PHY126 HW Set 2 (Chap.11)

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1 PHY16 HW Set (Chap.11) Problem 11.4 Descrpton:A physcs student s standng on an ntally motonless, rctonless turntable wth rotatonal nerta.31. She's holdng a wheel o rotatonal nerta. spnnng at 13 about a vertcal axs, as n the gure. When she turns the wheel upsde down, student and turntable begn rotatng at 7. What s the student's mass, consderng her to be a cylnder 3 Express your answer usng two sgncant gures. n dameter? How much work dd she do n turnng the wheel upsde down? Neglect the dstance between the axes o the turntable and wheel. Express your answer usng two sgncant gures. Solutons (a) Snce the turn table s rctonless, there are no external torques about ts axs, and the z component o angular momentum s conserved. Intally ths s all spn o the wheel, L = (. kg m )(13 rpm). When the wheel s nverted, the student and turrable acqure anugular momentum, such that L = L or = L. I I tot s the total moment o nerta about the z-axs, = I tot (7 rpm) = (. kg m )(13 rpm), or I tot =.817 kg m. Now I tot

2 ncludes the nerta o the turntable,.31 kg m, the student (a cylnder, (1/)M(.5 m), and CM o the wheel, mh 9whch can be neglected the axes o the wheel and turntable concde, and the spn axs s rctonless). Then (1/)M(.15 m) =( ) kg m, or M=45.1 kg. (b) The work done by the student s muscle equals the change n knetc energy, W = K = ( 1/ ) ω (the knetc energy o the spnnng wheel s unchanged, as s the gravtatonal nc I tot potental energy.) Thereore: W nc = (1/ )(.817 kg m )(7π / 3 s) =. J. Problem Descrpton: A thn rod o length Land mass M s ree to pvot about one end.. I t makes an angle wth the horzontal, nd the torque due to gravty about the pvot pont. (You'll need to ntegrate the torques on the ndvdual mass elements composng the rod.) Fnd the value o such that when the rod s released rom rest at ths angle, the vertcal component o the ntal acceleraton o the rod's tp s equal to (the acceleraton o gravty). Solutons (a) The rod has mass M and length L, and pvots at one end. The angle t makes wth the horzontal s θ. We use an ntegral or the torque, although we mght expect that ths torque was equal to the torque actng r r r r d τ on a mass element dm s dτ = df on the center o mass o the rod. The torque, so d τ = xgdm cosθ. We ntegrate ths over the length o the rod, where dm = ( M / L) dx. L M Mg cosθ L Mg cosθ 1 1 Thereore τ = xg cosθ dx = xdx L = mgl cosθ. L L = L τ = Iα and a = Lα = g, and the moment o nerta o a rod rotatng around one end s (1/ ) mgl cosθ 3 g cosθ I = (1/ 3) ML, α = τ / I = =. Settng a = Lα = g, (1/ 3) ML L (b) Snce g = ( 3/ ) g cosθ, so θ = cos 1 ( / 3) = 48.. Problem Descrpton: A sold ball o mass and radus s spnnng wth angular velocty about a horzontal axs. It drops vertcally onto a surace where the coecent o knetc rcton wth the ball s (see the gure). Fnd expressons or (a) the nal angular velocty once t's acheved pure rollng moton and (b) the tme t takes untl t's n pure rollng moton.

3 Solutons A sold spnnng ball drops onto a rctonal surace. At rst t sldes, but due to rcton t wll slow down ts spn and ncrease ts lnear moton untl t s purely rollng wthout sldng. The rctonal orce s F = µ F = Mg. There s a torque on the ball due to rctonal orce, actng on the edge. Ths torque n µ τ = µmgr serves to slow the ball s rotaton: we can use τ = Iα to nd the angular acceleraton α and then use ω = ω + αt to nd the resultng angular speed. The rctonal orce on the ball also accelerates the ball, so we can use = Ma and v = v + at to nd the speed o the ball. α = τ / I = ( µ MgR) /[( / 5) MR ] = 5µ g /(R), so ω = ω 5µg /(R). Snce a = F / M = µ Mg / M = µ g, v = v + µ gt. We set t R ω = v = Rω ( 5/ ) µ gt = v ( = ) + µ gt, so solvng ths or t gves R ω = µ gt(1 + 5/ ). So t = 7Rω /( µ g), and pluggng ths nto the equaton or ω gves ω = ω [ 5µ g /(R)][Rω /( µ g)] = ω (5/ 7) ω = ( / 7) ω. (a) ω = ( / 7) ω (b) t = ( 7 / )[ Rω /( µ g)].

4 Problem Descrpton: The gure shows the dmensons o a 88 wooden baseball bat whose rotatonal nerta about ts center o mass s.48. I the bat s swung so ts ar end moves at 5. Hnt: Remember the parallel-axs theorem. Express your answer usng two sgncant gures., nd ts angular momentum about the pvot pont Fnd the constant torque appled about to acheve ths angular momentum n.5. Express your answer usng two sgncant gures. Solutons (a) The rotatonal nerta o the bat about pont P s I P = Mh + I cm = (.88 kg)(.43 m) +.48 kg m =.11 1 kg m (see Eq.1.17.) Its angular velocty (drecton parallel to the axs o swng) about P s -1 ω P = v / r = (5 m/s)/(.74 m) = 67.6 s. Thereore, the angular momentum about P (also drected along the axs o swng) s LP = I P ω P = 14. J s. (b) For constant torque along the axs o swng, τ = L / t mples τ = ( 14. J s)/(.5 s) = 56.9 N m. ( we assume that the bat s ntally at rest.)

5 Problem Descrpton: A unorm, sold, sphercal asterod wth mass and radus 1. s rotatng wth a perod o 4.3 equator at 8.4. A meteorod movng n the asterod's equatoral plane crashes nto the. It hts at a 58 angle to the vertcal and embeds tsel at the surace. Ater the mpact the asterod's rotaton perod s 3.9. Fnd the meteorod's mass. Express your answer usng two sgncant gures. Soluton I we neglect external gravtatonal torque, the total angular momentum o the asterod and meteorod s conserved. Snce the meteorod s movng n the asterod s equatoral plane, ts angular momentum about the center o the asterod s n the same drecton as the rotatonal angular momentum o the asterod. r r L = Iω + R mv = ( / 5) MR ω + mvrsn 58 and L = ( I + mr ) ω = (M / 5 + m) R ω. Settng L =, solvng or m, and usng the L perod o rotaton nstead o the angular speed, we nd:.4mr( T T ) 7 m = =.81 1 kg. vrsn 58 RT

Rotation Kinematics, Moment of Inertia, and Torque

Rotation Kinematics, Moment of Inertia, and Torque Rotaton Knematcs, Moment of Inerta, and Torque Mathematcally, rotaton of a rgd body about a fxed axs s analogous to a lnear moton n one dmenson. Although the physcal quanttes nvolved n rotaton are qute