QJEEM23Phbce8: JEE Main 2023 Physics - Class 11 JEEM23Phbce8 Solution: A solid sphere is rolling on a horizontal plane without…

Given that the solid sphere is rolling without slipping, we have:

Angular momentum $L = \left(I_{\text{com}}\right)(\omega)$

Kinetic energy $K = \frac{1}{2}(I_{\text{com}})(\omega^2) + \frac{1}{2}MV_{\text{com}}^2$

For a solid sphere, the moment of inertia is $I_{\text{com}} = \frac{2}{5}MR^2$ , and the relationship between linear and angular velocity is $V_{\text{com}} = R\omega$ .

Substituting these values into the expressions for $L$ and $K$ :

$L = \frac{2}{5}MR^2 \frac{V_{\text{com}}}{R} = \frac{2MRV_{\text{com}}}{5}$

$K = \frac{1}{2}\left(\frac{2}{5}MR^2\right) \frac{V_{\text{com}}^2}{R^2} + \frac{1}{2}MV_{\text{com}}^2 = \frac{7}{10}MV_{\text{com}}^2$

Now, the given ratio of $\frac{L}{K}$ is $\frac{\pi}{22}$ :

$\frac{L}{K} = \frac{4}{7} \frac{R}{V_{\text{com}}} = \frac{\pi}{22}$

Since $V_{\text{com}} = R\omega$ , we can substitute this relationship into the equation and solve for $\omega$ :

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

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

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

Thus, the value of the angular speed is $4 \ \text{rad/s}$ .