PHYS 1443 – Section 003 Lecture #18AnnouncementsTorque and Vector ProductRotational Kinetic EnergyTotal Kinetic Energy of a Rolling BodyKinetic Energy of a Rolling SphereExample for Rolling Kinetic EnergySlide 8Work, 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 MotionsMonday, Nov. 10, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #18Monday, Nov. 10, 2002Dr. Jaehoon Yu1. Torque and Vector Products2. Rotational Kinetic Energy3. Work, Power and Energy in Rotation4. Angular Momentum 5. Angular Momentum and Torque6. Conservation of Angular MomentumMonday, Nov. 10, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu2Announcements•2nd term exam changes–Problem #7, the correct answer is c, 0.6. Those who answered b and c get credit–Problem #21, there is no correct answer All of you received credit for this problem•Class average: 52.1 (term 1: 53)–The two exams are identical in averageMonday, Nov. 10, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu3xyzOTorque and Vector ProductThe magnitude of torque given to the disk by the force F isLet’s consider a disk fixed onto the origin O and the force F exerts on the point p. What happens?sinFrBAC The disk will start rotating counter clockwise about the Z axisThe above quantity is called Vector product or Cross productFrxFrpBut torque is a vector quantity, what is the direction? How is torque expressed mathematically? Fr What is the direction? The direction of the torque follows the right-hand rule!!What is the result of a vector product?Another vectorWhat is another vector operation we’ve learned?Scalar productcosBABAC Result? A scalarsinBABAC Monday, Nov. 10, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu4Rotational Kinetic EnergyWhat do you think the kinetic energy of a rigid object that is undergoing a circular motion is? Since a rigid body is a collection of masslets, the total kinetic energy of the rigid object isSince moment of Inertia, I, is defined asKinetic energy of a masslet, mi, moving at a tangential speed, vi, isrimiOxyviiKRKiiirmI22IKR21The above expression is simplified as221iivm2221iirmiiK2iiirm221iiirm221Monday, Nov. 10, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu5Total Kinetic Energy of a Rolling BodyWhere, IP, is the moment of inertia about the point P.Since it is a rotational motion about the point P, we can write the total kinetic energySince vCM=R, the above relationship can be rewritten as221PIK What do you think the total kinetic energy of the rolling cylinder is?PP’CMvCM2vCMUsing the parallel axis theorem, we can rewriteK222121CMCMMvIK What 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 CM221PI 2221MRICM2222121MRICMMonday, Nov. 10, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu6Kinetic Energy of a Rolling SphereSince vCM=RLet’s consider a sphere with radius R rolling down a hill without slipping.KRxhvCM222121CMCMCMMvRvI 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?K2221CMCMvMRI2221CMCMvMRIMgh2/12MRIghvCMCM2222121MRICMMonday, Nov. 10, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu7Example for Rolling Kinetic EnergyFor solid sphere as 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.CMIThe moment of inertia the sphere with respect to the CM!!Since h=xsin, one obtainsThus using the formula in the previous slideWhat must we know first?RxhvCM2/12MRIghvCMCMsin7102gxvCMUsing kinematic relationshipxavCMCM22The linear acceleration of the CM issin7522gxvaCMCMWhat do you see?Linear acceleration of a sphere does not depend on anything but g and .5/212ghgh710dmr2252MRMonday, Nov. 10, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu8Example for Rolling Kinetic EnergyFor solid sphere as 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.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 torqueMxhRaCMCMWe know that What are the forces involved in this motion?MgfNewton’s second law applied to the CM givesFrictional Force, Normal Forcenxy252MRICMWe obtain f Substituting f in dynamic equationsCMMaMg57sin fMg sinCMMayFcosMgn 0fRCMIRICMRaRMRCM252CMMa52sin75gaCMMonday, Nov. 10, 2003 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu9Work, Power, and Energy in RotationLet’s consider a motion of a rigid body with a single external force F exerting on the point P, moving the object by ds.The work done by the force F as the object rotates through the infinitesimal distance ds=rd is What is Fsin?The tangential component of force F.dWSince the magnitude of torque is rFsin,FOrddsWhat is the work done by radial component Fcos?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. The work put in by the external force thendWsdF rdF sinddtdWdtdIdtdIdtdddIddIWfdfdI222121ifII drF
View Full Document
Unlocking...