Moment of inertia of a disc of mass M and rad | vedclass

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(B) $1$. The moment of inertia of a disc about its diameter is $I_d = MR^2/4$.$2$. The moment of inertia of a disc about an axis passing through its c

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$3$

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(B) $1$. The moment of inertia of a disc about its diameter is $I_d = MR^2/4$.$2$. The moment of inertia of a disc about an axis passing through its center and normal to the disc (perpendicular axis theorem) is $I_{cm} = I_d + I_d = MR^2/4 + MR^2/4 = MR^2/2$.$3$. To find the moment of inertia about an axis passing through a point on its edge and normal to the disc, we use the parallel axis theorem: $I = I_{cm} + Md^2$, where $d = R$ is the distance between the center and the edge.$4$. Substituting the values: $I = MR^2/2 + M(R)^2 = MR^2/2 + MR^2 = (3/2)MR^2$.$5$. Comparing this with the given expression $(x/2)MR^2$, we get $x/2 = 3/2$, which implies $x = 3$.

Moment of inertia of a disc of mass $M$ and radius $R$ about any of its diameter is $MR^2/4$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be $(x/2)MR^2$. The value of $x$ is

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