Question

A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is π : 22 then, the value of its angular speed will be ______ rad/s.

Hint:

For rolling motion, always use the rolling condition \( v = R \omega \) to simplify relationships between energy, angular momentum, and angular speed.

Answer 1

Correct Answer: 4

For a solid sphere rolling without slipping:
 

The total energy is the sum of translational kinetic energy and rotational kinetic energy: \[ E_{\text{total}} = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 \] where \( I = \frac{2}{5} M R^2 \) is the moment of inertia of the sphere about its axis of rotation.

Using the rolling condition \( v = R \omega \), the total energy becomes: \[ E_{\text{total}} = \frac{1}{2} M v^2 + \frac{1}{2} \left( \frac{2}{5} M R^2 \right) \omega^2 = \frac{7}{10} M v^2 \]

The angular momentum about the axis of rotation is: \[ L = I \omega = \frac{2}{5} M R^2 \omega. \] Given that the ratio of angular momentum to total energy is \( \pi : 22 \): \[ \frac{L}{E_{\text{total}}} = \frac{\pi}{22} \] Substitute the expressions for \( L \) and \( E_{\text{total}} \): \[ \frac{\frac{2}{5} M R^2 \omega}{\frac{7}{10} M v^2} = \frac{\pi}{22} \] Simplify using \( v = R \omega \): \[ \frac{\frac{2}{5} M R^2 \omega}{\frac{7}{10} M (R \omega)^2} = \frac{\pi}{22} \] Cancel terms: \[ \frac{\frac{2}{5}}{\frac{7}{10} R \omega} = \frac{\pi}{22} \] Solve for \( \omega \): \[ \omega = \frac{22 \times \frac{2}{5}}{\pi \times \frac{7}{10} R} \] Simplify: \[ \omega = \frac{4}{\pi R} \quad \text{rad/s} \]