The ratio of radii of gyration of a ring to a | vedclass

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(A) The moment of inertia of a body is given by $I = Mk^2$,where $k$ is the radius of gyration.For a ring of mass $M$ and radius $R$ about a tangentia

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

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(A) The moment of inertia of a body is given by $I = Mk^2$,where $k$ is the radius of gyration.For a ring of mass $M$ and radius $R$ about a tangential axis perpendicular to its plane,using the parallel axis theorem: $I_{	ext{ring}} = I_{	ext{cm}} + MR^2 = MR^2 + MR^2 = 2MR^2$.Thus,$Mk_{	ext{ring}}^2 = 2MR^2 \implies k_{	ext{ring}} = \sqrt{2}R$.For a disc of mass $M$ and radius $R$ about a tangential axis perpendicular to its plane,using the parallel axis theorem: $I_{	ext{disc}} = I_{	ext{cm}} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.Thus,$Mk_{	ext{disc}}^2 = \frac{3}{2}MR^2 \implies k_{	ext{disc}} = \sqrt{\frac{3}{2}}R$.The ratio of the radii of gyration is $\frac{k_{	ext{ring}}}{k_{	ext{disc}}} = \frac{\sqrt{2}R}{\sqrt{3/2}R} = \frac{\sqrt{2}}{\sqrt{3}/\sqrt{2}} = \frac{2}{\sqrt{3}}$.

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

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