Question

A solid cylinder of mass \( m \) and radius \( R \) rolls down an inclined plane without slipping. The speed of its C.M. when it reaches the bottom is:

• A. \( \sqrt{2gh} \)
• B. \( \frac{\sqrt{4gh}}{3} \)
• C. \( \frac{\sqrt{3}}{4}gh \)
• D. \( \sqrt{4gh} \)

Hint:

For rolling motion without slipping, the kinetic energy is split between translational and rotational energies.

Answer 1

The Correct Option is A

For rolling motion, the energy is shared between translational kinetic energy and rotational kinetic energy. The final speed of the center of mass of the cylinder is: \[ v_{\text{cm}} = \sqrt{2gh} \]
Final Answer: \[ \boxed{\sqrt{2gh}} \]