The ratio of the radii of gyration of a circu | vedclass

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(C) For a circular disc of mass $M$ and radius $R$,the moment of inertia about a tangential axis in its plane is given by $I_{	ext{disk}} = I_{	ext{

Vedclass · Accepted Answer

$\sqrt{5} : \sqrt{6}$

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(C) For a circular disc of mass $M$ and radius $R$,the moment of inertia about a tangential axis in its plane is given by $I_{	ext{disk}} = I_{	ext{cm}} + MR^2 = \frac{1}{4}MR^2 + MR^2 = \frac{5}{4}MR^2$.Since $I = MK^2$,the radius of gyration $K_{	ext{disk}} = \sqrt{\frac{5}{4}}R = \frac{\sqrt{5}}{2}R$.For a circular ring of mass $M$ and radius $R$,the moment of inertia about a tangential axis in its plane is given by $I_{	ext{ring}} = I_{	ext{cm}} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.Since $I = MK^2$,the radius of gyration $K_{	ext{ring}} = \sqrt{\frac{3}{2}}R = \frac{\sqrt{3}}{\sqrt{2}}R$.The ratio of the radii of gyration is $\frac{K_{	ext{disk}}}{K_{	ext{ring}}} = \frac{\sqrt{5}/2}{\sqrt{3}/\sqrt{2}} = \frac{\sqrt{5}}{2} 	imes \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{5}}{\sqrt{2} 	imes \sqrt{3}} = \sqrt{\frac{5}{6}}$.

The ratio of the radii of gyration of a circular disc about a tangential axis in the plane of the disc and of a circular ring of the same radius about a tangential axis in the plane of the ring is

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