Rolling — Worked Solutions

These solutions belong to Rolling. Each step states why it is useful. Try the questions first, then use this page to check your reasoning.

Question 1

For ideal rolling without slipping on a stationary surface, which statement is correct?

1. For no slipping, v_cm = R|ω|. Why: The translational and rotational surface motions match at contact.

2. At the bottom they cancel in the stationary-surface frame. Why: The center still moves at v_cm and the top adds the two motions, giving 2v_cm.

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Question 2

A wheel spins clockwise with its center initially at rest on a rough floor. Its bottom slips left. Which way does kinetic friction on the wheel act?

1. The bottom surface slips left relative to the floor. Why: Kinetic friction opposes relative motion at contact, not necessarily center motion.

2. Friction acts right, increasing center speed while reducing clockwise spin. Why: Its force accelerates the center right and its torque opposes the existing clockwise rotation.

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Question 3

A hoop and disk roll without slipping from rest through the same height. Which has greater center speed at the bottom?

1. Use Mgh = ½Mv² + ½I(v/R)². Why: Both start from rest with the same height change and roll without slipping.

2. The hoop has greater I/(MR²), so less energy is available for translation at a given total. Why: The disk reaches a larger center speed; this is an energy partition effect.

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Question 4

A hoop of mass 1.5 kg rolls without slipping at 2.0 m/s. Its provided inertia is I_cm = MR². Find translational, rotational, and total kinetic energy without needing its radius.

1. Compute K_trans = ½(1.5)(2.0²) = 3.0 J. Why: The center-of-mass motion contributes the translational term.

2. Insert I_cm = MR² and ω = v/R into K_rot. Why: The supplied hoop inertia and no-slip constraint apply together.

3. K_rot = ½MR²(v/R)² = ½Mv² = 3.0 J; K_total = 6.0 J. Why: Radius cancels in this particular model, and the two distinct contributions add.

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Question 5

A disk and hoop, each of radius 0.25 m, move right on a rough level stationary floor with clockwise spin. At one instant the disk has center speed 3.0 m/s and angular speed 8.0 rad/s; the hoop has 2.5 m/s and 10 rad/s. Identify which is instantaneously consistent with rolling without slipping. For the slipping object, predict the direction of kinetic friction and its qualitative effects on translation, spin, and mechanical energy. Can these data alone rank the two objects’ total kinetic energies? Explain what else is needed.

1. For the disk, R|ω| = 0.25(8.0) = 2.0 m/s, below v_cm = 3.0 m/s; its bottom slips right at 1.0 m/s. For the hoop, R|ω| = 0.25(10) = 2.5 m/s, matching v_cm. Why: Translation and clockwise surface motion subtract at the bottom. Matching values support instantaneous no slipping; the disk’s unequal values establish slipping.

2. Kinetic friction on the disk acts left. With other horizontal forces negligible, it reduces the rightward center speed and produces a clockwise torque that increases clockwise spin; contact sliding dissipates mechanical energy. Why: Friction opposes the contact point’s motion relative to the floor, and its force and torque have different effects on center translation and rotation.

3. The data cannot rank total kinetic energies without masses and rotational inertias about the centers. Why: K_total = ½Mv_cm² + ½I_cmω² depends on those quantities as well as the measured speeds. Radius and shape names alone do not supply all the needed values.

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