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: Plan:
Use the fact that the rotational share of the energy is $\frac27$ of the total.

Step 2: Steps:
For a solid sphere, translational KE : rotational KE $= 5:2$. Translational part $= \frac57K = \frac57\times7\times10^{-3} = 5\times10^{-3}$ J.
$\frac12mv^2 = 5\times10^{-3}$, so $v^2 = 10^{-2}$, and $v = 0.1$ m/s $= 10$ cm/s.

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