1Conservation of Angular Momentum8.01W11D2Rotational and Translational Comparison0fWd=⋅∫Frrot SPτω=0fSWdθθτθ=∫PowerWorkMomentumAng MomentumRotationKinetic EnergyTorqueForceTranslationQuantity2rot cm(1 / 2)KIω=2trans cm(1 / 2)KmV=2rot cm cm/2KLI=Lcm= Icmω2trans/2Kpm= p = mVcmext sys total/cmddtm==Fp Acm cm/ddt= LτP =⋅FvTorque and the Time Derivative of Angular MomentumTorque about a point S is equal to the time derivative of the angular momentum about S .SSddt=LτAngular Momentum for a System of ParticlesTreat each particle separatelyAngular momentum for system about S,,Si Si i=×Lrpsys,,11iN iNSSiSiiii======×∑∑LLrp2Angular Momentum and Torque for a System of ParticlesChange in total angular momentum about a point S equals the total torque about the point Ssys,,,11iN iNSiSiSi i Siiiddddt dt dt====⎛⎞== ×+×⎜⎟⎝⎠∑∑rLpLprsystotalSSSddt=Lτ()systotal,,,111iN iN iNSiSi Si i SSi SSiiidddt dt======⎛⎞=×=×==⎜⎟⎝⎠∑∑∑LprrFττInternal and External TorquesThe total external torque is the sum of the torques due to the net external force acting on each element The total internal torque arise from the torques due to the internal forces acting between pairs of elementsThe total torque about S is the sum of the external torques and the internal torquestotal ext intSSSSSS=+τττext ext ext,,11iN iNSSSiSiiii======×∑∑rFττint int int,,,,,111 11NiNiNiNiNSSSSj Sij Si ijjji ijji ji== ===== ==≠≠== = ×∑∑∑∑∑rFττ τInternal TorquesWe know by Newton’s Third Law that the internal forces cancel in pairs and hence the sum of the internal forces is zeroDoes the same statement hold about pairs of internal torques?By the Third Law this sum becomesint int,, , , , , , ,Si j S ji Si i j S j ji××rFrFτ+τ= +,,ij ji=−FF,11iNiNijijji====≠=∑∑0Fint int,, , , , , ,)Si j S ji Si S j i j−×rr Fτ+τ=(The vector points from the jthelement to the ithelement.,,Si S j−rrCentral Forces: Internal Torques Cancel in PairsIf the internal forces between a pair of particles are directed along the line joining the two particles then the torque due to the internal forces cancel in pairs. This is a stronger version of Newton’s Third Law than we have so far used requiring that internal forces are central forces. With this assumption, the total torque is just due to the external forcesHowever, so far no isolated system has been encountered such that the angular momentum is not constant.int int,, , , , , ,)Si j S ji Si S j i j−×=rr F0τ+τ=(sysextSSSddt=Lτ,,,)ij Si Sj−Frr (3Angular Impulse and Change in Angular MomentumAngular impulseChange in angular momentumRotational dynamics()()sys sys sysSS SfiΔ≡ −LL L ()()ext sys sysfitSSSSfitdt =−∫LLτext sysave int()SSSt=Δ=ΔJLτextfitSStdt=∫JτConcept Question: Change in Angular MomentumA person spins a tennis ball on a string in a horizontal circle with velocity (so that the axis of rotation is vertical). At the point indicated below, the ball is given a sharp blow (force ) in the forward direction. This causes a change in angular momentum in the 1. direction 2. direction 3. direction ˆrˆθˆk FΔL vConservation of Angular MomentumRotational dynamicsNo external torquesChange in Angular momentum is zeroAngular Momentum is conservedSo far no isolated system has been encountered such that the angular momentum is not constant.sysextSSSddt=Lτ()()sys sys sys0SS SfΔ≡ − =LL L 0 sysextSSSddt==L0τ()()sys sys0SSf=LLConcept Question: Twirling PersonA woman,holding dumbbells in her arms, spins on a rotating stool. When she pulls the dumbbells inward, the moment of inertia changes and she spins faster. The magnitude of the angular momentum about the vertical axes passing through her center is 1. the same. 2. larger.3. smaller. 4. not enough information is given to decide.4Concept Question: Figure SkaterA figure skater stands on one spot on the ice (assumed frictionless) and spins around with her arms extended. When she pulls in her arms, she reduces her rotational moment of inertia and her angular speed increases. Assume that her angular momentum is constant. Compared to her initial rotational kinetic energy, her rotational kinetic energy after she has pulled in her arms must be 1. the same. 2. Larger.3. smaller. 4. not enough information is given to decide.Table Problem : Rotating Chair and WheelA person is sitting on a chair that is initially not rotating and is holding a spinning wheel. The moment of inertia of the person and the chair about a vertical axis passing through the center of the stool is IS,p, and the moment of inertia of the wheel about an axis, perpendicular to the plane of the wheel, passing through the center of mass of the wheel is Iw= (1/4)IS,p. The mass of wheel is mw. Suppose that the person holds the wheel as shown in the sketch such that the distance of an axis passing through the center of mass of the wheel to the axis of rotation of the stool is d and that md2= (1/3)Iw. Suppose the wheel is spinning initially at an angular speed ωs. The person then turns the spinning wheel upside down. You may ignore any frictional torque in the bearings of the stool. What is the angular speed of the person and stool after the spinning wheel is turned upside down?Demo: Rotating WheelBicycle Wheel and Rotating StoolDemo: Train(1) At first the train is started without the track moving. The train and the track both move, one opposite the other. (2) Then the track is held fixed and the train is started. When the track is let go it does not revolve until the train is stopped. Then the track moves in the direction the train was moving.(3) Next try holding the train in place until the track comes up to normal speed (Its being driven by the train). When you let go the train remains in its stationary position while the track revolves. You shut the power off to the train and the train goes backwards. Put the power on and the train remains stationary.A small gauge HO train is placed on a circular track that is free to rotate.5Constants of the MotionWhen are the quantities, angular momentum about a point S, energy, and momentum constant for a system? • No external
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