Questions: Rotational Dynamics: Newton's Second Law for Rotation

Question 1

A solid cylinder and a hollow cylinder of the same mass and radius are released from rest at the top of an identical ramp. Which reaches the bottom first?

• A. The hollow cylinder, because it has more mass concentrated at the rim, generating more torque
• B. The solid cylinder, because its moment of inertia is smaller (I = ½mr²), so less of its energy is locked in rotation
• C. They tie — both experience the same gravitational force and normal force
• D. The hollow cylinder, because the angular velocity is higher when all mass is at the edge

Answer: (B)

For a rolling object, total energy is split between translational (½mv²) and rotational (½Iω²) kinetic energy. The hollow cylinder has I = mr², the solid cylinder has I = ½mr². Using the no-slip constraint v = ωr and energy conservation, the solid cylinder reaches the bottom faster because a smaller fraction of its energy is tied up in rotation — more is available for translational speed. The gravitational force is the same for both, but I determines how the energy is distributed between the two motion types.

Question 2

A mechanic applies force F to a wrench in a direction pointing directly toward the bolt (i.e., the force vector points straight at the rotation axis). What torque does this force produce?

• A. Torque = Fr, where r is the length of the wrench
• B. Torque = Fr sin(0°) = 0 — a force aimed at the axis produces no torque
• C. Torque = F/r, since the lever arm is inverted when the force aims inward
• D. Torque = Fr cos(0°) = Fr — the full force contributes because the angle is zero

Answer: (B)

Torque is τ = r × F with magnitude τ = rF sin θ, where θ is the angle between the position vector r and the force F. When the force is aimed directly at the rotation axis, θ = 0°, so sin(0°) = 0 and the torque is zero. Intuitively, this force has no 'twisting' component — it pushes along the wrench rather than perpendicular to it. Maximum torque occurs when the force is perpendicular to the lever arm (θ = 90°), which is why you push perpendicular to a wrench handle.

Question 3

When applying Στ = Iα to a rotating rigid body, the moment of inertia I and all torques τ must be computed about the same axis.

• T. True
• F. False

Answer: True

This is a fundamental consistency requirement. Both I (the distribution of mass relative to an axis) and τ (the rotational effect of each force) are axis-dependent quantities. If you compute I about the center of mass but compute a torque about a different point, the equation Στ = Iα becomes invalid. This is one of the most common errors in rotational dynamics problems — it's the rotational analog of mixing coordinate systems in linear mechanics.

Question 4

Applying a larger net force to a rigid body typically produces a larger angular acceleration.

• T. True
• F. False

Answer: False

Angular acceleration is governed by Στ = Iα — it depends on net TORQUE, not net force. A large force applied very close to or directly through the rotation axis produces little or no torque and thus little or no angular acceleration. A smaller force applied farther from the axis can produce a much larger torque. Additionally, I (moment of inertia) matters: the same torque produces a larger angular acceleration in a body with smaller I. The correct relationship is α = Στ/I, not α = ΣF/I.

Question 5

Why does a hollow cylinder roll more slowly down a ramp than a solid cylinder of the same mass and radius, even though gravity exerts the same force on both?

Model answer: Both cylinders have the same gravitational potential energy at the top of the ramp, but they distribute that energy differently between translational and rotational motion when rolling. The hollow cylinder (I = mr²) has a larger moment of inertia than the solid cylinder (I = ½mr²) because all of its mass is at the rim, far from the axis. By the no-slip constraint a = αr, more rotational inertia means more energy must go into spinning before any additional translational speed is gained. So the hollow cylinder reaches the bottom with the same total energy, but more of it is rotational — meaning less translational speed and a slower descent.

This problem illustrates a key feature of Στ = Iα: the same torque produces less angular acceleration when I is large. Since friction provides the torque that angularly accelerates the rolling cylinder, a larger I means friction must 'fight harder' to spin up the mass. The coupling constraint a = αr then limits the translational acceleration. The solid cylinder's mass is distributed closer to the axis (lower I), making it easier to spin and thus able to translate faster for the same energy input.