Question

A solid cylinder and a hollow cylinder, each of mass \( M \) and radius \( R \), are rotating with the same angular velocity \( \omega \). What is the ratio of their rotational kinetic energies \( \left( \frac{K_{\text{hollow}}}{K_{\text{solid}}} \right) \)?

• A. 1
• B. 2
• C. 3
• D. 3/2

Hint:

For rotating bodies, the moment of inertia depends on their mass distribution. The solid cylinder has less moment of inertia than the hollow cylinder, hence its rotational kinetic energy is half of the hollow cylinder's.

Answer 1

The Correct Option is B

Rotational kinetic energy \( K \) is defined as \( K = \frac{1}{2} I \omega^2 \), where \( I \) denotes the moment of inertia and \( \omega \) represents the angular velocity. Moment of inertia for a solid cylinder: The moment of inertia \( I_{\text{solid}} \) for a solid cylinder is given by \( I_{\text{solid}} = \frac{1}{2} M R^2 \). Moment of inertia for a hollow cylinder: The moment of inertia \( I_{\text{hollow}} \) for a hollow cylinder is given by \( I_{\text{hollow}} = M R^2 \). The corresponding rotational kinetic energies are: \( K_{\text{solid}} = \frac{1}{2} \times \frac{1}{2} M R^2 \omega^2 = \frac{1}{4} M R^2 \omega^2 \) \( K_{\text{hollow}} = \frac{1}{2} M R^2 \omega^2 \) The ratio of their rotational kinetic energies is calculated as: \( \frac{K_{\text{hollow}}}{K_{\text{solid}}} = \frac{\frac{1}{2} M R^2 \omega^2}{\frac{1}{4} M R^2 \omega^2} = 2 \) Therefore, the ratio of their rotational kinetic energies is 2.