Question

A solid sphere of mass \(1\) kg rolls without slipping on a plane surface. Its kinetic energy is \(7\times 10^{-3}\) J. The speed of the center of mass of the sphere in cm s\(^{-1}\) is

• A. \(1\)
• B. \(10\)
• C. \(100\)
• D. \(1000\)

Hint:

Total kinetic energy of a rolling sphere is \(\frac{7}{10}mv^2\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A rolling sphere has translational and rotational kinetic energy. For rolling without slipping, \(v = \omega R\), and for a solid sphere \(I = \frac25 mR^2\).

Step 2: Total energy:
\[ K = \frac12mv^2 + \frac12I\omega^2 = \frac12mv^2 + \frac12\cdot\frac25mR^2\cdot\frac{v^2}{R^2} = \frac12mv^2 + \frac15mv^2 = \frac{7}{10}mv^2 \]

Step 3: Solve:
\(\frac{7}{10}(1)v^2 = 7\times10^{-3}\), so \(v^2 = 10^{-2}\), and \(v = 0.1\) m/s \(= 10\) cm/s.

Final Answer:
The speed is \(10\) cm/s, option (B). \[ \boxed{10\ \text{cm s}^{-1}} \]