PHYS 1443 – Section 501Lecture #21AnnouncementsTotal Kinetic Energy of a Rolling BodyKinetic Energy of a Rolling SphereExample for Rolling Kinetic EnergyExample for Rolling Kinetic EnergyWork, Power, and Energy in RotationAngular Momentum of a ParticleAngular Momentum and TorqueAngular Momentum of a System of ParticlesExample for Angular MomentumAngular Momentum of a Rotating Rigid BodyExample for Rigid Body Angular MomentumConservation of Angular MomentumExample for Angular Momentum ConservationSimilarity Between Linear and Rotational MotionsPHYS 1443 – Section 501Lecture #21Wednesday, Apr. 14, 2004Dr. Andrew Brandt1. Rotational Kinetic Energy2. Work, Power and Energy in Rotation3. Angular Momentum 4. Angular Momentum and Torque5. Conservation of Angular MomentumWednesday, Apr. 14, 2004 PHYS 1443-501, Spring 2004Dr. Andrew Brandt1Announcements• HW#9 on Ch. 10 and is due 4/19. There will be two more homeworks after this. 4/26+5/3. It would help you to do them.• Updated grades are posted.Wednesday, Apr. 14, 2004 2PHYS 1443-501, Spring 2004Dr. Andrew BrandtTotal Kinetic Energy of a Rolling BodySince it is a rotational motion about the point P, we can write the total kinetic energyWhat do you think the total kinetic energy of a rolling cylinder is?Since vCM=Rω, the above relationship can be rewritten as221ωPIK =Where, IP, is the moment of inertia about the point P.PP’CMvCM2vCMUsing the parallel axis theorem, we can rewriteWednesday, Apr. 14, 2004 3PHYS 1443-501, Spring 2004Dr. Andrew BrandtK222121CMCMMvIK +=ωWhat does this equation mean?Rotational kinetic energy about the CMTranslational Kinetic energy of the CMTotal kinetic energy of a rolling motion is the sum of the rotational kinetic energy about the CMAnd the translational kinetic of the CM221ωPI=()2221ωMRICM+=2222121ωωMRICM+=Kinetic Energy of a Rolling SphereWednesday, Apr. 14, 2004 4PHYS 1443-501, Spring 2004Dr. Andrew BrandtSince vCM=RωLet’s consider a sphere with radius R rolling down a hill without slipping.=KRxhθvCMω222121CMCMCMMvRvI +⎟⎠⎞⎜⎝⎛=Since the kinetic energy at the bottom of the hill must be equal to the potential energy at the top of the hillWhat is the speed of the CM in terms of known quantities and how do you find this out?K2221CMCMvMRI⎟⎠⎞⎜⎝⎛+=2221CMCMvMRI⎟⎠⎞⎜⎝⎛+=Mgh=2/12MRIghvCMCM+=2222121ωωMRICM+Example for Rolling Kinetic EnergyFor the solid sphere shown in the figure, calculate the linear speed of the CM at the bottom of the hill and the magnitude of linear acceleration of the CM.The moment of inertia of the sphere with respect to the CM!!=CMIWednesday, Apr. 14, 2004 5PHYS 1443-501, Spring 2004Dr. Andrew BrandtSince h=xsinθ, one obtainsThus using the formula in the previous slideWhat must we know first?RxhθvCMω2/12MRIghvCMCM+=θsin7102gxvCM=Using kinematicrelationshipxavCMCM22=The linear acceleration of the CM isθsin7522gxvaCMCM==What do you see?Linear acceleration of a sphere does not depend on anything but g and θ.5/212+=ghgh710==∫dmr2252MRExample for Rolling Kinetic EnergyFor the solid sphere shown in the figure, calculate the linear speed of the CM at the bottom of the hill and the magnitude of linear acceleration of the CM. Solve this problem using Newton’s second law, the dynamic method.Wednesday, Apr. 14, 2004 6PHYS 1443-501, Spring 2004Dr. Andrew Brandt∑xFGravitational Force,Since the forces Mg and n go through the CM, their moment arm is 0 and do not contribute to torque, while the static friction f causes torqueMxhθαRaCM=CMτWe know that What are the forces involved in this motion?MgfNewton’s second law applied to the CM givesFrictional Force, Normal Forcenxy252MRICM=We obtain fSubstituting f in dynamic equationsCMMaMg57sin =θfMg−=θsinCMMa=∑yFθcosMgn−=0=fR=αCMI=RICMα=⎟⎠⎞⎜⎝⎛=RaRMRCM252CMMa52=θsin75gaCM=Work, Power, and Energy in RotationLet’s consider a motion of a rigid body with a single external force F exerted at the point P, moving the object by ds.The work done by the force F as the object rotates through the infinitesimal distance ds=rdθ is What is Fsinφ?dWThe tangential component of force F.Wednesday, Apr. 14, 2004 7PHYS 1443-501, Spring 2004Dr. Andrew BrandtSince the magnitude of torque is rFsinφ,FφOrdθdsWhat is the work done by the radial component Fcosφ?Zero, because it is perpendicular to the displacement.dWThe rate of work, or power becomesPHow was the power defined in linear motion?The rotational work done by an external force equals the change in rotational energy. ∑τThe work put in by the external force thendWsdF ⋅=()θφrdF sin=θτd=dtdW=dtdθτ=τω=αI=⎟⎠⎞⎜⎝⎛=dtdIω⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛=dtdddIθθωθτd∑=ωωdI=W∫∑=fdθθιθτ∫=fdIωωιωω222121ifIIωω−=()θφdrFsin=⎟⎠⎞⎜⎝⎛=θωωddIAngular Momentum of a ParticleIf you grab onto a pole while running, your body will rotate about the pole, gaining angular momentum. We’ve used linear momentum to solve physical problems with linear motion, angular momentum is useful for rotational motion.φsinmvrL=Let’s consider a point-like object ( particle) with mass m located at the vector location r and moving with linear velocity vprL ×≡The instantaneous angular momentum L of this particle relative to origin O is What do you learn from this?If the direction of linear velocity points to the origin of rotation, the particle does not have any angular momentum.What is the unit and dimension of angular momentum? 2/kgms⋅Because r changesNote that L depends on origin O. Why? The direction of L is +zWednesday, Apr. 14, 2004 8PHYS 1443-501, Spring 2004Dr. Andrew BrandtWhat else do you learn? Since p is mv, the magnitude of L becomesIf the linear velocity is perpendicular to position vector, the particle moves exactly the same way as a point on a rim.21[]MLT−Angular Momentum and TorquedtpdF =∑Can you remember how net force exerted on a particle and the change of its linear momentum are related?Total external forces exerting on a particle is the same as the change of its linear momentum.∑τThus the torque-angular momentum relationshipThe same analogy works in rotational motion between torque and angular momentum. Net torque acting on a particle is dtLddtLd=∑τxyzOpφL=rxprmWhy is this?Because v is parallel to the linear momentum()dtprd ×=dtpdrpdtrd×+×=dtpdr ×+= 0∑×= Frdtpdr ×=Wednesday, Apr. 14, 2004 9PHYS 1443-501, Spring 2004Dr. Andrew BrandtThe net torque
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