A disc of mass M and radius R is rolling with | vedclass

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(C) The angular momentum of a rolling disc about the origin $O$ is given by the sum of the angular momentum about its center of mass and the angular m

Vedclass · Accepted Answer

$\frac{3}{2} MR^2\omega$

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(C) The angular momentum of a rolling disc about the origin $O$ is given by the sum of the angular momentum about its center of mass and the angular momentum of the center of mass about the origin $O$.$L = L_{cm} + L_{orbital}$$L = I_{cm}\omega + M(v_{cm} 	imes r)$Since the disc is rolling without slipping,the velocity of the center of mass is $v_{cm} = R\omega$.The moment of inertia of the disc about its center is $I_{cm} = \frac{1}{2}MR^2$.The distance of the center of mass from the origin $O$ along the $y$-axis is $R$.$L = \left(\frac{1}{2}MR^2ight)\omega + M(R\omega)(R)$$L = \frac{1}{2}MR^2\omega + MR^2\omega$$L = \frac{3}{2}MR^2\omega$

List-$I$	List-$II$
$(i) P$	$(A) (0, 0)$
$(ii) Q$	$(B) (6, 0)$
$(iii) R$	$(C) (-2, 1)$
$(iv) S$	$(D) (-6, 0)$
	$(E) (-6, -3)$
	$(F) (-6, 3)$

$A$ 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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