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

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(A) The moment of inertia $(I)$ of a circular disc about its central axis is $I_{disc} = \frac{1}{2}MR^2$. The radius of gyration $(k)$ is defined by

Vedclass · Accepted Answer

$1 : \sqrt{2}$

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(A) The moment of inertia $(I)$ of a circular disc about its central axis is $I_{disc} = \frac{1}{2}MR^2$. The radius of gyration $(k)$ is defined by $I = Mk^2$. Thus,$k_{disc} = \sqrt{\frac{I}{M}} = \sqrt{\frac{R^2}{2}} = \frac{R}{\sqrt{2}}$.The moment of inertia $(I)$ of a circular ring about its central axis is $I_{ring} = MR^2$. Thus,$k_{ring} = \sqrt{\frac{I}{M}} = \sqrt{R^2} = R$.The ratio of the radius of gyration of the disc to that of the ring is $\frac{k_{disc}}{k_{ring}} = \frac{R/\sqrt{2}}{R} = \frac{1}{\sqrt{2}}$.Therefore,the ratio is $1 : \sqrt{2}$.

The ratio of the radii of gyration of a circular disc to that of a circular ring,each of same mass and radius,around their respective axes is

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