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A ring and a disc having the same mass roll without slipping with the same linear velocity. If the kinetic energy of the ring is 8J, then kinetic energy of the disc must be equal to_____ J.

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answer is 6.

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Detailed Solution

$\begin{array}{l} K . E_{ring} = \frac{1}{2} (2 {mr}^{2}) \frac{v^{2}}{r^{2}} = {mv}^{2} \\ K . E_{disc} = \frac{1}{2} (\frac{3}{2} {mr}^{2}) \frac{v^{2}}{r^{2}} = \frac{3}{4} {mv}^{2} \\ = \frac{3}{4} \times 8 = 6 J \end{array}$

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