91Ó°ÊÓ

The body is rolling without slipping, which means it has both translational and rotational kinetic energy. The total kinetic energy (TKE) is the sum of the translational kinetic energy (TKE_trans) and rotational kinetic energy (TKE_rot). Mathematically, it is expressed as:\[ TKE = TKE_{trans} + TKE_{rot} \] where\[ TKE_{trans} = \frac{1}{2}mv^2 \]and\[ TKE_{rot} = \frac{1}{2}I\omega^2 \]

According to the question, rotational energy constitutes 40% of the total kinetic energy. Therefore, \[ TKE_{rot} = 0.4 imes TKE \]

Let's analyze each option to find which body fits this condition:- For a **ring** rolling without slipping: - \( I = mr^2 \) - \( v = r\omega \) - \( TKE_{rot} = \frac{1}{2}(mr^2)\omega^2 = \frac{1}{2}mv^2 \) - \( TKE = TKE_{trans} + TKE_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2 \) - \( TKE_{rot} = \frac{1}{2}mv^2 = 0.5 imes TKE \)- For a **hollow cylinder** rolling without slipping (same as ring): - Similar derivation, resulting in \( TKE_{rot} = 0.5 \times TKE \)- For a **solid cylinder** rolling without slipping: - \( I = \frac{1}{2}mr^2 \) - \( v = r\omega \) - \( TKE_{rot} = \frac{1}{2}(\frac{1}{2}mr^2)\omega^2 = \frac{1}{4}mv^2 \) - \( TKE = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2 \) - \( TKE_{rot} = \frac{1}{4}mv^2 = \frac{1}{3} \times TKE \)- For a **hollow sphere**: - This body is not commonly included in rotational inertia comparisons, but hollow cylinders and hollow spheres usually have similar derivations to hollow rings.

Based on the energy distribution, only the hollow cylinder satisfies the condition: - \( TKE_{rot} = 0.5 imes TKE \), which matches a 50% contribution. - Other bodies either have higher or lower ratios, except these usually approximate similar contributions. - For exploratory purposes, the solid sphere could also fit certain understanding, but must compare with exact derivation.