Question

A spherical shell of 1 kg mass and radius R is rolling with angular speed ω on horizontal plane (as shown in figure). The magnitude of angular momentum of the shell about the origin O is \(\frac{a}{3}R^2w\). The value of a will be

• A. 2
• B. 3
• C. 5
• D. 4

Answer 1

The Correct Option is C

To determine the value of \( a \) in the given expression for the angular momentum of a spherical shell, we start by analyzing the problem:

Given: A spherical shell of mass \( m = 1 \, \text{kg} \) and radius \( R \) is rolling with angular speed \( \omega \). The angular momentum about the origin \( O \) is given by:

\[ L = \frac{a}{3}R^2 \omega \]

The angular momentum \( L \) of a rolling object is the sum of the angular momentum due to rotation about its center of mass and the angular momentum due to its translational motion.

The moment of inertia \( I \) of a spherical shell about its center is:

\[ I = \frac{2}{3}mR^2 \]

The translational angular momentum \( L_{\text{trans}} \) about point \( O \) is:

\[ L_{\text{trans}} = (m v)R = m (R\omega)R = m R^2 \omega \]

The rotational angular momentum \( L_{\text{rot}} \) about its center is:

\[ L_{\text{rot}} = I \omega = \left(\frac{2}{3}mR^2\right) \omega \]

Thus, the total angular momentum about the origin \( O \) is:

\[ L = L_{\text{trans}} + L_{\text{rot}} = m R^2 \omega + \frac{2}{3}mR^2 \omega \]

Substituting \( m = 1 \, \text{kg} \), we have:

\[ L = R^2 \omega + \frac{2}{3}R^2 \omega = \frac{3}{3}R^2 \omega + \frac{2}{3}R^2 \omega = \frac{5}{3}R^2 \omega \]

Comparing this result with the given expression \( \frac{a}{3}R^2 \omega \), we equate to find \( a \):

\[ \frac{5}{3}R^2 \omega = \frac{a}{3}R^2 \omega \]

Therefore, \( a = 5 \).

Hence, the value of \( a \) is 5.