A disc and a ring of the same mass are rollin | vedclass

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(A) The total kinetic energy of a rolling body is given by $K = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$.For a disc,the radius of gyration $k$ satisfies

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$\sqrt{4} : \sqrt{3}$

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(A) The total kinetic energy of a rolling body is given by $K = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$.For a disc,the radius of gyration $k$ satisfies $\frac{k^2}{R^2} = \frac{1}{2}$. Thus,$K_{disc} = \frac{1}{2}mv_d^2(1 + \frac{1}{2}) = \frac{3}{4}mv_d^2$.For a ring,the radius of gyration $k$ satisfies $\frac{k^2}{R^2} = 1$. Thus,$K_{ring} = \frac{1}{2}mv_r^2(1 + 1) = mv_r^2$.Given that $K_{disc} = K_{ring}$,we have $\frac{3}{4}mv_d^2 = mv_r^2$.Rearranging the terms,we get $\frac{v_d^2}{v_r^2} = \frac{4}{3}$.Therefore,the ratio of their velocities is $\frac{v_d}{v_r} = \sqrt{\frac{4}{3}} = \frac{\sqrt{4}}{\sqrt{3}}$.

$A$ disc and a ring of the same mass are rolling. If their kinetic energies are equal,then the ratio of their velocities will be:

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