A disc and a ring of same mass are rolling an | vedclass

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Question

${K_{disc}} = \frac{1}{2}mv_d^2\left( {1 + \frac{{{k^2}}}{{{R^2}}}} \right)\,$$ = \frac{3}{4}mv_d^2$  $\left[ {As\frac{{{k^2}}}{{{R^2}}} = \frac{

Vedclass · Accepted Answer

$\sqrt 4 :\sqrt 3 $

Vedclass · Answer

${K_{disc}} = \frac{1}{2}mv_d^2\left( {1 + \frac{{{k^2}}}{{{R^2}}}} \right)\,$$ = \frac{3}{4}mv_d^2$  $\left[ {As\frac{{{k^2}}}{{{R^2}}} = \frac{1}{2}\,\,\,\,\,\,{\rm{for disc}}} \right]$

${K_{ring}} = \frac{1}{2}m{v_r}\left( {1 + \frac{{{k^2}}}{{{R^2}}}} \right)$$ = mv_r^2$               $\left[ {As\frac{{{k^2}}}{{{R^2}}} = 1\,\,\,\,\,\,\,{\rm{for ring}}} \right]$

According to problem ${K_{disc}} = {K_{ring}}$

$\frac{3}{4}mv_d^2 = mv_r^2$ 

$\frac{{{v_d}}}{{{v_r}}} = \sqrt {\frac{4}{3}} $.

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

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