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(A) The total kinetic energy $(E_{total})$ of a rolling body is the sum of its translational kinetic energy $(E_t)$ and rotational kinetic energy $(E_

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$1/3$

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(A) The total kinetic energy $(E_{total})$ of a rolling body is the sum of its translational kinetic energy $(E_t)$ and rotational kinetic energy $(E_r)$.$E_{total} = E_t + E_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since $I = mk^2$ and $v = r\omega$,we have $E_r = \frac{1}{2}mk^2(\frac{v^2}{r^2}) = \frac{1}{2}mv^2(\frac{k^2}{r^2})$.For a disc,the radius of gyration $k^2 = \frac{1}{2}r^2$,so $\frac{k^2}{r^2} = \frac{1}{2}$.$E_{total} = \frac{1}{2}mv^2(1 + \frac{k^2}{r^2}) = \frac{1}{2}mv^2(1 + \frac{1}{2}) = \frac{3}{4}mv^2$.The rotational kinetic energy is $E_r = \frac{1}{2}mv^2(\frac{1}{2}) = \frac{1}{4}mv^2$.The fraction of total energy in rotational form is $\frac{E_r}{E_{total}} = \frac{\frac{1}{4}mv^2}{\frac{3}{4}mv^2} = \frac{1}{3}$.

$A$ disc rolls without slipping on an inclined plane. What fraction of its total energy is in the form of rotational kinetic energy?

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