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

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Question

(C) The angular momentum of a rolling body about a point on the surface is the sum of the angular momentum due to translation and the angular momentum

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The angular momentum of a rolling body about a point on the surface is the sum of the angular momentum due to translation and the angular momentum due to rotation about its center of mass.$L = L_{	ext{translation}} + L_{	ext{rotation}}$$L = MvR + I_c\omega$Since the disc is rolling without slipping,the linear velocity of the center of mass is $v = R\omega$.The moment of inertia of a disc about its center is $I_c = \frac{1}{2}MR^2$.Substituting these values:$L = M(R\omega)R + (\frac{1}{2}MR^2)\omega$$L = MR^2\omega + \frac{1}{2}MR^2\omega$$L = \frac{3}{2}MR^2\omega$

$A$ disc of mass $M$ and radius $R$ is rolling on a horizontal surface as shown in the figure. What is the magnitude of the angular momentum of the disc about the origin $O$?

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