Experiment 08: Physical PendulumExperiment 08: Physical Pendulum8.01tNov 10, 2004Goals Investigate the oscillation of a real (physical) pendulum and compare to an ideal (point mass) pendulum. Angular frequency calculation: Practice calculating moments of inertia, using them, and solving the τ = I a equation of motion.Equipment setup Suspend 1m ruler so it can swing over edge of table. Measure the period of oscillation with the DataStudio motion sensor. Set motion sensor on narrow beam, aim it to just miss support rod and hit ruler about 25 cm away. Place a chair about 40-50 cm from motion sensor to intercept ultrasound beam when ruler swings out of beam.Understanding the graphsPosition vs. time data from the motion sensor.What is happening:1. Along the top plateaus marked by A?2. At the downward peaks marked by B?How do you use this graph to find the period of oscillation of the pendulum?Starting DataStudio Create a new experiment. Plug motion sensor into the 750 and drag their icons to inputs in the Setup window. Double-click the Motion Sensor icon, set trigger rate to 120. Plot position vs. time.Ruler pendulumDelayed Start = None. Automatic Stop = 10 sec.Pull ruler aside and release it to swing at the same time you start DataStudio.Measure periods filling in the table below.Click 0.520.50 m0.250.25 m0.100.10 mPeriodθ0DisplacementModified ruler pendulumClip a 50g brass weight to the ruler at positions in table in order to change the moment of inertia. (Clip is 8.6 g.)Measure the period of oscillation filling in the table below:58.6 g58.6 g58.6 gWeight0.90 m0.20 m0.50 m0.20 m0.25 m0.20 mPeriodPositionDisplacementAngular Momentum and Fixed Axis Rotation8.01tNov 10, 2004Dynamics: Translational and Rotational MotionTranslational Dynamics• Total Force• Momentum of a System• Dynamics of TranslationRotational Dynamics of point mass about S• Torque • Angular Momentum about S• Dynamics of RotationtotalextFGtotaltotalextddt=pFGGtotalpG,SSm=×Lr pGGGtotalSSddt=LτGG,SSmm=×τ rFGGGAngular Velocity Vector and Angular Acceleration Vectorfor Fixed Axis Rotation• Fixed axis of rotation: z-axis• Angular velocity vector• Angular acceleration vector ˆddtθ=ω kG22ˆddtθ=α kGAngular Momentum of a Point Particle• point particle of mass m moving with a velocity • momentum • Fix a point S• vector from the point to the location of the object • angular momentum about the point SvGm=pvGG,SmrG,SSm=×Lr pGGGCross Product: Angular Momentum of a Point ParticleMagnitude:a) moment arm b) Perpendicular momentum,SSm=×Lr pGGG,sinSSmθ=LrpGGG,sinSmrθ⊥= rGSr⊥=LpGGsinpθ⊥= pG,SSTp⊥=LrGGCross Product: Angular Momentum of a Point ParticleDirection Right Hand RuleAngular Momentumfor Fixed Axis Rotation• Fixed axis of rotation: z-axis• Angular velocity• angular momentum about the point S• z-component of the angular momentum about S, ˆrω=× =vrω kGGG,SSm Sm=×=×Lr pr vGGGGG2,ˆˆˆSSmmrmvrmr mrωω=× = = =Lr v k k kGGGFixed Axis Rotation• Angular Momentum about z-axis• Rotational Dynamics2,totalSz SLmr Iωω==,,totalSzSz S SdLdIIdt dtωτα===PRS QuestionA person spins a tennis ball on a string in a horizontal circle (so that the axis of rotation is vertical). At the point indicated below, the ball is given a sharp blow in the forward direction. This causes a change in angular momentum dLin the 1. x direction 2. y direction 3. z directionPRS QuestionA dumbbell is rotating about its center as shown. Compared to the dumbbell's angular momentum about its center, its angular momentum about point B is 1. bigger. 2. the same. 3. smaller.Time Derivative of Angular Momentum for a Point ParticleTime derivative of the angular momentum about S:Product ruleKey Fact:Result:(),totalSSmdddt dt=×LrpGGG(),,,itotalSmSSm Smdddddt dt dt dt=×=×+×rLrp pr pGGGGGGG,,itotalSSm Sm Sdddt dt=×=×=LrprFτGGGGGG,,Sm Smddmmdt dt=⇒×=×=rrvvvv0GGGGGGGTorque and the Time Derivative of Angular Momentum for a Point ParticleTorque about a point S is equal to the time derivative of the angular momentum about S .totalSSddt=LτGGAngular Momentum for a System of Particles• Treat each particle separately• Total Angular Momentum for System about S,,iSi Sm i=×LrpGGG,,11iiN iNtotalSSiSmiii======×∑∑LLrpGGGGAngular Momentum and Torque for a System of Particles• Total torque about S is the time derivative of angular momentum about S,,,11 1itotaliN iN iNSitotalSSm i Si Sii idddt dt== === ===×==∑∑ ∑LLrF ττGGGGGGAngular Momentumof a Rigid Body for Fixed Axis Rotation• Fixed axis of rotation: z-axis• angular momentum about the point S• z-component of the angular momentum about S, ,, ,Si Si i Si i im=×=×∆Lrpr vGGGGG(),,iSi O i i izm=×∆Lr vGGG,, ,iiSi SO O i=+rr rGG GZ-component of the Angular Momentum about S • Mass element• radius of the circle• momentum • z-component of the angular momentum about S • Velocity• Summary:,ir⊥iimv∆(),,Si i i izLrmv⊥=∆,iivrω⊥=()()2,, ,Si i i i i izLrmvmrω⊥⊥=∆ =∆im∆Z-component of the Angular Momentum about S• Sum over all mass elements• Continuous body• Moment of Inertia• Main Result()()()2,,totalSSiiizziiLLmrω⊥==∆∑∑()()2totalSzbodyLdmrω⊥=∫2,()SzbodyIdm r⊥=∫(),totalSSzzLIω=Torque and Angular Momentum for Fixed Axis Rotation• torque about S is equal to the time derivative of the angular momentum about S• resolved in the z-direction totaltotalSSddt=LτGG()()()2,,, ,2totalSSztotalzSSzSzSzzdLdIddII Idt dt dt dtωωθτα=====Conservation of Angular Momentum about a Point S• Rotational dynamics• No external torques• Change in Angular momentum is zero• Angular Momentum is conservedtotaltotalSSddt=LτGGtotaltotalSSddt==L0 τGGG()()0total total totalSS Sf∆≡−=LL L 0GGGG()()0total totalSSf=LLGGPRS QuestionA figure skater stands on one spot on the ice (assumed frictionless) and spins around with her arms extended. When she pulls in her arms, she reduces her rotational inertia and her angular speed increases so that her angular momentum is conserved. Compared to her initial rotational kinetic energy, her rotational kinetic energy after she has pulled in her arms must be 1. the same. 2. larger because she's rotating faster. 3. smaller because her rotational inertia is smaller.Conservation Principles• Change in mechanical energy• No non-conservative work• Change in momentum• No external forcestotalnc mechanicalWE KU=∆
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