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(C) For a rolling hoop,the total kinetic energy $(KE_{roll})$ is the sum of translational and rotational kinetic energy: $KE_{roll} = \frac{1}{2}mV^2

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$\sqrt{2} : 2$

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(C) For a rolling hoop,the total kinetic energy $(KE_{roll})$ is the sum of translational and rotational kinetic energy: $KE_{roll} = \frac{1}{2}mV^2 + \frac{1}{2}I\omega^2$. Since $I = mr^2$ and $\omega = V/r$,we get $KE_{roll} = \frac{1}{2}mV^2 + \frac{1}{2}(mr^2)(V/r)^2 = \frac{1}{2}mV^2 + \frac{1}{2}mV^2 = mV^2$.For a sliding hoop,the kinetic energy $(KE_{slide})$ is purely translational: $KE_{slide} = \frac{1}{2}mv^2$.Given that the kinetic energies are equal,we equate them: $mV^2 = \frac{1}{2}mv^2$.This simplifies to $V^2 = v^2 / 2$,or $V^2 / v^2 = 1 / 2$.Taking the square root of both sides,the ratio of their speeds is $V / v = \sqrt{1/2} = 1 / \sqrt{2} = \sqrt{2} / 2$.

Two identical circular loops are moving with the same kinetic energy; one rolls and the other slides. The ratio of their speed is:

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