1443-501 Spring 2002 Lecture #16Kinetic Energy of a Rolling SphereExample 11.1Example 11.2Torque and Vector ProductProperties of Vector ProductMore Properties of Vector ProductExample 11.3Angular Momentum of a ParticleAngular Momentum and TorqueAngular Momentum of a System of ParticlesExample 11.4Angular Momentum of a Rotating Rigid BodyExample 11.6Conservation of Angular MomentumExample 11.81443-501 Spring 2002Lecture #16Dr. Jaehoon Yu1. Torque and Vector Product2. Angular Momentum of a Particle3. Angular Momentum of a Rotating Rigid Object4. Conservation of Angular MomentumToday’s Homework Assignment is the Homework #7!!!Mar. 27, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #152Kinetic Energy of a Rolling SphereSince vCM=RLet’s consider a sphere with radius R rolling down a hill without slipping.2222121MRIKCMRxhvCM2222212121CMCMCMCMCMvMRIMvRvIKSince 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?222/1221MRIghvMghvMRIKCMCMCMCMMar. 27, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #153Example 11.1For 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.2252MRdmrICMThe 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?RxhvCMghghMRIghvCMCM7105/212/122sin7102gxvCMUsing kinematic relationshipxavCMCM22The linear acceleration of the CM issin7522gxvaCMCMWhat do you see?Linear acceleration of a sphere does not depend on anything but g and .Mar. 27, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #154Example 11.2For 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.0cossinMgnFMafMgFyCMxGravitational 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 torqueMxhRaCMCMCMIfRWe know that What are the forces involved in this motion?MgfNewton’s second law applied to the CM givesFrictional Force, Normal Forcenxy252MRICMWe obtain CMCMCMMaRaRMRRIf52522 Substituting f in dynamic equationssin75 ;57sin gaMaMgCMCMMar. 27, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #155xyzOTorque 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?sinFrsinBABACBACThe 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 scalarMar. 27, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #156Properties of Vector Product dtBdABdtAddtBAdVector Product is Non-commutative What does this mean?If the order of operation changes the result changesABBA ABBA Following the right-hand rule, the direction changesVector Product of a two parallel vectors is 0.00sinsin BABABAC0AAThus,If two vectors are perpendicular to each otherABBABABABA 90sinsinVector product follows distribution law CABACBA The derivative of a Vector product with respect to a scalar variable isMar. 27, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #157More Properties of Vector ProductThe relationship between unit vectors, kji and , ,jkiikijkkjkijjikkjjii 0 kBABAjBABAiBABABBAAkBBAAjBBAAiBBBAAAkjiBAxyyxxzzxyzzyyxyxzxzxzyzyzyxzyxVector product of two vectors can be expressed in the following determinant formMar. 27, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #158Example 11.3Two vectors lying in the xy plane are given by the equations A=2i+3j and B=-i+2j, verify that AxB= -BxA ijjijijiBA 34232Since (2,3)(-1,2)ABkji kkkijkBA 73434 ijjijijiAB 43322Using the same unit vector relationship as above kkkAB 743 Therefore, AxB= -BxA Now prove this using determinant methodMar. 27, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #159Angular 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 motions, angular momentum will do the same for rotational motions.sinmvrL xyzOpL=rxprmLet’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? 22/ smkg Note that L depends on origin O. Why? Because r changesThe direction of L is +zWhat 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.The point O has to be inertial.Mar. 27, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #1510Angular Momentum and TorqueTotal external forces exerting on a particle is the same as the change of its linear momentum.Can you remember how net force exerting on a particle and the change of its linear momentum are related?dtpdrFr Thus the torque-angular momentum relationshipThe same analogy works in rotational
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