Question

A solid cylinder of radius $ R $ is at rest at a height $ h $ on an inclined plane. If it rolls down, then its velocity on reaching the ground is:

• A. \( \sqrt{\dfrac{5gh}{3}} \)
• B. \( \sqrt{\dfrac{2h}{3g}} \)
• C. \( \sqrt{\dfrac{2gh}{3}} \)
• D. \( \sqrt{\dfrac{4gh}{3}} \)

Hint:

For rolling motion, include both translational and rotational energy: \( KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \)

Answer 1

The Correct Option is D

For a solid cylinder rolling without slipping, total mechanical energy is conserved. Potential energy at height \( h \): \[ PE = mgh \] At the bottom, total energy is translational + rotational: \[ KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \] For a solid cylinder, \( I = \frac{1}{2}mR^2 \) and \( \omega = \frac{v}{R} \), so: \[ \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{1}{2}mR^2 \cdot \frac{v^2}{R^2} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2 \] Equating energies: \[ mgh = \frac{3}{4}mv^2 \Rightarrow v^2 = \frac{4gh}{3} \Rightarrow v = \sqrt{\dfrac{4gh}{3}} \]