1443-501 Spring 2002 Lecture #15Mid Term DistributionsSimilarity Between Linear and Rotational MotionsRolling Motion of a Rigid BodyMore Rolling Motion of a Rigid BodyTotal Kinetic Energy of a Rolling BodyKinetic Energy of a Rolling SphereExample 11.1Example 11.21443-501 Spring 2002Lecture #15Dr. Jaehoon Yu1. Mid-term Results2. Mid Term Problem Recap3. Rotational Motion Recap4. Rolling Motion of a Rigid BodyToday’s Homework Assignment is the Homework #6!!!Mar. 25, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #152Mid Term DistributionsCertainly better than before. You are getting there. Homework and examples must do good for you.46.419.7Mar. 25, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #153Similarity Between Linear and Rotational MotionsAll physical quantities in linear and rotational motions show striking similarity.Similar QuantityLinear RotationalMass Mass Moment of InertiaLength of motion Distance Angle (Radian)SpeedAccelerationForce Force TorqueWork Work WorkPowerMomentumKinetic Energy Kinetic Rotational dmrI2dtdrv dtddtdva dtdmaF IfixxFdxWvFP P221mvK 221IKRLMfidWvmp IL Mar. 25, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #154Rolling Motion of a Rigid BodyWhat is a rolling motion?To simplify the discussion, let’s make a few assumptionsLet’s consider a cylinder rolling without slipping on a flat surfaceA more generalized case of a motion where the rotational axis moves together with the objectUnder what condition does this “Pure Rolling” happen?The total linear distance the CM of the cylinder moved isThus the linear speed of the CM isA rotational motion about the moving axis1. Limit our discussion on very symmetric objects, such as cylinders, spheres, etc2. The object rolls on a flat surfaceR ss=RRs RdtdRdtdsvCMCondition for “Pure Rolling”Mar. 25, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #155More Rolling Motion of a Rigid BodyAs we learned in the rotational motion, all points in a rigid body moves at the same angular speed but at a different linear speed.At any given time the point that comes to P has 0 linear speed while the point at P’ has twice the speed of CMThe magnitude of the linear acceleration of the CM isA rolling motion can be interpreted as the sum of Translation and RotationRdtdRdtdvaCMCMWhy??PP’CMvCM2vCMCM is moving at the same speed at all times.PP’CMvCMvCMvCM+PP’CMv=Rv=0v=R=PP’CM2vCMvCMMar. 25, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #156Total 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 writ 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 rewrite2222212121MRIIKCMP222121CMCMMvIK 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 CMMar. 25, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #157Kinetic 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. 25, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #158Example 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. 25, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #159Example 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
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