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UT Arlington PHYS 1443 - Rolling Motion of a Rigid Body

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PHYS 1443 – Section 003 Lecture #15Similarity 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 11.2Torque and Vector ProductProperties of Vector ProductMore Properties of Vector ProductExample 11.3Wednesday, Nov. 6, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #15Wednesday, Nov. 6, 2002Dr. Jaehoon Yu1. Rolling Motion of a Rigid Body2. Total Kinetic Energy of a Rolling Rigid Body 3. Kinetic Energy of a Rolling Sphere4. Torque and Vector Product5. Properties of Vector Product6. Angular MomentumToday’s homework is homework #15 due 12:00pm, Wednesday, Nov. 13!!Wednesday, Nov. 6, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu2Similarity Between Linear and Rotational MotionsAll physical quantities in linear and rotational motions show striking similarity.Similar QuantityLinear RotationalMass Mass Moment of InertiaLength of motion Displacement Angle (Radian)SpeedAccelerationForce Force TorqueWork Work WorkPowerMomentumKinetic Energy Kinetic Rotational dmrI2dtdrv dtddtdva dtdmaF IfixxFdxWvFP P221mvK 221IKRrMfidWvmp IL Wednesday, Nov. 6, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu3Rolling 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=RRs dtdsvCMCondition for “Pure Rolling”dtdRRWednesday, Nov. 6, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu4More 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 RotationCMaWhy??PP’CMvCM2vCMCM is moving at the same speed at all times.PP’CMvCMvCMvCM+PP’CMv=Rv=0v=R=PP’CM2vCMvCMdtdvCMdtdRRWednesday, Nov. 6, 2002 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 writ the total kinetic energySince vCM=R, the above relationship can be rewritten as221PIK 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 CM221PI 2221MRICM2222121MRICMWednesday, Nov. 6, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu6Kinetic Energy of a Rolling SphereSince vCM=RLet’s consider a sphere with radius R rolling down a hill without slipping.2222121MRIKCMRxhvCM222121CMCMCMMvRvI 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Wednesday, Nov. 6, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu7Example 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.2252MRdmrICMThe 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?RxhvCM2/12MRIghvCMCMsin7102gxvCMUsing kinematic relationshipxavCMCM22The linear acceleration of the CM issin7522gxvaCMCMWhat do you see?Linear acceleration of a sphere does not depend on anything but g and .5/212ghgh710Wednesday, Nov. 6, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu8Example 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.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 torqueMxhRaCMCMWe know that What are the forces involved in this motion?MgfNewton’s second law applied to the CM givesFrictional Force, Normal Forcenxy252MRICMWe obtain f Substituting f in dynamic equationsCMMaMg57sin fMg sinCMMayFcosMgn 0fRCMIRICMRaRMRCM252CMMa52sin75gaCMWednesday, Nov. 6, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu9xyzOTorque 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?sinFrBAC The disk will start rotating counter clockwise about the Z axisThe above quantity is called Vector product or Cross productFrxFrpBut 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 productcosBABAC Result? A scalarsinBABAC Wednesday, Nov. 6, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu10Properties of Vector Product dtBAd Vector Product is Non-commutative What does this mean?If the order of operation changes the result changesABBA ABBA


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UT Arlington PHYS 1443 - Rolling Motion of a Rigid Body

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