Which of the following (assuming mass and rad | vedclass

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(C) The total kinetic energy $(K_{total})$ during pure rolling is the sum of translational kinetic energy $(K_{trans})$ and rotational kinetic energy

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Ring

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(C) The total kinetic energy $(K_{total})$ during pure rolling is the sum of translational kinetic energy $(K_{trans})$ and rotational kinetic energy $(K_{rot})$.$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Since $v = R\omega$,we have $K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2 = \frac{1}{2}mv^2(1 + \frac{I}{mR^2})$.The ratio of rotational kinetic energy to total kinetic energy is:$	ext{Ratio} = \frac{K_{rot}}{K_{total}} = \frac{\frac{1}{2}I\omega^2}{\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2} = \frac{\frac{1}{2}I(\frac{v}{R})^2}{\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2} = \frac{I}{mR^2 + I} = \frac{1}{1 + \frac{mR^2}{I}}$.For a Disc $(I = \frac{1}{2}mR^2)$: $	ext{Ratio} = \frac{1}{1 + 2} = \frac{1}{3} \approx 0.33$.For a Solid Sphere $(I = \frac{2}{5}mR^2)$: $	ext{Ratio} = \frac{1}{1 + 2.5} = \frac{2}{7} \approx 0.28$.For a Ring $(I = mR^2)$: $	ext{Ratio} = \frac{1}{1 + 1} = \frac{1}{2} = 0.5$.For a Hollow Sphere $(I = \frac{2}{3}mR^2)$: $	ext{Ratio} = \frac{1}{1 + 1.5} = \frac{2}{5} = 0.4$.Comparing the values,the Ring has the maximum percentage of rotational kinetic energy. Thus,option $C$ is correct.

Which of the following (assuming mass and radius are the same) has the maximum percentage of total $K.E.$ in rotational form during pure rolling?

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