Rigid Body Dynamics chapter 10 continuesRigid Body RotationCORRESPONDENCE:Rotational Kinetic EnergyImportant Concept: Moment of InertiaMoment of Inertia, contQuestion – WHAT IS THE MOMENT OF INERTIA OF THIS OBJECT??Let’s Look at the possibilitiesSlide 9Remember the Various DensitiesMoment of Inertia of a Uniform Thin HoopMoment of Inertia of a Uniform Rigid RodMoment of Inertia of a Uniform Solid CylinderMoments of Inertia of Various Rigid ObjectsParallel-Axis TheoremHowcome??From the top:Remember the Center of Mass?So:Parallel-Axis Theorem ExampleMoment of Inertia for a Rod Rotating Around One EndSlide 22Torque (Another Vector): tTorqueMore TorqueingSlide 26Net TorqueTorque vs. ForceTorque UnitsTorque and Angular AccelerationMore Associations:SO?Torque and Angular Acceleration, ExtendedTorque and Angular Acceleration, Extended cont.Torque and Angular Acceleration, Extended finalTorque and Angular Acceleration, Wheel ExampleProblemAnudder oneTorque and Angular Acceleration, Multi-body Ex., 1Torque and Angular Acceleration, Multi-body Ex., 2Slide 41Work in Rotational MotionPower in Rotational MotionWork-Kinetic Energy Theorem in Rotational MotionWork-Kinetic Energy Theorem, GeneralEnergy in an Atwood Machine, ExampleSummary of Useful EquationsRolling ObjectPure Rolling MotionRolling Object, Center of MassRolling Object, Other PointsRolling Motion Cont.Total Kinetic Energy of a Rolling ObjectTotal Kinetic Energy, ExampleTotal Kinetic Energy, Example contRigid Body Dynamicschapter 10 continuesaround and around we go …Rigid Body RotationRotation Axis 222222222121)(2121)(21)(IrmKErmrmKErvvmKEiiitotaliiiiiiiiiiMoment of Inertia:22iiirmICORRESPONDENCE:ImvImv222121RotationalRotational Kinetic EnergyWe there have an analogy between the kinetic energies associated with linear motion (K = ½ mv 2) and the kinetic energy associated with rotational motion (KR= ½ I2)Rotational kinetic energy is not a new type of energy, the form is different because it is applied to a rotating objectThe units of rotational kinetic energy are also Joules (J)Important Concept:Moment of InertiaThe definition of moment of inertia is The dimensions of moment of inertia are ML2 and its SI units are kg.m2We can calculate the moment of inertia of an object more easily by assuming it is divided into many small volume elements, each of mass mi2i iiI r m=�Moment of Inertia, contWe can rewrite the expression for I in terms of mWith the small volume segment assumption,If  is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be knownlim 2 20im i iiI r m r dmD �= D =��2I r dVr=�Question – WHAT IS THE MOMENT OF INERTIA OF THIS OBJECT??Let’s Look at the possibilitiesc dTwo balls with masses M and m are connected by a rigid rod of length L and negligible mass as in Figure P10.22. For an axis perpendicular to the rod, show that the system has the minimum moment of inertia when the axis passes through the center of mass. Show that this moment of inertia is I = L2, where = mM/(m + M).Remember the Various DensitiesVolumetric Mass Density –> mass per unit volume:  = m / VFace Mass Density –> mass per unit thickness of a sheet of uniform thickness, t : tLinear Mass Density –> mass per unit length of a rod of uniform cross-sectional area:  = m / L = Moment of Inertia of a Uniform Thin HoopSince this is a thin hoop, all mass elements are the same distance from the center222I r dm R dmI MR= ==� �Moment of Inertia of a Uniform Rigid RodThe shaded area has a mass dm =  dxThen the moment of inertia is/ 22 2/ 22112LLMI r dm x dxLI ML-= ==� �Moment of Inertia of a Uniform Solid CylinderDivide the cylinder into concentric shells with radius r, thickness dr and length LThen for I( )2 222 12zI r dm r Lr drI MRpr= ==� �dV=L(2rdr)Moments of Inertia of Various Rigid ObjectsParallel-Axis TheoremIn the previous examples, the axis of rotation coincided with the axis of symmetry of the objectFor an arbitrary axis, the parallel-axis theorem often simplifies calculationsThe theorem states I = ICM + MD 2 I is about any axis parallel to the axis through the center of mass of the objectICM is about the axis through the center of massD is the distance from the center of mass axis to the arbitrary axisHowcome??The new axis is parallel to the old axis of rotation.Assume that the object rotates about an axis parallel to the z axis.The new axis is parallel to the original axis.miLriNot insameplaneFrom the top: NEW OLD miLri-----iiiirLrLrLLLaarLiioldnewiiiinewiiinewiinewmIMLImrmMLIrmIanotemI22)2( :)(2212222Center of MassRemember the Center of Mass?01CMiiiCMmMrrrSince for our problem the sum is ABOUT the center ofmass, rCM must be zeroSo:-irLiioldnewmIMLI 22ZEROInew = ICM + ML2Parallel-Axis Theorem ExampleThe axis of rotation goes through OThe axis through the center of mass is shownThe moment of inertia about the axis through O would be IO = ICM + MD 2Moment of Inertia for a Rod Rotating Around One EndThe moment of inertia of the rod about its center is D is ½ LTherefore, 2112CMI ML=2CM22 21 112 2 3I I MDLI ML M ML= +� �= + =� �� �Many machines employ cams for various purposes, such as opening and closing valves. In Figure P10.29, the cam is a circular disk rotating on a shaft that does not pass through the center of the disk. In the manufacture of the cam, a uniform solid cylinder of radius R is first machined. Then an off-center hole of radius R/2 is drilled, parallel to the axis of the cylinder, and centered at a point a distance R/2 from the center of the cylinder. The cam, of mass M, is then slipped onto the circular shaft and welded into place. What is the kinetic energy of the cam when it is rotating with angular speed about the axis of theTorque (Another Vector): FTorqueTorque, , is the tendency of a force to rotate an object about some axis Torque is a vector = r F sin  = Fd = rXFF is the force is the angle the force makes with the horizontal d is the moment arm (or lever arm)More TorqueingThe moment arm, d, is the perpendicular