Find the moment of inertia of a solid cylinde | vedclass

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(C) The moment of inertia of a solid cylinder about its central axis is given by $I_{cm} = \frac{1}{2}MR^2$.According to the parallel axis theorem,the

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$ \frac{3}{2}M{R^2} $

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(C) The moment of inertia of a solid cylinder about its central axis is given by $I_{cm} = \frac{1}{2}MR^2$.According to the parallel axis theorem,the moment of inertia $I$ about an axis parallel to the central axis at a distance $d$ is given by $I = I_{cm} + Md^2$.Here,the axis is on the surface of the cylinder,so the distance $d$ from the central axis is equal to the radius $R$.Substituting the values,we get $I = \frac{1}{2}MR^2 + M(R)^2$.$I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.Therefore,the correct option is $C$.

Find the moment of inertia of a solid cylinder of mass $M$ and radius $R$ about a line parallel to the axis of the cylinder and on the surface of the cylinder.

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