Moment of inertia of ring of mass M and radius R about a tangent to the circle of the ring is\

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Professor PC Thomas Classes · Answer

Moment of inertia of a ring of mass M and radius $R$ about the diameter $A B$ is\ $\mathrm{I}_{\mathrm{AB}}=\frac{1}{2} \mathrm{MR}^{2}$\XY is the tangent to the circular ring in the plane of the ring.\ According to theorem of parallel axes.\ $\mathrm{xy}=\mathrm{I}_{\mathrm{AB}}+\mathrm{MR}^{2}=\frac{1}{2} \mathrm{MR}^{2}+\mathrm{MR}^{2}=\frac{3}{2} \mathrm{MR}^{2}$\