Question

A solid sphere of mass \( M \) and radius \( R \) is rolling without slipping with speed \( v \). Its total kinetic energy is:

• A. \( \frac{1}{2} M v^2 \)
• B. \( \frac{7}{10} M v^2 \)
• C. \( \frac{3}{10} M v^2 \)
• D. \( \frac{9}{10} M v^2 \)

Hint:

For rolling objects, total kinetic energy includes both translational (\( \frac{1}{2} M v^2 \)) and rotational (\( \frac{1}{2} I \omega^2 \)) components. Use \( \omega = \frac{v}{R} \) for rolling without slipping.

Answer 1

The Correct Option is B

- For a solid sphere rolling without slipping, the total kinetic energy is the sum of translational and rotational kinetic energy.
- Translational kinetic energy: \[ KE_{\text{trans}} = \frac{1}{2} M v^2 \]
- Rotational kinetic energy: The moment of inertia of a solid sphere about its axis is \( I = \frac{2}{5} M R^2 \). The angular velocity \( \omega = \frac{v}{R} \) (since it rolls without slipping). \[ KE_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{2}{5} M R^2 \right) \left( \frac{v}{R} \right)^2 = \frac{1}{2} \cdot \frac{2}{5} M R^2 \cdot \frac{v^2}{R^2} = \frac{1}{5} M v^2 \]
- Total kinetic energy: \[ KE_{\text{total}} = KE_{\text{trans}} + KE_{\text{rot}} = \frac{1}{2} M v^2 + \frac{1}{5} M v^2 = \frac{1}{2} M v^2 + \frac{2}{10} M v^2 = \frac{5}{10} M v^2 + \frac{2}{10} M v^2 = \frac{7}{10} M v^2 \]
- This matches option (B).