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6.System of Particles and Rotational Motion
hard
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
A
$\sqrt 4 :\sqrt 3 $
B
$\sqrt 3 :\sqrt 4 $
C
$\sqrt 3 :\sqrt 2 $
D
$\sqrt 2 :\sqrt 3 $
Solution
${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}} $.
Standard 11
Physics
