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(A) For a hollow sphere rolling without slipping,the moment of inertia about its axis of symmetry is $I = \frac{2}{3} mR^2$.The rotational kinetic ene

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$2$

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(A) For a hollow sphere rolling without slipping,the moment of inertia about its axis of symmetry is $I = \frac{2}{3} mR^2$.The rotational kinetic energy is $K_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{2}{3} mR^2) \omega^2 = \frac{1}{3} mR^2 \omega^2$.The total kinetic energy is $K_{total} = K_{rot} + K_{trans} = \frac{1}{2} I \omega^2 + \frac{1}{2} mv^2$.Since $v = R\omega$,we have $K_{total} = \frac{1}{2} (\frac{2}{3} mR^2) \omega^2 + \frac{1}{2} m(R\omega)^2 = \frac{1}{3} mR^2 \omega^2 + \frac{1}{2} mR^2 \omega^2 = (\frac{2+3}{6}) mR^2 \omega^2 = \frac{5}{6} mR^2 \omega^2$.The ratio of rotational kinetic energy to total kinetic energy is $\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{3} mR^2 \omega^2}{\frac{5}{6} mR^2 \omega^2} = \frac{1}{3} 	imes \frac{6}{5} = \frac{2}{5}$.Comparing this with $\frac{x}{5}$,we get $x = 2$.

$A$ hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is $\frac{x}{5}$. The value of $x$ is . . . . . . .

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