A spherical shell of 1 , kg mass and radius R | vedclass

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(C) Let $L_{O}$ be the angular momentum of the shell about the origin $O$.Since the shell is rolling without slipping on the horizontal plane,the velo

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$5$

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(C) Let $L_{O}$ be the angular momentum of the shell about the origin $O$.Since the shell is rolling without slipping on the horizontal plane,the velocity of the center of mass is given by $V_{cm} = \omega R$.The total angular momentum about the origin $O$ is the sum of the angular momentum due to the motion of the center of mass and the angular momentum about the center of mass:$L_{O} = m V_{cm} R + I_{cm} \omega$For a spherical shell,the moment of inertia about its center of mass is $I_{cm} = \frac{2}{3} m R^{2}$.Substituting the values $(m = 1 \, kg)$:$L_{O} = (1) (\omega R) R + (\frac{2}{3} (1) R^{2}) \omega$$L_{O} = \omega R^{2} + \frac{2}{3} R^{2} \omega$$L_{O} = (1 + \frac{2}{3}) R^{2} \omega = \frac{5}{3} R^{2} \omega$Comparing this with the given expression $\frac{a}{3} R^{2} \omega$,we get:$\frac{a}{3} = \frac{5}{3}$Therefore,$a = 5$.

$A$ spherical shell of $1 \, kg$ mass and radius $R$ is rolling with angular speed $\omega$ on a horizontal plane (as shown in the figure). The magnitude of the angular momentum of the shell about the origin $O$ is $\frac{a}{3} R^{2} \omega$. The value of $a$ will be ..............

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