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(C) For a solid sphere rolling without slipping,the angular momentum $L$ about the axis of rotation is $L = I_{com}\omega = (\frac{2}{5}MR^2)\omega$.S

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$4$

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(C) For a solid sphere rolling without slipping,the angular momentum $L$ about the axis of rotation is $L = I_{com}\omega = (\frac{2}{5}MR^2)\omega$.Since $V_{com} = R\omega$,we have $L = \frac{2}{5}MR^2(\frac{V_{com}}{R}) = \frac{2}{5}MRV_{com}$.The total kinetic energy $K$ is $K = \frac{1}{2}I_{com}\omega^2 + \frac{1}{2}MV_{com}^2 = \frac{1}{2}(\frac{2}{5}MR^2)(\frac{V_{com}}{R})^2 + \frac{1}{2}MV_{com}^2 = \frac{1}{5}MV_{com}^2 + \frac{1}{2}MV_{com}^2 = \frac{7}{10}MV_{com}^2$.The ratio $\frac{L}{K} = \frac{\frac{2}{5}MRV_{com}}{\frac{7}{10}MV_{com}^2} = \frac{2}{5} 	imes \frac{10}{7} 	imes \frac{R}{V_{com}} = \frac{4}{7} 	imes \frac{R}{R\omega} = \frac{4}{7\omega}$.Given $\frac{L}{K} = \frac{\pi}{22}$,we have $\frac{4}{7\omega} = \frac{\pi}{22}$. Assuming $\pi \approx \frac{22}{7}$,we get $\frac{4}{7\omega} = \frac{22/7}{22} = \frac{1}{7}$.Therefore,$7\omega = 28$,which gives $\omega = 4\,rad/s$.

$A$ solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about the axis of rotation of the sphere to the total energy of the moving sphere is $\pi: 22$,then the value of its angular speed will be $...........\,rad/s$.

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