Three bodies, a ring, a solid disc, and a sol | vedclass

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Question

(D) When a body of mass $m$ and radius $R$ rolls down an inclined plane of height $h$ without slipping, its velocity at the bottom is given by $v = \s

Vedclass · Accepted Answer

$V_S > V_D > V_R$

Vedclass · Answer

(D) When a body of mass $m$ and radius $R$ rolls down an inclined plane of height $h$ without slipping, its velocity at the bottom is given by $v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$, where $k$ is the radius of gyration.For a ring, $k^2 = R^2$, so $v_R = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$.For a solid disc, $k^2 = \frac{R^2}{2}$, so $v_D = \sqrt{\frac{2gh}{1+0.5}} = \sqrt{\frac{4gh}{3}} \approx 1.15 \sqrt{gh}$.For a solid sphere, $k^2 = \frac{2R^2}{5}$, so $v_S = \sqrt{\frac{2gh}{1+0.4}} = \sqrt{\frac{10gh}{7}} \approx 1.19 \sqrt{gh}$.Comparing the values, we find $V_S > V_D > V_R$.

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