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(A) The moment of inertia of a thin circular ring of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane i

Vedclass · Accepted Answer

$10 \,cm$

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(A) The moment of inertia of a thin circular ring of mass $M$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I_z = MR^2$. The radius of gyration $k_z$ is given by $I_z = Mk_z^2$, so $k_z = R = 10 \sqrt{2} \,cm$.The moment of inertia of the ring about its diameter is $I_d = \frac{1}{2} MR^2$. The radius of gyration $k_d$ about the diameter is given by $I_d = Mk_d^2$.Thus, $Mk_d^2 = \frac{1}{2} MR^2$, which implies $k_d = \frac{R}{\sqrt{2}}$.Substituting the value of $R = 10 \sqrt{2} \,cm$, we get $k_d = \frac{10 \sqrt{2}}{\sqrt{2}} = 10 \,cm$.

If the radius of gyration of a thin circular ring about an axis passing through its centre and perpendicular to its plane is $10 \sqrt{2} \,cm$, then its radius of gyration about its diameter is

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