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(C) The total kinetic energy of a body rolling without sliding is given by $K = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$,where $k$ is the radius of gyrat

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$8$

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(C) The total kinetic energy of a body rolling without sliding is given by $K = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$,where $k$ is the radius of gyration.For a circular disc,the moment of inertia is $I = \frac{1}{2}mR^2$,so $k^2 = \frac{1}{2}R^2$. The kinetic energy is $K_{disc} = \frac{1}{2}mv^2(1 + \frac{1}{2}) = \frac{3}{4}mv^2 = 6 \ J$.From this,$\frac{1}{2}mv^2 = \frac{6 	imes 2}{3} = 4 \ J$.For a thin circular ring,the moment of inertia is $I = mR^2$,so $k^2 = R^2$. The kinetic energy is $K_{ring} = \frac{1}{2}mv^2(1 + 1) = mv^2$.Since $\frac{1}{2}mv^2 = 4 \ J$,then $mv^2 = 8 \ J$.Therefore,the total kinetic energy of the ring is $8 \ J$.

$A$ thin circular ring and a circular disc of equal mass are rolling without sliding. If their linear velocities are equal and the total kinetic energy of the disc is $6 \ J$,then the total kinetic energy of the ring is (in $J$)

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