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(A) The angular momentum of a body rolling on a surface with respect to a point $O$ on the surface is given by the sum of the angular momentum due to

Vedclass · Accepted Answer

$\frac{14}{15}$

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(A) The angular momentum of a body rolling on a surface with respect to a point $O$ on the surface is given by the sum of the angular momentum due to the motion of the centre of mass and the angular momentum due to rotation about the centre of mass.$L_O = L_{	ext{linear}} + L_{	ext{rotational}} = MvR + I\omega$Since the body is rolling without slipping, $v = R\omega$, so $\omega = \frac{v}{R}$.$L_O = MvR + I\left(\frac{v}{R}ight) = vR \left(M + \frac{I}{R^2}ight)$For a solid sphere, $I_1 = \frac{2}{5}MR^2$. Thus, $L_1 = vR \left(M + \frac{2}{5}Might) = \frac{7}{5}MvR$.For a solid cylinder, $I_2 = \frac{1}{2}MR^2$. Thus, $L_2 = vR \left(M + \frac{1}{2}Might) = \frac{3}{2}MvR$.The ratio is $\frac{L_1}{L_2} = \frac{\frac{7}{5}MvR}{\frac{3}{2}MvR} = \frac{7}{5} 	imes \frac{2}{3} = \frac{14}{15}$.

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