1443-501 Spring 2002 Lecture #19The PendulumExample 13.5Physical PendulumExample 13.6Torsional PendulumSimple Harmonic and Uniform Circular MotionsExample 13.7Damped OscillationMore on Damped OscillationRotational KinematicsExample 10.1Rotational EnergyExample 10.4Calculation of Moments of InertiaExample 10.6Parallel Axis TheoremExample 10.8TorqueTorque & Angular AccelerationExample 10.10Work, Power, and Energy in RotationSimilarity Between Linear and Rotational MotionsRolling Motion of a Rigid BodyTotal Kinetic Energy of a Rolling BodyExample 11.1Torque and Vector ProductAngular Momentum of a ParticleAngular Momentum and TorqueExample 11.4Example 11.6Example 11.8Conditions for EquilibriumMore on Conditions for EquilibriumExample 12.1Example 12.1 ContinuedExample 12.3Example 12.4Example 12.71443-501 Spring 2002Lecture #19Dr. Jaehoon Yu1. The Pendulum2. Physical Pendulum3. Simple Harmonic and Uniform Circular Motions4. Damped Oscillation5. Review Examples Ch. 10-13No Homework Assignment today!!!!!2nd term exam on Wednesday, Apr. 10. Will cover chapters 10 -13.Apr. 8, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #192The PendulumA simple pendulum also performs periodic motion.The net force exerted on the bob is 22sin0cosdtsdmmamgFmgTFAtArALsAgain became a second degree differential equation, satisfying conditions for simple harmonic motionIf is very small, sin~Since the arc length, s, is sin2222gdtdLdtsdsin22LgdtdresultsmgmLTs222LgdtdLggiving angular frequencyThe period for this motion isgLT22The period only depends on the length of the string and the gravitational accelerationApr. 8, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #193Example 13.5Christian Huygens (1629-1695), the greatest clock maker in history, suggested that an international unit of length could be defined as the length of a simple pendulum having a period of exactly 1s. How much shorter would out length unit be had this suggestion been followed?Since the period of a simple pendulum motion isgLT22The length of the pendulum in terms of T is 224gTL Thus the length of the pendulum when T=1s is mgTL 248.048.914222Therefore the difference in length with respect to the current definition of 1m ismLL 752.0248.011 Apr. 8, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #194Physical PendulumPhysical pendulum is an object that oscillates about a fixed axis which does not go through the object’s center of mass.Therefore, one can rewriteThus, the angular frequency isThe magnitude of the net torque provided by the gravity is sinmgdImgdAnd the period for this motion ismgdIT22By measuring the period of physical pendulum, one can measure moment of inertia.OCMddsinmgConsider a rigid body pivoted at a point O that is a distance d from the CM.sin22mgddtdII Then 2ImgdImgddtdsin22Does this work for simple pendulum?Apr. 8, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #195Example 13.6A uniform rod of mass M and length L is pivoted about one end and oscillates in a vertical plane. Find the period of oscillation if the amplitude of the motion is small.Moment of inertia of a uniform rod, rotating about the axis at one end isSince L=1m, the period isLOPivotCMMg231MLI The distance d from the pivot to the CM is L/2, therefore the period of this physical pendulum isgLMgLMLMgdIT322322222Calculate the period of a meter stick that is pivot about one end and is oscillating in a vertical plane.sgLT 64.18.9322322 So the frequency is161.01 sTfApr. 8, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #196Torsional PendulumWhen a rigid body is suspended by a wire to a fixed support at the top and the body is twisted through some small angle , the twisted wire can exert a restoring torque on the body that is proportional to the angular displacement.Applying the Newton’s second law of rotational motionThus, the angular frequency isThe torque acting on the body due to the wire is IAnd the period for this motion isIT 22This result works as long as the elastic limit of the wire is not exceeded22dtdII is the torsion constant of the wire 2Idtd22OPmaxThen, again the equation becomesApr. 8, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #197Simple Harmonic and Uniform Circular MotionsUniform circular motion can be understood as a superposition of two simple harmonic motions in x and y axis.When the particle rotates at a uniform angular speed , x and y coordinate position becomeSince the linear velocity in a uniform circular motion is A, the velocity components are tAAytAAxsinsincoscost=0xyOPAxyOPAQxyt=t=t+xyOPAQvvx tAvvtAvvyxcoscossinsinxyOPAQaaxSince the radial acceleration in a uniform circular motion is v2/A=2, the components are tAaatAaayxsinsincoscos22Apr. 8, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture #198Example 13.7A particle rotates counterclockwise in a circle of radius 3.00m with a constant angular speed of 8.00 rad/s. At t=0, the particle has an x coordinate of 2.00m and is moving to the right. A) Determine the x coordinate as a function of time.Since the radius is 3.00m, the amplitude of oscillation in x direction is 3.00m. And the angular frequency is 8.00rad/s. Therefore the equation of motion in x direction isSince x=2.00, when t=0However, since the particle was moving to the right =-48.2o, Using the displcement tmAx 00.8cos00.3cos 2.4800.300.2cos ;cos00.300.21m 2.4800.8cos00.3 tmxFind the x components of the particle’s velocity and acceleration at any time t. 2.4800.8sin/0.242.4800.8sin00.800.3 tsmtdtdxvxLikewise, from velocity 2.4800.8cos/1922.4800.8cos00.80.242 tsmtdtdvaxApr. 8, 2002 1443-501 Spring 2002Dr. J. Yu, Lecture
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