12-Dimensional Motion: Rotation and Translation8.01W13D1 Fall 2006Introduction: Rotation and Translation of Rigid BodyTranslational motion: the total external force of gravity acts on center-of-massRotational Motion: object rotates about center-of-mass. Note that the center-of-mass may be acceleratingsyext total totalcmcmsddmmdt dt== =VpFAGGGGIntroduction: Rotation about the Center-of-Mass of a Rigid BodyThe total external torque produces an angular acceleration about the center-of-massis the moment of inertial about the center-of-massis the angular acceleration about the center-of-massis the angular momentum about the center-of-mass Icmextcmcm cm cmdIdtα==LGGGτ αcmcmLGTwo Reference Systems‘Laboratory reference frame’ is a coordinate system fixed to the lab(the center-of-mass of the rigid body is translating in this frame)‘Center-of-mass reference frame’ is a coordinate system that moves with the center-of-mass of the body which may or may not be accelerating with respect to the laboratory framecm=+rRrGGGcm=+vVvGGGcm=+aAaGGG2Questions: Torque about the Center of MassWhat forces do we use to calculate the torque about the center of mass?Do we use the same forces in the laboratory and center-of-mass reference frames?Where does the gravitational force act?Answer: The gravitational force acts at the center of mass as does the ‘fictitious force’ that is actually the result of the effect of the acceleration of the center of mass. Since they act at the center of mass, neither contributes to the torque. So we use the same forces that are observed in the laboratory frame to calculate the torque about the center-of-mass.extcm ,1Ncm i ii==×∑rFGGGτTorque due to Uniform Gravitational ForceThe total torque on a rigid body due to the gravitational force can be determined by placing all the gravitational force at the center-of-mass of the object.S,grav , grav, , ,111totat,totat1totatS,cm1NNNSi i Si i i SiiiiNiSiimmmmmm=====×= ×= ×⎛⎞=×⎜⎟⎝⎠=×∑∑∑∑rF r g rgrgRgGGGG GGGGGGGτTorque for Rotation and TranslationThe total torque about S is given by where the first term torque about S due to the total external force acting at the center-of-mass and the second term is torque about the center-of-mass is only due to the forces as seen in the laboratory frame.The total external torque produces an angular acceleration about the center-of-masscm cm,i1iNii===×∑rFGGGτ,cm cmSS=+GG Gττ τext,cm ,cmSS=×RFGGGτextcmcm cm cmdIdtα==LGGGτConcept Question: Rotation and TranslationTwo disks are separated by a spindleof smaller diameter. A string is woundaround the spindle and pulled gently.Which positions of the string causethe assembly to roll to the right?1) Only A2) Only B3) Only C4) A and B5) B and C6) A and B and C7) None of the configurations shown3Demo: Giant Yo-YoDemo: Descending and Ascending Yo-YoMwheel+axle= 435 gRouter≅ 6.3cmRinner≅ 4.9 cmIcm≅12MRouter2+ Rinner2()= 1.385× 104g ⋅ cm2Rotational Work-Kinetic Energy TheoremKinetic energy of rotation about center-of-mass Rotation and translation Wtotal=ΔKtrans+ΔKrot Wrot=12Icmωcm, f2−12Icmωcm,i2= Krot, f− Krot,i≡ΔKrot Wtotal=ΔKtrans+ΔKrot=12mvcm, f2−12mvcm,i2⎛⎝⎜⎞⎠⎟+12Icmωf2−12Icmωi2⎛⎝⎜⎞⎠⎟Concept Question: Rotation and TranslationTwo cylinders of the same size and mass roll down an incline, starting from rest. Cylinder A has most of its mass concentrated at the rim, while cylinder B has most of its mass concentrated at the center. Which has more total kinetic energy at the bottom? 1. A 2. B 3. Both have the same4Concept Question: Rotation and TranslationTwo cylinders of the same size and mass roll down an incline, starting from rest. Cylinder A has most of its mass concentrated at the rim, while cylinder B has most of its mass concentrated at the center. Which reaches the bottom first? 1. A 2. B 3. Both have the same Demo: Rolling CylindersConcept Question Problem:Cylinder Rolling Down an Inclined PlaneSimple PendulumSimple Pendulum: bob of mass m hanging from end of massless string string pivoted at S.Torque about SAngular accelerationMoment of inertial of a point mass about S,Rotational Law of MotionSimple harmonic oscillator equation,ˆˆˆˆ(sin cos) sinSSsmmlmg lmgθθ=×=×− + =−rgr r kGGGτθ22ˆddtθ= kGα IS= ml2SSSI=GGτα−lmgsinθ= ml2d2θdt25Simple Pendulum: Small Angle ApproximationAngle of oscillation is small Simple harmonic oscillatorAnalogy to spring equationAngular frequency of oscillationPeriod sinθ≅θ d2θdt2≅−glθ d2xdt2=−kmx ω0≅gl T0=2πω0≅ 2πlgPhysical PendulumPendulum pivoted about point S Gravitational force acts center of mass Center of mass distance from the pivot point lcmPhysical PendulumRotational dynamical equation Small angle approximation Equation of motionAngular frequencyPeriodSSSI=GGτα sinθ≅θ d2θdt2≅−lcmmgISθ ω0≅lcmmgIS T =2πω0≅ 2πISlcmmgRotation and TranslationThe center-of-mass of the physical pendulum is undergoing non-uniform circular motion.The body is also rotating about the center-of-mass.The physical pendulum is a special case (pivoted about a point) of a more general motion of a rigid body.6Concept Question: Physical PendulumA physical pendulum consists of a uniform rod of length l and mass m pivoted at one end. A disk of mass m1 and radius a is fixed to the other end. Suppose the disk is now mounted to the rod by a frictionless bearing so that is perfectly free to spin. Does the period of the pendulum1. increase?2. stay the same?3. decrease?Demo: Identical Pendulums, Different PeriodsDouble pivot: no rotation about center of mass.Single pivot: body rotates about center of mass.Appendices1. Torque About the Center-of-Mass2. Angular Momentum for Rotation and Translation3. Torque for Rotation and TranslationQuestions: Torque about the Center of MassWhat forces do we use to calculate the torque about the center of mass?Do we use the same forces in the laboratory and center-of-mass reference frames?Where does the gravitational force act?Answer: The gravitational force acts at the center of mass as does the ‘fictitious force’ that is actually the result of the effect of the acceleration of the center of mass. Since they act at the center of mass, neither contributes to the torque. So we use the same forces that are observed in the laboratory frame to calculate the torque about the center-of-mass.extcm ,1Ncm i ii==×∑rFGGGτ7Two Reference
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