A sphere of mass m and radius r rolls on a ho | vedclass

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Question

(C) The total kinetic energy of a rolling sphere is the sum of its translational and rotational kinetic energies: $K = \frac{1}{2} mu^2 + \frac{1}{2}

Vedclass · Accepted Answer

$7 u^2 / 10 g$

Vedclass · Answer

(C) The total kinetic energy of a rolling sphere is the sum of its translational and rotational kinetic energies: $K = \frac{1}{2} mu^2 + \frac{1}{2} I \omega^2$. For a solid sphere, the moment of inertia $I = \frac{2}{5} mr^2$ and the rolling condition is $\omega = u/r$. Substituting these, $K = \frac{1}{2} mu^2 + \frac{1}{2} (\frac{2}{5} mr^2) (u/r)^2 = \frac{1}{2} mu^2 (1 + \frac{2}{5}) = \frac{1}{2} mu^2 (\frac{7}{5}) = \frac{7}{10} mu^2$. By the law of conservation of energy, the total kinetic energy is converted into potential energy at the maximum height $H$: $K = mgH$. Therefore, $\frac{7}{10} mu^2 = mgH$. Solving for $H$, we get $H = \frac{7 u^2}{10 g}$.

$A$ sphere of mass $m$ and radius $r$ rolls on a horizontal plane without slipping with the speed $u$. Now if it rolls up an incline, the maximum height it would attain will be

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