TORQUE AND ANGULAR MOMENTUM IN CIRCULAR …

1 Project PHYSNETP hysics Bldg. Michigan State University East Lansing, MI MISN-0-34 TORQUEANDANGULARMOMENTUMINCIRCULARMOTION shaft1 TORQUEANDANGULARMOMENTUMIN CIRCULARMOTIONbyKirby Morgan,Charlotte,Michigan1. Introduction.. 12. Torqueand AngularMomentuma. De nitions.. 1b. Relationship:~ =d~L=dt..1c. MotionCon nedto a Plane.. 2d. CircularMotionof a Mass..33. Systemsof Particlesa. TotalAngularMomentum.. 4b. TotalTorque.. 4c. RigidBody MotionAbout a FixedAxis.. 5d. Example:Flywheel..5e. KineticEnergyof Rotation.. 6f. Linearvs. RotationalMotion.. 64. Conservationof AngularMomentuma. Statement of the Law .. 7b. If the ExternalTorqueis not Zero.. 7c. Example:Two Flywheels.. 7d. KineticEnergyof the Two Flywheels..85. NonplanarRigidBodies.. 9 Acknowledgments..9 Glossary.. 92ID Sheet: MISN-0-34 Title:Torqueand AngularMomentumin CircularMotionAuthor:Kirby Morgan,HandiComputing,Charlotte,MIVersio n:4/16/2002 Evaluation:Stage0 Length:1 hr; 24 pagesInputSkills:1. Vocabulary:kineticenergy(MISN-0-20),torq ue,angularacceler-ation(MISN-0-33),angul armomentum(MISN-0-41).

2 2. Solve constant angularaccelerationproblemsinvolvingtorq ue,moment of inertia,angularvelocity, rotationaldisplacement andtime(MISN-0-33).3. Justifythe use of conservationof angularmomentumto solve prob-lemsinvolvingtorquelesschangefromon e stateof uniformcircularmotionto another(MISN-0-41).OutputSkills(Knowledg e) nethe torqueand angularmomentumvectorsfor (a) a singleparticle(b) a systemof 's2nd law, derive its rotationalanalogandstatewhenit can be writtenas a equationfor linearkineticenergyandderive thecorrespondingone for conservationof angularmomentummay not holdinone systembut may if the systemis (ProblemSolving):S1. For massesin circularmotionat xedradii,solve problemsrelatingtorque,moment of inertia,angularacceleration,rotationalki neticenergy, work,and Given a systemin which angularmomentumis changingwithtimedue to a speci edappliedtorque,reconstructthe minimum ex-pandedsystemin which totalangularmomentumis the reactiontorquewhich producesthe compensatingchangein A DEVELOPMENTAL-STAGE PUBLICATIONOF PROJECTPHYSNETThe goalof our project is to assista network of educatorsand scientistsintransferringphysicsfromone personto support manuscriptprocessingand distribution,alongwithcommunicationand alsowork withemployers to identify basicscienti cskillsas well as physicstopicsthatare neededin scienceandtechnology.

3 Anumber of our publicationsare aimedat assistingusersin designed:(i) to be updatedquickly in responseto eldtestsand newscienti cdevelopments; (ii) to be usedin both class-room andprofessionalsettings;(iii)to show the prerequisitedependen-ciesexistingamongth e variouschunksof physicsknowledgeandskill,as a guideboth to mental organizationand to use of the materials;and(iv) to be adaptedquickly to speci cuserneedsrangingfromsingle-skillinstruc tionto ,reviewers and eldtestersare DirectorADVISORY COMMITTEED. AlanBromleyYale UniversityE. LeonardJossemTheOhioStateUniversityA. A. StrassenburgS. U. N. Y., Stony BrookViewsexpressedin a moduleare thoseof the moduleauthor(s)and arenot necessarilythoseof otherproject 2002,PeterSignellfor Project PHYSNET,Physics-Astronomy Bldg.,Mich. StateUniv.,E. Lansing,MI 48824;(517) our liberaluse policiessee: CIRCULARMOTIONbyKirby Morgan,Charlotte,Michigan1. IntroductionJustas for translationalmotion(motionin a straight line),circularorrotationalmotioncan be separatedinto kinematicsand coveredelsewhere,1the discussionherewill centeron to derive the rotationalanalogofNewton'ssecondlaw and thenapplyit to the circularmotionof a singleparticleandto systemsof particularwe wishto developthe relationshipbetween torqueand angularmomentumand discussthecircumstancesunderwhich angularmomentumis Torqueand AngularMomentum2a.

4 De de nedasvectorproductsof position, a force~Factson a particlewhosepositionwithrespect to the originOis thedisplacement vector~r. Thenthe TORQUE \about the point 0 and actingonthe particle,"is de nedas:2~ =~r ~F :(1)Now suppose the particlehas a linearmomentumPrelative to angularmomentumof the particleis de nedas:~L=~r ~p:(2)Thedirectionsof~ and~Lare given by the right-handrulefor crossproducts(seeFig. 1). :~ =d~L= de nitionsof torqueandangularmomentum,we can derive a usefulrelationshipbetween \Kinematics:CircularMotion"(MISN-0-9)and \Torqueand AngularAccel-erationfor RigidPlanarObjects:Flywheels"(MISN-0-33) .2 See \Forceand TORQUE "(MISN-0-5).5 MISN-0-34200(a)(b)F`p`r`qqr` :(a) TORQUE (b) angularmomentum(both directedout of the page).StartingfromNewton'ssecondlaw, writtenin the form~F=d~pdt;(3)the torqueis:~ =~r ~F=~r d~pdt:(4)Thiscan be rewrittenusingthe expressionfor the derivative of a crossproduct:Help:[S-1]~ =ddt(~r ~p) d~rdt ~p=d~Ldt ~v ~p:(5)Now~p=m~vso~v ~p= 0 (becausethe vectorproductof parallelvectorsis zero),so the torqueis:3~ =d~Ldt:(6)Thus the timerateof changeof the angularmomentumof a particleisequalto the torqueactingon MotionCon nedto a ~ =d~L=dtfor aparticletakes on a scalarappearancewhenthe motionof the particleiscon nedto a particleconstrainedto move onlyin thex-yplane,asshown in Fig.

5 2. Thetorqueon the particleis always perpendicularto3 Thisequationis valid onlyif~ and~Lare measuredwithrespect to the `p`r`xFigure2.~r,~F, and~pareall coplanarfor motionin `r` aparticlein circularmotion,~rand~ planeas is the angularmomentum[work this out usingEqs.(1) and(2)].Equivalently, we say that~ and~Lhave onlyz-components. Sincetheirdirectionsremainconstant, : =dLdt(motionin a plane):(7)Thisequationholdsonlyif~Fand~p are in the sameplane;if not (andtheywon'tbe for non-planarmotion),the full Eq. (6) must be CircularMotionof a torqueand angularmomentumfor the specialcaseof a singleparticlein circularmotioncan be easilyrelatedto the particle' a particleof massmmoves about a circleof radiusr withspeed v (notnecessarilyconstant)as shown in Fig. 3. Theparticle'sangularmomentumis:~L=~r m~v;(8)but since~rand~vare perpendicular,5the magnitudeof~Lis:L=mvr(9)andthe directionis out of the (9) may be rewrittenintermsof the angularvelocity (sincev=!)

6 R) as:L=mr2!:(10)4 Thecomponent equationsare x=dLx=dt, y=dLy=dt, z=dLz= \Kinematics:CircularMotion"(MISN-0-9).7 MISN-0-344 Similarly, the torqueis: =dLdt=mr2d!dt=mr2 ;(11)where is the particle' Systemsof systemof particlesis simplythe sumof the angularmomenta of theindividualparticles,addedvectorially. Let~L1,~L2,~L3,: : :,~LN, be therespective angularmomenta, about a given point, of the particlesin the point is:7~L=~L1+~L2+: : :=NXi=1~Li:(12)As timepasses,the totalangularmomentummay rateofchange,d~L=dt, will be the sumof the ratesd~Li=dtfor the particlesinthe ~L=dtwill equalthe sumof the torquesactingon a systemof particlesis justthe sumof the externaltorquesactingon the tointernalforcesis zerobecauseby Newton'sthirdlaw the forcesbetweenany two particlesare equaland oppositeand directedalongthe line torquedue to each such action-reactionforcepairis zeroso the totalinternaltorquemust alsobe totaltorqueon the systemis just equalto the sumof the externaltorques:~ =NXi=1~ i.

7 Ext(systemof particles):(13)For the system,then:~ =d~Ldt:(14)In words,the timerateof changeof the totalangularmomentumabout agiven point, for a systemof particles,is equalto the sumof the externaltorquesabout thatpoint and actingon the \TorqueandAngularAccelerationfor RigidPlaneObjects:Flywheels"(MISN-0-33). 7 For continuousmassdistributionsthe summationbecomesan RigidBody MotionAbout a rigidbody is asystemof particleswhosepositionsare all xedrelative to each !for each particlein the body, we may writeforthe totalangularmomentum:L= Ximir2i!!;(15)wherewe have assumedthatthe body is rotatingabout a xedaxiswithangularvelocity!. Thequantity in parentheses,I Ximir2i;(16)is calledthe \moment of inertia"of the we writeL=I!(17)and,sincethe axis of rotationis xed,equationswitha scalarappearanceholdfor the TORQUE : =dLdt=Id!dt=I :(18) a visualexample,we calculatethetorqueandangularmomentumfor the specialcaseof a ywheelwherethe entire massis uniformlydistributedaroundthe \dm" bethe massof an in nitesimalsegment of the rim as shown in Fig.

8 4. TheangularmomentumdLof the massdmis:dL=dm r2!:(19)Sincerand!are the samefor all points on the rim,the totalangularmomentumfor the ywheelis:L=ZdL=r2!Zdm=Mr2!=I!;(20)whereM is the totalmassof the identicalto Eq. (17)for a singlemassMin also the same, , =dLdt=Mr2 =I :(21)8 See \UniformCircularMotion:Moment of Inertia,Conservationof AngularMo-mentum,KineticEnergy, Power" (MISN-0-41). the rim of a y-wheel(heavy line). particles' KineticEnergyof a sys-tem can be writtenin termsof the system'smoment of inertiaand angularvelocity. We startwiththe statement thatthe totalkineticenergyof thesystemof particles,each of which is in circularmotionabout a xedaxisof rotation,is equalto the sumof the kineticenergiesof the individualparticleof massmimoves in a circleof radiusriabout the axisof the positionsof the particlesare all xedrelative to each other(as in a rigidbody),thenthe angularvelocity!isthe samefor all each particleis thus:Eik=12miv2i=12mir2i!

9 2 Thetotalkineticenergyof the rotatingbody is thereforeEk=12m1r21!2+12m2r22!2+: : :=12"NXi=1mir2i#!2By Eq. (16)the sum,Pmir2i, is just the moment of inertiaIof the body,so the rotationalkineticenergyfor a rigidbody can nallybe written:Ek=12I!2:(22) See if you can determinethe moment of inertiaand kineticenergyofour ywheelin :[S-2]3f. a comparisonof theequationsof dynamicsin \normal"and \rotational"form:10 MISN-0-347 Table1. ~F=d~P =dt~ =d~L=dtEk=MV2=2Ek=I!2=2 For a rigidbody rotatingabout a xedaxis,the Rigid-body !F=dp=dt=Ma =dL=dt=I Someof the equationsshown in the Table2 can be rewrittenin vector/ten-sor formso theyhave validity beyond linearand Conservationof AngularMomentum4a. Statement of the totalangularmomentumof a systemof particlesis conserved if thereare no externaltorquesactingon :~ = 0 =d~Ldt(23)so~Lis a constant. For a rigidbody rotatingabout a xedaxis,L=I!,so thatifIchanges,theremust be a compensatingchangein!in orderforLto If the ExternalTorqueis not the externaltorqueona systemis not zerothenangularmomentumis not conserved for , all is not lost,for if the systemis expandedto includewhatever is causingthe externaltorqueon it (thereforechangingit into aninternaltorque),angularmomentumwill be conserved for the Example:Two whentwoidentical ywheels,one spinningand one at rest,are suddenlybrought ywheelhas totalangularmomentumL0=Mr2!

10 0beforeit is allowed to comein contact withthe second ywheelscometogether,the frictionbetween themcausesa ywheels,one spinning,oneat rest,are brought be exertedon the spinningone and,sincethis torquecomesfromanexternalsource,the angularmomentumof the spinning ywheelis notconserved. If, however, our systemconsistsof both ywheels,the torqueseach exerton the otherare internalto the systemand so thereis no longeran for the expandedsystem,containingboth y-wheels,angularmomentumis conserved. Thistellsus that:Mr2!0=Mr2!f+Mr2!f= 2Mr2!f;(24)where!fis the nalangularvelocity at which the two together|thespinningone slowed down (!0!!f), theotherhavinggonefromzeroto!f|thetwo act as a single ywheel,withmass2M, rotatingwithangularvelocity!f=!0= :[S-3]. the Two ywheelsro-tatingat di erent speedsare brought into contact,it can be expectedthatkineticenergyis not conserved. Thisis becausethe nonconservativefrictionalforceactson the two initiallyEk0=12Mr2!