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 $\pi: 22$ then, the value of its angular speed will be ____________ $\mathrm{rad} / \mathrm{s}$.

Given that the solid sphere is rolling without slipping, we have: <br/><br/> Angular momentum $L = \left(I_{\text{com}}\right)(\omega)$ <br/><br/> Kinetic energy $K = \frac{1}{2}(I_{\text{com}})(\omega^2) + \frac{1}{2}MV_{\text{com}}^2$ <br/><br/> 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 $V_{\text{com}} = R\omega$. <br/><br/> Substituting these values into the expressions for $L$ and $K$: <br/><br/> $L = \frac{2}{5}MR^2 \frac{V_{\text{com}}}{R} = \frac{2MRV_{\text{com}}}{5}$ <br/><br/> $$K = \frac{1}{2}\left(\frac{2}{5}MR^2\right) \frac{V_{\text{com}}^2}{R^2} + \frac{1}{2}MV_{\text{com}}^2 = \frac{7}{10}MV_{\text{com}}^2$$ <br/><br/> Now, the given ratio of $\frac{L}{K}$ is $\frac{\pi}{22}$: <br/><br/> $\frac{L}{K} = \frac{4}{7} \frac{R}{V_{\text{com}}} = \frac{\pi}{22}$ <br/><br/> Since $V_{\text{com}} = R\omega$, we can substitute this relationship into the equation and solve for $\omega$: <br/><br/> $\frac{4}{7} \frac{R}{R\omega} = \frac{\pi}{22}$ <br/><br/> $\frac{4}{7\omega} = \frac{\pi}{22}$ <br/><br/> $\omega = \frac{4}{7} \times \frac{22}{\pi} \times 7 = 4$ <br/><br/> Thus, the value of the angular speed is $4 \ \text{rad/s}$.