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

The rotational kinetic energy \( K \) is given by: \[ K = \frac{1}{2} I \omega^2 \] Where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. ### Moment of inertia for the solid cylinder: The moment of inertia \( I_{\text{solid}} \) for a solid cylinder is: \[ I_{\text{solid}} = \frac{1}{2} M R^2 \] ### Moment of inertia for the hollow cylinder: The moment of inertia \( I_{\text{hollow}} \) for a hollow cylinder is: \[ I_{\text{hollow}} = M R^2 \] Thus, the rotational kinetic energy for each cylinder is: \[ K_{\text{solid}} = \frac{1}{2} \times \frac{1}{2} M R^2 \omega^2 = \frac{1}{4} M R^2 \omega^2 \] \[ K_{\text{hollow}} = \frac{1}{2} M R^2 \omega^2 \] Now, the ratio of their rotational kinetic energies is: \[ \frac{K_{\text{hollow}}}{K_{\text{solid}}} = \frac{\frac{1}{2} M R^2 \omega^2}{\frac{1}{4} M R^2 \omega^2} = 2 \] Thus, the ratio of their rotational kinetic energies is 2.