A ring and a disc roll on horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is 6 J then total kinetic energy of the disc is

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TestHub · Accepted Answer

Total kinetic energy of a rolling body is given by $K = \frac{1}{2} m v^{2} (1 + \frac{k ^{2}}{R ^{2}})$

For a ring, $\frac{k ^{2}}{R ^{2}} = 1$

$K_{ring} = \frac{1}{2} m v^{2} (1 + 1) = m v^{2}$

Given $K_{ring} = 6 J$ , we get $m v^{2} = 6 J$

For a disc, $\frac{k ^{2}}{R ^{2}} = \frac{1}{2}$

$K_{disc} = \frac{1}{2} m v^{2} (1 + \frac{1}{2}) = \frac{3}{4} m v^{2}$

Substituting the value of $m v^{2}$ :

$K_{disc} = \frac{3}{4} (6) = \frac{9}{2} J$

Answer: $\frac{9}{2} J$

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