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UT Arlington PHYS 1443 - PHYS 1443 Lecture Notes

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PHYS 1443 – Section 003 Lecture #16Similarity Between Linear and Rotational MotionsAngular 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 11.8Monday, Nov. 11, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu1PHYS 1443 – Section 003Lecture #16Monday, Nov. 11, 2002Dr. Jaehoon Yu1. Angular Momentum2. Angular Momentum and Torque3. Angular Momentum of a System of Particles4. Angular Momentum of a Rotating Rigid Body5. Angular Momentum ConservationToday’s homework is homework #16 due 12:00pm, Monday, Nov. 18!!Monday, Nov. 11, 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 Monday, Nov. 11, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu3Angular 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 xyzOpL=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.Monday, Nov. 11, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu4Angular 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?Thus the torque-angular momentum relationshipThe same analogy works in rotational motion between torque and angular momentum. Net torque acting on a particle is The net torque acting on a particle is the same as the time rate change of its angular momentumdtpdF dtLddtLdxyzOpL=rxprmWhy does this work?Because v is parallel to the linear momentum dtprd dtpdrpdtrddtpdr 0 Frdtpdr Monday, Nov. 11, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu5Angular Momentum of a System of ParticlesThe total angular momentum of a system of particles about some point is the vector sum of the angular momenta of the individual particles LLLLLn......21Since the individual angular momentum can change, the total angular momentum of the system can change.dtLdextThus the time rate change of the angular momentum of a system of particles is equal to the net external torque acting on the systemLet’s consider a two particle system where the two exert forces on each other.Since these forces are action and reaction forces with directions lie on the line connecting the two particles, the vector sum of the torque from these two becomes 0.Both internal and external forces can provide torque to individual particles. However, the internal forces do not generate net torque due to Newton’s third law.Monday, Nov. 11, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu6Example 11.4A particle of mass m is moving in the xy plane in a circular path of radius r and linear velocity v about the origin O. Find the magnitude and direction of angular momentum with respect to O.rxyvOLUsing the definition of angular momentumSince both the vectors, r and v, are on x-y plane and using right-hand rule, the direction of the angular momentum vector is +z (coming out of the screen)The magnitude of the angular momentum isLSo the angular momentum vector can be expressed askmrvL Find the angular momentum in terms of angular velocity LUsing the relationship between linear and angular speed pr vmr vrm vrm sinmrv90sinmrvmrvmrvmrvmrvvrmL 90sinsinkmrv kmr22mrIMonday, Nov. 11, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu7Angular Momentum of a Rotating Rigid BodyLet’s consider a rigid body rotating about a fixed axisiiiivrmL Each particle of the object rotates in the xy plane about the z-axis at the same angular speed, IdtdLzextThus the torque-angular momentum relationship becomesWhat do you see?Since I is constant for a rigid bodyMagnitude of the angular momentum of a particle of mass mi about origin O is mivirixyzOpL=rxprmSumming over all particle’s angular momentum about z axisiizLL iiizrmL2dtdLzis angular accelerationThus the net external torque acting on a rigid body rotating about a fixed axis is equal to the moment of inertia about that axis multiplied by the object’s angular acceleration with respect to that axis.2iirm iiirm2IdtdIIMonday, Nov. 11, 2002 PHYS 1443-003, Fall 2002Dr. Jaehoon Yu8Example 11.6A rigid rod of mass M and length l pivoted without friction at its center. Two particles of mass m1 and m2 are connected to its ends. The combination rotates in a vertical plane with an angular speed of . Find an expression for the magnitude of the angular momentum.IThe moment of inertia of this system is First compute net external torquecos21lgm1m1 gxyOlm1m2m2 gIf m1 = m2, no angular momentum because net torque is 0. If  2 at equilibrium so no angular momentum.212314mmMlILFind an expression for the magnitude of the angular acceleration of the system when the rod makes an angle  with the horizon.21extThus


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UT Arlington PHYS 1443 - PHYS 1443 Lecture Notes

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