PHYS 218sec. 517-520ReviewChap. 9Rotation of Rigid BodiesWhat you have to know• Rotational kinematics (polar coordinate system)• Relationship & analogy between translational and angular motions• Moment of inertia• Rotational kinetic energy• Section 9.6 is not in the curriculum.Analog between translation and rotation motionTranslation Rotationposition angle velocity angular velocity acceleration angular accelexvaθω22ration mass moment of inertia 11kinetic energy rotational kinetic energy 22force torque momentum mImv IFpαωτ21 21222 angular momentum 111222LFma IWEE WEEEmvU Emv IUτωω=== − = −=+ =++∑∑Angular velocity and accelerationθslength of the arc : , where is in radiansssrθθ=rAngular velocityratio of the angular displacement to tddtθθω∆∆=Angular velocityratio of the angular velocity to tddtωωα∆∆=The angular velocity and angular acceleration are vectors.Follow the right hand rule.0counterclockwiseω > 0clockwiseω <Rotation with constant angular accelerationAll the formulas obtained for constant linear acceleration are valid for the analog quantities to translational motion()()0200220000angular motion (fixed-axis rotation)constant12212ttttαωω αθθ ω αωω αθθθθ ωω==+=+ +− = −− =+()()0200220000linear motion (straight-line motion)constant12212avv atxvtatvv axxxx vvtθ==+=+ +− = −− =+Polar coordinate systemrGxyˆrˆθˆiˆjˆˆunit vectors in polar coordinate system: ,ˆˆunit vectors in Cartesian coordinate system: ,ˆPosition vector of : rijPr rrθ=GP()ˆˆˆcos sin ,ˆˆˆˆsin cos ,ˆˆˆˆˆsin cosrijdrijddr dr dijdt d dtθθθθθθθθθωωθθ=+= − +===− +=()ˆˆˆsin cos ,ˆˆˆˆcos sin ,ˆˆˆˆˆcos sinijdijrddddij rdt d dtθθθθθθθθθθθθωωθ= − += −−= −==−− = −()tanThenˆˆˆˆ ˆ.dr d dr dr drvrrrrrrdt dt dt dt dtvrωθω== =+=+ ∴=GGTherefore, this is valid in general.Polar coordinate system222222222tanˆˆˆˆ2 For a circular motion, 0 and we getˆˆ,raddv d dr d r drarrrrrdt dt dt dt dtdr d rdt dtarrrvvarr arrrωθ ω ω α θωαθωα⎛⎞⎛⎞ ⎛ ⎞⎟⎜⎟⎟⎜⎜⎟== + = − ++⎟⎟⎜⎜⎜⎟⎟⎟⎜⎜ ⎜⎟⎜⎝⎠ ⎝ ⎠⎝⎠⇒ === − +⎛⎞⎟⎜⇒ = − = − = − =⎟⎜⎟⎜⎝⎠GGGEnergy in rotational motionRotational motion of a rigid bodyirω()()()222 2222 22 2Kinetic energy of the -th particle11 122 2Then the total kinetic energy of a rigid body reads11 122 2iii ii iiiii iiiKmvmr mrKKmr mrIωωωωω== === = =∑∑ ∑2This defines the moment of inertia;iiImr=∑Depends on1. How the body’s mass is distributed in space,2. The axis of rotationMoment of inertia2If the mass distribution is continuous,Irdm=∫Moments of inertia for various rigid bodies are given in section 9.6Rotational kinetic energy is obtained by summing kinetic energies of each particles.Each particle satisfies Work-Energy theoremWork-Energy theorem holds true for rotational kinetic energy21WKK= −includes rotational kinetic energyParallel-axis theoremMoments of inertia depends on the axis of rotation.There is a simple relationship between Icmand IPif the two axes are parallel to each other.xyCMPTwo axes of rotation()()2222Position of , and ; choose the CM as the origin 0CM i i i cm cmPab d abImxy xy==+=+ ==∑()()(){}()22222Then2Pii ii iii i iiImrP mxaybmx y a mx= − = − + −=+−∑∑∑∑GG2iibmy−∑()222iCMma bIMd++=+∑2Parallel-axis theorem: PCMIIMd=+1. If you know ICM, you can easily calculate IP.2. IPis always larger than ICM. Therefore, ICMis smaller than any IP, and it is natural for a rigid body to rotate around an axis through its CM.Ex 9.8Unwinding cable IFFinitialfinal2mmass of the cylinder: 50 kg, radius of the cylinder: 0.06 m9.0 N, final angular speed and the final speed of the cable?mRF===2211 2 2122speed1Moment of inertia of the cylinder in this motion: 2Use Work-Energy theorem to obtain the 10, , 02Then, where 18 J102220 rad/s,other otherotherotherImRKU K I UEW E W FdWIWvRIωωω=== = =+= ==+=⇒ == =1.2 m/sω =Ex 9.9Unwinding cable IIcylinder with mass and radius and block of mass what is ?MRmvmhinitial finalv1122220,11,022KUmghKmvIUω===+ =Kinetic energy of mRotational kinetic energy of M;I=MR2/2, ω=v/R222 2By energy conservation,111 110222 2222 : speed when we neglect the rotational motion of the cylinder12vmgh mv MR m M vRghvghMm⎛⎞⎛⎞ ⎛ ⎞⎟⎟ ⎟⎜⎜ ⎜=+ +=+⎟⎟ ⎟⎜⎜ ⎜⎟⎟ ⎟⎜⎜ ⎜⎝⎠⎝⎠ ⎝ ⎠⇒
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