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(C) The angular momentum of a body about a point $P$ is given by $\vec{L}_P = \vec{L}_{cm} + \vec{r} 	imes \vec{p}_{cm}$,where $\vec{L}_{cm} = I_{cm}

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$\frac{7}{5} mv_0R$

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(C) The angular momentum of a body about a point $P$ is given by $\vec{L}_P = \vec{L}_{cm} + \vec{r} 	imes \vec{p}_{cm}$,where $\vec{L}_{cm} = I_{cm} \vec{\omega}$ is the angular momentum about the center of mass and $\vec{r} 	imes \vec{p}_{cm}$ is the angular momentum due to the motion of the center of mass.For a solid sphere,the moment of inertia about the center of mass is $I_{cm} = \frac{2}{5} mR^2$.In pure rolling motion,the angular velocity is $\omega = \frac{v_0}{R}$.The angular momentum about the center of mass is $L_{cm} = I_{cm} \omega = (\frac{2}{5} mR^2) (\frac{v_0}{R}) = \frac{2}{5} mv_0R$.The angular momentum due to the motion of the center of mass about the point of contact $P$ is $L_{trans} = m v_0 R$.Since both vectors point in the same direction (into the plane),the total angular momentum is $L_P = L_{cm} + L_{trans} = \frac{2}{5} mv_0R + mv_0R = \frac{7}{5} mv_0R$.

$A$ 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 $P$.

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