(C) For pure rolling,the velocity of the center of mass is $v = R \omega$.The angular momentum of a rolling body about a point on the ground (origin $

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(C) For pure rolling,the velocity of the center of mass is $v = R \omega$.The angular momentum of a rolling body about a point on the ground (origin $O$) is given by the sum of the angular momentum of the center of mass about $O$ and the angular momentum about the center of mass.$L_O = L_{cm} + L_{spin}$$L_O = M v R + I_{cm} \omega$Since $v = R \omega$ and the moment of inertia of a disc about its center is $I_{cm} = \frac{1}{2} M R^2$,we have:$L_O = M (R \omega) R + (\frac{1}{2} M R^2) \omega$$L_O = M R^2 \omega + \frac{1}{2} M R^2 \omega$$L_O = \frac{3}{2} M R^2 \omega$

Class 11 Physics System of Particles and Rotational Motion Conservation of angular momentum (combined translation and rotational motion)

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$A$ disc of mass $M$ and radius $R$ is rolling with angular speed $\omega$ on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origin $O$ is

A
$\frac{1}{2} M R^2 \omega$
B
$M R^2 \omega$
C
$\frac{3}{2} M R^2 \omega$
D
$2 M R^2 \omega$

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System of Particles and Rotational Motion

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Conservation of angular momentum (combined translation and rotational motion)

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$A$ disc of mass $M$ and radius $R$ is rolling with angular speed $\omega$ on a horizontal plane as shown. The magnitude of angular momentum of the disc about the origin $O$ is

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