The moment of inertia of a semicircular ring | vedclass

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Question

(D) Let the center of the semicircular ring be $O$. The moment of inertia of a complete ring of mass $2M$ and radius $R$ about an axis passing through

Vedclass · Accepted Answer

$2MR^2$

Vedclass · Answer

(D) Let the center of the semicircular ring be $O$. The moment of inertia of a complete ring of mass $2M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is $I_{cm} = (2M)R^2$. By symmetry,the moment of inertia of the semicircular ring of mass $M$ about the axis passing through $O$ and perpendicular to the plane is $I_O = MR^2$. Now,we use the parallel axis theorem to find the moment of inertia about an axis passing through point $A$ and perpendicular to the plane. The distance between the center $O$ and point $A$ is $R$. The parallel axis theorem states $I_A = I_O + Md^2$,where $d = R$. Substituting the values,we get $I_A = MR^2 + M(R)^2 = 2MR^2$.

The moment of inertia of a semicircular ring of radius $R$ and mass $M$ about an axis passing through point $A$ (one end of the ring) and perpendicular to the plane of the paper is:

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