Nov. 5, 2009**Lecture** 17

Outline• First hour– Some examples about moment of inertia– Angular momentum– Torque• Second hour– Atwood’s machine– Friday’s experiment– Pure roll: Rolling without slipping & skidding

P. 260Moment of Inertia

• Group problem: A crucial part of a piece of machinerystarts as a flat uniform cylindrical disk of radius and mass M.It then has a circular hole of radius drilled into it. The hole’scenter is a distance h from the center of the disk. Find themoment of inertia of this disk (with -off-center hole) whenrotated about its center, C. [Hint: Consider a solid disk and“subtract” the hole; use the parallel-axis theorem.]

SolutionI = MR = I + I → I = MR − I1 2 1 2total 2 0 remainder cutout remainder 2 0 cutout22 2 M12 R1I = m R + m h ; m = π R = M →cutout 2 cutout 1 cutout cutout 2 12π R R0 0RI = MR − m R + m h = MR − M R + hR21( ) ( 1 )1 2 1 2 2 1 2 1 2 2remainder 2 0 2 cutout 1 cutout 2 02 2M R R R h1= ( 4 −4 −22 22 2 0 1 1 )R00

flywheelFrom W. Lewin classnote at MIT

Angular MomentumLQ = r × p=m r × vQmagnitudedirection( )QL = mvr sinθLQQ ,⊗QCrQ⊥,rQ,⊥mθr QQv p Angular momentum depends on the reference point you choose(Not like momentum). It is not an intrinsic property of a particle.e.g.LC= 0This special property makes angular momentum is more difficult to learn

Case Ivt ()r CCangular momentum relative to point Ct = 0, LC= 0t, L ≠ 0angular momentum changes all the timeC

Case II: constant circular motionvt () L = mr × v = mrvC C cwhich does NOT changeCr Cr CQQuestion: how about other reference point?Point Q: it does notOnly at point C , angular momentum is conserved

More general: L = r × pQQ dLQdrd Q dp= ( rQ× p)= × p+ rQ×dt dt dt dt0 dp= mv × v + rQ×dt = × FQrQτ ≡ × Fr Q: torque

Case II: constant circular motion L = mr × v = mrvC C cvt ()r CrCwhich does NOT change dLC× F = 0 → = 0dtQCr CQuestion: how about other reference point?Point Q: it does notrQ dLQ× F ≠ 0 → ≠0dt

L=mr,v sin90Ci , iCi iL= mrv = mCdisk ,0r ω2i c, i i i C,i= ω∑mr=iI ωCvi2i C,i=rωQuestion: is this angular momentum different for some otherpoint? e.g. some point “A”Answer: No, an object rotates about an axis passing its CM, theangular momentum is independent of referenced point –named spin angular momentum.iACMmir i

• Orbital angular momentum:– The value will depend on the reference point– e.g. the earth rotates around the sun. Theangular momentum (orbital) does not changeonly when your reference point is on Sun• Spin angular momentum:– an object rotates about its center of mass.This angular momentum is independent of thereference (you got the same answer in thecenter of mass or other point)– an intrinsic property of an object– e.g. the Earth is spinning about its center ofmass, so the Earth has an intrinsic spinangular momentum.

Experiment: ice-skater's delight.

Experiment: ice-skater's delight.1 2I = MR21I = MR + 2 × mL22 2

Summary L ( ) Q= rQ × p= m rQ× v (1)LQ = IQωQ(2)τ ≡ r × FQ Q(3)τQ≡ IQαQ(4)dLQ = τQ, ext= rQ × Fext(5)dtrQ,⊥mθr QQv p Rotation about a stationary axis through the center of massL = I ωCM CM CM(6)(Spin) angular momentum