Question

A solid cylinder of radius \( R \) and mass \( M \) rolls down an inclined plane of height \( h \). When it reaches the bottom of the plane, its rotational kinetic energy is
\textit{(g = acceleration due to gravity)}

• A. \( \frac{Mgh}{3} \)
• B. \( Mgh \)
• C. \( \frac{Mgh}{2} \)
• D. \( Mgh \)

Hint:

For rolling motion, part of the potential energy is converted into rotational kinetic energy, which for a solid cylinder is \( \frac{1}{3} \) of the total energy.

Answer 1

The Correct Option is A

Step 1: Energy conversion.
The total mechanical energy at the top of the inclined plane is potential energy \( Mgh \), which is converted into rotational kinetic energy and translational kinetic energy at the bottom. The rotational kinetic energy for a rolling object is \( \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
Step 2: Rotational kinetic energy of a solid cylinder.
For a solid cylinder rolling without slipping, the moment of inertia is \( I = \frac{1}{2} MR^2 \), and the rotational kinetic energy at the bottom is: \[ E_{\text{rot}} = \frac{1}{3} Mgh \]
Step 3: Conclusion.
Thus, the correct answer is (A) \( \frac{Mgh}{3} \).