Questions: Rolling Motion Without Slipping: Equations and Analysis

Question 1

A hollow cylinder (I = mR²) and a solid cylinder (I = ½mR²) of identical mass and radius are released from rest at the top of the same incline. Which reaches the bottom first, and why?

• A. The hollow cylinder, because its mass concentrated at the rim gives it more rotational momentum
• B. They arrive at the same time, because they have identical masses and the same gravitational force acts on both
• C. The solid cylinder, because its smaller moment of inertia means less gravitational energy goes into rotation, leaving more for translational acceleration
• D. The hollow cylinder, because distributing mass at the rim increases the total kinetic energy available

Answer: (C)

The acceleration is a_cm = g sin θ / (1 + I/mR²). For the solid cylinder (I = ½mR²): a_cm = (2/3)g sin θ. For the hollow cylinder (I = mR²): a_cm = (1/2)g sin θ. The solid cylinder accelerates faster and wins. The rolling constraint forces some gravitational PE into rotational KE; the more mass concentrated at the rim, the larger I/mR², the more energy goes into spinning rather than translating. Total mass is irrelevant — it cancels in the formula.

Question 2

Why does static friction do no work on an object rolling without slipping?

• A. Static friction is too small to transfer significant energy in rolling motion
• B. The contact point is instantaneously at rest, so the friction force acts through zero displacement and does zero work
• C. Static friction acts perpendicular to the direction of motion, so its work component is zero by the dot product
• D. Energy conservation only applies when there is no friction; static friction means energy is not conserved

Answer: (B)

Work is force times displacement in the direction of the force. The no-slip condition v_cm = Rω ensures the contact point has zero velocity — it is instantaneously stationary. Since the contact point undergoes zero displacement during an infinitesimal time interval, the static friction force does zero work regardless of its magnitude. This is why energy is fully conserved in rolling without slipping: no mechanical energy is converted to heat.

Question 3

For an object rolling without slipping down an incline, the acceleration of the center of mass is independent of the object's total mass.

• T. True
• F. False

Answer: True

The formula a_cm = g sin θ / (1 + I/mR²) looks like it depends on m, but for uniform objects, I is proportional to mR² — so I/mR² is a pure geometric factor depending only on mass distribution, not total mass. For a solid disk, I/mR² = 1/2 regardless of m; for a ring, I/mR² = 1 regardless of m. Total mass cancels, and acceleration depends only on mass distribution geometry and incline angle. This is analogous to how all objects fall at the same rate in free fall.

Question 4

Static friction causes energy loss in rolling without slipping, similar to how kinetic friction converts mechanical energy to heat in sliding.

• T. True
• F. False

Answer: False

Static friction does NO work in rolling without slipping because the contact point is instantaneously at rest — work = force × displacement, and displacement is zero. Therefore no mechanical energy is converted to heat, and total mechanical energy (translational + rotational KE + gravitational PE) is conserved. Kinetic friction in sliding does convert energy to heat because the contact point IS moving relative to the surface. The contrast between static and kinetic friction in this context is one of the key insights of rolling motion.

Question 5

Explain why a hollow cylinder rolls more slowly down a ramp than a solid cylinder of the same mass and radius. Connect your answer to the rolling constraint and mass distribution.

Model answer: The rolling constraint v_cm = Rω links translational and rotational motion, so all gravitational PE released as the object descends must split between translational KE (½mv_cm²) and rotational KE (½Iω²). For the hollow cylinder (I = mR²), the total KE is ½mv_cm²(1 + 1) = mv_cm² — the energy is split evenly. For the solid cylinder (I = ½mR²), total KE is ½mv_cm²(3/2) — less goes to rotation. Since the same mgh is available, the hollow cylinder must allocate more energy to spinning, leaving less for translational acceleration. Mass concentrated at the rim (large I) is 'expensive' to spin under the constraint ω = v_cm/R, which is why it trades translational speed for rotational speed compared to the solid cylinder.

The key is understanding that the rolling constraint is a coupling — you cannot increase v_cm without proportionally increasing ω. The hollow cylinder's large moment of inertia makes spinning energetically expensive, which is why it must slow its translational acceleration to satisfy the constraint. Students who think heavier objects roll slower miss this: mass cancels in the formula, and only the geometry of mass distribution (I/mR²) matters.