Rapproximately,whereRistheradius ofthe 146 PLANE MECHANICS [SEC. 5.4 earth inthe firstcrude model. Hence (5.319) maybewritten (5.320) g'=g+RWcos2 X, (5.321)6= sinXcosX. Thus wecan find thegravitational intensity gfinterms of measurable quantities, gbeing measured bymeans ofapendulum (cf.Sec. 6.3). Theoretically, gisgiven byameasurement of thetension inaplumb line,butthis isnotapractical method. AttheNorth andSouth Poles, g=983cm.sec~2 ;attheEquator, g=978cm. sec.""2Equation (5.321) gives thedeviation ofthe plumb linefrom thedirection ofthegravitational force;it isamaximum atalatitude of45. Other effects oftheearth's rotation willbetreated inSec. 13.5. 5.4.SUMMARY OFMETHODS OFPLANE DYNAMICS I.Equationsofmotion ofaparticle. (5.401)raf=P (vector form); (5.402) mx Z, myY(Cartesian coordinates) ; (5.403) mq=Ptj^-=Pn (resolution along tangent and pnormal) ; (5.404) m(f-rtf2 )=R,r*&=const, (central force). litPrinciple ofangular momentum foraparticle* (5.405) h=N, where h=m(xy yx)=mr*6. III.Principle ofenergy foraparticle. (5.406) T-W, where T=img2=$m(xz+y2 ),W=work done; (5.407) T+V=E (conservation ofenergy). IV.Principle oflinearmomentum forasystem. (5.408) M=F, SBC. 5.4] METHODS OFPLANE DYNAMICS 147 where nM=Vm,q ; ^T (5.409) mfF(motionofmass center). V.Principle ofangular momentumfor_a system. (5.410) h**N, where n t(i2/ 2A&), N=moment ofexternal forces. (This holds with respect toafixed point andwith respect tothe mass center.) VLPrinciple ofenergy forasystem. '(5.411) T=W, where n J?7=5}Wiflf , IF=work done; ti (5.412) T+V=E(conservation ofenergy). VII.D'Alembert's principle. The reversed effective forces(mtf)andthe real forces together give statical equilibrium. VIII.Moving frames ofreference. (i)Aframe ofreference having atranslation with constant velocity relative toaNewtonian frame isalsoNewtonian. (ii)Aframe ofreference having atranslation with constant acceleration frelative toaNewtonian framemaybetreated as Newtonian ifafictitious force mfisapplied toeach particle. (iii)Aframe ofreference rotating with constant angular velocityo>relative toaNewtonian framemaybetreated asa Newtonian frame iftoeach particle there areapplied two fictitious forces: Coriolis force withcomponents (2mwy, Centrifugal force withcomponents (mrfx, 148 PLANE MECHANICS [Ex.V EXERCISES V 1.Atacertain instant, aparticle ofmass w,moving freely inavertical plane under gravity,isataheight habove theground andhasaspeed q. Usetheprinciple ofenergy tofind itsspeedwhen itstrikes theground. 2.What istheleastnumber ofrevolutions perminute ofarotating drum, 2feetininternal diameter,inorder thatastone placed inside thedrummay becarried right round? Assume that thecontact between thestone and thedrum isrough enough toprevent sliding. (The reaction ofthedrum on thestonemust bedirected inward.) 3.Askier, starting fromrest,descends aslope117yards longandinclined atanangle ofsin"1fatothehorizontal. Ifthe coefficient offriction between theskisandthesnow is$,findhisspeed atthebottom oftheslope. Iftheskierwith hisequipment weighs 200lb.,howmuch energy isdissipated inovercoming friction? 4.Abead ofmassmslides onasmooth wire intheform ofaparabola with axis vertical andvertex downward. Ifthebead starts from rest atanendofthelatusrectum (oflength 4a),findthespeed withwhich it passes through thevertex. Find alsothereaction ofthewireonthebead atthispoint. 6.Aheavy particle restsontopofasmooth fixed sphere.Ifitisslightly displaced,findtheangular distance from thetopatwhich itleaves the surface. 6.Two barges ofmasses mi,ra2atadistance dfrom each other are connected byacable ofnegligible weight. Onebarge isdrawn uptothe otherbywinding inthecable. Ifneither barge isanchored, nndthedistance through which eachbarge moves. (Neglect anyfrictional effects duetothe water.) 7.Anairplane withanairspeed of120miles perhour starts fromA togotoBwhich isnortheast ofA. Ifthere isawind blowing from the north at20miles perhour,inwhat direction must thepilot point theair- planeifhewishes togoinastraight linefromAto5? 8.Aheavy particle issuspended from afixed pointbyalight string of length a.Ifthestring would break under atension equal totwice the weight oftheparticle, findthegreatest angular velocity atwhich thestring andparticle canrotate asaconical pendulum without thestring breaking. 9.Asteamer sailing eastat24knots is1000 feettothenorth ofalaunch which isproceeding north at7knots. Find theshortest subsequent dis- tance between them ifthese courses aremaintained. Draw rough dia- grams, showing (i)thetracks relative tothewater, (ii)thetrack ofthesteamer relative tothelaunch, (iii)thetrack ofthelaunch relative tothesteamer. 10.Anautomobile travels round acurve ofradius r.Ifhistheheight of thecenter ofgravity above theground and2athewidth between thewheels, show that itwilloverturn ifthespeed exceeds-\/gra/h, assuming noside- slipping takes place. Ex.V] METHODS OFPLANE DYNAMICS 149 11.Every second ngasmolecules, each ofmass m,strike thesideofa box. Usetheprinciple oflinearmomentum tofindtheforce required to holdthesideoftheboxinplace, assuming thateachmolecule hasthesame speed qbefore andafter hitting thesideandthatthemolecules move atright angles totheside. Explain precisely whatdynamical "system"youuse. 12.Explain howamanstanding onaswing canincrease theamplitude oftheoscillations bycrouching andstanding upatsuitable times. 13.Alight string isattached toafixed point andcarries atitsfreeend aparticleofmass m.Theparticleisdescribing complete revolutions about under gravity, andthestring isjusttautwhen theparticleisvertically above 0.Find thetension inthestring when inahorizontal position. 14.Show that,ifanairplane ofmassMinhorizontal flight drops abomb ofmass m,theairplane experiences anupward acceleration mg/M. 16.Achain ofanynumber oflinkshangs suspended fromoneend. The suspension andtheconnections between thelinks aresmooth. Thechain is displacedinavertical plane andreleased from rest. Show that inthe resulting oscillations thecenter ofgravity ofthechain never rises higher than itsinitial position, andthat ifitdoes ever risetothatsame height, thewhole chain isatrestatthat instant. 16.Awheel spinsinahorizontal plane about avertical axisthrough its center; thebearings aresupposed frictionless, andthere isamass clipped on onespoke. During themotion themass slipsalong thespoke outtotherim. Does thiscause anincrease ordecrease intheangular velocity ofthewheel? 17.Assuming thataskierkeops hislegsandbody straight andneglecting friction and airresistance, show thathemust keep hisbody perpendicular totheslope ofahillinorder thathemay preserve hisbalance without support from theforward orrearends ofhisskis. Giveageneral discussion oftheproperdirection forhisbodywhen friction and airresistance* are taken intoaccount. 18.Anicefloewithmass 500,000 tons isneartheNorth Pole,moving west attherateof5miles aday. Neglecting thecurvature oftheearth,findthe magnitude anddirection oftheCoriolis force. Express themagnitude in tonswt. 19.Aparticle moves inasmooth straight horizontal tubewhich ismade torotate with constant angular velocity about avertical axiswhich intersects thetube. Prove thatthedistance oftheparticle from theaxis isgivenby r=Aeat-fBe'"', whereAandBareconstants depending ontheinitial position andvelocity oftheparticle. Ifwhen t=theparticleisatadistance rafrom theaxis,what velocity must ithave along thetube inorder that after averylong interval oftime itmaybevery close totheaxis? 20.Aloop ofstring isspinningintheform ofacircle about adiameter of theloopwith constant angular velocity. Neglecting gravity, prove that themass perunit length ofthestring must beproportional tocosec3 0, 6being measured from thediameter. 160 PLANE MECHANICS [Ex.V 21.Aparticle moves with constant relative speed qround therimofa wheel ofradius a;thewheel rollsalong afixed straight linewithuniform velocity F.Taking thewheel asframe ofreference,findtheCoriolis force andthecentrifugal force. Indicate them inadiagram. 22.Asystem ofparticles moves inaplane. Prove that, provided the mass center isnotatrest,there exists attime tastraight lineLsuchthatthe angular momentum about anypoint onLiszero. Further, show that,if noexternal forces actonthesystem, thelineLisfixed forallvalues oft. CHAPTER VI APPLICATIONS INPLANE DYNAMICS MOTION OFA PARTICLE 6.1.PROJECTILES WITHOUT RESISTANCE Thescience ofballistics isconcerned withthemotion ofprojec- tiles. Thetheory oftheexplosion ofthecharge andthemotion oftheprojectile inthebarrel ofthegunbelong tointerior ballis- tics,withwhich weshall notbeconcerned. After theprojectile leaves thebarrel ofthegun, itmoves under theinfluence of gravity andtheresistance ofthe air;thepurpose ofexterior ballistics istopredict, from given muzzle velocity andangle of elevation ofthegun,thepath ortrajectory oftheprojectile. Onaccount ofthecomplicated nature oftheresistance ofthe air,anaccurate mathematical prediction isnotpossible. The greatest difficulties arisefrom thefactthattheprojectileisof finite size.Toavoid these,weregard theprojectile asaparticle. Inthepresent section, weshallmake afurther andmuchmore drastic simplification; weshall assume thatnoresistance is offered bythe air.Wecannot claim that thetheory based on thishypothesis gives results ofmuch practical value inballistics, except inthecase ofprojectiles thrown with small velocities.* Theparabolic trajectory. LetOxyberectangular axes,Oxbeing horizontal andOy vertical, directed upward. Theequations ofmotion ofaparticle under theinfluence ofgravity are (6.101) mx=0,my=-mg. Integration gives (6.102)x=uQ,&-VQ gt, (6.103) x=XQ+u<>t, y=yQ+vrf ifltf2 , where XQ,yo,u^VQareconstants ofintegration. Itisevident *Cf .C.Cranz andK.Becker, Handbook ofBallistics (H.M.Stationery Office, London, 1921), Vol.I,p.17. 151 152 PLANE MECHANICS [Sec. 6.1 that(a?o, 2/0)istheposition and(t*o, VQ)thevelocity, both at time t=0.Theequations (6.103) givethepath, ortrajectory, of theparticle. Obviously, atacertain instant(t=vQ/g),wehavey=0, sothatatthatinstant thevelocityishorizontal. Now theorigin Oandtheinstant from which tismeasured maybechosen as weplease. Letuschoose atthepoint where thevelocity is horizontal andmeasure tfrom that instant. Thenwehave (6.104) for t=0, x=y=0, y=0, Substituting in(6.102) and(6.103), weobtain XQ=?/o=0, VQ=0, andsotheequations ofthetrajectory become (6.105) x=uQt, y=-ijf. Elimination oftgives theequation ofthetrajectory intheform (6.106) =-g, aparabola with itsvertex attheorigin (Fig. 71). ThefocusFoftheparabolaissituated atadistance a=^u\]g below thevertex, andthedirectrix Lisatthesame height above thevertex. Atanypoint onthe trajectory thespeed qisgivenby (6.107)2=x*+y*=ul+gH*=u\-2gy=2g(a-y); thus thespeedatanypointPonthe trajectory isequaltothespeed ac- pxquired infree fall toPfrom restat FIG. 71. Parabolic trajectory,thedirectrix L. with focus F,vertex o,and From this itfollows that, whert directrix L.,.,._,.' aprojectileisfiredfrom apointPwith speed g,thedirectrix ofitsparabolic trajectoryis atthegreatest height reached byasecondprojectile, fired straight upfromPwith thesame speed q.Inparticular, we notethat alltrajectories obtained byfiring projectiles atvarious inclinations inonevertical plane, butwith a-common initial speed g,have ageometrical propertyincommon, namely, a common directrix. SEC. 6.1] MOTION OFAPARTICLE 153 Theaxesshown inFig.71arethesimplest forthediscussion of general properties ofthetrajectory. But inballistic problems itispreferable totaketheorigin at thepoint ofprojection andmeasure thetimefrom theinstant offiring (Fig. 72). Ifa.istheinclination oftheinitial velocity tothehorizontal, wehave, asin(6.102) and(6.103), x= cosO FIG. 72. Parabolic trajectory referred tothepoint ofpro- jection. The projectile strikes theground when y=0,thatis,when (6.109) then (6.110)sina; x=2Psin2<x. Q Wenotethat itisamaximum This istherange oftheprojectile, (forgiven qQ)whena=45. Thegreatest height attained bytheprojectileisobtained by putting y=0.Then,by(6.108) , (6.111) t= sina, y=^ Limits ofrange. Letussuppose thatagungives toaprojectile amuzzle velocity #o-Theguncanbepointed in any direction. What region in space canbereached bythepro- jectile? The question maybeanswered analytically, butthemost elegant solution isgeometrical. Wemay confine ourattention toonevertical plane through thegun,which isat inFig. 73. Firstweask:Where isthe focus ofthetrajectory passing through anassigned pointP? Theanswer isgivenbythefollowing construction:Co FIG. 73.Construction forthe fociofthetwoparabolic trajectories passing through P. 154 PLANE MECHANICS [SEC. 6.2 FIG. 74. Paraboloidal region within rango.Draw thedirectrix L,ataheight %ql/g above 0.Draw the circleCwith center 0,touching LatA.Since isapoint on theparabolic trajectory, thefocus must lieatadistance OA from 0,andsoitmust lieon Co. Similarly,ifthecircleC isdrawn with centerPto touch L,thefocus must lie onCalso. Hence thefocus must lieatanintersection of thecircles CoandC.Ingen- eral, there willbeeither two points ofintersection, FIand Ft,ornone. Intheformer case,Piswithin range andF\, Fzarethefociofthetwotrajectories throughit.Inthelatter case,Pisoutofrange. IfPisatthelimit ofrange, thecirclesCandCtouch. Then Pisequidistant from andahorizontal lineLI,drawn ataheight twice that ofL,i.e.,ataheight q\/g. Thus thelocus ofPin spaceisaparaboloid ofrevolution, having forfocusandAfor vertex (Fig. 74). Allpoints inside thisparaboloid arewithin range, and allpoints outside itareoutofrange. 6.2.PROJECTILES WITH RESISTANCE General equations. Weturnnow tothemore practical problem inwhich theair exerts ontheprojectile (still regarded asaparticle) aforceR acting inadirection opposite to thevelocity (Fig. 75). Inresolving forces andaccel- eration inorder toobtain equa- tions ofmotion inscalar form, wehave achoice oftwopro- cedures:(i)resolution along horizontal and vertical direc- tions, and(ii)resolution along thetangent andnormal totheR mg FIG. 75. The forces acting ona projectile. trajectory asin(5.103). Denoting by6theinclination tothehorizontal ofthetangent tothetrajectory, weobtain by (i)theequations SBC. 6.2] MOTION OFAPARTICLE 155 (6.201) mx~-Rcos0,my=Rsin-mg. Denoting byptheradius ofcurvature ofthetrajectory, we obtain by (ii)theequations (6.202) mq^=-R-mgsin0,^=wgrcos0, where qisthespeed anddsanelement ofarcofthetrajectory. Theequations (6.202) are, ofcourse, onlyadifferent mathe- matical expression of(6.201). Theresistance experienced byagiven projectile depends onits speed andonthedensity ofthe air.Regarding the airas stratified intohorizontal layers each ofconstant density, sothat thedensity isafunction ofyonly,wemayexpressRintheform (6.203) R=R(y, q)t toshow that itisafunction ofyandqonly. Since (6.204) cos6=-; sin=t q Q wemay write (6.201) intheform (6.205)=-$.r, y=-$y-g, where<isafunction ofy,q,namely, (6.206) *(y, q)= The mathematical problem ofthedetermination ofthe trajectoryismademuchmore difficult bythefactthatthere isno physicallyvalid formula expressing Rasafunction ofq.For small values ofq,Rvaries asq]butthissimple lawbreaks down before wereach those velocities which areofinterest inballistics. Forlow ballistic velocities, Rvaries asq2 ;but thislawagain breaks downwhen thevelocity oftheprojectile approaches the velocityofsound. Thelawofdependenceisthencomplicated andcanberepresented only graphically orbytables ofvalues obtained experimentally. Hence, wemust notexpect tofindany simple formulas fortrajectories. Ingeneral, (6.205) must be integrated byatedious process ofstep-by-step numerical integra- tion. The differential equations areintegrated approximately over small intervals oftime; theerrors duetoapproximation become insignificant when theintervals arevery small. 156 PLANE MECHANICS [SBC. 6.2 The difficulties involved intheabove method leadustodo whatwesooften doinapplied mathematics replace thecom- plicated physical problem byonethat issimpler mathematically. Thus, weshall confine ourattention below tothecasewhereRis independent ofyanddevote particular attention tothecases whereRvaries asg2orasq. Resistance independent ofheight. Iftheresistance depends onthespeed only, sothatR=R(#), theproblemismost easily attacked bymeans of(6.202). Letus write (6.207) R=mg<t>(q) forconvenience. Noting that decreases asthearclengths increases, wehave p=ds/dOj and elimination ofdsfrom (6.202) gives ldg=</>(<?)+ sin0. qdd(6.208)v ' (6.209)dx ~TZdd dy-^dd, . ~TZ=~J~TB=^~Pcos=-~' y ys .n qan -^=-J^-JZ=psm= %---ycos6 This iscalled theequation ofthehodograph, since #,6arethepolar coordinates ofapoint onthehodograph. Ifwecansolvex (6.208), alldesired information about the trajectory maybeobtained byquadratures. By(6.202), we have * ' 9l g #2 jtan g _*?sec q g Letussuppose thattheprojectile isfiredfrom theorigin attime t=withspeed qQatanangle ofelevation .Let (6.210) q=/(0) bethesolution of(6.208), supposed known. Then integration of (6.209) gives (6.211)_ dddxds ~dsd6 dycte 3*dd dtds dsdO Bf(B)dB. These equations express x,y,tasfunctions ofoneparameter andsodetermine thetrajectory. SEC.6.2] MOTION OFAPARTICLE 157 Butwecannot use(6.211) until thefunction /(0)isknown, i.e.,until thedifferential equation (6.208)isintegrated. Ifthe function<f>(q)isgeneral, theintegration of(6.208) cannot evenbe reduced toquadratures. Forsome special forms of<f)(q)the integration canbereduced toquadratures, andinsome cases the solution /(0)canbeexpressed interms ofelementary functions. Weshall consider bolow thecase<t>(q)=C#2 ,butbefore making thisspecial choice of <weshall transform (6.208) bychanging toanewindependent variable $,defined by tanh^=sin 0. Itiseasily seenthat (6.208) transforms into (6.212)i=tanh* Resistance varying asthesquare ofthevelocity. Letustakethelawofresistance tobe (6.213) R=m<7<K<?), 4>(q)=Cq\ whereCisaconstant. Division of(6.212) by %q*gives (6.214)=-tanh*~2C> ^ }dt\q2/ q2 this isastandard type ofequation, with solution (6.215)~=sech2^(A-2CJcosh2^dtf =cos2e[A-Ctanh-1(sin 0)]-Csin0, whereAisaconstant ofintegration, tobefixedbythe initial conditions. Theoretically atleast, theequations (6.211) nowdetermine thetrajectory, /(0)being thereciprocal ofthesquare root ofthe right-handsideof(6.215). But itisevident thatthecalculations involved arevery complicated. When theprojectile moves inaverticalline, theproblem ismuch simpler.Ifiisaunitvector directed vertically upward andtheposition vector oftheprojectileist/i,theacceleration is i/i.Theforce ofgravityismgi. The resistance ismgCy2i formotion upward (y>0)andmgCy2iformotion downward 158 PLANE MECHANICS [SBC. 6.2 (y<0).Hence theequation ofmotion is (6.216a) g=gCy2g formotion upward; (6.2166) g=*gCy2g formotion downward. Since wehave (6.217a) 1^^., gdy formotion upward;l+ (6.2176).2=gdy formotion downward. Thus, onintegration, (6.218a) y=-log(1+C#2 )+A formotion upward, (6.2186) y=~log(1-Cy2 )+A' formotion downward, where A,A1areconstants ofintegration. Letusconsider twospecial cases, corresponding to(a)ashell fired verti- callyupward, and (b)abomb dropped vertically downward. (a)Letgbetheinitial velocity ofthe shell, firedfrom y=0.Then, by(6.218a), wehave Theheight htowhich theshell rises Isfound byputting y 0;thus, (6.220) h=^log(1+CflD. Iftheresistance issmall, thisgives, approximately, (6.221)/i=*~-i^l Q Q inwhich thefirstterm isthewell-known expression forheight attained under noresistance. (6)Letthebomb bedropped from y with novelocity. Then (6.2186) gives (6.222) y2^1og(l-W ybeing, ofcourse, negative. When thebomb hasdropped adistance h, wehave log(1-CM-- SBC. 6.3] MOTION OFAPARTICLE 159 andhence itsspeed is (6.223) q.yjl"*"***- Asktends toinfinity, gtends toC~*. This isthelimiting velocity;itsvalue is (6.224) C~*-qVw/R, whereRistheresistance atanyspeed q. Relations connecting yand tmaybeobtained from (6.216) byintegration. Thus, (6.216a) maybewritten Oneintegration gives yinterms oft,andasecond integration gives y. Resistance varying directly asthevelocity. Although thelawofresistance (6.225) R=mgCq isnotaccurate physically,itissosimple totreat mathematically that itisauseful approximation atleastanimprovement over theassumption ofnoresistance atall. Turning back to(6.206), wenote that<isnowaconstant ($=gC),andtheequations (6.205) areeasy tohandle, because thevariables xandyarcseparated. Weobtain, onintegration, (6.226) where x,y<>arethecoordinates and UQ,VQthecomponents of velocity for t=0. When 3>issmall, these equations yield approximately {x=x+uti$u2 , y=y.+Vot-fa*- inwhich theterms independent of3>correspond totheparabolic trajectory. 6.3.HARMONIC OSCILLATORS Thesimple pendulum. Asimple pendulumconsists ofaheavy particle attached tooneend ofalight rodorinextensible string, theother end oftherodorstring being attached toafixed point. Wecon- 160 PLANE MECHANICS [SBC. 6.3 aider onlymotions ofthependulum inwhich thestring remains inadefinite vertical plane. InFig. 76,Bisthepoint ofattach- ment andAistheparticle (ofmassra),drawn aside from its position ofequilibrium 0.Oxyarerec- tangular axes intheplane ofmotion, Ox being horizontal. The particle moves under the.influ- ence oftwoforces:(i)itsweight mg,and (ii)thetension Sinthestring. SinceS actsalong thenormal tothecircular path oftheparticle,itisclear that thedy- namical problem presented bythesimple pendulumisprecisely thesame asthat of _ __themotion ofaparticle onasmooth '*circular supporting curve, fixed inaver- tical plane. rmgLetAB= Z,OSA=B]then FIG. 76.Thesimple pen-,Qfm fi_Iy. fiX ' fi_. fidulum.. COS-- j') Sin= j> where x,yarethecoordinates ofA.Theequations ofmotion are (6.302) mx=Ssin0,my=Scos6-mg. Letusinvestigate small oscillations about theposition of equilibrium, assuming xand itsderivatives tobesmall. Then yand itsderivatives aresmall, ofthesecond order, andcos6 differs from unity byasmall quantityofthesecond order. Thus, tothefirstorder inclusive, thesecond of(6.302) gives (6.303) S=mg, andsubstitution from thisandfrom (6.301) in(6.302) gives (6.304) x+p*x=0, p=J2. Thegeneral solution ofthis differential equation is (6.305) x=Acospt-hBsinpt, where A,Bareconstants, tobedetermined bytheinitial condi- tions.Amotion given byanequation ofthisform iscalled simple harmonic. Differentiation of(6.305) gives (6.306) x=-Ap sinpt+Bpcospt. SBC. 6.3] MOTION OFAPARTICLE 161 Wenote thatxandxhave thesame pair ofvalues attimes ti, ti+r,h+2r, i+3r, where hisarbitrary, and (6.307) r=^= P When theposition andvelocity ofaparticle arerepeated over andoveragain atequal intervals oftime,wesaythatthemotion isperiodic; theinterval iscalled itsperiodic time. Hence the above motion ofasimple pendulumisperiodic, with periodic timegivenby(6.307). InSec. 13.2 the finite oscillations ofapendulum willbe discussed, and itwillbefound that the finite oscillations are periodic butnotsimple harmonic; theformula fortheperiodic time isdifferent. Theharmonic oscillator. The simple pendulum, executing small oscillations,isonly onephysical example ofatype ofdynamical system offrequent occurrence. Many problemsofoscillation canbediscussed inasingle mathematical form; sowecreate asingle mathematical model forthem all.This model2 iscalled theharmonic oscillator , (^ , x (Fig. 77). Jf,/ . .,, ,.,FIG.77.Theharmonic oscillator.Theharmonic oscillator consists ofaparticle which canmove onastraight line,which weshall take forz-axis. Itisattracted toward theorigin byacontrolling force varying asthedistance. Ifiisaunitvector inthepositive direction ofthez-axis, thecontrolling forcemaybewritten mp2 ai,wheremisthemass oftheparticle andpisaconstant. Theacceleration isai,andhence theequation ofmotion is (6.308)mxi=mp2 xi, or,inscalar form, (6.309)*+P2x=0. Thegeneral solution ofthisequation maybewritten intheform (6.305), thatis, (6.310) x=Acospt+Bsinpt, sothatthemotion issimple harmonic. 162 PLANE MECHANICS [Sue. 6.3 Letusnowdefine constants a,ebytheequations (6.311) a=V^2+B\ A . B cos=j sin e= VA*+B* Then (6.310) maybewritten (6.312) x=acos(pt+e). Weobserve thatxcovers therange(a, a)andthatthemotion isperiodic with periodic time2ir/p. Thefollowing terminology isused inconnection with theharmonic oscillator: amplitude=a, period orperiodic time=T=2fr/p, frequency=number ofoscillations perunittime =1/r=p/27r, phase=pt+c. Effect ofadisturbing force. Letusnowsuppose that,inaddition tothecontrolling force, there actsontheharmonic oscillator aforce whose component inthepositive direction ofthe ar-axis ismX. Theequation of motion (6.308)isnowmodified totheform (6.313) rnxi=-mp*xi+mXi, or,inscalar form, (6.314) *+p*x=X. First letussuppose thatXisconstant. Thegeneral solution of(6.314)isthen (6.315) x=~+acos(pt+ ). Thismotion issimple harmonic, butthecenter oftheoscillations isdisplaced totheposition x=X/p2 ,asisseenbywriting (6.315) intheform (6.316) x-~=acos(pt+e). When theoscillation orvibration hasahighfrequency, sothatp islarge, thedisplacement ofthecenter issmall. SBC. 6.3] MOTION OFAPARTICLE 163 Letusnowsuppose thatXisitself simple harmonic, varying according totheformula (6.317) X=kcosct, where k,careconstants. Then theequationofmotion (6.314) reads (6.318) x+p*x=kcos ct. Inthegeneral casewhere cjp,thesolution is k (6.319) x=acos(pt+e)+-, Tzcosc*>pc where a,eareconstants, tobedetermined bytheinitial condi- tions. Thus themotion isasuperpositionofanundisturbed simple harmonic motion andasecond motion; thelatter hasthe same period astheforce, andanamplitude which becomes large when thedifference between theperiodsofthefreeoscillator and thedisturbing force becomes small. Thegreat increase inthe amplitudeoftheoscillations under this lastcondition iscalled resonance. We shall discuss itagain below, taking resistance intoconsideration. Dampedoscillations. Letusnowsuppose that, inaddition tothecontrolling force, there actsontheparticleofaharmonic oscillator aforce of resistance proportionaltothevelocity, called thedamping force. Thecomponentofthisforce inthepositive direction ofthez-axis maybewritten 2rapt:r, where\Lisapositive constant. The equationofmotion (6.308)ismodified to (6.320) inn=-rnp'2xi2mnxi, or,inscalar form, (6.321)x+2fjix+p*x=0. Now, (6.322)x=Cent willsatisfy thisequation providedthatnsatisfies thecharacter- isticequation (6.323)n*+2/m+P2=0. 164 PLANE MECHANICS [Sac. 6.3 This equation hastwo roots, whichmayberealorcomplex; they are (6.324) ni,n2=-/*VV-P2 - Letusdefine arealpositive number /by (6.325)I=V|M2~P2 |, thesign | |indicating absolute value. Wehave thentwocases todiscuss:(i)lightdamping, ju<p;(ii)heavy damping, ^>p. CASE(i)Li^fadamping (/*<p). Inthiscase, (6.326) ni,n2= /* #, andthegeneral solution of(6.321)is (6.327) x=<7ie<-"+'I)<+tV""-'. Thismaybewritten (6.328) x=<r*(A cos ft+Bsintt), where (6.329) A=Ci+C2,=z(Ci-C2). Since, forphysical reasons, weareinterested onlyinrealvalues ofxyitisevident that, although the differential equationis satisfied by(6.327) with Ci,C2complex constants, weshould adopt asour final solution (6.328) with realconstants A,B. Thesemaybechosen tosatisfy theinitial conditions, i.e.,togive assigned values toxandxfort=0. Themotion given by(6.328) may alsobewritten (6.330)x=a*-*' cos(It4-c), wherea,eareconstants chosen tofitthe initial conditions. This isnotasimple harmonic motion, notbeing oftheform (6.312). Thefactor e~M*indicates ageneral decayoftheoscilla- tions, xtending tozeroasttends toinfinity. Bychanging theinstant fromwhich tismeasured, wecanget ridofein(6.330), obtaining thesimpler expression (6.331) x=oerM* cosU. SEC. 6.3] MOTION OFAPARTICLE 165 Onplottinga;asafunction oft,weobtain thegraph shown in Fig. 78.(Thegraph ofthesimple harmonic motion x=acos It isacosinecurve, resembling theabove curve, except forthe latter's tendency todieaway.) Thecurve of(6.331) should be compared with thecurves (6.332a) (6.3326) FIG. 78. Position-time graph foralightly damped harmonic oscillator. alsoshown inFig.78. Itiseasily seen that (6.331) touches (6.332a) and(6.3326) at (6.333a) (6.3336)I' (2nH I respectively, nbeing anyinteger. Wemay regard (6.331) asaharmonic motion with decaying amplitude, given byae~M< .Butthefollowingisamore accurate description. Themaxima andminima of(6.331) occur atinstants t tn, where (6.334)ltn=n-K a, tan a.V 166 PLANE MECHANICS fSBC. 6.3 nbeing anyinteger. Thecommon interval is (6.335)r= ; thus 2ir/lmay bereferred toastheperiodoftheoscillations. Thevalues ofxcorresponding to(6.334) are (6.336)xn=(l)nae-^ cosa. Thus successive values areconnected by (6.337)=-e-"+1-< )=-cr**". #n The ratio ofsuccessive swings toopposite sidesis,inabsolute value, (6.338)Xn+l Itisusual todefine thelogarithmic decrement ofthedamped oscillation asthelogarithm ofthereciprocal ofthisratio tobase 10;itsvalue is(TT/Z/Q logice. CASE(ii)Heavy damping (/*>p). Inthecase ofheavy damping, wehave,by(6.324), (6.339) rci,ri2=-M+I, andsothegeneral solution of(6.321)is (6.340) x=e-^(Aclt4-Be~lt ). Then, (6.341) x=er[A(l-n)e-B(l This vanishes onlyif Since e2"isasteadily increasing function oft,there canbeat mostonesolution. Hence thevelocity oftheoscillator vanishes atoneinstant atmost. Themotion isnon-oscillatory, or deadbeat, xtending tozeroasttends toinfinity, since/u>L Forced oscillations. Finally, letussuppose thataharmonic oscillator issubject to (i)acontrolling force(mpzx), (ii)adamping force(2mp,x), (iii)adisturbing force(mkcos ct). SBC. 6.3] MOTION OFAPARTICLE 167 Theequation ofmotion isnow (6.343) x+2t*x+p*x kcos ct. Thegeneral solution is (6.344) x=xl+x, } where x\satisfies (6.345) x,+2Mzi+p'zi=0, andcontains twoconstants ofintegration, while x2isanypartic- ularsolution of (6.346) x24-2^2+p*x t=kcos ct. Then x\corresponds tothemotion ofthedamped oscillator without disturbing force;itisgiven by(6.330) or(6.340), according asthedampingislight orheavy. Asfor 2,(6.346)issatisfied bytheexpression (6.347) x*=Ecosct+Fsinct, where ._ k(p2c2 ) (b.d4S) 4- (p2_C2)2 This solution ismost easily found byreplacing cos ctbyelct in(6.346) andfinding acomplex constant GsothatGelctisa solution; x%willthenbetherealpart ofthis solution. Alter- natively, wemay substitute (6.347) directly in(6.346) tofindthe constants. Ifwedefine _ ,+(2MC) 2__C2 00817 wemaywrite (6.347)intheform (6.350)*2=bcos(d4-rj). Thus thegeneral motion ofthedamped oscillator under the influence ofthedisturbing force is (6.351) x xi+6cos(ct+77). 168 PLANE MECHANICS (Sfic. 6.4 As$>oo,zi-0; thus after alongtime themotion tends totheforced oscillation (6.352) x=bcos(ct+r?). This isasimple harmonic motion ofamplitude 6andperiod equal tothat ofthedisturbing force, butthere isadifference inphase. Iftheperiod ofthedisturbing force isequal tothefreeperiod oftheoscillator, wehave thecase ofresonance. Then c=p, 77=^?r,andso k k (6.353) x= cos(pt-fr)= sinpt, thedisturbing force beingmkcospt.The difference of TTin phaseisinteresting. 6.4.GENERAL MOTION UNDER ACENTRAL FORCE Cartesian equations andthelawofdirect distance. Consider aparticle ofmassmattracted toward afixed point byaforcemP.(Wemayinclude thecase ofrepulsion bytaking Pnegative.) Forrectangular Cartesian coordinates Oxy, the equations ofmotion are ,AM \.mPx.mPy (6.401) mx=-- ,my=---> where r2=x2+y2 .Werefer tothisasmotion under acentral force, because theline ofaction oftheforce passes through a fixed center 0. Asanapplication of(6.401), letusconsider thelawofdirect distance, meaning thereby thatPisproportional tor.Letus confine ourattention toanattractive force, putting (6.402) P=jfcr, where &isaconstant. Then (6.401) read (6.403) x+kzx=0, y+k*y=0; these equations have thegeneral solutions (x=Acoskt+Bsinkt, \y=Ccoskt+Dsinkt, SBC. 6.4] MOTION OFAPARTICLE 169 where thecoefficients areconstants, tobefixedbythe initial conditions. Ifwesolve (6.404) forcosktand sinkt,andeliminate tby theidentity cos2kt+sin2kt=1, weget (6.405) (Cx-Ay)2+(Dx-By)2=(BC-AD)\ This isacentral conicandnecessarily anellipse since x,yremain finite, asweseefrom(6.404). Thus theorbit described under a central attractive force varying directly asthedistance isanellipse havingitscenter atthecenter offorce. This motion iscalled elliptic harmonic. Toillustrate thesignificance oftheconstants in(6.404), letussuppose that attime t theparticleisatxa,y=0, moving with velocity VQinthedirection ofthe t/-axis, sothat x=0,y=VQ.Putting thisinformation into theequations (6.404),first asthey stand andthen intheform obtained by differentiation, weget a=A,=0,= ,VQ=Dk, andsothemotion isgivenby (6.406) x=acoskt, 2/=T~sm^-K Polar coordinates. Returning tothegeneral problem ofmotion under acentral force, weshallnow obtain equations easier tohandle than (6.401). Itisonly inthecase ofthelawofdirect distance that (6.401) areconvenient. Since theforce ontheparticle passes through 0,ithasno moment about 0.Hence, bytheprinciple ofangular momentum (5.214), theangular momentum about isconstant. Weshall change slightly thenotation ofChap. V,now letting hdenote angular momentum perunitmass; then,by(5.106), (6.407)h=r*6=constant, wherer,arethepolar coordinates oftheparticle. Analter- 170 PLANE MECHANICS [SBC. 6.4 native expression forhispq,where pistheperpendicular from onthevelocity vectorq. Ifwefollow theradius vector, drawn from totheparticle, weobserve thatwhen itturns through aninfinitesimal angle d6 itsweeps outanarea^r2d&.Thus thearcal velocity Amaybe defined as (6.408) A=r2 0, Abeing,infact, thetotal areaswept outfromsome initial instant. Weobserve thathistwice thearcal velocity, andby (6.407) wehave thefollowing important result: Inmotion under a central force,thearcal velocityisconstant. This fact isus&l tosimplify theproblemofdetermining the orbit.Wedefine (6.409) u=~, thereciprocaloftheradius vector. Then, since (6.407) maybe written (6.410)6=hu*, wehave 1du A,du (6.411) Thevector equation ofmotion is (6.412) mi=mP, wherePistheattractive force perunitmass. Resolution along theradius vector gives, by(4.107), (6.413)f-r6*=-P, wherePistheinward componentofP.By(6.411), weobtain (6.414)gf-f SBC. 6.4] MOTION OFAPARTICLE 171 This isthedifferential equation oftheorbit ofaparticle moving under anattractive central forcePperunitmass. Ifwecansolve thisequation, obtaining uasafunction of0,wehave theequation oftheorbit inpolar coordinates. Henceforth, letussuppose thatPisafunction bfronly. It iseasy toseethattheworkdone inpassing fromoneposition to another isthenindependent ofthepath described. Thus the systemisconservative, andwemayusetheprincipleofenergy. LetT,V,Eberespectively thekinetic energy, potential energy, andconstant total energy,allperunitmass. Then (6.415) T+V-E. Now, by(6.411), (6.416) T=i(f*+r*)-\ Thepotential energy perunitmass issuchthat P=-grad F, sothat, sincePistheinward component ofP, (6.417) p=,v where roissome constant. Hence, (6.415) gives Thisequationisreally equivalent to(6.414), asmaybeseenon differentiation. Weare,ofcourse, toremember thatV,being a function ofr,isalsoafunction ofu. Apsides andapsidal angles. Anapseisapoint onanorbit atamaximum orminimum distance from thecenter offorce. Thecondition foranapseis r=0,or,equivalently, (6.419)*f-0. From (6.418) itfollows that, atanapse, 172 PLANE MECHANICS [SEC. 6.4 SinceVisafunction ofu,this isanequation todetermine the values ofuattheapsides, supposing theconstants Eandh known. Asanillustration,letusreturn tothecaseP=k2r.Then (6.421) F=P2r2=-5, andso(6.420) maybewritten (6.422) w4-p(#^2-P2 )=0. This quadratic equation inu2yields rootswf,u\,which will bethesquares ofthereciprocals ofthesemiaxes oftheelliptical orbit. Byastudy ofapsides, wemayobtain ageneral description of anorbitwithout actually solving thedifferential equation (6.414). Toestablish animportant feature, letustemporarily forget the dynamical problem andthink ofadifferential equation (6.423) ^=f(y), tobesolved under theinitial conditions y=i/o,dy/dx for x=0.These initial conditions de- termine aunique solution. Since the transformation x x'leaves the form ofthedifferential equation and the initial conditions unaltered,itis evident thatthecurve y=F(x) ywhich satisfies (6.423) andtheinitial condi-^ tions, issymmetric with respect tothe SincePisafunction ofu,itistjlear that (6.414) and(6.423) areequations Fm.79.-Symmetry ofaofthesame form>withtheCOrrespond- central orbitwithrespect toanenCCU*y,6*X.IfW6measure 6 apse me.from anapse, the initial conditions for(6.414) arethesame asthose for(6.423). Hence, acentral orbit issymmetric with respecttothelinedrawn fromtheforce center toanapse. This result throws much lightonthegeneral structure ofa central orbit. LetAandB(Fig. 79)beconsecutive apsides, and SEC. 6.4] MOTION OFAPARTICLE 173 letussuppose thattheportionABoftheorbit isknown. The orbit issymmetrical about OB;hence, wemayobtain somemore oftheorbitbyfolding theportion ABoverthelineOB,obtaining BC,withanapse atbysymmetry. Theapsidal distance 00 isequal totheapsidal distance OA. Again, foldingBCover00, wegetCDwithanapse atD,andOD=OB. Thefollowing facts arenow clear: (i)Any central orbit hasonlytwoapsidal distances. The orbit isacurve touching twoconcentric circles, theradii ofwhich arethetwoapsidal distances. (ii)Once theorbitbetween twoconsecutive apsidesisknown, thewhole oftheorbitmaybeconstructed byoperations offold- ingover apsidal radii. (iii)The angle subtended atthecenter bythearcjoining consecutive apsidesisaconstant. Itiscalled theapsidal angle. Insome cases, (i)maybeviolated. Theradius oftheinner circlemaybezero, orthat oftheouter circlemaybeinfinite. Butthese aretoberegarded asexceptional cases. Letusnow consider how theapsidal angleistobefound. When uisincreasing, (6.418) gives (6.424)^ where (6.425) F(u)- Thus d=_*L and so,ifMI,uzarethereciprocalsoftheapsidal distances (with Ui<u%),theapsidal angle ais (6.426) Asanillustration,letusconsider thelawofdirect distance, whereFisgivenby(6.421). HereF(u)isoftheform 174 PLANE MECHANICS [Sue. 6.4 whereAandBareconstants. ButF(u)=atanapse; hence u=1*1,u=uzareroots ofF(u)=0,andso F(u) Thus, by(6.426), aswealready knew from thefactthat theorbit isacentral ellipse. Stability ofcircular orbits. Inacircular orbit uisconstant. Hence by(6.414) the possible radii ofcircular orbits aredetermined by (6.428) =I- Withanattractive force (P>0),wecanobtain acircular orbit ofanyradius, byprojecting theparticle atright angles tothe radius vector with thatvelocity which makes (6.429) h* VI Butthequestion arises: Arethese circular orbits stable or unstable fInother words,ifslightly disturbed, willtheresulting orbit lieclose totheoriginal circular orbit, orwill itdeviate far from it?This questionisimportant physically, because in nature small disturbances arealways present, andthey will destroy anunstable circular orbit. Theonly circular orbitswe r*nhope toobserve arethose that arestable. Itshould beclearly understood thatweshallnotconsider the effects ofdisturbing forces which continue toactontheparticle; weassume that theposition andvelocity oftheparticle have been disturbed andinvestigate theresulting motion under the original central force. Letu=uQandh=hQinthecircular orbit. Then,by(6.429), (6.430) hi: Tostudy thedisturbance, weput (6.431) u=-Uo+{, SEC. 6.4) MOTION OFAPARTICLE 175 where Jand itsderivatives areassumed tobesmall. Wealso assume thath hoissmall. Substitution in(6.414) gives (6.432) +..+ Now, onexpansion inpowers of{, <6-433>Ffsrl-K,('+&" where P'=dP/du, andthesubscript indicates evaluation foru=w . Then, tothefirstorder insmall quantities, (6.432) becomes (6.434) g+At-- , where (6.435) A-1-^(~J--)=3-?*, hlu*\P UQ/ Po by(6.430) ;Bisanother constant whose value doesnotinterest us.Thesolution of(6.434)is (6.436a) J=~+Cicos(\/3 (?)+C2sin(\/I 0), (6.4366) -+Cicosh(v^^Z ^)+C2sinh(x/11^* 0), (6.436c) f-i^^2+CJ+C2, according asA>0,A<0,orA=0. Ofthese solutions, only (6.436a) remains permanently small; theothers increase indefinitely with 0.Hence thecircular orbit of radius I/UQ isstable if,andonly if, (6.437) Inparticular,letusconsider thecase ofaforce varying inversely asthenthpowerofthedistance sothat (6.438) P=^=ku\ Then uP' 176 PLANE MECHANICS [Sue. 6.5 andsocircular orbits under theattractive force (6.438) arestable if,andonly if, (6.439) n<3. Thus thelawofdirect distance (n=1)andthelawofthe inverse square (n=2)give stable circular orbits; thelawof theinverse cube (n=3)gives unstable circular orbits. 6.5.PLANETARY ORBITS Thelawoftheinverse square. Asalready remarked inSec. 3.1,Newton's lawofgravitational attraction states thattwoparticles ofmasses w,w',atadistance rapart, attract oneanother with equal andopposite forces of magnitude (6.501) whereGisthegravitational constant. Coulomb's law ofelectrostatic attraction states thattwo particles carrying electric charges e,e'(inelectrostatic units), atadistance rapart, repel oneanother with equal andopposite forces ofmagnitude PP' (6.502)^. Ifeand e'have opposite signs, thisforce isaforce ofattraction. Herewehavetwoexamples ofthelawoftheinverse square. Thelaw(6.501) governs astronomical phenomena inparticular, themotion ofaplanet round thesun.Thelaw(6.502) governs atomic phenomena inparticular, themotion ofanelectron in anatom about thecentral nucleus. Inthis case, ofcourse, the charges e,e'have opposite signs, sothat theforce isoneof attraction, asinthegravitational case. Itisremarkable that thesame form forthelawofattraction should holdonsuch different scales. Theexpressions (6.501) and (6.502), combined withNewton's lawofmotion, constitute twohypotheses regarding phenomena ingravitational and electrostatic fields. Foralong time, they wereaccepted ascompletely validfromaphysical point ofview, butthat isnolonger thecase. Themodern astronomer knows Sue. 6.5] MOTION OFAPARTICLE 177 that gravitational attraction should bediscussed interms ofthe general theoryofrelativity, andthephysicistinsists thatprob- lemsontheatomic scale belong toquantum mechanics. It would, however, create acompletely false impressionifwewere tosaythatthelawoftheinverse square hasdisappeared from modern science. Nearly allthecalculations ofastronomers are stillbased on(6.501) andgive results inexcellent agreement with observation. Moreover, thephysicist often fallsbackon thesimple atomicpicture based on(6.502) andNewton's law ofmotion. Inwhatfollows, weshall discuss themotion ofaplanet attracted bythesun. Obviously, byamere change ofconstant, thesame reasoning willapply tothemotion ofanelectron in anatom. Determination oftheorbit. Thesunandaplanet arercgaidcd asparticles,ofmassesMandm,respectively. Theattraction ofthesunontheplanet, given by(6.501), produces anacceleration GM/r2 ]andthe attraction oftheplanet onthesunproduces anacceleration Gm/r2 .These acceleration?* are intheratioM/m, which is actually avery large number. Hence, without any serious departure from reality, wemay neglect theacceleration ofthe sunandtreat itasifitwere atrest. Laterweshall seehowto treat theproblem exactly. Weconsider then thecase ofaparticle attracted toward a fixed center byaforcePperunitmass, where (6.503) P= /*being some positiveconstant. The differential equation (6.414) fortheorbitnowreads (6.504) +- fr Thegeneral solution is (6.505) u=^+Ccos(0- ), whereCand 0oareconstants ofintegration. Thisis,inpolar coordinates,theequation ofthemost generalorbit described under acentral force varying astheinverse square ofthedistance. 178 PLANE MECHANICS [SEC. 6.5 Thepotential energy perunitmass is (6.506) F=JPcJr=-=-/m, theconstant ofintegration being chosen tomakeVvanish at infinity. Letusnow substitute from (6.505) in(6.418), theequation ofenergy, inorder toexpress theconstant Cinterms ofEandh (the total energy andangular momentum perunitmass^. We get Cf+j|+2C^2cos(9- )=~ [#+Jj+nCcos(6- sothat (6.507) C>=g+ Byrotating thebase line=0,wecanmake =and C>in(6.505) ;thisweshallsuppose done. Then theequation (6.505) fortheorbit reads e\ (6.508) u=pl+-l+rcose From thefocus-directrix property ofaconic, weknow that itsequation inpolar coordinates maybewritten (6.509) u=i(1+ecos0), where Iisthesemi-latus-rectum(i.e., halfthefocalchord parallel tothedirectrix) and etheeccentricity; 6ismeasured from the perpendicular dropped from thefocus onthe directrix. The conicmaybeofanyofthefollowing types: ellipse (e<1), parabola (e=1), hyperbola (e>1). Inthecase ofthehyperbola, (6.509) gives onlythebranch adja- centtothefocus. * Comparing (6.508) and (6.509), wenote that itisalways possible tobring theequations intocomplete agreement by choosing for Iand ethevalues SBC. 6.5) MOTION OFAPARTICLE 179 (6.510) I=, e-J /* V1+2M2 Accordingly, wemay say:The orbit described byaparticle, attracted toafixed center byaforce varying astheinverse square ofthedistance, isaconic having thecenter offorce forfocus. The semi-latus-rectum and theeccentricity aregiven by(6.510) interms oftheangular momentum andenergy perunitmass. The orbit may beofthefollowing types: ellipse (E<0), parabola (E=0), hyperbola (E>0). Thefactthat orbitsmaybeclassified thus interms ofthetotal energyisremarkable. Themostimportant orbits inastronomy (those oftheplanets) are ellipses. Recurring comets describe orbits which areelon- gated ellipses, approximating toparabolas. Abody with a parabolic orhyperbolic orbit would pass outfrom thesolar system, never toreturn. Constants oftheelliptical orbit. Letusnowconfine ourattention totheelliptical orbit. Since theorbits oftheplanets areofthistype, agreat wealth of technical detail hasbeen developed about the elliptical orbit. Weshall heregiveonlyabrief treatment. Itisevident from (6.509) thattheshape and sizeofanorbit (butnot itsorientation inspace) aredetermined bythetwo constantsI,e.These arerelated totheconstants E,hby (6.510). Thus, ofthevarious constants which appearinour equations, wearetoregard /x(theintensity oftheforce center) asgiven once forall,whereas theconstantsI,e,E,htake differ- entvalues fordifferent orbits. Onaccount of(6.510), onlytwo ofthese constants areindependent. Wemay useasaninde- pendent pairanytwowhich prove convenient. Instead ofusing (I,e)asfundamental constants,itisbetter touse (a,e),where aisthesemiaxis major oftheorbit. Now, (6.511)I--a(l-e2 ), 6being thesemiaxis minor. We shall refer to(a,e)asthe geometrical constants ofanorbit and (E,h)asitsdynamical 180 PLANE MECHANICS [SBC. 6.5 constants. Theformulas oftransformation from one setto theother areasfollows: (6.512)a= E=-2E 2a2Eh* There isasimple formula giving thespeed qatanypoint oftheorbit interms oftheradius vector. Bytheequation of energy, wehave to*- J-B. Substituting forEfrom (6.512), weobtain (6.513) Theperiodic time. Wenow ask:How longdoestheparticle take todescribe the elliptical orbit? Thistime iscalled theperiodic time(T).We seekanexpressionforrinterms ofthefundamental constants. The periodic time cannot beobtained from(6.414), because thetime hasbeen eliminated from this equation. We refer pinstead to(6.408), which gives for theareal velocity A=ifc. IfFisthefocus atwhich thecen- terofforce issituated,itfollows at once thattheparticle describes an arcVP,starting from thevertex* Vnearer toF,inatime 2A/h, whereAisthearea ofthesector subtended atFbythisarc (Fig. 80). Following thepointPrightround theorbit,wegetfor theperiodic timeFia. 80.-Therateofincrease ofA isconstant. (6.514)2A *ThevertexViscalled perihelion thepoint closest tothesun the other vertex being called aphelion thepointaway from thesun. SBC. 6.5] MOTION OFAPARTICLE 181 whereAisnow thetotal area ofthe ellipse. Wemight sub- stitute A=7ra6; but, tobesystematic, weshould express r interms ofeither thegeometrical constants orthedynamical constants. Since 6=\/l e2 , weobtain, bysome easy calculations, <' '- (Weremember thatE<fortheelliptical orbit.) Itisremarkable thattheformula involves onlyonegeometrical constant oronedynamical constant. Allorbits with thesame semiaxis major have thesame periodic time; soalsohave all orbits with thesame total energy. Kepler's laws. Before themathematical theory given above hadbeen devel- opedbyNewton, Kepler deduced thefollowing laws ofplanetary motion from acareful study ofthe results ofastronomical observations: I.Each planetdescribes anellipse with thesuninonefocus. II.Theradius vector drawn from thesuntoaplanet sweeps outequal areas inequal times. III.The squares oftheperiodic times oftheplanets are proportionaltothecubes ofthesemiaxes major oftheir orbits. Starting from Newton's lawofgravitation, wehaveshown that allthese statements aretrue. But itisinteresting toadopt thehistorical point ofviewandfacetheproblem asitpresented itself toNewton: Given Kepler's laws asastatement offact,what isthelawofgravitationalattraction? Law IItellsusthaththeangular momentum perunitmass isconstant, andhence that theforcemust bedirected toward thesun.FromLaw I,weknow that theequation ofanorbit maybewritten u=-T(1+ecos0). Thenby(6.414) theforce perunitmass is (MID " 182 PLANE MECHANICS [SEC. 6.5 Thus foreach planet theforce varies inversely asthesquare of thedistance. But itremains toprove that theforce isofthe form (6.517) mP2, wheremisthemass oftheplanet and/*aconstant, thesame for alltheplanets. Toshow this,weappeal toLaw III.We know that foranelliptical orbit, described under acentral force directed toafocus, 2irab r=-/T' andso a3h*a h*' But,byLaw III, this isaconstant, thesame for allplanets. Thus h*/listhesame forallplanets; and so,by(6.516),P=/m2 , where /*isthesame forallplanets. Hence (6.517) istrue,and Newton's lawofgravitationisthusdeduced asaconsequence ofKepler's laws. IfKepler's laws were accurately true,weshould have to regard thesunasfixedandtheplanets asattracted onlyby thesun.More precise measurements show that Kepler's laws areonlyanapproximation andthat theinverse-square lawof attraction holds forevery pair ofbodies. Itisfortunate that theobservations ofKepler's timewere crude, because otherwise thesimplicity ofthelawofgravitation would havebeen obscured. Thetwo-body problem. Anaccurate dynamical treatment ofthesolarsystem involves complexities fargreater than those encountered intheabove dis- cussion. First, thesun isaccelerated bytheattractions ofthe planets; secondly, themutual attractions oftheplanets influenqe their motions. Afulltreatment oftheproblem belongs tothe subject ofcelestial mechanics, andwemake noattempt todiscuss ithere. Wemayhowever ask:What isthebehavior oftwobodies which attract oneanother accordingtothelawoftheinverse square? This SEC. 6.5] MOTION OFAPARTICLE 183 problem presents itself innature inthecase ofadouble starand intheproblem ofthemoon's motion relative totheearth, the attraction ofthesunbeing neglected.Weshowed inSec. 5.2that,ifthere arenoexternal forces, themass center ofasystem moves inastraight linowith con- stant velocity, relative toaNewton- ianframe ofreference. Wecanthen take another Newtonian frame in which themass centerCisatrest. Wesuppose thisdone forthetwo-body problem inFig.81. Letm,m'bethemasses ofthepar- ticles, r,r'their position vectors rela- tivetoC,and iaunitvector drawn paralleltothelinejoiningmr tom.Then theequations ofmotion are,invector form,FIG. 81.Thetwo-body problem. (6.518)mi=- m'r"Gmm' Gmm' . Now, from thedefinition ofmass center, /m+m' m 7(6.519) hence (6.518) maybewritten GmM'r+r" m' (6.520)(m+m')*' fm'r'Gm'M . ' !<>^Mm3 (m+ Butthese equations have theform ofequations ofmotion under central forces varyingastheinverse square ofthedistance. Therefore, eachbodymoves about thefixedmass center asifattracted toitbythegravitational forcedue toamassM'inthefirst case andamassMinthesecond case. Tofindthemotion ofmrelative tom',wenotethattherelative position vector is (6.521) R-r-r'. 184 PLANE MECHANICS [SBC. 6.6 Thus, by(6.518), (6.522) mR=m*-Sn>V=-Gm(m R+m/) i, sinceR=r+r'. This resultmaybeexpressed asfollows: TT&emotion ofoneof thebodies (m)relative totheother(mf )takes place precisely asif thelatter werefixedand itsmass increased fromm'tom+mf . Itisevident fromsymmetry that,when aparticleisattracted byafixed center 0,itsorbit liesinaplane, i.e.,theplane contain- ingandtheinitial velocity vector. Inthecase ofthetwo-body problem, theorbits both lieinoneplanewhen viewed inaframe ofreference inwhich themass centerCisfixed. Butinanyother Newtonian frame themotion appears very complicated. 6.6.SUMMARY OFAPPLICATIONS INPLANE DYNAMICS MOTION OFAPARTICLE I.Ballistics. (a)Noresistance; thetrajectoryisaparabola. (b)Resistance independent ofheight [R=mg4>(q)]] the trajectory maybefound byquadratures when thefollowing differential equation ofthehodograph hasbeen integrated: (6.601) orv 'cos (c)Resistance proportional to#2 ;(6.601) maybeintegrated interms ofelementary functions, butthecomplete determination ofthetrajectoryisvery complicated. Motion inavertical line iseasily determined. (d)Resistance proportional tog;thetrajectoryiseasily found. II.Harmonic oscillators. (a)Simple harmonic oscillations: (6.602) x+p*x=0, (6.603) x=Acospt+Bsinpt, or x=acos(pt+ ); (T==2ir<J-forsimple pendulum). r ifv / (b)Oscillations with disturbing forcemX: (6.604) X=constant; center ofoscillation displaced. SBC. 6.6] MOTION OFAPARTICLE 185 k (6.605) X=kcosct\ x-acos(pt+c)+2_ 2cos ct. (c)Oscillations withdamping ( (i)Light damping (/*<p)oroscillatory: (6.606) x=ae-*' cos(It+e), (Ratio ofsuccessive swings toopposite sides=e~rft/l.) (ii)Heavy damping (/x>p)ordeadbeat: (6.607) x=Ae-^-w+Be~^+l)i yI=vV-P2 . (d)Forced oscillations (periodic disturbing forcemkcos ct): after alongtime themotion approximates to (6.608) x=6cos(ct+17), where &and17areindependent oftheinitial conditions. III.General motion under acentral force(mPtoward center). (a)Equations ofmotion: (6.609) *=-^y=-^; d*u P (6.610) <d6*' (h=r26=pg=twice theareal velocity ; (6.611) (-JT)+u2 -TO;(^~constant total energy). \cLB/hu (b)Orbit symmetric with respect toapse line. (c)IfP=k2 r,theorbit isacentralellipse. IV.Planetary orbits. (a)Theorbitunder anattraction mn/r2isaconic section with onefocus atthecenter offorce: ellipse forE<0,parabola for E=0,hyperbolaforE>0. (b)For ellipticalorbit (6.612)-=1+ecos (6.613) periodic time=r=2?r^/ 186 PLANE MECHANICS [Ex.VI V.Two-body problem. Relative motion isthesame asifonebody were held fixedand itsmass increased tothesum ofthetwomasses. EXERCISES VI 1.Aparticleisprojected upward inadirection inclined at60tothe horizontal. Show that itsvelocity when atitsgreatest height ishalf its initial velocity. (Neglect theresistance ofthe air.) 2.Aparticle ofmass ramoves onastraight lineunder theinfluence ofa force directed toward theorigin onthelineandproportional tothedis- tance from 0]theforce atunitdistance isofmagnitude wfc2 .The particle passesOwithavelocity M.Ifxisitscoordinate attime tandvitsvelocity atthat instant, show that r2+fc2s2-u*. 3.Prove byageneral argument, notinvolving anyparticular lawof resistance, thatabodythrown vertically upward inaresisting medium will return tothepoint ofprojection withavelocity lessthan thatwithwhich it wasprojected. 4.Agunismounted onahillofheight habove alevel plain. Show that,iftheresistance oftheairisneglected, thegreatest horizontal range for given muzzle velocity Visobtained byfiring atanangle ofelevation such that coscc2-2(1+gh/V2 ). 6.Find thegreatest distance thatastone canbethrown inside ahori- zontal tunnel 10feethighwithavelocity ofprojection of80feetpersecond. Find alsothecorresponding time offlight. 6.Inaresisting medium twoidentical bodies areletfallfrom thesame position atinstants separated byaninterval t.Show thatthedistance between them tends tothelimit vt,where visthelimiting velocity. 7.Calculate therateoflossofenergy (kinetic+potential) foradamped harmonic oscillator vibrating asin(6.330). 8.Aspring withcompression modulus Xsupports amass m.Show that theperiod ofvertical oscillations under gravity is2ir\/ml/\, where Iisthe natural length ofthespring. (The compression modulus istheratio* ofthe force producing compression tothecompression perunit length.) 9.Aparticle moves inaplane, attracted toafixed center byaforce varying astheinverse cube ofthedistance. Findtheequation oftheorbit, distinguishing thethree different caseswhichmay arise. 10.Aparticle isattracted toward afixed center byaforce /i/r2perunit mass, Mbeing aconstant and rthedistance fromthecenter. Itisprojected fromaposition Pwithavelocity ofmagnitude q,making anangleawith OP. Assuming thatOP<2///0J,show thattheorbit isanellipse; determine (interms ofAC,qQ)a.andthedistance OP)theeccentricity oftheorbitand theinclination ofthemajor axistoOP. 11.Aparticle moves under theinfluence ofacenter which attracts witha force [(6/r2 )+(c/r4 )],band cbeing positive constants and rthedistance from thecenter. Theparticle moves inacircular orbit ofradius a.Prove thatthemotion isstableif,andonly if,a*b>c. Ex.VI] MOTION OFAPARTICLE 187 12.Aparticle ofmassmmoves inacentral field ofattractive force of which theintensity is where kisaconstant. Prove thatacircular orbit ofradius risstableif,and only if,r2<\. 13.Deduce thefollowing relations foranelliptical orbitunder theNew- tonian lawofattraction: ra(l ecosE),ME-esinE, where ristheradius vector drawn from thecenter ofattraction, athe semiaxis major oftheorbit,Etheeccentric anomaly, andMthemean anomaly. (These anomalies areangles, defined asfollows. LetObethe geometrical center oftheorbit,Vitsperihelion, andPtheposition ofthe particle attime t.ThenEistheeccentric angle ofPvanishing whenP isatV.Todefine M,weconsider apointmoving ontheorbit with con- stant angular velocity about 0,starting fromVwithPandcompleting the circuit intheactual periodic time. IfQistheposition ofthispoint attime t,thenthevalue ofAfcorresponding toPistheangleQOF.) 14.Asimple pendulum ofmassmandlength aishanging inequilibrium. Attime t=asmall horizontal disturbing forceXcomes intooperation and continues toact,varying withtimeaccording totheformula Xmbsin2pt, where p2g/a. Findaformula giving theposition ofthependulum atany time. 15.Abodyofmassmisprojected vertically upward inamedium for which theresistance ismk*v*. Ifthe initial velocity is t>,show thatthe body returns tothepointofprojection withavelocity v\suchthat . 1 g+k*vl 16.Mud isthrown offfrom thetireofawheel (radius a)ofacartraveling ataspeed F,whereV2>ga.Neglecting theresistance ofthe air,show thatnomudcanrisehigher thanaheight ,V* ,ga* +25+2F* above theground. 17.Ashell isfired vertically upward withspeed 90.The resistance is mgCq*. Show that itattains itsgreatest height attimeI,givenby tan(gtv/C)- floVU. Deduce that,nomatter howlarge gomay be,tcannot exceed ^""'C""*. 18.Aparticle ofmassmdescribes anelliptical orbit ofsemiaxis major a under aforce mp/r* directed toafocus. Prove thatthetime average of 188 PLANE MECHANICS [Ex.VI reciprocal distance is 1rdt I-i =5- rJra anddeduce thatthetimeaverage ofthesquare ofthespeed is Theintegrals areevaluated foracomplete revolution. 19.Abomb isdropped fromanairplane flying horizontally ataheight h with speed U.Assuming thelinear law ofresistance R=mgCq asin (6.225), andfurther assuming that thisresistance issmall, show that the time offall isapproximately Show alsothat thehorizontal distance through which thebomb falls is approximately 20.Intheproblem oftwobodios attracting according totheinverse square law,there arefour orbits: theorbits ofcither body relative totheotherand theorbits ofeither body relative tothemass center. Show that allfour orbits have simultaneous apsides andthesame eccentricity. 21.Two particles ofmasses m,m'distant aapart areprojected with velocitiesq,q',respectively; thedirections ofprojection andthelinejoining theparticles aremutually perpendicular. Find thecondition that the relative orbits under theirmutual attraction maybeellipses; assuming the condition tobesatisfied,findtheperiodic time. 22.Themotion ofanoscillator mayberepresented graphically inaplane, xbeing shown asabscissa andxasordinate. Thehistory oftheoscillator isthen acurve. Show that foranundamped harmonic oscillator this curve isanellipse, andforalightly dampedoscillator itisacurve spiraling intotheorigin. Investigate thecurve foraheavily damped oscillator, and show that itwillbeastraightlinethrough theorigin forspecialinitial conditions. CHAPTER VII APPLICATIONS INPLANE DYNAMICS- MOTION OFA RIGIDBODYANDOFASYSTEM 7.1.MOMENTS OFINERTIA. KINETIC ENERGY ANDANGULAR MOMENTUM Definition ofmoment ofinertia andsome direct calculations. Themoment ofinertia ofaparticle about aline isdefined as/=mr2 ,wheremisthemass oftheparticle and ritsper- pendicular distance from the line. Themoment ofinertia ofa system ofparticlesisdefined asthesum ofthemoments of inertia oftheseparate particles. Thus, (7.101) /=2}m'r*<> i-i ifthesystem consists ofnparticles ofmasses mi,m2,mnt situated atdistancesri,r2, rnfrom thelineabout which the moment ofinertia istaken. Itisevident that themethod ofdecompositionisapplicable tomoments ofinertia. Thus,ifasystemissplit intotwoparts withmoments ofinertia /iand /2,themoment ofinertia ofthe complete systemis (7.102) /=/i+Ii. Itisconvenient todefine alength kcalled theradius ofgyration. Ifasystem oftotalmassmhasamoment ofinertia /,theradius ofgyration kisdefined bytheequation (7.103) mk*=/. When thesystem consists ofasingle particle,itisevident that theradius ofgyration about any line issimply thedistance from the line. Inthecase ofacontinuous distribution ofmatter, thedefinition (7.101) passes over into (7.104)/=JY2dm, where theintegration sign indicates thelimit ofaprocess in which thesystemisdivided intoagreatnumber ofvery small parts, andthesumtaken; dm isthemass ofaninfinitesimal 189 190 PLANE MECHANICS [SEC. 7.1 element, and risitsdistance from the lineabout which the moment ofinertia istobefound. Moment ofinertia hasthedimensions [ML*] and ismeasured ingm.cm.a inthe c.g.s. system andinIb.ft.2inthe f.p.s. system. Thesquare ofthe radius ofgyration hasthedimensions [moment ofinertia]/[mass], or[ML*]/ [M], i.e., [L2 ].Thus radius ofgyration hasdimensions [L]andsoisa length. Letusnowcalculate some simple moments ofinertia. Hoop. Itisevident that themoment ofinertia ofathin hoop ofmassmandradius aabout alinethrough itscenter perpendicular toitsplaneisma2 . Rod. Letuscalciilate themoment ofinertia ofauniform rod ofmassmandlength 2aabout alinethroughitscenter per- pendicular toitslength. Taking therodforor-axis, with the origin atthecenter oftherod,themass ofanelement dxis ,mdxdm=^2a Hence, by(7.104), wehave mdx ., (7.105)2a Rectangular plate. Consider nowauniform rectangular plate ofmassmandedges oflengths 2a,2b.Wewish tocalculate themoment ofinertia about thelineinitsplane passing through thecenter and parallel totheedge 26.Weimagine theplate split into thin strips parallel totheedge2a.Letdmbethe mass ofastrip. Then themoment ofinertia ofthestripis Jo2dm,by(7.105), andhence themoment ofinertia ofthewhole plateis-ywa2 . Circular disk.Wewish tofindthemoment ofinertia ofa uniform circular disk ofmassmandradius aabout alinethrough itscenter perpendicular toitsplane. Weimagine thedisksplit upintothinringsbyagreatnumber ofcircles concentric withthe boundary.Ifr,r+draretheinner andouter radii ofaring, thearea ofthering is2irrdr,and itsmass is , 2irrdrdm=m5ira2 Themoment ofinertia is (7.106) /= r2dm= r9dr SEC. 7.1]MOTION OFARIGIDBODYANDOFASYSTEM 191 Circular cylinder. Themoment ofinertia ofasolid circular cylinder about itsaxisfollows immediately from (7.106). For wemayimagine thecylinder splitupbyplanes perpendicular toitsaxis intoagreatnumber ofthin disks. When weadd together theirmoments ofinertia, we get7=iwa2 ,wheremisthemass of thecylinder andaitsradius. The length ofthecylinder doesnotappear explicitly intheformula. Sphere. Tofindthemoment ofin- ertia ofasolid sphere ofmassmand radius aabout adiameter, weimagine itsplit into thin circular disks by planes perpendicular tothediameter inquestion. Figure 82shows thesec- tion ofthesphere byaplane through thediameter (Ox)about which themoment istobecalculated. Ifpisthedensity ofthematerial, themass 'ofthediskbetween planes atdistances #,x+dxfrom thecenter is piry2dx, where yistheradius ofthedisk.By(7.10G), themoment of inertia ofthedisk is dl=^Trpy4dx. Buty2=a2#2 ,andsoFIG. 82. Solid sphere split intothin circular disks forthe calculation ofmoment of inertia. But=ITTPfa (-X2 Ja m= andsothemoment ofinertia ofthesphereis (7.107) I=|ma2 . Exercise. Showbythetheory ofdimensions, without calculations, that theabove moments ofinertia ofthehoop, therod,thecircular disk, the circular cylinder, andthesphere areallnecessarily oftheformCmaz ,where Cisapurenumber. Theorem ofparallel axes. Thetheorem ofparallel axes gives usaneasymethod ofcal- culating themoment ofinertia ofasystem about any line, when themoment ofinertia about aparallel linethrough the mass center isknown. Figure 83shows aprojection onto a 192 PLANE MECHANICS [SEC. 7.1 plane perpendiculartothetwo lines, 0'being theprojection of thelinethrough themass center and theprojection ofthe other line. Introducing parallel co- ordinate axes asshown, let(a,6)be thecoordinates of0'relative to0. Ifx,yarethecoordinates ofany point relative totheaxesOxy,and ,x',y'thecoordinates ofthesameO a FIG. 83.Coordinates fortho proof ofthetheorem ofparallel axes.point relative toO'x'y', then (7.108)a, b. With thenotation used atthe beginning ofthissection, themoments ofinertia about thelines through and0'are,respectively, (7109) mt(z?+</?),7'= i-+yft. Then, by(7.108), (7.110) I'+m(az wheremisthetotalmass ofthesystem, since Li n -2a FIG. 84.Rod and sphere: themoment ofinertia aboutL isrequired.from thedefinition ofmass center. Wemay state (7.110) inwords as follows: Themoment ofinertia ofa system about anaxisLisequal tothe moment ofinertia ofthesame system about anaxisthrough themass center parallel toL,together with themoment ofinertia aboutLofaparticle witha mass equaltothetotalmass ofthesys- tem,placed atitsmass center. Asanapplication ofthistheorem ofparallel axes,wenotethat by(7.105) themoment ofinertia ofarodoflength 2aabout a linethrough oneend,perpendicular totherod,is SBC. 7.1]MOTION OFARIGIDBODYANDOFASYSTEM 193 Asanother application, suppose wewish tofindthemoment of inertia about the lineLoftheapparatus shown inFig. 84, consisting ofarodofmassmandlength 2a,withasphere ofmassMandradius battached totheendoftherod.From (7.105) and (7.107), combined with thetheorem ofparallel axes,weobtain atonce (7.111) /=ma2+M[ffe2+(2a+fc)2 ]. Theorem ofperpendicular axes. Thetheorem ofperpendicular axes isuseful forthecalculation ofmoments ofinertia ofplane distributions ofmatter. Let Oxyz berectangular axes, and letthere beadistribution of matter intheplane=0.Denoting byA,B,Cthemoments ofinertia about thethree axes,wehave (7.112) A Obviously, (7.113) C-A+B; thisresult constitutes thetheorem ofperpendicular axes. Asanapplication, suppose wewish tofindthemoment of inertia ofarectangular plate ofedges 2a,26,about alinethrough itscenter perpendiculartoitsplane. Taking theorigin atthe center, theaxesOxy parallel totheedges, andtheaxisOzper- pendicular totheplate,wehave, asalready established, (7.114) A=%mb\ B= Hence therequired moment ofinertia is (7.115) C=A+B=m(a2+62 ). More information about moments ofinertia willbefound in Sec. 11.3. Exercise. Themoment ofinertia ofahoop ofmassmandradius a about adiameter is$nia2andthat ofacircular disk ofthesamemass andradius about adiameter isimo2 .Verify these statements. Kinetic energy andangular momentum. Aswehave seeninSec. 5.2,kinetic energy andangular momen- tumplayanimportant partinthedynamics ofsystems. We 194 PLANE MECHANICS [SBC. 7.1 shallnowshowhow these quantities aretobecalculated when thesystemisarigidbodymoving parallel toaplane. Letusfirstsuppose that therigidbodyisrotating about a fixed axis. Let o>betheinstantaneous value oftheangular velocity. Then thekinetic energy ofaparticle ofmassmv situated atadistance r-from theaxis israT2a>2 ,andsothekinetic energy oftherigidbody (supposed toconsist ofnparticles)is (7.116) T=|a where Iisthemoment ofinertia about theaxis. Theangular momentum ofaparticle about theaxis ofrotation iswtr2 o>,and sotheangular momentum oftherigidbodyis (7.117) h=< Letusnowsuppose thattherigidbodynolonger rotates about afixed axisbutmoves inageneral manner parallel toafixed fundamental plane (cf.Sec. 4.2). Letusimagine anobserver traveling withone oftheparticles (A)oftherigidbody and observing themotion oftheparticles relative tohim.Hecan compute arelative kinetic energy andarelative angular momen- tumabout alinethrough Aperpendicular tothefundamental plane, using inthese computations thevelocities oftheparticles relative tohim. Since relative tohimthebody rotates withang- ularvelocityo>about afixed axisthrough A,theformal calcu- lations areprecisely asabove andlead toformulas (7.116) and (7.117) fortherelative kinetic energy andangular momentum. Although this istrue foranyparticle Aofthebody, theresults aremost useful whenAisthemass center. Letusrestate them: The kinetic energy andangular momentum, both relative tothe masscenter, ofarigid bodymoving paralleltoaplane are (7.118) T=i/co2 ,h=Jw, where coistheangular velocity ofthebodyandIitsmoment of inertia about anaxisthroughthemass center perpendiculartothe plane ofmotion. Thefollowing theorem ofKonig enables ustocomplete the calculation ofthekinetic energy ofarigidbodymoving parallel toaplane. Weshallproveitingeneral three-dimensional form. SEC. 7.1]MOTION OFARIGIDBODYANDOFASYSTEM 195 THEOREM OFKONIG. Thekinetic energy ofamoving system is equal tothesum of(i)thekinetic energy ofafictitious particle moving with themass center andhaving amass equaltothetotal mass ofthesystem and(ii)thekinetic energy ofthemotion relative tothemass center. LetOxyzbefixed axesandO'x'y'z' parallel axesthrough the mass center. Let x,y,zbethecoordinates ofthemass center referred toOxyz. Then, foranyparticle, wehave (7.119) xt=x+x't,.yt=y+y'i} z,=z+zj. Thekinetic energy ofthesystem is (7.120) T= Letusdifferentiate (7.119) andsubstitute forz,yifztin(7.120). Certain terms vanish onaccount oftherelations (7.121) Ym&=2)mtf %= ff\ ff\ i which areconsequences ofthedefinition ofmass center given in Sec. 3.1.Weobtain (7.122) T=im(x2+#2+I2 )+^i)mt(x?+y?+z?), wherem=Vmt,thetotal mass ofthesystem. Thus the theorem isproved. Itisconvenient torefer tothekinetic energy ofthefictitious particle asthe"kinetic energyofthemass center," sothat ourresult reads (7.123) T=T+T', whereTQisthekinetic energyofthemass center andT'the kinetic energy relative tothemass center. This general result holds eventhough thesystemisnotarigidbody. Inthecase ofarigidbody,wehave theimportant formula (7.124) T 196 PLANE MECHANICS [Sic. 7.2 wherem=mass ofbody, q=speed ofmass center, /=moment ofinertia about mass center,* co=angular velocity. Exercise. Find thekinetic energy ofadisk ofmassmandradius o, rolling along theground withspeed q. 7.2.RIGIDBODY ROTATING ABOUT AFIXED AXIS General methods. InSec. 5.2wedeveloped theprinciple ofangular momentum (5.214) andtheprinciple ofenergy (5.223). Letusinsert inthese equations thevalues ofhandTgiven in(7.117) and (7.116); thenwehave (7.201)/co=N (principle ofangular momentum), (7.202) l/co2+V=E (principle ofenergy). These equations represent thetwofundamental methods of finding themotion ofarigidbody which turns about afixed axis. Letusrecall themeanings oftheterms: /=moment ofinertia about thefixed axis, w=angular velocity,N=moment ofexternal forces about thefixedaxis, ortorque. V=potential energy, E=total energy (aconstant). Itmust beremembered that (7.201) isalways valid; (7.202), ontheother hand, holds onlywhen thesystemisconservative. Itwould nothold, forexample,ifthere wereafrictional torque. Flywheels. Letusconsider aflywheel rotating about afixed axiswhich passes throughitsmass center. Gravity contributes nothing to themoment about theaxisandsodoesnotinfluence themotion. Wesuppose thetorqueNsupplied byamotor orbrakes. Since theforces involved here willnot,ingeneral, beconservative, we use(7.201) astheequation ofmotion; sowewrite (7.203)Jco=N. Wemaynote theresemblance between thisequation andthe *More precisely, themoment ofinertia about thelinethrough themass center perpendiculartotheplane ofmotion. SEC. 7.2]MOTION OFARIGIDBODYANDOFASYSTEM 197 equation ofmotion ofaparticle moving onastraight line, mu=X; mass corresponds tomoment ofinertia, linear velocity toangular velocity, force totorque. Thismathematical similarity maybe used tosolve aproblem inthedynamics ofarotating flywheel, when thesolution oftheanalogous problem foraparticle moving onastraight line isalready known. IfthetorqueNisconstant, (7.203) gives (7.204)co=yt+A, whereAisaconstant ofintegration; hence,if istheangle turned through, wehave 6=coand (7.205)=^*2+At+B, whereBisanother constant ofintegration. This motion is analogous tothemotion ofaparticle under aconstant force. Theimportance oftheflywheel inmachinery liesinitscapacity tosmooth outmotion. Inasteam orgasoline engine thetorque isnotuniform, andwithout aflywheel (orsomething equivalent) themotion would bejerky. Asanillustration, letuswork out thecasewhere aflywheel ofmoment ofinertia 7isunder the action ofatorque withafluctuating part, N=NQ+Nicosct, (No,Niconstants), andaload proportional toangular velocity. Theequation of motion is (7.206) 7w=No+Nicosct-Lo>, thelastterm corresponding totheload; thus (7.207) w+-ju=~(No+Nicosct). This isastandard type ofdifferential equation, which hasthe integrating factor eu/I ;thesolution is (7.208)o)=e~Lt/I IA+jfeLt"(N Q+Nicos ct)dt\, whereAisaconstant ofintegration. UsingRtodenote "real part of,"wehave 198 PLANE MECHANICS [Sec. 7.2 (7.209)J*ew"cosctdt=RJ* _.e(L//+tc) =72 =eLtR^7^72(cos*+*sin ct} _cn//cos (ct+ ) ((L/iy+c']*' where isaconstant; itsvalue isofnopresentinterest. Hence, (7.210)=Ac++c, ]tcos (ct+.). The firstterm diesaway astincreases. The ratio oftheampli- tude ofthethirdterm tothesecond term is (7.211)~ Byincreasing themoment ofinertia oftheflywheel, wecanmake thisratio assmall asweplease andsoapproximate tothesteady motion co=N/L,eventhough thefluctu- ating partNicos ctmaybegreater inmag- nitude than thesteady torque No. Thecompound pendulum. InSec. 6.3,wediscussed thesmall oscil- lations ofasimple pendulum, consisting of aheavy particle attached toafixed pointby alight string. Weshallnow discuss the compound pendulum, which isarigidbody freetooscillate under theinfluence ofgrav- ityabout afixed horizontal axis.Weshall usetheequationofenergy (7.202), butthe results maybeobtained with equal ease from (7.201). InFig.85theplaneofthepaperistheplane through the mass centerCperpendiculartotheaxis ofsuspension. Theaxis cuts itat0,which iscalled thepoint ofsuspension. Weshall usethefollowing notation: a=OC,m=mass ofpendulum, ke=radius ofgyration about(7, fco=radius ofgyration about 0.Fio. 85.Acompound pendulum. SEC. 7.2]MOTION OFARIGIDBODYANDOFASYSTEM 199 If6denotes theinclination ofOCtothevertical, thepotential energy is V=mgacos0, and(7.202) gives (7.212) $mk* Q6*-mgacos6=E, whereEisaconstant. This equation gives theangular velocity atany position, whenEhasbeenfound from theinitial conditions. Forexample,ifthependulum starts withCdirectly below andwithangular velocity w,wehave E=Tfmkfa* mga] equation (7.212) gives ffyO1O\ ^i2 2 "J/**1*91/1(7.213)2=cog -TJ-sin2 TfO. This willvanish when takes thevalues a,where andsothependulum oscillates through therange (,). Since sin2 -^0cannot exceed unity,itisevident from (7.213) that 6never vanishes if ifstarted with suchanangular velocity, thependulum travels right around. Differentiation of(7.212) gives (7.214) klS+gasin=0, asanalternative form fortheequation ofmotion ofacompound pendulum. Hadweused (7.201), weshould have obtained this equation directly without differentiation. Forsmall oscillations, wereplace sin6by$andobtain the solution (7.215)=acos(pt+e), where a,areconstants ofintegration andp2*ga/k\. This 200 PLANE MECHANICS [SEC. 7.2 isasimple harmonic motion with periodic time (7.216) T=?*= P The simple pendulumisaspecial case ofthecompound pendulum; forasimple pendulum oflength Z,wehave ko= Z, a= Z,andso(7.214) gives (7.217)16+gsin6=0, asthegeneral equation ofmotion ofasimple pendulum. The equation (6.304) wasvalid only forsmall oscillations. Ifwecompare themotion ofacompound pendulum, givenby (7.214), withthemotion ofasimple pendulum, given by(7.217), wenote that thetwoequations aremathematically identical provided that (7.218)I= ^- Thus, corresponding toanycompound pendulum, wecancon- struct asimple pendulumoflength givenbythisformula, which willoscillate inunison withthecompound pendulum;itiscalled theequivalent simple pendulum. Exercise. Show thatasquare plate ofside26suspended fromonecorner oscillates inunison withasimple pendulum oflength (4\/2/3)6=1.896. Letusnowsuppose thatarigidbodyisgiven, withanumber ofthin parallelholes drilled throughit.Wecanform acom- pound pendulum bypassing anaxis ofsuspension through any oneoftheholes. Howdoestheperiodic time ofsmall oscillations depend onthepositionoftheholechosen? Toanswer this question, wenote that,bythetheorem of parallel axes (7.110), (7.219) kl-a2+/c?. Hence theformula (7.216)fortheperiodic timemaybewritten (7.220) r2=- Ifwechange thepositionofthepointofsuspensioninthebody, achanges butkcdoes notchange. Weseethat Ttends to SBC. 7.3]MOTION OFARIGIDBODYANDOFASYSTEM 201 infinityifatends tozero orifatends toinfinity. Thus, we canobtain very slow oscillations bymoving thepoint ofsus- pension close tothemass center orfarfrom it. Ondifferentiating (7.220) with respect toa,weget df2, 47T2A *J\j~(T)==-11--1rdav'g\ a2/ Thus theperiodic time isaminimum when thepoint ofsuspension isatadistance fromthemass cen- terequaltotheradius ofgyration about themass center. Wenowaskwhether itispOS-FIG. 86.Theperiodic time isthe sible toshiftthepointofsuspen-m'toU8pd "8 sionfrom apositiontoanew position Of (Fig. 86)onthelineOCwithout changing theperiodic time. WithOC=a,CO'=b,thecondition forequality of periodic timesis,by(7.220), fr2If2 IW/C J,L C+^=b+r which issatisfied if (7.221)ab=fc2- Thepoint 0',related inthisway tothepointofsuspension 0, iscalled thecenter ofoscillation. 7.3.GENERAL MOTION OFARIGIDBODY PARALLEL TOAFIXED PLANE General methods. Probably themost useful principleavailable forthesolution ofproblemsinmechanics istheprincipleofenergy intheform (5.223), namely (7.301)T+V=E. Onemust ofcourse make sure, before attempting toapply this principle,that thesystemisconservative, i.e.,that ithas apotential energy V. When thesystem consists ofasingle rigidbodymoving parallel toafixed plane, wemay write (7.301)intheform (7.302) im<Z2+$Io>2+V=E, 202 PLANE MECHANICS [Sac. 7.3 wherem=mass ofbody, q=speed ofmass center, /=moment ofinertia about mass center, a)=angular velocity ofbody. However, (7.302)isonlyoneequation. Sometimes werequire more equations, andthenwemayemploy theprinciples of linear andangular momentum intheforms (5.209) and (5u219). IfOxyarefixed axes inthefundamental plane, wehave (7.303) mx=X,my=7,/ci=N, where x,yarethecoordinates ofthemass center, X,Yarethe totalcomponents ofexternal forces inthedirections oftheaxes, andNisthetotalmoment ofexternal forces about themass center. Ofthefour equations (7.302) and (7.303), atmost three areindependent. Cylinder rolling down aninclined plane. Consider acylinder ofmassm andradiuso,rolling down aplane inclined atanangleatothehori- FIG.87-^^^down anzontal (Fig. 87).Wewish tode- termine themotion, andasanil- lustration weshalldosobytwomethods, firstusing theprinciple ofenergy andthen theprinciples oflinear andangular momen- tum.Weassume themass center tobesituated onthegeo- metrical axisofthecylinder. Letxbethedisplacement attime tofthecenter ofthecylinder from itsinitial position atrestatt=0,and 6theangle through which ithasturned. Then, bythecondition ofrolling, (7.304) x=aO. IfA;istheradius ofgyration ofthecylinder about itsaxis,its kinetic energyis (7.305) T=\m&+ or,by(7.304), (7.306) SBC. 7.3]MOTION OFARIGIDBODYANDOFASYSTEM 203 Thepotential energy is (7.307) V=mgx sina. Hence, by(7.301), (7.308) & 2)x*mgx sina whereEistheconstant total energy. Actually E=0,since x== for t=0.Differentiating (7.308) with respect tot, weobtain (7.309)gsin a. Thus thecylinder rollsdown theinclined plane withaconstant acceleration. Aparticle would slidedown asmooth plane ofinclination a withanacceleration gsina.Thevalue given by(7.309) is alwayslessthan gsin#,except inthelimiting casek=0, which corresponds toaconcentration ofallthemass ofthe cylinder onitsaxis. Ifthecylinder isathin shell,wehavek=a,and hence (7.310) x=sna. Ifthecylinderissolidanduniform, wehave k2=a2 ,andhence (7.311) x=|0sina. FIG. 88. External forces acting Themethod ofenergy doesnotoncyhnder - tellusthereaction between thecylinder andtheplane, orhow rough theplane must beinorder that slipping maybeavoided. Tofindoutthese things, weturn totheprinciples oflinear and angular momentum, using (7.303). Thereaction oftheplane onthecylinder mayberesolved into anormal component Nandacomponent Fintheplane (Fig. 88). These forces, with theweight mg,form thecomplete system of external forces. Thus, wehave (7.312)'mx=mgsinaF,=mgcosaN, [mk*'6=Fa, 204 PLANE MECHANICS [Sue. 7.3 thesecond equation coming from resolution perpendicular tothe plane. Using (7.304) andeliminating F,weget gsina asin(7.309). Hence thecomponents ofthereaction are /^o^\ rfc2 .mgk2sina ,, (7.314) F=m-5x=%,iN=mgcosa. fl O~t~ AC Inorder that rollingmayoccur without slipping, wemusthave F/N^M,or /^oie\ -^fc2tana (7.315) p> - , where/*isthecoefficient ofstatic friction asin(3.202). Self-propelledvehicle. Consider anautomobile (Fig. 89).Theexternal forces acting onitare (i)gravity, (ii)thereactions oftheground onthewheels, (iii)resistance ofthe air. Without knowing anyfurther details, wecanapply theprinciple oflinearmomentum intheform (5.209) tothecomplete auto- Fio. 89.Automobile. mobile. Ifitsmass ismand itistraveling onahorizontal road with acceleration f,thenmiequals thetotal horizontal component ofground reactions and airresistance. The total vertical component ofgravity, ground reactions, and airresistance is zero. Application oftheprinciple ofangular momentum inthe form (5.219) requires alittle care. For simplicity, weshall suppose that thewheels havenomassandhence noangular momentum. Theangular momentum oftheautomobile about SEC. 7.3]MOTION OFARIGIDBODYANDOFASYSTEM 205 itsmass center isthen zero. Hence thetotalmoment about the mass center ofground reactions and airresistance iszero. Wecanusetheabove results tofindthegreatest possible acceleration ofanautomobile onastreet forwhich thecoefficient offriction between ground and tire isju.Since, byhypothesis, themass ofeach wheel iszero, therate ofchange ofangular momentum ofawheel about itscenter iszero; hence thetotal moment ofexternal forces onawheel iszero. These forces consist ofaforce exerted bytheaxle, acouple duetoengine or brakes, andaground reaction. Ifthecoupleisabsent, the ground reaction canhavenomoment about thecenter ofthe wheel. Infact, inthecaseofawheel without mass, undriven and unbraked, theground reaction hasnofrictional component. No FIG. 90. External forces acting onautomobile. Figure 90shows theexternal forces acting onanautomobile, driven through therearwheels ontheright, airresistance being neglected. Leth=height ofmass center above ground, 61=distance offront axleinfront ofmass center, 62=distance ofrearaxlebehind mass center, Ni resultant ofvertical reactions attwofront wheels,#2=resultant ofvertical reactions attworearwheels, F=resultant offrictional forces attworearwheels. Then, (7.316)tntf=F, )=tfi+N (=bzNz--mg, iJfi-hF. Solving forNi,Nz,F,weobtain (7W^ N.-mgb*"^ NV/.oi/; ivim-r ,T>M F=mf. Bythelawofstatic friction (3.202)F/N 2^M,andso 206 PLANE MECHANICS [Site. 7.3 or (7.319)+62 This fraction represents the greatest acceleration possible without slipping. Themaximum negative acceleration obtainable bytheapplica- tionofbrakes, without slipping between thetiresandtheground, maybefound inasimilar way. Internal reactions. Theprinciplesoflinear andangular momentum donotinvolve theinternal reactions between theparticles ofarigid body. Nevertheless these reactions exist, andwhen theybecome excessive thebodymay break. Asweshallnow see,theprin- ciples oflinear andangular momentum maybeused tofindthe reactions. D'Alembert's principle (cf.Sec. 5.2)may alsobeused. o o FIG. 91. (a)Arodrotating about oneond. (6)Reactions ontheportion BA. The essential point tonote isthat,when internal reactions aresought, thedynamical system considered isonly partofthe rigidbody.Weshall illustrate themethod withanexample. Figure 91ashows auniform rodOArotating aboutOwith angular velocity w,which weshall firstsuppose tobeconstant. Bisanypoint intherod.Weseekthereaction across thesection oftherodatB.AsinSec. 3.3,thereaction exerted byOBon BAconsists ofatension T,ashearing force S,andabending momentM(Fig. 916). Letusregard gravity asnon-existent. Then T,S,Mconstitute thewhole system ofexternal forces acting onBA. LetOA= Z,OB r.The acceleration ofthemass center ofBA isofmagnitude $(l+r)w2 ,directed along AB. Ifm SBC. 7.4]MOTION OFARIGIDBODYANDOFASYSTEM 207 isthemass oftherod,themass ofBA ism(l r)/L Hence the principle oflinear momentum, applied toBAasadynamical system, gives (7.320) T-*^f-(I2-r), 5=0. Theangular momentum ofJ5Aabout itsmass center isconstant. Hence, bytheprinciple ofangular momentum, M=0.Thus, when wisconstant, thereaction intherodatBisatension, as given by(7.320). Asacheck, wenote thatT=forr= I, andT=$mo)H forr=0. Letusnowconsider themore general casewhere wisavariable function of t.By(4.107) theacceleration ofthemass center of BAhascomponents (7.321) %(l+r)w2alongAB, %(l+r)ciperpendicular toAB. Hence, (7.322) T=*^(Z2-r2 ),8=i^ (J2-r2 ). Iffcistheradius ofgyrationofBAabout itsmass center, the angular momentum ofBAabout itsmass center ismk*u(l r)/L Hence, bytheprincipleofangular momentum, (7.323) M-iS(l_r)-(I-r). Thus, (7.324) M=OTti>a z~r) [**+W-r*)]. But fc2=Ad-r)f , andso (7.325) M=i?y(I-r)*(2J+r). Exercise. Findwhere therod ismost likely tobreak, assuming that this occurs where thebending moment isgreatest. 7.4.NORMAL MODES OFVIBRATION Degrees offreedom. Theposition ofasimple pendulumisdetermined bythevalue ofonevariable, namely,itsinclination tothevertical, orthe horizontal component ofthedisplacement ofthebob.Asystem 208 PLANE MECHANICS [SBC. 7.4 whose position maybespecified byonevariable orcoordinate issaidtobeasystem with onedegree offreedom. Arodwhich canmove inaplane, withoneendconstrained to move onafixed line,canbedescribed astoposition bytwovari- ables, namely, thedistance oftheconstrained endfromafixed point ontheconstraining lineandtheinclination oftherodto the line. Each ofthese variables cantake arbitrary values. Asystem whose position maybespecified bytwoarbitrary and independent variables orcoordinates issaidtobeasystem with twodegrees offreedom. Similarly, there aresystems withndegrees offreedom, where n=3,4,--; InSec.6.3wediscussed theoscillations orvibrations ofasimple pendulum andinSec.7.2those ofacompound pendulum. Each ofthese isasystem withonedegree offreedom. Wenowproceed todiscuss systems withtwodegrees offreedom. Particles onastretched string. Letthere bealight elastic string oflength 3a,stretched between points A,B.Lettwo particles, each ofmass m,be a a FIG. 92.Loaded string vibrating. attached tothestring atthepointsoftrisection. Forsimplicity, weshall neglect gravity; or,equivalently, wemaysuppose the particles supported onasmooth horizontal plane. Initially theparticles areatrestandthetension inthestring isaconstant (S)throughout. Theparticles aregiven small dis- placements perpendicular tothestring andthen released. We wish toinvestigate theresulting oscillations. Figure 92shows thesituation attime t.The particles are atCandD,their displacements from thepositions ofequilibrium being denoted byxand y,which aresmall quantities. The inclinations oftheportionsofthestring toABaresmall, ofthe same order asxand y.Hence, since thecosine ofasmall angle differs from unitybyasmall quantityofthesecond order, itisseenthatthelengths AC,CD,DBareeach equal toa,to SBC. 7.4]MOTION OFARIGIDBODYANDOFASYSTEM 209 thefirstorder ofsmall quantities inclusive. Hence thetensions inthese portions areequal toStothisorder. Resolving forces inthedirection perpendicular toAB,we obtain thefollowing equations ofmotion: (7.401) mx=-2-s5JZJ>, my=S5-H2-Sa a a a Writing (7.402) k*=,'ma wesimplify these equations to (7.403) *+2k2x-kzy=0,-k2x+y+2k2y=0. Wetrysolutions oftheform (7.404) x=Acos(nt+e), y=Bcos(TI+*), where-4,#,n,eareconstants. The equations (7.403) are satisfied provided A,B,nsatisfy theequations (7-405)l_Ai*+*(-+2fc)=0. Elimination ofAandBgives thedeterminantal equation n2-2k2k* k2n2-2k2 (7.406) or (7.407) n4-4fc2n2+3fc4=0; thesolutions aren\,n2,where (7.408) n\=A;2 ,n-3A;2 . When nisknown, either oftheequations (7.405) gives, forthe ratioB/A, (7.409) I=2~F Thus forn=m,5/-A=1;andforn=n2,J5/A=1. Hence ifAi, 1arearbitrary constants, thefollowingisa solution of(7.403): (7.410) x=Aicos(fa+ i), y=Aicos(fa+ 1). 210 PLANE MECHANICS [SBC. 7.4 Also,ifA2,2arearbitrary constants, thefollowingisasolution of(7.403): (7.411) x=A*cos(ktV3+*2), 2/=*-4 tcos(fa\/3+ 2). Thus, (7.410) and (7.411) represent possible vibrations ofthe particles. Themost general vibration isgivenbyadding these expressions, thus: (7.412)x=Aicos(kt+ 1)+Azcos(kt\/3+ 2), 2/=AiCOS(fa+ i)^2COS(fa\/3+ 2). Weknow that this isthegeneral solution, because itcontains four constants ofintegration whichmaybechosen tosatisfy initial conditions corresponding togiven positions and velocities of theparticles att=0. Letussuppose, forexample, thatwhen t=wehave x=y=x=0,v. This corresponds tothecasewhere themotion isstarted by giving ablow toD.Puttingt in(7.412) andtheequations obtained bydifferentiating (7.412), weobtain (7.413)AiCOS 1+AzCOS 2=0, AiCOS 1AzCOS 2=0, sin 1A*\/3sin e2==0, -Aisin 1+A2\/3 si which arefourequations forAi,A2,i,c2.vsm c2=T> A/ Thesolution is (7.414) l=62=fa Ai--^^2 sothatthemotion oftheparticlesisgivenby (7.415)rsin%2-- ^=sin (Art\/3) h ~sin A;+^sin(A;\/3) J- Thismotion iscomplicated, butthesimple harmonic motions ofwhich itiscomposed areeasy todescribe. These simple harmonic vibrations arecalled normal modes ofvibrations. They SBC. 7.4)MOTION OFARIGIDBODYANDOFASYSTEM 211 areexecuted byavibrating system when theinitial conditions areproperly chosen. Thevibration given in(7.415)isnotanormal mode ofvibra- tion,noringeneralisthatgivenby(7.412). But iftheinitial conditions arechosen sothatA2=0,wehave thenormal mode ofvibration (7.410), whereas,iftheinitial conditions arechosen sothatAi=0,wehave thenormal mode ofvibration (7.411). Theperiodic times andfrequencies ofnormal modes arecalled normal periods andnormalfrequencies. Intheabove problem thenormal periods are 27T 27T fc' kV5 When asystemisexecuting anormal vibration,itsconfigura- tions areusually simple todescribe. Thus,inthemode (7.410) wehave x=y,and in(7.411) wehave x=y.Typical configurationsforthenormal modes areshown inFigs.93aand 6. (6) Fio. 93. (a)Loaded string vibrating infirstnormal mode, vibrating insecond normal mode.(6)Loaded string Wehavebeen discussing thevibrations ofaparticular system twoparticles onataut string. Allproblems ofvibration have certain features incommon; andalthough weshallnotattempt here toprove these facts,itwillbeuseful tosumthemup: (i)Avibration mayberegarded asasuperposition, oraddi- tion, ofsimple harmonic vibrations. (ii)Each simple harmonic vibration iscalled anormal mode of vibration. Itispossible tomake asystem vibrate inanormal modebystarting with suitable initial conditions. (iii)The periods andfrequencies ofthenormal modes are called thenormal periods andfrequencies. 212 PLANE MECHANICS [SEC. 7.4 (iv)Thenumber ofnormal modes isequal tothenumber of degrees offreedom ofthesystem. (v)Thenormal periods arefound bysolving adeterminantal equation, e.g., (7.406). Vibrations ofaparticle inaplane. Asanother exampleofavibrating system withtwodegreesof freedom, letusconsider aparticle which moves inaplanein afield offorce such thatthepotential energy perunitmass is (7.416) V=i(az2+2hxy+by*), where a,hjbareconstants. The force components perunit mass arethen (7.417) X=-(ax+hy), Y=-(hx+by). These vanish attheorigin, which istherefore aposition of equilibrium. Theequationsofmotion are (7.418) x=-ax-hy, y=-hx-by. Trying asolution (7.419) x=Acos(nt+e), y=Bcos(nt+c), weseethat (7.418) aresatisfied provided A,B,nsatisfy (7420) (A(n*-a)-Bh=0, ^ }\-Ah+B(n*-b) =0. Hence, nmust satisfy thedeterminantal equation 2-a -h -h n2-b or (7.422) n4-n\a+6)+ah-W=0; thesolutions aren\,HI,where(7.421)=0, 6)-V(a-6)2+4J. Ifoneofthese values should benegative, thecorresponding n would beimaginary, andthesolution (7.419) would contain hyperbolic instead oftrigonometrical functions. This case will SEC, 7.4]MOTION OFARIGIDBODYANDOFASYSTEM 213 bediscussed inSec. 7.5;forthepresent, weassume that ni,na arereal. Then thenormal periods are2rr/ni, 2ir/n 2,andthenormal modes ofvibration are vft*ct (7.424) x=Aicos(nit+ 1), y=-^r AIcos(nrf+ i), and (7.425) x=Azcos(nzt+ 2), y=^-^A2cos(nzt+e2), where Ai,A2,ei,c2arearbitrary constants. Thegeneral motion isfound byadding thesolutions (7.424) and(7.425), justaswe added (7.410) and (7.411). Thepreceding discussion isactually more general thanmight appear. Letusagain suppose thataparticle moves inaplane under aconservative force system, with potential energyVper unitmass; butinstead ofassuming thesimple expression (7.416) fory,weshallmerely assume that itisafunction which canbe expanded inaTaylor series. LetXo, 2/0beapositionofequilibrium. Since thecomponents offorcemust vanish there, wehave ().-* thesuffix zero indicating evaluation atx=XQ,y=y$. Ifwe expand VinaTaylor series about x,2/0,twoterms intheexpan- sionvanish onaccount ofthese equations, andso (7.427)F-F.+ (y-,). theterms notwritten being ofahigher order ofsmallness if xxQ,y 2/0aresmall. NowVisalways undetermined towithin anadditive con- stant; there istherefore nolossofgenerality inputting V=0. Ifweshift theorigin totheposition ofequilibrium (z ,^o)and define a,A,6by 214 PLANE MECHANICS [Sic. 7.5 theprincipal partof7forsmall values ofxand#is (7.429) V=i(az2+2hxy+fey2 ), which isformally thesame as(7.416). Thedeductions based on (7.416) were exact; thesame formal deductions hold approxi- mately forsmall vibrations about anypositionofequilibrium. Thenormal modes ofvibration aregivenby(7.424) and(7.425), where ni,n2aregivenby(7.423), a,ft,6having thevalues (7.428). 7.6.STABILITY OFEQUILIBRIUM Apositionofequilibrium foranysystemissaid tobestable when anarbitrary small disturbance doesnotcause thesystem to depart farfrom thepositionofequilibrium. Otherwise,itis unstable. By"small disturbance" wemean that, attheinitial instant, theparticlesofthesystemaredisplaced from their positionsofequilibrium through small distances and their velocities aresmall. Thesystemisstable ifintheresulting motion theparticles remain atsmall distances from their positions ofequilibrium. Thus acompound pendulum hanging from itsaxisofsupport isinstable equilibrium.Ifitisbalanced with itsmass center above theaxisofsupport, theequilibriumisunstable, because a small disturbance willcause thependulum tomove rightaway from thepositionofequilibrium. Condition ofminimum potential energy. Letasystem have apotential energy V.Weknow bythe principleofvirtual work (cf.Sec. 2.4) that, forasystem in equilibrium, nowork isdone inasmall displacement. Thus 5V=foranysmall displacement fromapositionofequilibrium, andsoVhasastationary value there. Stationary values areofvarious kinds; thequestion ofstability turns onthecharacter ofthestationary value of7.Wemake thefollowing statement: //,inaposition ofequilibrium,the potential energy isaminimum, then theequilibriumisstable. Toprove this,letusrecall theprinciple ofenergy, (7.501) T+VjB, SBC. 7.5]MOTION OFARIGIDBODYANDOFASYSTEM 215 whereEisaconstant. Since potential energyisalways undeter- mined towithin anadditive constant, there isnolossofgenerality inassuming V=attheposition ofequilibrium. Then, since Visaminimum there,wehaveV>forallpositions nearthat ofequilibrium. Theconstant Eisfound from thesmall initial disturbance. Let To,VQbethe initial kinetic and potential energies. ThenE=To+F,which issmall andpositive. In thesubsequent motion, (7.502) V=E-T<E, sinceTcannot benegative. ThusValways remains lessthan thesmall positive constant E,andsotheequilibriumisstable, since toescape toafinite distance fromtheposition ofequilibrium thepotential energy would have tobecome finite. Asanillustration, consider asimple pendulum ofmassmand lengtha.Ataninclination tothedownward vertical, the potential energyis V=mga(l cos6), ifwechooseV=attheposition inwhich thependulum hangs vertically. ThenVisaminimum for=0.Suppose thatthe pendulumisdisturbed toanangle and isgiven akinetic energy To.Inthesubsequent motion, asin(7.502), mga(l cos 6)<To+mga(l cos), where theright-hand side issmall. Thus cos6must remain nearly equal tounity, or,inother words, must remain small. If,ontheother hand,weconsider thatposition ofequilibrium inwhich thependulumisbalanced directly above thepoint of support (thestring being replaced byalight rod)andmeasure from thisposition, wehave V=mga(cu8 -1). ThenVisamaximum for=0.Ourinequality (7.502) isstill valid;itreads w0ra(cos 1)^To+mgra(cos 1), where0o,TQrefer totheinitial disturbance. But thisinequality isnotviolated as increases from to?r,andsoitdoesnot 216 PLANE MECHANICS [SBC. 7.5 restrict themotion. The position ofequilibriumisactually unstable, butthisinequality shows onlythat itmaybeso. Thefollowing statement istrue forasystem withanynumber ofdegreesoffreedom, butweshallproveithereonly forsystems withonedegree offreedom: //the potential energy ataposition of equilibrium isnotaminimum, then theequilibrium isunstable.** Letxbethevariable which fixes thepositionofthesystem. Con- sider thegraph ofthepotential xenergyVagainst x(Fig. 94). FIG.94-Graph ofpotential Btheprincipleofvirtual WOrk,energy against position; stable^f r- j equilibrium atAandD,unstable WChave8V= foranmfillltesi- equiiibrium atBandc.maldisplacement dxfromaposition ofequilibrium. Infact, ataposition ofequilibrium (7.503)-Q, sothat thepositions ofequilibrium correspond tothose points onthegraph where thetangentisparallel tothez-axis, i.e.,the pointsAyB,C,D.AtAandD,Visaminimum, andhencewe know thatequilibrium atAorDisstable. Letusnowconsider theposition corresponding toB.Suppose thesystemisdisplaced toaneighboring position B'and isthen released from rest. Since thetangent atB'isnotparallel tothe #-axis, thesystem cannot remain inequilibrium atB'. Itmust start tomove; andsince itskinetic energy (being positive) must increase incomparison with itsinitial zerovalue,Vmust decrease, andsothesystem mustmove stillfarther away from B.Itcan come torestonlywhenVtakes thesame value asatB' .Thus it cannot stopmoving until ithaspassed theposition corresponding toA. Clearly, this isacase ofinstability. AtCthepotential energy hasastationary value, but itis neither amaximum noraminimum. Byconsidering aninitial displacement inthedirection ofD,itisseenthattheequilibrium *Intheparticular casewhereVisconstant (asforasphere resting ona horizontal table), theequilibriumisoften called neutral. Actually,itis unstable inthesense ofourdefinition. SEC. 7.5]MOTION OFARIGID BODYANDOFASYSTEM 217 isunstable. Thiscompletes theproofoftheitalicized statement onthepreceding page. Tosumup:Aposition ofequilibriumisstableif,andonly if,the potential energyisaminimum. Hence, inthecase ofasystem withonedegree offreedom, a sufficient condition forstability is (7.504) >0, attheposition ofequilibrium. This condition isalsonecessary, unless d*V/dx* = attheposition ofequilibrium;inthat exceptional case,wehave toexamine thehigher derivatives. Itisclearfrom Fig.94thatbetween anytwopositionsofstable equilibriumtheremust beatleastoneposition ofunstable equi- librium. Points such asC,where apoint ofinflection onthe graph coincides withatangent parallel tothez-axis, areexcep- tional. Ingeneral, positionsofstability andinstability alternate. Stability ofequilibriumofaparticleinaplane. Itwasshown in(7.429) that, nearapositionofequilibrium, the potential energy perunitmass ofaparticle inaplanemaybe written (7.505) V=$(ax*+2hxy+by*), where a,h,bareconstants. Itisknown, from theanalytical geometryofconies, thatwemay choose newrectangular axes Ox'y' such that (7.506) ax*+2hxy+by*-aV2+b'y'*, wherea',b'arenewconstants. Letusrecallhowa',bfarefound. Foranyconstant value ofX, (7.507) ax*+2hxy+by*-\(x*+y*)=a'x'*+b'y'*-X(z'2+y'*). IfX=a'orX=&',theright-handside isaperfect square. For either ofthese values ofX,theleft-hand sidemust alsobea perfect square. Thus,ifX=a'orX=&', (7.508) (a-X)(6-X)-h*=0, or,indeterminants! form, aX h h b-X(7.509)0. 218 PLANE MECHANICS [Sue. 7.5 Infact, a'andVaretheroots ofthisquadratic equation. Suppose thetransformation carried out, sothat, near the position ofequilibrium, (7.510) V-i(aV+6V2 ). Theequations ofmotion are Wxdx7*~aV' (7.511) f '' or (7.512)x'+aV=0, #'+Vy'=0. Various caseshavenowtobedistinguished: (i)a!>0,x'=Acos(Vo7+Bsin(Vo7-0, (ii) a'=0,x'=A*+5, (in) a'<0,a;'=Ae^'^'-' +Ber'S^''*. These arethesolutions ofthe first of(7.512), according tothe sign ofa! .Thesolutions ofthesecond equation fory'maybe similarly classified according tothesign ofb'. Thesolution forcase(i)indicates that x'remains permanently small, sothat there isstability asfarasxrisconcerned. The solutions forcases(ii)and (Hi)indicate instability. Thus, there isstability if,andonly if,both a'and b'arepositive; since a',b1 aretheroots of(7.509), wemay state ourresult asfollows: When theorigin isaposition ofequilibrium,thepotential energy foradjacent positions isgiven by(7.505). Theequilibrium is stableif,andonly if,thetworoots ofthedeterminantal equation (7.509) arepositive. Itisclearfrom (7.510) thatVisaminimum attheorigin if,and only if,theroots of(7.509) arepositive. Hence, wehaveadirect proof inthiscasethatminimum potential energyisthecondition forstability, both necessary and sufficient. Inthe case ofstability, oscillations along theaxes of x'and y'arenormal modes andthenormal periodstiro where Xi,X2aretheroots of(7.509). SBC, 7.5JMOTION OFARIGIDBODYANDOFASYSTEM 219 Problems ofbalancing. Theoreticallyitispossible tobalance aneedle onitspoint, but inpracticeitisextremely difficult todoso.There isaposition ofequilibrium with theneedle vertical, but itisunstable. The instabilityisobvious inview ofthegeneral testgiven above, because theheight ofthecenter ofgravityisdecreased asthe needle isdisplaced from theverticalposition, andsothepotential energyisamaximum fortheverticalposition. (a) FIG. 95. Cylindrical body rolling onahorizontal piano: (a)position ofequi- librium, (b)displaced position. If,instead ofaneedle, wetrytobalance abody witharounded base,itisnotimmediately evident whether theequilibrium inthe position ofbalancingisstable ornot.Butthecondition of minimum energy gives usaneasy test.Weshall confine our attention tocases where thepossible motion ofthebodyis two-dimensional. Figures 95aandbshow endviews ofacylindrical body in contact with arough horizontal plane; thelower part ofthe section isacircular arcofradius a.Inequilibrium (Fig. 95a) thecenter ofgravity Cmust lievertically above thepoint of contact, since thebodyisinequilibrium under twoforces (the weight andthereaction) andtheir lines ofaction must coincide. Lethbetheheight ofthecenter ofgravity above thepoint of contact. Figure 956shows adisplaced position, inwhich thebody has beenturned through anangle0.IfWistheweight ofthebody, thepotential energyis (7.513) V=W[a-(a-h)cos0]. Hence, <7-514> w(0~ 220 PLANE MECHANICS [SEC. 7.5 ifthe Thus theequilibriumisstableif,andonly if,a>h,i.e. center ofgravityliesbelow thecenter ofthecircle. Letusnowconsider amore general problem, which includes the preceding asaspecial case. LetAbeacylinder ofanysection, balanced onafixed cylinder A',thecontact being rough andthe common tangent horizontal (Fig. 96a). Cisthecenter of gravityofA.D,D'arethecenters ofcurvature ofthesections ofthecylinders atthepoint ofcontact, andp,p'aretheradii of curvature; theheight ofCabove thepoint ofcontact ish. (a) (6) FIG. 96.One cylindctioiling onanother: (a)position ofequilibrium, (b)dis- placed position. Figure 96&shows adisplaced position, inwhich thepoint of contact hasmoved through asmall angle0'about D',andthe lineDCnowmakes asmall angle6withDD'. Sincewearecon- cerned onlywithsmall values ofand0',itislegitimate toregard thesections intheneighborhood ofthepoint ofcontact ascircular arcs ofradii pand p'.Then, bythecondition ofrolling, (7.515) p0=p'0'. IfWistheweight ofA,thepotential energyis (7.516) V=W[(p'+p)cos 0'-(p-h)cos(0+0')], SEC. 7.5]MOTION OFARIGIDBODYANDOFASYSTEM 221 or,since0,0'aresmall, (7.517) V-W[(p-h)(9+O'Y-(p'+p)0'2 ]+C, approximately, whereCisaconstant. By(7.515), thismaybe written (7.518) V=iW [(P- /i)(l+p ^)2- (P'+ P)]+C. Thecondition that thisshallbeaminimum for 6'= is (7.519) (p- Ji) (l+^y-(p'+p)>0, or,equivalently, (7.520) h<-p ^~,-PT-P This isthecondition ofstability. Onletting p'> ,itreads /t<p,agreeing with theresult established earlier. Ifonthe otherhandweletp<*> ,wegeth<p'asthecondition forthe stability ofabody withaflatbasebalanced onacylinder with radius ofcurvature p'. Bymeans oftheprinciple ofenergy,itiseasy tofindtheperiod ofsmall oscillations ofastable balanced system when disturbed. Forexample, thebodyshown inFig.95hasthepotential energy given by(7.513). Toconvert todynamical units, weputW mg,wheremisthemass ofthecylinder. Thus, when is small, thepotential energyisapproximately (7.521) V=mgh+$mg(a- Since, atany instant, thecylinderisturning about thelineof contact, thevelocity ofthemass center isapproximately hi;the angular velocity ofthecylinderis6.Hence (7.302) gives (7.522) \mWfr+i/02+mgh+mg(a-h)0*=E, where 1isthemoment ofinertia about alinethrough Cparallel tothegenerators.Ifwewrite (7.523) F=h*+^, anddifferentiate (7.522), weget (7.524) W+g(a-K)B=0; 222 PLANE MECHANICS [SEC. 7.6 thisgives asimple harmonic motion with period (7.525)r=2irk/Vg(a-h). 7.6.SUMMARY OFAPPLICATIONS INPLANEDYNAMICSMOTION OFARIGIDBODYANDOFASYSTEM I.Moments ofinertia. (a)Definitions: r?or/=/r2dm=J7J"p(z2+t/2 )dxdydz] (7.601) /= (6)Devices forcalculation: (i)Theorem ofparallel axes (allmoments ofinertia follow immediately when those foraxesthrough mass center areknown). (ii)Theorem ofperpendicular axes (foraplane distribution, themoments ofinertia about axesperpendicular tothe plane follow immediately when those foraxes inthe plane areknown). (c)Standard results: II.Kinetic energy (T)andangular momentum (K). (a)Rigidbody turning about fixed axis: (7.602) T=i/co2 ,h=/o>. Ex.VII]MOTION OFARIGIDBODYANDOFASYSTEM 223 (6)Rigidbody ingeneral plane motion: (7.603) T=|mg2+i/co2 ,h-/(h,Iabout mass center). III.Motion ofatigid body. (a)Rotation about fixed axis: (7.604)7o>=N (angular momentum), (7.605) i/o>2+V=E (energy). (b)Compound pendulum: (i)Exact equation ofmotion: (7.606) P0+gasin=0, (krelative toaxis), (ii)Periodic time forsmall oscillations: 2irk (7'607)T= VTa (111)Equivalent simple pendulum: (7.608)I=~ (c)General motion parallel toplane: (7.609) mx=X,my=7, /w=N (momentum); (7.610) img2+i/w2+V=E (energy). IV.Normal modes ofvibration. (a)Avibration isingeneral notperiodic;itiscomposed of simple harmonic vibrations with different frequencies. These are thenormal modes. Asystem vibrates inanormal mode if started under specialinitial conditions. (6)Thenormal frequenciesarefound byassuming asimple harmonic solution oftheequationsofmotion andsolving a determinantal equationobtained onthisassumption. V.Stability ofequilibrium. Necessary and sufficient condition forstability: thepotential energyisaminimum. EXERCISES VH 1.Auniform rodoflengthIandmassMisfreetorotate inavertical plane about anaxisatadistance afrom itscenter. Ifitisreleased froma 224 PLANE MECHANICS [Ex.VII horizontal position,find itsangular velocity when passing through the vertical position. 2.Abucket ofmassMisfastened tooneendofalight rope; therope iscoiled round awindlass intheform ofacircular cylinder (radius a)which isleftfreetorotate about itsaxis. Prove thatthebucket descends with acceleration g 1+(//Jl/a2 )' where /isthemoment ofinertia ofthecylinder about itsaxis. 3.Three uniform rodrf, each ofmass w,formanequilateral triangle of side 2o.The triangleissuspended from one corner. Find thelengths oftheequivalent simple pendulumsforoscillations under gravity (i)when thetriangle oscillates initsownplane; (ii)when theplaneofoscillation isperpendiculartotheplane ofthe triangle. 4.Awheel consists ofathinrimofmassMandnspokes each ofmassm, whichmaybeconsidered asthinrodsterminating atthecenter ofthewheel. Ifthewheel isrolling with linear velocity v,expressitskinetic energy in terms ofM,m,n,v. Withwhat acceleration will itrolldown arough inclined planeofinclina- tiona? 5.Abuoyisformed byjoining theedge ofathinmetal conical shell to theedge ofahemispherical shell ofthesame material andthickness. The radii ofthehemisphere andofthemouth ofthecone areeachequal to5ft., andtheslant height ofthecone is10ft.Thebuoyisplaced withthehemi- sphere incontact with therough horizontal surface ofadock sothat the axis isvertical. Ifslightly disturbed,determine whether ornot itwill return tothevertical position. 6.Oneendofaheavy chain isattached toadrum andthechain is wrapped around thedrum, making ncomplete turns, withasmall piece of chain hangingfree. Thedrum ismounted onasmooth horizontal axle, andthechain isallowed tounwrapitself. Apply theprincipleofenergyto findtheangular velocity ofthedrum attheinstant when thechain iscom- pletely unwrapped,interms ofthemass ofthechain (m},theradius ofthe drum(r),andthemoment ofinertia ofthedrum (7). 7.Arectangular plate swings inavertical plane about oneofitscorners. Ifitsperiodis1sec.,findthelengthofthediagonal. 8.Aparticle ofmassmmoves inaplane under theaction ofaforce withcomponents X--k*(2x+y),Y--*(x+2y), where kisaconstant. What isthepotential energy? Find thenormal periodsofoscillation about theposition ofequilibrium. 9.Auniform circular plate ofradius aandmassMisdragged along a smooth sheet oficebymeans ofalong string attached toapointAonthe Ex.VII]MOTION OFARIGIDBODYANDOFASYSTEM 225 rimoftheplate. Thetension Tinthestring iskeptconstant throughout. Ifinitially theplate isatrestandthediameter through Amakes asmall angle withthestring, show that thisdiameter oscillates about thedirection ofthestring with aperiod equal to 27'" 10.Aparticle Ahangs from afixed pointbyalight string, andanother particle Hofthesamemass hangs fromAbyasecond light string ofthe same length. Find thenormal periods ofoscillation, andsketch thenormal modes 11.Ahomogeneous solid cylinder, whose section isasemicircle ofradius a, restswith itsflatfacehorizontal andincontact withafixedrough circular cylinder ofradius6,thegenerators ofthetwo cylinders being parallel. Find thegreatest value ofa/6forwhich there isstability. 12.Find theradius ofgyration ofauniform semicircular plate about a linethrough themass conter perpendicular totheplate. 13.Aladder (length 2a)rests against asmooth vertical wallandasmooth horizontal floor, theinclination tothofloor boing initially.Find the inclination oftheladder tothefloor attheinstant when theupper end leaves thewallasitslidesdown under theaction ofgravity. 14.Apendulum consists ofabobofmassmattheendofalight rodof length 3a. Itissuspended from thepoint oftheroddistant 2afrom the bob.Ahorizontal forcembcosnt(where bandnareconstants and6is small) isappliedtotherodatitsupper end. Find theangular amplitude oftheforced oscillations ofperiod 2ir/n. 15.Auniform solid ellipsoid ofrevolution ofsemiaxesa,b(the axisof revolution boing 2a)iscutintwobyaplane through theconter perpendicular totheaxisofrevolution. Ifeither half willbalance instable equilibrium with itsvertex onahorizontal piano, prove that 16.When aship rollsthrough asmall angle, theupward thrust ofthe water intersects thecentral plane oftheshipatapoint called themetacenter. Find aformula fortheperiodic time ofrolling ofaship interms of/t,the height ofthemetacenter above themass center oftheship,and/r,theradius ofgyration oftheshipabout afore-and-aft axisthrough themass contor. Istheperiodic time increased ordecreased byshifting cargo horizontally from thecenter oftheship tothesides, thisshiftbeing done symmetrically with respect tothecentral planeoftheship? 17.Asquare frame, consisting offourequal uniform rods oflength 2a rigidly joined together, hangs atrest inavertical plane ontwosmooth pegsPjQatthosame lovel. IfPQ candthepegsarenotboth incontact with thesame rod,show that there arcthree positions ofequilibrium, pro- vided a<c\/2. 226 PLANE MECHANICS [Ex.VII Ofthese positions, show that theonly unstable one isthesymmetrical position. If,however, a>c-\/2,show thattheonly possible position of equilibriumisstable. 18.Aparticle issuspended byalight string oflength afrom thelower endofarodofthesamemassandlength 2a,which isfreetoturnabout its upper end. Forvibrations about equilibrium inavertical plane, show that thetwonormal frequencies arcgivenby where psatisfies 4p2-25p+9=0. 19.Arodoflength 2ahangs from asupport which isgiven asmall hori- zontal displacement varying with time according totheequation=6sinpt,where 6andpareconstants. Find theequation ofmotion forsmall oscillations. Integrate theequation, obtaining aresult withtwo arbitrary constants. Find these constants ontheassumption thatwhen t*therod ishanging vertically andhasnoangular velocity; hence show thattheinclination oftherodtothevertical isgivenby (nsnpt-psn 20.Two simple pendulums, each ofmassmandlength a,hang from a trolley ofmass ^l/which canrunwithout friction along horizontal rails. Asmall impulse, parallel totherails,isapplied tooneofthependulums and imparts toitanangular velocity o>,theother pendulum andthetrolley having novelocity atthat instant. Investigate theresulting motion, and express thedisplacement ofthetrolley andtheinclinations ofthependulums tothevertical asfunctions ofthetime. Show that,iftheratiom/M issmall, themotions ofthependulums relative tothetrolley mayberegarded assimple harmonic motions with slowly varying amplitudes, theamplitudes being given bytheabsolute values of where ,... CHAPTER VIII PLANE IMPULSIVE MOTION 8.1.GENERAL THEORY OFPLANE IMPULSIVE MOTION Theconcept ofanimpulsive force. Foraparticle moving inaplane under theaction ofaforce withcomponents X,Y,theequationsofmotion are (8.101) mx=X, my=Y. Multiplying bydtandintegrating from ttot\,weobtain (8.102) AM)-Fxdl, A(my)=f"Ydt, /fo Jt9 whereAdenotes anincrement during thetime interval(fo,ti). Thevector withcomponents (8.103) rXdt, CtlYdt JlQ /'o iscalled theimpulse ontheparticle during thetime interval (to,ti).Wemay state (8.102) inwords asfollows: Theincrement inmomentum isequaltotheimpulse. Letusnowsuppose thataparticle ofmassmcanmove along the x-axis. Attime t=0,itisatrest atx 0.Atthis instant aforce (8.104) X=Asin^ commences toactandactsuntil t=T.(Aandrareconstants.) During thistimetheequationofmotion oftheparticle is (8.105) mxAsin > andso ArA ;rAU=X=(1COS I irm\ r) Art AT* .irt_sm_, irm ?T2m r 227 (8.107) X=228 PLANE MECHANICS [SBC. 8.1 theconstants ofintegration having been chosen tofittheinitial conditions. Thus inthetime interval(0,T)theparticle receives increments invelocity andposition givenby /Om*N A 2^TAAT* (8.106) Aw=>Az= v 'TTW TTW Themean value oftheforce is frA' fit TJo r TT andso(8.106) maybewritten V" V^2 (8.108) AM=,Az= v m m Theexperiment mayberepeated using different values of Aand T.Wenote that aslong astheproduct XTremains unchanged thevalue ofAwremains thesame. IfweletA(and therefore X)tend toinfinity and letTtend tozeroinsuchaway thatXTmaintains afixed value(7,wehave (8.109) AM-> ,Ax-0. Wenote that, inthislimiting case of"an infinite force acting for aninfinitesimal time," there isaninstantaneous change in velocity butnochange inposition. Returning tothegeneral equations (8.102), wemay letthe forcecomponents (X,Y)tend toinfinity andthetime interval ti fotozeroinsuchawaythattheintegrals remain constant or approachfinite limits. Under these circumstances aparticle moving inaplane experiences (inthelimit) aninstantaneous change ofvelocity. Since thevelocity remains finite during this change, thedisplacementiszero inthe limit. The instanta- neous change inmomentum isgivenby (8.110) A(mz)=limP1Xdt, A(m#)=limP1Ydt. Havewehereintroduced anew ingredient orconcept into mechanics? Itmust beadmitted thatwehave, because no force, however large, canproduce aninstantaneous change in momentum. Toplace ournewideasonasecure foundation, we admit theconcept ofanimpulsive force, withcomponents denoted SEC. 8.1] IMPULSIVE MOTION 229 byX,P:itissuch that,when applied toaparticle, theimpulsive force causes aninstantaneous change inmomentum given by (8.111) A(mx)=X, A(my)=f. Wemust, however, regard theimpulsive force, notassomething absolutely new,butasconnected withtheordinary force (X,Y) bytherelations (8.112) X=limrXdt,Y=limPYdt, ti-*toJtoti-+toJh obtained bycomparison of(8.110) and (8.111). Onaccount ofthisconnection,itisunnecessary torepeatfor impulsive forces results already obtained forordinary forces. Wedraw attention tothefactthatimpulsive reactions between theparticles inarigidbodyobey thelawofaction andreaction. Thetheory ofmoments applies toimpulsive forces, andwemay speak ofanimpulsive couple. Theideaofequipollence may also beused. Itisclear from (8.112) that forces which remain finite as ti to(e.g., gravity) contribute nothing totheimpulsive force. Animpulsive force istheproduct ofanordinary forceandatimeand is equal toachange inmomentum. Hence, impulsive forcehasthedimensions [MLT*1 ];itsmagnitude isexpressed indyne sec.orgm.cm.sec."1inthe c.g.s. system andinpoundal sec.orIb.ft.sec."1inthef.p.s. system. Principles oflinear andangular momentum. Weexpressedin(5.206) thelawthat, foranysystem, therate ofchange oflinearmomentum isequal tothesum oftheexternal forces. IfMx,Myarethecomponents oflinearmomentum in thedirections ofaxesOx,Oy,andX,Ythetotalcomponents of external force inthose directions, then (8.113) A,=X,Mv=Y. Letusmultiply bydt,integrate from t=fcto t= ti,andthen proceed tothelimit t\>fo,supposing theforces totend to infinity. Then (8.114) AM*=X,AMtf=Y, where X,Yarethesums ofthecomponents oftheexternal impulsive forces. Inwords,thesudden change inthelinear momentum ofasystem isequaltothetotal external impulsive force. 230 PLANE MECHANICS [Sue. 8.1 These equations may alsobewritten invector form: (8.115) AM=F, where Fisthevector sum oftheexternal impulsive forces. Similarly, weobtain from (5.209) thevector equation (8.116) mAq=F, wheremisthemass ofthesystem andAqthesudden changein thevelocityofthemass center. Ifthere arenoexternal impulsive forces, wehaveF=0,and hence Aq=0.Thus, when theimpulsive forces arepurely internal, there isnosudden change inthevelocity ofthemass center. This result isofinterest inconnection with collisions and explosions. Here, inphysical reality, wefindlarge forces acting forshort intervals oftime, andwemay treat thephenomena mathematically bymeans ofimpulsive forces. Thus,ifa shunting locomotive strikes acar,themass center ofthesystem "locomotive +car"hasthesame velocity justbefore andjust after thecollision. Theburstingofashell intheairproduces a setoffragments, themass center ofwhich hasthesame velocity asthemass center oftheshell before bursting. Consider nowthechange inangular momentum about afixed lineduetotheaction ofimpulsive forces. Equation (5.214) applies. Multiplying bydt,integrating over therange (to,ti) andproceeding tothelimit asusual,wefind (8.117) M=limftlNdt=ft. i-*fo Jt* Inwords,thesudden change inangular momentum about thefixed axis isequaltothemoment oftheexternal impulsive forces about theaxis.* Wemay treat similarly theequation (5.219) which concerns motion relative tothemass center. Wefindthat thesudden change inangular momentum relative tothemass center isequal tothemoment oftheexternal impulsive forces about themass center. Ifthesystemisarigidbody withafixedaxis, (8.117) maybe written (8.118) 7A-#, *Itiseasily proved that,$,defined asthelimit ofthetime integral ofthe moment, isequal tothemoment oftheimpulsive forces. SBC. 8.2] IMPULSIVE MOTION 231 where 7isthemoment ofinertia about thefixed axis,Awthe change inangular velocity, andNthemoment oftheimpulsive forces about theaxis. Forarigidbody which canmove parallel toaplane, (8.118) holds, provided weunderstand 7tobethemoment ofinertia about themass center andNthemoment ofimpulsive forces about themass center. Wehavenowconverted theprinciplesoflinear andangular momentum intoforms valid inthecasewhere impulsive forces act. Since theapplication ofimpulsive forces leads tosudden changes invelocity, wemay callthetheory ofimpulsive motion adiscontinuoustheory, reserving theword continuous forthose cases inwhich nosudden changes invelocity occur. 1Inthecontinuous theory theprinciple ofenergyisuseful for determining motions, either completely orinpart. But itmust beusedwith great caution inthediscontinuous theory, because wefind ingeneral thatwhen impulsive forces actthelawof conservation ofmechanical energy does nothold. Actually, theenergyisnotlost;itisconverted intoheat oremployed to deform thebodies onwhich theimpulses act.Butmechanical energy disappears, and itiswithmechanical energy alone that weareconcerned inthisbook. Exercise. Animpulsive forcePisapplied atoneendofabarofmassm andlength 2o,inadirection perpendicular tothebar. Find thevelocity imparted totheother endofthebar,assuming (i)that thecenter ofthe bar isfixed, (ii)thatthebar isfree. '8.2.COLLISIONS Asremarked above, acollision between twobodies gives rise (inphysical reality) tolarge reactions acting forashort time, andsowetreat theproblem ofcollision mathematically bymeans ofimpulsive forces. The collision ofspheres andthecoefficient ofrestitution. Asanillustrative example, weshall discuss theproblemof thecollision oftwospheres which aremoving along theline joining their centers (Fig. 97). Taking theaxisOxalong thelineofcenters,letususethe following notation: 232 PLANE MECHANICS [Sue. 8.2 mi,mz=masses ofthespheres, Ul)uz velocities ofcenters before collision, ui>u*^velocities ofcenters after collision, P=magnitudeofimpulsive reaction. Wehave then (8.201) mi(u{-ui)=-P,m2(u'2-u2)=P, andso (8.202) miu{+m2uf z=miUi+m2u2, asindeed wemight have deduced directly from thefactthat there isnoexternal impulsiveforce. Ourproblemistofindtheresult ofthecollision, i.e.,tofind u{, u'twhen ui,u2aregiven. But forthiswehave onlyone equation (8.202), andthat isnotenough togivetwounknowns. Wecanproceednofurther without anadditional hypothesis, andhereweintroduce theidea ofthecoefficient ofrestitution. FIG. 97. Collision oftwospheres. Consider theproblem ofcollision asitmight occur inreality, saybetween twotennis balls. Actually theballswould become distorted during thecollision andthenwould bound away from oneanother, regaining their spherical shapes. This isacom- plicated process whichwecannot follow through mathematically, andweareobliged tosubstitute some simple hypothesis based onexperimental results. Weintroduce theexpressions speed ofapproach qaandspeed of separation q8.Forageneral collision, these speeds arecalculated fortheparticles ofthetwobodies atthepoint ofcontact, com- ponents ofvelocity along thecommon normal atthatpoint being used. Inourproblem ofcolliding spheres, wehave (8.203) qa=HIu2, q8=14 u{. SEC. 8.2] IMPULSIVE MOTION 233 The following general hypothesisisadopted: Thespeeds of separation andapproach areconnected bytherelation (8.204) q8=eqa, where eisapositive number, called thecoefficient ofrestitution. Thevalue ofedepends onthematerials ofwhich thebodies are composed and alsoontheir shapes and sizes;itnever exceeds unity invalue. When e=1thebodies aresaidtobeperfectly elastic, andwhen e=they aresaidtobeperfectlyinelastic. Intheproblem ofthespheres, wenowhave (8205) 2=i 2. Hence, ,_mi-em*+(l+e}_J2_} (8.206) t /i i \ wii.mz em\u'2=(1+e)- :-uiH-- - :-uz,mi+m2 mi+mz' andsotheproblemissolved. Inthecase ofperfectly inelastic spheres (e=0),wehave HI=u'z;there isnorebound. Ifthespheres areofthesame mass (mi=m2)andthere is perfect elasticity (e=1),wehave (8.207) u{=uZfu'2=m; thismeans that thespheres exchange velocities. This case is particularly interesting because, inthekinetic theory ofgases, themathematical model represents themolecules byperfectly elastic spheres. Compression andrestitution. Thehypothesis (8.204) appears artificial; wegenerally prefer toadopt hypotheses which havesome plausibility. Thehypoth- esismayhowever beputinanother form, which suggests rather better itsconnection with physical reality. Todothis, wereturn tothephysical picture ofthecollision oftwotennis balls. Atfirst thecenters oftheballs areapproaching one another, andtheballs arebeing distorted. Then they start to regain their spherical shapes, pressing against oneanother until they separate. Thus thewhole period ofcollision isdivided 234 PLANE MECHANICS [Sac. 8.2 intoaperiod ofcompression andaperiod ofrestitution. We may adopt, instead of(8.204), thefollowing hypothesis: The impulse during restitution bears totheimpulse during com- pression adefinite ratio e.Or,passing tothelimit ofinfinite forces andvanishing time,wemaymake thefollowing formal statement ofourhypothesis: IfPiisthemagnitude oftheimpul- sivereaction ofcompression requiredtoreduce thespeed ofapproach tozero, then themagnitude P2oftheimpulsive reaction ofrestitution is (8.208) P2=cPi where eisthecoefficient ofrestitution. These twoimpulsive reactions actinthesame direction. Itisbynomeans obvious that (8.208)isequivalent to(8.204), but itcanbeproved without muchdifficulty. Weshall here merely establish theequivalence fortheproblem ofthespheres. Wehave, forcompression, (8.209) m\u m\u\=Pi,w2nw2w2=Pi, where uisthecommon velocity when thespeed ofapproachis zero. Forrestitution, wehaveby(8.208) (8.210) miu'i m\u=ePi, Wehaveherefourequations, which canbesolved foru(,u'2,u,PI. Toprove thatwegetthesame result asthatgiven by(8.204), weeliminate ufrom (8.209) andalsofrom (8.210) ;weobtain (8.211) /mim 2(ui-u)=(mi+m2)Pi, \WiW 2(i4~u()=e(mi+w2)Pi, fromwhich (8.204) follows atonce. Motion relative tothemass center. Themathematics ofdiscontinuous motions ismuch simpler than that ofcontinuous motions, because theequations tobe solved arealgebraic, not differential. Butthealgebra may become complicated, and itissometimes advisable tousea special Newtonian frame ofreference. Thus, inthecase ofthe twospheres considered above, wemayuseaframe ofreference inwhich themass center isatrestbefore collision. Itis,of course, atrestinthisframe after collision also.Wehavethen SBC. 8.3] IMPULSIVE MOTION 235 (o.ZiL^)' hence, (8.213) u{=-etii, t4= Asaresult ofthecollision, thecomponents ofvelocity are reversed insignandmultiplied bythecoefficient ofrestitution. Ifthekinetic energyisTbefore collision andTrafter collision, wehave Oandsothelossofkinetic energy is (8.214) T-T'=(1-e2)T. Since e^1,kinetic energyislost inevery caseexcept that of perfect elasticity (e=1). Wehave discussed thecollision ofspheres inthecasewhere their centers move along thelinejoining thecenters. Provided thespheres aresmooth, theextension tothecasewhere the spheres have general motions isimmediate; the componentsofmomentum (andhence velocity) indirections parallel tothecommon tangent planeofthespheres undergo nochanges, and thecomponentsofvelocity along thecommon normal change asdescribed above. 8.3.APPLICATIONS Weshallnow illustrate theapplication ofthe principlesoflinear andangular momentum by twoexamples. The ballistic pendulum. Consider arigidbody, hanging inequilibriumNfrom ahorizontal axis (Fig. 98).Abullet, traveling horizontally, strikes thebody atAandFlQp 93. Baiiis- becomes embedded init.Asaresult ofthe ticpendulum, impact, thebody swings asacompound pendulum, rising through anangular displacement abefore coming torest.Onaccount of itsimportance inballistics, theapparatusiscalled aballistic pendulum. From theangleaandtheconstants ofthesystem, wecancompute thevelocity ofthebullet, asweshallnowshow. 236 PLANE MECHANICS [SEC. 8.3 Letustake asdynamical system thebody andthebullet. Then theforces between thebodyandthebullet areinternal. During thebrief interval ofimpact theonly external forces acting are (i)gravity and(ii)thereaction atO.Theforce of gravityisafinite forceandsocontributes noimpulsive force. Since thereaction athasnomoment about 0,itisevident that theprinciple ofangular momentum enables ustostate that angular momentum ofsystem about before impact=angular momentum ofsystem about after impact. LetONbethevertical through 0,ANbeing horizontal. Letmbethemass ofthebullet, qitsspeed, /themoment of inertia ofthebody about 0,andwtheangular velocity immedi- ately after impact. Then theangular momentum ofthesystem about is before impact: mqON, after impact: (m-AO2+^)w- Ifthemass ofthebullet isvery small incomparison withthat of thebody,wemay neglectmAO2incomparison with /;thuswe have (8.301) mql=Io>, where I=ON. Although wecannot apply theprinciple ofenergy during impact, wecanapplyitinthesubsequent motion. Thus, again neglecting themass ofthebullet incomparison with that ofthebody,wehave (8.302) i/o>2=Mgh(l-cosa), whereMisthemass ofthebodyandhthedistance ofitsmass center from 0.Hence, from (8.301), (8.303) 9=-, which gives thespeed ofthebullet interms ofaandconstants ofthesystem. Linked rods. Twouniform rodsAB,BC,each ofmassmandlength 2o, areconnected byasmooth joint atBand lieinonestraight line SEC. 8.3] IMPULSIVE MOTION 237 onasmooth horizontal table (Fig. 99a).Ahorizontal blowP isstruck atC,inadirection perpendicular toBC.Wewish to findthemotion generated. Letusdraw aschematic diagram (Fig. 996), separating the rods inorder torepresent thereactions without confusion.ABC r FIG. 99o.Apairofrods, linked atB,receive ablow atf. o FIG. 99&. Diagram ofvelocities andimpulsive forces. Taking rectangular axesOxy,withOxparallel toABCandOy inthesense ofP,weshall usethefollowing notation : u\,Vi=components ofvelocity ofcenter ofAB, u2,vz=components ofvelocity ofcenter ofBC, a)i angular velocity ofAB, co2=angular velocity ofBC, X,F=components ofreaction onBCatB, X,Ycomponents ofreaction onABatB. Since therods arejoined atB,thispointmusthave thesame velocity whether considered asapoint ofABorofBC. Thus, (8.304) u\=u2, Vi+ctcoi=Vz ttoj2. Theprinciple oflinearmomentum applied toeachrodgives /_v v /r>O/\K\ JWM/i -~".A. frlUs .A. (8.305) <=_y =y4-P- theprincipleofangular momentum gives (8.306) mk2o>i=-aY, mWu* -aF+aP, where kistheradius ofgyration ofeach rodabout itscenter, sothat k*=a2 . 238 PLANE MECHANICS [Sac. 8.4 From the firstequation of(8.304) andthefirsttwoof(8.305), weseethat (8.307) ui=u2=0,5=0. Therenowremain in(8.304), (8.305), and(8.306) fiveequations forthefollowing fiveunknowns: 01>02,Wl, &>2jY. Itismost symmetrical tofindffirstbysubstitution in(8.304) from theother equations; wefind (8.308) Y-iP, andhence (8.309)iP p0i=-i-, *-*S^PP___o_f .^maTma Thevelocity of isP/w,downward inthediagram. 8.4.SUMMARY OFPLANE IMPULSIVE MOTION I.Componentsofimpulsive force. (8.401) X=limftl Xdt, Y=limfl7<B. <i-*fo^atv-*kJk II.Instantaneous change inmotion, (a)Particle: (8.402) A(mx)=X, A(my)=f. (6)Anysystem: (8.403) AM*=X,&MV=K, A^=N. (X,Y,N=totalcomponents andmoment ofexternal impulsive forces.) (c)Rigid body with fixed axis: (8.404) 7Ao>=N. (d)Rigidbodymoving parallel toafixed plane: (8.405) mAw= ,mA0f,/Ao>& (w,v=components ofvelocity ofmass center;N=impulsive moment about mass center.) Ex.VIII) IMPULSIVE MOTION 239 III.Collisions. Either (8.406) q. eqa or (8.407) P2=ePi. (e=coefficient ofrestitution; e^1.) EXERCISES Vin 1.Abar2ft.long, ofmass 10lb.,liesonasmooth horizontal table. It isstruck horizontally atadistance of6in.fromoneend,theblow being perpendicular tothebar;themagnitude oftheblow issuch that itwould impart avelocity of3ft.persec.toamass of2lb.Findthevelocities of theends ofthebarjust after itisstruck. 2.Auniform rodofmassmandlength 3ahangs from apinpassing throughitatadistance afrom theupper end. Find interms ofm,a,gthe magnitude ofthesmallest blow, struck atthelower endoftherod,which willmake theroddescribe acomplete revolution. 3.Aball isdropped onthefloorfrom aheight h.Ifthecoefficient of restitution ise,findtheheight oftheballatthetopofthenthrebound. 4.Abar,6ft.long,isswinging about ahorizontal axlepassing through itatadistance of1footfrom oneend. Atwhat point must ablow be struck tobringittorestwithout causing anyimpulsive reaction onthe axle? (This pointiscalled thecenter ofpercussion.) 5.Aparticle moving withaspeed of30feetpersecond inadirection making anangle of60with thehorizontal strikes asmooth horizontal plane andrebounds, thecoefficient ofrestitution being $.Find thespeed andthedirection ofmotion oftheparticle immediately after impact. 6.Auniform square plate ofmassMand side2arestsonasmooth horizontal table. Ahorizontal impulsive force ofmagnitude Pisapplied atacorner inadirection perpendicular tothediagonal atthat corner. Show thattheangular velocity generated bythisimpulsive force is 3 2Ma 7.Atugofmassmtons isattached toabarge ofmassMtonsbyacable themass ofwhichmaybeneglected. Thecable isslack. Thetugmoves andhasacquired aspeed ofvft.persec.when thecablebecomes tautand thebargeisjerked intomotion. Assuming thatthecable hasacoefficient ofrestitution andneglecting theimpulsive resistance ofthewater, find (i)thespeed imparted tothebarge; (ii)themean tension (intons wt.)inthecable during thejerk,supposing thistotake tsec. 8.Abilliard ball ofradius aandmassMrestsonahorizontal table. Inavertical plane through thecenter oftheballthere isapplied ahorizontal 24Q PLANE MECHANICS [Ex.VIII impulsiveforce ofmagnitude P. Ifthelineofaction oftheimpulseisata height habove thetable, findthe initial velocityofthat pointoftheball which isincontact withthetable. 9.Two gearwheels ofradii ai,a2and axialmoments ofinertia Ji,/2, respectively, canrotate freely about fixed parallelaxles. Initially the wheel ofradius aiisrotating withangular velocity w,while theother wheel isatrest. Ifthegearwheels aresuddenly engaged,findtheangular velocity ofeachwheel afterward. 10.Abeam ofmass 100 Ib.andlength 6ft.hangs fromana*xlepassing throughitatadistance of1ft.fromoneend. Itisdrawn aside through anangle of30andthen released. Itisstopped dead atthelowest point ofitsswing byahorizontal blowwhich strikes itataheight of2ft.above itslower end,What istheimpulsive reaction ontheaxle,expressedinIb. ft.sec.-1? 11.Twouniform rodsAB,BC,each ofmassMandlength 2a,aresmoothly jointed together and restonasmooth horizontal plane, theangle between therodsbeing 45.Ahorizontal impulsiveforcePisapplied atAina direction atright angles toABandaway from therodBC. Findtheinitial angular velocity ofBC. 12.Asmooth rodoflength 2aandmassMrestsonahorizontal plane. Asmallbodyofmassmmoves inthispiano with velocity vinadirection inclined totherodatanangle of45;itstrikes therodatapoint distant c from thecenter. Ifthecoefficient ofrestitution between therodandthe bodyise,findtheangular velocity oftherodandthevelocity ofthebody after collision. 13.Aflywheel whose axialmoment ofinertia is200 Ib.ft.2rotates with anangular velocity of300revolutions porminute. Find inft.Ib.wt.sec. theangular impulse which would berequired tobring theflywheel torest. Hence findthefrictional torque atthebearingsiftheflywheel romes to restin10minutes under friction alone. 14.Oneendofeach offourequal uniform rods issmoothly jointed tothe circumference ofauniform disk ofradius aandmassM.Thelength ofa rod is2aand itsmass ism.Thepoints ofattachment areatequal angular intervals. Initially thesystemisatrestonasmooth horizontal plane with eachrodlying along aradius ofthediskproduced. Ahorizontal impulsive force ofmagnitude Pisapplied totheouter endofonerodinadirection perpendicular toit.Show thattheinitial angular velocityofthedisk is a(M+2m)' inasense opposed tothedirection oftheimpulse. 15.Auniform rodofmassmandlength 2aismoving onasmooth hori- zontal plane. Atacertain instant, itscenter hasvelocity components u along therodandvperpendiculartoit,andtherodhasanangular velocityo>. What impulsive forcemust beapplied toapoint oftherodatadistance b from thecenter inorder tobring thatpoint torestinstantaneously? 16.Twouniform circular platesAandBteach ofradius aandmaso m, areconnected byarodoflength 2aandmassm,eachendofwhich islinked Ex.VIII] IMPULSIVE MOTION 241 smoothly toapointonthecircumference ofoneoftheplates. Thesystem isatrestonasmooth horizontal plane withthecenters oftheplates inthe lineoftherodproduced. Animpulsive couple$actsontheplate A. Determine theinitial motion oftheplate B. 17.ABtEC,CDarethree equal rods, smoothly hinged tooneanother atBandC.Theylieonasmooth horizontal plane, forming three sides of asquare. ABcanturn freely about A,which isfixed. Animpulsive force applied toDsetsDinmotion with avelocityvdirected away from A. Prove thattheinitial velocity ofBisoppositeindirection tothat ofDand equal inmagnitude to-f^v. 18.Forthecollision oftwosmooth latninas moving inaplane, prove that theassumption thattheimpulsive reaction ofrestitution isequal toetimes theimpulsive reaction ofcompression loads totheresult thattheratio of thespeeds ofseparation andapproachise. 19.Onastraightline/,there aresituated nparticlesallofthesame mass. Initially the particles areatthepoints A\,A2t- -Anwhere OAi<OAz<-<OAn,Obeing afixed pointofL,andthevelocity of therthparticle isinthedirection OArandofmagnitude, ur. Ifui>HZ>->unandtheparticles arcallperfectly elastic, findthe final velocity ofeachparticle. What would betheresult ifalltheparticles were perfectly inelastic,? 20.Anumber ofequal uniform rods aresmoothly jointed together to formachain which hangs atrestunder gravity. Theupper endAofthe chain isfreetoslideonasmooth horizontal axis. Ifaiiimpulsive force is applied toAalong theaxis,show thattheinitial angular velocities ofthe lastthree rods areintheratios 11 :3:1. PART II MECHANICS INSPACE CHAPTER IX PRODUCTS OFVECTORS Uptothepresent, ourdevelopment ofmechanics hasbeen restricted, forthemost part, totwodimensions. Wenowcome tothesystematic treatment ofmechanics inspace. Herewe mustmake adecision astonotation. Ontheonehand, wehave theordinary notation ofcoordinates; ontheother hand, the vector symbolism. Each has itsadvantages, butonthewhole thevector notation hasproved more useful onaccount ofits compactness. We shall therefore use itextensively (butnot exclusively) throughout the rest ofthebook. The present chapter, together with Sec.1.3, explains themathematical language tobeemployedlater. 9.1.THESCALAR ANDVECTOR PRODUCTS Indeveloping thetheoryofvectors, wetrytoextend to vectors theoperations ofordinary (scalar) algebra, asfaras possible. InSec.1.3,thiswasdone successfully fortheaddition andsubtraction ofvectors andforthemultiplication ofavector byascalar. Wenowconsider themultiplication ofvectors by oneanother, andherethemethods ofordinary algebra arenot soeasy togeneralize. Actually, wedefine twotypes ofproduct thescalar product andthevector product. AsinSec. 1.3,weusePi,P2,Patodenote thecomponents of avectorPonrectangular axesOx,Oy,Oz,andPtodenote its magnitude. Scalar product. The scalar product oftwovectors PandQ,written PQ, isdefined by (9.101) P-Q=PQcosfl, where istheangle between PandQ.SinceQcos6isthe component ofQinthedirection ofP(cf.Sec.1.3),itisclear thatPQisequal tothemagnitude ofPmultiplied bythe componentofQinthedirection ofP. 245 246 MECHANICS INSPACE [Sue. 9.1 Inparticular,3tPisthecomponent ofavectorPinthe direction ofaunitvector ^.Thus theworkdonebyaforceP inaninfinitesimal displacement3.5sis,by(2.401), 6W=P&8s. Since thedirection cosines ofParePi/P, P*/P, P*/P and those ofQareQi/Q,Q2/Q,Qs/Q, wehave Hence, using (9.101), wehave thefollowing expression forthe scalar product oftwovectors interms oftheircomponents: (9.102) PQ=PiQi+P.Q Z+P3QS. From thedefinition,itisclear that thescalar product oftwo perpendicularvectors vanishes. Either from thedefinition (9.101) orfrom (9.102), itfollows that theorder ofthefactors inascalar productisimmaterial. Thus, infact, scalar multiplicationiscommutative. Itisalso dis- tributive; thatis, P.(Q+R)=P.Q+P-R. Toshowthis,werecall that thecomponents ofQ+Rare Qi+Ri,Q2+#2,Qs+Rs',andtherefore, by(9.102), P(Q+R)=/MQx+R,)+P2(Q2+ ,)+P3(Q3+fi.) =(PxQi+P2Q2+P3Q3)+(Pi#i+P2#2+P8#3)=PQ+PR. Athird lawgoverning theoperation ofmultiplication in ordinary algebra, namely, theassociative law,doesnotconcern ushere sinceweattach nomeaning toPQR.However, we have defined such quantities as(PQ)RandP(Q R),each being theproduct ofavector byascalar. These quantities are,ofcourse, quite different, onebeing avector with thedirec- tionofRandtheother avector with thedirection ofP. Exercise. Avector hascomponents (1,3, 2)inthedirections ofrec- tangular axesOxyz. What isitscomponent along thelinexy z,the positive sense being that inwhich xincreases? SBC. 9.1) PRODUCTS OFVECTORS 247 Positive rotations. Before defining thevector product, weshall introduce a convention concerning the ^^ signofarotation. Arotation about adirected lineLissaidtobepositiveif itbears tothedirection ofL thesame relation astherota- tion ofaright-handed screw ^ bears toitsdirection oftravelFlG '100-Apositive rotation ' (Fig. 100). Thus arotation from south tocast isapositive rotation about theupward vertical; theearth's rotation about itsaxisdrawn from south tonorth isalsopositive. Right-handed triads. Consider three non-coplanar vectors. These three vectors, taken insome order, formanordered triad. Since allthetriads ofwhichweshallspeak areordered, theadjective willbeunder- stood infuture, andatriad willmean anordered triad. Let P,Q,Rbeanorthogonal triad, theorder being asindicated. This triad issaid toberight-handediftherotation through a right angle fromPtoQisapositive rotation about R.Any other triad issaid toberight-handedifitcanbedeformed con- tinuously into aright-handed orthogonal triad without its vectors becoming coplanar atanystage inthedeformation. Ifthetriad P,Q,Risright-handed, then thetriad Q,P,R issaidtobeleft-handed. Ifthetriad ofunitcoordinate vectorsi,j,k,introduced in Sec. 1.3,isaright-handed triad, theaxesOxyz aresaid tobe right-handed. Weshall always useright-handed axes forthe sake ofconsistency. Vector product. Given twovectors PandQ,wedraw theunitvector nper- pendicular tobothPandQ,such thatthetriad P,Q,nisa right-handed triad.Wedefine thevector product ofPandQ, written PXQ,by (9.103) PXQ=PQsin0n, where istheangle between PandQ(Fig. 101). 248 MECHANICS INSPACE [Sue. 9.1 Itisclearfrom thedefinition thatarotation fromPtoQ, through ananglelessthantworight angles,isapositive rotation aboutPXQ.Wenote that the magnitude ofPXQisPQsin6}this isequal tothearea oftheparallelo- gramwhose adjacent sides arePand PxQQ. Thevector product PXQoftwo non-zero vectors vanishesif,andonly if,PandQarecodirectional oroppo- site; inparticular, Q PXP=0. FIQ. 101.Thevector product. T, / ij.i_ ,Letusnow findthecomponents ofPXQ.Ifwedenote thisvector byR,thenRisperpendic- ulartobothPandQ,andwehave Therefore, (9.104) Rz=fc(P 3Qi-PiQ 3),-P2Qi), where kisanundetermined factor. Now, R*=fij+R\+Rl =kWl+PI+PD(Ql+Ql+Ql)-(Pid+P2Q2+P30s)2 ]=A;2P2Q2sin20. But,bydefinition, R=PQsin0, andhence k=1. From considerations ofcontinuity,itisevident that kisto have thesame sign inallcases. This signmay therefore be determined byconsidering theparticular casewherePandQ areunit vectors directed along thepositive axes ofxand y, respectively; then, P!=1,P2=0,P8=0, Qi=0,Q2=1,Q3=0, SBC. 9.1] PRODUCTS OFVECTORS 249 andhence, by(9.104), Ri=0,#2=0,#8=k. ButPXQis,inthiscase,aunitvector directed along thepositive axis ofz,sothatRs=1.Hence, k=+1here,andsoinall cases. Thus, quite generally, thecomponents ofR=PXQare (9.105)Ri=P2Q3-P3Q2, R*=P3Qi-PiQ 3, R3=PiQ,-P2Qi. Note thatthenumber describing thecomponent andthesub- scripts intheleading term oftheexpression forthatcomponent areacyclic permutation ofthenumbers1,2,3. From thedefinition (9.103),itisevident that (9.106) PXQ=-QXP. Again, using (9.105),itiseasily shown that (9.107) PX(Q+R)=PXQ+PXR. Thus, vector multiplicationisnotcommutative butdoesobey theusual distributive lawformultiplication. Fortheunitcoordinate vectorsi,j,k,itiseasily seenthatthe following relations hold : (9.108) 1 jXk= i,kXi=j,iXj Ifweassume thedistributive lawforscalar andvector products andtheformulas (9.108), wecanestablish (9.102) and(9.105) directly. Thus, P.Q=(Pii+P2j+P3k)(Qa+Q2j+Q3k) and PXQ=(Pii+P2j+P3k)X(Qii+Q2j+<?*k)=(P2Q3-P3Q2)i+(P3Qi-PiQ 3)j+(PiQ 3-P2Qi)k. Ifwemultiply avector byascalar m,wedonotalter its lineofaction; wemerely changeitsmagnitude, andreverse its 250 MECHANICS INSPACE [Sue. 9.2 direction ifmisnegative. From thisfactandthedefinitions ofthescala,r andvector products, weseethat (i) (mP)-Q=P-(wQ)-m(P. Q); (ii) (mP)xQ=PX(mQ)=m(PXQ). Hence,ifascalar factor appears inaproduct ofvectors, itspositionisactually ofnoimportance;itmayb#shifted toanyposition without altering thevalue oftheproduct asa whole. Exercise. IfPXQRandPXRQ,then thevectorsQandR both vanish. Differentiation ofproducts ofvectors. Thederivative ofavector with respect toascalar hasbeen defined inSec. 1.3.Wesawthere that thederivative ofthe sum oftwovectors isequal tothesum oftheir derivatives, as inordinary calculus. The ordinary rule holds also forthe derivatives ofthescalar andvector products. This isshown as follows: (9.109)/(P.Q)=lim(P+AP).(Q+AQ)-P.Q v 'du^^'AU->O Aw rAP*Q+P*AQ+APAQ"~ =... du^du Similarly, writing "cross" for"dot," weobtain (9.110)jjL(PxQ)- XQ+Pxg- Itisimportant topreserve theorder ofPandQin(9.110), butnotin(9.109). 9.2.TRIPLE PRODUCTS Mixed triple product. Letusconsider three vectors P,Q,andR.Promthemwecan form theproduct P(QXR),called theirmixed triple product. This isthescalar product ofPandthevectorV=QXR,and soisascalar. Weshallnow expressitinterms ofthecom- SEC. 9.2] PRODUCTS OFVECTORS 251 ponents ofthethree vectors. From (9.102) and (9.105), we have P(QXR)=P-V or,indeterminantal form, (9.201) p.(QXR)=PiQi \Q\ From therulegoverning theinterchange ofcolumns ina determinant,itfollows that P(QXR)=Q(RXP)=R(PXQ), and P(QXR)=-P(RXQ)=-Q (PXR). Thus amixed triple productisnotchanged byacyclic permuta- tion ofthevectors;itssignisreversed when twoofthevectors areinterchanged. Wemay interpret themixed triple product geometrically as follows: Aswehave seen, themagnitude ofQXRisequal tothe area oftheparallelogram whose adjacent sides represent QandR. NowP(QXR)istheproduct ofthemagnitudeofQXRby thecomponent ofPinthedirec- tion ofQXR(that is,per- pendicular totheplaneofQand R).Hence themagnitudeof P.(QXR)isequal tothevol- ume oftheparallelepiped whose adjacent edges represent P,Q,andR(Fig. 102). The sign ofP(QXR)isalso significant;itispositive ornegative according astheangle between PandQXRisacute orobtuse, i.e.,according asP,Q,Rform aright- orleft-handed triad. From thegeometrical interpretation,itisobvious that P-(QXR)=0 ifthevectors P,Q,Rarecoplanar.QxK Fjo102._ ixed triple product. 252 MECHANICS INSPACE [Stec. 9.3 Vector triple product. From thevectors P,Q,R,wecanform another product, namely, PX(QXR) ;this isevidently avector and iscalled thevector triple product. Weshallnow express thisproduct asthedifference oftwo vectors. DenotingitbyUandwriting V=QXR,wehave U=PXV; hence, using (9.105), C/i=P273-P3F2 -(P< Similarly, 17,=(P.R)Q S- 17,=(P.R)Q- These three expressions for/i,C72,C73canbecombined intothe vector equation (9.202) U=PX(QXR)=(PR)Q-(PQ)R. The following remark isanaidinremembering thisexpres- sion: sinceQXRisperpendicular totheplane ofQandR,the vectorPX(QXR)must bointhisplane; hence, PX(QXR)=qQ+rR, where qandrarescalars. Exercise. Evaluate allthevector triple products oftheunitcoordinate vectorsi,j,k,including those inwhich oneofthevectors isrepeated. 9.3.MOMENTS OFVECTORS Themoment ofavector about alinewasdefined inSec. 2.3 asascalar. There wespoke also of"themoment ofavector about apoint A,"butonly asanabbreviation for"themoment about alinethrough Aperpendicular totheplane containing A andthevector." Now thatweareinpossession ofthepowerful vector notation, weshallmake afresh start.Weshall define thevector moment ofavector about apoint, and (interms ofit) SEC. 9.3] PRODUCTS OFVECTORS 253 thescalarmoment ofavector about aline; thislatter definition willbeshown toagree with thatgiven inSec. 2.3. Moment ofavector about apoint. LetPbeavector withorigin atB,andAanypoint inspace (Fig. 103).Wedefine thevectormoment ofPaboutA(orbriefly themoment ofPabout A)asavector M,givenby (9.301) M=rXP, where r=AB,theposition vector ofBrelative toA.ThusM D C B FIG. 103.Themoment ofvector about apoint. fFIG. 104.Themoment ofP aboutAisrequired. isavector perpendicular totheplane ofrandP,withmagnitude (9.302) M=rPsinB=aP, where istheangle between randP,andatheperpendicular fromAonthelineofaction ofP. Asanillustration, letuscalculate themoment ofagiven forcePabout a pointA.InFig. 104,Pisaforce ofknown magnitude applied atHand acting along thediagonal HFofoneface ofthecubeABC H. Ifi,j,k isatriad ofunitorthogonal vectors atA(asshown), wehave AH-6(i+k), P--^(J~ ] where 6denotes anedge ofthecube. ThemomentMofPaboutAisnow easily calculated; itis M-6(1+k)X^= (j-k). 254 MECHANICS INSPACE [SBC. 9.3 Hence, by(9.108), Thus themoment ofPaboutAisavector withcomponents (b 6P/\/2 WVV2)inthedirections oftheedgesAEtAB,AD,respectively. Returning tothesituation shown inFig. 103,letusinvestigate the effect ofsliding Palongitsline ofaction. It"becomes B P B' P Fio. 105.Themoment ofavector isunchanged whenweslidethevector along itslineofaction. (Fig. 105)avectorPatB',whereAB'=r+kP(kbeingsome scalar). Themoment aboutAisnow M'=(r+fcP)XP. ButPXP=0,andhence M'=rXP=M. Thus themoment ofavector about apointisunaltered bysliding thevector alongitslineofaction. Weshallnowmake animportant deduction from theabove fact. LetPatBandPatB1betwovectors withacommon line ofaction L,and letAbeany point. Sliding Palong Luntil itsoriginisatB,wedonotalter itsmoment about A. Thus,ifAB=r,thesum ofthemoments aboutAofPatB and-Pat B'is rXP+rX(-P)=rX(P-P)=0. Inwords, fortwovectors inthesameline,withequal magnitudes butopposite senses,thevector sum ofmoments about anypoint SBC. 9.3) PRODUCTS OFVECTORS 255 iszero. Inparticular, bythefundamental lawofaction and reaction(cf.Sec.1.4),wehave (9.303) The vectorsum ofmoments about anarbitrary point of theforces ofinteraction between twoparticles ofanysystem iszero. Bythedistributive lawforvector multiplication, wehave, foranyvectors, (9.304)rXP+rXQ+rXR+-- LetP,Q,R, bevectors withcommon origin B,and letr betheposition vector ofBrelative toapoint A.Then, for vector moments about apoint, wehave thetheorem ofVarignon (cf.Sec. 2.3):Thesum ofthevector moments about apointAof vectors P,Q,R, withcommon origin B,isequaltothevector moment aboutAofthesinglevectorP+Q+R+ withorigin B. Moment ofavector about aline. LetMbethemoment ofavectorPabout apoint A,and letLbeanylinethrough A.Ofthetwosenses onL,wechoose oneaspositive anddistinguishitby aunit vector 3.lying onL.We define thescalarmoment ofPaboutLas thecomponent M\ofMintheposi- tivesense ofI/;expressedinsymbols, (9.305) MX=3t.M. Weshallnowshow thattheabove > definition isequivalent tothatgiven inSec. 2.3.Wetake special axes Oxyz asshown inFig.106;theorigin coincides withA,andOzliesalong the lineLinthe positivesense. Relative tothese axes,Phascomponents (X,F,Z)andactsata pointBwithcoordinates(x,y,2).ThemomentMofPaboutA (or0)is (9.306)M-(xi+y]+*k)X(XI+Yj+Zk) =(yZ-zY)l+(zX-zZ)j+(xY-/*R 256 MECHANICS INSPACE [Sac. 9.3 wherei,j,karetheunit coordinate vectors. Since &=k, wehave (9.307) MX=kM=xY-yX. But thisquantityisprecisely themoment asgiven by(2.303); thetwo definitions ofthemoment ofavector about alineare nowcompletely reconciled. Itmight appear thatthevalue ofthemoment MXofPabout L, asgiven by(9.305), depends onthechoice ofapointAonthis line. This isnotactually thecase. For letA'beanyother point onL,sothatAA' fa,where kissome scalar. The moment about A'ofPatBis M'=(-ASt+r)XP, where r=AB.ThecomponentofM'inthepositive sense of Listherefore M(=3i'[(-Wi +r)XP] =-A&.(3L XP)+3i-(rXP)=31M=MX, sinceA(3iXP)=0. Thetheorem ofVarignon forscalar moments ofvectors about alinefollows directly from (9.304) ;wehave merely totakethe scalar product ofeach sidewithXThisvery simple proofby vector methods should becompared with that ofSec.2.3,where onlyelementary methods were used. There areoccasions, however, where scalar methods aremore direct than vector methods. Onsuch occasions, werequire formulas forthemoments ofavector about theaxes ofcoordi- nates. These are,by(9.306), (9.308) yZ-zY, zX-xZ, xY-yX, where (X,Y,Z)arethecomponentsofthevector applied at (z,2/, z). Exercise. Avector withcomponents (1,2,3)actsatthepoint (3,2,1). What isitsmoment about theorigin, andwhat areitsmoments about the coordinate axes? SBC. 9.4] PRODUCTS OFVECTORS 257 9.4.SUMMARY OFPRODUCTS OFVECTORS I.Scalar product. (9.401) PQ=QP-PQcosB=PjQj+P2Q2+P8Q3. II.Vector product. (9.402) PXQ=-QXP=PQsin0n. (naunitvector perpendicular toPandQ;triad P,Q,nright- handed.) III.Usual rules ofalgebra andcalculus apply toproducts of vectors,iforder invector productsispreserved. IV.Themixedtriple product. PiOn (9.403) P.(QXR)=P2Q2J PaQaJ V.Thevector triple product. (9.404) PX(QXR)=(P-R)Q-(PQ)R. VI.Moment ofavector about apoint. (9.405)M=rXP =(yZ-zY)i+(zX-rcZ)j+(xY- VII.Moment ofavector about adirected lineO). (9.406) MX=*(rXP). EXERCISES IX 1.Solve theequations 2A+B=M, A+2B-N, MandNbeing given vectors. 2.Three vectors arcrepresented bythediagonals ofthree adjacent faces ofacube, allpassing through thesame corner anddirected away from it.Find theirsum. 3.What isthemoment about there-axis ofaforce ofmagnitude 3applied atapoint with coordinates (2,3,5),inadirection making angles of60 with theaxes ofyand zandanacute angle with theaxisofx? 4.A,B,C,Dareanyfour vectors. Prove thatthere exist scalarsa,6, c,d(not allzero), such that oA+&B+cC+dD=0. 258 MECHANICS INSPACE [Ex.IX 5.IfAXBAXC,show thatBC-H&A,where kissome scalar. 6.IfAandBareanytwounit vectors, prove that themoment ofA aboutBisequal tothemoment ofBabout A. 7.Find themoments, about acorner ofacube, ofthree unitvectors converging ontheopposite corner along three edges. Show thatthesum ofthemoments iszero.How could you obtain this result without calculation? 8.Aforce withcomponents (X,Y,Z)actsatthepoint (a,6,c).What isitsmoment about alinethrough theorigin with direction cosines(I,m,n)? 9.Aforce ofmagnitude Pactsalong thelinejoining opposite corners ofacube ofedge 2a.Find themoment oftheforce about alinewhich is adiagonal ofaface ofthecubeandwhich doesnotcutthelineofaction oftheforce. 10.Adirected lineLpasses through thepoint (a,6,c)with direction cosines(I,m,n).Prove thatthemoment aboutLofaunitvector pointing along thez-axis isbn cm. 11.Prove thatthemoment ofavector about alinevanishesif,andonly if,thevector cutsthelineorisparallel toit. 12.Prove theidentities (i) AX(BXC)+BX(CXA)+CX(AXB)-0, (ii) AX[BX(CXD)J=(BD)(AXC)-(BC)(AXD), (iii) (AXB)X(CXD)=B[A (CXD)]-A[B (CXD)]. 13.Oxyzj Ox'y'z' aretwosetsofrectangular Cartesian axes.Pisavector withcomponents X,Y,ZonOxyzandcomponents X',Y'tZ'onOx'y'z1 . Show that X'-anX+aiZY+OnZ, where an, ais,AHarethedirection cosines ofOx'with respect toOxyz. Develop similar formulas forY'and Z'. 14.Solve thedifferential equation where aisaconstant vector. 15.Show thatthedifferential equation where aandbareperpendicular constant vectors, hasthegeneral solution r-/(fla +fc+e-gj(aXb); here/(i)isanarbitrary function andc,earearbitrary constant vectors. CHAPTER X STATICS INSPACE 10.1.GENERAL FORCE SYSTEMS Before proceeding toconditions ofequilibrium, letusdevelop some results valid foranysystem offorces, whether theyproduce equilibrium ornot. The total forceandthetotalmoment. Letthere beasystem ofparticles with position vectorsTI, r2,Tnrelative toapoint 0,and letforces PI,P2,Pn actonthem.Wedefine the totalforceFofthissystem asthe vector sum oftheforces, i.e., (10.101) F=2)P.. Themoment oftheforcePaboutOisr,XP,by(9.301). Wedefine the totalmoment Goftheforce system about the basepoint asthesum ofthesemoments, i.e., n (10.102) G= ]rfXP*. Thescalar components ofthetotal forceandthetotalmoment onaxesOxyz areeasy towritedown. Letxi,yt,zlbethecoordi- nates ofthetthparticle andXitYiyZithecomponents ofPt. Then thecomponentsofFare (10.103) x=5)xt,Y=5)Yifz=2)zt, andthecomponents ofGare (10.104) L- N 259 260 MECHANICS INSPACE [Ssc. 10.1 Since thescalarmoment aboutOxisthecomponent alongOxof thevector moment about 0,itisevident that L,M,Narethe total scalar moments about theaxesOxyz ;thus, forexample, L isthesum ofthescalar moments ofalltheforces about Ox. Change ofbase point. Itisclear thatFdoesnotdepend onthechoice ofbasepoint0. Ontheother hand,Gdoesdepend onthis choice. Letussee howGchanges whenwechange thebase point from to0', where 00'=a. Ifrj, ig,r'naretheposition vectors oftheparticles relative to0',wehave (10.105) Ti=r(+a. Then,ifG'isthetotalmoment about 0',wehave G'=riXP, t=l =J)(r,-a)XP,; 1=1 and so,by(10.101) and(10.102), (10.106) G'=G-aXF. Thisequation showshowthetotalmoment changes withchange ofbase point. Equipollent force systems. InSec.2.3,wegave thegeneral definition ofequipollence butused itonly intherestricted sense ofplane equipollence. Werecall thattwoforce systems areequipollentif(inthelan- guage usedabove) thetotal forces ofthetwosystems areequal, andalsotheir total scalarmoments about anarbitraryline.We shallnow establish thefollowing fundamental theorem: //twoforce systems areequipollent, they have thesame total forceand thesame totalmoment about anarbitrary base point thesame forboth systems. Conversely, iftwoforce systems have thesame totalforceandthesame totalmoment about some onebase point tthen theyareequipollent. SBC. 10.2] STATICS INSPACE 261 LetSiandS2betwoforce systems and anybase point. The total forces ofthetwosystems willbedenoted byFI,F2, andtheir totalmoments about byGi,G2,respectively. If/Siand&areequipollent, thenFI=F2from thedefinition ofequipollence. Further, thescalar moments ofSiandS2 about anylinethrough areequal, andsothevectors GIandG2 have thesame component along any linethrough 0;hence Gi=G2.Thus,foranarbitrary basepoint 0,wehave (10.107) F!=F2,G!=G2, which establishes the firstpart ofthetheorem. Toprove theconverse, wemust show that SiandSzare equipollentif(10.107) hold forsome onebase point O.The first condition ofequipollence, namely, theequality oftotal forces,isevidently satisfied;itremains toprove thatthescalar moments ofSiand$2about any lineareequal. Iftheline passes through 0,theequality ofscalar moments follows at oncebyprojecting theequal vectors GI,G2onthe line. Ifthe linedoesnotpassthrough 0,let0'be-anypointonit.Thetotal moment GJofSiabout 0'isexpressed interms ofFIandGi asin(10.106); there isasimilar expression forthetotalmoment G2ofSzabout 0'.Those vectors areobviously equalbyvirtue of(10.107), andsothescalarmoments inquestion arealsoequal. Theproof ofthetheorem isnowcomplete. Ifthetotal forceFofasystemiszero,and alsothetotal moment Gabout some one.base point,itfollows thatFandG arezero for allbase points. Wesaythen that thesystemis equipollenttozero. 10.2.EQUILIBRIUM OFASYSTEM OFPARTICLES InChaps. IIandV,wedeveloped thegeneral principles ofstatics anddynamics inaplane. Inestablishing these principles, wesometimes gave theresults inthree-dimensional form, where there wasnoparticular difficulty involved. Now wehave todevelop general principles inthree dimensions, and itmight bethought thatthenewworkwould have tobe builtontopoftheold.That isnotthecase. Sinceweare now inpossession ofthepowerful vector method,itisonthe whole simpler toestablish thegeneral principles directly from thebasic laws ofSec. 1.4.That iswhatweshall do,except 262 MECHANICS INSPACE(SEC. 10.2 inthose cases where thevector method offers noadvantage. Inthemain, therefore, therest ofthebook islogically inde- pendentofPartI,except forthelaws ofSec. 1.4. Necessary conditions ofequilibrium. Forasingle particle, thecondition ofequilibriumis (10.201) P=0, wherePisthevector sum oftheforces acting ontheparticle. Letusnow consider asystem ofnparticles inequilibrium. LetPdenote theresultant oftheexternal forces acting onthe ithparticle. Inaddition totheexternalforces, there acton each particle anumber ofinternal forces duetotheother particles ofthesystem. LetP#denote theforceontheithparticle due tothejthparticle; bythelawofaction andreaction, these inter- nalforces satisfy (10.202) Pt/+P,t=0. Since each particleisinequilibrium,itfollows from (10.201) thattheexternal andinternal forces satisfy theequations P!+04-Pi2+Pis+'+Pm=0, (10.203)p2+Psi++P23++P2n=0, Pn+Pnl+Pn2+' ' '+P,n-l+=0. When weaddthese equations, theinternal reactions cancel on account of(10.202), andso (10.204) F=0, whereFisthetotal force oftheexternal force system, viz.,]P- Let rdenote theposition vector oftheithparticle relative toabase point 0. Ifwemultiply theequations (10.203) vec- torially byti,r2,rninorder, weobtain XPi++riXPi2+nXPis+ +riXPm=0, r2XP2+r2XP++r2XP2s+ (10.205) +r2XP2n-0, rnXP+rXPi+rXP2+ +rnXPn,-i+=0. SBC. 10.2J STATICS INSPACE 263 Now, riXPi2+r2XP=(ri-r2)XPi2 0, since ri r2andPi2lieinthesame line [cf.(9.303)]. Hence, onaddition oftheequations (10.205), theterms symmetrically placed with respect tothelineofzeros cancel inpairs, andwe get (10.206) G=0, whereGisthetotalmoment about oftheexternal force n system, viz.,J)rXP.. Thus, forasystem inequilibrium, FandGboth vanish, andso (inthelanguage ofSec. 10.1)wehave thefollowing general result :* //asystem ofparticles isinequilibrium, then theexternal force system isequipollenttozero. Interms ofthetotal forceFandthetotalmoment Gabout anybase point 0,theabove statement isequivalent tothetwo vector equations (10.207) F=0,G=0. Resolving vectors along rectangular axesOxyzandusing the notation of(10.103) and(10.104), wegetthefollowing sixscalar conditions ofequilibrium: (10.208) X=0,Y=0,Z=0; (10.209) L=0,M=0,N=0. Inthisform theconditions appear asgeneralizations of(2.308). Thewhole theoryofstatics restsontheequations (10.207). Most frequently, these equations areapplied toarigid body, treated asawhole. Butthey arevalid foranysystem, which maybeapart ofarigidbody orapiece ofanelastic material, or even avolume offluid. Asanexample, weshall presently discuss theequilibrium ofaflexible cable inspace (cf.Sec.3.4 fortheplane case). Theequilibrium ofarigidbodywillbe considered inmore detail inSec. 10.4. *Thiscondition isequivalent totheconditions obtained inSec. 2.3. 264 MECHANICS INSPACE [SBC. 10.2 Curves inspace. Werequire some elements ofthegeometry ofcurves inspace; they areofimportance apart from thepresent connection and willbeusedagain later. LetCbeacurve inspace andAanypointonC.LetPbeany other point onC,distant sfromA(sbeing measured along the curve). Theunitvectori,tangent tothecurve atP,isclearly avector function ofs.Since iiremains equal tounity along C,wehave (10.210)i'Ts"' Itfollows thatthevector di/dsisnormal toCateach point P. Let1/p(pistheradius ofcurvature ofCatP)denote themagni- tude ofthisvector. Thenwemay write (10.21D I-i, wherejisaunit vector normal toC;itistheunit principal normal vector. Theplane ofiandjiscalled theosculating plane. The unitbinormal vectorkatPisdefined asfollows: Itis normal toboth iandjand issodirected that(i,j,k)isaright- handed triad. Theequation (10.211)isthefirst oftheFrenet-Serret formulas. Thecomplete setofformulas is dij djk idk_ j -T" > ~7~=~~ >T~==' as p~ds Tp ds T where risacertain scalar, called theradius oftorsion.*These formulas areeasily proved. Since j.-0, k.-0, Jds'ds' itfollows that (10.213) g=ai+6k,*=ai+ffl, *Wenotethatpisnecessarily positive, since itisdefined asthereciprocal ofthemagnitude ofdi/ds} rmaybepositive ornegative. Sac. 10.21 STATICS INSPACE 265 where a,6,a,ftarescalars. Ondifferentiating therelations (10.214) i-j=0, j-k=0,k-i-0 andusing (10.211) and(10.213), wefind (10.215) a=-^ a=0, 6+=0. Hence, writing b=1/r,weobtain (10.212). Foracurve C,drawn onasurface,thevector iisnecessarily atangent toS.Butthevectorjisnotnecessarily normal to S;itmayevenbeatangent toS,asinthecasewhereSisa plane.Ifjisnormal toSateach point ofC,thecurve iscalled ageodesic onS. Flexible cables. Letusnowconsider aflexible cable inequilibrium under the action ofknown external forces andthetensions atitsends. -Ti k FIG. 107. Forces onanclement ofcable. Figure 107shows aninfinitesimal portion PQ,Pbeing ata distance sfromoneendofthecable. Leti,j,kdenote theunit tangent, principal normal, andbinormal vectors atP.The forces acting ontheelement PQ(length ds)maynowbedescribed asfollows: (i)aforce TiatP,whereTisthetension atP; (ii)aforce (T+dT)(i+di)=(T+dT)[i+(j/p) da]atQ, whereT+dTisthetension atQ; (iii)aforceRds (Rii+R%j+72gk)ds,whereRisthe external force perunitlengthofthecable. (This force actsat some unspecified pointoftheelement PQ.) Theelement PQ isasystem inequilibrium under these forces, andsowemayapply theconditions (10.207) toit.From the first ofthese conditions, wehave 266 MECHANICS INSPACE [Sao. 10.3 (10.216)-Ti+(T+dT)(i+i da)+Rtf+flak)da=0. Thesecond oftheconditions (10.207)issatisfied identically tothe firstorder inds.From (10.216), weatonce obtain thescalar equations (10.217) R*=0. These arethegeneral equations ofequilibrium. They enable us tofindtheform ofthecable and alsothevariation intension alongit.Inparticular, thelastofthese equations tellsusthat theosculating plane ateach point contains theexternal force vector. Example. Alight cable rests incontact withasmooth surface S,under no forces except thereaction ofSand thetensions atitsends. Itisrequired to find thecurveCinwhich thecable reststandalso thetension ateach point. Theexternal force vectorRdsisthereaction ofthesurface Sonthe element. Since thisreaction isnormal toS,itisalsonormal toC,andso Ri-0.But fls 0,bythelastof(10.217), andso R-#aj. Itfollows thattheprincipal normal vectorjisnormal toSateachpoint ofC. Hence,Cisageodesic onS.Thus, toconstruct ageodesic joining twogiven points onasurface, wehavemerely tostretch alightthread between these points. IfSisasphere, Cisanarcofagreat circle; ifSisacylinder, Cisa curve onthecylinder which maps intoastraight line,when thecylinder is cutalong agenerator andunrolled onaplane. Again, sinceR\ 0,thefirstequation in(10.217) gives - Thus thetension Tisconstant; inparticular, thetensions attheends are equal. 10.3.REDUCTION OFFORCE SYSTEMS Ifwesucceed infinding asimple forcesystem Sf ,equipollent toagiven system S,wesaythatwehave reduced thesystem S tothesystem S'.Weshall presently consider, insome detail, the SEC. 10.3] STATICS INSPACE 267 reduction offorce systems; but,before doing this,itisconvenient tohaveavector description oftheparticular forcesystem known asacouple. Moment ofacouple. AsinSec.2.3,acoupleisdefined asapair ofparallel forces, equal inmagnitude butopposite insense. Figure 108shows a couple consisting oftheforcesPandPapplied atthepointsAandB,respectively;isanypoint inspace. Thevector -PG=pxP FIG. 108.Themoment ofacouple. moment Gofthiscouple about iseasily found; denoting OBby randBAbyp,weobtain (10.301) G=rX(-P)+(r+p)XP=pXP. Thisvalue ofGisindependent oftheposition ofthepoint 0. Inother words, acouple has thesamemoment about allpoints inspace. Thus thevectorGmayberegarded asafreevector; itisperpendicular totheplane determined bytheforces P,P ofthecouple;itsmagnitudeisp'P,where p'istheperpendicular distance between thelines ofaction ofthese forces (Fig. 108). Sincetwocouples which have thesamemoment areequipollent force systems, acoupleiscompletely specified (asfarasequi- pollenceisconcerned) byitsfreemoment vector, or,briefly, its moment. Thus, whenwespeakofacouple G,wehave inmind anyone ofaninfinite number ofcouples, each ofwhich has moment G. Toavoid confusion indiagrams, thearrowheads indicating couples maybemarked withacrossbar, asinthefigure. 268 MECHANICS INSPACE [Sue. 10.3 Compositionofcouples. Letthere beasystemofforces consisting ofanumber of couples Gi,G2,.Thetotal force ofthissystemiszero,and thetotalmoment about anypointisclearly (10.302) G=Gi+G2+- . Thus, asystem consisting ofcouplesisequipollenttoasingle couple;itsmoment isequaltothevectorsumofthemoments ofthe individual couples. Inother words, couples arecompounded bytheparallelogramlaw. Exercise. Forces withcomponents (2,0,0), (1,0,0), (1,0,0)actat thepoints (0,0,0),(0,1,0),(0,0,1),respectively. Show thattheycanbe reduced toacouple, and find itsmagnitude anddirection. Reduction ofaforce system toaforceandacouple. Consider ageneral forcesystem S,with total forceFandtotal moment Gwith respecttoabase point 0.Consider alsoa Qsecond force system 8',consisting only ofa Fsingle forceFapplied atOandasingle couple G.Obviously, 8'isequipollent toS.There- foreageneral force system canalways bereduced toasingle force applied atanarbitrary base point, together withacouple. O Just aswerepresent asingle force byan FIG. 109. Repre-arrow, sowecanrepresent ageneral force sentation ofagen- , ,,. , ji .TV i/w\ eralforce system bysystem byadiagram such asthat in*ig.109; aforce Fand athisshows aforceFacting atabasepointcoupe'andacouple G.Although thecoupleGis afree vector,itisconvenient todraw itoutfrom thebase point 0. > Ifwechange thebasepointfrom to0',where 00'=r,wedo notalterF,butthemoment about 0'isnotG;itisfound by adding toGthemoment ofFabout 0';thisgives [cf.(10.106)] G'=G-rXF. Hence, under achange ofbase point from to0',theforceF andthecoupleGbecome F'andG',respectively, where (10.303) F'=F, G'=G-rXF. SBC. 10.3] STATICS INSPACE 269 WenotethatF'=FandF'G'=FG;inwords, thescaJars FandFGareinvariant under achange ofbase point. Ifeither ofthese invariants vanishes foronechoice ofbase point, then it vanishes forallchoices ofbase point. Reduction toawrench. Awrench consists ofaforceFandacoupleGwith parallel representative linesegments. This relation between FandGis expressed bythevector equation (10.304) G=pF, where pissome scalar having thedimensions ofalength. The quantities pandFarccalled thepitchandintensity ofthewrench, respectively. The lineofaction oftheforceFiscalled theaxis ofthewrench. Ageneral force system canalways bereduced toawrench. Weshallnowshowhow this isdone. Letusfirstreduce the system inquestion toaforceFatabasepoint andacouple G. Changing thebasepoint to0',where 00'=r,weobtain aforce F'andcouple G'.These constitute awrench if (10.305) G'=pF'. Since, by(10.303), F'=F, G'-G-rXF, theequation (10.305)issatisfied ifrandpsatisfy (10.306) G-rXF=pF. Thisis,infact,avector equation forr(theposition vector of0') andp(thepitch ofthewrench). Letustake asorigin ofrectangularCartesian coordinates, and let(Fi,F2yFZ), (Gi,G%,(73)denote thecomponents ofF,G, respectively. Thevector equation (10.306)isequivalent tothe scalar equations gi~yFt+zFz_G2-zFi+xF9_OB-xFz+yFl_ 7[ Fl F,-P> where x,y,zarethecoordinates of0'.These equations show that F',G'constitute awrench provided 0'liesonthestraight 270 MECHANICS INSPACE [SEC. 10.3 linewithequations Gl~VF*+zF*G*~zFl+xF*G*~xF*+^Fl --- ft--p-- r1 r2 T3 Wenote that,if(x,y,z)isanypointonthisline,then (x+kFi, y+kFz, z+kFs),where kisanyscalarfactor,isalsoonit.It follows that thislinehasthedirection ofF;itistheaxisofthe wrench towhich thesystemisreduced. Thepitchpisfound bytaking thescalar product ofFandthe vectors onthetwosides of(10.306). Wefind F-G=pF2 ; therefore, (10.308) p= Itis,ofcourse, notaccidental that thepitch oftheresulting wrench isafunction oftheinvariants FandFG. Ifp=0,thewrench degenerates intoasingle force; inthis caseFG=0.Ifpisinfinite, thewrench degenerates intoa couple; inthiscaseF=0.Ineach ofthese special cases, we have aforce system equipollent toaplane system offorces. Conversely,iftheforce systemisequipollent toaplane system offorces, thenoneorother ofthese special casesmust arise. Exercise. Inthereduction ofagiven forcesystem toaforceandacouple, thecoupleGdepends onthebase point. Forwhat base pointsisGleast? Reduction ofasystem ofparallel forces. Asetofparallel forces isasystem ofparticular importance inmechanics, e.g.,theweights ofanumber ofparticles. Anyftparallel forcesmaybedenoted by/bJP,&2P, knP, wherefci,&2,*knarescalars. Selecting abase point 0,we firstreduce thissystem toaforceFatandacouple G.Let r,(s=1,2, n)denote theposition vectors, relative to0,of thepointsofapplication oftheseveral forces. Then, (10.309)F-(W-fcP> (r-X*-p)-(2fc'r ')XP=rXF, Sac. 10.3] STATICS INSPACE 271 where (10.310) k=]gk,,r=(Vk,T.)/k. i \-i' Thisreduced system isclearly equipollent tothesingle force F, applied atthepointCwith position vector r. Thus asystem ofparallel forces canbereduced toasingle force, unless A;=0.Ifk=0,itcanbereduced toacouple. The point C,with position vector rgiven by(10.310),is called thecenter ofthesystem ofparallel forces. Itspositionis determined solely bythevectors r8andtheratios ofthenumbers ki,k%, kn.Itisunaltered byturning theseveral forces about their points ofapplication, provided they retain their magnitudes andremain parallel tooneanother. Iftheforces inquestion aretheweights oftheparticles ofa system, thepointCisthecenter ofgravity ofthesystem (cf. Sec. 3.1). Thereduction ofsome special force systems. Theforce systems encountered inpractical problems areoften extremely complicated. However, thedetails ofsuch aforce system arerelatively unimportant when itactsonarigidbody; ifweknow thetotalforceandthetotalmoment, weknow allthat isessential forthediscussion ofequilibrium. The total force andthetotalmoment playanequally important part indynamics (seeChap. XII). 1.Analysis offorces onanairplane. Figure 110shows anairplane inflight. Cisthemass center. The orthogonal right-handed triad ofunitvectorsi,j,kisfixed intheairplane; jisperpendicular totheplane ofsymmetry andpoints totheright;iandk lieintheplane ofsymmetry. The direction ofiisfixed insome conven- tionalway (e.g., paralleltothoairscrew axes), soastobenearly horizontal andpoint forward when theplaneisinnormal flight; kwillthenbedirected nearly vertically downward. Theforces acting ontheairplane areasfollows: (i)Theweights ofthevarious parts. These constitute asystem of parallel forces andcanbereduced toasingle forceWacting vertically down- ward through C\Wisthetotalweight oftheairplane. (ii)Thethrust, ordriving force,Pduetotheairscrews. This force isin thedirection ofthevector iornearly so. (iii)Forces arising from theaction ofthe air.These forces aredue mainly tovariations inpressureoverthewing surface andinalesser degree 272 MECHANICS INSPACE [SEC. 10.3 tofrictiftn;theyformaverycomplicated system. Reducing thissystem to aforceFatCandacouple G,weresolve asfollows: F=Xi+Yj+Zk, G-Li+Mj+Nk. Fornormal flight,Xisthedrag, Zisthelift,andMisthepitching moment; YtL,Narezero. Theprecise terminology ofaerodynamic theory isnotquite sosimple asthis. Under normal flight conditions, however, the differences aresmall. Thetheoretical determination oftheforcesystem (F,G)isa*very difficult problem inhydrodynamics; and, inpractice, experimental methods areused. Amodel oftheairplane (ortheairplane itself) ismounted inawind tunnel. Direct measurements arethenmade oftheforcesystem required tokeep the model atrestinastream ofair. k FIG. 110. Reference vectors foranairplane. Ananalysis oftheforces onabullet orshell follows thesame lines. In thiscase, however, thedriving forcePisabsent. 2.Analysis ofstresses inabeam. Consider abeam inequilibrium and letOxbealineinthedirection ofits length. Weimagine thebeam cutintwobyaplaneIIperpendicular toOx atA(Fig. 111).Weshalldenote byRthepart ofthebeam totheright ofn, andbyLthepart tothe left. Theforces onLareasfollows: (i)Applied forces, such asgravity orexternal loads. These areequi- pollent toasingle forceFatAandacouple G. (ii)Forces exerted acrossnbyRonL.These areinternal forces forthe whole beam, butexternal forces forthesystem L.They arecalled the stresses across theplane sectionnandareequipollent toaforceSatAanda couple M.Introducing theorthogonal triad ofunitvectorsi,j,k,asshown, wewrite S*Sd+Sd+S8k,M-Mil+M2j Thefollowing terminologyisused: Si=*tension, 2,$3=shearing forces, Mi twisting couple,M2,M3bending moments. SEC. 10.4] STATICS INSPACE 273 Iftheapplied forces areknown, then F,Gareknown andwecanfindS,M. Wehave merely toapply theconditions ofequilibrium (10.207) toL, obtaining S--F,M--G. Ifwechange thesection IIbyvarying thedistance xofAfrom 0,FandGare known functions ofx\hence, SandMareknown functions ofx.Inother words, there aretwovector functions S(z),M(x) [orsixscalar functions Si(x),Sz(x),-Ms(x)]which give, foreach value ofx,aforce system equipollent tothestresses across thecorresponding cross section ofthebeam. Failure inanengineering structure, such asabridge,isduetoexcessive stress. Theengineer mustknow inadvance ifanygivenbeam orgirder islikely tofailunder theloads which itwillbecalled ontosupport. FIG. 111. Reference vectors forreactions inabeam. Although thevalues ofSandMdonotgive acomplete picture ofthe internal stresses, they arethequantities which theengineer calculates inorder toseewhether ornotastructure issafe. Theabove analysis also applies innaval architecture. Regarding the hullofashipasabeam, subject toknown applied forces, wecandetermine a force system (S,M)equipollent totheinternal stresses across anysection perpendicular toitslength. Ashipmustbesoconstructed that itwillwith- stand theaction ofstresses (S,M)arising from theapplied forces ofweight andbuoyancy. Inastorm theshipmaybesupported bywaves underbow andstern; thentheforces ofbuoyancy areconcentrated there, andtheshipis indanger of"breakingitsback." 10.4.EQUILIBRIUM OFARIGIDBODY Necessary and sufficient conditions ofequilibrium. Inourmathematical model, arigidbody isasetofparticles whose mutual distances areinvariable. Letusnow consider a rigidbody acted onbyexternal forces. Reducing theforce 274 MECHANICS INSPACE [SEC. 10.4 system toasingle forceFatabasepoint andacouple G,we knowby(10.207) thattheconditions (10.401) F=0,G= Rarenecessary forequilibrium. Wenowmake useoftherigidity ofthebody toprove thatthese conditions arealsosufficient, so thatthebodymust beinequilibriumifthey aresatisfied. Letussuppose thatarigidbody acted onbyexternal forces, satisfying (10.401),isnotinequilibrium; then theparticles ofthebody willbeonthepoint ofmoving. Thismotion willbe prevented byintroducing the following constraints (Fig. 112): (i)Thepoint (taken inthe body)isfixed; thisleaves the body freetoturnabout 0. (ii)With origin 0,wedraw a unitvectori;theparticle Aat itsextremityisconstrained to slide inasmooth tubewith axis inthedirection ofi.Thetwo constraints nowintroduced fix allpoints ofthebody onthe lineOA,but stillpermit thebody toturnabout this line. (iii)With origin 0,wedraw aunit vectorj,perpendicular toi;theparticle Batitsextremityisconstrained tomove between twosmooth planes parallel totheplaneOAB. Ifthese planes areclose toeach other, thisconstraint willprevent the motion ofB. These three constraints together prevent anymotion ofthe body, andsoitmust remain atrest. Itistherefore inequi- librium under theaction ofthegiven external force system and thereactions ofthese constraints. Now thereactions ofconstraint areequipollent toaforce F'atandacouple G'. Since thebodyisinequilibrium, F+F'-0,G+G'=0; and so,by(10.401), (10.402) F'-0, G'-0.FIG. 112. Constraints preventing mo- tionofarigid body. SBC. 10.4] STATICS INSPACE 275 Inview ofthesmoothness oftheconstraints atAandJ5,we seethattheforces ofconstraint are (i)aforceP=PJ+P2j+Pskapplied at0; (ii)aforceQ=Q2j+Qakapplied atA(position vector i relative to0) ; (Hi) aforceR=RJs.applied atB(position vectorjrelative to0). Thevector kisaunitvector completing theorthogonal triad i,j,k.Oncalculating F,G'interms ofPi,P2,P3,Qz,Qs,Rz,we findthatthesixscalar equations contained in(10.402) imply Pi=P2=P3=Q2=Q3=#3=0. Hence theconstraints introduced actually exertnoreactions, and sothebody remains inequilibrium even ifthey areremoved. The sufficiency oftheconditions (10.401)isnow established. Itispossible tostate theconditions ofequilibrium informs other than (10.401). Forexample,itiseasy toseethat,ifthe external forces havenomoment about each ofthree non-collinear points, then theconditions (10.401) aresatisfied. Conversely, iftheexternal forces satisfy (10.401) forsome particular base point, theyhavenomoment about anypoint. Thus,ifG,G',G" denote thetotalmoments oftheexternal forcesystem about each ofthree non-collinear points, theconditions (10.403) G=G'-G"- areboth necessary and sufficient forequilibrium. Occasionally theconditions (10.403) areeasier toapply than (10.401). Applications. Theconditions (10.401) willnowbeapplied tosolvetwoproblems. Example1.Figure 113shows apulley, withradius randcenter A,rigidly attached toahorizontal shaftBCD. This shaft isfreetoturn insmooth bearings atBandC;theendDprojects beyond thebearing atC,andtoitis rigidly fastened acrankDEwithhandle EH. TheanglesBDEandDEH areright angles. AweightWisattached tothelower endofacordpassing round thepulley, theother endofthecordbeing fixed tothepulley. To raiseW,aman applies aforcePat//inadirection perpendicular toBCD andmaking anangle <f>withthehorizontal. Itisrequired tofindthemagnitude ofPandalsothedirections andmagni- tudes ofthereactions RandR'atBandC,respectively. 276 MECHANICS INSPACE [SEC. 10.4 Leti,j,kbeanorthogonal triad ofunitvectors atA,ilying alongAC andkpointing vertically upward. Lengths aredenoted asfollows: BA=AC=a,CD6,DE=c,EH=d.Then,ifDEmakes anangle withthevertical, theforces acting onthewhole system canbedescribed as follows (position vectors being taken relative toA): (10.404)aforceP=Pcos</>jPsin <k at(a+6-f aforceWkatrj, aforceR=R2j+Rfcatai, aforceR;=#JjH-/e'3katai.,+csinj+ccos6k, t-Wk FIG. 113. Pulley andshaft turned byacrank. Reducing thisforcesystem toaforceFatAandacouple G,wefind F=(Pcos</+R2+R'2)j+(-P sin<t>-W+#3+ G=[(a+b+d)i+csinj-fccos k]X[Pcos^j-Psin k] +rjXWk-aiX(flJ+#3k)+aiX(/&+ =[JTr-PCcos(0-*)]i+[P(a+*>+d)sin*+a,K 3-afl'8]j cos<*>-aR 2 Forequilibrium, these vectors must vanish. Equating them tozeroand performing some simple calculations, wefind (10.405)--"~P sm ,Tf C3=--+P sin*' These equations constitute thesolution ofourproblem. Thevalue ofPgiven above isleastwhen<f>=0,i.e.,when theforce at//is perpendicular tothecrankDE; thismay alsobeseenquite simply bytaking moments about thelineBD. Thus, toraisetheweight withtheleast effort, themanshould push atright angles tothecrank. Asfarastheman iscon- cerned, theactual position ofthepoint //inthehandle isofnoimportance; achange indmerely alters thereactions atBandC. SBC. 10.7] STATICS INSPACE 297 thisis,therefore, theonly position ofequilibrium, observe thatAsforstability, we SinceV=forqi q$ and ispositive forallother values,itfollows thatVisaminimum attheposition ofequilibrium; theequilibriumis stable. Example 2.Weshallnow discuss adevice known asHooke's joint. This isused totransmit atorque or couple from one axis toanother, inclined tothe first. Figure 119shows the essential features ofthejoint.AB isashaft oraxis, branching into thefork BCD] A'B' isanother axis, with fork B'C'D'. These forks arecon- nected byarigidbodycomposed of twobarsCD,C'D', joined perpendic- ularly attheircommon center O. TJie linesAB,A'B'meet atwhen produced andareperpendicular to CD, C'D', respectively. There are smooth bearings atC,D,C',D',and theaxesAB,A'B' arefreetoturn insmooth bearings atAandA'. Let Ibeaunitvector inthedirec- tionABand I'aunitvector inthe direction B'A'. When acouple A/I isapplied toABCD, thesystem will move unless motion isprevented byother forces. Wepropose tocalculate thecouple M'V applied toA'B'C'D', which (together withMIandthe reactions atthebearings) gives equilibrium. Tosolve suchaproblem, thegeneral planisasfollows:(i)select general- izedcoordinates qi,q*,- -forthesystem; (ii)calculate anexpression of theform (10.708) forthework done inadisplacement; (iii)equate the generalized forces tozero,andsolve. Wenotethatthefixedelements ofthesystem arethelinesAB,A'B'and thepoint O.SinceABCD canmerely turnabout AB,asingle coordinate 6(theangle turned through)issufficient tofixit.WhenABCD isfixed, thelineCD isfixed. Now thelineC'D'mustbeperpendicular tobothCD andA'B' (from theconstruction ofthejoint) ;hence C'D' isfixed,andsothe angle suffices tofix,notonlyABCD, butA'B'C'D' also. Thesystem has onedegree offreedom, and isageneralized coordinate. When 8increases to6-f50,A'B'C'D' turns through some small angle 50'; thedisplacementsofABCD, A'B'C'D', CDC'D' areasfollows: ABCD-. arotation 501, A'B'C'D': arotation 50'I', CDC'D': arotation dn=5wii+5nij+5nk,FIG. 119. Hooke's joint. 298 MECHANICS INSPACE [Sue. 10.7 wherei,j,kareunit vectors along CD,C'D'andperpendicular tothem. Since nowork isdonebythereactions atthebearings andtheinternal reactions atC,C",D,/>',wehave [cf.(10.704)] (10.715) 6WmMl dO1+MT 50'I' =MSB+M'60'. Wemustnow find 69'interms of60.ThepointDbelongs totworigid bodiesABCD andCDC'D'. Equating thetwoexpressions for itsdis- placement fcf(10.501)), wehave (10.716) 601Xai-6nXai, where a=OD OD'. Similarly, byconsidering thedisplacement ofD't wefind (10.717)60'I'Xaj-5nXaj. Now, resolving along i,j,k,wehave I= sin <j+cos</>k,I' sin tf>'i+cos#'k, where<f>, <J>'aretheangles between kandAB,B'A', respectively. Sub- stituting these values forIand I'in(10.716) and(10.717), weobtain lcos*80=dn*>~sin*5*"5n |cog^^,_fi^_gm^5(?/_gn^ Equatingtheexpressions for5^a,weobtain (10.719) 60f-cos sec</ ^. Substituting thisvalue in(10.715), weget (10.720) 6W-(Af-fM'cos sec </>')5^, andsothesingle generalized force is 6M-fJf'cos sec <'. Forequilibrium, thismust vanish, and sothecouple required tohold A'B'C'D' isMT,where M'--3fsec cos '. Thefraction ofthetorqueMtransmitted through thejoint issec<t>cos0'. Example3.InFig. 120,ABrepresents ashaft freetoturnabout ahori- zontal axisL,perpendicular toABatA;DErepresents aheavy barthreaded ontheshaftABandperpendicular toit.Thepitch ofthethread isp,so that,whenDEturns through anangle about AB,Emoves alongAB through adistancep<f>.Thepoints C,C'arethecenters ofgravity ofAB, DE,respectively. Wewish tofindthepossible positions ofequilibrium of thissystem under gravity,allfriction being neglected. Westart with astandard configuration inwhichAB isvertical andDE liesinthevertical planeIIperpendicular toLatA.LetXQdenote the distance AEinthisposition. Wepass toageneral configuration bynwing- SBC. 10.7] STATICS INSPACE 299 ingABthrough anangle andturningDEtomakeanangle 4>with II.The angles and<t>aregeneralized coordinates. Interms ofthem, thepotential energyis (10.721) V--wa cos9-W[(XQ 4-p<)cos+6sin9cos <], where aAC, 6=EC'and10,Wdenote theweights ofAB,DE, respec- tively. Theconditions ofequilibrium are (10.722)wasm 6+W(XQ -fP^)sin Wbcos cos <=0, dV_ =,_PFpcos+Wbsin sin<j> iO<f>0. With numerical values fortheconstants, thisequation canbesolved graph- ically orotherwise. Interms of<,6isgivenbyElimination ofgives for <theequation (10.723) Wb2sin20-2wpa -f (10.724) tan0 Some interesting results canbededuced without solving (10.723). Ifpis small, apositionofequilibrium occurs forsome small value of<,i.e.,with Fio. 120.--AbarED isthreaded onashaftAB,which canturnabout ahorizon- talaxisL. DEclose toII.Forthisposition, tan6isfinite, sincepand <arcsmall of thesame order. Another position occurs near =$TT;forthis, tan is small, andsoAB isnearly vertical. Ifpislarge, theright-hand sideof (10.723) mayexceed Wb*forall4>;inthiscase, there isnoposition ofequilib- rium, andthebarDEsimply runsdown theshaftAB,turning asitgoes. Example 4.The lasttwoexamples considered above arethree-dimen- sional incharacter. Weconclude withanapplication ofthemethods of workandenergy toatwo-dimensional system withmany degrees offreedom. 300 MECHANICS INSPACE [SEC. 10.7 Consider achain ofnequal uniform rods,smoothly jointed together and suspended fromoneendA\. (Figure 121shows thecasen=5).Ahori- zontal forcePisapplied totheother endAn+iofthischain. Itisrequired tofindtheequilibrium configuration. FIG. 121.Achain ofrods pulled byahorizontal force. Asgeneralized coordinates, wetake theinclinations (tothodownward vertical) 0i, 2,Onoftheseveral rods inorder. Ifeachrodhaslength 2a andweight w,thepotential energy inageneral configurationis V=wacos 0iw(2a cos 0\+acos 2) w(2a cos0iH-2acos 2+ =-wa[(2n-1)cos 0i-f(2n-3)cos 22acos n.j-facos0) -f3cos0_j+cos0]. Inasmall virtual displacement theworkdonebygravityis -87=-t0a[(2n-1)sin 0i50i+(2n-3)sin 2802+sin 50n], Thework donebytheforcePistheproduct ofPandthohorizontal dis- placement of*4nfi, i.e., P5(2a sin0i+2asin 2+ +2asin n). Adding thesetwoexpression, wofind, fortho tot.ilwork done, dW=[2Pcos 0i (2n-\}wsint]a50i (10.725) +[2Pcos 2-(2n-3)ivsin 2]a502 +-- -fI2/3cos tt^sin n]a50B. Thismust vanish forarbitrary values of50i,602,50n,ifthodisplace- ment isfromaposition ofequilibrium. Hence, equating tozerothebrackets ontheright of(10.725), wefind SEC. 10.8] STATICS INSPACE 301 2P /tan 0i= - 2P (10.726) * fl2/>tan U=w These equations givetheinclinations oftherods tothedownward vertical intheequilibrium configuration, thetangents form aharmonic progression. 10.8.SUMMARY OFSTATICS INSPACE I.Conditions ofequilibrium. (a)Forasingle particle (necessary and sufficient): (10.801) P=0, or * (10.802) X=0,Y=0,Z=0. (b)Foranysystem (necessary), orforarigidbody (necessary and sufficient): (10.803) F=0,G=0. (F=totalforce,G=totalmoment.) (c)Foranysystem with workless constraints (necessary and sufficient): (10.804) dW=0, (5W=workdonebyapplied forces). II.Equipollence. (a)Conditions ofequipollence: (10.805) F=F,G=G'. (6)Anysystem offorces canbereduced toaforceFatan assigned point, together with acouple G. IfG=pF,the reduced systemisawrench. III.Displacementsofarigid body. (a)Finite displacements: (i)Anydisplacementofarigidbody with afixed pointis equivalent toarotation n(Euler's theorem). 302 MECHANICS INSPACE [Ex.X (ii)Ageneral displacementisequivalent toatranslation s, followed byrotation n. (6)Infinitesimal displacements: (i)Infinitesimal rotations compound vectorially. Theorder ofapplicationisimmaterial (ii)Forarigidbody withafixed point, thedisplacement ofaparticle ofthebodyis (10.806) 5nXr. (iii)Ingeneral, thedisplacement ofaparticle ofthebodyis (10.807) Ss+6nXr. IV.Work andpotential energy. (a)Work doneonaparticle: (10.808) BW=P-5s=X8x+YBy+ZSz. (b)Work doneonarigidbody: (10.809) 8W=F5s+G5n. (c)Work doneonageneral system: (10.810) 8W=Qifyi+Q2fy2+-+Qn8qn. (d)Work doneonaconservative system: n*y (10.811) dW=-57=-J)Sp Sqr. EXERCISES X 1.Aforce withcomponents (-7, 4,-5)actsatthepoint (2,4, 3). Find itsmoment about theorigin. Find also itsmoment about theline x=y-z, thepositive senseonthelinebeing that inwhich xincreases. 2.Arigidbody isacted onbyaforce withcomponents (1,2,3)ata point (3,2,1)andbyaforce withcomponents (1, 2, 3)atapoint (-3, 2, 1).Givethecomponents oftheequipollent forceandcouple attheorigin. 3.Aparticle ofweightwisplaced onarough plane inclined tothehori- zontal atanangle a.Ifthecoefficient offriction is2tana,findtheleast horizontal force across theplane which willcause theparticle tomove. Determine thedirection inwhich theparticle moves. Ex.X] STATICS INSPACE 303 4.Atripod consisting ofthree uniform rigid legs, each oflength 2a andweight w,supports acamera ofweight W,thelegsbeing smoothly jointed together atthetop.Thetripod stands onarough horizontal plane (coefficient offriction/*),thefeetforming anequilateral triangle. Find anexpression forthegreatest length ofasideofthistriangle consistent with equilibrium. 5.Prove, bytheprincipleofvirtual work, that foraninextensiblc cable (either freeorincontact withasmooth surface) TV constant, whereTisthetension andYdsthepotential oftheexternal force acting on anelement dsofthecable. 6.Aforce withcomponents (3,5,6)actsatapoint with coordinates (1,2,3),andaforce withcomponents (8, 2,Z)actsatapoint with coordinates(4,6, 7). Ifthepairofforces hasnoresultant moment about thex-axis, findZ. 7.Determine thepitch ofthewrench equipollent totwoforces ofmagni- tudes P,Q,inclined tooneanother atanangle a,theshortest distance between their lines ofaction beingc. 8.Asquare gateABCD, ofweightWandedge a,hashinges atBandC, thelineBCbeing vertical withBontop.Thehinge atCcansupport a downward thrust, butthat atBmerely supplies avertical axisofrotation. Thewind blows onthegate, exerting auniform pressure p.Thegateis kept inposition byalightropeattached totheouter upper cornerAandtoa pointEontheground, whereCE aandCEisaperpendicular tothegate. Find interms ofW,p,athetension intheropeandthemagnitude ofthe reaction ateach ofthehinges. 9.Three identical sphereslieincontact withoneanother onahorizontal plane. Afourth identical sphere restsonthem, touching allthree. Show thatthecoefficient offriction between thespheres isatleast (-\/3 \/2) andthatthecoefficient offriction between eachsphere andtheplane isat least(V3-V2)/4. 10.Denoting by 0,$theusual polar angles,findthepolar angles ofthe axisofaninfinitesimal rotation equivalent tothree infinitesimal rotations, allofthesame magnitude, with axeswhose polar angles are (0-60, 4>-45), (0=120,$=135), (0=60,=225). 11.Arigidbody receives insuccession three rotations about three mutually perpendicular intersecting lines fixed inspace, each rotation being through aright angleandthesenses being cyclic. Find theaxisandmagni- tude ofthesingle equivalentrotation. 12.Arigidbody receives afinite translation andafinite rotation (through anangle B)about anaxisDperpendicular tothetranslation. Show thatthe resultant displacement isequivalent toarotation (through anangle 0) about anaxisparallel toD,theposition ofthisaxisdepending ontheorder inwhich thetranslation androtation areapplied. 13.Show thatanyfinite displacement ofarigidbody isequivalent toa screw, i.e.,atranslation andarotation about anaxisparallel tothetransla- tion(Chasles' theorem). 304 MECHANICS INSPACE [Ex.X 14.Incoining torestonaslippery road, thewheel ofacartravels 10foot forward and2feetsideways, atthesame time turning through anangle of 180about itsaxle. Locate theaxisoftheequivalent screw displacement. 15.Show that, ingeneral, aforcesystem maybereduced toaforce acting along anygiven linetogether withanother force. (The lines ofaction ofthe twoforces aresaid tobeconjugate ) 16.Arigidbody isacted onbyaforceFatandacouple G.Pisan assigned point, with position vector rrelative to0.Show that there is asingle infinity oflines through Pabout which theforce system hasno moment; show thatthese lines lieinaplane andfind, inCartesian coordi- nates, theequationofthisplane. (The lines arecalled nullhnes, and theplane anullplane.} 17.Show that,ifarigidbodyisinequilibrium under theaction offour forces, theinvariant (FG)ofanytwo isequal totheinvariant (FG)of theother two. Show alsothat theinvariant (FG)ofanythree ofthe forces iszero. 18.Aheavy uniform inextensible cable hangs incontact with asmooth right-circular cone ofsemiverticul angle a,theaxisoftheconebeing vertical. Prove thatthecable hangs inacurve satisfying theequation (ID*+z*sin2a=z*(A+Bz}2 > where zisthedepth below thevertex ofthecone, <f>istheazimuthalangle, andA,Bareconstants. 19.InaHooke's joint (Fig 119)forcesystems (F,G)and (F',G')(includ- ingthereaction ofthebearings atA,A')actonthepartsABCD, A'B'C'D', respectively, being taken forbase point. Forequilibrium, show that couples G,G'must both beperpendicular totheplane CDC'D*. 20.There aretwoidentical rough stones, eachbeing anoblate spheroid ofserniaxes a,b(a>b).One islaidonahorizontal iloorandtheother balanced ontopofit,theaxes ofsymmetry being vertical. Confining atten- tiontodisplacements inavertical plane through theaxisofsymmetry, show thattheequilibriumisstable ifa2>362 . 21.Discuss thestability ofanyfour successive positions ofequilibrium forthesystem shown inFig. 120. Consider onlythecasewhere pissmall incomparison withaand 6. 22.Aforce1system isequipollent toaforceFatandacouple G;another force systemisequipollent toaforce F'atOandacouple G'.Prove that, iftheaxes oftheequivalent wrenches intersect, then FG'+F'G-(p+/>')F F', wherep,p'arethepitchesofthewrenches. 23.InaHooke's joint (Fig. 119) lettheangle 0,through whichABCD isturned about AB,bemeasured from zerowhenCD liesintheplane of ABandA'B'. Denoting byatheangle between ABandB'A', findthe angles <f>and'interms ofand a.Check youranswers byconsidering thespecial cases:(i) 0,(ii)6 %*. CHAPTER XI KINEMATICS. KINETIC ENERGY ANHANGULAR MOMENTUM Wenowapproach thestudy ofdynamics inspace. Weshall require (i)asimple way ofdescribing themotions ofparticles andof rigid bodies; (ii)methods ofcalculating kinetic energy and angular momentum. These items belong tokinematics(ifweunderstand theword to include mass aswellasmotion) andform thesubject matter of thepresent chapter. 11.1.KINEMATICS OFAPARTICLE LetOxyzberectangular axes fixed inaframe ofreference and I,J,Kunitvectors along them. Forany particle, with coordi- nates x,y,z,wedefine thefollowing vectors(cf.Sec. 1.3): (11.101)Position vector: r=xl+yj+zK, Velocity: q=-r=.rl+yj+zK, Acceleration: f=~=jcl+yj+zK. at Thesimplest wayofdescribing themotion ofaparticleisto givethevector function r(J). For,when risknown asavector function ofthetime t(i.e.,whenx,y,zareknown asscalar functions oft),wecantrace thepath oftheparticle and find itsvelocity andacceleration atanyinstant bydifferentiation. Wefrequently require expressions forthecomponentsof velocity andacceleration indirections other thanI,J,K.Two particular resolutions ofthese vectors willnowbeconsidered. Tangential andnormal components ofvelocity andacceleration. Figure 122shows thepathCofamoving particle A\AQis afixed point onC.ThearclengthAA Qisdenoted bys.From 305 306 MECHANICS INSPACE [SBC. 11.1 (11.101), weseethat thevector di/ds hascomponents dx/ds, dy/ds, dz/ds along I,J,K;itistheunittangent vector toC atAand willbedenoted byi. Forthevelocity ofAwehave K<"t--si. Hence, thevelocity ofa.particle is directed alongthetangenttoitspath, andhasmagnitudes. Fortheacceleration wehave xdq ..,. f=Tt=sl+s But,by(10.211),di FIG. 122.Aparticle moving in space.-_-==,as p wherejistheunit principal normal vector andptheradius of curvature ofCatA.Hence, *2 (11.103)f=6:i+-j, andsowemay state: The acceleration ofaparticle liesinthe osculating planetoitspath;thecomponents inthedirections of thetangent andprincipal normal aresand s2/p,respectively. These results should becompared withthose forthecorresponding two- dimensional case (cf.Sec. 4.1). Componentsofvelocity andacceleration incylindrical coordi- nates. InFig. 123,Aistheposition ofaparticle attime tandM thefoot oftheperpendicular fromAontheplane Oxy. The polar coordinates (R, <j>)ofM,together with the^-coordinate ofA,arethecylindrical coordinates (R, <j>,z)ofA.Leti,j,k beunitvectors atAinthedirections oftheparametric lines of these coordinates(i.e., those directions ineach ofwhich just one ofthethree coordinates R,<,zincreases, theother two remaining constant). Wewish tofindthecomponents ofthe velocity andacceleration ofAalong i,j,k. SEC. 11.1] KINEMATICS 307 Thevector kisconstant inmagnitude and direction. The directions ofiandjdonotdepend onRand z;they are,however, dependent on<t>.AsinSec. 4.1(where r,correspond to R,4>),wehave (11.104) Sinced<t> i FIG. 123. Cylindrical coordinates. weobtain, ondifferentiatingrwithrespect totandusing (11.104), (11.105) q=~=Ri+Rfo+*k, Asecond differentiation with respect totgives (11.106)f=(R-RW\+jjj t(#2 <ttJ+*k- From these equations, wecanread offthecomponentsofthe velocity (q)andtheacceleration(f)inthedirections ofi,j,k, when required. Compositionofvelocities andaccelerations. Weoften need toconnect thevelocities (oraccelerations) ofaparticle relative totwodifferent frames ofreference, Sand S'.Weshall here think only ofthecasewhere there isno relative rotation oftheframes. Let beapoint fixed inSand 308 MECHANICS JNSPACE [Sue. 11.2 Orapointfixed in8'.Aparticle Ahasposition vectors r=OA and r'=0'A;they areconnected by (11.107) r=r+r', > where r=00'. Differentiation gives (11.108) q=qo+q',f=fo+f, whereq,f=velocity andacceleration ofArelative toS, q',f=velocity andacceleration ofArelative toS', q ,fo=velocity andacceleration ofS'relative toS. Theequations (11.108) givethelaws ofcomposition ofvelocities andaccelerations. 11.2.KINEMATICS OFARIGIDBODY Motion ofarigidbody withafixedpoint. Consider arigidbody constrained torotate about afixed point O.Lett\,fabetwoinstants;inthetime interval fa t\ thebody receives adisplacement which isequivalent (cf.Sec. 10.5) toarotation nabout 0. Ifwekeep tifixedand let fa approach ti,thedirection ofnwillapproach some limiting direction, which wedenote bytheunitvector i.The ratio of theangle ofrotation ntothetime interval fa fawillapproach alimiting value co.The vector G>=coiiscalled theangular velocity ofthebody attheinstant t\.Atthisinstant thebodyis rotating about alinethrough inthedirection of<o;this line iscalled theinstantaneous axis ofrotation. The rate ofturning iscoradians perunittimeand isarotation inthepositive sense about theinstantaneous axis. Inaninfinitesimal time dtthebody receives aninfinitesimal rotation <odt;andsothedisplacement ofaparticle ofthebody is,by(10.501), dr=(odtXr, where ristheposition vector relative to0.The velocity of thisparticleis (11.201)' q=J=Xr. Thisformula gives thevelocity ofany particle ofthebody in terms oftheangular velocity vector <o.Thus,ifo>isknown as SBC. 11.2] KINEMATICS 309 avector function ofthetime,wecanfindthevelocity ofany particle atanytime; inother words, thesingle vector function o() suffices todescribe themotion. Asthebody turns about 0,theinstantaneous axis(determined by <o)willoccupy different positionsinthebody. Since this axisalways passes through 0,itslocus inthebodyisaconewith vertex 0;itiscalled thebody cone (orpolhode cone). Similarly, thelocus oftheinstantaneous axisinspaceisanother conewith vertex 0;itiscalled thespace cone (orherpolhode cone). Arigidbodymoving parallel toafundamental planemaybe regarded asabody turning about apoint atinfinity. Inthis case thebody andspace cones become cylinders; their inter- sections with thefundamental plane arethebody andspace centrodes ofourearlier theory (cf.Sec. 4.2). WesawinSec.4.2that, inthe?motion ofarigidbody parallel toaplane, thebody centrodo rollsonthespace centrode. Simi- larly,inthemotion ofarigidbody withafixed point, thebody cone rollsonthespace cone. Toestablish this result wemust show that: (i)thebody conetouches thespace cone; (ii)thoparticlesofthebody onthelineofcontact ofthecones areinstantaneously atrest. LetOAbetheposition oftheinstantaneous axisofrotation at some instant. Itisagenerator ofthefixed space coneandalso ofthemoving body cone. After aninfinitesimal timedt, another generator OBofthebody conecomes intocoincidence with agenerator OB' ofthespace cone. Butthedisplacement intime dtisaninfinitesimal rotation ofmagnitudecodtabout OA, andsotheangle between theplanesOAB,GAB' isaninfinitesimal angle. Since these planes represent thetangent planes tothetwo cones along thegenerator OA,itfollows thatthetangent planes cannot cutatafinite angle; thecones must therefore touch. Since allparticles ofthebodyontheinstantaneous axisOAare instantaneously atrest,thesecond oftheabove conditions isalso satisfied, andtheresult isestablished. Thecomponents ofangular velocity interms oftheEulerian angles. InSec. 10.6wedefined theEulerian angles 0,<,^;they describe (relative toafixed triadI,J,K)thepositionofatriad 310 MECHANICS INSPACE [Sue. 11.2 ofunitorthogonal vectorsi,j,k,fixed inarigidbody turning about apoint (Fig. 118). Themotion ofthebodyisdeter- mined when6, <t>,$areknown asfunctions ofthetimet;but thismotion canalsobedescribed bytheangular velocity o>().Wewrite andseek expressions fori, 2,w3interms of0,<,t,andtheir rates ofchange. Inaninfinitesimal time dtthebody receives therotation odt.But,by(10.607), thisrotation is (sin^d6 sin6cos^d<t>)\-t-(cos^d0+sin sin^d<t>)j +(cos d<+ where d0,d<, d\fraretheinfinitesimal increments in0,<,^in time dt.Equating thisexpression to<*dianddividing bydt, wehave !coi=sin^6sin6cos^<, <o2=cos^+sin6sin^<, w3=cos<^+ \l/. These equations givethecomponents ofangular velocity when themotion isknown, i.e.,when9,<, \f/areknown asfunctions of thetime. Conversely, when thecomponents of<>areknown at anytimet,wecansolve theabove equations for0,<,^asfunc- tions oftandsodetermine themotion. Exercise. Find thecomponents of <*>onthefixed triadI,J,K(Fig. 118) interms of9, <t>,tyandtheir rates ofchange. General motion ofarigidbody. Letusconsider arigidbodymoving inageneral manner. We select aparticle Aofthebody asabase point anddenote its velocity byq^.Inaninfinitesimal timedt,thedisplacement ofthebodyisequivalent toatranslation q^dt,andarotation dnaboutA(cf.Sec. 10.5). By(10.502), thedisplacement of anyparticle Bofthebodyis q^dt+dnXr, where r=AB. Hence,forthevelocity ofBwehave (11.203) q=q^+*>Xr, SBC. KINEMATICS 311 where co=dn/dt. Weobserve that thisvelocity consists oftwo parts: (i)thevelocity q^ofthebase point, and(ii)thevelocity ofBrelative toA,viz.,<oXr.Itisclear thatthevelocity ofB relative toAisprecisely thesame asifthebody were turning aboutA(asafixed point) withangular velocity<o. Ifwealter thebase point A,thetranslation qAdtischanged, buttherotation daremains thesame. Itfollows that the vector o>pertains tothemotion ofthebody asawhole;itisthe angular velocity ofthebodyand istoberegarded asafreevector, since itdoes notdepend onourchoice ofbase point. The equation (11.203) gives thevelocity ofanypoint ofthebody when theangular velocity<oandthevelocity q^areknown; thus, thetwovectors o>andqAcompletely describe themotion. When CDandqAareknown asvector functions ofthetime, wehave apicture ofthemotion atany instant. From the velocity qofanyparticle J5,asgiven by(11.203),itsacceleration fcanbe found bydifferentiation; thus, Fia. 124.Awheol rolling ona straight track. WehaveHere d^A/dtistheacceleration fAof thebasepointA;itdepends solely onthemotion ofAandnotonthe angular velocity. Asforthelast term, di/dtisthevelocityofB relative toA]therefore, by(11.201), itequals c*Xr. then (11.204) f=iA+~Xr+<oX(o>Xr). Example 1.Asasimple illustration, letusconsider acircular wheel rolling withconstant speed along astraight level track (Fig. 124).Wetake asbasepoint thecenterCofthewheel anddenote itsvelocity byV.This vector isconstant;itliesintheplane ofthewheel and ishorizontal. The angular velocity<>ofthewheel isavector perpendicular toitsplane;italso isaconstant vector. By (11.203) ,aparticle Bofthewheel hasvelocity V+oXr, > where r=CB. Since <*Xrisavector perpendicular to*>,thisvelocity liesintheplaneofthewheel afactwhich isintuitively obvious. Since 312 MECHANICS INSPACE [SBC. 11.2 andVarcconstant vectors, theacceleration ofBis,by(11.204), f=<aX(<*Xr)=r)-r2=-r2 . Thus, each particle ofthewheel hasanacceleration ofmagnitudero>2 , directed toward C. Example 2.Asasecond illustration,letusconsider themotion ofthe propeller ofanairplane making aturn. Inparticular,letusseehowthe velocity andacceleration ofthetipofthepropeller maybefound. Forsimplicity, weshallsuppose thatthecenter ofthepropeller describes ahorizontal circleCwith constant speed F;let&betheradius andAthe center ofC.Figure 125shows theposition ofthepropeller when theline fromthecenter tothetipBmakes anangle withthevertical. Flo. 125. Motion ofanairplane propeller. Leti,j,kbeatriad ofunitorthogonal vectors atO;ipoints alongAO, kpoints vertically upward, andjcompletes thetriad. Thevectorjis clearly theunittangent vector toCat0.Asabasepoint forthedescrip- tion ofthemotion, wetake thepoint 0;itsvelocityisVj.Theangular velocity oofthepropeller consists oftwoparts: (i)anangular velocity orspin sj(where s=6),imparted bytheengine; (ii)anangular velocity (V/6)k, duetotheturning oftheairplane. Hence, Thesecond part ofG>arises from thefactthat, intime 2irb/Vj theairplane (andtheaxis ofthepropeller) would turnthrough anangle 2?rabout the vertical. From (11.203), wehave, forthevelocity ofanypoint ofthepropeller (position vector rrelative to0), q=Vj+uXr. SBC. 11.3] KINEMATICS 313 Inparticular, thevelocity ofBis (11.205) qfl=Vj+(sj+~k\X(asin i+acos k) =vcos i+(l+jsinWj-t;sink, where a=OBand t>=sa,thespeed ofthetiprelative totheairplane. Hence theabsolute speedisgivenby 9fl= 2+V2 (l+|sin 0)2 . Actually, a/6willbesmall, andsog|=w2 -+-F2 ,approximately. Ifthe trigonometrical term isretained, qntakesmaximum andminimum values when thepropellerishorizontal. Theacceleration ofBmaybefoundbydifferentiating (11.205). Weshall assume that sisconstant; then thescalars,V,bandthevector karecon- stant, whereas thescalar andthevectors iandjarevariable. Tofind di/dtanddj/dt,wenotethat iandjmayberegarded astheposition vectors ofpoints fixed inabody, turning aboutOwith angular velocity (F/&)k. Hence, by(11.201), Itisleftforthereader toverify thattheacceleration ofBis (v2 V2V2a \ 2vV v2 sin H---h sinBJi+-r-cosj--cos k. a b b2/ b a Under normal circumstances, theterms inv2/afarexceed theother terms in magnitude, andsotheacceleration isduealmost entirely tothespin ofthe propeller. 11.3.MOMENTS ANDPRODUCTS OFINERTIA Themoment ofinertia ofasystem wasdefined inSec. 7.1. Foraparticle ofmassmdistant pfrom alineL,themoment of inertia aboutLismp2 .Forasystem ofparticles, themoment ofinertia isthesum ofthemoments ofinertia oftheseveral particles. Wenow define products ofinertia. LetP,Qbetwo planes, and letp,qdenote theperpendicular distances fromthem ofa particle ofmass m.Thedistance iscounted positive ornegative according astheparticleliesononesideortheother ofthe corresponding plane. Theproduct mpqiscalled theproduct ofinertia oftheparticle with respect totheplanes P,Q.Fora system ofparticles, theproductofinertia isthesum ofthe products ofinertia oftheseveral particles. 314 MECHANICS INSPACE flc.11.3 LetOxyz berectangular axes; themoments ofinertia ofa systemofparticles about theaxesOx,Oy,Ozare,respectively, (11.301) A=2m(?/2+z2 ),B=Sm(z2+z2 ) C= Heremisthemass ofatypical particle, x,y,zare itscoordi- nates, and thesummation extends over allparticles ofthe system. Theproductsofinertia with respect tothecoordinate planes, taken inpairs, are (11.302) F=Swyz, G=2mzx, H=Zmxy. Foracontinuous distribution ofmatter thesummations are replaced byintegrations, themassmbeing replaced bythemass pdr(p=density) ofasmallvolume element dr. Itisaremarkable factthat, when A,B,C,F,(7,Hareknown, wecanfindthemoment ofinertia /ofthesystem about any line through 0.Toseethis,we recall that,bydefinition, FIG. 126.Themoment ofinertia about thelineLisrequired. ofthevector productwhere pistheperpendicular distance ofatypical particle P (mass m)from thelineL(Fig. 126).NowpOPsin0,where Qistheangle between OPand T^i ixi -j. iL]thus,pequals themagnitude Xr,where 3*isaunit vector alongL and r=OP.Thecomponentsof^.arethedirection cosines a*)3,7ofL,andthecomponents ofrarethecoordinatesx,y,z ofP.Hence, thecomponents of^Xrare $z yy, yx az, ay fix. Thus, since pisthemagnitude ofthevector with these com- ponents, wehave (11.303) 7=2m[(0*-T2/)2+(7*-**)2+(0-0*)2 1* SEC. 11.3] KINEMATICS 315 This gives7interms ofA,B,C,F, ,H,andthedirection cosines ofL. When A,B,C,F,6r,//areknown foranysetofrectangular axesthrough themass center, wecanfindthemoment ofinertia / ofthesystem about any lineLvery easily. This isdone in twosteps: (i)use(11.303) tofindthemoment ofinertia 7about aline through themass center parallel toL; (ii)apply thetheorem ofparallel axes(cf.Sec. 7.1)tofind /. IfA,B,C,F,G,Hareknown forapoint other than themass center, wecanfindIinasimilar manner, buttwoapplications ofthetheorem ofparallel axesarcrequired. Themomentalellipsoid. Byvarying a,0,7in(11.303), weobtain themoments of inertia about alllines through 0.Letusmeasureoff,along each linethrough 0,adistance OQ=l/-\/I, where /denotes themoment ofinertia about thelineinquestion. Thelocus of Qhastheequation (11.304) Ax2+By*+Cz*-2Fyz-2Gzx-2Hxy=1. This istheequation ofaquadric surface with center 0;in general,itisaclosed surface, since /doesnotvanish forany line.* Hence, (11.304)istheequation ofanellipsoid;itiscalled themomental ellipsoid at0. When theequation ofthemomental ellipsoid atapointis known, wefind themoments andproducts ofinertia with respect totheaxes ofcoordinates byinspecting thecoefficients inthisequation. Under arotation ofaxesfromOxyz toOx'y'z', theequation ofthemomental ellipsoid changes from (11.304) to AV2+B'y'z+C'z'2-ZF'y'z'-2G'z'x'-ZH'x'y'=1. The coefficients A1 ,B',C',F',G',//'give themoments and productsofinertia forthenew axes. Thequadric represented bytheequation (11.304) sumsup theinertial propertiesofthesystem with respect toaxesthrough the origin. Theform oftheequation changes (inthesense that thevalues ofthecoefficients change) whenwerotate the *There isonlyoneexceptionalcase. Ifallparticles ofthesystem lieon alineL,then/-forL;thequadricisthenacircular cylinder with axisL. 316 MECHANICS INSPACE [SEC. 11.3 coordinate axes, butthequadricitself remains aninvariant model oftheinertial properties. Therepresentation ofphysical properties bymeans ofaquadric surface isoffrequent occur- rence itisused inelasticity, hydrodynamics, andother branches ofapplied mathematics. Since thecoefficients intheequation ofthequadric change whenwechange theaxes, theycannot bo called scalars,inthesense thatmass isascalar. Norarethey components ofavector. Thewhole setofsixcoefficients, or more precisely thearray A-H -G (11.305) -// B-F -0 -F C iscalled atensor. This isthesimplest example ofthecon- ceptwhich hasplayed suchanimportant part inthetheory of relativity. Anarrayisalsocalled amatrix. Just asweuseasingle letter todenote avector (which mayberegarded asamatrix with three elements), sowemaydenote amatrix byasingle letter. Theoperations ofalgebra maybeapplied tomatrices, yielding acompact andpowerful notation inmechanics.* Existence ofprincipal axesandmoments ofinertia. Equation (11.303) gives usthemoment ofinertia about any lineLthrough interms ofthedirection cosines ofLandthesix coefficients shown inthearray (11.305). We shallnowshow thatthenumber ofcoefficients maybereduced from sixtothree bymaking asuitable choice oftheaxesOxyz. LetOxyzbeanyaxes. Then7,asgivenby(11.303), attains itsmaximum value forsome lineLI.Letthismaximum beI\. Letustakeaxes Ox'y'z' suchthatOxrcoincides with LI;weleave thedirections oftheothertwoaxes unspecified forthepresent, except fortheconditions thatthey shallbeperpendicular toOx' and tooneanother. Then,forany lineL,themoment of inertia is I=AV2+B'p">+CV2-2F'p'y'-2G'y'a'-277V/3', *SeeR.A.Frazer, W. .1.Duncan, andA.R.Collar, Elementary Matrices (Cambridge University Press, London, 1938). SEC. 11.3] KINEMATICS 317 where A',B'',C",F',G',7T,arethemoments andproducts of inertia fortheaxes Ox'y'z' ',anda', /3',7'arethedirection cosines ofLrelative toOx'y'z'. Ifweputa'=1,0'=7'=0,thenL coincides withLI,andsoA'=7i,themaximum moment of inertia. Weshallnowshow thatthevanishingofG'andHrisaneces- saryconsequence ofthefactthat/isamaximum forof=1, p=y=0.Since a'2+0'2+7/2=1,wecanwrite 7-7X=-2a'(G'y' +H'ff)+(B'-7i)/3'2 +(G"-7ih'2~2F'0Y. IfwetakealineLnear LI,a/willbenearly unity and/3',7'will besmall. IfatleastoneofG',Hfisdifferent fromzero,wecan choose/3',7'(reversing oneorboth signsifnecessary) sothat (G'y'+H'0')isnegative. But, since/3'and 7'arcsmall, the signoftheright-hand side oftheabove equationisdetermined bythe first term. Therefore, 7 7imay bemade positive. But this isimpossible since I\isthemaximum of7.Therefore thehypothesis wemade about G'andH'isfalse,andweconclude thatGfII'=0,sothat foranylineL, 7=7i'2+B'p*+CV2-2*V0Y- This istrue forallaxes Ox'y'z' such thatOx'coincides with LI. Letusnow subject Oyrtotheconditionthat, ofalllines per- pendicular toOx',Oy'hasthemaximum moment ofinertia, say 72.Thenwehave B'1^andsoforany lineL, ForlinesLperpendicular toOx'wehave a'=0,0'2+7/2=1, andso /-/2=-2F'/3Y +(C'-72)T/2 . IfwetakealinenearOy1 ',ft'willbenearly unity and 7'willbe small. Itisevident that,ifF'docsnotvanish, wecanmake 7 greater than72,which isimpossible since 72isamaximum. Therefore Ff 0,andwehave foranylineL, 7=7!'2+72/3'2+CY2 . Substituting 7/2=1-a'2- /2 ,weget /-C"=(/i-<7>'2+(72-C")0/2^0, 318 MECHANICS INSPACE [SEC. 11.3 andsothethirdmoment ofinertia C"istheleast ofallmoments ofinertia forlinesthrough 0. Wemaysumupasfollows, simplifying thenotation: Itis always possibletochoose rectangular axesOxyz such thatthemoment ofinertia Iofasystem about alineLthroughisgiven by (11.306) I=Aa*+Bp*+Cy*, wherea,/?,7arethedirection cosines ofLrelative toOxyz. These axes arecalled principal axes ofinertia at0,andthemoments of inertia A,B,Cabout them arecalled principal moments of inertia. The planes defined bytheprincipal axes arecalled principal planes;foranypairofprincipal planes, theproductof inertia vanishes since F,G,andHareabsent from (11.306). For principal axes, theequation ofthemomental ellipsoidis (11.307) Ax2+By*+Cz*=1. Exercise. Find principal axes ofinertia forathin straight uniform rod atitsmiddle point. General method offinding principal axesandmoments ofinertia. Theprincipal axesandmoments ofinertia havebeenshown to exist.We shallnowshowhow tofindthem, starting from general axesOxyz withmoments andproducts ofinertia A,B, C,F,G,H.LetOx'y'z' betheprincipal axesandA1 ,B',C'the principal moments ofinertia. Any pointPhastwo sets of coordinates, (x,y,z)and(a;', y'',z'),according totheaxeswhich areused. One setofcoordinates arelinear functions ofthe other, such that x*+y*+z2=x'*+y'*+z'2 forevery point P.Also,forevery point P,wehave Ax*+By*+Cz*-2Fyz-2Gzx-2Hxy=A'x'* +By*+cf z'*, since each side represents themoment ofinertia about OP, multiplied byOP2 .Therefore, nomatter howtheconstant K ischosen, wehave theidentity Ax*+By*+(Jz*-2Fyz-2Gzx-2Hxy-K(x*+y*+z*)=AV2+By*+C'zr*-K(xf*+y'*+z'*). Letusdenote each side ofthisidentity by$.Consider the SEC. 11.3] KINEMATICS 319 equations Explicitly, these equations read (A'-K)x'==0,('-K)y'=0, (C"-K)z'=0. Rejecting thetrivial solution x'=?/'=z'=0,wemust choose /fequal toA'r ,Z?7 ,orC".Wehave then thefollowing three solutions : K=A',a/arbitrary, y'=0,zr=0; 7f= ',x'-0,y1 arbitrary,z'=0;K=C",a:'=0,y'=0,2:'arbitrary. Thus (11.308) have nontrivial solutions provided Kisequal to oneoftheprincipal moments ofinertia; thecorresponding values ofx'yy'yz'givetheprincipal axes. Now _ dx dxfdx dy'dx dz'dx' and similar equations could bewritten ford$/dy andd$/dz. Therefore, (11.308) imply that IfK",x',y',z'arechosen asabove, (11.308) aresatisfied, and therefore (11.309) aresatisfied. Explicitly, (11.309) read ((4-K)x-Hy-Gz=0, (11.310){-Ex+(B-7f)2/-Fz=0, (-Ox-Fy+(C-K)z=0. Since these equations have asolution other than#=^=3=0, itfollows that (11.311)A-K-H-G -HB-K-F -G-^ C~0. This isacubic equation forK,and, aswehave seen,itsthree roots arethethree principal moments ofinertia. Tosumup: 320 MECHANICS INSPACE [SEC. 11.3 Starting with general axesOxyz withmoments andproducts of inertia A,B,C,F,G,H,thethree principal moments ofinertia at arethevalues ofKsatisfying thecubic determinantal equation (11.311), and thedirections ofthethree principal axes aregiven by theratios x:y:z determined by(11.310) when theabove values ofK aresubstituted. Theproblem offinding principal axesandmoments ofinertia isessentially thesame asthegeometrical problem offinding the directions andmagnitudes oftheprincipal axes ofanellipsoid from itsgeneral equation.* Theuseoftheequations (11.309) ismost naturally suggested bytheproblem toffinding station- aryvalues (including maximum andminimum values) ofthe expression Ax*+By*+Cz*-2Fyz-2Gzx-2IIxy, subject tothecondition x2+y2+z2=1. Method ofsymmetry. Forabody which exhibits symmetry,itisoften possible to findprincipal axes ofinertia very simply. InSec. 3.1theidea ofsymmetry wasused inconnection with mass centers. Amore thorough discussion requires theconcept ofacovering operation, which wenowproceed todefine. Ifwerotate abody ofrevolution about itsaxisthrough any angle, wedonot alter thedistribution ofmatter thewhole body appears exactly asbefore. Similarly,ifweturn athrce- bladed propeller about itsaxisthrough anangle 2ir/3, thefinal distribution ofmatter isthatwith which westarted. These rotations areexamples ofcovering operations. Ingeneral, a covering operation forabodyisatransformation which does not alter thedistribution ofmatter asawhole, although theindividual particles aremoved. Inthecase ofacurve orsurface, where no distribution ofmatter isinvolved, acovering operationisa transformation which leaves thecurve orsurface unchanged as awhole. Thecovering operations which weshall consider are (i)arotation about alineoraxis,and(ii)areflection inaplane. *Cf.D.M.Y.Sommerville, Analytical Geometry ofThree Dimensions (Cambridge University Press, 1934), Chap. VIII. tCf.R.Courant, Differential and Integral Calculus (Blackie &Sono, Ltd., Glasgow, 1936), Vol.II,pp.18&-191. SEC. 11.3] KINEMATICS 321 Whenever there exists acovering operation* forabody, the bodyissaidtopossess symmetry.Ifthecovering operationisa rotation through anangle 2ir/n about anaxis(where nisa positive integer other than unity),thisaxis iscalled anaxis of n-gonal symmetry; forn=2,3,4thesymmetryisdigonal, trigonal, tetragonal, respectively. Thus theaxis ofathree- bladed propellerisanaxis oftrigonal symmetry; foratwo- bladed propeller theaxis isofdigonal symmetry.Ifthecovering operationisareflection inaplane, then thatplaneisaplane of symmetry forthebody. When wespeak ofanaxisofsymmetry, without qualification, weunderstand thatarotation through anyarbitrary angleisa covering operation. Asurface ofrevolution has thistypeof symmetry. Wenow return totheproblem offinding principal axes of inertia forabody possessing symmetry. Inthisconnection wehave thefollowing theorem: Acovering operation forabody, which leaves apoint ofthebodyunchanged,isacovering operation forthemomental ellipsoid at0.The proof ofthistheorem depends onthefollowing facts, which hold foranydistribution ofmatter whether symmetrical ornotandareeasily proved: (i)when abodyisrotated about aline,themomental ellipsoid atanypoint onthelineturns with thebody; (ii)when abodyisreflected inaplane, themomental ellipsoid atanypoint ontheplaneisalsoreflected inthisplane. When therotation (orreflection) isacovering operation forthe body, thedistribution ofmatter isunaltered, andthemomental ellipsoid atapoint ontheaxis ofrotation (orintheplane of reflection)isthesame asbefore. The rotation (orreflection) istherefore acovering operationforthemomental ellipsoid also,andsothetheorem isproved. Nowweknowfrom thegeometryoftheellipsoid that,when the axesareunequal, there areonlyvery special covering operations; these are(i)arotation through anangleTTabout aprincipal axis and(ii)areflection inaprincipal plane. Iftheellipsoid has more general covering operations,itmust necessarily beof revolution or,inparticular,asphere. Thus,forexample,ifa rotation through anangle 2^/3isacovering operation, the *Other than arotation through four right angles; this isatrivial opera- tion, since itleaves every particle ofthebodyback initsoriginal position. 322 MECHANICS INSPACE [Sue. 11.3 ellipsoid must beofrevolution. Ifarotation through anangleir about alineLisacovering operation, thenLmust beaprin- cipal axis. Ifareflection inaplanenisacovering operation, then IImust beaprincipal plane. We shallnowapply these facts tothemomental ellipsoid. Thetruth ofthefollowing statements willbeobvious: (i)Anaxis ofn-gonal symmetryisaprincipal axis ofinertia atanypoint ofitself. (Example: atwo-bladed propeller.) (ii)Atanypoint onanaxis oftrigonal ortetragonal sym- metry, themomental ellipsoid hasthisaxisforaxis ofrevolution, andtwooftheprincipal moments ofinertia areequal. (Exam- ple:athree- orfour-bladed propeller.) (iii)Thenormal toaplane ofsymmetryisaprincipal axis ofinertia atthepoint where itcuts theplane ofsymmetry. (Example: thehull ofaship.) Principal axes ofinertia foranumber ofbodies aregiven in thetable onpage 324. Ineach caseanargument, based onthe ideas ofsymmetry, canbeused toverify thattheprincipal axes aregiven correctly. Themomental ellipse. Letusnow consider adistribution ofmatter inaplane n, and letOx,Oyberectangular axes inthisplane. SinceHis aplane ofsymmetry,itsnormal at isaprincipal axis ofinertia, andthesection ofthemomental ellipsoid atbytheplaneII isaprincipal section;itiscalled themomental ellipse atO. IfAandBdenote themoments ofinertia about Ox,Oy, respectively, andHdenotes theproduct ofinertia with respect toplanes through Ox,Oy,perpendicular ton,theequation of this ellipseis (11.312) Ax2-2Hxy+By2=1. (Weseethisbyintroducing thethird axisOzandputting2= intheequationofthemomental ellipsoid.) Itisclear that the principal axes ofthis ellipse areprincipal axes ofinertia at0. Tofindthemweproceed asfollows. LetOx',Oy'benewaxes at0,Ox'making anangle with Ox. If(xf' ,y'\ (x,y)denote thecoordinates ofapoint referred totheaxes Ox'y', Oxy, respectively, then x=x'cos y'sin6, y=x'sin+yfcos 0. SEC. 11.3] KINEMATICS 323 Theequation oftheellipse (11.312) referred totheaxes Ox', (Vis A(x' cosB-yfsin0)2 2H(x' cos y'sin0)(x' sin+y'cos6) +B(x' sinB+y'cos0)2=1, or,equivalently, (11.313) AV2-ZH'x'y'+B'y'*=1, where A'=Acos2-2//sin cos6+Bsin2 0, //'=(A-J3)sin cos+#(cos2-sin2 0), B'=Asin26+2Hsin cos+Bcos20. Now (11.313) represents theequationofanellipse referred to principal axes atitscenter ifHf=0.Hence, Ox',Oyrare principal axes ofinertia at if 2/7 (11.314) tan20=_* Thetwovalues of intherange (0, TT)satisfying thisequation givethedirections ofthetwoprincipal axes. Thecomplete set ofprincipal axes at areOx',Oyf ,andalineperpendicular to them. % Thismethod offinding principal axes ofinertia atapoint can beapplied toanycasewhere oneprincipal axistE^vthe pointis known;itneed notberestricted, ashere, tothecase ofaplane distribution ofmatter. Inparticular,itapplies toanybody with aplane ofsymmetry oranaxis ofdigonal symmetry. Moments ofinertia ofsome simple bodies. The table onthefollowing page gives theprincipal axesand moments ofinertia atthemass center forsome simple bodies. Themoments ofinertia about theaxes Ox,Oy,Ozaredenoted (asusual) byA,B,C,respectively. Inallcases thebodies are homogeneous, i.e.,ofconstant density. Some ofthemoments ofinertia given inthetablehave already been calculated inSec. 7.1.Weshall givethecalculations for theellipsoid andleave thereader toverify theothers forhimself. Theequation ofanellipsoid Ewithsemiaxesa,b,c,referred toprincipal axesatitscenter,is 324 MECHANICS INSPACE [SEC. 11.3 Themoment ofinertia about the#-axis isgivenby ACCC (V*+22 )dxdydz, where pisthedensity^ jindtheintegration extends throughout theellipsoid E.Weputxr=x/a, y'=y/>b,z'=2/candobtain SEC. 11.3] KINEMATICS 325 ^4.err Pjjj where therange ofintegrationisnow theinterior ofaunit sphere S.From thesymmetry ofS, fffdx' dyfdz'= z'2dx' dyfdz' (S) Butthislastintegral hasalready been calculated inSec. 7.1;itis themoment ofinertia ofasphere (ofunit radius anddensity) about adiameter andhasthevalue 8?r/15. Hence, A=(b*+c*) asgiven inthetable. Thevalues forBandCfollow inexactly thesame way. Thefollowing rule,known asRouth'srule,summarizes most of theresults given inthetable onpage 324:Forsolid bodies ofthe cuboid, elliptical cylindrical, andellipsoidal types,themoment of inertia about aprincipal axisthrough thecenter (and paralleltothe generators,inthecaseoftheelliptical cylinder)isequalto m(a2+b2 ) wheremisthemass ofthebody, a,barethesemiaxes perpendicular totheprincipal axis inquestion, andn=3,4,or5according asthebody belongstothecuboid, elliptical cylindrical,orellipsoidal type. Themethods ofdecomposition anddifferentiation. Ifwewish tocalculate amoment ofinertia, wecanalways do sobyevaluating amultiple integral. But inmany cases there aresimpler methods. Onemethod istodivide thebody intoa number ofparts, foreach ofwhich themoment ofinertia is known; byadding themoments ofinertia ofthese parts, we obtain therequired result. This isthemethod ofdecomposition andhasbeenused already inSec. 7.1. Another method, known asthemethod ofdifferentiation, can 326'MECHANICS INSPACE [SEC. 11.3 beused tofindthemoment ofinertia ofashellwhen thecor- responding moment ofinertia forasimilar solid isknown. Asanexample, letusfindthemoment ofinertia ofaspherical shellabout adiameter. We firstconsider auniform solid sphere ofdensity pandradius r.Itsmoment ofinertia about adiameter is(87r/15)pr5 .Iftheradius ofthissphereisincreased tor+dr, themoment ofinertia isincreased by dl=(87r/3)pr4 dr-, this isthemoment ofinertia ofaspherical shell ofradiusr, thickness dr,andmass 4?rpr2dr.Hence themoment ofinertia ofaspherical shell, ofradius aandmass m,about adiameter isfwa2 . Similarly, byconsidering theellipsoid _Fa2 A-262 andincreasing ktok4-dk,wecanfindtheprincipal moments of inertia atthecenter ofathin shellbounded bytwosuchellipsoids. Thereader willhavenodifficulty inshowing that,for A*=1,the results areW+c2 ),im(c2+a2 ),im(a2+62 ), where raisthemass oftheshell. Equimomental systems. Two distributions ofmatter which have thesame totalmass andthesame principal moments ofinertia atthemass center aresaid tobeequimomental systems. Forexample, ahoop of massmandradiusa/V2isequimomental withacircular plate ofmassmandradius a. Such systems areinteresting onaccount ofthefollowing fact:Two rigid bodies* which areequimomental have thesame dynamical behavior. Bythiswemean thattwosuch bodies, when acted onbyidentical force systems, willbehave inthe sameway;ifthebodies were fixed inside twoidentical boxes, weshould notbeabletodistinguish between them. This result willbeevident whenwehavedeveloped thegeneral principlesof dynamics inChap. XII. SBC. 11.4] KINEMATICS 327 11.4.KINETIC ENERGY Thekinetic energy ofarigidbody withafixedpoint. Consider arigidbody turning about afixed point with angular velocity<>.Aparticle Pofthisbody, with velocity qandmass 5m,haskinetic energy ^5m g2 (cf.Sec. 5.1); the kinetic energy ofthebodyis (11.401) T=iSdm- q*, where thesummation extends over allparticles ofthebody. Weseekanalternative expression forT7 ,involving theangular velocity<>andtheprincipal moments ofinertia at0. LetOxyzbeanyrectangular axes at0,andi,j,kunitvectors along them. Resolving vectors inthedirections oftheseaxes, wewrite r=xi+yj+2k, w coj+co2j+w3k, > where r=OP. Forthevelocity qofP,wehave,by(11.201), (11.402)q=wXr=(cdaz-w37/)i+(usz-i0)j+(any-a>2)k. Hence, bysubstitution from (11.402) in(11.401), weobtain 2T=Sdm[(ci>23 w3?/)2+(ws wiz)2+(wi?/ w2.r)2 ] =co2 .2dm-(if+z2 )+col26m(z2+a*2 )+co26w-(x2+?/) 2a>2co3S 5m2/2 2a>3Wi25m z# or (11.403) T=i(Aa>2+Bui+Cul-2Fco 2co3-2(7co 3a>i 2//C01C02), where-4,#,C,F,(j,Harethemoments andproducts ofinertia forOxyz. Iftheaxes areprincipal axes ofinertia, then F=G=H=0, andweobtain, astherequired expression forthekinetic energy, (11.404) T=KAco?+B<*\+Cw|), where A,B,Carenowprincipal moments ofinertia. The expression (11.404)isvalid onlywhen theaxesOxyz areprincipal axes ofinertia at0;forother axes,itisevident 328 MECHANICS INSPACE [SEC. 11.4 from (11.403) thatTinvolves both products andmoments of inertia. Ifweuseaxeswith directions fixed inspace, notonly willTinvolve both products andmoments ofinertia worse still, these willvary with thetime. Toavoid these complica- tions,itiscustomary touseaxeswhich arepermanently principal axes ofinertia at0,sothatthesimple formula (11.404) holds at anytimeandA,J5,Careconstants. Ingeneral theprincipal axes at arefixed inthe*body. But ifthemomental ellipsoid at isofrevolution, onlyoneofthem needbesofixed; theothertwomaybeanyperpendicular lines intheplane perpendicular totheaxis ofrevolution. Itmight appear thattheuseofsuchaxes, fixed neither inspace norinthe body, would introduce aneedless complication. Butactually itsimplifies considerably thetheoryoftopsandgyroscopes. Forarigidbody turning about afixed lineLthrough 0,it iseasily seenthattheformula (11.403) simplifies totheformula (7.116), given inthetwo-dimensional theory. Wehave merely totake Ozalong L;then on=co2=0. co3= co,and (11.403) gives T=iGV, whereCisthemoment ofinertia about L. Thekinetic energy ofarigidbody ingeneral. Letusnow findthekinetic energy Tofarigidbodymoving quite generally inspace. Applying thetheorem ofKonig (cf.Sec.7.1),wehave (11.405) T=%mql+T', wherem=mass ofbody, #o=speed ofmass center, Tr=kinetic energy ofmotion relative tomass center. Butthemass centermayberegarded asabasepoint inthebody; andso,asexplained inSec. 11.2,themotion relative tothemass center isthat ofarigidbody turning about afixed point. Thus, T'isgiven by(11.404) with aproper interpretation ofthe symbols. Wetherefore have (11.406) T=%mql+i(A?+Bu\+Cco2 3), SEC. 11.5] KINEMATICS 329 where A,B,C=principal moments ofinertia atthemasscenter, i,w2,ws=components oftheangular velocitycointhe directions ofprincipal axes ofinertia atthe mass center. Inapplying theprinciple ofenergy, provedinSec. 5.2,to particular systems, weneed expressions forkinetic energy.. Foraparticle thekinetic energyissimply w</2 ;forarigidbody, wehave theformulas (11.404) and (11.406). With theaidof these fundamental formulas, wefindnodifficultyincalculating thekinetic energy ofanysystem. 11.5.ANGULAR MOMENTUM Theangular momentum ofaparticle about alinewasdefined inSec. 5.1asthemoment ofthelinearmomentum vector about thelineinquestion. Now indealing withmoments ofvectors inthree dimensions,itisthevectormoment about apoint which isfundamental, rather than thescalar moment about aline. Accordingly, wedefine theangular momentum ofaparticle asa vector; thescalar angular momentum defined inSec. 5.1is, ofcourse, merely onecomponent ofthevector defined here. Angular momentum ofaparticle andofasystem ofparticles. Consider aparticle ofmass m,moving with velocity qrelative tosome frame ofreference S.The linearmomentum iswq (cf.Sec. 5.1).Wedefine theangular momentumh,about any point 0,asthemoment ofraqabout 0;hence, by(9.301), (11.501) h=rXmq, where ristheposition vector oftheparticle relative to0. It isclear thathdepends ontheframe ofreference used inthe measurement ofq. Forasystem ofparticles, theangular momentum isthevector sum oftheangular momenta oftheseveral particles. LetTW,, rt,qtdenote themass, position vector (relative toapoint 0), and velocityofthezthparticle, respectively. The angular momentum about is n (11.502) h=2)(r<Xmtqt), =i where nisthenumber ofparticlesinthesystem. 330 MECHANICS INSPACE [SEC. 11.5 Wenote that,if isfixed intheframe ofreference, then qt=ft.Inthat case thecomponents ofhalong rectangular axes fixed intheframe are (11.503) i Letusnowconsider theeffect ofchanging theframe ofrefer- ence. LetS'beanewframe, having avelocity qoftranslation relative toS.Then thevelocities qt,q(ofaparticle relative to S,S',respectively, areconnected by (11.504) q<=q+q5, according to(11.108). Theangular momenta about arethen h=i)(r.Xmtqt)forS;h'=Y (r<Xm>q' z)forS'. t=i =i Substituting from (11.504) intheexpression forh,wefind (11.505) h=(^mtrt)Xq+h7 . If isthemass center,^mr*=0*an^s^efirs*termonthe rightvanishes. This gives thefollowing remarkable result: Angular momentum about themass center isthesame forallframes ofreference inrelative translational motion. Generallyitismost convenient touseaframe ofreference inwhich themass center isfixed. Inspeakingofangular momentum about apoint 0,weshall in future always understand aframe ofreference inwhich is fixed. Angular momentum ofarigidbody. Themost interesting application of(11.502)istothecase ofarigidbody turning about 0.Theformulas which weare about todevelop arefundamental ingyroscopic theory. SEC. 11.51 KINEMATICS 331 Inaslightly different notation, wehave,fortheangular momentum about0, (11.506) h=S(rX8m-q), where 8misthemass ofatypical particle,ritsposition vector, andqitsvelocity; thesummation extends over allparticlesin thebody. But,by(11.201), q=<oXr, where CDistheangular velocity ofthebody. Hence, (11.507) h=Sdm[rX(oXr)]=2dm- [or2-r(co r)]. Letusresolve thisvector along anorthogonal triadi,j,k at0.Intheusual notation, wewrite r=xi+y]+zk,co=wii+w-j+cojs anddenote byA,B,C,F,G,Hthemoments andproducts of inertia with respect tothetriadi,j,k.Thecomponent ofh inthedirection ofiis hi=25m[o>i(z2+2/2+z2 )x(=coiS8m (y2+z2 )co2S8mxyco3S5m22 =A&I 7/OJ2 G&Z. Similar expressions forthecomponents7i2and /i3arefound in thesameway; thecomplete setofcomponentsis hi= /i2= Thestructure ofthese formulas should becompared with the array (11.305). Ifi,j,karoprincipal axes ofinertia at0,thenF=G=H=0, andthese formulas aregreatly simplified. Theybecome (11.509) hi=Acoi,h=w2,/i3=Cw3, where A,B,Carenow principal moments ofinertia at0. 332 MECHANICS INSPACE [SEC. 11.6 Asinthecase ofkinetic energy,itisusual tochoose the coordinate vectorsi,j,kindirections which arepermanently principal axes ofinertia at0.With such achoice forthese vectors, thesimple formulas (11.509) hold atany time, and A,ByCareconstants. Ifthebodyisconstrained torotate about afixed axis,wemay takekalongthis axis. Then coi=0,w2=0,o>3=w,and (11.508) gives (11.510) hi=-G, h*=-Fw, h,=GV Thus theangular momentum vector doosnot liealong theaxisof rotation, unless thelatter isaprincipal axisofinertia. However, thecomponent/&3along theaxis ofrotation isequal totheproduct ofthemoment ofinertia about that axisandtheangular velocity. This isinagreement with (7.117). 11.6.SUMMARY OFKINEMATICS, KINETIC ENERGY, ANDANGULAR MOMENTUM I.Kinematics ofaparticle. Velocity: (11.601) q=-r= i, (i=unittangent vector). Acceleration: (11.602)f=-77=.si+ j, (j=unitprincipal normal at p vector). II.Kinematics ofarigid body. (a)Rigid body withafixed point: Motion described byvectoro>;velocity ofanyparticle ofthe bodyis (11.603) q-QXr. (6)Rigid body ingeneral motion: Motion described byvectors q^, <>;velocity ofanyparticle of thebodyis (11.604) q=q^+co Xr. Ex.XI] KINEMATICS 333 III.Moments andproducts ofinertia. (a)General formulas: (A=Sra(2/2+z2 ),B=Sm(z2+a:2 ), C=2 F=Zroys,Cf=2wz.r,//=S (11.606) /=Aa2+B(3*+6V-2F(3y-2Gya- (b)Momental ellipsoid (r=\/\/l): General form: (11.607) Ax*+By*+Cz*-2Fyz-2Gzx-2Hxy=1. Form forprincipal axes: (11.608) Ax*+By2+Cz*=1. (A,B,Careprincipal moments ofinertia; FGH=0.) IV.Kinetic energy. (a)Particle: (11.609) T=%mq*. (b)Rigidbody withafixed point (principal axes): (11.610) T=i(Awf+Bu\+CVO- (c)Rigid body ingeneral motion (principal axes atmass center): (11.611) T=Imql+i(Ao>2+#co2+Ca,2 3). V.Angular momentum. (a)Particle: (11.612) h=rXmq. (b)Rigidbody turning about apoint (principal axes): (11.613) h=Awii+J5co 2j+Co>3k. EXERCISES XI 1.What isthekinetic energy ofahomogeneous circular cylinder,of massmandradiusa,rolling onaplane with linear velocity? 334 MECHANICS INSPACE [Ex.XI 2.Foracertain orthogonal triad ofaxesatthemoments ofinertia ofa body are3,4,5,andtheproducts ofinertia vanish. What isthegreatest moment ofinertia ofthebody about any linethrough 0? 3.Arod, oflength 2aandmass m,turns about oneend0,describing a conewith semivertical angle a.Itcompletes arevolution intime T.Find themagnitude anddirection oftheangular momentum about 0. 4.Arigidbody isturning about afixed point 0,andOxyz arerectangular axes. Ifthecomponentsofvelocity oftheparticle withcoordinates(1,0,0) are(0,2,5),findthecomponentinthedirection ofthex-axis ofthevelocity oftheparticle with coordinates(0,0,1). 6.Find thelength ofahomogeneous solid circular cylinder ofradius<z, given thatthemomental ellipsoid atthemass center isasphere. 6.Abody turns about afixed point. Prove thattheangle between its angular velocity vector and itsangular momentum vector (about thefixed point) isalways acute. Showthat,iftheprincipal moments ofinertia AyB,Carealldifferent, thentheangle vanishes only ifthebody isturning about aprincipal axis. 7.Find themoment ofinertia ofasolidhomogeneous cubeabout an arbitrary linethrough itscenter. What aretheprincipal axes ofinertia atacorner? 8.Find themoment ofinertia ofarectangular plate 3ft.by4ft.,ofmass 20lb.,about adiagonal. 9.Find thecomponents ofvelocity and acceleration along thepara- metric lines ofspherical polar coordinates r}6, <f>,foraparticle moving in space. Check yourformulas byapplying them tothefollowing special cases: (i) <j>=constant; (ii)?r. 10.Acardrives round acurve ofconstant curvature atconstant speed. What isthemagnitude anddirection oftheinstantaneous acceleration ofthe highest point ofatire? 11.Auniform circular diskofradius aandmassmisrigidly mounted on oneendofathin light shaftCD,oflength6.Theshaft isnormal tothe disk atitscenter C.Thedisk rollsonarough horizontal plane,Dbeing fixed inthisplanebyasmooth universal joint. Ifthecenter ofthedisk rotates about thevertical through Dwithconstant angular velocity n,find theangular velocity, thekinetic energy, andtheangular momentum ofthe diskabout D. 12.Acar isturning acorner, themiddle point oftheback axledescribing acircle ofradius r.Ifthelength oftheaxle is2aandthewheels areregarded asuniform disks, each ofradiusb,prove thattheratio ofthekinetic energies oftheback wheels is 6(r+a)2+ft2 6(r-a)3+&*' Ex.XI] KINEMATICS 335 13.Anellipsoid ofrevolution with fixed center rollswithout slipping ona fixed plane. Describe thespace andbody cones. 14.Aplaneisfixed inspace. Coordinate axesOxyz rotate about 0. Their angular velocity hascomponents i,w2,waalong them. Iftheequa- tionoftheplane atanyinstant is Ax+Dy-fCz=1, prove that dA dB dC .R-I--saJ[5a>3 CC02,-T7-=C/C01 /IC03, -yT=/1W2~-DCOl. 15.Agovernor consists oftwoequal spheres, ofmassmandradius a. They arefixed totheends ofequal light rods, oaoh oflength ca,which are hinged toacollar onavortical axle. Bymeans ofalight linkage andsliding collar, theequality oftheinclinations tothevortical ofthetworods is ensured. Ifthisangle ofinclination is aridtheangular velocity ofthe governor about itsvertical axle iso>,show thatthekinetic energyis +co2sin0}+Cw2cos2 0, where A=m(|a2+c2 ),C=lmaz . 16.Prove thattheangular momentum cfamoving system about apoint Oisthesumofthefollowing parts: (i)theangular momentum about ofaparticle moving with themass center andhaving amass equal tothetotalmass ofthesystem; (ii)theangular momentum ofthesystem about themass center. 17.Asteel ball isplaced between twohorizontal planes, which rotate with angular velocities w,'about vertical axes L,Z/.Assuming thatno slipping takes placo, show thatthecenter oftheballdescribes ahorizontal circle withcenter intheplane containing LaridL'.Show thatthedistances ofthiscenter fromLandL'areintheratio ':w. 18.Forarigidbody ingeneral motion, show thatthere isnopoint atrest. Show also that,ingeneral,there isonepoint, andonly one,withno acceleration. 19.Forasystem ofparticles, prove that thekinetic energy ofmotion relative tothemass centermaybeexpressedintheform wherem=totalmass ofsystem,mt=mass oftypical particle, ,-=magnitude ofvelocity ofm^relative tom,, andthesummation contains oneterm foreach pairofparticles. 336 MECHANICS INSPACE [Ex.XT 20.OAisalightrodoflength bwhich turns withangular velocityftabout anaxisOBperpendicular toit.Aisthemiddle point ofarodCDofmassm andlength 2a,hinged toOAatAinsuchawaythatCD isalways coplanar withOB. If6denotes theangleOAC, provethatthecomponents ofthe angular momentum ofCDabout inthedirections OA tOBandadirection perpendiculartothemare,respectively, sin cos0, w!2(62 -f-a2cos2 0), Jma20. CHAPTER XII METHODS OFDYNAMICS INSPACE Thefollowing three principles arefundamental inNewtonian mechanics: (i)theprinciple oflinearmomentum, (ii)theprinciple ofangular momentum, (iii)theprinciple ofenergy. General forms forthefirstandlastofthese principles have already beengiven inChap.V;weshallmerelyrecallthem here, stressing their applications todynamics inspace. Thetreatment ofthe principle ofangular momentum, given inthischapter,isinde- pendent ofthatgiven inChap. V;there isreason forthis, since intwodimensions angular momentum isascalar, whereas in three dimensions itisavector. Webegin ourdiscussion oftheabove principles byconsidering thesimplest ofallsystems asingle particle. 12.1.MOTION OFAPARTICLE* Equations ofmotion. Foraparticleofmassmacted onbyaforce P,wehave,by thefundamental law(1.402), (12.101) mf=P, where fistheacceleration relative toaNewtonian frame of reference. This vector equation canalsobewritten intheform (12.102)|(mq)=P, where qisthevelocityoftheparticle. Inthisform,itisoften referred toastheprinciple oflinear momentumforaparticle: Therateofchange oflinearmomentum ofaparticle isequaltothe applied force. Byresolving thevectors fandPinthedirections ofrectangular axesOxyZj fixed intheframe ofreference, weobtain, asinSec. 5.1,theequations (12.103) mx-X,my-7, m&=Z, 337 338 MECHANICS INSPACE [Sic. 12.1 where X,F,Zarethecomponents ofPalong theaxes. These aretheequationsofmotion ofaparticle inrectangular Cartesians; other forms oftheequations ofmotion areobtained below. Leti,j,kbeunitvectors along thetangent, principal normal andbinormal tothepath oftheparticle. By(11.103), where sdenotes arclength along thepathandpistheradius of curvature. Writing P=P!i+P2j+P3k, weobtain from (12.101) thefollowing intrinsic equations of motion: (12.104) ms=P1;=P2,=P3. P Now leti,j,kbeunitvectors inthedirections ofthepara- metric lines ofcylindrical coordinates (R, <#>,z).By(11.106), wehave f=(&-RWi+i~ Thus,if weobtain, asequations ofmotion incylindrical coordinates, (12.105) m(R-Rp)=P,m~~(R^)=P ,mz=P,. Equations (12.103), (12.104), and (12.105) arcprobably the most useful forms oftheequations ofmotion ofaparticle. Other formsmaybeobtained byfollowing thesame general plan, namely, resolution ofvectors along asuitably chosen orthogonal triad. When thepath ofaparticleistobefound,itisbetter touse someformsuch as(12.103) or(12.105), rather than theintrinsic equations (12.104). However,ifthepathisknown beforehand, theequations (12.104) areparticularly convenient. Forexam- ple,consider aparticle sliding down asmooth curve under SBC. 12.1] METHODS OFDYNAMICS INSPACE 339 gravity. InthiscasePIissimply thecomponent oftheparticle's weight inthedirection ofthetangent and istherefore known. Integration ofthe firstequation in(12.104) gives s(and s)in terms ofthetime. The other twoequations then give the reaction ofthecurve ontheparticle without further integration. There isapoint ofinterest inconnection with (12.104). From thelast ofthese equations,itisclear that thepathis such that theosculating plane contains theapplied force.We recall that, foraflexible cable inequilibrium [cf.(10.217)], the osculating plane alsocontains theexternal force; there isaclose analogy between thetwoproblems. Exercise. Aparticle moves onasmooth surface under noforces except thereaction ofthesurface. Show that itspathisageodesic onthesurface. Isthisresult truewhen thesurface isrough? Integration oftheequations ofmotion. Toobtain theequations ofmotion ofaparticleisonequestion, buttosolvethem isanother. Thesecond task ismuch harder than the first. Indeed, wemaysaythatonlyaveryfewofall possible problems indynamics canbecompletely solved,ifby solution wemean theexpressionofthecoordinates aseasily calculable functions ofthetime t.However,itisalways possible toobtain solutions intheform ofpower series in t.Considera- tion ofthisprocess leads tothefollowing important general theorem: Themotion ofaparticleisdetermined when itsinitial position and velocity aregiven. Letussketch theproof ofthistheorem inthecase ofafree particle, moving inaccordance with theequations (12.103). Weshallsuppose thatX,F,Zaregiven functions ofx,yyzand perhaps ofx,y,z,t,also. (Consider,forexample, aprojectile under theaction ofgravity and airresistance, asinSec. 6.2.) Then theequations (12.103) giveXQ,yQ,ZQinterms of (12.106) X,7/o,ZC,XQ, t/o, , thesubscript indicating evaluation att 0.Ifwedifferentiate (12.103) andconsider theresulting equations att=0,wesee thatthey give thethird derivatives ofx,y,zwith respect to tatt= interms ofthequantities (12.106), since XQ,yQ,ZQ have already been found. Proceeding inthisway,wecan determine allderivatives ofx,y,zatt=interms ofthequan- 340 MECHANICS INSPACE [&BC. 12.1 titles (12.106). Thus,wehave allthecoefficients inthefollowing Taylor expansions forx,y,z: x=xQ+xt y=2/o+Vot z=z+tot+\z<p+' ' . Theabove series provide aformal solution oftheequations ofmotion and, aswehave seen, thissolution depends onlyon o,2/o,z,#o,#o,2o-Forthecompletion oftheproof,itisnecessary todiscuss theconvergence ofthe series; thisbelongs tothe theoryofdifferential equations, andweshall merely remark that theconditions ofconvergence (forsome rangeofvalues for/)aresatisfied inalltheproblems weshall consider. Inorder thatasolution forthemotion ofafreeparticle (not necessarily expressed inpower series) maybemade tofitthe stated initial conditions, theremust beavailable sixconstants ofintegration. Forasetofparticles moving under forces depending ontheir positions andvelocities,itmay beshown byanargument similar tothatgiven above thatthemotion isdetermined when the initial positions and velocities aregiven. (Thisismost easily seenbymeans ofLagrange's equations;cf.Chap. XV.) Thenumber ofconstants ofintegrationisdouble thenumber of degrees offreedom. Ifweregard theuniverse ascomposed ofparticles, thisresult leads toarather surprising conclusion. Ifweknew atthe present moment theposition andvelocity ofevery particle in theuniverse andcould solve thedifferential equations ofmotion, weshould beable topredict thewhole future oftheuniverse. Even more surprising, since themotions ofalltheparticles could befollowed backward intime aswellasforward, weshould beinaposition touncover thehistory oftheuniverse from its beginning. Isthispractical science? Itisnot,forsuchacomplete knowl- edge ofpresent conditions isquitebeyond ourpower. From a philosophical point ofview, however, thequestionisofinterest thequestion astowhether thepastandfuture aredetermined bythepresent. That they aresodetermined isimplied in Newtonian mechanics, and itishere thatquantum mechanics SEC. 12.1] METHODS OFDYNAMICS INSPACE 341 introduces anewandrevolutionary idea: Nothingiscertain, only probable. Principle ofangular momentum. By(11.501), theangular momentum ofaparticle about a fixed pointis (12.107) h=rXwq. Letuscalculate therateofchangeofh.Differentiating (12.107), wefind (12.108) h=fXwq+rXmq=qXwq+rXraf =rXP, wherePistheforce acting ontheparticle. Inwords,therate ofchange ofangular momentum ofaparticle about afixed point isequaltothemoment oftheapplied force about thatpoint. Ifwetake asorigin ofrectangular axesOxyzandresolve vectors inthedirections ofthese axes,weobtain !m(yzzy)=yZ zY, m(zx-xz)=zX-xZ, m(xy-yx)=xY-yX, where X,Y,Zarethecomponents ofP.The last ofthese equationsisthesame asthatobtained inSec. 5.1foraparticle moving intheplane Oxy,moments being taken about (orOz). Principle ofenergy. Foramoving particle wehave, asinSec.5.1, (12.110) f=W, wheretistherate ofincrease ofthekinetic energy andWis therateatwhich theapplied forces dowork. This general form oftheprinciple ofenergyisoflittle use,except inthecasewhere theworking forces areconservative. ThenW=F,whereV isthepotential energyoftheparticle and (12.110) gives, on integration, (12.111) T+V=E, whereEisaconstant, thetotal energy. 342 MECHANICS INSPACE [SEC. The reader may ask: Seeing that theequations ofmot (12.103) arethree equations forthreeunknowns (and therei mathematically complete), whydowetrouble todevelopf more equations (12.109) and (12.111)? Theanswer is:r . latter equations often give directly pieces ofinformation wh canbeused inconjunction with (12.103) tosimplify theworl Example. Letusconsider themotion ofaparticle ona.smooth sph Weusecylindrical coordinates (72,<,z)with origin atthecenter of sphere, theaxis ofzbeing directed vertically upward. Then theequa ofthesphereis (12.112) R2=a2-z*. Theforces acting ontheparticle areitsweight mgandthenormal reac Nofthesphere. Byresolving these forces along theparametric line R,<,andz,wefindPR, P<t,,andPein(12.105); these equations, togel with (12.112), arefourequations fromwhichwecanfindN,R,<,zasft tions ofthetime. This plan ofdealing with themotion, though strail forward, isnotsosimple asthatgiven below. We firstnote that, sinceNdoesnowork, theprinciple ofenergy app Now thepotential energy oftheparticle ismgz,and itskinetic energ %mqz ,where qisthevelocity, withcomponents given by(11.105). Hei by(12.111), (12.113) $m(R*+R*<i>*+z*)-fmgz=mE, whereEishereused todenote theconstant energy perunitmass. Ag sinceNandtheweight havenomoment about Oz,theangular moment about Ozisconstant. Thecomponents oflinearmomentum intheR-an directions havenomoments about Oz;the^-componentismR$ and moment ismR2 <f>.Hence, (12.114) R*<t>=h, where hisaconstant. This result follows alsofrom thesecond equal of(12.105), sinceP$=inthiscase.When theinitial position and veloi areknown, theconstants Eandhcanbefound, andtheequations (12.1 (12.113), and(12.114) provide three equations todetermine R,<,and z. From (12.112), wehave,bydifferentiation, When thisvalue ofttandthevalue ofj>from (12.114) aresubstitutec (12.113), weget (12.115) *,= SEC. 12.2) METHODS OFDYNAMICS INSPACE 343 This isasingle equation forzasafunction of/;when thisequation hasbeen solved, (12.112) givesRinterms of tdirectly, and <canbefound bya quadrature from (12.114). The solution of(12.115) isgiveninthenext chapter. 12.2.MOTION OFASYSTEM Principleoflinearmomentum; motion ofthemass center. Wenow recallsome results established inSec. 5.2. Ifmlandqt denote themass andvelocity ofthezthparticle ofasystem, thenthelinearmomentum is (12.201) M= Jm,qt, where nisthenumber ofparticles. By(5.206), wehave (12.202) M:=F, whereFisthevector sum oftheexternal forces. This isthe principle oflinearmomentum initsgeneral form. Itmaybe stated asfollows: The rateofincrease ofthelinear momentum ofasystem isequaltothevectorsumoftheexternal forces. Ifqdenotes thevelocity ofthemass center andmthetotal mass, thelinearmomentum Misraqand(12.202) gives (12.203) wq=F. This istheequationofmotion forasingle particle ofmassm under aforce F,and sowehave thefollowing result, already stated inSec. 5.2:Themass center ofasystem moves likeaparticle , having amass equaltothemass ofthesystem, acted onbyaforce equaltothevectorsum oftheexternal forces acting onthesystem. This alternative statement oftheprinciple oflinearmomentum isparticularly useful;itreduces thedetermination ofthemotion ofthemass center ofanysystem under known external forces toaproblem inparticle dynamics. Asillustrations, wemay consider themotion ofahigh-explosive shell oroftheearth in itsorbitround thesun.Todetermine themotion ofthemass center oftheshell,weneedknow only thesum oftheforces exerted bytheairontheelements ofitssurface and, ofcourse, theweight ofthe shell. Similarly, inthecase oftheearth,its mass center moves likeaparticle subject tothegravitational fields ofthesun,moon, andother bodies inthesolar system. 344 MECHANICS INSPACE [SEC. 1 Principle ofangular momentum. By(11.502), theangular momentum ofasystem ofpartic about apointis (12.204) h=J(rtXmtq<). Herew=mass ofithparticle, rt=position vector ofzthparticle relative to'0, qt=velocity oftthparticle relative to0, n=number ofparticles insystem. Inwhat follows, weshall consider tobeeither afixed poi inaNewtonian frame ofreference orthemass center oft system. Therate ofchange ofhis n ^ h=5)(ftXm^i+rXfn.q<). t-i Since ft=qt,thefirstvector product vanishes. Thuswehave (12.205) h=(rtX where ftistheacceleration oftheithparticle relative toO. If isafixed point, then ftisacceleration relative toaNe tonian frame, and*so iiiA-Pi+P{, wherePt,P^are,respectively, theexternal andinternal fore ontheithparticle. Hence, by(12.205), (12.206) h=Vr,XP<+Vr,XPJ. Thesecond summation vanishes, since theinternal forces have ] moment about anypoint (cf.Sec. 10.2). Hence, (12.207) h=G, whereGisthetotalmoment oftheexternal forces about tl fixed point 0. SBC. 12.2] METHODS OFDYNAMICS INSPACE 345 If isthemasscenter, theacceleration oftheithparticle relative toaNewtonian frame is fo+ft, where fistheacceleration of relative toS[cf.(11.108)]. Hence, theequation ofmotion oftheithparticleis mt(f+ft)=Pt+Pi. Substitution formtin(12.205) gives (12.208) h=VrXPi+2rtXP't- (Vmtrt)Xf. i=i t=ivi=i' Thesecond summation vanishes asbefore, andthelastvanishes n since2)m^i=0.Hence, weobtain again anequation ofthe t=i form (12.207), whereGisnowthetotalmoment oftheexternal forces about themass center. Wemaysumuptheprinciple ofangular momentum asfollows: The rateofchange oftheangular momentum ofasystem about a point, either fixed ormoving with themasscenter, isequaltothetotal moment oftheexternal forces about thatpoint; insymbols, (12.209) h=G. The equations (12.203) and (12.209) arefundamental in dynamics and, indeed, instatics aswell. Like thegeneral conditions ofequilibrium F=0,G=0,they hold forany systemitmaybethewhole, oranypart, ofagiven distribution ofmatter. When thesystemisasingle rigid body, (12.203) and (12.209) provide twovector equations forqand<o,the velocity ofabase point (themass center) andtheangular velocity ofthebody (cf.Sec. 11.2). Moreover, when applied toasystem inequilibrium, forwhich qandhvanish, theyreduce totheconditions (10.207), thebasic equations instatics. Example. Asasimple illustration, letusconsider acylinder rolling down aninclined plane. Themass center moves inavertical plane, andsothe vectors qandFlieinthisplane. Resolving them alongandperpendicular totheinclined plane, weobtain the firsttwoequations in(7.312). The angular velocityisparallel totheaxisofthecylinder, andtheangular momentum about themass center is 346 MECHANICS INSPACE [SBC. 12.3 where /isthemoment ofinertia about theaxisofthecylinder. Sincehhas afixed direction, (12.209) gives asingle scalar equation thethirdandlast oftheequations (7.312). Principle ofenergy. Inaddition totheprinciples oflinear andangular momentum, there isathird general principle theprinciple ofenergy. This principle, established inSec. 5.2,isvery useful when theworking forces areconservative. Then itleads tothelawofconservation ofenergy, (12.210) T+V=E, whereTandVarethekinetic andpotential energies andEis theconstant total energy. Thetwomostcommon systems inmechanics aretheparticle andtherigid body. Foreach ofthese systems, theprinciple of energyisnotindependent oftheprinciples oflinear andangular momentum;itmay, however, beused inplaceofanyoneofthe scalar equations deduced from (12.203) and(12.209) byresolution ofvectors. When itisused inthisway, thevalue ofthelawof conservation ofenergy liesinitssimplicity;itinvolves only positions and velocities, notaccelerations. Exercise. Arigidbody turns about afixed axiswith constant kinetic energy. Show thatthemagnitude oftheangular momentum h(about a point ontheaxis) isalsoconstant. Isthedirection ofhnecessarily fixed? 12.3.MOVING FRAMES OFREFERENCE InSec. 5.3thequestion wasraised: Ifthelawsgoverning the motion ofabody inaNewtonian frame ofreference areknown, howdoes itmovewhen viewed from aframe ofreference moving relative totheNewtonian frame? This question hasbeen answered foraparticle moving inaplane; weshallnowconsider three-dimensional motion. Frame ofreference with translational motion. LetSbeaNewtonian frame ofreference andS'aframe of reference which has, relative toS,amotion oftranslation only. Foramoving particle, wehave, asin(11.108), (12.301) f=fo+f, SEC. 12.3J METHODS OFDYNAMICS INSPACE 347 where foistheacceleration ofS'relative toS.SinceSisNew- tonian, thelawofmotion is mf=P, wheremisthemass oftheparticle andPtheforce acting onit. InS'thelawofmotionis,by(12.301), (12.302) mf=P-mf . Thus themotion ofS'gives risetothefictitious force wf . Thismeans thatwecanregard S'asNewtonian, provided we addtotheactual forces afictitious force mfoneach particle. Rotating frames; rate ofchange ofavector. Leti,j,kbeatriad ofunitorthogonal vectors inaframe of reference $',which rotates withangular velocity Qrelative toa Newtonian frame S.Any vectorPmaybeexpressed inthe form (12.303) P=Pii+P2j+Psk. Weshallnowcalculate therate ofchange ofPasestimated byan observer inS. Incalculating dP/dt, wemustremember thatnotonlydo PI,P2,P3vary, but alsothevectorsi,j,k.Straightforward differentiation of(12.303) gives .1 . Now iisavector fixed inarigidbody (/S'),which rotates with angular velocityii.Wemay think ofiastheposition vector ofaparticle Bofthisbody relative toabasepoint A,theorigin ofi.Then di/dtisthevelocity ofBrelative toA ;andso,by (11.203), di/dt=aXi.Thesame reasoning applies toj andk;thuswehave (12.305)^=OXi,jj=QXj,*=flXk. Substituting these results in(12.304), weobtain (12.306)?*=^+QXP, 348 MECHANICS INSPACE [Sac. 1 where (12 -307>f-ir'+TrJ+Tr* Weusethesymbol 8/8t todenote apartial differentiation whichi,j,kareheld fixed. Wenote thatdP/dt consists oftwo parts. The firstpa $P/dt,istherate ofchange ofPasmeasured by*anobsen moving with S';itmaybecalled theraleofgrowth, since, calculating it,wethink ofthevectorPaschanging orgrowii whereasi,j,kremain constant. Thesecond partofdP/dt,vi ftXP,isduetotherotation ofthetriadi,j,k;itmaybecall the rate oftransport. Thus, forarotating frame, therate change ofavector equals rateofgrowth plus rateoftransport. Motion ofaparticle relative toarotating frame. LetS'beaframe ofreference which rotates with anguj velocityftabout apoint O,fixed inaNewtonian frame Relative to/S,thevelocity qofamoving particle Ais,by(12.30< (12.308) q^J^I+oxr, where r=OA.Theacceleration is (12.309) f,g_j+axq. Substitution from (12.308) gives (12.310)f-+ Xr+Ox+O Letq'and i'denote, respectively, thevelocity andaccelerate oftheparticle relative to',sothat (12.311) q'-f,f-g- Now, (dO=SO 50 (12.312) )dtS<^"*" fit' IOX(OXr)=O(O r)-rfis , SBC. 12.3] METHODS OFDYNAMICS INSPACE 349 andso(12.310) maybewritten (12.313) f=f+f,+fc, where (12.314) ft=^Xr+a(a r)-rJ22 , fc=2QXq'. Foraparticle fixed inS',q'=0;thenf=and fe=0, sothat freduces toft.Forthisreason ftmay, inthegeneral case,becalled theacceleration oftransport. The acceleration fc iscalled thecomplementary acceleration oracceleration ofCoriolis. Wenote that theacceleration ofCoriolis isperpendicular to bothQandq'. Foraparticle ofmassmacted onbyaforce P,thelawof motion inSis mi=P; in/S',thelawofmotion is (12.315) mf'=P-mft-mfc. Thus therotation ofS'gives risetotwo fictitious forces, mtt and mfc.The lastofthese istheCoriolisn force; the first isintimately related tothe force usually known ascentrifugal force. When these two forces areadded tothe actual force P,thelaw ofmotion ofa particle in8'isprecisely theNewtonian lawwesaythatS'isreduced torestbythe introduction ofthese fictitious forces. Frames withconstant angular velocity. Letusnow consider thecasewhere the angular velocity oftherotating frame S'is constant. SinceQisaconstant vector,it determines afixed axisofrotation through 0.LetANbetheperpendicular from the position oftheparticle Atothis axis (Fig. 127). ThenN Fio. 127.Thevectors Randrforaparticle A. ftr=Qrcos0, where 6istheangleAON. The acceleration oftransport is, therefore, 350 MECHANICS INSPACE [Sac. 12 ft=QQrcos6-rfi2 =ONW-(ON whereR=NA. Inthiscasetheequation (12.315) becomes (12.316) mf=P+mRfl2-2mftXq'; thefictitious forcewRfi2isthecentrifugal force, asordinari understood. Foraparticle atrest inS',q'=0,and sotheonly fon required toreduce S'torest isthecentrifugal force. Thecond tionforrelative equilibriumis (12.317) P+mRn2=0. This isactually thecondition used inSec. 5.3,indiscussingtl equilibrium ofaparticle onornear theearth's surface. Exercise. Show that inaframe with constant angular velocityj reduced torest, thecentrifugal force perunitmass isgrad V,whe V=-* Frames ofreference ingeneral motion. Theabove results havebeen obtained onthesupposition th, thepointOisfixed inaNewtonian frame ofreference. If moving, theformulas (12.308) and(12.309) forqarid fgivemere thevelocity andacceleration ofArelative to0.Thecomple expressionsforvelocity andacceleration areobtained byaddii thevelocity qoandtheacceleration fof tothese expressioi forqand f,respectively. Thus, relative toaframe S'moving inageneral manner,tl motion ofaparticle takes place inaccordance withtheequatic (12.318) mf=P-mf-ndt-mfc, wheref=acceleration ofparticle relative to', P=force applied toparticle, fo=acceleration ofbase point in' (relative toaNei tonian frame), f,fc=acceleration oftransport andacceleration ofCorio) [cf.(12.314)]. Forslow-moving frames, forwhich fandtheangular velocity aresmall, thefictitious forces mf,mft,mfcmaynot 1 SEC. 12.4] METHODS OFDYNAMICS INSPACE 351 noticeable. However, asremarked inSec. 5.3,theybecome importantforanairplane making asharp turnorpulling outofa power dive; theformula (12.318) enables ustoestimate the force which interferes with themotion ofthepilot's hands in manipulating thecontrols. 12.4.MOTION OFARIGIDBODY Thegeneral principles ofSec. 12.2govern themotion ofany system. Inthissection, they areused tofind explicit equations ofmotion forarigid body. Rigid body withafixed point. Consider arigidbody constrained torotate about afixed point 0.By(11.509), theangular momentum aboutOis (12.401) h=Auj.+Bwzj+Cco3k, wherei,j,k=unit vectors inthedirections ofprincipal axes ofinertia at0, A,B,C=principal moments ofinertia at0, i,w2,o>3=components oftheangular velocity<oofthe body inthedirectionsi,j,k. Aspointed outinSec. 11.4, ingeneral theprincipal axes at arefixed inthebody; inthatcasethetriadi,j,khastheangular velocity<a.But ifA,B,Carenot alldifferent, wemayusea principal triad which isfixed neither inthebody norinspace. Toallow forallpossibilities, weshalldenote theangular velocity ofthetriadbyQ,noting that&=wifthetriad isfixed inthe body. Writing weapply (12.306) andobtain (12.402) h=+aXh +Cw 3k+(1 X (Cco 3 352 MECHANICS INSPACE [SEC. 12.4 Now,by(12.209), h=G, whereGisthetotalmoment oftheexternal forces about 0;hence (12.402) gives, astheequations ofmotion ofarigidbodywitha fixed where G\}(r2,Gzarethecomponents ofGalong i,j,k. Ifi,j,karefixed inthebody, sothat 2=o,theequations (12.403) become !Ai-(B-C)o) 2o)3=Gi, #C02-(C-4)0)30)!=G2, Co>3 (A #)0)10) 2=#3. These areEuler's equations ofmotion forarigid body withafixed point. When theworking forces areconservative, wecan use, in place ofanyoneofthethree equations in(12.403) or(12.404), thefollowing equation, deduced from theprinciple ofenergy (12.210): (12.405) iGlwJ+#o)|+Ccof)+V=E. Example1.Arectangular plate spins with constant angular velocity w about adiagonal. Find thecouple which must actontheplate inorder totnaintain thismotion. InFig. 128,Oisthemass center oftheplate andi, j,kareunit vectors along theprincipal axes of inertia at0;kisnormal totheplate, and iandjlieinitsplane,ibeing parallel tothelength. The princi- palmoments ofinertia at areFIG. 128.Arectangular plate spin- ningabout adiagonal. (12.406) B=|ma2 ,C=\m(a*+&2 ), wheremisthemass oftheplate, 2aitslength, and26itsbreadth. Ifaistheangle between iandtheaxisofrotation, sothattana theangular velocity oftheplateis <o=cocosai-fcosinaj.6/0, SEC. 12.4] METHODS OFDYNAMICS INSPACE 353 Thecomponents ofuinthedirectionsi,j,kare,therefore, coi=coCOSor, o>2=cosina, coj 0. Substituting these values ofcoi,co2,co3andthevalues ofA,B,Cfrom (12.406) intheequations (12.404), weget (12.407) Gi=0,G2=0,G3=|w(a2-62 )a>2sinacosa These arethecomponents ofthecoupleGwhich must actontheplate; we observe thattheaxis ofthecoupleisnormal totheplateandturns with it. Ifwesuppose theplato toturn inbearings attheends ofthefixed diagonal andtobesubject only tothereactions atthese bearings, then clearlyitis these reactions which supply thecouple G.p]ach reaction, liesintheplane oftheplateand isofmagnitude a2-62 i(a2+62)* Astheplate turns, these reactions turnwith it. Fluctuating reactions ofthissortmust beavoided inthecaseofflywheels androtors. Itisnotenough tomake sure that themass center liesontheaxis of rotation. The rotating body must be balanced sothat theaxis ofrotation isa principal axis ofinertia. Itislefttothe reader toprove, bymeans of(12.404), that thefluctuating reactions vanish for any rotating body if,andonly if,this condition issatisfied. Example2.Acircular diskofradius a andmassmissupported onaneedle point atitscenter; itissetspinning withangular velocitycoabout alinemaking ananglea with thenormal tothedisk. Find theangular velocity ofthedisk atanysubsequent time. InFig. 129,kisaunitvector normal to thediskatthecenter 0,andi,jarefixed intheplane ofthedisk;wesuppose j chosen sothat the initial angular velocity liesintheplane ofkandj. Theangular velocityofthediskatanytime is <0=C0]l ~\"COgl "{"COski att=0, o>i=0, coz=toosinor, cos cooCOSa., Theprincipal moments ofinertia at are A=B \ma*. C $ma*.FIG. 129. Disk spinning about itscenter,iandjfixed in thedisk. 354 MECHANICS INSPACE [SBC. 12.4 Since theexternal forces (thereaction atandtheweight ofthedisk) havenomoment about 0,theequations (12.404) give AOl~(A-C)W 2W8=0, Aw 2-(C-A)wao)i=0, CW3 =0. From thelast ofthese equationsitfollows thatw3isconstant; hence, w3=cocos <*.Multiplying thesecond equation in(12.408) byi(=\/ 1) andadding theresult tothe firstofthese equations, weget A\ i(C A)UQ cosa=0, where =01+tw2.SinceC2A,thisequation canbewritten toocosa=0; thegeneral solution is where isaconstant. From theinitial conditions, wehave o=*osina, andhence (12.409) o>i=wosinasin(w<cosa), w2=wosin cos(w<cosa), 03=woCOSa. This isthesolution oftheproblemasstated, but itdocsnottellusatoncehow thediskmoves inspace. Tofindthis,we must either introduce theEulerian angles defining thepositions ofi,j,krelative to fixed axes, oruseadifferent method. Example3.Find themotion inspace of thedishconsidered inExample2. InFig.130,histheangular momentum vector,o)theangular velocity vector, and kaunit vector normal tothedisk as before;iandjareunit vectors inthe plane ofthedisk,butnotfixed init.The vectorjistaken intheplane determined byhand k.Wenote thefollowing jfacts: (i)Since theexternal forces have no moment about 0,then,by(12.209), his ithasafixed direction inspace determined bythe initialFIG, 130. Diskspinning about itscenter; jcoplanar with haconstant vecto] and k. conditions. (ii)Sinceh=A<i)\i -f-^4w 2j4-Cwskand intheplane ofjand k. (iii)Since thetriadi,j,kisnotfixed inthedisk, itsangular velocity ais different from u.However, kisfixedboth inthetriadandinthedisk. As0,then wi=and <olies SBC. 12.4] METHODS OFDYNAMICS INSPACE -355 apoint ofthetriad, theextremity ofkhasvelocity OXk;asapoint ofthe disk,ithasvelocity uXk.Hence, (Oil+G2j+flak)Xk=(wii+ 2J+w3k)Xk, andsofti= o>i,122=w2.Since wi- 0,wehave fii=0. Applying thegeneral equations (12.403) andmaking useoftheabove facts, weget Ca> 3w2-0, Ad>2=0, Cd> 3-0. .Thus,a>2,ws,hi,h$areconstants, and S23/fi 2=fti/wj=Cw3/^la>2=h3/hz' f theangular velocity Qofthetriad hasconstant magnitude and liesalong the fixed direction h.The following facts concerning themotion arenow obvious: (i)Thediskspins about itsnormal kataconstant rate wa. (ii)Theangle between kandh,givenbyhcos/3=hs,isconstant; the normal tothediskmoves onaconewith axish,turning abouthatthecon- stant rate ft. (iii)Theangleabetween <oandk,givenbytocosa=ws,isconstant; the angular velocity vector todescribes aconeabout thenormal tothedisk. This isthebody cone (ef.Sec. 11.2). Theangle /3between oandhis alsoconstant, andsothespace conehasconstant sernivertical anglea3 andaxish;itliesinside thebody cono. Theabove problemisaspecial case ofthemotion ofarigidbody witha fixed point under noforces, considered inChap. XIV. General motion ofarigidbody. Wenowconsider arigidbodymoving quite generally. LetF denote thetotal external force andGthetotalmoment ofthe external forces about themass center. By(12.203), theaccelera- tion fofthemass center (relative toaNewtonian frame)is given by* (12.410) mi=F, wheremisthemass ofthebody. Forthemotion relative tothe mass center wehave,by(12.209), (12.411) h=G, where histheangular momentum about themass center. This lastequationisexactly thesame asifthemass center were fixed, andsocanbetreated bythemethods given above. *Forsimplicity, wedrop thesubscripts fromqand fo,thevelocity and acceleration ofthemass center. 356 MECHANICS INSPACE [SEC. 12.5 Letusresolve thevectorsf,F,6,Galong aprincipal triad i,j,katthemass center. Asbefore, thistriad issupposed to bepermanently aprincipaltriad. Itsangular velocity willbe denoted byft;ifthetriad isfixed inthebody,Q=<a,theangular velocity ofthebody. Now,by(12.306), where q=ui+vj+wk isthevelocity ofthemass center. Substituting forfin(12.410) andnoting that (1-2.411) leads toequationsoftheform (12.403), weobtain thefollowing scalar equations ofmotion: m(u vQ3+ m(b wQi m(w u&i <j)i #0)2^3+(12.412) Here theconstants A,B,Caretheprincipal moments ofinertia atthemass center. Theequations (12.412) aresixequations forthecomponents ofvelocity ofthemass center andthecomponents ofangular velocity ofthebody. Foranyoneofthese sixequations, we cansubstitute thelawofconservation ofenergy, (12.413) T+V=E, provided theexternal forces areconservative. Inamore explicit form, (12.413) reads (12.414) im(w2+v2+w2 )+|G4 o>?+B<*\+CwJ)+V=E. Exercise. From thebasic equations (12.410) and(12411), deduce the principle ofenergy forarigidbody intheform T=Fq-hG <o. 12.6.IMPULSIVE MOTION Theprinciples ofdynamics, thus farconsidered inthischapter, dealwith ordinary orcontinuous motion. Bythiswemean thattheforcesacting, andtheaccelerations produced, arefinite. SBC. 12.5] METHODS OFDYNAMICS INSPACE 357 Sometimes wehave todeal with problems inwhich sudden changesinvelocity occur. Fortwo-dimensional problems of thistype,weusethemethods ofChap. VIII; similar methods formotion inthree dimensions willnowbedeveloped. General equations ofimpulsive motion. Integration oftheequations (12.203) and(12.209) from time ttotime t\gives* (12.501) A(roq)=f'Fctt,Ah=P1Gdt,Jt$ I/to whereAdenotes anincrement inthetime interval t\ to.These equations express theprinciples oflinear andangular momentum inintegrated form. Inwords, theyread asfollows: (i)theincrement inthelinearmomentum ofasystemisequal tothetotalimpulse oftheexternal forces; (ii)theincrement intheangular momentum about apointO (either afixed pointinaNewtonian frame orthemass center of thesystem inquestion)isequal tothetime integralofthetotal moment about oftheexternal forces, i.e.,the totalangular impulse about 0. Inthisform theprinciples oflinear andangular momentum can easily beapplied toproblems where sudden changes in velocity occur. Themethod ofprocedureisessentially that given inChap. VIII, andsoweshall giveonlyabrief outline here. Thevery short time interval t\ tQinwhich thechanges occur isregarded aninfinitesimal. Any finite force willthen contribute nothing tothetotalimpulsive force F=limftlFdt. <1-><0 J** Ontheother hand, aforce P,forwhich P=limrPdt isfinite, contributes theimpulsive forcePtoF. Ifrdenotes theposition vector ofthepointofapplication ofsuchaforceP *Forsimplicity, wedrop thesubscript from q ,thevelocity ofthemass center. 358 MECHANICS INSPACE (SEC. 12.5 (the position vector being relative tothepoint about which theangular momentum iscalculated), then rdoesnotchange byafiniteamount intheinfinitesimal time t\ ta.Hence, limf*(rXP)dt=rXlimJP1Pdt=rXP, andtheforcePcontributes theimpulsive moment r%XPtothe totalimpulsive moment 6=limJo"Gdt. Then, from (12.501), wehave (12.502) A(mq)=F, Ah=G, whereF=total impulsive force=vector sum ofexternal impulsive forces, G=total impulsive moment =totalmoment ofexternal impulsiveforces. These arethegeneral equations cfimpulsive motion. Forarigidbody ingeneral motion, the first oftheabove equations gives thechange inthevelocityofthemass center; the second gives thechange intheangular momentum (andhence thechange inangular velocity) about themass center. Fora rigidbody with afixed point, thesecond equation in(12.502) alone suffices todetermine thechange inangular velocity. Example. Asquare plate, ofmassmandedge 2a,issuspended from one corner 0. Itisstruck atacorner inahorizontal direction perpendicular to theplane oftheplate. About what linedoes theplate begintoturn? Leti,j,kbetheprincipal triad ofinertia at (Fig. 131);ipoints upward along thediagonal through 0,andjisahorizontal vector intheplane ofthe plate. Before theplateisstruck, theangular momentum about iszero;imme- diately afterward,itis h=4oni -f whereon, cos,wsarcthecomponents ofangular velocity andA,B,Cthe principal moments ofinertia at0. IfPisthemagnitude oftheblow, tho external impulsive forces arePkatthepoint r--oV2 (i+J), SBC. 12.5] METHODS OFDYNAMICS INSPACE 359 nomoment about 0,the andanimpulsive reaction Qat0.Since second oftheequations (12.502) gives +Bu zj+Co>,k--oV2 (i+j)X Hence, 3-0; theplate begins toturnabout aline initsplane passing through 0.The angle 0,between iand this axis of rotation,isgivenby __2_Atan&~~*~~~~~"75*Wl /> Since B= wefindra2+2mo2= tan0=-J; theaxis ofrotation isindicated in Fig.131bythevectors. Theimpulsive reaction Qat isFlQ ,m.Square pkte, suspended easily found. The velocity ofthefrornoandstruck byablowpk> mass center, immediatelyafter the plate hasbeenhit,is Thus, from the firstof(12.502), weget -k=Pk+0, andso =-* 360 MECHANICS INSPACE [SEC. 12.6 12.6.SUMMARY OFMETHODS OFDYNAMICS INSPACE I.Motion ofaparticle. (a)Equationsofmotion: (12.601) wf=P (vector form) ; (12.602) mx=X, my=Y, mz=Z (Cartesian coordinates) ; (12.603) ms=Pi,m-=P2,=P3 P (intrinsic equations). (6)Principleofangular momentum: (12.604) h=rXP. (c)Principleofenergy: (12.605) f=TF, (T7=w<72 ,Tf=work done); (12.606) T+F=E (conservation ofenergy). II.Motion ofasystem. (a)Principle oflinearmomentum: (12.607)Hi!=F,(M=2)mtqt);v 1=17 (12.608) mq=F (motion ofmass center). (6)Principle ofangular momentum: (12.609) h=G(fixed point ormass center), (c)Principle ofenergy: (12.610) t=F; (12.611) T+V=E (conservation ofenergy). III.Motion ofarigid body. (a)Rigid body withafixed point: (12.612) 6=~+QXh =G;ot SAui(B C)w 2w3=Gi, Bws-(C-4)ttjtti=(J2, (7d>3 (A J5)cOiOJ2=CrsJ (12.614) i(Af+Bwi+CJ)+V=E (conservative forces). Ex.XII] METHODS OFDYNAMICS INSPACE 361 (6)Rigid body ingeneral: (mf=F (motion ofmass center), \i=G (motion relative tomass center) ; (12.616) im?2+i(A!+Bu\+C!)+V=E (conservative forces). IV.Rotating frame ofreference. Rate ofchange ofanyvector: (12'617) f-Tt+QxP" V.Impulsive motion. General equations: (12.618) A(mq)= ,Ah=i. EXERCISES XII 1.Aheavy particle moves onasmooth surface. Show that itsspeedis thesamewhenever itspath cutsagiven horizontal curve onthesurface. 2.Show directly from Euler's equations (12.404) that,ifG=and A=B,then o>isconstant. 3.Aparticle isattracted toward afixed linebyaforce, perpendicular to thelineandvarying asthedistance from theline. Show that itspathisa curve traced onanelliptical cylinder. 4.Asolid ofrevolution rotates withconstant angular velocity wabout a fixed axiswhich passes throughitsmass center and isinclined totheaxisof symmetry atanangle a.Prove thatthereactions oftheaxisonthesolid areequipollent toacouple ofmagnitude (C 4)cu2sinacosa, whereCisthemoment ofinertia about theaxis ofsymmetry andAthe other principal moment ofinertia atthemass center. 6.Explain howaman, standing onasmooth sheet ofice,canturn round bymoving hisarms. 6.Abaroflength 2oisfitted atitsmiddle point withanutwhich moves without friction onafixed vertical screw ofpitch p;thebarremains hori- zontal andturns withthenut. Find theacceleration ofthenut. 7.Two particles, ofmasses m,w',attract oneanother according tothe inverse square law. Attime t=0,misattheorigin andhasavelocity u along thex-axis, andm'isatthepoint (a,6,c)andhasvelocity components (u'j v'jwf ).Determine (i)thecoordinates ofthemass center attimet, (ii)theconstant areal velocity ofthemotion ofm'relative tom, (iii)theconstant areal velocity ofthemotion ofmrelative tom'. 8.Twomensupport auniform pole ofmassmandlength 2ainahori- zontal position. They wish tochange endswithout changing their positions 362 MECHANICS INSPACE [Ex.XII ontheground, bythrowing thepoleintotheairandcatchingit. Ifthepole istoremain horizontal throughout itsflightandthemagnitude oftheimpul- siveforce applied byeachman istobeaminimum, findthemagnitudes and directions oftheimpulsive forces. 9.Anequilateral triangleisformed ofthree rods, each ofmassmand length 2a. Ithangs fromonevertex, about which itisfreetoturn.Ablow Pisstruck ononeofthelower vertices inadirection perpendicular tothe planeofthetriangle. Prove that theimpulsive reaction onthepoint of support hasamagnitude If*. 10.Asolidhomogeneous ellipsoid ofmassmandsemiaxesa,b,cspins with constant angular velocityo>about anaxiswhich isfixed inspace and makes constant angles a,0,ywith theaxes oftheellipsoid. Show thatthe components (along theaxes oftheellipsoid) ofthecouple thatmust acton itinorder tomaintain thismotion are |ra<o2(&2c2 )cos cos7 andtwosimilar expressions. 11.Aparticle ofmassmmoves inaplane under theaction oftwo forces. Oneforce isanattraction mk*rtoward theorigin ;theother isperpendicular tothevelocity qandhasmagnitude mk'q. Show thatthemotion isgiven byanequation oftheform x+iy-e^k/t(Aeict+Be~ict ). Howmany arbitrary constants (tofitinitial conditions) arepresent inthis solution? 12.Aninsect runswithconstant relative speedvround therimofawheel ofradius awhich rollsalong astraight roadwithuniform velocity V.Find themagnitude anddirection of(i)theacceleration relative tothewheel, (ii)theacceleration oftransport, and(iii)theCoriolis acceleration. Indi- catethese accelerations inadiagram. 13.Afree rigidbodyisatrest. Find three linear scalar equations to determine thecomponents ofanimpulsive force which, applied atan assigned point ofthebody, imparts tothat point anassigned velocity. Solve these equations inthecasewhere theassigned pointliesononeofthe principal axes ofinertia atthemass center. 14.Athinrodofmassmandlength 2aismade torotate with constant angular velocity wabout anaxiswhich passes through oneendoftherod andcuts itataconstant angle a.Reduce theforcesystem exerted bythe axisontherodtoaforce atthefixedendandacouple. 16.Arigid triangular targetisfixed atthecorners, andabullet isfired normally into it.Find theregion inwhich thebullet must strike inorder thatnosupport mayexperience animpulsive reaction normal tothetarget greater than halfthemomentum ofthebullet. 16.Auniform circular diskofmassMandradius aissomounted that it canturn freely about itscenter, which isfixed. Itisspinning withangular velocity about theperpendicular toitsplane atthecenter, theplane being horizontal. Aparticle ofmass m,falling vertically, hitsthedisknear the Ex.XII] METHODS OFDYNAMICS INSPACE 363 edgeandadheres toit.Prove thatimmediately afterward theparticleis moving inadirection inclined tothehorizontal atanangle a,givenby .m(M+2m) vtan a.=4TtffTtf ,,(M(M -f-4m)aw where visthespeed oftheparticle justbefore impact. 17.Ahomogeneous ellipsoidofsemiaxes a,6,c, (a>b>c) istobemounted onahorizontal axisLinsuchawaythat itmay oscillate asacompound pendulum with thesmallest possible periodictime. What positionofLrelative totheaxes oftheellipsoid should beselected? 18.Acrankshaft ofmass m,intheform ofaletter Sformed outoftwo semicircles, each ofradiusa,spins withangular velocity<oinbearings atits ends. Find themagnitudes ofthenvictions exerted onthebearings, and show thedirections ofthese reactions inadiagram. 19.Asystemissotinmotion byimpulsive forcesappliedtocertain pre- scribedparticles.IfPistheexternal impulsive forceonatypical particle, qitsvelocity, and &ianarbitrary infinitesimal displacementconsistent with theconstraints (assumed workless), show that S(mq-5r)=S(P 6r), where thesummation ontheleftextends over allparticlesofthesystem and thesummation ontheright overtheprescribed particles. LetTbethekinetic energy oftheactual motion andT'that ofanyother motion (q')consistent with theconstraints andmaking q'=qforthe prescribed particles; prove thatT<T'(Kelvin's theorem). 20.Arhombus ABCD isformed offouruniformrods, each ofmassmand length 2a,smoothly jointed atthevertices. Prove that iftherhombus isin motion initsplane,insuchawaythatAandCaremoving along thediagonal AC,thekinetic energy maybeexpressedintheform T=2m(v-2awsin0)2+jJwaV, whcro visthevelocityofA,wtheangular velocity ofAB,and 6theinclina- tion ofACtoAB. Hence, prove byKelvin's theorem (seeExercise 19^that,iftherhombus isatrest intheform ofasquare and isjerked intomotion byanimpulsive force applied atAinthedirection AC,then theangular velocity imparted totherods is 3\/2 v 10'a where visthevelocity impartedtoA. CHAPTER XIII APPLICATIONS INDYNAMICS INSPACE MOTION OF APARTICLE 13.1.NOTEONJACOBIAN ELLIPTIC FUNCTIONS Sofar,wehave usedonlytheelementary functions, alone or incombination polynomial, trigonometrical, exponential, and logarithmic. Wenow find itnecessary tointroduce theelliptic function. Definition ofafunction bymeans ofadifferential equation. Avariable yissaid tobeafunction ofxwhen tovalues ofx there correspond values ofy.Infact,afunction isdetermined byarulewhich assigns ywhen xisgiven. Usually thisrule isa formula admitting direct calculation ofy(e.g., x2 ,3sin2x),but wemay alsouseadifferential equation todefine afunction; we must, however, assigninitial conditions tomake thesolution unique. Consider, forexample, thedifferential equation with theconditions y=0,dy/dx=1forx=0. Ifwehad never previously heard ofthefunction sinxythisequation andthe initial conditions would serve todefine it.Another wayof defining sinxisbythedifferential equation (13.101) feY=1-y\ with theconditions (13.102) y=0,j|>0,forx=0. Atypeofdifferential equation with periodic solutions. Inconnection withelliptic functions, wehave tostudy the differential equation 364 SEC. 13.1] MOTION OFAPARTICLE 365 (13.103) where kisaconstant suchthat<k<1.Itisreally simpler, however, totakeamore general pointofviewandstudyfirstthe differential equation where f(y)isageneral function. Wecanfindoutasurprising amount about thesolutions ofthisequation without specifying thefunction f(y).Weshall, however, assume that f(y)is continuous; that itvanishes fory=aandy=b(a<6),but its derivative doesnotvanish foreither ofthese values; and finally that f(y)>fora<y<b.(The right-hand side of(13.103) hasthese properties,ifwetakea1,6=1.) Letustake ageometrical point ofview, regarding xandyas rectangular Cartesian coordinates inaplane. The equation (13.104)isthen arelation between theslope andtheordinate onacurve, andasolution, orintegral curve,isacurve forwhich thisrelation issatisfied. Anumber ofstatements canbemade regarding theintegral curves of(13.104). These willnowbegiven, followed bytheir proofs. (A)Ifweknow anintegral curve, then thatcurve translated through anydistance parallel tothe a*-axis isalsoanintegral curve. (B)Ifanintegral curve starts inthefundamental strip a<y^6,itcannot passoutofthatstrip. (C)Every integral curve inthefundamental strip touches the bounding linesy=a,y=bandhasatnoother point atangent parallel tothez-axis. (D)There isone,andonly one, integral curve touching a bounding lineatagiven point.* (E)Allintegral curves maybeobtained from oneintegral curvebytranslation parallel tothex-axis. (F)Anintegral curve issymmetric with respect toitsnormal atapoint ofcontact withabounding line. *Thebounding linesy=a,ybsatisfy (13.104), butwedonotregard them asintegral curves. They aresingular solutions. 366 MECHANICS INSPACE SEC. 13.1 (G)Thez-distance between successive contacts ofanintegral curve with thebounding lines is (13.105) (H)Any solution of(13.104), y=<t>(x),isaperiodic function with period 2P,wherePisgiven by(13.105); thismeans that (13.106) 2P)= forallvalues ofx. (I)Ify=<f>(x)isanyonesolution of(13.104), thenthegeneral solution isy=4>(x+c),where cisanarbitrary constant. Some ofthese properties areshown inFig. 132. Proofs : (A)Neither slope norordinate ischanged bythetranslation; iftherelation (13.104)issatisfied bythecurve before translation, itwillbesatisfied after translation. y=a Fia. 132. General character ofasolution ofthedifferential equation (13.104). (B) Ifthecurve passed outofthestrip, /(?/)would become negative anddy/dx imaginary. (C)Byhypothesis, /(a)=/(&)=0;hence, dy/dx=onthe boundinglines. Further, f(y)> fora<y<6,and so dy/dx cannot vanish between thebounding lines. (D)Letx=zo,y=abeapoint onaboundingline. Ifwe invert(13.104), takethesquare root,andintegrate, weget (13.107) x-zo--dr,x SEC. 13.1] MOTION OFAPARTICLE 367 according asdy/dx< ordy/dx>0.These twoequations together give (intheneighborhoodofx=x)theunique integral curve satisfying thecondition oftangency. (E)LetCandC'beanytwointegral curves. Wehave to show that C'maybemade tocoincide withCbyatranslation. LetCtouch yaatx=XQ.Translate C'until italsotouches y=aatx=XQ.By(A),itisstillanintegral curve after translation; by(D),itcoincides with C. (F)Thetwoequations (13.107) give thetwoparts ofan integral curve, meeting atapoint oftangency withy=a.To agiven yythere correspond equal values ofxXQ,except for sign. This establishes thesymmetry fortheparts ofthecurve running upfrom y=atoy=b.Butthere isthesamesym- metry with respect tothenormal atacontact withyb.Itis notdifficult toseethat thisimplies symmetry ofthewhole curve with respect tothenormal atanypoint ofcontact withabound- ingline. Ifthepart ofthecurve totheright ofsuch anormal isfolded over thenormal,itwillcoincide with thepart ofthe curve onthe left. (G)This isobvious from (13.107). (H)This follows from (G)andthesymmetry ofthecurve. (I)This merely expresses (E)inanalytic form. Since the differential equationisofthe first order, weexpect justone constant ofintegration. TheJacobian elliptic functions. Letusnowapply ourgeneral results tothedifferential equation (13.108)(jy=(1- 2/2)(l-*V), (0<fc<1). Thefundamental stripis 1^y^1.Wedefine theJacobian elliptic function snxtobethatsolution of(13.108) which satisfies theconditions (13.109) y=0,^>0, forx=0.ax Itisevident thatsnxdepends onthevalue ofJfc,which iscalled themodulus ofthefunction, andwemay write itsn(x,k) ;but itisusual tosuppress theexplicit dependence onk.(Inspeaking ofthefunction, wecall it"ess-en-ex."). 368 MECHANICS INSPACE [SEC. 13J From theresult(I),stated onpage 366,weknow thatthemost general solution of(13.108)is y=sn(x+c), where cisanarbitrary constant. Further, by(H),weknow thatsnxisaperiodic function. Thus, (13.110) sn(x+IK)=snx, whereKistheelliptic integral (13.111) K=^f1 %L dy Kis,ofcourse, afunction ofk. By(F)thegraph ofthefunction snxhassymmetry with respect toeachnormal atacontact withthelinesy=1.But, since theright-hand sideof(13.108)isaneven function ofy,the graph hasafurther symmetry. Theequation (13.108) andthe conditions (13.109) areunchanged whenwechange xinto x andyintoy,andsothecurve isunaltered byareflection in theorigin. Forthegeneral case,shown inFig. 132,thewhole curve canbeconstructed bymeans ofthesymmetry whenwe know ahalfwave, running from y=atoy=b.Inthecase of snx,weneed merely know thecurve from y=toy=1or, equivalently, from x=tox=K. Theproperties ofsnxmaybesummed upasfollows: =(1 sn2 j:)(l k2sn2 x), (13.112)(0<*<!), snO=0,f-T-snzj=1, sn(x+4K)=snx. Wenow define other elliptic functions, enxanddnx,bythe equations n311^ /cn2x^^~~sn2X) cno=i, (16.116) \dn2Z=1-/b2sn2 x, dn=1, SEC. 13.1] MOTION OFAPARTICLE 369 with thefurther condition that the functions andtheir derivatives shall becontinuous. Since k<1,dnxis always positive. Itisclear that enxhastheperiod4Kanddnx theperiod 2K. Ifwetakethesquare roots ofthe two sides ofthe firstequation in (13.112), wegetanambiguous sign. However, forcontinuity, onesign must betaken throughout, andthat signisfixedbyconsidering x=0. Thuswefind (13.114) -j-snxcnxdnx.ax Differentiation of(13.113) gives dsnx-j-snxdx =snxenxdnx,d denx-j-enx snx-p-snxdx dx andso (13.115) -T-enx snrcdna;. Similarly, (13.116) -T-dnx=fc2snzcnz. Just as(13.108)isageneralization of(13.101) andreduces toitifk=0, sotheelliptic functions aregeneral- izations ofthetrigonometric func- tions. Infact,ifk0,wehave (13.117)snx sinxy enx=cosx, dnx=1, and(13.114), (13.115) reduce to familiar formulas. 370 MECHANICS INSPACE [SEC. 13.2 Thetheory ofelliptic functions isextensive, but thisvery brief presentation contains enough toenable ustosolve certain dynamical problems. Fornumerical tables ofthefunctions snx,enx,dnx,seeL.M.Milne-Thomson, Die elliptischen Funktionen vonJacobi (Verlag Julius Springer, Berlin, 1931). Tables ofelliptic integrals may alsobeused tofindtheelliptic functions;cf.J.B.Dale, Five Figure Tables ofMathematical Functions (Edward Arnold, London, 1903), orE.Jahnke and F.Emde, Tables ofFunctions (B.G.Teubner, Leipzig, 1938). Figure 133shows graphs ofthefunctions, drawn fork*=0.7; thismakesK=2.07536. Exercise. Show that, inthelimit k=1,weget snx=tanhxyenxdnx=scch x. 13.2.THESIMPLE PENDULUM Themotion interms ofelliptic functions. Wecannowgivetheexact solution forthemotion ofasimple penduluminterms ofelliptic functions.* Letmbethemass ofthebobandathelength ofthependulum. Theequationof energy (12.111) gives (13.201) \mtffr-mgacosB=E, where 6istheinclination ofthestring tothedownward vertical, andEtheconstant total energy. Weshallsuppose themotion tobeoscillatory withamplitude a,sothat 6=for= a. ThenE=mgacosa,and(13.201) maybewritten (13.202) 62=2p2(cos-cosa)=4p2(sin2%a-sin2$0), where p2=g/a. Letusdefine<j>by (13.203) sin|0=sin\asin , sothat |cos\B6=sin\acos <<. *Since themotion ofacompound pendulumisidentical withthat ofthe equivalent simple pendulum (cf.Sec.7.2), thesolution nowgiven applies alsotothecompound pendulum; in(13.202) andthesubsequent equations, wearetoputp*=ga/k*, where aisthedistance ofthemass center from the axisofsuspension andktheradius ofgyration about that axis. SBC. 13.2] MOTION OFAPARTICLE 371 Multiplying (13.202) byicos2|0,weget sin2v&cos2 <t> (j>2=p2sin2%acos2 <j>cos2 0, or (13.204) tf=P2-sin2 \OLsin20). Ifwemultiply thisequation bycos2 <andput (13.205) y=sin <=Sm? ,k=sinia,sinTct weget (13.206) i/2=p2 (l- ?/2 )(1-k*y2 ). Except fortheconstant p2ontheright, thishastheform ofthe equation (13.108). Togettheexact form, wedefine anew independent variable by (13.207) x=pt andobtain (13.208)=(1-*)(!-*V). Thegeneral solution ofthisequationis (13.209) y=sn(x+c), where cisaconstant ofintegration. Honce, wehave thefollow- ingresult: Thegeneral oscillatory motion ofasimple pendulum, withamplitude a,isgiven by (13.210) sin$0=sin$ani\p(t- )1, where toisaconstant cfintegration, p2=g/a,and themodulus oftheelliptic functionisk=sin-Jar. Weusually findindynamical problems that,ifsnappears, the other elliptic functions en,dnhave simple physical meanings. From (13.210) wegetatonce,by(13.113), (13.211) cos$0=dn[p(t- Jo)], anddifferentiation of(13.210) gives (13.212)6=2psin$aen[p(t-<)]. 372 MECHANICS INSPACE [SEC. 13.2 Theperiodic time. Aswehave seen, theperiods ofsnxandenxare4X. Thus, by(13.210) and (13.212), themotion repeats itself after atime 4K'/p, andsotheperiodic time ofthependulumis (13213) r==- PP Putting y sin<,weget (13.214)r-i"-** PJVl-*2si Now, flT Jos1.3 andsowehave thefollowing infinite scries fortheperiodic time ofthependulum: Forvery small amplitude a,weget,asafirstapproximation, a agreeing with (6.307), where Iwasused todenote thelength. Thenextapproximationis (13.216) r. Itisevident from (13.215) that theperiodic time increases steadily with theamplitude. SEC. 13.3] MOTION OFAPARTICLE 373 13.3.THESPHERICAL PENDULUM Aparticle ofmassmisattached toafixed pointbyalight string orrodoflength aandoscillates under theaction ofgravity. Since theparticleisthus constrained tomove onasphere, thissystemis called aspherical pendulum. Under specialinitial conditions, aspherical pendulumwillmove inavertical plane; then themotion isthat ofasimple pendulum, discussed inSec. 13.2. Although weshall beable todeter- mine thegeneral motion ofaspherical pendulum interms ofelliptic functions, there aretwoparticular motions which canbediscussed quite simply. The first isamotion inwhich theparticle performs small oscillations near the lowest point ofthesphere, andthe second ismotion inahorizontal circle.O / FIG. 134. Spherical pendu- lum Small oscillations (first approximation). LetOxyz berectangular axes,Obeing atthelowest point ofthesphere andOzbeing directed vertically upward (Fig. 134). Ifi,j,kisaunitorthogonal triad along theaxes, theposition vector oftheparticleis (13.301)r=xi+yj+zk. TheforcePontheparticleismadeupofgravity andthetension (S)inthe string. Now thedirection cosines ofthestring (running from theparticletothepoint ofsupport) are x al/?a andso (13.302) P=--[xi+yj+(z-a)k]- Sofartheexpressions areexact. But ifxandyarcsmall,z isasmall quantity ofthesecond order, since theplane2= touches thesphere. Hence, wehave asequation ofmotion, 374 MECHANICS INSPACE [Sue. 13.3 omitting small quantities ofthesecond order, (13.303) m(A+yj)=-^i-^j+(S-m?)k. Comparing thecoefficients, weseethatS=mg,and (13.304) x+p*x=0, if+P2 */=0,U2=|Y These aresimple harmonic equations, asin(6.403). Asfarasits projection onthehorizontal planeisconcerned, thebob ofthe pendulum moves likeaparticle attracted toward byaforce proportionaltothedistance from 0.Asshown inSec.6.4,the pathisanellipse with center at0.(When theellipse degener- atestoastraight line,wegetthemotion ofasimple pendulum, performing small oscillations.) Wehave idealized theproblem byleaving out allconsideration offrictional resistance. The effect ofthis istocause thebob ofthependulum tospiralintoward 0,instead ofcontinuing for ever intheelliptical path. However, theapproximation (neglect ofz)isperhaps amore serious oversimplification. Weshall see theeffect ofthislater,whenweconsider thesecond approximation. Theconical pendulum. Anyprescribed motion ofaparticle willtake place under tile action ofasuitable force, namely, aforce equal totheacceleration multiplied bythemass ofthe particle. Thus thebob of*a spherical pendulum maybemade tomove inanywayonthe sphere defined bythelength ofthependulum; toproduce this motion,itisingeneral necessary toaddasuitable force tothe weight ofthebobandthetension inthestring. But ifwecan findamotion inwhich nosuch additional force isrequired, then thatmotion isapossible motion ofthependulum under weight andtension alone. Consider amotion inahorizontal circle ofradiusRatconstant speed q.Theacceleration isofconstant magnitude qz/Rand is directed inalong theradius ofthecircular path. Resolving along this radius, along thetangent tothecircular path, and vertically, wefindthatnoadditional force isrequired provided that Ssin=-~; Scos=mg, SEC. 13.3) MOTION OFAPARTICLE 375 wheremisthemass ofthebob,Sthetension, and6theinclina- tion ofthestring tothedownward vertical (Fig. 135). Since sin6R/a, elimination ofSgives gR* (13.305)--- If6denotes thedepth ofthehorizontal circle below thecenter ofthesphere, sothatb2=a2R2 , wehave (13.306) This gives thespeed qatwhich ahorizontal circle atdepth bmay(^ bedescribed bythebob ofthe- pendulum. When behaving in thisway, thependulumiscalled FIG. 135. Conical pendulum.aconical pendulum, since thestring describes aright circular cone. Exercise. Find thetension inthestring ofaconical pendulum moving atadepthb.Examine thelimits b*a,b 0. Thegeneral motion ofaspherical pendulum. Toinvestigate thegeneral motion ofaspherical pendulum, we take cylindrical coordinates R,<,z,theorigin being atthe center ofthesphere andtheaxisofzdirected vertically upward. Wehave already obtained theequations ofmotion in(12.114) and(12.115); theymaybewritten (13.307) (a2-z2 )<=h, (13.308)z2=/(z), where U3.309,*,.![[<,_.,(.-)_*} Werecall thataisthelength ofthestring (i.e.,theradius ofthe sphere onwhich theparticle moves), andhandEareconstants, thevalues ofwhich depend ontheinitial conditions. Weshall, fordefiniteness, assume hpositive, sothat<t>increases; there isno lossofgenerality here, sincewecanreverse atwillthesense in which 6ismeasured. 376 MECHANICS INSPACE [SEC. 13.3 Ourplanistosolve (13.308) forzasafunction oft;then (13.307)willgive byaquadrature. Wenote that f(z)isacubic. Itispositive forlarge positive values ofz;itisnegative forz=a;itispositive forvalues ofz occurring during themotion, asweseefrom (13.308). These lastvalues must ofcourse lieintherange (a,a).Since a cubic cannot havemorethan three changes ofsign,itfollows that -a Fm. 136. Graph of/(z). thegraph of/(z)isofthegeneral nature shown inFig.136;the function f(z)hasthree real zeros, Zi,z2,23,such that (13.310) a<Zi<Z2<CL< (Inexceptional cases,wemayhaveoneormore signs ofequality instead ofinequality.) Since /(z)cannot benegative during the motion, weseethat zoscillates between thevaluesZi,z2.We notethat theequation (13.308)isofthetype (13.104), sothat allthegeneral results established for(13.104) apply to(13.308). Toobtain zasafunction oft,weproceed asfollows: Since Zi, 2,23arethezeros of/(z),wehave (z-zi)(z-z2)(z- that z=2uu. Then (13.308)(13.311) /( Letusdefine u gives (13.312) This suggests thedifferential equation (13.108) forthe elliptic function sn.LetusdefinejT-j (Z2~Zi~M2)(Z3 ZiW2 ). SEC. 13.3] MOTION OFAPARTICLE 377 (13.313) k= Then (13.312) maybewritten (13.314)*>2=p2(l-v*)(l- andso (13.315) v=sn[p($- fe)], where Joisaconstant ofintegration. Hence wehave !z-Zi=(z2-21)sn2 [p(t-*)], 22-z=(22-20cn2[p(t- to)], 23-*=(z3-zOdn2[p(*-*)]. Anyoneofthese three equations giveszasafunction of t.We notethat zhastheperiod z 9JT (13.317) r=, where jf^isasin(13.111). Since, by(13.307), </>increases steadily throughout themotion/the path of theparticle onthesphereisas shown inFig. 137. Wohave already seenthatthe particle oscillates between thetwo levels z=Ziand z=z2.We shallnowshow thatthearithmetic mean ofthose levels liesbelow theFIG. 137.Thepath ofthebobofa spherical pendulum. center ofthesphere,thisstatement being equivalent to (13.318) zi+z2<0. Wehave two different expressions forf(z), (13.309) and (13.311) ;theymust, ofcourse, beidentically equal, andtherefore (13.319) + 378 MECHANICS INSPACE [SBC. 13.3 From thesecond ofthese, wehave (13.320) 2l+ ,=-L*!?. 23 Since 3iand 22areeach lessthanainabsolute value and z3is positive, (13.318) follows atonce. Thepathonthesphereisrepresented analytically byarelation connecting zand<j>.Elimination oftfrom (13.307) and(13.308) gives therequired relation intheform ofadifferential equation Ifwelookdown onthependulum fromagreat distance above, thebobappears todescribe aplane curve, withRand <aspolar coordinates. Themotion resembles that ofaparticle attracted toward acenter offorce, theareal velocity (^R^<j>) being acon- stant forboth motions. Just asweconsidered theapsides ofthe orbit ofaparticleinaplane, sowecanconsider theapsides ofthe horizontal projection ofthepath ofthespherical pendulum. These points correspond tostationary values ofR,i.e.,toR=0. Hence z=atthese points, sinceR*+z*=a2 .Astheactual path oscillates between thecircles atheights z=z\,z=z%on thesphere, sothehorizontal projection ofthepath oscillates between circles ofradii\A2 z?and\/azz2 ,. Theapsidal angleaistheincrement in <corresponding tothe passage from z=zitoz=z2.Thus, by(13.321), (13.322)<x=hf" 2_d* ha m (a2z2 )\/(z Zi)(z z2)(z z3) Exercise. FindZi,z2,ZBforaconical pendulum withthebobatadepth 6 below thecenter ofthesphere. Small oscillations (second approximation). Theexpression (13.322)istoocomplicated asitstands tobe ofmuch interest. But ityields adefinite simple resultwhenwe suppose theoscillations tobesmall. Thereasoningisdelicate, because, asZiand Zztend toa(thelowest point onthesphere), theextent oftherangeofintegration tends tozero,andthe SEC. 13.3] MOTION OFAPARTICLE 379 integrand tends toinfinity. Themethod ofapproximation is important; thesamemethod maybeused infinding therotation oftheperihelion ofMercury inthegeneral theory ofrelativity. Before making anyapproximation, however, weshall firstput (13.322) intoaform inwhich a,z\,z%aretheonly constants occurring explicitly. Todothis,werefer to(13.319). Wehave h2 / a2+2iZ2 ^(zi+z2+*3), 23=~ ifweeliminate z3from thesetwoequations andsubtract zfrom thesecond equation, weobtain (13.323) where23~Z= -r[z(Zi+2o)+O2+ZiZZ\, z\H-22 (13.324) S=V(a+zi)(a+za),D=V(o~*i)(o-22). Substitution in(13.322) gives fortheapsidal angle therequired expression (13.325) a-aSDf"F(z)dz, j&\. where F(z]=l (a2-22 )Vfe-z)(z-zl)(z(z,+22)4-a2+ ia] Wecannot usethebinomial theorem toexpand negative powers ofterms which vanish whenz,z\,z*tend to a.How- evertheterm inthesquare bracket remainsfinite, andwemay expand anegative powerofit.Putting (13.326)z=-a+f, where fissmall, wegetapproximately, i.e.,neglecting f2 , 1 1 380 MECHANICS INSPACE [SEC. 13.3 Hence, wecanwrite (13.327) a=aSI-t^(z,+zz)J, where dz (a2- =pJ*i(a-dz(13.328) ./=r (a z)v(z2z)(z These integrals areevaluated without difficulty bymeans ofthe substitution z=2sin2+zicos2 0, andwefind Substitution in(13.327) gives (13.330)=frTl+8D- Itisevident from (13.324) thatSissmall; thus thelastterm issmall, andweintroduce onlyanegligible error (ofthesecond order)ifwesubstitute inthefraction D=2a, Zi+zz=-2a. This gives (13.331) a=i Now ifRi,Rzarethedistances from thecentral vertical tothe apsides, wehave accurately 2n2__ 2 02_ 2_ 2 Of) 7?7?Jtju-^j, 7t2t*^2, Oxx/ti/t2, andsotheapproximate formula (13.331) becomes (13.332) a= Wesaw, inconnection withtheequations (13.304), that inthe firstapproximation thepath oftheparticle isacentral ellipse. SEC. 13.4] MOTION OFAPARTICLE 381 Forthat curve theapsidal angleisfar.Nowwesee,from (13.332), thattheapsidal angle isalittle greater thanfar.This means thattheapseadvances; thepathisapproximately acentral ellipse, butthisellipse turns slowly forward (i.e., inthesame sense asthat inwhich theparticle describes thepath). Inonerotation oftheparticle, theapseadvances through anangle (13.333) 4-27r= 4a24a2 whereAisthearea oftheellipse. Theadvance disappears when A=0,i.e.,when theorbit isflattened intothetrack ofasimple pendulum. Thisadvance oftheapsecanbeshown byfitting alight writing device tothebob ofthependulum. This traces therotating elliptical pathonasheet ofpaper, placed underneath. 13.4.THEMOTION OFACHARGED PARTICLE INANELECTROMAGNETIC FIELD Much ofourknowledge ofthestructure ofmatter isderived from thestudy ofthemotion ofcharged particles (electrons or ionized atoms) inelectromagnetic fields. Further interest has beenadded totheproblem bytheinvention oftheelectron microscope andother devices,inwhich streams ofelectrons produce images inmuch thesameway asimages areformed by raysoflight inanoptical instrument. Electrostatic andmagnetostaticfields. Weshall consider only staticalfields, i.e., fields which donot change with time. Such fields areproduced byelectric charges atrestincondensers orbysteady currents; permanent magnets may alsobeused. Inanelectrostatic field, there exists ateach point ofspace an electric vector E.Itisthenegative ofthegradient ofanelectric potential V,sothat (13.401) E=-grad V. Thepotential Vcannot take arbitrary values throughout space; itmust satisfy Laplace's partial differential equation 382 MECHANICS INSPACE [SBC. 13.4 Similarly, inamagnetostaticfieldthere isateachpoint ofspace amagneticvector H,such that (13.403) H=-grad Q; 12isthemagnetic potential, and italso satisfies Laplace'sequation Both fieldsmay bepresent atthesame time. The force exerted onaparticle carrying anelectric charge c,moving with velocity q,is (13.405) P=eE+eqXH, iftheunits aresuitably chosen. Weaccept these basic formulas ofelectromagnetic theory as thefoundation forourdynamical deductions. Before proceeding todiscuss special fields, weshall obtain an equationofenergy from (13.405). Ifmisthemass ofthe particle,itsequation ofmotion is* (13.406) mq=E+eqXH. Taking thescalar product ofeach sidewithq,weget ft(?<q2 )=mqq=eEq=-e(grad V)q=- -jg" Hence, wehave theequation ofenergy (13.407) ira?2+eF=constant. Ifthefield ispurely magnetic, sothatVdisappears, thespeed of theparticle remains constant. Exercise. Thepotential duetoachargeeattheorigin isV=e/r,where r*=a;2 -f-y2+z*.Verify that this satisfies Laplace's equation, andshow thattheforcebetween twocharges atrest satisfies theinverse square law (6.502). *This isthenonrelativistic equation ofmotion and isagoodapproxima- tion ifthevelocity oftheparticleissmallcompared withthevelocity oflight. Intheaccurate relativistic equation, wereplace theleft-hand sideof(13.406) SEC. 13.4] MOTION OFAPARTICLE 383 Motion inauniform field. Asimple solution of(13.402) is V=ax+by+cz+d, where a,6,c,dareconstants. This gives auniform electric field, inwhich Eisaconstant vector. Similarly, wemayhave a uniform magnetic field, inwhichHisaconstant vector. Letusnowsuppose thataparticle, ofmassmandcarrying a charge ,moves inauniform electric andmagneticfield. Ifr istheposition vector oftheparticle, wehave asequationof motion (13.408) mi=cE+efXH. Letusnowchoose ouraxes sothatOzisparallel toH.The vector equation (13.408) gives thethree scalar equations (with theusual notation forcomponents) (13.409);*+** v-*-- m' Tocomplete thesolution most conveniently, weintroduce the complex quantities f=x+iy,F=Ei+iE2. Then the firsttwoequationsof(13.409) maybewritten together inthecomplex form (13.410) ^+'If^* m"" This isadifferential equation withconstantcoefficients, andthe characteristic equation forsolutions oftheform entis n(n+)=0. \m/ Thus thegeneral solution of(13.409)is 4-iy=f=A4- =C+D<+!f(13.411) 384 MECHANICS INSPACE [SEC. 13.4 where p=eH/m andA,B,C,Dareconstants ofintegration; AandBarecomplex, whereas Cand Z)arereal. These equations givethemotion ofacharged particle inauniformelectric andmag- netic field. Letusexamine thismotion inthecasewhere theelectric and magneticfields areperpendiculartooneanother. Then E&=Q andthe^-velocityisconstant. Letus,forsimplicity, assume that thiscomponent ofvelocity vanishes andthat z=through- outthemotion. Then thetrajectoryisdescribed bythecomplex position vectorf,asgivenbythe first of(13.411). Letuswrite thisequation intheform /7-/rA (13.412) r-M- jfJ=Ber**. Werecall thatanycomplex number Zmaybewritten intheform IfZisacomplex position vector, \Z\istheradius vector, and argZ theazimuthal angle. Equating moduli andarguments in (13.412), wehave (13.413) [f- (A- )]=arg f-A- =tagS- pt. IfFwere zero, the firstequation would indicate motion ina circle with center Aandradius\B\,aridthesecond equation would tellusthat thecircle isdescribed with constant angular velocity p.(The signshows thesense.) The effect ofthe F-term issimply toimpose anadditional motion inwhich the center ofthecirclemoves withconstant complex velocity (13.414)- j=1(JB,-iEl). This velocityisofmagnitude E/Hand isperpendicular tothe electric vector. Wesumupourdescription ofthemotion ofacharged particle inperpendicular uniform electric andmagnetic fields asfollows: //started withavelocity perpendiculartoH,theparticle moves asif SEC. 13.4] MOTION OFAPARTICLE 385 itwere attached totheedgeofacircular diskwhich moves inaplane perpendicular toH;thediskspins with constant angular velocity eH/m, and itscenter hasaconstant velocity E/H perpendicular toE(Fig. 138). Thesurprising part ofthisresult isthat,onthe whole, theparticle docsnotmove inthedirection oftheelectric field, butperpendicular toit. Fio. 138. Motion ofacharged paiticle inperpendicular uniform electric and magnetic fields. Exercise. Suppose thecharged particle starts from restattheorigin at time t=0.Starting from (13.412) prove thefollowing facts concerning the motion: (i)Attime t=2-jr/p,itwillbeatrestagain atadistance 2irE/(pH) from theorigin. (li)IfHisvery small, and iftheparticleisallowed totravel foradefinite finite time titthen at t=tiitscomplex positionisapproximately ^eF/J/m, and itscomplex velocityisapproximately eFti/m. Motion inapurely electric fieldandinapurely magnetic field. Wehaveworked out(13.411) forthegeneral case inwhich both electric andmagneticfields arepresent. Inthecase ofa purely electric field(H=0)wereturn to(13.408), which 386 MECHANICS INSPACE [SBC. 13.4 becomes (13.415) mi=E. Ifthefield isuniform, theacceleration isconstant, andsothe particle describes aparabolic trajectory likeaprojectile under gravity (cf.Sec. 6.1). Theplaneofthetrajectoryisdetermined bythevectorEandtheinitial velocity. Inthecase ofauniform purely magnetic field(#=0),the equations (13.411) read r=A+Be-1 ,z=C+Dt,(p= Bymoving theorigin, wecanmakeAC 0;thenwehave (13.416) f=Be~lpt ,z=Dt. Hence in=i*i,D_ PB since these values areconstant,itisclear that thetrajectory isacircularhelix, with axis paralleltothemagnetic field. The azimuthal angular velocityisp=eH/m. The simplest motion inauniform magnetic field isone in which the initial velocityisperpendicular tothe field. Then D=in(13.416), andthetrajectoryisacircle described with constant speed. Thedetermination ofthechargecandthemassmofan electron isaproblem ofgreat physical interest. Letusseehow the results wehave established help inthat determination. The firstthingwenotice isthat andmappear inourequations only intheform e/m,andtherefore itisonly thisratio thatwe canhope tofind. Itwould seem asimple matter tofinde/m from thecircular motion described inthepreceding paragraph. Theangular velocityiseH/m, andsowehave, onequating two different expressions fortheangular velocity, . R*m2' where qistheconstant speed andRtheradius ofthe circle. (Wehave squared thetwoexpressions toavoid thequestion of sign, which isofnoimportance here.) Ifwecould measure q,R,andH,weshould atoncehave e/m.NowRcanbemeas- SEC. 13.4] MOTION OFAPARTICLE 387 uredfromaphotograph ofthetrack oftheelectron, andHcan alsobemeasured; butqpresents adifficulty electrons move too fast forustofind their speeds directly. Wehave therefore to find qindirectly. Before entering themagnetic field, the electron isaccelerated from restbyanelectric field. Ifitstarts from restatpotential VFandenters themagnetic field withspeed qatpotential V=Vi,then bytheprinciple ofenergy (13.407). Elimination ofqbetween thetwoequations gives This isasuitable expression forthedetermination ofe/m, since allthequantities ontheright aremeasurable. This isthemethod ofKaufmann. Theelectromagnetic units are such that (13.405) holds. Axially symmetricfields. LetR, <j>,zbecylindrical coordinates. Afield issaidtobe axially symmetric with respect tothe2-axis ifthepotentialisa function ofRand zonly (i.e.,independent of<).Anelectric field ofthistypeisproduced byasystem ofcharged plates perpendicular tothe z-axis, acircular holewith center onthe 2-axis being cutfrom each plate. Anaxially symmetric mag- netic field isproduced bycurrents flowing incircular coils arranged inplanes perpendicular tothe 2-axis, thecenters of thecoilsbeing onthe2-axis. LetVbeanaxially symmetric electric potential. Weassume thatVcanbeexpandedinapower series inxand?/,thecoeffi- cients being functions ofz.Onaccount oftheaxialsymmetry, xandycanbeinvolved onlyintheform xz+y2(=R2 ),andso theexpansionisoftheform 7?2 7?4 (13.417) F=Fo(z)+g-Fi()+~7,()+. Then, byaneasy calculation, (13.418)a+ 388 MECHANICS INSPACE [SEC. 13.4 Inorder thatLaplace's equation (13.402) maybesatisfied, the functions F,FI, must satisfy thesequence ofordinary differential equations (13.419) FJ'OO+27i(*)=0, yj'W+|7 2(s)=0,. Itisevident that VQ(Z) (thepotential ontheaxis ofsymmetry) may bechosen arbitrarily, theburden ofsatisfying Laplace's equations being placed onVi(z), Vz(z\. Wemay treatanaxially symmetric magneticfield inexactly thesame way. Assuming forthemagnetic potential anexpan- sion oftheform P2 7?4 (13.420) a=Goto+~rQi()+a2(*)+, wededuce therelations (13.421) OJ'CO+2QiGs)=0, ttJ'Cs)+iQ2(2)=0,-- - . Motion ofacharged particle near theaxis ofsymmetryofan electromagneticfield.* Ifweintroduce thepotentials from (13.401) and(13.403), the general equations ofmotion (13.406) read,when written outin full, (13.422)e,,dV, .dfl .c.. a;=A-1-5-+y-; 2-5- xto*dz dy ,,'dV. .50 .dti\ y=k(-^+*te-x -te)' -k(a-~k \dz*-^-y-^ where k=e/m. Inthemost interesting applications, the charged particleisanelectron carrying anegative charge;inthis casekispositive. Letusassume thattheelectromagnetic fieldhasthe2-axis for axis ofsymmetry, sothatwehave theexpansions (13.417) and (13.420) forVand0,respectively. We shall consider only motion neartheaxisofsymmetry, sothat #,y,andtheir deriva- tives aresmall. Then, neglecting terms oforder higher than the *Reference maybemade toN.Chako andA.A.Blank, Supplementary Note No.I,inR.K.Luncberg, Mathematical Theory ofOptics (Brown University, 1944) SEC. 13.4] MOTION OFAPARTICLE 389 first,werewrite (13.422) intheform y z=*FJ, where theprime denotes d/dz. The last ofthese equationsis equivalent,inourapproximation, totheequation ofenergy (13.407), whichmaybewritten (13.424) s2=2k(V<>- C7), whereCisaconstant. Thisconstant maybedetermined when theinitial values ofzandzaregiven. Wenote that (13.424) determines zasafunction ofz,to within asign. Letusassume that zispositive throughout the motion. Thenwemaywrite (13.425) z=w(z)>0, w*=2k(V Q-C). Thefunction wistheaxialcomponentofvelocity. Sinceweare neglecting xzand?/2 ,itisclear that, toourorder ofapproxima- tion,walsorepresents themagnitudeofthevelocity vector. By(13.419) and(13.421) wehave (13.426) fVl=-4*7---L^(w*)--~(urn/'+^'2 ), IOi=-40?. Itisconvenient tointroduce complex notation, writing f=x+zy.Wemultiply thesecond equation of(13.423) byi andadd ittothefirst;thisgives, onmaking useof(13.426), (13.427) f=-\(ww"+u/2)f-ftOjf-4#ti>ni/ r. Weshallchange theindependent variable from ttozbythe equations (13.428) f=fz=wf, f-u>sf"+tir. Substitution in(13.427) gives fr+JT+or=o, (13.429) _ti/ "w"W "25"' This isthedifferential equation fromwhich thepath ofthe particleistobedetermined byfinding fasafunction ofz. 390 MECHANICS INSPACE [Sflc. 13.4 Thecomplex variable frepresents thevector displacement ofthe particle perpendicular totheaxis ofsymmetry (z-axis). The coefficients PandQarefunctions oftheindependent variablez> and aredetermined bytheaxial potentials F(z)and tt(z) andbythe initial conditions, which areneeded toobtain the value ofCin(13.425); wenotethatwdepends onC. Inthecase ofanelectrostaticfield,weputQ=0;wenote thatthenPandQarereal. Inthecase ofamagnetostatic field, weput7=0;then,by(13.425), wisaconstant andtheterms inPandQinvolving derivatives ofwdisappear, leaving purely imaginary expressions. There isnosimple general method ofsolving (13.429). How- ever, theequation maybesimplified byusing astandard device toeliminate thefirst-order derivative byachange inthedepend- entvariable. Tocarry thisout,wesubstitute (13.430) f(z)=u(z)v(z) in(13.429) andobtain (13.431) u"v+u'(2v'+Pv)+u(v"+Pv'+Qv)=0. Wenowchoose vsoastomake thecoefficient ofu'vanish. We dothisbywriting (13.432)*>-expT-i fP)L Jzo where zistheinitial value of2.When wesubstitute (13.432) in(13.431) weget,aftersome easy calculation, (13433)(W.4A Also,by(13.430), (13.434) f=exp[-* P(f)d*]. Whenwesubstitute forPandQtheexpressions given in(13.429), weobtain amuch simpler expression forSthanwemight expect, and(13.433) reads (13.435)"+,- The relation between theactual displacement vector fandthe SBC. 13.4] MOTION OFAPARTICLE artificial displacement vector uis (13.436) f=391 whereWQisthevalue ofwwhen z=ZQ. Itmaybeconvenient forreference towrite theresults sepa- rately forthecases ofelectrostatic andmagnetostaticfields: Electrostatic field: u"+S(z)u=0, S(z) (13.437)W 16 w(z)=V2/f[Fo(z)-C],A-=--,m Vo(Zo), WQ=W(Z ). Magnetostatic field: u"+S(z)u=0, wo=constant velocity, (13.438)jj.20'2=it' 6 m ft/r 1 =wexp^-(fioo-flo)i There aresomeremarkable features inthepreceding work. In (13.429) thecoefficients PandQwere complex; butwhen we transform to(13.435), wegetarealcoefficient S.NotonlyisS real,itisalso positive;thishasanimportant bearing onthefor- mation of"images" byanaxially symmetric electromagnetic field, asweshall seelater. Themathematical difference between theelectrostatic case andthemagnetostatic case islessthanwemight expect. In each casethecoefficient Sispositive. The chief difference lies intherelation between fand u.Intheelectrostatic case, the connection isreal,andthecomplex vector fhasthesame direc- tion asthecomplex vector u.Inthemagnetostatic case the 392 MECHANICS INSPACE [SEC. 13.4 connection iscomplex; themagnitudes ofthetwovectors are equal, andfisobtained fromubyrotation through anangle 5<"-*> Tosumup,inthegeneral electromagnetic case, thedetermina- tion ofthepath oftheparticle involves thesolution of(13.435). Asinitial conditions, wemayassume that theparticle starts from thepoint (20,fo)with velocity WQinadirection giving tof' thevaluefo-Thus, (13.435)istobesolved, with the initial conditions forz=ZQ, (13.439) u=f,'-ft+iA- where no=n(*o), a=oj(*o). Intheelectrostatic case, (13.437) replaces (13.435), andweput floo=in(13.439);inthemagnctostatic case, (13.438) replaces (13.435), andweputFH=in(13.439). Exercise. Show that thesmall angle between theinitial velocity vector andtheaxisofsymmetryis|fj|. Theelectromagnetic lens. Inanoptical instrument, such asamicroscope, camera, or telescope, rays oflight arebentbyasystem ofglass lenses, thesystem usually having anaxis ofsymmetry. Thefunction oftheinstrument istoproduce animage, theraysfrom each point oftheobject being brought toafocus atanimage point. Inrecent times, there hasbeen aremarkable development of electromagnetic devices analogous totheimage-forming optical instrument. Instead ofrays oflightbentbyglass lenses, there arestreams ofelectrons whose trajectories arecurved bymeans ofelectromagneticfields.*When anaxially symmetric electro- magnetic field isused, theequation (13.435)isthefundamental equation from which the trajectories oftheelectrons are determined. LetOz(Fig. 139)betheaxisofsymmetry ofanelectromagnetic field;letnbeaplane perpendicular tothis axis,with theequa- *Cf.L.M.Myers, Electron Optics (Chapman &Hall, Ltd., London, 1939), p.100. SEC. 13.4] MOTION OFAPARTICLE 393 tion z=ZQ.LetPbeanypoint onnwithacomplex position vector x+iy f.Wesuppose thatfrom thepointPQthere areprojected anumber ofidentical charged particles (electrons). Their velocities have acommon magnitude w,buttheir direc- tions aredifferent; thedirections are,however, nearly parallel totheaxis ofsymmetry, sothatourmethods apply. Thetrajectory ofeach electron satisfies (13.435). Thefunc- tionw(z)isgivenby(13.425). Since alltheelectrons have the same charge e,thesamemassm,andthesame initial velocity w , FIG. 1JJ9 Formation ofanimage theconstant Chasthesame value forallthetrajectories. Hence w[andconsequently 8in(13.435)]isthesame forallthetrajec- tories. Thisis,ofcourse, amathematical idealization. Asfar asweknow,allelectrons have thesame charge andthesame mass, butwecannot secure accurately acommon initialvelocity. Hence,inpractice, theconstant Candthefunctions wandS willnotbequite thesame for allthetrajectories. This leads towhat iscalled, from theoptical analogue, Achromatic aberra- tion," thevelocity oftheelectron corresponding tothecolor of thelight. Butforourpurposes weshall neglect this effect and regardwandSasthesame forallthetrajectories. Wenote that,from (13.439), forz=ZQwehaveu=fforall theelectrons; tmttheinitial value ofufdepends ontheparticular electron since fJisnotthesame forthem all. Itisknown from thetheory oflinear differential equations thatthegeneral solution of(13.435)isoftheform* (13.440) u=af(z) *Cf.E.L.Ince, Ordinary Differential Equations (Longmans, Green& Co.,Ltd.,London, 1927), p.119. 394 MECHANICS INSPACE [SBC. 13.4 where a,/?arearbitrary constants ofintegration (which maybe complex) and/(z), g(z) areindependent particular solutions. Since thecoefficient Sin(13.435) isreal,wecanobtain two real independent particular solutions bytaking theinitial conditions (13.441) /(z )=0, /'(z )=1; g(z )=1, g'(z Q)=0. With thischoice of/andg,itfollows from (13.440) that a=UQ,=uQ, where UQ,u'arethevalues ofu,u'when z=z;then (13.440) maybewritten (13.442) u=u'J(z)+uQg(z). LetHIbethevalue ofuatthepoint where thetrajectory cutsthe planez=z\]then (13.443) u,=nJ/OsO+ Inthefamily oftrajectories which weareconsidering, i.e.,a family starting from apoint (z ,fo),uhasacommon value but UQchanges from trajectory totrajectory. Thus, ingeneral, (13.443) willgiveanareaontheplanez=z\whenwesubstitute thevarious values ofu'Qcorresponding tothevarious initial directions. Theequation (13.443)willdefine asingle point on theplanez=Ziif,andonly if, (13.444) /fa)=0. Letusrecall that thefunction/(z)isdefined bythefollowing differential equation and initial conditions: (13445)(13.445) /(*)=0, /(*,)-!. Canwefindaplanez=z\(other than z=z)suchthatthewhole family oftrajectories cut itinasingle point? This isequivalent toasking whether theequation (13.444) hasasolution other than Zl=Z . Although wecannot giveadefinite answer tothisquestion in general, wecandiscuss itqualitatively. Consider thegraph of /(z). Thisgraph starts from thez-axis atz,sloping upat45. Thus/(z)ispositive atfirst;henceby(13.445), since$ispositive, /"(z) isnegative, and itremains negative aslong as/(z)ispositive. SEC. 13.4] MOTION OFAPARTICLE 395 Thismeans thatthegraphisconvex when viewed from above. Either oftwothings happens. Thecurvemay turndown and cutthe2-axis atsome point Zi(Fig. 140a); oritmay turn so slowly that itreaches z= >before coming down tothe2-axis (Fig. 1406). Intheformer case, theequation (13.444) hasa solution;inthelatter case,ithasnosolution (atleast notfor values ofz\greater than z ,andweareinterested onlyinsuch values). Ingeneral terms, wemaysaythatthelargerSis,the more chance there isthat there willbeasolution; because the largerSis,themore rapidly doesthegraph off(z)turndownward. f(z) f(z) oz (a) FIG. 140.Graph off(z):(a)inthecasewhere animage isformcrl, (b)inthecase wheie animageitsnotformed. If(13.444) hasasolution, thenthefamily oftrajectories start- ingoutfrom apoint POmeet again inapoint PI,asshown in Fig. 139. Borrowing thelanguage ofoptics, wemaycallPthe object point andPItheimage point. Inthis sense, anaxially symmetric electromagnetic fieldmayform images. Wemight go further andsaythat itwillformimagesifitisstrong enough, because Sisincreased byanincrease inthestrength ofthefield. Itwillbenoted thattheequation (13.444), which determines theplane HIonwhich theimageisformed, doesnotinvolve f. Consequently wemay state thefollowing important result: // anobject point POontheplane IIohasanimage PIontheplane HI,thenevery object pointontheplanen(near theaxisofsymmetry, tomake theapproximatemethod valid) hasanimage ontheplane Hi. Infact,wehaveanobject planeandanimage plane, justasinthe 396 MECHANICS INSPACE [SEC. 13.4 optics ofalens. Hence, byanalogy, wemayspeak ofanaxially symmetric electromagneticfield asanelectromagnetic lens. Suppose thatontheobject planenthere issomeminute struc- turewhich wewish tophotograph. Wesetupaphotographic plate attheplane HIandbombard theplanenfrom theleft with astream ofelectrons. Each point ofnbecomes asource ofelectrons travelling ontowards IIiandconverging toanimage point onHI.Thus thestructure onnisreproduced point for point onHI.Apoint onIItransparent toelectrons gives a "bright" point onHi,andapoint onHOopaque toelectrons gives a"dark" point on IIj. Essentially, this ishowanelectron microscope works. Since magnificationisthemost important function ofamicroscope, letusnowlookintothequestion ofthe magnification mproduced byanelectromagnetic lens. Theimage ofapoint (z ,fo)is(z\,fi),where Ziisgivenby (13.444) and fiby(13.436) and(13.443); wehave (13.446) ui=u<*g(zi)= and (13.447) fl=f(*) exp [-*' where wi=w(zi). Magnificationisdefined by (13.448) m andso,by(13.447), themagnification ofanelectromagnetic lens is (13.449) m= Werecall that g(z)isdefined bythefollowing differential equa- tionand initial conditions: (13.450) (jw-lf^^-O. Itisarealfunction, sinceSisreal;andthemodulus sign in (13.449)isneeded only totake care ofthepossibility that g(zi)isnegative. Inthecase ofanelectrostaticfield, theexponential disappears from (13.447); thevector fihasthesame direction asforthe SEC. 13.4] MOTION OFAPARTICLE 397 opposite direction, according asg(zi)ispositiveornegative. In thecase ofamagnetostatic field, (13.447) reads (13.451) ft=fo<7(zi) cxp^~(floo- Qoi)], where QOI isthevalue ofQatz=z\.Themagnificationis m=\g(zi)\. Theimage vector fiisobtained byfirstapplying thismagnification totheobject vector fandthen rotatingit about thes-axis through anangle (13.452)=^(Goo-QOI). Figures 141aandbshow theprojections ofobject point PO,image point PI,andtrajectories ontheplanez=0.They aredrawn form=2anda=7r/4. Inactual electron microscopes the magnification maybeashigh as200,000. Approximations forelectromagnetic lenses. Thedetermination ofthefocal properties ofanelectromagnetic lensdepends, aswehave seen,onthesolution ofthedifferential equation (13.453) /"(z)+S(*)/CO=0, with theinitial conditions, asin(13.441), (13.454) /(z )=0, /'(so)=1. There isnosimple wayofsolving thisequation, andwehave to fallbackonapproximate methods. Weshall consider thecase where theelectromagneticfield isconcentrated onashort length ofthez-axis, sothatthere ispractically nofieldoutside thisshort range. Making amathematical idealization, weshallassume that there isafield for h<z<handnofield outside that range. Intheabsence ofelectric field, theaxial potential VQiscon- stant, and so,by(13.425), wisconstant andw'=0.Inthe absence ofmagnetic field, theaxial potentialftisconstant and 8=0.Thus, by(13.435), S=forthatrange ofvalues ofz forwhich theelectromagnetic field vanishes. Byhypothesis, this isthecase outside therange h<z</?,andthen the differential equation (13.453) becomes very simple: }"(z)0. 398 MECHANICS INSPACE y[SEC. 13.4 FIQ.14la.Formation ofanimage byanelectrostatic lens. Fio. 1415. Formation ofanimage byamagnetostatio lens. SEC. 13.4] MOTION OFAPARTICLE 399 The function f(z)istherefore alinear function of z.This corresponds tothefact that,intheabsence ofelectromagnetic field,anelectron travels inastraight linewithconstant velocity. The projections oftrajectories inFigs. 14laand 6aredrawn forsuch acase; each projection consists oftwostraight lines, connected byacurve. Thecurve isproduced bytheaction ofaconcentrated electromagneticfield. Ifthe fieldextended \45yy/A/,A Iyy i ^1//VV IIS(z> z -h h zt Fia. 142. Graphs ofS(z)andf(z)foraconcentrated electromagnetic field. from zto2i,theprojectionsofthetrajectories would becurved alltheway. Figure 142shows graphsofS(z)and/(z)forthecase ofacon- centrated electromagnetic field,drawn ontheassumption that /(z)vanishes forsome value Ziofz,sothatanimageisformed. Combining the initial conditions (13.454) with thefactthat /(z)islinear forz< h,weobtain (13.455) /(-fc)--h-ZQ,f(-h) =1. Thusweknow thevalue of/and itsfirstderivative onentering 400 MECHANICS INSPACE [SEC. 13.4 theconcentrated field.Wenowtrytofindoutwhathappens as wegothrough the field. Transferring thesecond term of(13.453) totheright-hand sideandintegrating from htoz,weget,remembering (13.455), (13.456) /'(z)=1-S(QJ(Q Another integration gives (13.457) /()=z-z- Puttingz=hinthese twoequations, weobtain f/(/O=h-Z-f\dr,f\8(&f(& dk, (13.458) {J J~h Atfirst sightitmayappear thatwehavefound thevalues of/ and itsderivative onleaving thefield, but ofcourse this is illusory, because wedonotknow thefunction /occurring inthe integrals. However, wecanusetheabove equations asabasis forapproximation. IfS(z)isfiniteandhissmall,itisevident from (13.458) that thechanges in/and /'inpassing from z htoz=harcsmall. Buttogetanimage, thegraph of/must bebentthrough an angleofmore than 45onpassing throughthe field. Infact, theremust beafinite change in/',andconsequently wemust use astrongfield. Itisclear that S(z)must belarge oftheorder h"1 .Then theintegral inthesecond of(13.458)isfinite; the doubleintegralinthefirst of(13.458)issmall oforderh,showing thatalthough thechangein/'isfinite, thechange in/issmall. Before introducing theapproximation, letusgetanexpression forZi,thecoordinate oftheimage point. From thelinearity of thegraph of/outside thefield,wehave (13.459) f(h)=- or ^-- This willgiveusz\ifwecanevaluate f(h)and/'(/0 from (13.458). Wenowtakeupthemethod ofapproximate solution bythe method ofiteration. Thekeyequationis(13.457). For/under SEC. 13.4] MOTION OFAPARTICLE 401 thesign ofintegration, wesubstitute /asgivenbytheequation itself. Thisdoesnotgetridof/ontheright-hand side,but it pushesitunder more signs ofintegration andthus reduces its importance. Thisprocedure gives (13.461) /(*)-_.- dr, 8(& dt-*- dp]. Themultiple integrals aretobeevaluated starting from theright- hand side.Wecanrewrite thisintheform (13.462) /()=2-0+2of'hdr,J_'A8(&d| hrdr,rs(&<*JhJh J Thisexpressionisaccurate. Forzintherangeh<z<h, the first integralissmall oforder/i,andtheremaining integrals are small oforder hz .Differentiation of(13.462) gives (13.463) /'(z)=1+z S(Qdt--z h&S(Q dl; f.S(Qdff dqFS(p)f(p) dp.n JhJti Here thefirstintegralisfinite, andtheremaining integrals small, oforder h.Weobserve that in(13.462) and (13.463), only the lastintegrals areunknown. Ifwerequired ahigher approximation, wecould substitute again for/under thesign ofintegration; then theintegrals con- taining/would besmall oforder A3intheexpression for/(z),and small oforder h2intheexpressionfor/'(z). This process could becontinued indefinitely. Let us,however, content ourselves with approximations for f(h)andf(h)which retain terms oforder hbutreject terms of order ft2 .Weshallcommit anerror only oforder /i2ifwesub- stitute /(p)= zinthelastintegral in(13.463). Accordingly, puttingz=h,wegetthefollowing approximate expressions: (13.464)f(h)=h-z +z hdrj /'(/*)=1+z* S(S)d{-_$8(0 dq S(p) dp. 402 MECHANICS INSPACE [SEC. 13.4 Towrite these results more neatly, weintroduce the finite constants fA=fS(Q dfcB=A-'/* (13.465){J-h J~h [D-*-'f$({)df*(f- v%/~~n %/~n Wenote that,byinversion oforder ofintegration, (13.466) Consequently, (13,464) read "z+/i[1+(A~dq* S(p)dp Theconstants A,B,Dmaybeevaluated numericallyifthefield isgiven;itshould benoted however thatthey involve alsothe initial velocity MO,sinceSinvolves WQ(cf.equation (13.435)). Exercise 1.Show that ifS=K/h, aconstant, forh<z<h, then A-2K, B=0,D-ftf 2. Exercise 2.Evaluate 4,B,andZ),if,for-h<z<h, S(z)-a/i-1cos2 1|, where aisaconstant. Using (13.467)in(13.460), theimage z\corresponding toan objectzisgivenby__ Zi_h~ ZQ^h[i+(A_ ormore symmetrically,tothesame order ofapproximation (i.e. neglecting/i2 ), (13.469) --A =-h(-+-+ Z\ ZQ \Zi ZQ IfweletZQoo ,weget (13.470)-A=-h(-+D\ z\ \z\ / SEC. 13.5] MOTION OFAPARTICLE 403 Thevalue ofz\soobtained gives theimage ofanobject at infinity. Ifwearesatisfied with therougher approximation inwhich h isneglected, (13.469) becomes (13.471)---*,A. Z\ ZQ IfweletZQ> oo 9thecorresponding value ofz\iscalled the focal lengthFoftheelectromagnetic lens;by(13.471), wehave (13.472) =A Exercise. Show that intheroughest approximation, thefocal length ofaconcentrated electric lens isgivenby andthefocal length ofaconcentrated magnetic lensby whereHisthemagnitude ofthemagnetic vector ontheaxisofsymmetry 13.6.EFFECTS OFTHEEARTH'S ROTATION The effect oftheearth's rotation onaplumblinewasfound inSec. 5.3. This isastatical phenomenon relative totherotating earth, andonly thecentrifugal force isinvolved. Indynamical problems ontherotating earth theCoriolis force alsoenters, and theeffects arehard topredict without acareful mathematical analysis. Equationsofmotion ofaparticle relative totheearth's surface. Weaccept themodel oftheearth used inSec. 5.3anoblate spheroid turning about itsaxis ofsymmetry with constant angular velocityQ.The axis issupposed fixed inaNewtonian frame ofreference. The vertical atanypoint ontheearth's surface isdefined bytheplumb line,andthehorizontal planeis perpendicular tothevertical. The latitude Aistheangle of elevation oftheearth's axisabove thehorizontal plane. 404 MECHANICS INSPACE [SEC. 13,5 InFig. 143,SN istheearth's axis,drawn fromsouth tonorth; isapoint onorneartheearth's surface, andBthefoot ofthe perpendicular dropped from onSN]Iisaunitvector alongBO andKaunitvector parallel toSN.The triad ofunitvectors i,j,kisfixed relative totheearth anddirected asfollows: iishorizontal andpoints south; jishorizontal andpoints east; kisvertical andpoints upward. ]'io. 143. Vectors used indiHrussing theeffects oftheeaith'ts rotation. LetusputBO=aanddenote byrtheposition vector ofa moving particle relative to0.Then theposition vector ofthe particlerelative toBis (13.501)rB=a+r. SinceBisafixed point inaNewtonian frame ofreference, the absolute acceleration oftheparticleis (13.502) IB=+r. Here aistheacceleration of0;since moves inacircle with constant angular velocity 12,wehave (13.503) a==-aQ2I=-a!22(sinXi+cosXk). Letmbethemass oftheparticle. The force ofgravityis proportional tomandmaybewritten mF.Wedenote byP SEC. 13.5] MOTION OFAPARTICLE 405 theresultant ofallother forces. Theequation ofmotion is (13.504) miB=wF+P, or,by(13.502) and(13.503), (13.505) mr=mF+P+mafl2(shiXi+cosXk). Letusapply thisequation toaplumb line,hanging inequilib- riumwith thebobat0.ThenPisthetension intheplumb line andpoints inthedirection k.AsinSec.5.3,wedefine gtobe thistension, divided bythemass ofthebob,sothat P=mgk. Since r=0,(13.505) gives (13.506) Fo+afl2(shiXi+cosXk)=-0k, whereFistheforce ofgravity perunitmass at0. NowFinthegeneral equation (13.505) andFin(13.500) arenotequal vectors unless theparticleisatO,fortheearth's gravitationalfieldchanges from point topoint. Butweshall assume thattheparticle always stays soclose tothatvariations intheearth's field arenegligible. Soweintroduce our first approximation, putting (13.507) F=Fo intheequationofmotion (13.505). When, further, wesubstitute forFfrom (13.506), wegetforanymoving particle (13.508) mr=P-mgk. Ournext task istoresolve thisequation intocomponents along i,j,k.This triad turns withaconstant angular velocity (13.500)ft=OK= cosXi+ftsinXk, and so,by(12.310), (13.510)?=S|+2QxJ +QX(OX f)' ot ot Wenow introduce asecond approximation, dropping the last term onaccount ofthesmallness of0.Substitution from (13.510) in(13.508) gives thevectorform oftheequations ofmotion relative totheearth's surface, 406 MECHANICS INSPACE [SEC. 13.5 (13.511) m~f=P-mgk-2wQX~ Here 8*x/Bt* and dr/Bt are,respectively, therelative acceleration andvelocity; thelastterm istheCoriolis force. Thecentrifugal force hasbeen eliminated intwosteps,firstby(13.506) and secondly byneglect ofthelastterm in(13.510). Itwillbe noticed that (13.511)isessentially thesame differential equation as(13.406) or(13.408) theequation ofmotion ofacharged particle inanelectromagnetic field. Letusnowintroduce axesOxyz coincident indirection with (i,j,k),sothatOxpoints south andOyeast. LetX,F,Zbe thecomponents along these axes oftheforce P,which,itwill beremembered,istheforce other than gravity. SinceQis given by(13.509), wegetfrom (13.511) thescalar form ofthe equations ofmotion, Imx=X+2wQ sinXy, my=Y-2mO(sin Xx+cosXz), mz=Zmg+2m$l cosXy. Motion ofafreeparticle. Bya"free particle" wemean hereaparticle onwhich there actsnoforce butgravity. Asremarked inSec. 6.1,this isan idealization difficult toapproach inpractice. The resistance oftheairisalways present andproduces discrepancies between mathematical predictions andobserved motions. Itistherefore notsurprising thattheminute effects duetotheearth's rotation arehard todetect. Forafree particle, weputX=Y=Z=in(13.512). The resulting equations areeasy tointegrate, especially asfurther approximations, based onthesmallness of8,arepermissible. Theequations nowread (x2ftsinXy, y=-2Q(sin X : z=-.0+28GOJ(13.513) \y=-2Q(sin X+cosXz), tcosXy. Each ofthese equations canbeintegrated once. Without loss ofgenerality, wemaysuppose that theparticle starts from the origin att=with velocity (u ,t>o,Wo),andsoweget SEC. 13.5] MOTION OFAPARTICLE 407 Ix~2J2sinXy+UQ, y=-2Q(sin Xx+cosXz)+ t>, z=gt+212cos\y+WQ. Ifwesubstitute from the firstandthird ofthese equations in thesecond of(13.513) andneglect122 ,weobtain (13.515) y=212(^0 sinX+WQcosX gtcosX), andhence, byintegration, (13.516) y=vtlM2(wsinX+wcosX)+i%*3cos X. Then the firstandthird equations of(13.514) give,onneglecting x**Uot+ Vot*sinx> z=wot- \gt^+QvQt*cos X. Two cases areofparticular interest, aparticle dropped from rest,andaparticle representing aprojectile fired with high velocityinaflattrajectory. Inthecase ofaparticle dropped from rest,weput UQ=VQ=Wo 0, andget (13.518) x=0, y=$Qgt* cosX,z--$gt*. Thepathisasemicubical parabolaintheeast-west vertical plane, (13.519) ,Z=-.i.z. Itisevident from (13.518) that thedeviation from thevertical is toward theeast.From (13.519), thedeviation for fallfrom a height his flcosX-2h. \y This iszeroatthepoles (X=^r),asweshould expect. Foraprojectile with large UQandVo,weneglect theterm in WQandalsothelastterm in(13.516). Thus theprojectionofthe trajectory onthehorizontal plane hastheequations (13.520) x=Uot+Itoo*2sinX, y=vQt-Quo? sinX. 408 MECHANICS INSPACE [SEC. 13.5 Thesemaybeexpressedincomplex form inthesingle equation (13.521) x+iy=(UQ+ivQ)(t-ilM*sinX). Letusput x+iy=Re*+, UQ+iv=qoeia andwrite, aswemay since ftissmall, 1HitsinX=exp(Hit sinX).* Then (13.521) takes theform qlexp(iaiQ,tsinX), andso (13.522) R=got, <t>=a ftsinX. Themagnitude oftheposition vector grows ataconstant ratego, andatthesame timethevector turns ataconstant rate sinX. IntheNorthern Hemisphere, Xispositive and thisrotation is clockwise when viewed fromabove;intheSouthern Hemisphere, itiscounterclockwise. Thismeans that theprojectile experi- ences, onaccount oftheearth's rotation, aslight deviation totherightintheNorthern Hemisphere, andtotheleftinthe Southern Hemisphere. This isknown asFereVs law. Foucault's pendulum. Letussuppose apendulum setupattheNorth Pole. If started properly,itmay vibrate asasimple pendulum ina vertical plane which isfixed intheNewtonian frame ofreference. Astheearth turns under thependulum withangular velocity 12, theplane ofvibration ofthependulum appears toanobserver ontheearth toturnwithanangular velocityft.Foucault wasthe first topoint outthatapendulum could beused to demonstrate theearth's rotation. Itisnotnecessary thatthe pendulum should besituated atone oftheearth's poles; an apparent rotation, duetotherotation oftheearth, may be observed atanylatitude except ontheequator. Weshallnowapply (13.512) tothemotion ofapendulum. Thependulum consists ofaparticle ofmassmattached bya light string oflength atoapoint with coordinates(0,0,a). Thus, inequilibrium theparticle rests attheorigin. Weshall discuss small oscillations about thisposition, aproblem already SEC. 13.5J MOTION OFAPARTICLE 409 solved [cf.(13.304)] forthecase 12=0.Thequestionofinterest now istofindhowthesimple motion there described ismodified bytherotation oftheearth. We recall thatX,Y,Zarethecomponentsofforce other than gravity. Forthependulum,thisconsists ofthetension 8 inthestring; asin(13.302) thecomponents are (13.523) Z=--S,7-^S,z=5LZ^ s. CL a QI Wehave totake care oftwoseparate approximations. The first, based onthesmallness of12,hasalready been used in obtaining (13.512);itconsists inneglecting122 .Thesecond approximationisthat arising from thesmallness oftheoscilla- tions. Thismeans thatx,y,and their derivatives aresmall; zand itsderivatives aretherefore small ofthesecond order and consequently willbeneglected. The lastequation of(13.512) gives, sinceZ=Sapproxi- mately, (13.524) S=mg 2ml2 cosXy, andsothe firsttwoequations become ~20sinX'^+p'x~ > +212sinXz+p*y=0, where p* g/a. Multiplying thesecond equation byiand addingittothefirst,wegetthesingle complex equation (13.526) f+2tl2sinXf+p^=0, (f=x+iy). Thegeneral solution is (13.527) f=A&*<+Ben , whereAandBarecomplex constants depending onthe initial conditions and HI,n^aretheroots oftheequation (13.528) n2+2iQsinXn+p*=0. These roots are ni,n2= t'OsinX i\/ti2sin2X+ Neglecting122 ,wemay write (13.527) intheform (13.529) f=fiexp(-i$lt sinX), 410 MECHANICS INSPACE [SEC. 13.5 where (13.530) fi=A#+Be-*". Tointerpret thisresult, wesuppose forthemoment that 12=0, sothatf=fi.Onseparation oftherealandimaginary parts, itiseasily seen that thepathisanellipse with center atthe origin themotion being acomposition ofperpendicular simple harmonic motions[cf.(6.405) and(13.304)]. The effect ofthe second factor in(13.529)istorotate thecomplex vector fi through anangletitsinX,proportional tothetime.Wemay sumupourresult asfollows: Theeffect oftheearth's rotation ontheelliptical path ofaspherical pendulum istocause theellipse torotate withanangular velocityftsin X.This rotation is clockwise intheNorthern Hemisphere and counterclockwise in theSouthern Hemisphere. IfweputX=far,sothat thepen- dulum isattheNorth Pole, theangular velocity becomes ft andmay beregarded asdue directly totheearth's rotation beneath thependulum. When wediscussed thespherical pendulum inSec.13.3,with- outtaking theearth's rotation into consideration, westarted withafirstapproximation andobtained anelliptical orbitfrom equations (13.304). Wethen proceeded toasecond approxi- mation andfound, in(13.333), anexpression fortherateatwhich theelliptical orbit advances. Inthecase ofFoucault's pendu- lum, thesituation ismore involved. Inthe first place, the angular velocityftoftheearth issmall(cf.page 143),andthat factwasused inobtaining (13.512). Secondly, wehave con- sidered onlysmall oscillations (first approximation). Ifwewere toproceed toasecond approximation, wewould findtwosuper- imposed rotations oftheelliptical orbit onedepending, asin (13.333), onthearea oftheorbit (area effect), andtheother an angular velocity12sinXduetotheearth's rotation (Foucault effect). Unless special precautions aretaken, thearea effect is likely tobemuch larger than theFoucault effect andtoconceal it.Toprevent this,itisusual todraw thependulum aside with athread and start themotion byburning thethread. This means that, for t=0,wehave f=andf=fo(say). Then, by(13.529), (13.531) A+B=fo, p(A-JB)-(A+7?)ftsinX-0. SBC. 13.6) MOTION OFAPARTICLE 411 Now (13.530) maybewritten (13.532) fi=(A+B)cospt+i(A-B)sinpt, or,by(13.531), (13.533) fi=ft(cospt+isinpt-sinX/p). Itiseasy toseethat thisrepresents motion inanellipse with semiaxes|fo|and\fo\QsinX/p. Foranellipse with these semi- axes, described inaNewtonian frame ofreference, theadvance oftheapse inoneperiod (2ir/p) is,by(13.333), (13.534) ~^- |fol'ftsinX, andsotheangular velocity oftheellipse duetothearea effect is (13.535) ^sinX. Since|fo|/aissmall, thisangular velocityismuch smaller than theFoucault angular velocityftsinXandmayberegarded as negligible. Hence,ifthependulumisstarted byburning a thread, therotation oftheorbitmayberegarded asduetothe Foucault effect alone. Itshould benoted that, foranysmall elliptical orbit, there isadistinction between thearea effect andtheFoucault effect. Thearea effect isalways arotation inthesense inwhich the ellipseisdescribed andreverses when thatsense isreversed, but theFoucault rotation takes place inadefinite sense (clockwise in theNorthern Hemisphere andcounterclockwise intheSouthern). 13.6.SUMMARY OFAPPLICATIONS INDYNAMICS INSPACE MOTION OFAPARTICLE I.Jacobian elliptic functions. (a)Differential equation: (13.601)gay=(i-ifld-*V), (0<*<i). (b)General solution: (13.602) y=sn(a:+c). 412 MECHANICS INSPACE [SEC. 13.6 (c)Other elliptic functions: (13.603) en2x=1-sn2 x, dn2x=1-Psn2x\ (13.004) -7-snx=enxdnx,j-enx= snxdnz, -r-dn#= A12sn a:en a;.dx (d)Periodicity: (13.605) sn(x+4/0=snre, en(x+4/0=ena;, dn(x+2/0=dnx; jcr_^=JL===,=p- ,d* .. JO^/(i_y2)(!_^.2^2)JO^ J2gin2^ II.Simple pendulum. (a)Motion: Isin?6 sin^asn[p(t to)], ,g .-i^2, /,.=gm^a (b)Periodic time: (13.608) r--A-2sin2 -(1+iVa2 ),approximately. III.Spherical pendulum. (a)General motion: thependulum oscillates between two levels, found bysolving acubic equation; theanalytical solution is It-zi=(22-Zi)sn2[p(t- <<,)], =?,F=g2~g^ a 3-21 (6)Small oscillations: inthe firstapproximation thepathis anellipse; inthesecond approximation theellipse turns ata rateproportional toitsarea. IV.Motion ofacharged particle inanelectromagneticfield. (a)Uniform electric field: thetrajectoryisaparabola. (6)Uniform magneticfield: thetrajectoryisahelix. HEC. 13.6] MOTION OFAPARTICLE 413 (c)Axially symmetric electromagneticfield: thetrajectory satisfies u"+S(z)u=0, 3F'2 16 (13.610)-*+%=exp- > iw.=speed ofparticle= ^/2ArfFFo-fr Fo=axial electric potential, Fo=Vo(z ), fio=axialmagnetic potential, Prime indicates d/dz. (d)Electromagnetic lens: (i)Image planez z\ofobject planez=2determined by (13.611)=0, =0, (ii)Magnification givenby=0, (13.612) o, o, w= (e)Magnet ostatic lens :rotation ofimage givenby (13.613) 414 MECHANICS INSPACE [Ex.XIII V.Effects oftheearth's rotation. (a)Equations ofmotion: (13.614)mx=X+2mQ sinXyy my=Y 2wl2(sin Xx+cosXz), mz=Zmg+2mfl cosXy, X,Y,Z=force other than gravity, X=latitude. (6)Afalling body deviates totheeast. (c)Aprojectile deviates totheright intheNorthern Hemis- phere. (d)Foucault's pendulum turns withangular velocity12sinX, clockwise intheNorthern Hemisphere. EXERCISES XIII 1.Asimple pendulum ofmassmandlength aperforms finite oscillations, thegreatest inclination ofthestring tothevertical being 30. Find the tension inthestring when thebob isinitshighest position. 2.The string ofaspherical pendulum isheldouthorizontally andthe bobstarted with ahorizontal velocity perpendicular tothestring. Find to thenearest footpersecond themagnitude ofthisvelocity iftheminimum inclination ofthestring tothevertical inthesubsequent motion is45. Thelength ofthestring is54inches. 3.Aparticle carrying achargeeisprojected from theorigin with a velocity UQinthedirection ofthez-axis. There isauniform magnetic field ofstrengthHparalleltothez-axis. Iftheparticle crosses theplane x= atadistance afrom theorigin,find itsmass. (Isotopes areseparated ina mass spectroscope byamethod such asthis.) 4.Aheavy bead isfreetomove onasmooth circular wire ofradiusa, which rotates withconstant angular velocity Qabout afixed vertical diam- eter. Find thepossible positions ofrelative equilibrium. If122>g/a t findtheperiod ofsmall oscillations about aposition ofstable equilibrium. 5.Astone isthrown straight up,rises toaheight of100ft.,and falls to earth. Estimate thedeviation duetotherotation ofearth, thelatitude of theplace being 45North. (Neglect airresistance.) 6.Aparticle moves under gravity onasmooth surface ofrevolution with axis vertical. Theequation ofthesurface incylindrical coordinates is given intheformR=*F(z). Ifthevelocityishorizontal andofmagnitude q\ataheight z\,andagain horizontal andofmagnitude q<iataheight Zi, determineqiand#2interms ofziand ^^.(Usetheprinciples ofenergy and angular momentum.) 7.Inasimple pendulum thebob isconnected byalight string oflength a tothefixed point ofsupport. Thebobstarts inthelowest position with speed g.Show that if Ex.XIII] MOTION OFAPARTICLE 415 thestringwillslacken during themotion, sothatthebob fallsinward from thecircular path. 8.Show that,onaccount oftherotation oftheearth, atrain traveling south exerts aslight sideways forceonthewestern railofthetrack. Give anapproximate expressionforthisforce interms ofthemass ofthetrain,its speed, thelatitude, andtheangular velocity oftheearth. 9.Aparticle moves onasmooth surface ofrevolution with axisvertical. Theequation ofthesurface incylindrical coordinates isR F(z). Usethe principlesofangular momentum andenergy toshow that R*4>=h, %(z*+&+RW) -hgz-E, where handEareconstants. Deduce that zsatisfies adifferential equation oftheform &<=/(*). 10.Aparticleslides onasmooth cycloid inaverticalplane, thecusps of thecycloid being upward. Show thattheperiodic time ofoscillations under gravity isindependent oftheamplitude. 11.Aspherical pendulum oflength aandmass raoscillates between two levels which areatheights band cabove thelowest point ofthesphere. Expressitsconstant totalenergy interms ofmta,6,c,andg,taking thezero ofpotential energy atthelowest point ofthesphere. Check youranswer by putting b=c. 12.Acharged particle moves inauniform electric andmagnetic field, the electric andmagnetic vectors being perpendicular tooneanother. Show that,ifproperly projected, thepath oftheparticle isacycloid. 13.Thebobofaspherical pendulum, 10feetlong, justclears theground. Apegissetup1footduesouth from theequilibrium position ofthebob,and thebob isdrawn outtotheeastthrough adistance of2feet. Find (approxi- mately) thedirection andmagnitude ofthevelocity withwhich thebob should bestarted from thisposition,inorder tohitthepegandmake itfall overtoward thewest. 14.Forasimple pendulumoflength amaking complete revolutions, show thattheperiodic time is 4afl : I qoJOdy - 2/2)U- where qQisthespeed atthelowest position andk*=*ql/4ga. 16.Asmooth cup isformed byrevolution oftheparabola za4axabout theaxisofz,which isvertical. Aparticle isprojected horizontally onthe inner surface ataheight zwithaspeed -\/2kgz Q.Prove that,ifA;=i,the particlewilldescribe ahorizontal circle; alsothat,ifkFJ 0jitspath will lie between twoplanes z=zand z=|z. 16.Aspherical pendulum oflength aisheldouthorizontally, andthebob isstarted withagreat horizontal velocity qQ.Show that itfallstoadepth below itsinitial position given approximately by2gra2 /c5. 416 MECHANICS INSPACE [Ex.XIII 17.Aparticle moves onsmooth surface ofrevolution, theaxisofsymmetry being vertical. Show thatmotion inahorizontal circle ofradiusRisstable if d*z 3dz where z=z(R) istheequation ofthesurface incylindrical coordinates. Deduce thatthemotion ofaconical pendulumisstable. 18.Aparticle moves under gravity onarough vertical 'circle. Itstarts from restatoneendofthehorizontal diameter andcomes torestatthe lowest point ofthecircle. Findanequation todetermine thecoefficient of friction. 19.Aheavy bead starts from restatapointAandslidesdown asmooth wire in.theform ofahelixhaving theparametric equations xacos0, y=asin0,z=aOtana, theaxisofzbeing vertical. When itisvertically under A,asecond bead starts from restatA.Show thatthetangential acceleration ofeachbead is gsina,anddeduce aformula determining allthesubsequent instants at which onebead isvertically underneath theother. 20.Aheavy particleisconstrained tomove ontheinner surface ofa smooth right circular cone ofsemivcrtical angle a,theaxisoftheconebeing vertical andthevertex down. The particleisinsteady motion inacircle atheightbabove thevertex. Find theperiodic time forsmall oscillations about thissteady motion. 21.Aparticle slides inasmooth straight tubewhich rotates with con- stant angular velocity about avertical axiswhich doesnotintersect the tube. Thetube isinclined atananglo atothevertical. Initially thepar- ticle isprojected upward along thetubewithspeed <?o,relative tothetube, from thepoint where theshortest distance between theaxisandthetube meets thetube. Show that,nomatter what thelength ofthetubemay be, theparticlewillescape attheupper endprovided cotana 22.Consider anaxially symmetric electric field inwhich theaxial poten- tial isoftheform Fo=az+&Show thatthe field isuniform through- outspace andparallel tothe2-axis. Verify directly from (13.429) thatthe trajectory ofacharged particleisparabolic. Consider alsothecase ofanaxially symmetric magnetic fieldwith the axial potential oftheformQ=az+b.Verify from (13.438) thatthe trajectory isahelix. 23.Aparticle moves under gravity onasmoothsurface, theprincipal radii ofcurvature atitslowest point being a,b(a>ft).Thesurface rotates with constant angular velocity wabout thenormal atthelowest point. Show that,ifOxyarehorizontal axesattached tothesurface atthelowest point anddirected along thelines ofcurvature atthat point, thentheequa- Ex.XIII] MOTION OFAPARTICLE 417 tionofmotion forsmall vibrations nearthelowest point are Considering solutions ofthefoimx=Aent ,y-Bent ,deduce thatthere will beinstabilityifw*liesbetween g/aandg/b. 24. Ifinamagnetostatic lenstheaxialcomponentftofthemagnetic vector isconstant, show thatanobject point onthoaxisatz=willgive animage at z=__, where WQistheinitial velocity. 25.Foramagnetostatic lens,with thefieldconcentrated inh<z<ht prove that,totheorder hinclusive, themagnification ofanobject inthe planez ZQis |l+ylzo-k(B+Dzo)h, where A,#,Daieconstants defined by(13405) Evaluateexplicitly,if theaxialcomponentofmagneticfieldHisconstant inh<z<h. CHAPTER XIV APPLICATIONS INDYNAMICS INSPACEMOTION OFA RIGIDBODY 14.1.MOTION OFARIGIDBODYWITH AFIXED POINT UNDER NOFORCES Ifarigidbodyisconstrained toturnabout asmooth fixed axis,under noforces other than thereaction ofthe axis, the motion isextremely simple: thebody spins withconstant angular velocity. Butif,instead offixing alineinthebody, wefix onepoint only, themotion under noforces ismuch more com- plicated. Theproblem offinding thismotion isofwider interest thanmight appear atfirst sight, for themotion ofarigidbody relative toitsmass center isthesame asif themass center were fixed(cf.Sec. 12.4). Themounting ofabody soasto fixonly onepointismuch more complicated than that required to giveitafixed line. Itmaybedone byanarrangement oflight rings, known as"Cardan's suspension" (Fig. 144). Thebodyisrepresented bytheinner circle. Thepoints A,B arefixed. Rotation oftheringRiFIG.144. Cardan's suspension. aboutABgives onedegree offreedom. Rotation ofthe ringR2aboutCDgives asecond degree offreedom. Rotation ofthebody itself aboutEFgives thethird. Thebody cantake upallpositions inwhich thepoint ofthebody isfixed inspace, being thecommon intersection ofAB,CD,andEF. Allthe apparatus, except thebody itself,istoberegarded asmassless inthemathematical theory; thiscannot, ofcourse, beachieved inpractice, butthemasses ofRiandR%aremade assmall as possible compared withthemass ofthebody. 418 SEC. 14.1] MOTION OFARIGIDBODY 419 There aretwoways oftreating theproblem ofthemotion of abody with afixed point under noforces thedescriptive and theanalytic. The descriptive method, ormethod ofPoinsot, gives agood qualitative idea ofthemotion. Inthecasewhere thebody hasanaxis ofdynamical symmetry, thedescriptionis particularly simple; weshall consider thatcaseindetail later. Themethod ofPoinsot. Let bethefixed point inthebody, andAtB,Ctheprincipal moments ofinertia at0.Leti,j,kbeunitvectors fixed inthe bodyanddirected along theprincipal axes at0.Fortheangular velocity andangular momentum wehave,by(11.509), (14.101)to=cjii+co2j+w3k, h= When wesaythatthebodyisunder noforces, wemean more precisely thattheforces acting on thebodyhavenomoment about 0. (Thus ourargument applies toa heavy body under theaction cf gravity, provided that isthe center ofgravity.) Since theex- ternal forces donoworkandhave nomoment about 0,wehave the following facts toassist usindis- cussing themotion : /\ ji_ i j. m FIG. 145. Tho invariable line (l)thekinetic energy TISandtheinvariable plane. constant; (ii)theangular momentum hisaconstant vector. From thefirst ofthesewehave,by(11.404), (14.102) Aco?+Bu\+Cul=2T=constant; from thesecond, weknow thathhasadirection fixed inspace andalsoaconstant magnitude, sothat (14.103) A2 co?+2 co|+C2 l=h*=constant. Letusdraw through alineOPinthefixed direction ofh (Fig. 145); this iscalled theinvariable line. LetOQrepresent theangular velocity<>atanyinstant. Drop theperpendicular QNonOP;thenON=o>h/h. But,by(14.101) and(14.102), (14.104)<*h-2T, 420 MECHANICS INSPACE [SEC. 14.1 andso 2T (14.105) ON=~=constant. ThusNisafixed point during themotion, andsotheplane through N,perpendicular totheinvariable lineOP,isafixed plane;itiscalled theinvariable plane. Theextremity Qofthe angular velocity vector comoves ontheinvariable^plane. Letusnowtake thepoint ofview ofanobserver whomoves withthebody. (This iswhatwedoinourdaily lives, forwelive onarotating earth butregard apoint ontheearth's surface as " fixed.") Tosuchanobserver, thevectorsi,j,karefixed, but boththevectors hand CDarechanging. Ifi,j,karetaken as coordinate axesandtheextremity ofthevector coisgiven coordinates x,y,z,then x=on, y=o)2, z=o)3. Byvirtue of(14.102) and (14.103), wehave (14.106) Ax2+By2+Cz2=2T, A*x*+B2y2+C2z*=h2 . Infact,toanobserver moving with thebody, theextremity Qof theangular velocity vectorodescribes acurve which istheinter- section ofthetwoellipsoids (14.106), fixed inthebody. The first ofthese two ellipsoidsissimilar totherhomental ellipsoid andhasthesame axes;itiscalled thePoinsotellipsoid. The invariable planeisfixed inspace, buttotheobserver moving with thebodyitisamoving plane. Ittouches asphere ofradius ON,but ithasanother remarkable property: the invariable plane touches thePoinsot ellipsoidattheextremity of theangular velocity vector. Toseethis,wenote that thetangent plane tothePoinsot ellipsoid atthepoint (on,o>2,o>3)is (14.107) AU&+B^y+CW=2T. Sothedirection ratios ofthenormal totheellipsoid atthis point are A&I, Buz, C3. Butthese areprecisely thecomponents ofangular momentum; hence thenormal tothePoinsot ellipsoid attheextremity ofthe angular velocity vector isparallel totheangular momentum vector, i.e.,parallel toOP. Thisproves theresult. SEC. 14.1) MOTION OFARIGIDBODY 421 Aswehave indicated, there aretwodifferent points ofview: (i)thepoint ofview ofanobserver Sfixed inspace; (ii)thepointofview ofanobserver Srfixed inthebody. Itisconfusing totrytolookatthings simultaneously from the twopoints ofview.Weshall clarify thesituation bytaking themupseparately. Theobserver/S,fixed inspace, cutsaway (inhisimagination) allthebody except anellipsoid thePoinsot ellipsoid. He fixes hisattention onthismoving ellipsoid andonafixed plane (theinvariable plane). Asthebody moves, theellipsoid always touches theplane. Itactuallyrollsontheplane, since ithas anangular velocity vector which passes through thepointof contact oftheellipsoid andtheplane. This isafairly com- plicated type ofmotion;itbecomes quite simple, however, when thePoinsot ellipsoidisasurface ofrevolution, asweshall seelater. But itmay inanycasebevisualized bythinking of theinvariable plane asasheet ofpaper andthePoinsot ellipsoid asaninked surface. Inthecourse ofthemotion acurve isthus drawn ininkontheinvariable plane. Onjoining thefixed point tothepoints onthiscurve, wegetthespace cone(cf.Sec. 11.2). Ontheother hand, theobserver/S',fixed inthebody, turns hisattention tothetwo ellipsoids (14.106), fixed asfarasheis concerned, andinparticular totheir curve ofintersection. The angular velocity vector traces outacone (thebody cone), formed byjoining thefixed point tothiscurve. Thetwopoints ofview arebrought intocontact bythegeneral result: thebody cone rollsonthespace cone. The difference between thetwo isthis:Sregards thespace cone asfixed, but S'regards thebody cone asfixed. Theabove method gives aqualitative, rather thanaquantita- tive, description ofthemotion. Foraquantitative description, wemust useananalytic method. Thecase ofageneral body; analytic method. Since theexternal forces havenomoment about 0,Euler's equations (12.404) give !Ai-(B-C)w 2o>3=0, Ba2-(C-A)co 3o>i=0, Cw3-(A-#)!,=0. 422 MECHANICS INSPACE [SBC. 14.1 Wehave also, asin(14.102) and (14.103), (14109)U4.iuy; whereTandhareconstants, whichmaybefound byinserting thevalues ofi,w2,w3at=0.[The equations (14.109) may bededuced from (14.108) directly.]Weshallassume thatA,B,Carealldistinct. Wemaysuppose thetriadi,j,kchosen sothatA>B>C.Then itfollows from (14.109) that 2AT-h*>0,2CT-h*<0. There arethree very simple particular solutions of(14.108). These are o>i=constant, W2=0, s=0, to2=constant, w3=0, coi=0, wa=constant, wi=0, W2=0. These three solutions correspond tosteady rotations about the three principal axes ofinertia. Itisaremarkable factthatthese aretheonly axes about which thebody willspin steadily under noforces; theequations (14.108) aresatisfied byconstant values ofi,o>2, o>aonlyiftwoofthese constant values arezero. Turning nowtotheproblem offinding themost general solu- tion of(14.108), wemust firsteliminate twooftheunknowns, soastogetadifferential equation involving justoneunknown. Itproves best toconcentrate ourattention onw2.Wesolve (14.109) forwf, J,obtaining (14.110) o>?=P-Qcol, wS=R-SJ, where P,Q,R,Sarepositive expressions involving A,B,C,T,h. Substitution inthesecond equation of(14.108) gives Thisequationisofthesameform as(13.312) andmaybetreated inthesame way. Thus, from (14.111), weget where (14.113) SEC. 14.1] MOTION OFARIGID BODY 423 theconstants 0,p,kbeing positive functions ofA,B,C,T,h,with k<I.Hence, (14.114)=sn[p(t- to)], where tQisaconstant ofintegration. Substitution in(14.110) gives either (14.115a) or (14.1156)adn[p(t-<)], aen[p(-a>37en[p(t- )], 7dn[p(l- )], where aand7arefunctions ofA,B,C,T,h,determined except forsign.When wesubstitute in(14.108), wefindthat ajfry is negative. Fordefiniteness, wemaymake apositive bysuitable choice ofthesense ofthevectori;then7isnegative. That istheoutline ofthemethod offinding coi,o>2,o>3asfunc- tions of t.Thecompletion oftheargument consists infilling inthealgebraic details. Caremust betaken inselecting the constants/?,p,k,sothatkislessthan unity;itbecomes necessary todistinguish between thetwo cases (a)ft2>2BT,and(6) h*<2BT. Intheformer case,wearrive at(14.115a), inthe latter at(14.1156). Weleave ittothereader toverify the following results. CASE (a):h'2>2BT. 424 MECHANICS INSPACE [SEC. 14.1 Inestablishing these results, thefollowing identityisuseful: (14.117) (B-C)(^2-2AT)+(C-A)(h*-2BT) +(A-B)(h*-2CT)=0. Butthedetermination ofi,co2,w3asfunctions oftdoesnot complete thesolution oftheproblem. Weshould beable to tell,from given initial conditions, theposition ofthebody atany time. Todothis,wespecify thedirections ofthetriadi,j,k, relative toatriadI,J,K,fixed inspace, bymeans oftheEulerian angles 0,*,^.Then, by(11.202), !o>i=sin\l/6 sin9cos^<, co2=cos$6+sh)sm^<, 0)3=COS+$. Ifwesubstitute forOH,co2,u3from (14.114) and (14.115), we obtain three differential equations for6,<,^.Thesolution of these equationsinthis general form presents aformidable problem.Itisgreatly simplifiedifwechoose thevectorKinthe direction oftheinvariableline, defined bytheconstant vector h. Then thecomponentsofhalong i,j,karefound bymultiplying hbythedirection cosines ofKrelative toi,j,k.Those direction cosines areeasily found(cf.Fig. 118,page 280)byprojecting K oni,j,k;they are sin cos^, sin sin^, cos 6. Hence, !Ao>i=hsin6cos^, Bwz=hsin6sinf, Co>3=hcos 6. From these equations, weget6and\l/asfunctions oftwithout anyintegration, thus: (14.120) cose=^p,tan*=- |^- Tofind<f>,wededuce, from the firsttwoof(14.118), (14.121) sin(j>=co2sin\f/ o>icos^, andso<f>isobtained byaquadrature, since0,^,wi,co2arealready known asfunctions of t. SEC. 14.1] MOTION OFARIGIDBODY 425 From theperiodic property ofthe elliptic functions, wesee that0,sin^,cos^,<areperiodic functions of/;ingeneral,< doesnotincrease byamultiple of2irinaperiod, andthemotion as awhole isnotperiodic. Thecase ofabody withanaxis ofsymmetry. When themomental ellipsoid atthefixed point hasanaxis ofsymmetry, twoofthethroemoments ofinertiaA ,B,Cbecome equal tooneanother. This willbethecase ifthebody isasolid ofrevolution ofuniform density, but allthat isactually required isthesymmetry ofthemomental ellipsoid. Themotion ofthe body under noforces isthen greatly simplified. Infact, the simplificationissogreat that itiseasier todiscuss theproblem afresh, rather than toapply theformulas ofthegeneral case. Themotion canbedetermined, both qualitatively andquantita- tively, bythemethod ofPoinsot. Letusdenote theprincipal moments ofinertia atbyAand C,Cbeing theaxial momentofinertia andAthetransverse moment ofinertia. (Thismeans thatCisthemoment ofinertia about theaxis ofsymmetry andAthemoment ofinertia about any perpendicular linethrough 0.)The casesA>Cand A<Cdiffer insome respects, butforthepresent wemay treat them together. The Poinsot ellipsoidisofrevolution. Since itscenter is fixedand itrollsontheinvariable plane, thefollowing facts are obvious : (i)Thebody coneandthespace cone areboth right circular cones. (ii)The angular velocity vector isofconstant magnitude (w=OQ)andmakes aconstant angle withtheinvariable lineOP (cf.Fig. 145). (iii)Theinvariableline,theangular velocity vector, andthe axis ofsymmetry arecoplanar atevery instant. (iv)Theaxisofsymmetry makes aconstant angle (a)withthe angular velocity vector andaconstant angle (/3)with theinvari- able line. Togetaclear idea ofthebehavior ofthebody, letusstart attheinstant t=withthebody insome definite position and 'withsome definite angular velocity <o,sayOQo. LetORobethe initial position oftheaxisofsymmetry. Then a.= 426 MECHANICS INSPACE [SEC. 14.1 LetOSobeperpendicular toORo intheplane RoOQo. We resolve o>along ORoandOS,obtaining componentso>cosa and o)sina.Since these areprincipal axes, theangular momen- tumvector hhascomponents CeocosaalongORoandAv>sina along 0$oJ ithas, ofcourse, nocomponent perpendicular tothe plane RoOQo. Wearenow inaposition toconstruct h,andhence theinvariable lineOP.Theangle (=RQOP)isgivenby (14.122) tan8=77tan a. Wehavenow todistinguish two cases, asshown inFigs. 146a and 6. N FIG.146.- (a)ThecasewhereA>C. (b)ThecasewhereA<C. CASE (a):A>C(asinthecase ofarod). Here/3>a;the angular velocity vector liesbetween theaxisofsymmetry andthe invariable line. CASE (b):A<C(asinthecase ofaflatdisk). Herej3<; theinvariable line liesbetween theaxis ofsymmetry andthe angular velocity vector. Wehavespoken oftheinstant t=0.Butasimilar construc- tionmaybemade atany instant, and, aswehave seen, the angles aand0,andthemagnitude oftheangular velocitycoare constants. Sothefigures wehave constructed represent the state ofaffairs atany instant, theplane containing thefigure rotating about theinvariable lineOP. This rotation isduetoan angular velocityo>ofconstant magnitude, inclined toOPata constant angle; hence theplane containing theaxis ofsymmetry andtheangular velocity vector rotates aboutOPwithaconstant angular velocity. Weshalldenote thisangular velocity byft. There isonemore constant ofimportance. Itistheangular velocity oftheinstantaneous axisabout theaxisofsymmetry, as SEC. 14.1] MOTION OFARIGIDBODY 427 judged byanobserver moving with thebody.Weshalldenote thisangular velocity byn. Wehave, inall,thefollowing constants: a,0,w,ft,n. Ofthese, aand coaredetermined byinitial conditions; /3isgiven by(14.122). Weshallnow setupequations tofind andn,and atthesame time getaclear picture ofthemotion byconsidering thespace andbody cones (Figs. 147aand6).OPistheinvari- able line,OQtheinstantaneous axis,ORtheaxisofsymmetry, andQN,QMaredrawn perpendicular toOP,OR,respectively. Ineach casethespace cone isfixed, andthebody cone rollson iace-Cone r-ConeSpace-Gone O (a) (6) FIG.147. (a)ThecasewhereA>C. (6)ThecasewhereA<C. it.Thismotion iseasy tofollow inFig.147a. Themotion in Fig.1476maybeunderstood bythinking ofwhat themotion looks likefrom above, orbymaking asimple model outofthick paper andworking thecones through thefingers inorder toreproduce thecondition ofrolling. CASE(a):A>C.The lineORturns aboutOPwith angular velocityft.Thus, intimedt,thepointMreceives adisplacement OMsinftftdt=cosasinOQftdt, perpendicular totheplanePOR. ButMisapoint fixed inthe body cone, which isturning aboutOQwith angular velocity w. Hence thedisplacement ofMisalso QMcosao)dt=cosasinaOQ o>dt, 428 MECHANICS INSPACE [Sue. 14.1 sinceQMcosaistheperpendicular distance ofMfrom OQ. Equating thetwoexpressions, weobtain or,by(14.122),__ (14.124)ft=co./sin2a+-pcos2a. Tofind n,wenote that, intimedt,Qtravels adistance QNtt dt onthespace cone. But,from thedefinition ofn,inthesame timeQtravels adistance QMn dtonthebody cone. From the condition ofrolling, these distances areequal tooneanother, and so QMn=QNil, or (14.125) n=Q5L<?JL!> =sin <?~ >. sina sin/3 Interms ofthebasic constants, wehave (14.126a) n=A~C cocosa. Thesense ofthisrotation nisobviously retrograde, when com- pared with co. CASE (b):A<C.The reasoning inthis case follows the same lines,andwegetthesame formula (14.124) forft,while (14.1266) n=jcocosa. Thesense ofthisrotation nisdirect, whencompared with co. Inthecase oftheearth, which isslightly flattened from the spherical form, wehaveA<C,andtheratio (CA)/A is small. Thus, case (6)applies, butinanextreme form, since the instantaneous axis isclose totheaxis ofsymmetry andais small. Theangular velocity nrepresents therateatwhich the celestial pole, oraxis ofrotation oftheearth, moves round the earth's axisofsymmetry. Fortheperiod, wehaveapproximately SEC. 14.2] MOTION OFARIGID BODY 429 Ifthesidereal dayistaken asunit oftime, then 27r/co=1,and calculation gives thevalue 305fortheperiod. This prediction is inpooragreement with observation; forthough arotation ofthis sort isobserved,itsperiodisabout 440days.*Themodel used (arigidbody) proves atfault here,onaccount oftheelasticity oftheearth. Exercise. Acircular disk ismounted sothat itcanturn freely about its center. Itsangular velocity vector makes anangle of45with itsplane andhasamagnitude of20revolutions persecond. Make arough sketch ofthespace andbody cones, andfind J2and n. 14.2.THESPINNING TOP Thespinning topisthemost familiar exampleofagyroscopic system. Theword "gyroscope" wasinvented todenote an instrument inwhich theearth's rotation produced aneffect which could beobserved. Buttheword"gyroscope" (or"gyrostat") isnowused foranysystem inwhich arapidly rotating body issomounted that itmaychange thedirection ofitsangular velocity vector.* Why doesaspinning topnot falldown? How does itsrapid rotation render itapparently immune totheforce ofgravity, which makes non-spinning bodies fall? Itisdifficult togivea simple answer tothisquestion. Theonlyway toexplain the phenomenonistoconstruct themathematical theory ofthetop. Forourpurposes, weshallunderstand a"top" tomean arigid body withanaxisofsymmetry, acted onbytheforce ofgravity. Apoint ontheaxis ofsymmetryisfixed. Thusweidealize the ordinary topbysupposingittoterminate inasharp point (or vertex) andtospinonafloorrough enough toprevent slipping. Steady precession ofatop. Themotion ofany rigidbody withafixed point satisfies the equation (14.201) h=G, where histheangular momentum about andGthemoment oftheexternal forces about 0.Inmostdynamical problems, we *Of .H.N.Russel, R.S.Dugan, and J.Q.Stewart, Astronomy (Ginn and Company, Boston, 1945), Vol.I,pp.118,131-132. Thesection ofthebody conebytheearth's surface isacircle withadiameter ofabout 26feet. 430 MECHANICS INSPACE [SEC. 14.2 mgK FIQ. 148. Vector diagram fortop insteady precession.think oftheforces asgiven andthemotion asunknown; inthat case,Gisgivenandhistobefound. Butwecanlookat(14.201) theotherwayround. Wemayregard themotion asprescribed, sothathisknown asavec- torfunction ofthetime. Then (14.201) shows directly themo- mentGwhich must beapplied tothebody inorder togive this motion. Letusnow describe asimple motion ofatop, called steady precession, andinquire what forces must actonthetopinorder that thismotion maytake place. Insteady precession, theaxis ofsymmetry ofthetopdescribes withconstant angular velocity a right circular conewiththevertical foraxis. Atthesame time thetopspins about itsaxis ofsymmetry withconstant angular velocity. Weshall usethefollowing notation (Fig. 148): a=distance ofmass centerDfrom fixed vertex 0,m=mass oftop, A=transverse moment ofinertia at0, C=axialmoment ofinertia at0,K=unitvector directed vertically upward, (i,j,k)=unitorthogonal triad, withkalongODand iinthe plane ofkandK,=inclination ofODtothevertical. Wenotethat (14.202) K=sin6i+cos6k. Theangular velocity vector <oofthetop liesintheplane (k,K). Itcanberesolved alongiandk;weshall write (14.203) ii+5k and call sthespin ofthetop.Thevelocity ofthepointDis 6>Xak=(ii+sk)Xak=coiaj. Weunderstand bytheprecession ptheangular velocity with SEC. 14.2] MOTION OFARIGIDBODY 431 whichODrotates about K.Thevelocity ofDisthen pKXak=pasinj. Equating thetwoexpressionsforthevelocity ofD,wehave (14.204) coi=psin 6. Inthesteady precession 0,s,andpareconstants. Theangular momentum is (14.205) h=Awt+Csk =Apsin0i+Csk. This vector liesintheplane (k,K),androtates rigidly with it. Thushisthevelocity ofapoint with position vector hinarigid body which turns withangular velocity />K. Therefore, (14.206) h-pKXh =p(sin 6i+cos k)X(Ap sin6i+Csk)=psin6(Ap cos6 Cs)j. Thesteady precession takes place, with assigned values of 0,p,ands,provided that themoment about ofallforces (including gravity)is (14.207) G=psinB(Ap cos-Cs)j. Now theweight ofthetopisaforce mgK atDandsohasa moment akX(mgK)=mga sin6j about 0. Ifthis isequal toG,asgiven by(14.207), noforce other than theweight ofthetopisrequired tomaintain the motion. Thus thesteady precession takes place undergravity aloneif (14.208) p(Cs-Apcos 6)=mga. This isasingle equation connecting thethree constants 6,p,s.Thereis,therefore, adoubly infinite setofsteady precessions corresponding toarbitrary values oftwooutofthe three constants. Itisnot,however, possible toassign com- pletely arbitrary values oftwo oftheconstants; these values must besuch that (14.208) yields arealvalue forthethird constant. 432 MECHANICS INSPACE [SEC. 14.2 In Ifweseeatopspinning, 8andpareeasy toobserve, terms ofthem, sisgivenby (14.209) .=V?+4~L?. Wenote that,iftheprecession issmall, thespin isgreat and isgiven approximately by (14.210)mga ~Cp' This isaverysimple anduseful formula. Exercise. Adisk, 6inches indiameter,ismounted ontheendofalight rod 1inchlongandspins rapidly. Itprocesses once in15seconds. Find approximatelythespin, inrevolutions persecond, andthevelocity ofa point ontheedge ofthedisk. General motion ofatop. Todiscuss thegeneral motion ofatop,weshall usethesame notation fortheconstants ofthetopasthatused above. FIQ. 149. Vector diagram fortopingeneral motion. LetI,J,K(Fig. 149)beafixed orthogonal triad,Kbeing directed vertically upward. Leti,j,kbeanorthogonal triad, withkpointing alongOD,theaxisofsymmetry ofthetop,and icoplanar withkandK;thusjishorizontal. Thetriadi,j,k isfixed neither inspace norinthetop,butkisfixed inthetop. Let6, <t>betheusual polar angles ofkrelative tothefixed triad. Variations inarereferred toasnutation, andvariations in<j>asprecession. Let (14.211)<>=d>ii -f-GJ2J ~~hcoak SEC. 14.2] MOTION OFARIGIDBODY 433 betheangular velocity ofthetop,and (14.212) a=ftii+ 2j+Qsk theangular velocityofthetriadi,j,k.Itiseasy toseethat (14.213) Q!=sin<, Q,=-0, U3=cos6<. Now therelative motion ofthetopandthetriadi,j,kconsists onlyofarotation about k.Hence, (14.214) coi=fli=sinB<,o>2=Q2=0. Theangular momentum is (14.215) h-Awii+Ao> 2j+Cwjc, and itsrate ofchange is,by(12.306), (14.216) h=Ahi+Aw 2j+Cwak+OXh. Themoment about oftheweight ofthetopis (14.217) G=akX(-nigK)=-mgasinj. Themotion ofthetopsatisfies thefundamental equation (14.218) h=G. When wesubstitute theexpressions given above, thisvector equation gives three scalar equations for0,<,andw3.However, anindirect method ofattack proves simpler, andweshallmake direct useonlyofthethirdcomponent of(14.218). Thecomponentof(14.218) inthedirection ofkgives Ctos=0, since, by(14.214) arid(14.215),12Xhhasnocomponent inthe direction k.Hence (14.219)<at=s, aconstant; thespin ofthetopisaconstant. Further, since the weight ofthetophasnomoment about K,thecomponent of angular momentum inthisfixed direction isconstant, andso (14.220) hK=a, aconstant. By(14.214) and (14.215), thisgives immediately (14.221) Aj>sin2+Cscos6=a, 434 MECHANICS INSPACE [SEC. 14.2 sinceK=sin6i+cos6k.Finally, wehave theequation of energy (14.222) T+V=E, or (14.223) iA(wf+wf)+iCw|+mgacos9=E, Ebeing aconstant. Substitution from (14.214) and (14.219) gives (14.224) A(62+&sin26)+Cs2=2(E-wgfacos0). Wehave in(14.221) and(14.224) twoequations todetermine 6and <asfunctions ofthetime. Itisconvenient towrite Cs=/?.Then ourtwoequations read A<j>sin2=apcos6 * 2sin2 61)+=2(#-m0cos0). Theplanisnow obvious. Wearetosubstitute for <inthe second equation from the first; this willgiveadifferential equa- tion for 6.When this issolved, the first equation, willgive< byaquadrature. Letusputxcos 0.Onmultiplying thosecond equation in(14.225) bysin26andsubstituting for<,weobtain forxthe differential equation (14.226). L-rx- jv/ =2(E-mgax)(l- Thisequation maybewritten (14.227) **=/(*), where (14.228) /(*)=I This isacubic inx,and,bythesameform ofargument asthat used inSec. 13.3forthespherical pendulum, weseethat ithasa SEC. 14.2] MOTION OFARIGIDBODY 435 graph ofthegeneral formshown inFig. 150;thefunction f(x) hasthree realzerosx\,#2,#3,such that 1<Xi<xz<1< 3. (Inspecial cases, wemayhave oneormore signs ofequality instead ofinequality.) Thus /(x)maybewritten (14.229) /(*)=(x- Again bythesame argument asinSec. 13.3, thesolution of (14.227)is (14.230)cos=x=xi+(x*-xi)sn2\p(t-/)1, f(x) Fia. 150. Graph off(x) foratopingeneral motion. where pandthemodulus koftheelliptic function aregivenby ._mga(x>-*,),2_JT,-Xi (14.231)2A 3~ Theconstants x\,xz,x^arefunctions oftheconstants occurring in(14.228), i.e.,theconstants ofthetopand a,0,E;thelatter areknown when theinitial position andangular velocity ofthe toparegiven. Thecomplete solution forthemotion oftheaxis ofthetopis givenby(14.230) and afa (14.232) Since xisknown asafunction of tythis lastequation gives byaquadrature. This analytic solution does notimmediately give aclear idea oftheway inwhich thetopbehaves. However, wecan 436 MECHANICS INSPACE [SEC. 14.2 construct theessential features ofthemotion, byfixing our attention ontheintersection oftheaxis ofthetopwithaunit sphere having itscenter at0. Itisinteresting tocompare the motion ofthispoint withthemotion ofaspherical pendulum. Inthe first place,itisclearfrom (14.230) thattherepresenta- tive point ontheunit sphere oscillates between two levels B=0iand=2,given by cos0i=#1, cos 2=#2. Thisbehavior islikethat ofthespherical pendulum; butwhile themean level forthespherical pendulum must liebelow the center ofsphere, that isnolonger necessarily true forthetop. Thereis,however, amore striking difference; inthecase ofthe a b FIG.151. Motion oftheaxisofatop. (a)without loops, (b)with loops. top,wemayhave loops onthecurve. Theabsence ofloops, asinFig.151a,orthepresence ofloops, asinFig.1516, depends ontheway inwhich themotion isstarted, i.e.,onthevalues of theconstants a,/?,E.The criterion fortheexistence ofaloop isthat<j>should sometimes increase andsometimes decrease, andthecondition forthis isthat<j>should vanish during the motion. By(14.232),<=when x a//3; since xoscillates between x\and#2,itisjustaquestion astowhether a/0 lies inthisrange ofoscillation. Ifitliesintherange, there are loops;ifnot,there arenoloops. Cuspidal motion ofatop. Aparticularly interesting case ariseswhen thetopisspinning with itsaxis fixed inposition andthen released. Itstarts to fallbutrecovers and rises toitsformer height, repeating this process overandover again. This casecanbediscussed interms ofthetheory justdeveloped. Thebehavior ofatopdepends essentially onthecubicf(x) of (14.228), andtoitwemust direct ourattention. SEC. 14.2] MOTION OFARIGIDBODY 437 First, letusnotethat initially xx(say), x=0,and$=0. Hence, by(14.232), a=#c ;also,by(14.226), Substitution in(14.228) gives (14.233) f(x)=^(Xo-z)(l-*2 )~J2(*o-*)2 . Obviously, onezero off(x)isx=#o;but isthiszero #1or Differentiation gives, forxXQ, f(xo )=_ Since thisvalue isnegative,itisclearfrom Fig.150thatx=z2, notXL The oscillation ofxisfrom xito#o(orx2),where Xiisthe smallest zero of/(x). Putting (14.234)v x weseethatthethree zeros off(x)are a:2=so, Z3=X+VX2-2Xx+1. Thus theaxis ofthetopfallsdown from aninclination 0o (where cos0o=#o)toaninclination0i,where (14.236)cos0i=xi=X-VX2-2Xx+1. Then itstarts toriseagain andswings upto6,where the axis isagain instantaneously atrest. IfXislarge, i.e.,ifthespinisgreat, binomial expansion ofthe radical in(14.236) gives approximately sin20ocos0i=cosH\ Thedifference 0i issmall, andsowemayusetheapproxima- tion cos 0i-cos=~(0i 0o)sin8 . 438 MECHANICS INSPACE [Sue. 14.2 The axis fallsonlythrough thesmall angle (14.237) 0i = ^!Tfa sin 0> The differential equation ofthepathoftherepresentative point ontheunitsphereis dx x4(1 #2)^/J(x) Since /(x)vanishes likexXQasx >o,itisclearthat d<j>/dx= atthehighest positions ofthe axis. Hence thepath ofthe representative point meets thecircle =0oatright angles; the pathhascusps atthese points, directed upward. Stability ofasleeping top. Anyone whohasseen atopspinningisfamiliar with the general nature ofthemotions discussed above. Sometimes thetopexecutes amotion ofsteady procession, andsometimes tho more general motion inwhich theaxis ofthetopnodsupand down asitprocesses. Athird typo ofmotion isoften seen, inwhich thetopspins with itsaxis vortical. The axisremains stationary andthere isnoapparent motion ofthetopasawhole- itisthen said tobeasleeping top.Asmall disturbance ofa sleeping topproduces only asmall oscillation when thespin isgreat; when thospinhasboon considerably reduced byfric- tional resistance, thetopbegins towobble andultimately falls down. Wenaturally ask:What isthecritical value ofthespin below which themotion ofasleeping topisunstable? Theanswer isfound byexamining thecubic f(x)given in (14.228). Since == forasleeping top,wehave, by (14.221) and(14.224), a==Cs,E=|Cs2+mga\ hence, (14.228) gives (14.238) Weobserve that x=1isadouble zero off(x).Two cases arise: either thethird zero off(x)isgreater than unity, orit islessthan unity. Theforms ofthegraph off(x)forthesetwo SEC. 14.2] MOTION OFARIGID BODY 439 cases areshown inFigs. 152aand 6;interms ofthenotation used forthezeros of/(#), Fig.152ashows thecasewhere the third zero isxs,andFig.1526thecasewhere itisx\. When thetopsuffers asmall disturbance, thegraphoff(x) forthedisturbed motion willnotbethesame asthat forthe f(x) f(x) FIQ. 162.- (a)Graph off(x) forastable sleeping top. (b)Graph off(x) for anunstable sleeping top. Ineach case, thebroken curve isthegraph fordis- turbed motion. sleeping top.The difference willbesmall, however, since only small changes intheconstants a,#,Ecanresult from asmall disturbance. Thebroken curves inFigs.152aandbindicate the way inwhich thegraphs off(x) aremodified byasmall dis- turbance. Ingeneral, thethree zeros off(x)willbecome dis- tinct two ofthem must, ofcourse,lieintherange (1,1), andthethirdmust exceed unity. 440 MECHANICS INSPACE [SBC. 14.2 Since, inthedisturbed motion, thevalue ofxliesbetween thetwosmaller zeros off(x),itisclear thatoneorother ofthe following descriptions applies: (i)The axis ofthetop,when disturbed, doesnotdepart far from itsoriginal vertical position themotion isstable. This corresponds toFig. 152a. (ii)The axis ofthetopfalls toaninclination0i,where cos0i=#1 themotion isunstable. Thiscorresponds toFig.1526. Tofindthecritical value ofthespin s,itremains todistinguish thetwocases analytically. Thetwotypesofcurve aredistinguished bythesignof/"(#) atx=1;itisnegative inFig.152aandpositive inFig.1526. Differentiating (14.238), wefind this isnegative and themotion ofasleeping topisstable, if (14.239)s>>- Inthelimiting case s2=4Amga/C2 ,itiseasy toseethat all three zeros off(x)coincide ata;= 1,andthemotion isstable. Exercise. Show that, foramotion ofsteady procession, thecubic f(x) hasadouble zerolying intherange (1,1).Hence show that thistype of steady motion isalways stable. Stability ofaspinning projectile. Itisawell-known factthatanelongated projectile acquires stability from thespinimparted toitbytherifling inthegun. Bythiswemean that itdoesnotturnbroadside ontoitsdirec- tionofmotion when inflight, nordoes ittumble asanonspinning projectile often does.We shallnow useourtheory ofthe motion ofatopinanattempt toexplain thisspin stabilization. InSec.6.2wediscussed themotion ofaprojectile inaresisting medium, theprojectile being treated asaparticle. Theproblem becomes much more complicated when theprojectileistreated asarigid solid ofrevolution, subject togravity andtotheaero- dynamic forcesystem duetothepressure ofthe air.Wecannot SBC. 14.3J MOTION OFARIGIDBODY 441 enter hereinto thisgeneral problem; instead, following theolder writers onballistics, weshallmake some drastic simplifications. Thus, weshallassume that theaerodynamic force systemis equipollent toasingle force (thedrag) with fixed direction and constant magnitude (R), intersecting theaxis oftheprojectile atafixed point (thecenter ofpressure). WecannowuseFig. 149,with suitable changes, forthediscussion ofthemotion ofthe projectile relative toitsmass center. Let bethemass center,Kafixed unitvector opposed tothedirection ofthedrag,andk aunitvector along theaxis oftheprojectile. ThepointDis taken tobethecenter ofpressure; andtheforce atD,i.e.,mgK inthecase ofthetop,isnow tobereplaced byRK.The weight oftheprojectile actsthrough 0,andsothetotalmoment about isduetotheaerodynamic forces alone;itis (14.240) G=-ftasin0j, where aisthedistance ofthecenter ofpressure infront ofthe mass center and 6thetingle between theaxis oftheprojectile andthedirection ofthedrag reversed. Since thefundamental equation (12.209) formotion relative tothemass center isofthesameform as(14.218) andtheexpres- sion (14.240) forGisofthesameform as(14.217) with thecon- stantmgachanged toRa,wecanapply tothemotion ofthe projectile thesame mathematical analysis asweapplied tothe motion ofthetop. Inparticular,iftheprojectileismoving alongitsaxis, thefixed direction ofthedrag willalso lieonthis line,andwehavewhat isessentially asleeping top.Then (14.239) yields thecondition forstability ofthespinning pro- jectile, i.e.,thecondition thattheprojectile,ifslightly disturbed, willnotdevelop large oscillations. This condition is (14.241) *>^, where sisthespin oftheprojectile, Athetransverse moment of inertia at(i.e., atthemass center), andCtheaxialmoment of inertia. 14.3.GYROSCOPES Thestability ofagyroscope. Letussuppose thatagyroscope (i.e., arigidbody withan axisofsymmetry)ismounted inaCardan's suspension (Fig. 144), 442 MECHANICS INSPACE [Sue. 14.3 sothat itsmass center isfixed. Itissetspinning about itsaxis ofsymmetry with agreat angular velocitys.Now letan impulsive coupleGbeapplied tothegyroscope. The instan- taneous change inangular momentum is[cf.(12.502)] (14.301) Ah=G. Wehavethenavector diagram asinFig. 153.ThevectorOA is h,theangular momentum before theimpulsive couple was B Fio. 153. Change inangular momentum duetoanimpulsive couple. applied.Itismade long, because s(and consequently /?,)is > > assumed tobelarge. ThevectorAB isAh,andOBrepresents thefinal angular momentum. Itisclear that theangleAOB issmall;infact,ittends tozeroasstends toinfinity. Thus, theapplication ofanimpulsive couple toarapidly spinning gyroscope makes only asmall change inthedirection ofthe angular momentum vector. Itiseasily seenthatthecorrespond- ingchangeinthedirection oftheangular velocity vector isalso small. This simple result illustrates the stability which arapid rotation imparts toabody. Thegyroscope shows, asitwere, anunwillingness toalter thedirection ofitsaxis.When itdoes yield,itdoes soinamanner which continues tocause surprise even tothose familiar withthetheory. Thegyroscopic couple. InFig. 154a,isafixed point ontheaxis ofagyroscope, andjaunitvector fixed inspace. Aspointed outearlier, any motion canbeproduced, provided suitable forces areapplied. Letusdemand that thegyroscopeshall spin with constant angular speed sabout itsaxis,andatthesame time thattheaxis shallturn (orprocess) withconstant angular speed pintheplane perpendicular toj.Ifkisaunitvector along theaxis ofthe gyroscope and icompletes thetriad, then theangular velocity ofthegyroscopeis (14.302)<o=pj+sk, SEC. 14.3] MOTION OFARIGIDBODY 443 andthetriad(i,j,k)hasanangular velocity (14.303)11=pj. IfAandCare,respectively, thetransverse andaxialmoments ofinertia, theangular momentum is (14.304) h=Apj+Csk, and itsrate ofchangeis (14.305) h=aXh-Cspi. Thus thegyroscopic coupleGrequired tomaintain thismotion is (14.306) G-Cspi, thatis,acoupleofmagnitude Csp,produced byapair offorces intherotating plane ofjand k. Cs Ap (a) FIQ. 154. (a)Angular momentum diagram foraprocessing gyroscope. (b)Relations between couple, precession, andspin. Therelations between thecouple, theprecession, andthespin areshown inFig. 1546. Itismore interesting herenottoshow thevectors intheusual way, buttorepresent thembyarcs in theplanes perpendicular tothem. The curiousfact, hard to understand intuitively,isthat theplane ofthecouple doesnot coincide with theplane oftheprecession, but isperpendicular toit.Instead ofyielding tothecouple, theaxisofthegyroscope turns atright angles totheplane ofthecouple. Itisevident from (14.306) thatwhen thegyroscope spins rapidly, avery great coupleisrequired toproduce even amod- erate rate ofprecession. Although thediagram ofFig.1546shows onlyasimple gyro- 444 MECHANICS INSPACE [SEC. 143 scopic phenomenon,itisvery useful fromapractical standpoint. Wenote that thethree quadrants form asingle closed curve. Aswetraverse itinthesense indicated bythearrows, wecover thefollowing quadrantsinorder: Couple, Precession, Spin. This iseasy toremember, since theletters C,P,Sareinalpha- betical order. Example Anairplane hasarotary engine, which rotates inaclockwise direction when viewed frombehind. Theairplane makes aleftturn. Does thegyroscopic effect oftherotating engine tend tomake thenose riseorfall? Firstwesuppose that thepilot sotsthe rudder andelevator insuchaway thatthe nose goes neither upnordown. Themass center oftheengine describes acircular arc Castheairplane turns (Fig. 155). The angular velocity oftheengine ismadeupof alargecomponent along thetangent to(7 andasmall vertical component, duetothe turning oftheairplane asawhole. The quadrants ofprecession andspinaretherefore asshown Hence, bytheruleofalphabetical order, thecouple quadrant comes down in front. Tomaintain themotion described, thepilotmust settherudder andelevator in suchawaythataerodynamic forces, acting onthem, produce therequired couple. Ifthepilotfliestheairplane witharotary engine inthesameway ashewould flyasimilar anplane withastationary engine, thecouple required tomaintain thesteady flight inahorizontal circle willnotbepresent. Since thecoupleisonewhich tends todepress thenose, initsabsence thenose will rise.What willhappen after theinitial lifttakes placeisacomplicated question, notcovered bythepresent simple theory. Thegyrocompass. Ifthespinning oftheearth onitsaxisweremuch faster than itactually is,itwould beasimple matter todevise amechanism bywhich thetruenorth could befound onashipatsea.How- ever, theearth's rotation issoslow thatanapparatus ofgreat delicacyisrequired,inorder that aminute effectmaynotbe wiped outbyfrictional resistances. Themodern gyrocompassFIG. 155. Airplane turning. SEC. 14.3] MOTION OFARIGIDBODY 445 issuchapiece ofapparatus. Thesimple system which weshall discuss ismuch lesselaborate than thegyrocompass asitis actually constructed.* However, thebasic factthataspinning gyroscope enables ustofindthenorth isdemonstrated bya discussion ofanideally simple gyrocompass. InFig. 156,PQ ispart oftheearth's axis,drawn from south tonorth. Thepointisontheearth's surface atlatitude A. Thus thehorizontalline,drawn duenorth from 0,isinclined totheearth's axisatanangle X;this line isOQinthediagram. TheunitvectorKisparallel toPQ. Agyroscopeismounted inaCardan's suspension (Fig. 144)sothat itsmass center liesat0.But itisnot leftfroototurn about 0}onepair ofthebearings inthe suspensionislocked,sothatthoaxis ofthe gyroscope canmove onlyinthohorizontal plane at0.Theunitvector kliesalong theaxis ofthegyroscope, making withOQ avariable angle 8;iisperpendicular tok inthehorizontal plane, andjisvertical (i.e.,coplamir withPQ, OQumlporpcndi- cular toOQ). Thogyroscopeitsolf isnot shown inthediagram; thecurve isaunit circle inthehorizontal plane. Theangular velocity ofthetriad(i,j,k)ismado upofthe angular velocity oftheearth, which wemay write 2K,andan angular velocity duetochange in0,infact, 0j.Since K= sin6cosXi+sinXj+cos 6cosXk, theangular velocity ofthetriad is (14.307)o>'=-ftsin cosXi+(6+ sinX)j +ftcos6cosXk. Theangular velocityo>ofthegyroscope differs from thisonly in *Cf.H.Lamb, Higher Mechanics (Cambridge University Press, 1929), p.144;R.F.Deimol, Mechanics oftheGyroscope (TheMacmillan Com- pany,New York, 1929); A.L.Rawlings, TheTheory oftheGyroscopic Compass (TheMacmillan Company, NewYork, 1929); E.S.Ferry, Applied Gyrodynamics (John Wiley&Sons,NewYork, 1932).P FIG. 156. Vector dia- gram forgyrocompass. 446 MECHANICS INSPACE [SBC. 14.3 thethirdcomponent; thus, (14.308)o>=-ftsin cosXi+ (6+ ftsinX)j+sk, where sistheaxial spin, including acomponent oftheearth's rotation. Theangular momentum is (14.309) h=-A 12sin cosXi+A(6+QsinX)j+Csk, where ^4andCare,respectively, thetransverse andaxialmoments ofinertia. Tokeep theaxis ofthegyroscopeinthehorizontal plane, thebearings ofthesuspension must exert acoupleGon it. Since thebearings areassumed tobesmooth, nowork isdone bythiscoupleinrotations about eitherjork.Hence,Gis perpendicular tothese vectors, andwemay write (14.310) G=Gi, whereGmaybeeither positive ornegative. Thefundamental equation h=Ggives (14.311) -AQcoa BcosX6i+A8j+6Y.sk+'Xh=Gi. Wenowpickoutthejandkcomponents ofthisequation andso obtain twoscalar equations forsand 0: , .I^ ~^~CsttcosXsin9AiT2sin9cos6cos2X=0, (14-.ol.ttJ1yry.,- Thesecond equation shows that thespinsisconstant. With 122neglected, the firstequation maybewritten (14,313)S+n*sin (9=0, where /CsiTcos Xn-i(14.314) n=J- Now (14.313)istheequation ofmotion ofasimple pendulum. Itispossiblefortheaxis ofthegyroscope togoright round thehorizontal circle, but iftheinitial values of6and 6aresmall, themotion willbeoscillatory. Theimportant fact isthis: Theaxisofthegyroscopeoscillates symmetrically about thedirection 6=0.Hence, bybisecting theangle ofswing, wemay find thenorth. The gyroscopetherefore acts asagyrocompass. SEC. 14.4] MOTION OFARIGIDBODY 447 indicating the true north; themagnetic compass, ofcourse, indicates themagnetic north. Forsmalloscillations, theperiodic time ofswing forthegyro- compassis (14.315)r==2wLAv v 'n \Csl2cosX Since 12issosmall(1revolution persidereal day=2r/86,164 radians persecond), thespin ofthegyroscope (s)must begiven alarge value inorder tomake rreasonably small. IfX= ?r, theperiodic timebecomesinfinite, andthegyrocompassfailsto function; butthat isonly tobeexpected, forthepoints inques- tionaretheNorth andSouth Poles. Exercise Assuming themass ofthegyroscope concentrated inathin ring, findthenumber ofrevolutions persecond requiredforaperiodic time of10seconds atlatitude 45. 14.4.GENERAL MOTION OFARIGIDBODY The general motion ofarigidbody consists of(i)motion ofthemass center and(ii)motion relative tothemass center. Theequations governing these have been given inSec. 12.4. But itwould bewrong tosuppose thatthedetermination ofthe general motion always splits intotwo parts aproblem in particle dynamics andaproblem inthedynamics ofabody with afixed point. Constraints make thetwoproblems interlock, andcomplicationsarise.Wecannot give ageneral plan for thesolution ofallsuchproblems butshalldetermine themotions oftwosystems asexamples. Themotion ofabilliard ball. Abilliard ball isstruck byacue.Attime t=0,wesuppose thattheball isincontact withthetable;itscenter hasahorizon- talvelocity qo,andtheballhasanangular velocity G>O.Wewish tofindthesubsequent motion oftheball. Ifthetable were perfectly smooth, thecenter oftheballwould retain thevelocity q ,andtheangular velocity wwould alsobe retained. Butweshallassume thetable toberough, with a coefficient ofkinetic frictionju. Atageneral timet,thecenter hasahorizontal velocity q, andtheballhasanangular velocity<o.LetKbeaunitvector 448 MECHANICS INSPACE [SEC. 14.4 drawn vertically upward (Fig. 157). The reaction ofthetable ontheballatthepoint ofcontact Pmaybewritten whereRisthemagnitude ofthenormal reaction andFtheforce offriction; F,ofcourse, actshorizontally. TheweightismgK, where raisthemass oftheball. Thus theequationofmotion ofthecenter is,by(12.410), (14.401) mq=F+(R-mg)JL. But, since theballremains incontact with thetable, theacceleration ofthe center hasnovertical component. ThusR=mg,andwehave (14.402) mq=F. FIG. 157. Ball slipping on atable.Since every axisat isaprincipal axis ofinertia, theangular momentum about Oish=7wA-2 co,where kistheradius of gyration about adiameter. Thus theequation formotion relative tois,by(12.411), (14.403)roA-%=-aKX(F+KK)=-aKXF, where aistheradius oftheball. Inthevector equations (14.402) and (14.403), there are actuallyfivescalar equations. There areseven unknowns, viz., twocomponents ofq,throecomponentsof<o,andtwocomponents ofF.Thus, twomore equations arerequired; they arefurnished bythelawofkinetic friction, aslongasthere isslipping betweon theballandthetable. Thislaw tellsusthatFactsinadirection opposite tothevelocityoftheparticleoftheballatP,andthat (14.404) F=nR Thus, (14.405) F-- whereq'isthevelocity oftheparticle atP,givenby (14.406) q'= c*X(-aK). SEC. 14.4] MOTION OFARIGIDBODY 449 Hence, using (14.402) and(14.403), weget (14.407) mq'=F+~(KXF)XK sinceKF=0.Thus, by(14.405) and(14.407), thederivative ofq'hasadirection opposed toq'.This implies that, aslong asslippingistaking place, thevectorq'hasafixed direction. LetIbeaunitvector inthisfixed direction. Then (14.408) q'=01, F=-/impl, and,by(14.407), themagnitudeofq'changes according tothe equation (14.409) <?'=- Hence, whereq'Qisthemagnitude ofqj,theinitial velocityofslipping, viz., (14.411) qj=q()-a<oXK. By(14.410), weshallhave</=when (14.412)t=-^--r-i^; v 'Ma2+ A-2' atthisinstant slippingceases. Itiseasy toseethat, once slip- ping ceases, themotion becomes asimple rolling inastraight line atconstant speed,for(14.402) and (14.403) aresatisfied by constant values ofqandw,withF=0. There isapoint ofinterest inconnection with themotion of thecenter before slippingceases. By(14.402) and (14.408), wehave (14.413) 4=-M0I- Thismeans that theacceleration ofthecenter oftheball is constant indirection andmagnitude, andsothecenter describes aparabolic path aslong asslipping persists. 450 MECHANICS INSPACE [Sac. 14.4 Theabove results hold forany ballinwhich there isaspheri- callysymmetric distribution ofmatter. Iftheball issolidand homogeneous, weput2=fa2 . Themotion ofarollingdisk. Everyone knows thatachild's hoop, orarolling coin, acquires stability from itsmotion. Ifthehoop orcoin rolls slowly,itwill start towobble violently, but ifitrollsfast,itcanpassoversmall obstacles without being upset. We shallnow discuss such motions, idealizing forsimplicity tothecasewhere thebody has asharp rolling edge.Wecantreat thehoopandthecoin (or indeed anybody with asharp circular edge, possessing anaxisandaplane of symmetry)inasingle argument byusing general symbols formoments ofinertia. Forpurposes ofreference, however, we shall usotheword "disk." Figure158shows thedisk inageneral position; Pisthepoint ofcontact with theground, whichweshallsuppose rough FIG. 158. Dibk rolling onenough toprevent slipping. Let 6be aplane.^mc iination oftheplane ofthedisk tothe vertical, and$theangle between afixed horizontal direction andthetangent tothediskatP.Let(i,j,k)beaunit orthogonal triad, kbeing perpendicular tothedisk atitscenter and ilying along theradius toward P] jistherefore horizontal and liosinthepianoofthedisk. Forthevelocityofthecenter andtheanguhir velocity ofthe disk,wemay write q=ui+v]+wk,o>=o>ii+oj2j+wsk. These twovectors arenotindependent, because theparticle at Pisinstantaneously atrest. This gives thecondition q+o>Xai=0, where aistheradius ofthedisk; or,inscalar form, (14.414) u=0, v+aojs=0,waui=0. These equations determine qwhen o>isknown. Now theangular velocity Qofthetriad arises solely from SEC. 14.4] MOTION OFARIGIDBODY 451 changes in6and <.Theformer gives anangular velocity 0j,andthelatter anangular velocity <j>about OQ,thevertical through 0.Thus, (14.415) a=-cos fa- 6j+sin6<k. Buttheangular velocities ofthediskandthetriad differ only inthekcomponent. Therefore (14.416) coi=-cosB<l>,o)3=-0. Forthereaction oftheground, wewrite (14.417) R=flii+#2j+ft3k. By(12.410) and(12.411), thetwovector equationsofmotion are (mq=R+?rc<7(cosisin6k), }h=aiXR, wheremisthemass ofthediskandhistheangular momentum about 0;wehave h=Acoii+.<4co 2j+Ow3k, AandCbeing transverse andaxialmoments ofinertia atO. Since, by(12.306), (14.419) q=-wi+ j+ibk+aXq, thefirst of(14.418) gives thethroe scalar equations (m(uOw sin <<>)=li\ -f-mgcos0, m(v+sin6fat-fcos<M=72a, m(?/?cos fa) \611) 7?3 ingsin 0. By(14.414) and (14.116), weeliminate?/, *>,i/Jandobtain (ma(02+sin<w3)=/?i+^.<7os0, -?/i<z(w3 +cos00<^)-K2, ma($ cos 0a?3)=/?3 7??^rsin 0. Turning now tothesecond of(14.418), wehave (14.422) h=Aciii+A^j+6'wjc+ftXh, andsowegetthethree scalar equations (Awi<70co 3Asin^w2=0, Aw 2+Asin <wi+(7cos <co3=-a/^ 3, Cd)3Acos <a>2+A0a)i= 452 MECHANICS INSPACE [SEC. 14.4 By(14.410), thesebecome (14.424)A-~(cosB<)+C0o>3-Asin 0<=0, Associating these equations with (14.421), wehave sixequations forthesixunknowns 6,<,cos,RitRz,Rz-They aretheequations (12.412) applied toourspecial problem. Before proceedingtodiscuss thestability ofthediskrolling straight ahead,letusconsider simple steady motjons satisfying theequations (14.421) and(14.424). Themost obvious solution is (14.425) f'"'* v IRl=-mg,=constant, 0)3=constant, This isthestraight-ahead motion,inwhich theplaneofthedisk isvertical. Another simple motion isgivenby (14.426)=constant, <#=constant,co3=constant. Thecorresponding reactions are,by(14.421), IHI=m(a sinB<j>&3gcos0), #2=0, /^s=m(acos6^>oj3+gsin0). When wesubstitute in(14.424), thesatisfaction ofthese equa- tions requires (14.428) (C+ma2 )cos6<a>3+mga sin6=Asin cose<2 ; thiscondition must besatisfied bytheconstant values of0,<, w3,inorder thatthesteady motion may exist. Itis,ofcourse, arolling inacircular path. Exercise.If,inthemotion givenby(14.426), and <aresmall, show that thetimetaken tocomplete thecircular path isapproximately Letusnow discuss thestability ofthe rolling disk.We suppose thedisktobeslightly disturbed from thesteady motion givenby(14.425). Inthedisturbed state thefollowing quanti- SEC. 14.5] MOTION OFARIGIDBODY 453 tiesareassumed tobesmall, since they vanish inthesteady motion: (14.429) 0,6,S,<,<,o>3,R1+mg,RZlR9. With only first-order terms retained, (11.421) and (14.424) become (0=#1+nig, A<t>+(70o>3=0, mau*=-#,, AS-Cfa*=aR 9, maS mafas /?a+mgO, Cu* aR>>. From thesecond and lastequations weseethat wa=constant. Elimination of <andRsfrom theother equations gives (14.431) A(A+ma*)8+[C(C+ma2 )o>S-Amga]0=a, where aisaconstant ofintegration. Obviously thecondition forstabilityis (14.432)C(C+ma2 ) 14.6.SUMMARY OFAPPLICATIONS INDYNAMICS INSPACE- MOTION OFARIGID BODY I.Rigid body with fixed pointunder noforces. (a)Forageneral body, themotion isgiven byrolling the Poinsot ellipsoid ontheinvariable plane; there isananalytic solution interms ofelliptic functions. (b)Forabody withanaxis ofsymmetry, thePoinsotellipsoid isofrevolution, andthemotion isgiven byrolling theright circular body coneontheright circular space cone ataconstant rate. II.Thespinning top. (a)Steady precession (p)with fastspin (s): (14.501)s=-y- (approximately).op (b)General motion expressible interms ofelliptic functions. (c)Sleeping topstable if (14.502)s*> 454 MECHANICS INSPACE [Ex.XIV III.Gyroscopes. (a)Afinite impulsive couple, applied toafast-spinning gyroscope,alters thedirection oftheaxisofrotation only slightly. (6)Gyroscopic coupleGrequired tomaintain precession p: (14.503) G=Csp. Couple>Precession Spin. (c)Gyrocompass: (14.504) r=27rT, v 7cosX IV.General motion ofarigid body. (a)Center ofaslippingbilliard balldescribes aparabola. (b)Rolling disk isstable ifitsangular velocity wsatisfies (14.505)o,*>- cT<f+^ EXERCISES XIV 1.Acircular disk, pivoted atitscenter,issetspinning with angular velocityo>about alinemaking anangle awith itsaxis. Find, interms of wanda,thetimetaken bytheaxisofthedisk todescribe acone inspace. 2.Agyroscope canturn freely about itsmass center which isfixed. Initially,itissetspinning about itsaxis,which isthen struck perpendicu- larly. Find theangular velocity immediately afterimpact interms ofthe initial spin, themoments ofinertri, themagnitude oftheimpulsive force,and itsdistance from 0.Draw adiagram showing thedirection oftheimpulsive force, theangular velocity, andtheangular momentum justafter impact. 3.The center ofasquare plateisfixed. Ifatacertain instant the angular velocity vector makes anangle of30with thenormal totheplate, findtheinclination oftheangular momentum vector tothenormal. 4.Arigidbody turn?? about afixed point under theaction ofasingle forceF(inaddition tothereaction atthefixed point).Iftheextremity oftheangular momentum vector, drawn from O,liesinafixed planeP throughout themotion, show thatFintersects orisparallel totheperpen- dicular dropped fromOonP. 5.Aheavy homogeneous right circular cone spins with itsvertex fixed. The axisofthecone is4in.long,andtheradius ofthebase is2in.The axismaintains aconstant inclination tothevertical andcompletes arotation about thevertical in5sec. Findapproximately thenumber ofrevolutions persecond oftheconeabout itsaxis. 6.Alamina turns freely under noforces inthree-dimensional motion about itsmass center, which isfixed. UseEuler's. equationstoprove that thecomponentofangular velocity intheplane ofthelamina isconstant in magnitude. Ex.XIVJ MOTION OFARIGIDBODY 455 7.Arigidbody turns about afixed point under noforces. Themomen- talellipsoid atthefixed pointihofrevolution, andtheaxialmoment of inertia Cisgreater than thetransverse moment ofinertia A.Show that ifaistheangle ofinclination oftheinstantaneous axistotheaxisofsym- metry, then thesemiangle ofthespace cone is tun-*(C~4Ltan_ C+Atan2a Noting theinequality C<A-f-tt,satisfied ingeneral bymoments of inertia, show thatthesemmnglo ofthespace conecannot exceed tan-1 {\/2. 8.Make arough estimate ofthespeed atwhich atwenty-five-cent piece must rollinorder that itsmotion maybestable. 9.Asolidhomogeneous cuboid ofedges 2a,2a,4acanturn freely under noforces about itscenter,\\hich isfixed. Itissetspinning with angular velocity uabout adiagonal. Kind thesemivertical angle ofthecono described inspace bythelinethrough thecenter parallel tothelonger edges, andshow thatthetime takenbythis linetomove onceround tho cone is10jr/(w \/ll). 10.Acircular disk ofmassmandradius aismade torollwithout slipping insteady motion onarough horizontal plane,itsplane being vertical and itstrack acircle ofradius b.Itcompletes acircuit intime r.Reduce toa force atthecenter ofthodiskandacouple,theforcesystem (including weight andthereaction oftheplane) which must actonthedisk inorder that thismotion maytake place. Thecomponents oftheforceandthecouple aretobeexpressedinterms ofa,&,r,M. 11.Arigidbody withanaxisofsymmetry canturnabout itsmass center. There actson itafnction.ilcouple, Xo,where o>istheangular velocity vector andXapositive constant Show thattheaxialcomponent ofangular velocity isreduced tohalf itsoriginal value inatime (Clog.2)/X,whereC istheaxialmoment ofinertia. If(/exceeds thetransverse moment ofinertia A,show alsothatthesemiangle ofthebody cone decreases steadily. 12.Anegg-shaped solid ofrevolution rolls insteady motion onarough horizontal plane, theaxisoffigure being horizontal. Establish therelation (a2-fk*)ns-abn*+bg, where aistheradius ofthegreatest circular section, bthedistance ofthe mass center from this section,Atheradius ofgyration about theaxisof figure, andn,sthevertical andhorizontal components ofangular velocity. 13.Prove thatatopcannot move insteady precession with spinsand inclination tothevertical, unless CY2s2>4Amga cos B. 14.Asolidcone ofheight bandsemivertical anglearolls insteady motion onarough horizontal table,thelineofcontact rotatingwithangular velocity 456 MECHANICS INSPACE [Ex.XIV il.Show thatthereaction ofthetableonthecone isequipollent toasingle forcewhich cutsthegenerator ofcontact atadistance '$cosaH---cota from thevertex, whore kistheradius ofgyration oftheconeabout agenera- tor.Deduce thatthegreatest possible value for 12is astheconewould overturn iiS2exceeded thisvalue. 15.Acircular disk ofradius aspinsonasmooth table about avertical diameter. Prove themotion isstable iftheangular velocity exceeds 2\/g/a* 16.Acircular disk ofradius arollsonarough horizontal planeinsteady motion. Thespeedofitscenter isqtt,and itsplaneisinclined tothevertical ataconstant angle0.Show thattheradius rofthecircle described bythe center ofthedisk satisfies theequation 4gr2Qqlrcot6qlacos 0; deduce that,when issmall, 3'/.1 .ir=approximately. 200 17.Arigidbodycanturn freely about asmooth axis, forwhich itsmoment ofinertia is7. Itisacted onbyacouple ofconstant magnitude G,applied in such away that,when thebody hasturned through anangle 0,thevector representing thecouple makes anangle withthoaxis. Ifthebody isini- tially atrest,find itsangular velocity when ithasturned through aright angle. Iftheaxis isanaxisofsymmetryofthebody,findthereaction exerted bythebodyontheaxiswhen ithasturned through anangle0. 18.Alight axleLcarries twogyroscopes; Listheircommon axisofsym- metry, andtheycanturn freely about it.Lissomounted that itcanturn freely about afixed point ^on it,halfway between themass centers ofthe gyroscopes. Find aquadratic equationtodetermine theangular velocities ofsteady precession under theaction ofgravity,interms ofthefollowing constants: m,w',themasses ofthegyroscopes, C,C",their axialmoments ofinertia, AjA',their transverse moments ofinertia attheirmasscenters, 8,s',their spins, 2a,thedistance between their centers, 0,theinclination ofLtothevertical. 19.Arigidbody turns about afixed point under noforces. Show that, relative tothebody, theextremity oftheangular momentum vector hmoves Ex.XIV] MOTION OFARIGIDBODY 457 onthecurve ofintersection ofthesphere *s+yz+s2=K andthecone theaxesbeing principal axes ofinertia, Sketch thecurves, taking A>R>Candconsideringallpossible values of22'A2 . 20.Atopisspinning about itsaxiswhich isvertical. Itisatthesame time sliding over asmooth horizontal plane with velocity qu.Thevertex strikes asmallsmooth ridgeontheplane,thedirection ofthemotion being inclined atanangle atothedirection oftheridge. Hthecoefficient of restitution fortheimpactise,findtheangle atwhich thevertex rebounds from theridge interms of</n,a,e,andtheconstants ofthetop. Find also thedirection ofmotion ofthemass center immediately afterimpnct 21.Athinelliptical plate ofsemiaxosa,b(a>6)canturn freely about its center, which isfixed;itissetinmotion withanangular velocity nabout an axis initsplane equally inclined totheaxes oftheellipse. Show thatthe instantaneous axis willagain beintheplane oftheplateafter atime where _A"" 22 22.Athinhemispherical bowl ofmassmandradius astands onasmooth horizontal table. Ahomontal impulsiveforce; ofmagnitude f*isapplied along atangent tothenm. Find themagnitude anddirection oftheveloc- ityinstantaneously impartedtothepointofthebowl incontact with the table. Show that,nomatter how largePmay be,therimofthebowl willnever come intocontact with thetable. 23.Arigidbody withanaxis ofsymmetry kismounted sothat itcan turn freely about itsmass center, which isfixed. Toagiven point onthe axisofsymmetry there isapplied aforceFK,whereKisafixed unitvector andFagiven function oftheangle between kandK.Show thatthe system isconservative, andobtain equations analogous to(14.227) and (14.228). Show thatkoscillates between tworight circular cones having Kfortheircommon axis. CHAPTER XV LAGRANGE'S EQUATIONS 16.1.INTRODUCTION TOLAGRANGE'S EQUATIONS Thequestion must have occurred tomany people:Ifscience keeps ongrowing atitspresent rate,how aresucceeding genera- tions ofstudents tokeepupwith it?Wemay findapartial answer bylooking back atwhat hashappened during thepast twohundred years orso. First, there hasbeen thedevelopment ofspecialization a broad specialization into subjects (pure mathematics, applied mathematics, astronomy, physics, chemistry), followed bya narrower specialization into branches (differential geometry, hydrodynamics, spectroscopy, tomention afew). Each branch isnow bigenough toprovide work foralifetime. Astillnar- rower specializationisnotapleasant prospect, forintensive work inarestricted range becomes intime uninteresting and sterile. But, sidebysidewiththegrowth ofspecialization, wefindan increasing tendency tousemathematical methods. Mathe- matics gives toscience thepower ofabstraction andgeneralization, andasymbolism thatsayswhat ithastosaywith thegreatest possible clarity andeconomy. Themathematician, penetrating deeply into thestructure oftheories,isoften able todetect common features, notobvious onthesurface. Inthisway, hebreaks down thebarriers between restricted fieldsandbrings the specialists into contact with oneanother. Further, the mathematician cancompress amass ofdescriptive theory into afew differential equations andsogreatly reduce thebulk of science. Long before science reached itsmodern state ofcomplexity, Lagrange invented auniform method ofapproach foralldynami- calproblems. Thismethod hasformed- thebasis fornearly all work onthegeneral theory ofdynamics and isthefoundation onwhich quantum mechanics isbuilt. Inthemore elementary 458 SEC. 15.1] LAGRANGE'S EQUATIONS 459 parts ofmechanics, ithasnotyetsupplanted themore direct andphysical approach, because itsrather abstract andgeneral character hasmade itappear difficult. However,itseems probable that astime passes themethod ofLagrange willwork itswayfrom theendtothebeginning oftextbooks onmechanics. Theeasier, butmore cumbrous, methods arebecoming aluxury forwhich wecannot afford thetime. Instead ofproceeding atonce toLagrange's equationsin their fullgenerality, weshall start with thecase ofaparticle inaplane. Much ofthedifficulty ofunderstanding themethod maybeovercome byastudy ofthiscomparatively simplecase. Lagrange's equations foraparticle inaplane. Consider aparticle ofmass m,moving inaplane. LetOxy berectangular Cartesian axes,and letX,Ybethecomponents oftheforce acting onthe particle. The usual equations of motion are (15.101) mx=X, my=Y. Now letqi,q%beany curvilinear coordinates(e.g., polar coordinates). Itwillbepossible toexpress xandyinterms of<7iandq^andsowemay write (15.102) x=x(qi, (72), y=y(qi, <?2). These relations hold forallvalues ofthetimet,andso The partial derivatives occurring here arefunctions ofq\,q^we cancalculate themwhen thefunctions (15.102) aregiven. Looking at(15.103)inaformal wayandforgetting thatqi isactually thederivative of</i,wemayregard them asequations expressing thetwoquantities JT,yasfunctions ofthefourquanti- tiesqi,q2,qi,qz]wemay expressthisbywriting (15.104) *=f(q ly?2,qi,&), y= flffai, 2,ffi,&) Ifwespeak ofthepartial derivatives dx dx dx dx dqi dq* dqi dq2 460 MECHANICS INSPACE [SEC. 15.1 orthecorresponding derivatives ofy,weunderstand thatthey arecalculated from (15.104),allthequantities q\,q^ ft,fa being treated asconstants, except theonewith respect towhich wedifferentiate. Butthefunctions in(15.104) arethesame asthose in(15.103), andso Furthermore, dx/dq\isafunction of</i,q2;so,following the motion oftheparticle, ddx' *x ' d2* . But,ontheother hand,ifwedifferentiate the first of(15.103) partially with respect toqi tweget /tr^rtpr\ (15.107) This isequaltotheexpressionin(15.106). Hence, assembling this result with theother results obtained byusing q%andy, wehave ^.^5.=^ ^L^L^dy. (15.108)dtdgi- dqi dtdqi'dqi ddx dx ddy dy dtdqz dq% dtdq% dq% Theequations (15.105) and (15.108) arefundamental inthe development ofLagrange's equations. The kinetic energy oftheparticleis (15.109) T= Ifwesubstitute forxandyfrom (15.103), weobtain afunction of<?i,72,q\,qz, (15.110) T=T(q l9g2,ft,ft). Actually thisfunction isoftheform (15.111) T=iH!+2/iftft 4 where a,A,6arefunctions ofq\9q%. SEC. 15.11 LAGRANGE'S EQUATIONS 461 Now, in(15.109), Tisexpressed asafunction ofx,y\by (15.103), x,yarefunctions ofqi,#2,qi,#2;hence weobtain, using (15.105), n*>119^ ~ 4-^L**$ (*^ dqi~ dxdqi+ dydji -dx ,.dy =mx~ hmy- ddT.dx ,.dy t.dxmx+my~+mxTherefore, by(15.108) and(15.109), wehavo (15.113) Subtracting andusing theequations ofmotion (15.101), wegeta?7 r=mx---\-rny--u Thereis,ofcourse, asimilar equation with<?2instead ofqi. Anysmall virtual displacement* oftheparticle corresponds to incrementsdqi,dq2inthecoordinates #1,q%.Thecorresponding increments inx,yare (15.115) te=|fe+g*,- g-^+|259, Theworkdone inthisdisplacementis (15.116) dW=Xdx+Ydy, or (15.117) dW=Qidqi+Q28qz, where (15.118)Q^X^ +Yg-,Q^xf- +Y^-.dqi dqi dq2 dq2 Hence, wehave thefollowing result: Themotion ofaparticle inaplane satisfiesthedifferential equations *Itshould beemphasized that thisvirtual displacementisarbitrary;it isnottobeconfused withthedisplacement actually occurring inthemotion. Tfwewant torefer tothelatter, wewrite dx,dy,dq\, dq*. 462 MECHANICS INSPACE [SEC. 15.1 ddTdT_nddTdT-n dtdjl~ d~Ul >A*f2~ 5^~y" whereq\,q%areanycurvilinear coordinates, Tisthekinetic energy (expressed asafunction ofq\, <?2,<h,#2)araZ Qi,Q2areobtained fromtheexpression dW,asin(15.117), forthework done inan arbitrary small displacement. These areLagrange's equations ofmotion. The curvilinear coordinatesq\,q*are, ofcourse, generalized coordinates, asdiscussed inSec. 10.6; thequantities Qi,Q2are thegeneralized forces [cf.(10.708)]. Itmust beclearly understood that theLagrangian method only provides thedifferential equations ofmotion;itdoesnot solvethem. Itistrue that themethod does givesome hints helpful forsolution, butthat isamatter intowhich wecannot gohere. Example. Consider aparticle ofmassmmoving inaplane, under an attractive force /xw/r2 ,directed totheorigin ofpolar coordinatesr,9. If wetake asgeneralized coordinates tfi=r, 92=0, anddenote thegeneralized forces by ft,O,theequations ofmotion (15.119) read Now T=%m(r2+r22 ), andso (15.121)=wir,=mrtf2 ,-^=rar2 0,-=0. TofindRandO,wehave (foranarbitrary displacement 5r,50) ft5r-fO5fl=dW==~ 5r, andso (15.122) R--^~, 6-0. Substituting from (15.121) and(15.122) in(15.120), weget (15.123) mf mrd*= ^-r--5-(mr2d)=0. rz fdt These equations arethesame as(5.104); thepresent method ofobtaining them issimpler than themethod used earlier. SEC. 15.2] LAGRANGE'S EQUATIONS 463 16.2.LAGRANGE'S EQUATIONS FORAGENERAL SYSTEM Lagrange's equations forasystem withtwodegrees offreedom. Wepassnowfrom aparticle moving inaplane toanysystem with two degrees offreedom, with generalized coordinates qit#2(cf.Sec. 10.6). LetNbethenumber ofparticles forming the system, and lettheCartesian coordinates ofaparticle (ofmass mt)beXi, 7/t,Zi(i=1,2, N).ThenXi,y^Ziarefunctions ofgi,#2,andwemay write (15.201) x%=Xi(qi, g2), yt=2/tfei, (72), 2t=ft(qi, #2). Herewehave 3ATequations likethetwoequations (15.102), and weobtain ondifferentiation 3JVequations like(15.103), (15.202) dZi Asamatter offact,thewhole argument forasystem withtwo degreesoffreedom follows very closely theargument fora particleinaplane; thecomplication introduced byhaving 3N Cartesian coordinates, instead ofonlytwo,isnotserious. Thus, ifweuseasymbol tostand foranyoneofthecoordinates %i, 2/t,2i,weobtain, exactly asin(15.105) and(15.108), thefollow- ingequations: n-9n~v a_a dt_d( (15.203)- - ^ ? q Thekinetic energy ofthesystemis N (15.205) T=i2*mffe2+t/? -i andthis isexpressibleintheform (15.206) T=rfei, (?2, tfi, 464 MECHANICS INSPACE [SEC. 15.2 Asinthecase ofthesingle particle, this isaquadratic expression (15.207) T=%(aq\+2A44,+&#), where a,^,barefunctions of#1,q%. Then, by(15.203) and(15.205), ri5208^ar-y/^^, (15.208)-2 + and so,by(15.204), The lasttermontherightisdT/dqi. LetXt,Y%,Z*bethecom- ponentsofforce (external andinternal) acting ontheithparticle, sothat (15.210) wiA=Xt,my,=Yit mzi=2f. Then (15.209) maybewritten Now,ifQi,(harethegeneralized forces, sothattheworkdone inageneral displacementis (15.212) BW=QiBqi+Q2tq*, itisclearfrom (10.707) that theexpression ontheright-hand side of(15.211)isprecisely thegeneralized force Qi. Thus, associating with (15.211) thecompanion equation in 2,wehave Lagrange's equations ofmotion forasystem with twodegrees offreedom, (15213)ddT-dT- ddT-W_~(15.216)QJ[-V>> Jt^ 3q2-V*' (SW=QiSqi+Q2tqj. SEC. 15.2] LAGRANGE'S EQUATIONS 465 Theform ofthese equationsisprecisely thesame asforaparticle inaplane; noadditional complexity hasbeenadded byconsider- ingthegeneral system withtwodegrees offreedom, ofwhich a particle inaplane is,ofcourse, aspecial case. When thesystemisconservative, with potential energy V(q\, #2),thegeneralized forces areconnected withVby(10.712). Thus, (15.213) maybewritten .([<W_dT =_37^^_i?! =_5?Z ( 'dtdtfi dql~ dqi dtdq>2dqz dqz' Example1.Woshallnow findtheequations ofmotion ofaspherical pendulum. Letrabethemass oftheparticle andatheradius ofthesphere onwhich theparticleisconstrained tomove. Wetake asgeneralized coordinates where istheangular distance from thehighest point ofthesphere and < theazimuthal angle. Then, T=$wa*(02+sin26<2 ),V=mgacos0, andso AT1dT r)V-=ma*&, ma2sin cos<j>z , ~^r=mgasin9,30 o0 o0 g-,n..*g=0, g=0. Thus (15.214) give, asequationsofmotions ofaspherical pendulum, ma?8 mazsin cos <2=mgasin0, 5T(mo2sin2^)=0. Example2.Consider auniform barhanging byoneendfrom asmooth horizontal rail. Itcanmove only inthevortical plane through therailand isunder theinfluence ofgravity andahorizontal force A'applied toits lowest point. Letusfindtheequations ofmotion. Forgeneralized coordinates, wetake QI=distance ofpoint ofsuspension fromsome fixed point onrail, qz=inclination ofbartovertical. Then, n |[5( (acosq wherem=mass ofbar, 2a length ofbar, kradius ofgyration about mass center. 466 MECHANICS INSPACE [Sue. 15.2 Theabove expression reduces to T=Mq\+2acosq,qfa+(a2+*;)$}]. Thegeneralizedforces aregiven by Qi%-hQ2$?2=X8(qi+2asing2)+ Thus Qi=X, <?2=2Xacos92 andsotheequations ofmotion are,by(15.213), m ~dt^l^~aCS<?2^=^' m-57[acos/?2#i-h(a2+&2 )(?2]+masinq%qifa-2Xacos72 wa0a sinqz. Ifthebarremains nearly vertical, sothat g2issmall, these equations simplify totheapproximate form Lagrange's equations forageneral system. Consider asystem withndegrees offreedom andgeneralized coordinatesqi,q^*qn.Themethod offinding Lagrange's equationsinthis general case differs from themethod given above only inaslightly greater complexity, duetothendegrees offreedom. Weshall givehereanargument complete inessen- tialsbutomitting details which canbesupplied bythetype of argument used earlier.* Letmlyxt,#t,zl(i=1,2, N)bethemassandcoordi- nates oftheithparticle. Then, forr1,2,-n, dTdXj dTdy,dTdzl\ ^\dq rdytdqrdZtdq r) ddT dtddr *AsinSec. 10.6,non-holonomic systems willnotbeconsidered./..dx< ..dyt ..dz> V'^'^*d SEC. 15.3] LAGRANGE'S EQUATIONS 467 This lastequation maybewritten intheform ddT_dT__/YdXivdyt7dz*\ dtdq rdqr- ft^'dfr+'Wr+ *'Wr)' whereX^Ft-,Zlarethecomponentsofforce acting ontheith particle. Thus, by(10.707), wehave Lagrange's equations ofmotion for asystem withndegrees offreedom, !=<2" fr-1,2, ), whereQrarethegeneralized forces, defined bythecondition that thework done inageneral displacement is (15.216) 3W=J)Qr8qr. r=1 //thesystem isconservative, (15.217) Qr=- |(r=1,2, n). Two features ofLagrange's equations should beemphasized. First, there isnounique setofgeneralized coordinates; however wechoose them, theequationsofmotion always have theform (15.215). Secondly, since onlyworking forces contribute to8W, reactions ofconstraint areautomatically eliminated.* 16.3.APPLICATIONS Components ofacceleration inspherical polar coordinates. Although thenormal useofLagrange's method istoobtain equations ofmotion,itmaysometimes beused indirectly togive information noteasy toobtain otherwise. Consider aparticle moving inspace. Letustake thespherical polar coordinates r,6,asgeneralized coordinatesqi,q%,#3.Then,iftheparticle isofunitmass, T=(r*+r*&*+r*sin26 <2 ). (Toobtain this,weneed only thecomponents ofvelocity along theparametric lines.) IfR,0,*arethegeneralized forces, the equationsofmotion are,by(15.215), *Except where forces offriction dowork. 468 MECHANICS INSPACE [SBC. 15.3 rrd2rsin2 <2=R, d -r(r26) r2sin cos <2=0, d/o o -x -E(r2sin26(p)=<. Let/r,/0,/^bethecomponents ofacceleration along thepara- metric lines. These areequal tothecompo,nents offorce in these directions. Hence, equating two different expressions forwork done inanarbitrary displacement, wehave dW=fr5r+far50+farsind<j>=Rdr+650+*5<, andso r=JK=rr02rsin2d>2 (15.301)=1e=i^(r2 0)-rsin cos < 7* 7*nC =_- . rsin rsin eft These arethecomponentsofacceleration along theparametric lines ofspherical polar coordinates. Normal frequencies ofvibration ofasystem withtwodegrees offreedom. LetCbeaposition ofequilibrium ofaconservative system withtwodegrees offreedom. Letuschoose generalized coordi- nates such that</i= <?2=atC.Thekinetic energyisexpres- sible intheform (15.302) T= where a,h,barefunctions of</i,qz.Letus,however, consider only small oscillations about C,sothatq\,qz,qi, (faaresmall. Then theprincipal part ofThastheform (15.302), where a,/i,b areconstants, viz.,thevalues ofthecoefficients forqi=q2=0. Consider now thepotential energy V.Wemay choose C asstandard configuration, sothatV=forq\=#2=0.The expansion ofVinaTaylor series reads (15.303) F-Zfc + t, SBC. 15.3] LAGRANGE'S EQUATIONS 469 where thepartial derivatives areevaluated forq\=q%=0. But,bytheprinciple ofvirtual work (10.714), for#1=#2=0.Hence theprincipal part ofVis (15.304) V=\(Aq\+2Hqiq2+Bq\), where A,H,Bareconstants. Thus, toourapproximation, TandVarehomogeneous quadratic forms withconstant coeffi- cients,Tbeing quadratic inthevelocities andVinthecoordinates. Wehave dT . ,,. dT dVA.u -offi+A*,g=0,-Aft+Hq* dT ,.,,. dT7 AdFy/ ,D =%x+6^2,=0,=Uqi+Bq2, andsoLagrange's equations read Weseekasolution oftheform </i=a.cos(tit+e), q2=/3cos(n+e). When wesubstitute in(15.305) andeliminate aand/3,weobtain fornthedeterminantal equation (15.306)J*I-^-^=0. IfTii,n2aretheroots ofthisequation, thenormal periods (cf. Sec. 7.4)are2ir/ni, 2ir/n^ andthenormal frequencies arcni/2ir, Itis,ofcourse, assumed that theequilibriumisstable. If itwere not,weshould discover thefactthrough theappearance ofazeroorimaginary value forn. The top. Consider atopwith fixed vertex 0.Thesystem hasthree degrees offreedom. Fortwogeneralized coordinates, wetake 0,0,thepolar angles oftheaxisofthetop, 6=being directed 470 MECHANICS INSPACE [SEC. 15.3 vertically upward. Forthethird coordinate, wetaketheangle ^between twoplanes, onefixed inthetopandpassing through itsaxis,andtheother containing thevertical through andthe axis ofthetop.Then theangular velocity hascomponents 0, sin6<,atright anglestooneanother andtotheaxisofthetop, andacomponent 4>+cos6 <along theaxis. Thus thekinetic energyis (15.307) T=%A(6*+sin2 <2 )+|C(^+cos <2 , whereAandCarethetransverse andaxialmoments ofinertia atthevertex. Thepotential energyis (15.308) V=mgacos0, where aisthedistance ofthemass center from thevertex. Lagrange's equations thenread A6-Asin cos <2+Csin<(^+cos <)=mgasin0, d~[Asin2 <+Ccos0(^+cos <)]=0, +cos]=0.(15.309) The lasttwoequations give atonce the firstintegrals /ieoim (Asin2e*+Ccos*<*+cos**)= (15.310) |^+CQS^_ft where aand areconstants. (These areactually integrals of angular momentum.) Whenwesubstitute inthefirstof(15.309), wegetadifferential equationfor (15.311)9+a~ *= sin 0. Ifwemultiply thisequation by 0,integrate once, andput cos=#,weget(14.226). Thedetailed theory ofthemotion thenproceeds asinSec. 14.2. Lagrange's equations forimpulsive forces. When impulsive forces act,there areinstantaneous changes invelocity, without instantaneous changes inposition. In terms ofgeneralized coordinates qr,there areinstantaneous SEC. 15.3] LAGRANGE'S EQUATIONS 471 changes inqr,butnotinqr.Asusual, weapproach impulsive forces byalimiting process, inwhich theforces tend toinfinity andtheinterval during which they acttends tozero.We multiply (15.215) bydtandintegrate over theinterval(o,ti). When ti Jo,thesecond term onthe leftdisappears, andwe have Lagrange's equations forimpulsive forces, (15.312) Aff=&' (r-1,2, -n); hereAdenotes asudden increment andQrarethegeneralized impulsive forces [cf.(8.112)] (15.313) Qr=limf11Qrdt. *i-Xo J** Thesemaybecalculated from aformula analogous to(15.216), (15.314) 8W= where 8W isthework which would bedone inageneral dis- placement bytheimpulsive forces iftheywereordinary forces. TheLagrangian method isparticularly useful forsystems of linked rods, because theimpulsive reactions areautomatically eliminated. Thus, totakeanexample, consider theproblem worked inSec. 8.3(Figs. 99aand 996). Asgeneralized coordi- nateswetakex,y,thecoordinates ofthejoint, and0i, 2,the inclinations oftherods totheir initial line. Then, forthegiven position (6i=02=0), T=$m[x*+(y- orf,)a+k*6\+x*+(y+a02)2+k*t}], where kistheradius ofgyration ofarodabout itscenter. Now ifX,F,61,62arcthegeneralized impulsive forces, wehave 25x+Yby+6150i+02502=P(8y+2a502). Thus X=0,f=P, 81=0, 82- Lagrange's equations give 2mx=0, m(y-a6i)+m(y+a02)=?, -ma(y- a^i)+mk26i=0,^ ma(y+a6*)+mk*6 z=2aP. 472 MECHANICS INSPACE [SEC. 15.4 Hence weobtain, with /c2=a2/3, P P P x=0, y=--> ^=-.-, 2= _.. 'm ma ma 16.4.SUMMARY OFLAGRANGE'S EQUATIONS I.Finite forces. Forsystem with kinetic, energy T,expressed asfunction of (15.401) 4^~%r=On (r=1,2, n). V 7 didQ'r d^r' V ' ' 7 n (15.402) dW=^Qr5gr. (15.403) Qr=~T > forconservative system. II.Impulsive forces. (15.404) A|?=Qr, (r=1,2, n). C/O'j' n (15.405) 5^= ]Qrfyr. EXERCISES XV (TobedonebyLagrange's equations) 1.Find theequation ofmotion ofasimple pendulum, taking inturnthe following generalized coordinates : (i)theangular displacement, (ii)thehorizontal displacement, (iii)thevertical displacement. 2.Find theequation ofmotion ofasphere rolling down arough inclined plane. 3.Find theequations ofmotion ofaspherical pendulum, taking as generalized coordinates thehorizontal Cartesian coordinates ofthebob. Reduce theequations totheir principal parts foroscillations neartheequi- librium position. 4.Four flywheels withmoments ofinertia/j,/2,Is,/4areconnected by light gearing sothat their angular velocities areinfixed ratios n\:nz'.n^n^. Driving torques #1,#2,N*,Ntareapplied totheflywheels. Find their angular accelerations. 5.Show that ifageneralized coordinate (q\)doesnotappear explicitly ineitherTorVfthendT/dqiisconstant throughout themotion. Ex.XV] LAGRANGE'S EQUATIONS 473 Arodhangs byauniversal joint from itsupper end. Foroscillations under gravity, usetheabove result andtheequationofenergy tofinda differential equation ofthe firstorder for0,theinclination oftherodtothe vertical, 6.Apendulum consists oftwoequal barsAB,BC,smoothly jointed atB andsuspended from A.Themass ofeachbar ism,and itslengthis2a. Find thenormal periods forsmall oscillations inavertical plane under gravity, intheform where Xisanumerical constant. 7.Thependulum described inKxercise 6hangs atrest.Ahorizontal impulse Pisapplied atitslowest point. Find theangular velocities imparted tothebars. 8.Theends ofaheavy uniform barofmass 120 Ib.aresupported by springsofequal strength, thebarbeing horizontal. Thestrength ofthe springsissuch thataweightWof100Ib,placed gently atthemiddle point ofthebar,causes ittodescend 1in.Find, totwosignificant figures, the normal frequenciesofsmall vibrations ofthebar(without theweight W), considering onlyvibrations inwhich thesprings move vertically 9.Onasphere, and <arepolar angles. Aparticle describes asmall circle =constant withconstant speed </.Find thegeneralized forces O,4 consistent with thismotion. 10. (\>. ,.vicradynamical system with kinetic andpotential energies where /isagiven function. Bychoosing suitable newcoordinatesq{,g, reduce theproblemofdetermining themotion totheevaluation ofan integral involving thefunction/.Determine qi}qzasfunctions of tif /(*)-x\ 11.Acarriage hasfourwheels, each ofwhich isauniform disk ofmassm. Themass ofthecarriage without thewheels isM.The carriage rolls without slipping down aplane slope inclined tothehorizontal atanangle a,the floor ofthecarriage remaining parallel totheslope. Aperfectly rough spherical ballofmass m'rollsonthefloor ofthecarriage along aline paralleltoalineofgreatest slope. Show thattheacceleration ofthecarriage down theplane is 7M+28m+2m' TFT- 42wT+2m''Sm"' andfindtheacceleration oftheball. 12.Arhombus ofequal rods,smoothly jointed, liesonaplaneintheform ofasquare. Animpulseisapplied toonecorner, along thediagonal through that corner. Find theangular velocities imparted totherods, interms of theimpulse (/*),themass (m)ofarod,andthelength (2a)ofarod. 474 MECHANICS INSPACE [Ex.XV IS.Asmooth circular wire carries abead. Thewire issuspended froma point on it.Find thenormal periods ofsmall vibrations under gravity when thewireswings initsownplaneandthebead slides onthewire. Show that,when thebead isfixed tothewire atitsposition ofequilibrium when freetoslide, theperiod coincides withoneofthesetwonormal periods. 14.Using thefactthatTisahomogeneous quadratic expression inthe generalized velocities gr,show thattheintegral ofenergy T-fV=constant maybeproved asamathematical deduction fromLagrange's equations. Note that,if/isahomogeneous function ofdegreeminx\,x% xn,then r=s1 16.Adynamical system haskinetic energy T- andpotential energy V- Additional generalized forces areapplied.Allthecoefficientsa,h,6,A,H,B,andthek'sareconstants. Show that theenergy sumT+Vdecreases steadily during anymotion, provided kn>0, ifciifc,,>(fci,+/b21)2 . 16.Agyroscope ismounted inalight Cardan's suspension (Fig. 144). Take Eulerian angles simply related tothesuspension, andfindtheequations ofmotion ofthesystem under theaction ofacoupleGapplied totheouter ring,Gbeinginthelineoftheouter bearings. 17.Asystemissaid tohave"moving constraints" when thoconfigura- tion ofthesystemisdetermined bythevalues ofgeneralized coordinates #i#2,* <?nandthevalue ofthetime t.Show that, forsuch asystem, Lagrange's equations hold inthesame form aswhen there arenomoving constraints, butthatthekinetic energy isnolonger ahomogeneous quadratic expressionin,, 2, #. Apply this result tofindtheequationofmotion ofaheavy bead ona smooth circular wire, thewirebeingmade torotate about thevertical diam- eterwith constant angular velocity. CHAPTER XVI THESPECIAL THEORY OFRELATIVITY 16.1.SOMEFUNDAMENTAL CONCEPTS Thehardest part ofasubjectisthebeginning. Once acertain stageispassed, wegainconfidence and feelthat,ifneed be,we could carry onbyourselves. Theprocess oflearningisvery much thesame whether inswimming orinmechanics aninitial feeling ofinsecurityisfollowed byafeeling ofpower. Thesimple things thatwelearn firstarethehardest tochange later. Whether they aremuscular actions ormental concepts, they areusedagainandagain untiltheybecome part ofus.Our bodies orminds have learned tofollow apattern, which canbe broken onlybyaconscious effort. Breaking uptheNewtonian pattern. Wearenowfaced with thetask ofbreaking upthepattern of Newtonian mechanics, tomakeway forthenew pattern of relativity, which weowe toEinstein. This would becom- paratively easyifitweremerely aquestion ofmaking changes inthelaterandmore elaborate parts ofthesubject. Butthat isnotthecase. Thechangeistobemade rightdown inthe foundations inourconcept oftime. Toshowhowfundamental thechange is,weshall describe an imaginary experiment, putting intoopposition thepredictions thatwould bemadebyafollower ofNewton ontheonehandand afollower ofEinstein ontheother. Two clocks stand sidebysideataplace P.They areof thevery finest construction and identical with oneanother. Their readings arethesame, andtheycontinue toruninperfect unison aslong asthey stand sidebysideatP.Oneclock is leftatP;theother isputinanairplane andflown with great speed onalong flight, being finally brought back toPandset upbeside theclock thathasstood thereunmoved. Willthere thenbeanydifference between thereadings ofthe twoclocks? 475 476 MECHANICS INSPACE [SEC. 16.1 The practical physicist will, before answering, make inquiries astothewayinwhich theclock wastreated onthe flight whether itwasknocked about, whether itwassubjected to extremes ofheatand cold,andsoon.Letussuppose thatthe greatest carehasbeen taken, sothat effects duetothese acci- dental causes may beruled out ofconsideration. Then the answers areasfollows: Newtonian theory:Theclocks willshow thesame reading. Relativity theory: Thereadings willnotbethesame. The clock thathasbeenontheflight willbeslow incomparison with theclock thathasstayed athome. Thefollower ofNewton reasons along these lines:Aperfect clock registers thetime.Aflight inanairplane doesnotalter this fact, provided that proper precautions aretaken. Since after theflight each clock registers thetime, theymust agree. Wecannot yetgive thereasoning ofthe relativist; that willcome later inthechapter. Forthepresent, wemust be satisfied with thewords withwhich therelativist would begin hisattack ontheargument oftheNewtonian: There isnosuch thing asthetime, inanyabsolute sense. Itwould beimpossible todecide between thetwopredictions bycarrying outtheexperiment wehave described. The rela- tivist would predict adifference between thetworeadings far toosmall todetect. Theairplane would have toflywithaspeed comparable with that oflight before theeffect would benotice- able. But itistheprinciple that isimportant. Other experi- ments canbecarried outinwhich thepredictions ofthetwo theories aredifferent andthe difference islarge enough to measure;inevery casetherelativistic prediction proves correct. There canbenodoubt thatthetheory ofrelativity gives usa mathematical model closer tonature than theNewtonian model. Wemust therefore payattention tothewords: There isnosuch thing asthetime, inanyabsolute sense. Once that pointis conceded, thebasis oftheNewtonian patternisbroken, andthe wayisopen forrelativity. Theingredients ofrelativity. Thetheory ofrelativityisdivided intotwoparts: (i)thespecial theory; (ii)thegeneral theory. SEC.16.1]THESPECIAL THEORY OFRELATIVITY 477 The special theory deals withphenomena inwhich gravitational attraction plays nopart, while thegeneral theory might becalled "Einstein's theory ofgravitation/* Inthisbook, weshallbe concerned solely with thespecial theory. Atthisstage thereader should glance overChap.Itoconcen- trate hisattention again onfundamental matters. Part, but notall,ofthecontents ofthatchapter willpassover intothe theory ofrelativity, andwemust understand clearly what passes overandwhat does not. Letustherefore start again with ablank sheet andput in,onebyone, theingredients ofthe theory ofrelativity. Firstweintroduce aparticle, understood inthesame sense asbefore. Nextweintroduce &frame ofreference andanobserver init.Theobserver hasameasuring rodwithwhich hecan measure thedistances between theparticles which form his frame ofreference. Ifthedistances between these particles remain constant, theobserver declares that hisframe ofreference isarigid body. Nowweprovide theobserver with aclock. AsinChap. I, this isanapparatus inwhich thesame processisrepeated over andover again, therepetitions defining equal intervals oftime. Theactual mechanism oftheclock doesnotmatter. Wemay think ofitasanordinary watch, driven byaspring andcontrolled byanescapement. Aswehave indicated above, thetransporting ofaclock isan operation whichmay lead tocurious consequences. Weshall therefore notexpect theobserver tocarryhisclock about but shall provide himwithagreatnumber ofclocks, allofidentical construction. These willbedistributed throughout hisframe ofreference andkept fixed init. Wemust notoverlook thefactthat thesynchronization of these clocks raises animportant and difficult question.Ifthere isnosynchronization, theobserver willnotesome strange things ashewalks among hisclocks. Forexample, hemay start at 2:15(bythelocal clock), walk amile,andfindthatthetime is 2:10(bythelocal clock). Under suchcircumstances, inordinary life,onewould putaclock inhispocket andwalk around, setting each local clock ashepassed toagree withtheclock inhispocket. But ifourobserver does thishefinds thefollowing strange result. The clocks which hesynchronizes inwalking outfrom 478 MECHANICS INSPACE [SEC. 16.1 hisbasenolonger agree with theclock inhispocket when heis walking back. This isthesamephenomenon asthat described earlier inthe case oftheclockandtheairplane, andthereason for itwillbe made clear later. The effects aresosmall astobenegligible inordinary life,butourobserver isexpected tobemathe- matically accurate. This description ofthe difficulties ofsynchronization may explain why thetheory ofrelativity hashad forthepopular mindmuch thesame appeal asAlice inWonderland. Familiar ideas areturned upside down. Why doestheobserver notsimply setalltheclocks toshow thecorrect time? Theanswer is:There isnosuch thing asthecorrect time. Werecall that,inChap. I,weintroduced theidea ofanevent something happening suddenly atapoint. Wecarry thisidea over into relativity, where weshallmake extensive useofit. Even though hisclocks arenotyetsynchronized, theobserver isprepared todescribe anyevent byassigning fourcoordinates toit.Ofthese coordinates, three arespatial (x,y,z),andthe fourth(t)isgiven bythelocal clock, i.e.,theclock situated at thepoint where theevent occurs. Galilean frames ofreference. Wehave already seen inNewtonian mechanics theimportance ofmaking aproper choice offrame ofreference. Thelaws of Newtonian mechanics take their simplest form only incertain special frames, which wecalled Newtonian. Similarly, inrela- tivity there areframes ofreference which areparticularly convenient touse. These arecalled Galilean frames ofreference.* They correspond innature torigid bodies situated inremote space, farfrom attracting matter, andwithout rotation rela- tivetothestars asawhole. We shallnowmake thefollowing hypothesis regarding a Galilean frame ofreference: I.AGalilean frame ofreferenceisarigid body, isotropic with respecttomechanical andoptical experiments. *This istheusualname, andnotaverygood one, forGalileo lived before Newton andofcourse hadnoidea ofthetheory ofrelativity. "Einstein frame ofreference" would beabetter name. SBC. 16.1]THESPECIAL THEORY OFRELATIVITY 479 Toexplain this,wenote that "isotropic" means "thesame inalldirections." Theneighborhood oftheearth isnot iso- tropic. Ifwedrop astone, itfalls inadefinite direction and theearth's rotation defines adirection which wecandetect by means ofagyrocompass. However, inapplying thetheory of relativity, wemay often regard theearth asaGalilean frame, foritsgravitational attraction maybesmallcompared withother forces involved and itsrotation maybeofnoimportance. Theassumption thatarigidbodyinremote spaceisiso- tropicisatleast plausible. Toassert that itwasnotisotropic would atonce raise thequestion: Why should anyonedirection beprivileged above another? Thehypothesis refers tomechanical andoptical experiments. Wemust provide theobserver withapparatus toperform these. Weshall therefore givehimmechanisms bywhich hecanexert forces, andlamps andmirrors bywhich hecansendoutflashes oflightand reflect them. InNewtonian mechanics, wehadnooccasion torefer tolight. Optics appeared tobeaseparate subject. Inrelativity, onthe other hand, wehave todiscuss optics andmechanics together. Thesynchronizationofclocks. Space doesnotpermit ustoattempt anaxiomatic treatment of thetheory ofrelativity. Toreach themost interesting deduc- tions quickly, weshall outline some steps inthedevelopment without proof. Thus, weshall only sketch themethod ofsynchronization of clocks inaGalilean frame ofreference. Thesynchronization isdonebymeans oflight signals. Taking theclock attheorigin asmaster clock, theobserver sends outflashes oflight tothe other clocks, fromwhich they arereflected bymirrors back to0. Let tiand 2bethetimes (asgivenbytheclock at0)atwhich a flash leaves andreturns toitafter reflection atapoint A. Inordinary life,weshould reason inthisway:Ifvisthevelocity oflightandrthedistance OA,thelightwould takeatime r/v togoandatime r/vtoreturn. Thus tz t\=2r/v,andthe time ofarrival atAis t=ti+r/v=ti+i(2- i)=i(i+it). Butwecannot usethisargument, because velocityisaderived concept, depending onthemeasurement ofboth distance and 480 MECHANICS INSPACE [Sac. 16.2 time.Weshould bearguing inacircle ifweused velocity to define time.Weshallmerely adopt asdefinition ofsynchroniza- tionthattheclock atAissynchronized when itissettoread s(ti+ 2)attheinstant when theflash strikes it.Bythisrule, alltheclocksmaybesynchronized with theclock at0. Weaskthereader toaccept thefact that, inconsequence of theassumption ofisotropy, thissynchronizationissatisfactory. Thatis,arepetition oftheprocess, withanother clock asmaster- clock, willfind allclocks reading justwhat theyought toread, sothatnochange inthesettingsisnecessary. Thismeans that there isnoconfusion such aswepredicted earlier,inthecase where thesynchronization wasattempted bycarrying aclock about. Theobserver nowhasaserviceable time system. Hecan measure velocities, andinparticular thevelocity oflight. By virtue oftheassumed isotropy, thisproves tobeaconstant, thesame foralldirections. Although wehavemetsomenew ideas inconnection with synchronization, there isnothing neworstrange about thefinal picture ofaGalilean frame ofreference andthetimesystem we have setupinit.Itdiffers innoessential wayfrom theconcept wehave used inNewtonian mechanics. Wedonotencounter therealpeculiar! tiosofrelativity untilweconsider twoGalilean frames ofreference andtherelations between them. 16.2.THELORENTZ TRANSFORMATION Theprinciple ofequivalence. Letussuppose thatwecanshoot arocket right outofthe solar system. Inthisrocket weplace anobserver. When the io'..ket haspassed farbeyond thesolar system,itforms aGalilean frame ofreference. Theobserver haslost.thesense ofmotion hehadwhen rushing past theplanets. Heseesaround him nothing butstars, andthey aresofaraway thattheyappear fixed. Asecond identical rocket isshotoutwithagreater speed and onsuchatrack that itovertakes the first. Initthere isalsoan observer. NowwehavetwoGalilean frames ofreference.' Imagine thatthetwoobservers leave their rockets andtravel independently inspace. One ofthem comes upon oneofthe rockets. How ishetotellwhether itistherocket heoccupied SBC. 16.2]THESPECIAL THEORY OFRELATIVITY 481 before ortheother one? Toanswer thisquestion, heisallowed toperform anymechanical oroptical experiments hechooses. Thesame question inadifferent form occurred tothephysicist Michelson in1881. What heasked might beputthus: Isit possible totelltheseason oftheyear (i.e., theposition ofthe earth initsorbitround thesun)bymeans ofoptical experiments performed onaclouded earth? Theearth atthetwoseasons corresponds tothetworockets (Galilean frames ofreference). Itwasfullyexpected thattheseason could bedetermined inthis way, for itwasthen believed that lightwaspropagated inan "ether," andthedifference between thevelocities oftheearth through theether atthetwoseasons should beameasurable quantity. The question wasput toexperimental testbyMichelson and laterbyMichelson andMorley in1887.* Theexpected result wasnotobtained. Asfarasthisexperiment wascon- cerned, thetwo seasons (Galilean frames ofreference) were indistinguishable. Generalizing from thenegative result oftheMichelson- Morley experiment, wemake ohefBlowing sweeping hypothesis: II.PRINCIPLE OFKQUIVA.LEN ,E.Two Galilean frames of reference arecompletely equivalent forALL physical experiments. This gives theanswer tothequestion raised earlier. The observer isnotable totellwhich rocket hehasfound. No experiment hecanperform will tellhimwhich itisthey are indistinguishable,likeidentical twins. Tothehypothesis already madeweaddanother: III.Any twoGalilean frames ofreference have, relative toone another, auniform velocity oftranslation. The relativevelocity islessthan thevelocity oflight. Wemayrecall that, inNewtonian mechanics, twoNewtonian frames ofreference aresimilarly related, butinthat casethere is norestriction ontherelative velocity. Theassumption thatthe velocity oflightisalimitwhich cannot beexceeded issomething essentially new. Tosumup,wehavemade three hypothesesinall.The first deals with asingle Galilean frame ofreference; thelasttwocon- cerntherelations between twoGalilean frames ofreference. *Foranaccount oftheMichelson-Morley experiment,seeL.Silberstein, TheTheory ofRelativity (Macmillan Company, Ltd.,London, 1924), p.71. 482 MECHANICS INSPACE [SBC. 16.2 TheLorentz transformation. LetSand S'betwo Galilean frames ofreference. (We may without confusion alsousethese letters forobservers in thetwoframes.) Consider anyevent, observed byboth observ- ers.Tothisevent, Sattaches coordinates(x,y,z,t),and S' attaches coordinates (z',y',2',t'}.Indoing this,each observer uses hisownmeasuring rodand clocks. The.event determines thecoordinates, andconversely thecoordinates determine the event. Thus, consideringallpossible events, fournumbers (x,y,z,t)determine anevent, andthat inturn determines the fournumbers (x1 ,y\z',tf ).The lastfournumbers aretherefore functions ofthe first four,andweexpress thisbywriting (16.201) x'=f(x, y,z,0, y'=9(x, y,z,t), z'=h(x, y,z,0, ?=l(x,y,z,t). Such relations, connecting thecoordinates oftwoGalilean observ- ers,constitute aLorentz transformation. Wehavenowtoinvestigate theforms ofthese functions. We shall not,however, suppose that theaxesOxyz andO'x'y'z' aregiven arbitrary directions intherespective frames. Weshall suppose them sochosen thatOxandO'x' lieonacommon line when viewed byeither observer, this linebeing parallel tothe relative velocity ofeither frame with respect totheother. We shall .consider theLorentz transformation only forevents occur- ringonthiscommon line. Thus y=z=yf=z'=0,andthe transformation isoftheform (16.202) x'-f(x,0,t'=l(x, t). Wemust carefully avoid theideathatthere isany"absolute" frame from whichSandSfmaybeviewed. Wemust look at things either astheyappear toSorastheyappear toS'. First, Sseestheparticles ofhisownframe. They arefixed asfar asheisconcerned, andthrough them there passtheparticles of theframe S'.These particlesallmove parallel toOxwith a constant speed 7,thespeed ofS'relative toS.Similarly, to S'theparticles ofhisframe appear fixed, andtheparticlesofS passwith aspeed V,directed inthenegative sense ofO'x'. Todojustice toboth observers,itisbesttodrawtwodiagrams, asinFigs. 159aand 6.InFig.159awetake theview ofSand SEC. 16.2J THESPECIAL THEORY OFRELATIVITY 483 regard Oxyz asstationary; inFig.1596wetake theview ofS' andregard O'x'y'z' asstationary. Thetwoobservers now fixtheir attention onaflash oflight traveling along thecommon lineOx,O'x1 .Itwilladdtothe complexity ofourwork ifweassume thattheunits ofspace and timeusedbySandS'arecompletely independent. Weshall therefore suppose thattheyhavebeen supplied withmeasuring rodsandclocks from acommon stock. Then, byvirtue ofthe principle ofequivalence, thespeed oflight* hasacommon value inthetwoframes. Thiscommon valueweshalldenote byc. ,/yIf TfZ ZrZ Z' (a) (&) Fio. 169. (a)Theframe S'moving relative totheframe S. (&)Theframe <S moving relative totheframe S'. AsSobserves theflash, herecords thetime tatwhich (by hislocal clock) theflash reaches thepositionx.Since thespeed oflightisc,xisafunction oftsatisfying (*)-- But similarly, fortheobservations of$', dt' Thus, forthesequenceofevents given bythepassage ofthe flash,wehave thetwoequations dt*-dx*/c*-0,dtf*-dx'z/c*=0. Now amotion satisfying either ofthese equations represents the passage ofaflash oflightandtherefore must satisfy theother *3.00X1010cm.seer1or186,000 mileseer1 484 MECHANICS INSPACE [SEC. 16.2 equation. Thus,ifoneoftheequationsissatisfied, soisthe other, andtherefore wehave theidentity (16.203)dt'2-dx'2/c2=k(dt2-dx2/c2 ), where kissomeunknown factor. Butbytheprinciple ofequiv- alence wemust alsohave (16.204)dt2-dx2/c2SEEk(dt'2-dx'2/c*). Comparing these two identities, weseethat k2=1,and so k=-f-1or 1.Toseewhich value totake,wefollow the particle 0',fixed in8'.Then dx'=0,andso,by(16.203), the history of0'satisfies But,byhypothesis III,(dx/dt)2<c2 ;hence, k=+1. Accord- ingly, (16.205) dt'2-dx'2/c2=dt2-dx2/c2 . TheLorentz transformation must besuchthat thisidentity holds. Weshallassume thatthetransformation islinear, andthatthe zeros oftime arechosen sothat t=t'=when 0'ispassing through 0.Thuswewrite, inplace of(16.202), (16.206) x'=ax+pt,t'=a'x+P't, where a,p,a',0'areconstants soconnected that theidentity (16.205)issatisfied. Wehave !dx'=adx+ dt,dt'=a'dx+p'dt, dt'2-dx'2/c2=(a'dx+p'dt)2-(adx+pdf)2/c2 =dt2-dx2/c2 , andso,equating thecoefficients ofdx2 ,dt2 ,anddxdt, (16.208) a2-cV2=1,P2-c2P'2=-c2 ,op-cV/3'=0. Letusdefine<t>}0'bytheequations (16.209) sinh=ca', sinh 0'=p/c. Then, bythefirsttwoequations in(16.208), wehave a=cosh 0, p'=cosh<#>', SEC. 16.2]THESPECIAL THEORY OFRELATIVITY 485 andthelastof(16.208) gives sinh (<'-0)=0. Thus<j>'=<,andthetransformation (16.206)is (16210}\xr=xcosh4>+ctsinh <, \ctr=xsinh^+ctcosh 0. Itiseasy toverify directly that thistransformation satisfies (16.205) foranyconstant<#>. Toidentify <,wetake theviewpoint ofSandfollow the particle 0'.For0'wehavexf=0,dx/dt=7,andsodifferen- tiation ofthe first of(16.210) gives (16.211) tanh* =-V/c; hence, (16.212) cosh = Sowehave theLorentz transformation (16.213) x'=y(x-70,*'=7[t- 1 7 Solving forj, ,weget (16.214) x=7(z'" Ifwenowtaketheviewpoint ofSrandfollow 0,wehavex=0, dxr /dt'=7',where 7'isthespeed of$relative toS'.But whenweputx inthe first of(16.214) anddifferentiate, we obtain dx'/dt'=7.HenceV 7,asindeed wemight have anticipated from theprincipleofequivalence. Nowwehave theexplanation whythetheory ofrelativity did notforce itself ontheattention ofmankind long ago. Apart from thehigh velocities ofelectrons, which were notobserved until comparatively recent times, physicists andastronomers havehadtodealonlywith relative velocities very small indeed compared withthevelocity oflight. IfV/cissmall, then7isvery 486 MECHANICS INSPACE [Sflc. 16.2 nearly unity; asF/c 0,theLorentz transformation (16.213) tends to (16.215)x'=x-F,*'= t, asinNewtonian mechanics (cf.Sec. 5.3). Immediate consequencesoftheLorentz transformation. Toaperson accustomed tothinking intheNewtonian way, some ofthepredictions ofthetheory ofrelativity arestartling. Outstanding among these arethecontraction ofamoving body andtheslowing down ofamoving clock. These apparently curious facts areconsequences oftheLorentz transformation (16.213). First, letusconsider ameasuring rodwhich 5'laysdown along hisaxis O'x'. Tohim itisafixedmeasuring rod. IfA,B areitsends, thehistory ofAisasequence ofevents forwhich x'=(X')AJ aconstant, andthehistory ofBisasequence of events forwhich x'=(xf )B,alsoaconstant; thelength ofthe rod is (16.216) L'=(x')B-(x')A. Viewed byS,therod isnotfixed. Aninstantaneous picture, taken bySattimet,showsAat(x)A,say,andBat(x)B.If asked what istheapparent length oftherod,Snaturally says that itis (16.217) L=(x)B-(x)A. Now,bythe firstequationof(16.213), n2itt /(*').=?[(*).-Fl, (16.218) |(x%=7[(x)A_yt]i andsubtraction gives, inview of(16.216) and(16.217), L'=tL, or (16.219; L=Z//T=L'VI-F2/c2<L'. ThusLislessthan Z/;therodappears toStobecontracted in theratio\/l-V2/c2 :1. Wemight expect that,ifSfviewed arodfixed in$,hewould seeanexpansion instead ofacontraction. But ifwecarry out SEC. 16.2]THESPECIAL THEORY OFRELATIVITY 487 thecalculation, nowusing (16.214) instead of(16.213), wefind thesame contraction again. Each observer considers that the measuring rodoftheother iscontracted. Now letusconsider aclock carried along inSr .Let (')A, (t')Bbetworeadings oftheclock. These aretwoevents, both with thesame#',andwith t'=(t')A,tr=(t')B,respectively. Viewed byS,theclock ismoving; letthetimes ofthetwoevents be(t)Al(t)n,respectively, asmeasured inhistimesystem. From thesecond equation of(16.214), wehave rv\ A=7 [')-.H-^-J- Bysubtraction, (0.-M>=T[')B or (16.221) T=yT> whoreTisthetime interval recorded bySand T'thetime interval recorded by8'.Since T'<T,theclock carried by S'appears to8toberunning slow. Just asinthecase of contraction oflength, this result works both ways. Each observer considers theclock oftheother toberunning slow. Space -time. Itdoesnotseem possible atfirst sight todojustice simul- taneously toeach oftwoGalilean observers, foritappears neces- sary totake thepointofview ofeither theoneortheother. This difficultyisovercome byusing aspace-time diagram. There isnothing peculiarlyrelativistic about aspace-time diagram. Wehave used theidea inNewtonian mechanics, asinFig.78,whenweplotted theposition ofadamped harmonic oscillator against thetime. But itisinrelativity thatweget fulladvantage from thisidea. Consider firstoneGalilean observer 8.Draw oblique Car- tesian axesonasheet ofpaper, andlabelthem Qx,fit(Fig. 160). (WeuseQinstead of fororigin, toavoid confusion with the origin oftheobserver 'saxes.) Anyevent which happens onthe 488 MECHANICS INSPACE [SEC. 16.2 axisofOxintheobserver's frame willhave attached toitvalues ofxand t.Itcanthenberepresented byapoint inFig. 160, which isourspace-time diagram. Thehistory ofaparticle moving along Oxwillappear asacurveCinthespace-time diagram. Ifitmoves withuniform velocity, dx/dtisaconstant, andthe curve becomes astraight lineC\. Consider now asecond Galilean observer S'.Instead of making anewspace-time diagram forhim,wesuperimpose his diagram onthat ofS.Butweusenew oblique axes $lx'tf (Fig. 161), sorelated tottxtthatthegeometrical lawoftransfor- mation ofcoordinates intheplane isprecisely theLorentz transformation (16.213). NowanyeventEhas, ofcourse, two Fia.160. Histories ofparticles inthespace-time diagram.Fio.161.-Anevent andthespace-time axes of twoGalilean observers. pairs oflabels (x, ),(#', t')\butsince these areconnected bythe Lorentz transformation, Eappears asasingle point inthespace- timediagram. Infact, thespace-time diagram gives usarepresentation of events independent oftheframe ofreference. Itisthesame situation aswehave ingeometry. Thesides ofapolygon drawn onaplane have equations which depend onthechoice ofaxes. Thepolygonitself issomething absolute. Wemust not,however, rushtotheconclusion thattheordinary methods ofgeometry canbecarried over completely into the plane ofthespace-time diagram. Forexample, inordinary geometry weareaccustomed tospeak ofthedistance between twopoints assomething independent oftheaxes used. Ifwe havetwosetsofrectangular axesOxy,Ox'yfinaplane, thenthe square ofthedistance between adjacent pointsis (16.222) dx2+dy2=dx/2+dy'\ SEC. 16.2]THESPECIAL THEORY OFRELATIVITY 489 Infact, thequadratic expression dx2+dy2isaninvariant. Butinthespace-time diagram dt2+dx2^dt'2+dx'2 . Theinvariant quantity is,by(16.205), (16.223) dt*-dx2/c2=dt'*-dx'2/c2 . Ingeometry, wedenote theinvariant (16.222) byds2and calldsthedistance between twoadjacent points intheplane. This suggests thatweshould giveaname tothesquare root of (16.223). However, theminus signintroduces acomplication, since theinvariant maybenegative, andhence itssquare root imaginary. Soweput (16.224)cds2=dt2-dx2/c2 , where e=+1or1,according astheexpression ontherightis positive ornegative. We calldstheseparation between the events (x }t)and (x+dx,t+dt). Inthecase oftwoevents which arenotadjacent, wedefine theseparation assfds,taken along thestraight linejoining thepoints inthespace-time diagram which correspond tothem. Since dx/dtisconstant along astraight line,weeasily find (16.225)es2=(t,-*2)2-(xl-o-2)2/c2 , where thetwoevents inquestion arc(xi, ti)and (z2,2).Wenote that,ifthetwoevents have thesame x,theseparationissimply thedifference between the^s;iftheyhave thesamet,thesepara- tion isthedifference between theZ'H,divided byc. The lines inthespace-time diagram satisfying oneorother oftheequations (16.226)dt=dx/c,dt=-dx/c, arecalled nulllines, because theseparation between anytwo points onsuch aline iszero. Clearly anull linerepresents thehistory ofaflash oflight traveling along theaxisOxofa Galilean frame,inonedirection ortheother. Sofarwehavemade nohypothesis regarding themotion ofa particle inaGalilean frame. Weshallpostpone thediscussion of motion under aforce toSec. 16.3,butnowaccept thelawthata free particle travels inastraight linewithconstant speed, justasin 490 MECHANICS INSPACE SEC. 16.2 Newtonian mechanics. Thismeans thatafreeparticle traveling alongOxmoves inaccordance with dx where uisaconstant. Itshistory appears inthespace-time diagram asastraight line,withequation (16.227)t=constant.u Weshallnowshowhowthecontraction ofamoving rodand t t' Fio.162. Space- timediagram forthe contraction ofamoving rodandthe slowing down ofamoving clock.*theslowing down ofamoving clock appear inthediagram. InFig.162 the linesa,b represent thehistories ofthe ends ofameasuring rodlying onOx'andfixed inS'.These lines aredrawn parallel to!2', because thex1ofeachendof therod isaconstant. The length //(judged bySr )is proportional totheseparation A'B'. Togettininstantane- ouspicture from theviewpoint of8,wedraw aline parallel toSix(i.e., with tconstant), cutting a,6atA,Byrespectively. Then thelengthL(asjudged byS)isproportional totheseparation AB. ThatL5*L'is evident from thediagram. The lineainFig.162may alsoberegarded asthehistory ofa clock fixed inS'.Initshistory, A',Aaretwoevents, andthe time interval T'between them (asjudged byS')isthesepara- tionA'A.Asjudged by8,however, theinterval between these events isT,theincrease intinpassing fromA1toA;itis,infact, theseparation A'C,whore A'C isdrawn parallel to Sit. Itis evident thatTf^T. There aresome factsabout thespace-time diagram which we leave totheconsideration ofthereader. Why dotheaxes Qx'fnotinterlace with SlxtlWhy doesTappear smaller than T'inFig. 162,whereas wehaveproved thereverse in(6.221)? SEC. 16.3]THESPECIAL THEORY OFRELATIVITY 491 Isittrue, foratriangle inthespace-time diagram, thatthesum oftheseparations represented bytwo sides isgreater than the separation represented bythethird side? 16.3.KINEMATICS ANDDYNAMICS OFAPARTICLE Composition ofvelocities. LetP,Qbetwo particles traveling with uniform velocities MI,uzalong theaxisOxofaGalilean frame ofreference S. What istherelative velocity ofthetwoparticles? InNewtonian mechanics, weshould answer: uzMI.Inrelativity, wesay that this isonly the difference between the velocities. The velocity ofQrelative toPisthevelocity ofQasestimated bya Galilean observer S'traveling along withP,i.e.,using aGalilean frame ofreference inwhichPisfixed. Between SandSrwehave theLorentz transformation [cf.(16.213)] (16.301)x'=yi(x-iii), f=7i*- =__71vT^fA"2" Consider nowthemotion ofQ;forit,dx/dt=uz,and itsvelocity asestimated bySfis /iAQH9\/dx'dx""Uldi-u*~~Ul (16.302) u== dt_Uldx/c*- i This isthelawwhich replaces theNewtonian law, (16.303)u'=Ma-UL Wenote that,ifu\and i/2aresmall compared withc,(16.302) differs verylittlefrom (16.303). Ifwesolve (16.302) foruzandthenmake achange innotation, wegettherclativistic lawofcomposition ofvelocities: Ifaparticle moves with velocity HIinS',andS'hasavelocity M2relative toS, then thevelocity oftheparticlerelative toSis Proper time. LetSbeaGalilean frame ofreference, and letPbeaparticle traveling along theaxisOxwithuniform velocity M.Itmaybe 492 MECHANICS INSPACE [SEC. 16.3 regarded asaparticle ofasecond Galilean frameSr .Consider twoadjacent events inthehistory ofP;thetime interval dt' between them (asestimated byaclock carried withP) is,by (16.213), dt'dtudx/c2 ' Butdx/dt=w,andso dt'=dt\A~u2\/dt2-dx*/c*=ds, where daistheseparation between thetwoevents. Thus the separation equalsthetime interval, asmeasured byaclock carried withtheparticle. Hitherto wehave considered only particles withuniform veloc- ities.Wenow think ofaparticle, traveling with accelerated motion. Itshistory appears inthespace-time diagram asacurve. How does a clock behave ifcarried with the accelerated particle? Ourprevious hypotheses donot tellus;wemust make anewassumption. This assumptionisasfollows: Thetime interval between twoadjacent events in thehistory ofanaccelerated clock is given bytheseparation between these events. This separationiscalled theinterval ofproper timebetween theevents. Thus theelement ofproper time foraparticle moving alongOxwithspeed uisn / x FIG. 163. Space-time dia- gram ofthe histories oftwo clocks. (16.305) where (16.306)ds=dt/y u, 1 VT^^ Nowwecangive theexplanation ofthecurious prediction made early inthechapter regarding thebehavior ofaclock taken onaflight. Figure 103 isaspace-time diagram; aisthehistory oftheclock thatstayed athome. Attheevent A(t=ti)the SBC. 16.3]THESPECIAL THEORY OFRELATIVITY 493 other clock leftanddescribed thespace- timecurve5,returning attheevent B(t=tz).Suppose both clocks read zero atA. Then atBtheclock thatstayed athome reads (since dx= throughout itshistory) (16.307) T=f*ds (along a) -JO Theclock that flewreads (16.308) r= JT*ds (along 6) Thus 7"<T7 ,which proves thevalidity oftheprediction. Equationsofmotion inabsolute form. Ithasbeenremarked thatgravitationliesoutside thescope of thespecial theory ofrelativity. Butthere areavailable other forces bymeans ofwhich accelerated motion maybeproduced. What weshallhave tosayistheoretically applicable tothe accelerated motions ofordinary life,butthedifferences between therelativistic andtheNewtonian predictions arethen fartoo small tomeasure. The differences become appreciable only inthedynamics ofatomic particles accelerated byforces of electromagnetic origin. However, thesame principle applies throughout, andwemay understand itbythinking ofany smallbodyunder theinfluence ofanyforce. Wehave accepted thehypothesis that allGalilean frames areequivalent. Thus, whatever form ofequations ofmotion oneGalilean observer adopts, asimilar formmust hold forany other Galilean observer. Infact, theequations ofmotion ofa particle must beinvariant under theLorentz transformation. IfwetaketheNewtonian equation (16.309) mS=P andapply theLorentz transformation (16.213) togetanequation 494 MECHANICS INSPACE [SBC. 16.3 inx'and',wefindanequation ofquite different form. Thus (16.309)isnotinvariant under theLorentz transformation. Toseewhatform ofequationissuitable, wehave toconsider space-time vectors. InFig.164,wehavetaken apointAinthespace-time diagram anddrawn adirected segment SlA. This isaspace- time vector; itscomponentsinthedirections &r, titarex,t,thespace-time coordinates ofA. Ifweuseother axestoY, the same vector hasdifferent components. Butbetween thetwo sets ofcom- ponents theLorentz transformation holds. Wedefine aspace-time vector as apair ofquantities (J,r)which trans- form, whenwechange axes inspace- time, just like(x,t),i.e.,according toAspace-time vec- tor. (16.310) Vi-v-/c2' Aswehave stated, ourproblemistobuild equations ofmotion invariant under aLorentz transformation. Thekey tothe solution isfound intheidea ofthespace-time vector. Weshall form equations inwhich aspace-timevector isequatedtoaspace- time vector. Consider aparticle moving alongOxwith ageneral motion. Thiscorresponds tosome curve inspace time. Theproper time, measured fromsome initial point,maybetaken asaparameter, andtheequations ofthecurve written x=X(S),t=t(s). Since dsisaninvariant, thepair ofquantities (dx/ds, dt/ds) isaspace-time vector. Weseethisbydifferentiating (16.213). Wecall(dx/ds, dt/ds) theabsolute velocity ofaparticle. Explic- itlywehave,by(16.305), (16.311)dx dsdt SBC. 16.3]THESPECIAL THEORY OFRELATIVITY 495 where u=dx/dt, thevelocity oftheparticle relative tothe Galilean frame ofreference S,corresponding toSlxt. Ifthe particle hasavelocity smallcompared withthat oflight, sothat u/c issmall, thecomponents oftheabsolute velocity areapproxi- mately (u,1). Similarly, (d2x/ds2 ,dH/ds2 )isaspace-time vector. We call ittheabsolute acceleration. Wenow accept, assatisfactory from thepoint ofview of invariance under theLorentz transformation, thefollowing equations ofmotion: t^T rl^t (16.312) m^=X,m,4i=r, as as where moisaconstant (theproper mass oftheparticle) and (X,T)isaspace-time vector, called theabsolute force. Inadifferent Galilean frame ofreference $',with velocity V relative to$,these equations read where (16.313) X'=y(X-VT), T'=y(T-Y =1 7VT^TVc2" Thismay beverified immediately byapplying (16.213) to (16.312). Equations ofmotion inrelative form. Letusnowputtheequations ofmotion (16.312) intoanother form inorder toshow therelation ofrelativistic toNewtonian mechanics. Since ds=dt/y u,these equations maybewritten (16.314)jt(myuu)-X/y uy~(mT)=T/yu. Letusdefine some terms, asfollows: (16.315)Relative mass=mmQyu= VI~u2 Relative momentum =mu. Relative force=P=X/y*=X Relative energy=E=me2 496 MECHANICS INSPACE [Sue. 16.3 Then the firstof(16.314) maybewritten (16.316)~(mi*)=P; inwords, rateofchange ofrelative momentum =relative force. This istheequation ofmotion ofaparticle moving onthez-axis under theinfluence ofaforceP. ItisoftheNewtonian form, but witharemarkable difference. The (relative) mass ofaparticle isnotaconstant;itvaries withthespeed oftheparticle according to(16.315). Ifu/cissmall, thevariation ofmfrom thevaluemisinsignifi- cant,butontheotherhandmtends toinfinity asthespeed ofthe particle (u)tends tothat oflight (c).Noparticle haseverbeen observed traveling withaspeed equal to,orgreater than, that of light. This physical factagrees with thetheory. Ifthespeed ofaparticle were toincrease uptoandbeyond thespeed oflight, therelative masswould become meaningless, passing through an infinite value toimaginary values. Todiscuss thesecond equation of(16.314), letusfirstreturn to (16.224). Thismaybewritten (16.317) since (dx/dt)2<c2foraparticle. Differentiation gives na<MttdtdH (lO.oIo) i;5 5-7 r~5=su-Ndsds2c2dsds2 Thus, by(16.312), (16.319) T~-\X~=0,ds c2ds or (16.320) c2T=Xu=Puy u. Infact, thesecond component ofabsolute force isclosely related tothe firstcomponent. Ifwemultiply thesecond of(16.314) byc2andsubstitute from (16.315) and(16,320), weget (16.321) -jj-=Pu. SEC. 16.3]THESPECIAL THEORY OFRELATIVITY 497 This istheequation ofenergy andjustifies thedefinition ofrelative energy asin(16.315). For(16.321) reads, inwords, rate ofchange ofrelative energy=rateofworking ofrelative force. ThishastheNewtonian form, buttheexpression forenergy (E) doesnotatfirstappear related totheNewtonian kinetic energy. However,ifweexpand bythebinomial theorem, weobtain <16 -322'B and ifu/cissmall, wehaveapproximately (16.323) E=mr2+%m<>u*. This differs from theNewtonian expression forkinetic energy onlybytheconstant mc2 ,which iscalled the restenergy or proper energy. Thequantity Woe2appears oflittle importance here, because Eisdifferentiated in(16.321), andsotheconstant disappears. Theequation (16.321) would stillhold ifwehadadopted the definition (16.324) E=.mc2 ..--mc2 forrelative energy. There are,however, good reasons for preferring (16.322) to(16.324) asadefinition ofenergy. Some of these areconnected with thedisintegration ofatoms andtheir structure. Theknown atomic weights oftheelements* arecon- sistent with theprinciple ofenergy onlyif(16.322)isregarded astheenergy ofaparticle. Inrelativity, massandenergy are nolonger distinct concepts. Evenwhen aparticleisatrest,it hasenergymc2 ,andwocannot convert thisenergy intoanother form without destroying oraltering themassm . Example. Consider aparticle moving onthex-axis under aconstant relative force P,starting from restattheorigin att=0.By(16.316) the motion satisfies 498 MECHANICS INSPACE [SEC. 16.4 Hence, (16.326) ,WoU -Pt. Solving forw,weget (16.327) ucPt 4-wjc* Wenotethatuislessthan cforallvalues oftandtends tothelimiting value casttends toinfinity. Thisbehavioris,ofcourse, quite different from the behavior ofaparticle under constant force inNewtonian mechanics. Since u=dx/dt, (16.327) gives, onintegration, (16.328) x Ifmcislargecompared with Pt,thisreduces approximately to (16.329) x-* *2 , thefamiliar Newtonian formula. 16.4.SUMMARY OFTHESPECIAL THEORY OFRELATIVITY I.There isnosuch thing asabsolute time. II.Lorentz transformation : (16.401) x1=y(x-Vt),t'=7ft-~ 7 III.Space -time diagram. (a)Anevent isrepresented byapoint. (6)Thehistory ofafreeparticleisrepresented byastraight line. (c)Thehistory ofaflash oflightisrepresented byanull line. (d)Theseparation oftwoadjacent events isds,where (16.402)6ds2=dtz-dx*/c* (t=1). IV.Kinematics anddynamics ofaparticle. (a)Element ofproper time formoving particle: (16.403) ds=dt Ex.XVI] THESPECIAL THEORY OFRELATIVITY 499 (6)Equation ofmotion: (16.404) (mow) =P, 7* (c)Energy: (16.405) E=rao7t*c2=mc2+wM2 ,approximately. (16.406) ^=P^ EXERCISES XVI 1.Anairplane setsouttoflyat500miles perhour. Show that itwould have toflyformore than athousand years inorder tomake adifference ofoneone-hundredth ofasecond between thetimes recorded byaclock in theairplane andaclock ontheground. 2.Showthat,ifx/cand taretaken ascoordinates inthespace-time diagram, thehistory ofaflash oflight isequally inclined totheaxes. Draw thehistory ofaflashwhich passes toandfrobetween amirror fixed atthe origin andamirror which moves along theobserver's axisOxwithconstant speed. 3.Prove directly from theformula (16.304) that,ifthemagnitudes of u\anduzareboth lessthanc,themagnitude ofthevelocity relative toSis lessthan c. 4.Two electrons move toward oneanother, thespeedofeachbeing 0.9c inaGalilean frame ofreference. What istheir speed relative toone another? 5.Forsuitably chosen axes intwoGalilean frames SandSr ,thecom- plete Lorentz transformation is x'=y(x-Vt), yf=y,z'=z,t'=y(t-Vx/c*), whereVistherelative velocity ofSandSf . Aparticle, asobserved by5',describes acircle x'z+y'*=a2 ,z'=0,with constant speed. Show that toStheparticle appears tomove inaellipse whose center moves with velocity V. 6.Allelectromagnetic waves travel with thefundamental velocity cin empty space. Aradio station fixed inaGalilean frame ofreference Ssends outwaves. Show that, toanobserver inanother Galilean frame S',these waves atanyinstant formafamily ofnonconcentric spheres. Isitpossible thattwoofthese spheres should intersect? 7.Show thattheLorentz transformation mayberegarded asarotation ofaxesthrough animaginary angle. 8.The history ofamoving particleisrepresented inthespace-time diagram bythehyperbola 500 MECHANICS INSPACE [Ex.XVI Show that d*x 19 dH 19. -3-^=&2 ,-r-i-Wt. ds2'ds* 9.Twoparticles, with proper masses mi,m2,move along theaxisOx ofaGalilean frame with velocitiesUi,M2,respectively. They collide and coalesce toform asingle particle. Assuming thelaws ofconservation of relativistic momentum andenergy, prove that theproper massm*and velocity MSoftheresulting single particle aregivenby ml f1--^ iUi-f?n272i/2 Wl7l T"W272 whereyf=1-w?/c2 ,7^=1-u\/c*. 10.Aparticle ofproper massmmoves ontheaxisOxofaGalilean frame ofreference, and isattracted totheorigin bya(relative) forcem^x. It performsoscillations ofamplitudea.Show thattheperiodic time ofthis relativistic harmonic oscillator is where Verify that,ifc > ,r *2ir/fc (theNewtonian result) ;andshow that if ka/c issmall, 27rA fc2o2\ .j^1+A-~T)'approximately. APPENDIX THETHEORY OFDIMENSIONS Two physicists areshipwrecked onadesert island. After making qualitative observations oftheir surroundings, theywish tomake measurements. Buthereadifficulty arises, forthey havenone oftheusual apparatus ofthelaboratory nometer scale, nosetofweights, noclock.* Everything they require theymust construct forthemselves. Iftheycanagree onthelength ofacertain stick asunit of length, themass ofacertain stone asunit ofmass, andthedura- tionofsome simple rcpeatable experiment asunit oftime,allwill bewell;both experimenters willassign thesamenumber tothe same measurable quantity. Butwhy choose one stick rather than another, onestone rather than another, oneexperiment rather than another? Ifthetwophysicists areobstinate, each inhisown preference ofunits, there isnovalid argument by which onecanpersuade theother toyield. Thisdisagreement concerns only physical measurements. In therealm ofpuremathematics, there iscomplete accord; both agree, forexample, that (1)2+2-4, (x+l)(s-1)-x2- 1, But inthematter ofthechoice ofunits, wemay wellimagine that neither physicistwillyield totheother. Sothey decide towork independently, each constructing hisown apparatus, measuring quantitiesintheunits heprefers, anddeveloping his own results. Iftheywish todiscuss their work,how isoneto interpret theresults oftheother? How fardotheir individual efforts contribute totheconstruction ofacommon science, independent ofthechoice ofunits? These arequestions which belong tothetheory ofdimensions. *Wemaysuppose theskyperpetually overcast, sothattherotation ofthe heavens cannot beused asaclock. 501 502 PRINCIPLES OFMECHANICS Itmayappear strange thatwehavebeen abletopostpone to anappendix thediscussion ofthese important questions. The explanationisthatthetheory ofdimensions becomes necessary onlywhenwewish tocompare results fortwodifferent systems ofunits. Inourwork,wehaveused arbitrary unitsand letters (algebra) instead ofactual numbers (arithmetic). Our results arevalid quite generally andcanimmediately beapplied inany particular system ofunits. Perhaps ananalogy with analytical geometry willbehelpful. Wemaydevelop results true forarbitrary Cartesian axes, e.g., propertiesofconies deduced from ageneral equation ofthe second degree. Wemeet thetheory oftransformations (the analogue ofthetheory ofdimensions) onlywhenweconsider two different setsofaxesatthesame time. Theinvariants ofanalyti- calgeometry areanalogous togeneral physical laws, true forall systems ofunits. Units anddimensions. Thetheory ofdimensions arises from thefactthat unitsmay bechosen arbitrarily. Instead ofarguing over therespective merits ofdifferent systems ofunits (e.g., centimeter-gram-second andfoot-pound-second), letusregard allsystems asequally valid. Each physicist may select hisown units. Thismeans thatheselects apiece ofmatter andsaysthat itsmass isunity, heselects arigid barandsays that itslengthisunity, andhe selects arepeatable experiment andsays that itsduration is unity. Hecannowmeasure masses, lengths, and times, and record them assomany units. Tofindavelocity, hemeasures distance traveled andtime taken, anddivides theonenumber bytheother. Hedeals similarly with acceleration, moment of inertia, kinetic energy, andsoon. Asforforce, there aretwo possible plans: (i)hemay use Newton's lawofmotion intheformP=mitodefine force in terms ofmassand acceleration, or(ii)hemay useaseparate arbitrary unit offorce. Thesecond planisgood instatics, but the first isfarsimpler indynamics andmaybeused instatics also.Weshall accept the firstplan forthepresent discussion. With thisunderstanding, allthequantities occurring inmechanics arebuiltupoutofmass, length, andtime; thephysicist canassign numerical values tothem all,once hehasselected hisfunda- mental units ofmass, length, andtime. THEORY OFDIMENSIONS 503 Two physical quantities mayhave different numerical values andyetbeofthesame type. Forexample, thelinearmomenta oftwo particles mayhave different numerical values, butthey areboth builtupoutofmass, length, andtime inthesame definite way; infact, (2) linearmomentum =masa*length. time Toexpress thismore compactly, weintroduce thesymbolsM,L,Tformass, length, andtime,andwrite symbolically (3) [linear momentum]=[MLT'1 ]. Thesquare brackets aretoremind usthat this isnoordinary equation connecting numbers butasymbolic shorthand toshow how linear momentum involves thefundamental quantities. This iscalled thedimensional notation; wesaythat linear momentum "has thedimensions [MLT~1}" Allthequantities occurring inmechanics maybeexpressed dimensionally inthe form where a, /3,7arepositive ornegative powers, notnecessarily integers. Thefollowinglistofdimensions iseasily verified: [velocity]=[LT~l ], [acceleration]=[LT~*\, [force]= [moment ofaforce] = [linear momentum] =[MLT~l ], [angular momentum]=[ML2jT-1 ], [energy]=[ML*T~*], [angular velocity]=[T7"1 ]* [momentofinertia] =[ML*]. Inwriting down thedimensions ofaphysical quantity, wepay noattention tonumerical factors. Thus thedimensions of andmfarethesame, viz.,[ML2T~2 ]. Wedonotaddorsubtract quantities having different dimen- sions, butwefrequently multiply such quantities byoneanother ordivide thembyoneanother. Therulebywhich weobtain the 504 PRINCIPLES OFMECHANICS dimensions oftheproduct orquotientisobvious from thedefini- tionofdimensions. Itisasfollows: LetQiandQ2bephysical quantities withdimensions [Qi] then ItAHSxaj te]Led IfQiandQ2have thesame dimensions, then andwesaythen thatQi/Qzisdimensionless. Forexample, the circular measure ofanangleisobtained bydividing alength (arc)byalength (radius), and soanangleisdimensionless. Itiseasy toverify that thefollowing combinations arealso dimensionless: forceXtime linearmomentum forceXlength energy moment ofinertiaXangular velocity angular momentum Exercise. Einstein's radiation formula isE=*hv,whereEistheenergy ofaphoton,visitsfrequency, and hisPlanck's constant. Show that Planck's constant hasthedimensions ofangular momentum. Change ofunits. Firstmethod. Letusnowconsider twophysicists Siand$2,whousedifferent units ofmass, length, andtime. When theymeasure thesame physical quantity, theyrecord different results. But,asweshall now see,itiseasy topassfrom onenumerical value totheother whenweknow theratios ofthetwosets ofunits. Forsymmetry, weintroduce athird physicist $o,using a third system ofunits; weshall call hisunits "absolute" for purposes ofreference, without meaning toimply thatthey are inanywaymore fundamental than theunits ofSior#2.Let theunits of*Sficontain m\tl\,andh,absolute units ofmass, length, THEORY OFDIMENSIONS 505 andtime, respectively; and lettheunits of$2containw2,1%,and tzabsolute units. Consider aphysical quantity Qwith dimensions [M*LPTi\. This quantityismeasured bySQ,Si,andSz,with numerical results asfollows: /So Oi >32 Qu Qi Q*. Nowevery unit ofmass recorded bySicorresponds tomiabsolute units, andsimilarly forlength and time. Hence, one /Si-unit ofthequantity measured corresponds tom-flfti* absolute units, andQi5i-units correspond toQimiali^absolute units. Butwe know thatQi/Si-units correspond toQoabsolute units, andso (4) Qo= similarly, (5) Qo= i Comparing (4)and(5),weseethat thelawoftransformation connecting theresults ofSiand$2is (6) or m(7) Ifweidentify theunits ofSwith those ofSi,sothat the absolute units arenowthe/Si-units, wehave mi=1, Zi=1, ti=1, andso m where mz,h, faarethenumbers of$i-units contained inthe $2-units. Thisformula gives thenumber Q2assigned by<S>2toa quantity,interms ofthenumber Qiassigned bySitothesame quantity andtheratios oftheunits. Exercise. Anenergyis362inc.g.s units. What isitsnumerical value inf.p.s. units? 506 PRINCIPLES OFMECHANICS Change ofunits. Second method. Theabove method islogical, butnotgood inpractice. Con- version from onesetofunits toanother isaprocess which we must beable tocarry outquickly andaccurately, andtherules should besimple andeasy toremember. Theformula(7)isbad because itinvolves athirdsystem ofunits, and (8)isbadbecause itisunsymmetrical andhard toremember. Themethod weare about todescribe isthat incommon use. Compare theequations (1)withthefollowing: (9)1meter=100cm.,1Ib.=453.6 gm., 22ft.persec.=15miles perhr. These aretruestatements, butthey differ from(1)inanimpor- tant respect: theequations (1)involve only pure numbers, whereas (9)involve measurable physical quantities. Todis- tinguish them, wemaycall (1)mathematicians' equations (or briefly M-equations) and (9)physicists' equations (orP-equa- tions). Weknow whatwecandowithM-equations according tothemethods ofalgebra andcalculus. There arecertain rules ofmanipulation, which weapply with confidence thatweshall never reach afalse conclusion. Letusboldly apply therules of algebraic manipulation toP-equations, treating suchwords as meter, cm.,Ib.asifthey were ordinary algebraic symbols. A word ofwarning, however thesigns =,+,and aretobe used toconnect only quantities ofthesame type, i.e.,ofthesame dimensions. Wethink againoftwophysicists Siand $2.LetSiname his units Mi,Li,Ti]and letSzname hisunitsM2,1/2,Tz.These arenames (likegm. orcm.), notnumbers. IfSimeasures a length, herecords theresult intheform Q=QiLi; this isaP-equation,inwhichQstands for"the quantity which ismeasured," aridQiisanumber. More generally,ifSimeas- uresaquantity withdimensions [MaUT^] therecords (10) Q=QiMfLJTf, where Q\isanumber. IfSzmeasures thesame quantity, he records (11) Q THEORY OFDIMENSIONS 507 There isnothing novel aboutthis;itiswhatwedowhenwewrite acceleration duetogravity=32ft.sec.~2 acceleration duetogravity=980cm.sec.~2 Nowwebring intooperation ourassumption thattheP-equa- tions (10)and(11)maybetreatod inthesamewayasweshould treat M-equations. Wegetatonce (12) QJlfLJTi* and Ifweinterpret Mi/Mz tomean theratio oftheunitMitothe unitMz,thenMi/Mzisapurenumber infact, themeasure ofMiinterms ofMz.Since Li/Z/2 andT\/T*may alsobe regarded aspurenumbers, (13)isanM-equation, although (12) (from which itwasobtained)isaP-equation. Equation (13)iswhatwohave been seeking aformula to giveQzwhen Qiandtheratios oftheunits areknown. Ifwe lackconfidence init,because ithasbeen obtained byasymbolic method, wecanreassure ourselves byturning back tothe first method; there onlyM-equations were used, andthededuction of(6)and(7)islogically sound. Weseethat(12)ismerely the P-equation corresponding totheM-equation ((>),and (13)isthe same as(7)both M-equations. When weactually carry outaconversion from onesystem ofunits toanother,itistheP-equation (12)rather than the M-equation (13) thatweuse. Itwould, however, bemore correct tosaythatweuseneither. Therein liesthesimplicity ofthesymbolic method; wetreat eachproblem onitsmerits, without having toremember anything, except that itispermis- sible tousethesymbolic method, inwhich words aretreated as algebraic symbols. Theformulas (12)and (13)were obtained only forpurposesofcomparison with (6)and (7). Toshow thesymbolic method inaction, letusconvert an acceleration of32ft.sec."2intomilehr.~2*Firstwewritedown *Itisconvenient towrite each unit inthesingular, toavoid waste of energy indeciding whether tousethesingular ortheplural. This isa mathematical symbolism, and initsimplicity ismore important than grammar. 508 PRINCIPLES OFMECHANICS 1mile=5280ft.,1hr.=3600sec., sothat 1-1 11^ lsec.=^hr. Then, (1ft.)32ft.sec.~2=32 (1sec.)2 =32^ ^3600 32X3600X3600 ., ,_2 5280milehr * =78,545fV mile hr.~2 Aphysicist would round offtheresult. Forhewould think of thenumber 32asobtained bymeasurement carried outonly to two-figure accuracy, andsohewould prefer towrite 32ft.sec."2=79,000 mile hr.~2 This question of"significant figures" hasnothing todowith the theory ofdimensions, andweshall notpursueitfurther. The discrepancy between thetwostatements arises from thetwoways ofthinking mathematical andphysical which wementioned inChap.I. Exercise. Work outtheexercise onpage505bytheabove method. Dimensionless quantities andphysical laws. Ifaquantityisdimensionless, thena=j3y in(13), and therefore Q\=Q2.Adimensionless quantity hasavalue independent ofthesystem ofunits employed. This factmakes such quantities particularly simple tohandle, because any possible confusion regarding units isautomatically eliminated. Anymathematical combination ofdimensionless quantitiesis itself dimensionless. Wecannowanswer thequestions raised inconnection with the twoshipwrecked physicists. Theformulas given above enable theonetointerpret theresults oftheother, i.e.,totransform them into hisown units. Asforthesecond question thebuilding up ofacommon science independent ofthechoice ofunits the THEORY OFDIMENSIONS 509 answer istobefound intheconcept ofthedimensionless quan- tity.Any equation connecting dimensionless quantitiesistruein allsystems ofunits, iftrueinone. Suppose,forexample, that aphysicist (prior tothetime of Galileo) made measurements onafalling body, using some system ofunits oflength andtime.Weassume thathewasable tomeasure theheight hfromwhich thebody fell,thespeed q withwhich itstruck theground, andthetime tittook tofall. Suppose hefound 2?-2h~2j forawhole setofexperiments inwhich hwasgiven different values. Hewould have been justified inregarding thisasa result ofgreat importance, because itholds inallsystems ofunits, since qt/handthepurenumber 2arcboth dimensionless. Asshown above, any equation connecting dimensionless quantitiesisaphysical law, inthesense that itstruth isinde- pendent ofthechoice ofunits. However,itisnotnecessary toexpress aphysical lawindimensionless form. Itismerely necessary that theequation should bedimensionally homo- geneous; i.e.,theterms equated tooneanother must have the same dimensions. This willensure that thelaw istrue inall systems ofunits,iftrue inone. Theconstants occurring inphysical laws usually havedimen- sions. Consider thelawofgravitational attraction (6.501) nGmm' wherePisthemagnitudeoftheforce between particles of mass m,m'atadistance rapart. Tomake thisdimensionally homogeneous, wemust assign suitable dimensions tothecon- stant G.This iseasily done ifwewrite theequation intheform mm Wehavethen [01- =[M-1L*T-2 ]. 510 PRINCIPLES OFMECHANICS Inthe c.g.s. system, G=G.67X10~8gm.-1cm.3sec.~2 Exercise. Form adimensionless combination ofthegravitational con- stant, density, andtime. Applications. Apart from itsuseinthechange ofunits, thetheory ofdimen- sions hasthree important applications: (i)Itsupplies uswithauseful check against slips incalculation. (ii)Itsuggests forms ofphysical laws. (iii) Itenables ustopredict thebehavior ofafull-scale system from thebehavior ofamodel. These applicationswillnowbeexplained. (i)Provided thatwedonotinsert numerical values, thedimen- sions ofevery combination ofsymbols occurring inourwork areobvious. Forexample,ifaisthelength ofapendulum and gtheacceleration duetogravity, then Thebasic lawofmotion (1.402)isdimensionally homogeneous, inthesense thatboth sides have thesamedimensions, viz., [MLT~2 ].The operations weperform onthisequation may change thedimensions ofthetwosides, butthey arebothchanged inthesame way. Thus, atallstages ofourdeductions wehave dimensionally homogeneous equations. Indeed,itisinevitable that thisshould beso,since otherwise thetwosides ofanequa- tionwould change differently onchange ofunits, and iftrue for onesystem ofunits would notbetrue foranother. This gives auseful check. Forexample, suppose weareengaged inworking outtheformula fortheperiodic time ofsmall oscillations ofa simple pendulum. Asaresult ofourworkwearrive, perhaps, attheresult r=2*. g Dimensionally, thisreads [T]=m which shows that theresult isincorrect. Such acheckwill, of course, never beofanyassistance asfarasanumerical coefficient THEORY OFDIMENSIONS 511 isconcerned; forexample, thetheory ofdimensions alone cannot tellusthat isincorrect. Exercise. Itissuggested thattheequation ofmotion ofaparticle ona line is d*x,dx where ahasthedimensions[L],and 6thedimensions [jT~2 ].Would you accept thisequation ascorrect? (ii)Toseehow thetheory ofdimensions suggests forms of physical laws,weshall consider thetransverse vibrations ofa heavy particle atthemiddle point ofastretched string. The quantities involved are ra=mass ofparticle, alengthofstring, S=tension, T=periodic time. Theperiodic timemust besome function ofthequantities m,a, S,andsowewrite T=/(m, a,S). The only combination ofm,a,Shaving thedimensions [T] isoftheformCmaa(3Sy ,where C,a,0,yarepurenumbers, at present unknown. Accordingly weassume andobtain thedimensional equation [T]= [ Hence, tt+7=0, /3+7=0,-27=1, or andsoourformula forTis Ima 512 PRINCIPLES OFMECHANICS Wecannot findthenumerical factorCfrom thetheory ofdimen- sions. Toobtain ittheoretically, wemust solve thedifferential equationofmotion. But,ifwearesatisfied withanexperimental result, oneexperimentwill suffice todetermine C. Thismethod isuseful inthecase ofacomplicated system, where thedirect solution ofthedifferential equationsisdifficult. Exercise. Consider thetransverse vibrations ofasystem consisting of 20equal particles, equally spaced onastretched string, Show thateach ofthetwenty normal periodsisoftheform r>\r=CV-^, wheremisthemass ofeach particle, athelength ofthestring, Sthetension init,andCanumerical constant which maydepend ontheparticular normal mode. (iii)Toshowhowthetheory ofdimensions enables ustousea model topredict full-scale phenomena, letusconsider theflow ofairpast thewing ofanairplane. The liftYonthewing obviously depends onthefollowing quantities: pthedensity oftheair, U=thespeed ofthewing relative totheair, I=alinear measurement ofthewing (e.g.,itswidth from back tofront, atsome definite position). The liftdepends, ofcourse, ontheshape ofthewing;weshall consider onlywings ofonedefinite shape, transformed intoone another bychanging thelengthI. Theproblemistocalculate the liftYonthefull-scale wing from themeasurement ofthe liftY'onamodel. NowYisa function ofp,U,1]andY'isthesame function ofp',Uf ,/',where theaccented quantities refer totheexperiment onthemodel, thesame units ofmass, length, andtime being used inboth cases. Sowewrite F=f(P,U,1), Y'=f(P',U', I'). Asinthepreceding example, wetake /(p,U,I)=CfV'f, whereCisapurenumber. Since[p]=[ML"8 ],[U]=[LT~l ], [1]=[L],and[F]=[MLT~*], weeasilyfind Y=CpUH\Y'=Cp'U'H'2 . THEORY OFDIMENSIONS 513 Hence thefull-scale liftis Ifthedensityoftheairisthesame forboth cases, thisbecomes 772/2VV..72yi/, When weinsert thenumerical values for7', /',Z',obtained from experiment onamodel inawindtunnel, andthevalues ofUand /appropriate tothefull-scale wing inflight, weareable toread offthevalue ofthe liftF. Exercise. Inorder tostudy the(Inflections inanelastic beam with con- tinuous andisolated loads (cf.Sec.3.3),anengineer builds amodel ofthe same material withalinear ratio 1:100.Show that,ifthedeflections inthe model aretobeoneone-hundredth ofthefull-scale deflections, thecon- tinuous loadperunitlength inthemodel-must beoneone-hundredth ofthe full-scale loud. Inwhat ratioshould theisolated loads bereduced? INDEX Thenumbers inheavy type refer totheSummaries attheends ofthe chapters. Absolute equations ofmotion, 493- 495 Absolute velocity, acceleration, and force, 494,495 Acceleration, 27,28,36,305,332 absolute, 495 ofautomobile, 205,206 complementary, 349 ofCoriolis, 349 incylindrical coordinates, 306,307 duetogravity (see </) radial andtransverse components, 120,126 inrelativity, 492, 493,495 inspherical polar coordinates, 407, 468 tangential andnormal compo- nents, 118, 119,126,305,306, 332 oftransport, 349 Accelerations, composition of,141, 307,308 Action and reaction, law of,32,36 Addition ofvectors, 19-22, 36 Air,resistance of,151,154-159, 184 Airplane, 271, 272, 444, 512, 513 Ames, J.S.,16 Amplitudeofoscillations, 162 Angle offriction, 87,88,114 Angular impulse, 357 Angular momentum,inimpulsive motion, 229-231, 238, 357, 358, 361 ofparticle, 128,129,146,329,333, 341,360Angular momentum, relative to mass center, 135, 136,147,330, 345, 355, 360,361 Angular momentum, ofrigid body, 193,194,196,222,223,330-332, 333 ofsystem, 134-136, 147,329,330, 344,345,360 Angular velocity, oftheearth, 143 ofrigid body, 122, 126,308-311, 332 Anomaly, 187 Aphelion, 180 Appell, P.,xi Applications,indynamics inspace, 364-411-414, 418-463, 464 ofLagrangc's equations, 467-472 inplane dynamics, 151-184-186, 189-222, 223 inpiano statics, 74-113-115 instatics inspace, 275-278, 296- 301 Applied force, 58,295 Approximations forelectromagnetic lenses, 397-403 Apse, 171-174, 186 advanceof,forspherical pendu- lum,381 Apsidal angle, 173, 174,378-381 Archimedes, 82 Areal velocity, 170, 180,186 Associative property ofvector addi- tion,22 Astatic center, 73 Astronomical frame ofreference, 31, 33,141,143 Astronomical latitude, 145,403 515 516 INDEX Attraction, electrostatic, 176 gravitational, 82-86, 114, 144- 146, 176, 177,404, 405,509 Automobile, 204-206 Axes ofinertia, principal, 316-324, 333 Axially symmetric electromagnetic field, 387-403, 413 Axis, ofrotation, instantaneous, 308 ofscrew displacement, 285 ofsymmetry, 78,321,322 ofwrench, 269,270 B Balancing, problems of,219-222 Ball slipping ontable, 447-450, 464 Ballistic pendulum, 235,236 Ballistics, 151,184 (See alsoProjectile) Bars inframe, 106,108 Base point, 61,259, 280,281 change of,260,283,284 Beam, internal reactionsin,92,93, 114,272,273 thin, 92-98, 114 Becker, K.,151 Bending moment, 92-98, 114, 272, 273 Billiard ball,447-450, 464 Binormal, 264 Blank, A.A.,388 Body centrode, 124,126,309 Body cone, 309,421,425,427,463 Bound vector, 18,19 Bridge, suspension, 100,114 Cable, flexible, 98-105, 114,116 incontact with curve, 104, 105, 116 inspace, 265,266 Cajori, F.,32 Calibration ofspring, 17 Campbell,J.W.,103 Cardan's suspension, 418Catenary, 100-104, 116 Celestial pole,motionof,428,429 Center, astatic, 73 ofgravity, 84-86, 114,271 instantaneous, 123-126 ofmass (seeMass center) ofoscillation, 201 ofpercussion, 2^9 ofsystem ofparallel forces, 271 Centimeter, 13 Central force, general, 128,168-176, 186 varying directly asdistance, lb'8, 169,186 varying asinverse square ofdis- tance, 176-184, 186,462 Central symmetry, 77 Centrifugal force, 143-145, 147,349, 350,406 Centrode, 124,126,309 Chain(seeCable) Chako, N.,388 Change ofbase point, 260,283,284 ofunits, 504-508 Charge onelectron, 386,387 Charged particle, inaxially sym- metric electromagnetic field, 388-403, 413 inelectromagnetic field, 176,381- 403,412,413 inuniform electromagnetic field, 383-387, 412 Chasles' theorem, 303 Circular diskandcylinder, moments ofinertia of,190, 191,222,324 Circular motion, 28 Circular orbit, stability of,174-176 Clock, 12,13,501 inrelativity, 475-480, 482, 483, 487,490, 492,493 Clocks, synchronization of,477-480 enx,368,412 Coefficient, offriction, 87,88,114 ofrestitution, 232-234, 239 Collar, A.R.,316 Collisions, 231-235, 239 INDEX 517 Commutative property invector operations, 19,21,246,257 Complementary acceleration, 349 Complex frame, 111,112 Components ofvector, 22-24, 32, 35,36,245 (SeealsoAcceleration; Velocity) Composition, ofaccelerations, 141, 307,308 ofcouples, 2G8 offinite rotations, 20,282 ofinfinitesimal displacements, 281-283, 302 ofvelocities, 140,307, 308,491 Compound pendulum, 198-201, 223, 370 Compression, 233,234 modulus, 186 Cone, body orpolhode, 309, 421, 425,427,463 offriction, 87 space orhorpolhode, 309,421,425, 427,453 Configurationofasystem, 64,286 Conical pendulum, 374,375 Conjugate lines, 304 Conservation ofenergy, 130, 131, 137,146,147, 196,201,223,231, 341, 346, 356,360,361 Conservative field, 66 Conservative system, 65-67, 294, 295,302 Constant ofgravitation (secGravi- tational constant) Constraints, 57-60, 285-292 moving, 474 workless, 54-57, 70 Contact, rolling, 55-57, 70,123,124 rough, 86-88, 114 smooth, 54,55,57,70,86 Continuity ofbodies, 14,15,76 Contraction ofmoving rod,486,487, 490 Coordinate vectors, 23 Coordinates, cylindrical, 306, 307, 338 generalized, 285-292, 302,462-472Coordinates, spherical polar, 467 Coriohs, accelerationof,349 force, 143, 144,147,349,406 Coulomb's law,176 Couple, 49 gyroscopic, 442-444, 464 impulsive, 229 momentof,49,50,267 twisting, inabeam, 272 work done by,64,293 Couples, composition of,268 Courant, R.,320 Covering operation, 320 Cranx, C.,151 Critical form foraframe, 113 Cuboid, moment ofinertiaof,324 Curvature, radius of,119,264 Curves inspace, 264,265 Cuspidal motion ofatop,436-438 Cylinder, balancing problem for, 219-222 moments ofinertia of,191, 222, 324 rolling down inclined plane, 202- 204 Cylindrical coordinates, 306,307,338 D Dale, J.B.,370 D'Alembert's principle, 138, 139,147 Damped oscillations, 163-168, 185 Deadbeat oscillations, 166,185 Decomposition, methodof,78,79, 114,325 Decrement, logarithmic, 166 Degreesoffreedom, 207, 208,287 Dcimel, R.F.,445 Density, 76,77 Determination ofpastand future, 340,341 Deviations duetoearth's rotation, 145,407,408,414 Differentiation, used tofindmoments ofinertia, 325,326 ofvectors andtheir products, 24- 26,35,250,257 518 INDEX Digonal symmetry, 321 Dimensional notation, 503 Dimensionless quantity, 504,508 Dimensions, theory of,501-513 Directed line,22 Discontinuity inbodies, 14,15 Discontinuous motion, 231 Disk,moment ofinertia of,190,222, 324 rolling onplane, 450-453, 454 Displacement,ofrigidbody, 61,62, 70,278-285, 290, 301, 302 reduced totranslation and rotation, 61,62,280,281,302 screw, 284,285 virtual, 53,58,461 Distributive propertyofvector operations, 20,246,249 Disturbing force, 162, 163,166-168, 184,186 dnx,368,412 Drag, 272 Dugan, R.S.,13,429 Duncan, W.J.,316 Dynamicalunit offorce, 34,502 Dynamics, plane, applications in, 151-184-186, 189-222, 223 methods of,127-146, 147 inrelativity, 491-498, 499 inspace, applications in,364-411- 414,418-453, 464 methods of,337-359, 360,361 (See alsoMotion; Particle; Rigid body; System of particles) Dyne, 34 E Earth, angular velocity of,143 attraction of,82,84r-86, 144-146, 404,405 models of,5,6,84,85,144, 403, 428,429 rotation of,13,143-146, 403-411, 414 Earth's axis,motion of,428,429Eccentric anomaly, 187 Effective force, 138 Einstein, A.,7,475,477,478 Elastic beam, 95-98, 114 Elastic bodies incollision, 231-235, 239 Electric field, axially symmetric, 387, 390-392, 396 uniform, 383-386, 412 Electric lens, focal length of,403 (See also Electrostatic lens) Electric potential, 381 Electric vector, 381 Electromagnetic field, 381-403, 412, 413 axially symmetric, 387-403, 413 uniform, 383-387, 412 Electromagnetic lens, 392-403, 413 approximations for,397-403 focal length of,403 magnification of,396,413 Electron, determination ofe/m for, 386,387 Electron optics (seeCharged particle) Electrostatic attraction, 176 Electrostatic field, 381,383-386, 387, 390-392, 396,412 (SeealsoElectric field) Electrostatic lens,398 Ellipse, momental, 322,323 Ellipsoid, momental, 315-322, 333 moments ofinertia of,323-325 ofPoinsot, 420,421,425,463 Ellipsoidal shell,moments ofinertia of,326 Elliptic cylinder andplate,moments ofinertia of,324 Elliptic functions, 364-370, 411,412 Elliptic harmonic motion, 169,374 Elliptic integral, 368 Elliptical orbit, 169, 179-182, 186, 374 Emde, F.,370 Energy, principle of,129-131, 136, 146,147,196,201,223,231,341, 342,346,352,356,360,361,382, 495-497, 499 INDEX 519 Energy inrelativity, 495-497, 499 total, 130, 137,341,346 (SeealsoKinetic energy; Poten- tialenergy) Equation ofthehodograph, 156,184 Equations ofmotion, impulsive, 229-231, 238, 357, 358, 361, 470-472 ofcharged particle, 382, 383, 386, 388,389 Lagrange's 458-472 ofparticle, incylindrical coordi- nates, 338 inaplane, 127, 128, 146, 461, 462 relative torotating earth, 403- 406,414 relative torotating frame, 142, 348-350 inrelativity, 493-498, 499 inspace, 337-342, 360 ofrigidbody, with fixed axis, 196, 223 with fixed point, 351, 352,360 ingeneral, 355,356,361 moving parallel toaplane, 202, 223 Equilibrium, ofparticle, 39,40,69, 70,262,301 ofrigidbody, movable parallel to afixed plane, 62-64, 70 inspace, 273-278, 301 stability of,214-222, 223,296 ofsystem ofparticles, 41-52, 261- 266,295-301 Equimomental systems, 326 Equipollent force systems, 47-52, 70,260, 261,266-273, 293,301 Equivalence, ofGalilean frames, 480, 481 mechanical, 9,10 Equivalent force systems, 47,63,64 Equivalent simple pendulum, 200, 223 Erg,54 Euler-Bernouilli, law, 96,114 theory ofbeams, 95-98, 114Eulerian angles, 288-290 angular velocity intermsof,309, 310 Eiiler's equations ofmotion, 352,360 Euler's theorem, 279,280,301 Event, 11,478,498 Ewald, P.P.,88 Extension, 96 External forces, 41,42 Ferel's law,408 Ferry, E.S.,445 Fictitious forces, 138, 139,141-144, 147,347-351,406 Field, scalar orvector, 28 Field offorce, 66 electromagnetic, 381-403, 412,413 electrostatic, 381, 383-387, 390- 392,396,412 gravitational, 82-86, 144,405 magnetostatic, 381,382,383-388, 390-392, 397,412 uniform, 67 Finite displacement ofrigid body, 278-282, 301,302 Flexible cable (seeCable) Flywheel, 196-198 Focal length,ofelectromagnetic lens,403 Foot, 13 Force, 15-17, 36 absolute, 495 applied, 58,295 central, 128,168-186, 462 centrifugal, 143-147, 349,350,406 oncharged particle, 176,382 Coriolis, 143, 144, 147, 349,406 effective, 138 external andinternal, 41,42 fictitious, 138, 139,141-144, 147, 347-351, 406 field of,66,67 offriction, 87,88 generalized, 293-301, 302, 462, 464-467, 472 520 INDEX Force ofgravity, 82-86, 144-146, 404,509 impulsive, 228-238, 357-359, 361, 470-472 inrelativity, 479,495^199 reversedeffective, 138,147 shearing, 92-97, 114,272,273 total, 259,285,301 transmissibiiity of,64 unit of,16,34,502 Force system, general, 259-261 invariants of,269,270,285 reduction of,50-52, 70,266-273, 301 Force systems, equipollent, 47-52, 70,260, 261,266-273, 293,301 equivalent, 47,63,64 Forced oscillations, 166-168, 186 Forces, parallelogram of,32,33,36 polygon of,40 triangle of,40 which donowork, 54-57, 70 Foucault's pendulum, 408-411, 414 Foundations ofmechanics, 3-35, 36 Frame ofreference, 11,12,35,477 astronomical, 31,33,141,143 Galilean, 478,479 moving, 139-146, 147, 346-351, 361 Newtonian, 32-34, 132, 134,147 reduced torest, 141-143, 147,347, 349,350 inrelative motion inrelativity, 480-491 rotating, 142, 143, 147,347-351, 361 Frames, 106-113, 116 analytical andgraphical methods, 113 critical forms, 113 just-rigid andover-rigid, 106,107 simple andcomplex, 111,112 summary ofmethods, 115 Frazer, R.A.,316 Freeparticle, inNewtonian mechan- ics,32 inrelativity, 489, 490,498Free vector, 18,19 Freedom, degrees of,207, 208,287 Frenet-Serret formulas, 264 Frequencies, normal, 211-214, 223, 468,469 Frequency ofharmonic oscillator, 162,163 (See alsoPeriodic time) Friction, 86-92, 114 angle of,87,88,114 coefficientof,87,88,114 coneof,87 limiting, 88 Function denned by differential equations, 364 Fundamental plane, 39 Future and past, determination of, 340,341 G 0,85,86,114,144-146, 405 Galilean frame ofreference, 478,479 General theory ofrelativity (see Relativity) Generalized coordinates, 285-292, 302,462-472 Generalized forces, 293-301, 302, 462,464-467, 472 Generalized impulsive forces, 470- 472 Geodesic, 265,266,339 Gradient vector, 28-31, 35,36 Gram, 13 Gravitation inrelativity, 477 Gravitational attraction, 82-86, 114, 144-146, 176-177, 404,405,509 Gravitational constant(<?),82-84, 176,509,510 Gravity, center of,84-86, 114,271 Growth ofvector, 348 Gyration, radiusof,189 Gyrocompass, 444-447, 454 Gyroscope, 429,441-447, 464 Gyroscopic couple, 442-444, 464 Gyroscopic effect ofrotary engine, 444 Gyrostat (seeGyroscope) INDEX 521 H Hamilton, W.R.,33 Harmonicoscillator, 159-168, 184, 186 with constant disturbing force, 162,184 damped, 163-168, 186 forced oscillationsof,166-168, 186 Hemisphere, mass center of,78,81 Hemispherical shell,mass center of, 81,82 Herpolhode cone, 309 Heterogeneous body, 76 Hodograph, 120, 121,125, 156,184 Holonomic system, 287 Homogeneous body, 76 Hooke's joint, 297,298 Hooke's law, 96,114 Hoop,moment ofinertia of,190,222 Horizontal plane,85 orrotating earth, 145,403 Horsepower, 54 Hyperbolic orbit, 179,186 Image, formed byelectromagnetic lens, 391,395-403, 413 Image plane, 395,413 Image point, 395 Impulse, 227,357 Impulsive couple, 229 Impulsive force, 228-238, 357-359, 361,470-472 Impulsive moment, 230, 231, 238, 358,361 Impulsive motion, 227-238, 239, 356-359, 361,470-472 Ince, E.L.,393 Inclined plane, 202-204 Indeterminate problems, 68,69,90, 278 Inertia, moments of(seeMoments ofinertia) productsof (see Products of inertia)Infinitesimal displacement ofrigid body, 61,62,70,281-285, 290, 302 Ingredients, ofmechanics, 8-17, 36 ofrelativity, 476-478 Instability (seeStability) Instantaneous axis,308 Instantaneous center, 123-126 Integration ofequations ofmotion inpower series, 339,340 Intensity ofwrench, 269 Internal forces, 41,42 Internal reactions, inbeam, 92,93, 114,272,273 inflexible cable, 98 inrigidbody, 56,57,70,206,207 Intrinsic equation ofcatenary, 102 Intrinsic equations ofmotion, 338 Invariable line,419 Invariable plane, 420,463 Invariant element inspace-time, 489 Invariants, offorce system, 269,270, 285 ofinfinitesimal displacement, 285 Inverse square law,176-186 Lsotropy ofGalilean frame, 478,479 Jacobianelliptic functions, 364-370, 411,412 Jahnke, K.,370 Joints, inaframe, 106,107 method of,108-110, 116 Just-rigid frame, 106,107 K Kaufmann, method of,387 Kelvin's theorem, 363 Kepler's laws, 181,182 Kinematics, ofparticle, 118-121, 126,305-308, 332 inrelativity, 485-495, 498 plane, 118-126 ofrigid body, 121-126, 308-313, 332 inspace, 305-313, 332 522 INDEX Kinetic energy, ofmass center, 195 ofparticle, 129, 146, 333, 460 ofrigid body, 193-196, 222, 223, 327-329, 333 ofsystem, 136, 137,147,463 Konig, theorem of,195 Lagrange's equations, 458-472 applications of,467-472 forgeneral system, 466,467,472 forimpulsive motion, 470-472 forparticleinaplane, 459-462 forsystem withtwo degrees of freedom, 463-466 Lamb, H.,16,113,281,445 Lamina, representative, 61 Lamy's theorem, 40 Laplace's equation, 381,382 Latitude, astronomical, 145,403 Law, ofaction andreaction, 32,36 ofmotion, 32,33,36,140-143, 147,347, 349,350 oftheinverse square, 176-185 oftheparallelogram offorces, 32, 33,36 Laws, offriction, 87,88 ofNewtonian mechanics, 31-34, 36 Left-handed triad, 247 Length, 11 unit of,11,13,14,501,502 Lens, electric, 403 electromagnetic, 392-403, 413 magnetic, 403 (See also Electrostatic lens; Magnetostatic lens) Level surface, 29 Lift, 272,512,513 Light, inrelativity, 479-481, 483, 485,489,496,498 speed of,27,483 Limiting velocity, 159 Line density, 77 Linear moment, 74Linear momentum, inimpulsive motion, 229,230,238, 357,358, 361 ofparticle, 128,337,495,496 ofsystem, 132-134, 146,147,343, 360 Linkedrods, 236-238, 471,472 Loaded string, vibrations of,208- 212,511,512 Logarithmic decrement, 166 Lorentz transformation, 480-491, 498 Luneberg, R.K.,388 M Mach, E.,16 Magnetic field, axially symmetric, 387,388,390-392, 397 uniform, 383-387, 412 Magnetic lens, focal length of,403 (See also Magnetostatic lens) Magnetic potential, 382 Magnetic vector, 382 Magnetostatic field, 381-388, 390- 392,397,412 Magnetostatic lens, 398,413 Mass, 9,10 ofelectron, 386,387 inrelativity, 495 unitof,10,13,14,501,502 Mass center, 74-82, 113,114 angular momentum relative to, 135, 136, 147, 330, 345, 355, 360,361 found byintegration, 77,81,82, 114 found bysymmetry anddecom- position, 77-79, 114 kinetic energy of,195 motionof,132-134, 147,230,238, 343,360,361 motion relative to,134-136, 147, 234, 235, 238, 344, 345, 360, 361 ofsolar system, 134 Mathematical models, 5,6 INDEX 523 Mathematical truth, 7 Mathematical wayofthinking, 4,5, 508 Mathematicians* equations, 506,507 Matrix, 316 Mean anomaly, 187 Measuring rodorscale, 11,477, 501, 502 relativistic contractionof,486, 487,490 Mechanical equivalenceofbodies, 9,10 Mechanics, foundationsof,3-35, 36 Mercury, 7,31,379 Metacenter, 225 Methods, ofdynamicsinspace, 337- 359, 360,361 ofplane dynamics, 127-146, 147 ofplane statics, 38-69, 70 Michelson, A.A.,481 Michelsori-Morlcy experiment, 481 Milne-Thomsoii, L.M.,370 Minimum ofpotential energy, 214 220,223,296 Mixed triple product, 250, 251,267 Model, mathematical, 5-7 used forpredictionoffull-scale phenomena, 512,513 Modes ofvibration, normal, 207- 214,223 Modulus, compression, 186 ofelliptic functions, 367 Young's, 96,114 Moment, bending, 92-98, 114, 272, 273 ofcouple, 49,50,267 impulsive, 230, 231,238,358,361 linear, 74 ofmomentum, 128 (See alsoAngular momentum) pitching, 272 total, 259,285,301 ofvector, about line,43-45, 255- 257 inplane mechanics, 44,70 about point, 252-255, 267Momentalellipse, 322,323 Momental ellipsoid, 315, 316, 318, 321,322,333 Moments ofinertia, 189-193, 222, 313-326, 333 foundbydecomposition and differ- entiation, 325,326 principal, 316,318-320, 322-325, 333 ofsimple bodies, 190-193, 222, 323-325 Momentum (secAngular momen- tum;Linear momentum) Morley, E.W.,481 Motion, defined, 14 ofcharged particle, 381-403, 412, 413 impulsive (seeImpulsive motion) ofmass center, 132-134, 147,230, 238, 343,360,361 ofparticle, under centralforce, 168-184, 186,462 determined by initial condi- tions, 339,340 inplane, 118-121, 126,127-131, 146, 151-184-186, 212-214, 459-462 relative tomoving frame of reference*, 139-143, 147,346- 351 inrelativity, 489-498, 499 inspace, 305-308, 337-343, 360, 364-411-414 relative tomass center, 134-136, 147, 234, 235,238, 344, 345, 360,361 ofrigid body, parallel tofixed plane, 121-124, 126,189-222, 223 with fixed point, 308-310, 332, 351-355, 360, 418r-444, 463, 454 general, 310-313, 332,355, 356, 361,447-453, 464 ofsystem, 131-139, 146,147,189- 222,223,343-346, 360 Moving constraints, 474 524 INDEX Moving frames ofreference, 139- 146,147,340-351, 361 Moving rod,contraction of,486,487, 490 Multiplication, ofvector and scalar, 19,20 ofvectors, 245-267 Murnaghan, F.D.,16 Myers, L.M.,392 N Necessary conditions ofequilibrium, 39,40,42,43,45-47, 59,60,69, 70,262, 263,273-275, 296,301 Neutral equilibrium,216 Newton, L,32,82,176,475 Newtonian frame ofreference, 32- 34,132, 134,147 Newtonian law ofgravitational attraction, 82 Newtonian mechanics, laws of,31- 35,36 Newtonian unit oftime, 12 w-gonal symmetry, 322 Non-holonomic system, 287,466 Normal componentsofvelocity and acceleration, 118, 119,125,305, 306,332 Normal frequencies and periods, 211-214, 223,468,469 Normal modes ofvibration, 207-214, 223 Normal reaction, 87 Normal vector, principal,264 Notation, dimensional, 503 forvectors, 19 Null lines inspace-time, 489 Null planes and lines instatics, 304 Nutation oftop,432 O Object plane, 395 Object point, 395 Observer inrelativity, 477 Optics, electron (see Charged particle)Orbit, central, 166-184, 185,186 circular, 174-176 elliptical, 169, 179-182, 186,374 (See alsoPlanetary orbit) hyperbolic orparabolic, 179,185 Ordered triad, 247 Orthogonal triad, 23 Oscillation, centerof,201 Oscillations, damped, 163-168, 186 deadbeat, 166,186 forced, 166-168, 185 harmonic, 160-162, 184 (See alsoPendulum; Vibration) Oscillator, harmonic, 159-168 Osculating plane, 264 Over-rigid frame, 106 Pappus, theoremsof,80,114 Parabola insuspension bridge, 100, 114 Parabolic orbit, 179,186 Parabolic trajectory ofprojectile, 151-154, 184 Parallel axes, theorem of,191-193, 222 Parallel forces, 270,271 Parallelepiped (seeCuboid) Parallelogram offorces, 32,33,36 Parallelogram law, forcouples, 268 forinfinitesimal rotations, 282 Particle, 8,9,36 angular momentum of,128, 129, 146,329,333,341,360 under central force, 128,168-184, 185,462 charged, 176,381-403, 412,413 dynamics of,127-131, 146, 151 184-186, 212-214, 337-343, 360, 364-411-414, 459-462, 491-198, 499 equilibrium of,39,40,69,70,262, 301 free, 32,489,490,498 kinematicsof,118-121, 125,SOS- SOS,332,485-495, 498 INDEX 525 Particle, kinetic energy of,129,146, 333,460 Lagrange's equations for,459-462 linear momentumof,128, 337, 495,496 inaplane, 64-66, 118-121, 126, 127-131, 146, 151-184-186, 212-214, 217, 218, 370-372, 459-462 potential energy of,65-67 principleofenergy for,129-131, 146, 341, 342, 360, 382, 496, 497,499 inrelativity, 477, 489-498, 499 onrotating earth, 144- J46,403- 411,414 inrotating frame, 142, 143, 147, 347-351, 361 inspace, 305-308, 329, 332, 333, 337-343, 346-351, 360,373- 411-414 onstretched stung, 511 Particles, onstretched string, 208- 212 system of(seeSystemofparticles) Pastand future, determinationof, 339,340 Pendulum, ballistic, 235,236 compound, 198-201, 223,370 conical, 374,375 equivalent simple, 200,223 Foucault's, 40&-411, 414 simple, 159-161, 184, 370-372, 412 spherical, 373-381, 412 Percussion, centerof,239 Perihelion, 180 Period (seePeriodic time) Periodic solutions ofadifferential equation, 364-367 Periodic time, 161 ofcompound pendulum, 200,201, 223,370 ofharmonic oscillator, 162, 163, 166,168 normal, 211-213, 223, 468, 469 ofplanet, 180-182, 186Periodic time ofsimple pendulum, 161,184,372,412 Perpendicular axes,theoremof,193, 222 Phase ofoscillator, 162 Philosophical ideas, 3-8 Physical lawsanddimensions, 508- 512 Physical truth, 7 Physical way ofthinking, 3-5,508 Physicists' equations, 506,507 Pitch,ofscrewdisplacement, 284, 285 ofwrench, 269,270,285 Pitching moment, 272 Plane, fundamental, 39 inclined, 202-204 invariable, 420,463 osculating, 264 ofsymmetry, 78,321 Plane dynamics, applications in, 151-184-186, 189-222, 223 methodsof,127-146, 147 Plane equipollence, 48 Plane impulsive motion, 227-238,239 Plane kinematics, 118-125 Piano mechanics denned, 39 Plane4statics, applications in,74- 113116 methodsof,38-69, 70 Planetary orbit, 176-184, 186 constants of,179,180,186 Kepler's laws for,181,182 periodic timeof,180, 181,186 Plumb line oillotating earth, 145, 146,403,405 Poinsot, methodof,419-421, 425- 429,463 Poinsot ellipsoid, 420,421,425,463 Polhode cone, 309 Polygon offorces, 40 Poschl, Th.,88 Position vector, 24,36,305 Positive rotation, 247 Potential, electric, 381 gravitational, 83 magnetic, 382 526 INDEX Potential energy, 64-67, 70,85,114, 130, 294-297, 299, 300, 302 forinverse square lawofattrac- tion, 83,178 aminimum forstability, 214-221, 223,296 Pound, 13 Poundal, 34 Power, 54 Prandtl, L.,88 Precession, 429-432, 438, 442-444, 463,454 Principal axes andmoments of inertia, 316-326, 333 Principal normal, 264 Principal planesofinertia, 318 Principleofangular momentum, in impulsive motion, 230,231,238, 357,358,361 forparticle, 128,129,146,341,360 relative! tomass center, 136, 147, 202,223,345,360,361 forrigid body, 196,202-204, 223, 352, 355,360,361 forsystem, 135,147,344,345,360 Principle ofenergy, forparticle, 129-131, 146,341,342,360,382 inrelativity, 496,497,499 forrigidbody, 196,201,223,352, 356, 360,361 forsystem, 136, 137,147,346,360 Principle ofequivalence, 480, 481 Principle oflinear momentum, in impulsive motion, 229, 230, 238,357,358,361 forparticle, 337,496 forsystem, 132,146,343,360 Principle ofvirtual work, 57-60, 70, 295,296,301 Procedure intheoretical mechanics, 6,7 Products ofinertia, 313-316, 333 Products ofvectors, 245-267 mixed triple, 250,251,267 scalar, 245,246,267 vector, 247-250, 267 vectortriple, 252,267Projectile, with resistance, 154-159, 184 without resistance, 151-154, 184 onrotating earth, 407, 408,414 stability of,440,441 Propeller, 312,313,321,322 Proper energy, 497 Proper mass, 495 Proper time, 491,492,498 Q Quantum mechanics, 7,8,177,340, 341 R Radial components ofvelocity and acceleration, 120,126 Radius, ofcurvature, 119,264 ofgyration, 189 oftorsion, 264 Range ofprojectile, 153,154 Rate, ofchange ofvector, 347, 348, 361 ofgrowth, 348 oftransport, 348 Rawlings, A.L.,445 Reaction, inbeam, 92-94, 114,272, 273 normal, 87 inrigidbody, 56,57,70 inrotating rod,206,207 atrough contact, 86-88, 114 atsmooth contact, 54,55,57,70 workless, 54-57, 70 Reactions ofconstraint, 57-59, 295 Rectangular cuboid, moments of inertia of,324 Rectangular plate, moments of inertia of,190,222,324 Reduction, ofdisplacement, to screw, 284,285 totranslation androtation, 61, 62,280,281,302 ofgeneral force system, 266-273, 301 INDEX 527 Reduction,ofplane forcesystem, 50-52, 70 ofsystem ofparallel forces, 270, 271 Relative energy, 495,497,499 Relative force, 495-499 Relative mass, 495 Relative momentum, 495,496 Relativistic contraction ofmoving rod,486,487,490 Relativistic slowing down ofmoving clock, 487,490 Relativity, fundamental concepts of, 475-480 general theory of,7,177,476,477 measurement oftimein,475-480, 483, 487, 490, 492, 493,498 motion ofaparticle in,489-498, 499 special theory of,475-498, 499 Representative lamina, 61 Resistance ofair,151,154-159, 184 varying asthesquare ofthe velocity, 157-159, 184 Varying directly asthevelocity, 159,184 Resonance, 163,168 Rest, 14 Rest energy, 497 Restitution, 233,234 coefficient of,232-234, 239 Resultant, offinite rotations, 20,282 offorces, 32,36 ofinfinitesimal displacements, 281-283, 302 Reversed effective force, 138,147 Right-handed triad, 247 Rigid body, 10,11,35 angular momentum of,193, 194, 222,223,330-332, 333 angular velocity of,122,126,SOS- SI1,332 displacement of,61,62,70,278- 285,301,302 dynamics of,189-207, 221-223, 351-356, 360, 361,418-463, 464Rigid body, equilibrium of,62-64, 70,273-278, 301 free,290 internal reactionsin,56,57,70, 206,207 kinematicsof,121-124, 126,SOS- SIS,332 kinetic energy of,193-196, 222, 223,327-329, 333 motion parallel toaplane, 121- 124, 126, 189-207, 221-223 motion inspace, 351-356, 360, 361,418-463, 464 inrelativity, 477 rotating about fixed axis,196-201, 223 workdonebyforces acting on,63, 70,292,293,302 Rigid body with afixed point, angular momentumof,330-333 angular velocity of,308-310, 332 displacement of,279-282, 290, 301,302 dynamics of,351-355, 360,418- 444,463,464 equations ofmotionof,352,360 Kulerian angles for,288-290, 309, 310 Eulor's theoremfor,279, 280,301 kinematics of,308-310, 332 kinetic energy of,327, 328,333 mounting of,418 under noforces, 418-429, 463 (See alsoGyroscope; Top) Rod,moment ofinertia of,190, 222-. Rolling contact, 55-57, 70,123,124 Rolling disk, 450-453, 464 Rotating frame ofreference, 142, 143,147,347-351, 361 Rotating rod,206,207 Rotation, oftheearth, 13,143-146, 403-411, 414 about fixedaxis, 196-201, 223 about fixed point, 279-282, 301, 302,308-310, 332 instantaneous axisof,308 inaplane, 61,62,121-126 528 INDEX Ilotation, positive, 247 Rotations, finite, resultant of,20,282 infinitesimal, resultant of,281, 282,302 Rough contact, 85-88, 114 Routh's rule,325 Russell, H.N.,13,429 S Scalar, 18 multiplied byavector, 19,20 Scalar field, 28 Scalar product, 245, 246, 249, 250, 257 Screw displacement, 284,285 Second, 13,14 Sections, methodof,110, 111,116 Semicircular plate and wire, mass centers of,80 Separationinspace-time, 489,498 Shearing force, 92-97, 114,272,273 Shell (seeProjectile) Significant figures, 508 Silberstein, L.,481 Simple frame, 111 Simple harmonic motion, 160-162, 184 Simple pendulum, equivalent, 200, 223 finite oscillations of,161,200,370- 372,412 small oscillations of,159-161, 184 Sleeping top,438-440, 463 Sliding vector, 18 Slowing down ofmoving clock, 487, 490 Small displacement (see Infinites- imaldisplacement) Smooth contact, 54,55,57,70 snx,367-370, 411,412 Solar system, dynamics of,182 mass centerof,134 Sommerville, D.M.Y.,320 Space centrode, 124, 126,309 Space cone, 309, 421, 425, 427,463 Space-time, 487-491, 498 vectorsin,494,495Special theory ofrelativity (see Relativity) Speed, 27 ofapproach, 232, 233,239 oflight, 27,479-481, 483,485,496 ofseparation, 232, 233,239 Sphere, mass center of,78 moment ofinertia of,191,222,324 Spheres, collision of,231-235 Spherical pendulum, 373-381, 412 apse of,378-381 general motionof,375-378, 412 small oscillations of,373,374,378- 381,412 Spherical polar coordinates, 467,468 Spherical shell, attraction of,83,84 moment ofinertiaof,326 Spinoftoporgyroscope, 430, 433, 442-444, 463,464 Spinning top (seeTop) Stability, ofcircularorbit, 174-176 ofequilibrium, 214-221, 223,296 ofgyroscope, 441,442,464 ofrolling disk, 450-453, 464 ofsleeping top,43&-440, 453 ofspinning projectile, 440, 441 Statically determinate problems for beams, 94,95 Statically indeterminate problems, 68,69,90,278 Statics, plane, applications in,74- 113-116 methods of,38-69, 70 inspace, 259-301;302 Stewart,,1.Q.,13,429 Stress, inbarofframe, 108 inbeam, 92,93,114,272,273 String (secCable) Subtraction ofvectors, 21 Sufficient conditions ofequilibrium, 39,40,59,60,62,63,69,70, 273-275, 301 Surface, level, 29 rough, 86-88, 114 smooth, 54,55,57,70 Surface density, 77 Suspension bridge, 100,114 INDEX 529 Symmetry, axis of,78,321,322 central, 77 ofcentral orbit, 172, 173,185 diagonal, trigonal, etc.,321 plane of,78,321,322 used tofindmass centers, 77,78, 114 used tofind principal axes, 320- 323 Synchronization ofclocks, 477-480 Systemofforces(seeForce system) System ofparticles, angular momen- tum of,134-136, 147, 329, 330, 344, 345,360 dynamics of,131-139, 146, 147, 189-222, 223, 343-346, 360 equilibrium of,41-52, 57-67, 70, 261-266, 295-301 kinetic energy of,136, 137, 147, 463 Lagrange's equations for,463-472 linear momentum of,132-134, 146, 147,343,360 potential energy of,64-67, 137, 294-297, 299, 300,302 T Tangential components ofvelocity andacceleration, 118, 119, 125, 305, 306,332 Tension, inbarofframe, 108 inbeam, 92-96, 114,272 incable, 98-105, 114,116,265,266 Tensor, 316 Tetragonal symmetry, 321 Tetrahedron, mass center of,79 Theory ofdimensions, 501-513 Theory ofrelativity (seeRelativity) Thinbeams, 92-98, 114 Thrust, 108 Time, inNewtonian mechanics, 12, 13 proper, 491,492,498 inrelativity, 476-480, 483, 487, 490-493, 498 unit of,12-14, 501,502 rimoshenko, S.,113Top, 429-441, 453 cuspidal motion of,436-438 general motion of,432-436, 463, 469,470 Lagrange's equations for,469,470 sleeping, 438-440, 463 insteady precession, 429-432, 453 Torque, 196 Torsion, radiusof,264 Total angular impulse, 357 Total energy, 130, 137,341,346 Totalforce, 259,285,301 Total impulse, 357 Total impulsive force, 357 Totalmoment, 259,285,301 Trajectory,ofcharged particle, 384, 386,389-392, 412,413 ofprojectile, 151-159, 184, 407, 408,414 Transformation, ofaxes inspace time, 488 Lorentz, 480-491, 498 Newtonian, 486 toprincipal axes ofinertia, 318- 320,322,323 Tninslation, 61,279 Transmissibili tyofforce, 64 Transport, acceleration of,349 ofvector, 348 Transverse* components ofvelocity andacceleration, 120,125 Triad, left-handed andright-handed, 247 ordered, 247 unitorthogonal, 23 Triangle offorces, 40 Triangular plate, mass centerof,79 Trigonal symmetry, 321 Triple products, 250-252, 257 Trusses (seeFrames) Truth, mathematical andphysical, 7 Twisting couple, 272 Two-body problem, 182-184, 186 U Uniform field offorce, 67 electromagnetic, 383-387, 412 530 INDEX Unitcoordinate vectors, 23,247 Unit, offorce, 16,34,36,502 oflength, 11,13,14,35,501,502 ofmass, 10,13,14,36,501,502 oftime, 12-14, 36,501,502 Unitorthogonal triad, 23 Units, arbitrariness of,36,502 c.g.s.and f.p.s., 13 change of,504-508 Varignon, theorem of,44,45,255, 26 Vector, 17-19 binormal, 264 bound, 18,19 components of,22-24, 32,36,36, 245 differentiation of24-26, 36,250 electric, 381 free, 18,19 gradient, 28-31, 36,36 magnetic, 382 momentof,43-45, 70,252-267 multiplied byscalar, 19,20 notationfor,19 position, 24,36,305 principal normal, 264 rateofchange of,347,348,361 sliding, 18 inspace-time, 494,495 zero, 21 Vectorfield, 28 Vector function, 24-26 Vector product, 247-250, 267 Vector triple product, 252,267 Vectors, addition of,19-22, 36 coordinate, 23,247 products of,245-267 subtractionof,21 Vehicle, self-propelled, 204^206 Velocities, composition of,140,307, 308,491 Velocity, 26-28, 36,305 absolute, 494 angular, 122,126,308-311, 332Velocity, areal, 170, 180,188 incylindrical coordinates, 306,307 oflight, 27,479-481, 483,485,496 limiting, 159 ofparticle ofrigidbody, 308-313, 332 radial andtransverse components of,120,126 tangential component of,118,126, 305,306,332 Vertical, 85 onrotating earth, 145,403 Vibration, normal modesof,207- 214,223 ofparticle inplane, 212-214 ofparticle onstretchedstring, 511 oftwo particles onstretched string, 208-212 (See alsoOscillations) Virtual displacement, 53,58,60,461 Virtual work, 57-60, 70,111, 116, 295, 296,301 W Ways ofthinking, 3-5,508 Weight, 17,86 onrotating earth, 145,404,405 Whittaker, E.T.,xi,16 Work, 53-67, 70,292-301, 302 donebycouple, 64,293 donebyforce, 53,70 donebyforces ongeneral system, 293,294,302 donebyforces onrigid body, 62, 63,70,292,293,302 Work, virtual, 57-60, 70,111, 116,. 295,296,301 Workless constraints, 54-.r) Wrench, 269,270, 285,30i Young, D.H.,113 Young's modulus, 96,114 Zero, force equipollent to,48,261 Zero vector, 21