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Question

$	herefore \quad \overrightarrow{\mathrm{L}}_{\mathrm{P}}=\overrightarrow{\mathrm{L}}_{\mathrm{cm}}+\overrightarrow{\mathrm{r}} 	imes \overrightarro

Vedclass · Accepted Answer

$\frac{7}{5}\,Mv_0R$

Vedclass · Answer

$	herefore \quad \overrightarrow{\mathrm{L}}_{\mathrm{P}}=\overrightarrow{\mathrm{L}}_{\mathrm{cm}}+\overrightarrow{\mathrm{r}} 	imes \overrightarrow{\mathrm{p}}_{\mathrm{cm}}=1_{\mathrm{cm}} \vec{\omega}+\overrightarrow{\mathrm{R}} 	imes \mathrm{m} \vec{v}_{\mathrm{cm}}$

here $v_{\mathrm{cm}}=v_0$

since sphere is in pure rolling motion hence $\omega=\mathrm{v}_{0} / \mathrm{R}$

$\Rightarrow \vec{\mathrm{L}}_{\mathrm{p}}=\left(\frac{2}{5} \mathrm{MR}^{2} \frac{\mathrm{v}_{0}}{\mathrm{R}}ight)(-\hat{\mathrm{k}})+\mathrm{M} \mathrm{v}_{\mathrm{o}} \mathrm{R}(-\hat{\mathrm{k}})$

$=\frac{7}{5} \mathrm{Mv}_{0} \mathrm{R}(-\hat{\mathrm{k}})$

A solid sphere rolls without slipping on a rough surface and the centre of mass has a constant speed $v_0$. If the mass of the sphere is $m$ and its radius is $R$, then find the angular momentum of the sphere about the point of contact

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