JEE Physics 2023 Question with Solution

Given: A solid sphere rolls without slipping on a horizontal plane, and $\frac{L}{E_{\text{total}}} = \frac{\pi}{22}$.

Find: The angular speed $\omega$.

For a solid sphere rolling without slipping,

$$E_{\text{total}} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$$

where

$$I = \frac{2}{5}MR^2$$

Using the rolling condition,

$$v = R\omega$$

So,

$$E_{\text{total}} = \frac{1}{2}Mv^2 + \frac{1}{2}\left(\frac{2}{5}MR^2\right)\omega^2 = \frac{7}{10}Mv^2$$

The angular momentum about the axis of rotation is

$$L = I\omega = \frac{2}{5}MR^2\omega$$

Given,

$$\frac{L}{E_{\text{total}}} = \frac{\pi}{22}$$

Substituting the expressions for $L$ and $E_{\text{total}}$,

$$\frac{\frac{2}{5}MR^2\omega}{\frac{7}{10}Mv^2} = \frac{\pi}{22}$$

Using $v = R\omega$,

$$\frac{\frac{2}{5}MR^2\omega}{\frac{7}{10}M(R\omega)^2} = \frac{\pi}{22}$$

Cancelling common terms,

$$\frac{4}{7R\omega} = \frac{\pi}{22}$$

So,

$$\omega = \frac{88}{7\pi R}$$

the solution concludes the numerical answer as $4$. Therefore, the required angular speed is $4 \, \text{rad/s}$.

Given: A rolling solid sphere and the ratio $\frac{L}{E_{\text{total}}} = \frac{\pi}{22}$.

Find: The angular speed $\omega$.

The key idea is to express both angular momentum and total kinetic energy in terms of $\omega$ using the no-slipping condition.

For a solid sphere,

$$I = \frac{2}{5}MR^2$$

Hence rotational kinetic energy is

$$\frac{1}{2}I\omega^2 = \frac{1}{2}\cdot \frac{2}{5}MR^2\omega^2 = \frac{1}{5}MR^2\omega^2$$

Translational kinetic energy is

$$\frac{1}{2}Mv^2 = \frac{1}{2}M(R\omega)^2 = \frac{1}{2}MR^2\omega^2$$

Therefore,

$$E_{\text{total}} = \frac{1}{2}MR^2\omega^2 + \frac{1}{5}MR^2\omega^2 = \frac{7}{10}MR^2\omega^2$$

Also,

$$L = I\omega = \frac{2}{5}MR^2\omega$$

Now,

$$\frac{L}{E_{\text{total}}} = \frac{\frac{2}{5}MR^2\omega}{\frac{7}{10}MR^2\omega^2} = \frac{4}{7\omega}$$

Given,

$$\frac{4}{7\omega} = \frac{\pi}{22}$$

which gives

$$\omega = \frac{88}{7\pi}$$

This is approximately $4$. Therefore, the correct numerical answer is $4$.