Unit 6 practice test: Energy and Momentum of Rotating Systems

Rolling without slipping needs v_cm = Rω. Rω = 0, 1.9, 3.9, 5.7, 5.7 m/s; v_cm = 8.0, 7.2, 6.5, 5.7, 5.7 m/s. They match by 1.2 s and stay matched.

• Correct: By 1.2 s, when v_cm equals Rω: Right: 0.11 m × 52 rad/s = 5.7 m/s = v_cm.
• At 0.0 s, because the ball is moving: At first ω = 0 while v_cm = 8.0 m/s: it slides without spinning.
• At 0.8 s, when ω reaches 35 rad/s: Rω = 3.9 m/s, still less than v_cm = 6.5 m/s.
• Not yet, because v_cm keeps falling: v_cm stops falling at 5.7 m/s, once it equals Rω.

The bottom of the ball slides forward on the lane, so kinetic friction on the ball points backward. That reduces v_cm, and because it acts at the bottom, a distance R below the center, its torque increases ω until v_cm = Rω.

• Correct: Backward: it slows v_cm and speeds up the spin: Right: one force, two effects.
• Forward: it speeds up v_cm and the spin: v_cm falls in the data, so the friction force points backward.
• Backward: it slows both v_cm and the spin: ω rises in the data; the backward force at the bottom gives a torque in the direction of rolling.
• No friction acts, because the lane is level: A skidding ball rubs on the lane, so kinetic friction acts.

While the contact point slides, kinetic friction dissipates mechanical energy as thermal energy. Total kinetic energy (translational + rotational) falls during the skid, then stays constant once the ball rolls without slipping.

• Correct: It decreases: kinetic friction dissipates energy: Right: sliding friction turns some kinetic energy into thermal energy.
• It stays the same: the spin gains what v_cm loses: The spin gains some energy, but kinetic friction during sliding removes energy overall.
• It increases, because the ball starts to spin: Nothing outside does positive work on the ball overall.
• It decreases to zero, because the ball stops skidding: The ball keeps rolling at 5.7 m/s.

Rolling without slipping, the bottom point has zero velocity relative to the lane, so nothing slides. On a level lane with no other horizontal force, no friction acts, the net force and net torque are zero, and v_cm and ω stay constant.

• Correct: Its contact point does not slide, so no friction acts: Right: no sliding, no kinetic friction; on a level lane no static friction is needed either.
• Static friction keeps pushing the ball forward: A forward force would speed the ball up; the data show constant v_cm.
• Kinetic friction balances the ball’s weight: Friction is horizontal and the weight vertical; they do not balance.
• The ball’s spin pushes it along the lane: Spinning does not push the ball; with no net force v_cm stays constant.

K = ½Iω², so K against ω² is a straight line through the origin with slope ½I.

• Correct: K against ω²: Right: K = (½I)ω² is proportional to ω².
• K against ω: K rises faster than ω; the points curve upward.
• K against 1/ω: K increases with ω, so it falls along a curve against 1/ω.
• √K against ω²: √K is proportional to ω, not ω².

K = ½Iω² = ½(0.80 kg·m²)(50 rad/s)² = 1000 J. The steps between rows grow (120, 200, 280 J), as a square law predicts.

• Correct: 1000 J: Right: K = ½(0.80 kg·m²)(50 rad/s)² = 1000 J.
• 800 J: This scales 640 J by 50/40, as if K were proportional to ω.
• 920 J: This adds the last step (280 J) again, but each step is larger than the one before.
• 2500 J: This forgets the ½I and uses ω² alone.

At a fixed ω, K = ½Iω² is proportional to I. Putting the same mass farther from the axle raises I, so a rim-heavy flywheel stores more energy.

• Correct: move more of its mass to the rim: Right: that raises I, and K = ½Iω² rises with it.
• move more of its mass toward the axle: That lowers I and so lowers the stored energy.
• make it a solid disk instead of a ring: A solid disk has less of its mass at the rim than a ring of the same mass and radius.
• leave the shape, since K depends on mass: K depends on I, which depends on where the mass is, not mass alone.