Q. A sphere of mass $m$ and radius $r$ rolls on a horizontal plane without slipping with the speed $u$. Now, if it rolls up vertically, the maximum height it would attain will be
A) $\frac{{3u}^{2}}{4g}$
B) $\frac{{5u}^{2}}{2g}$
C) $\frac{{7u}^{2}}{10g}$
D) $\frac{u^{2}}{2g}$
Explanation
(c) The rolling sphere has rotational as well as translational kinetic energy.
$\therefore$ Kinetic energy $= \frac{1}{2}mu^{2} + \frac{1}{2}I\omega^{2}$
$= \frac{1}{2}mu^{2} + \frac{1}{2}\left( \frac{2}{5}mr^{2} \right)\omega^{2}$
$= \frac{1}{2}mu^{2} + \frac{1}{5}mu^{2} = \frac{7}{10}mu^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because u = r\omega)$
From conservation of energy
$ie$, $mgh = \frac{7}{10}mu^{2}$
or $h = \frac{7u^{2}}{10g}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.