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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 $...........\,rad / s$.
$2$
$6$
$4$
$8$
Solution
$L =\left( I _{\text {com }}\right)(\omega) \text { and } K =\frac{1}{2}\left( I _{\text {com }}\right)\left(\omega^2\right)+\frac{1}{2} MV _{\text {com }}^2$
$L =\frac{2}{5} M^2 \frac{V_{\text {com }}}{R} \quad K=\frac{1}{2}\left(\frac{2}{5} M R^2\right) \frac{V_{\text {com }}^2}{ R ^2}+\frac{1}{2} M V_{\text {com }}^2$
$L =\frac{2 MRV _{\text {com }}}{5} \quad K =\frac{7}{10} MV _{\text {com }}^2$
$\text { Ratio } \frac{ L }{ K }=\frac{4}{7} \frac{ R }{ V _{\text {com }}}=\frac{\pi}{22}$
$\Rightarrow \omega=\frac{4}{7} \times \frac{22}{22} \times 7=4$
