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(C) Case $(i)$: When the hollow cylinder slides without rotating,it possesses only translational kinetic energy.$K_{trans} = \frac{1}{2}mv^2$Case $(ii

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$1:2$

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(C) Case $(i)$: When the hollow cylinder slides without rotating,it possesses only translational kinetic energy.$K_{trans} = \frac{1}{2}mv^2$Case $(ii)$: When the hollow cylinder rolls without slipping,it possesses both translational and rotational kinetic energy.$K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a thin hollow cylinder,the moment of inertia $I = mR^2$ and for rolling without slipping,$\omega = \frac{v}{R}$.$K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$Ratio of kinetic energies:$\frac{K_{trans}}{K_{total}} = \frac{\frac{1}{2}mv^2}{mv^2} = \frac{1}{2}$Thus,the ratio is $1:2$.

$A$ thin hollow cylinder open at both ends is moving with the same speed $v$ in two different cases:
$(i)$ It slides without rotating.
$(ii)$ It rolls without slipping.
Find the ratio of the kinetic energies in the two cases.

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$A$ thin hollow cylinder open at both ends is moving with the same speed $v$ in two different cases:$(i)$ It slides without rotating.$(ii)$ It rolls without slipping.Find the ratio of the kinetic energies in the two cases.