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

To solve the problem, we need to find the angular speed \(\omega\) of a solid sphere rolling without slipping. We know:

1. The ratio of angular momentum \(L\) about the axis of rotation to the total energy \(E\) of the sphere is \(\pi : 22\).
2. The angular momentum \(L = I\omega\), where the moment of inertia \(I\) for a solid sphere is \(\frac{2}{5}mr^2\) and \(\omega\) is the angular speed.
3. The total energy \(E\) consists of translational and rotational energies: \(E = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\). Since the sphere rolls without slipping, \(v = r\omega\).

Using \(v = r\omega\), we get:

• Translational energy: \(\frac{1}{2}mv^2 = \frac{1}{2}m(r\omega)^2 = \frac{1}{2}mr^2\omega^2\).
• Rotational energy: \(\frac{1}{2}I\omega^2 = \frac{1}{2}\left(\frac{2}{5}mr^2\right)\omega^2 = \frac{1}{5}mr^2\omega^2\).

Total energy \(E = \frac{1}{2}mr^2\omega^2 + \frac{1}{5}mr^2\omega^2 = \frac{7}{10}mr^2\omega^2\).

Angular momentum \(L = \frac{2}{5}mr^2\omega\).

Setting up the given ratio:

\[\frac{L}{E} = \frac{\frac{2}{5}mr^2\omega}{\frac{7}{10}mr^2\omega^2} = \frac{4}{7\omega}\]

Given this ratio equals \(\frac{\pi}{22}\), we equate and solve for \(\omega\):

\[\frac{4}{7\omega} = \frac{\pi}{22}\]

Cross-multiplying gives:

\[4 \times 22 = \pi \times 7\omega\]

\[88 = 7\pi\omega\]

Simplifying for \(\omega\):

\[\omega = \frac{88}{7\pi}\]

Approximately, \(\omega \approx \frac{88}{22} \times \frac{7}{7} \approx 4 \, \text{rad/s}\).

This value of \(\omega\) lies within the given range of \(4,4\), thus verifying it.

The value of the sphere's angular speed is \(\omega = 4 \, \text{rad/s}\).