The ratio of the radius of gyration of a ring | vedclass

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Question

(A) The radius of gyration $K$ is given by $K = \sqrt{\frac{I}{m}}$.For a ring of mass $m$ and radius $R$,the moment of inertia about its central axis

Vedclass · Accepted Answer

$\frac{2}{\sqrt{3}}$

Vedclass · Answer

(A) The radius of gyration $K$ is given by $K = \sqrt{\frac{I}{m}}$.For a ring of mass $m$ and radius $R$,the moment of inertia about its central axis perpendicular to the plane is $I_{	ext{ring, center}} = mR^2$.For a disc of mass $m$ and radius $R$,the moment of inertia about its central axis perpendicular to the plane is $I_{	ext{disc, center}} = \frac{1}{2}mR^2$.Using the parallel axis theorem,the moment of inertia $I'$ about a tangential axis perpendicular to the plane is $I' = I_{	ext{center}} + mR^2$.For the ring: $I'_{	ext{ring}} = mR^2 + mR^2 = 2mR^2$. Thus,$K_{	ext{ring}} = \sqrt{\frac{2mR^2}{m}} = \sqrt{2}R$.For the disc: $I'_{	ext{disc}} = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2$. Thus,$K_{	ext{disc}} = \sqrt{\frac{3}{2}}R$.The ratio of the radius of gyration of the ring to that of the disc is $\frac{K_{	ext{ring}}}{K_{	ext{disc}}} = \frac{\sqrt{2}R}{\sqrt{3/2}R} = \frac{\sqrt{2}}{\sqrt{3}/\sqrt{2}} = \frac{2}{\sqrt{3}}$.

List-$I$	List-$II$
$(A)$ Moment of inertia of solid sphere of radius $R$ about any tangent	$(I)$ $\frac{5}{3} MR^2$
$(B)$ Moment of inertia of hollow sphere of radius $R$ about any tangent	$(II)$ $\frac{7}{5} MR^2$
$(C)$ Moment of inertia of circular ring of radius $R$ about its diameter	$(III)$ $\frac{1}{4} MR^2$
$(D)$ Moment of inertia of circular disc of radius $R$ about any diameter	$(IV)$ $\frac{1}{2} MR^2$

The ratio of the radius of gyration of a ring to that of a disc (both circular) of the same radius and mass,about a tangential axis perpendicular to the plane is

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