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(A) For a disc rolling without slipping,the translational kinetic energy is $K_t = \frac{1}{2}mv^2$.The rotational kinetic energy is $K_r = \frac{1}{2

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$2/3$

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(A) For a disc rolling without slipping,the translational kinetic energy is $K_t = \frac{1}{2}mv^2$.The rotational kinetic energy is $K_r = \frac{1}{2}I\omega^2$. Since $I = \frac{1}{2}mR^2$ and $\omega = v/R$,we have $K_r = \frac{1}{2}(\frac{1}{2}mR^2)(v/R)^2 = \frac{1}{4}mv^2$.The total kinetic energy is $K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.The ratio of translational kinetic energy to total kinetic energy is $\frac{K_t}{K_{total}} = \frac{\frac{1}{2}mv^2}{\frac{3}{4}mv^2} = \frac{1/2}{3/4} = \frac{2}{3}$.

$A$ disc is rolling without slipping on a straight surface. The ratio of its translational kinetic energy to its total kinetic energy is

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