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) $3u^{2}/4g$
B) $5u^{2}/2g$
C) $7u^{2}/10g$
D) $u^{2}/2g$
Explanation
(c) The rolling sphere has rotational as well as translational kinetic energy
$\therefore$ Kinetic energy $= \frac{1}{2}mu^{2} + \frac{1}{2}I\omega^{2}$
$= \frac{1}{2}mu^{2} + \frac{1}{2}\left( \frac{2}{5}mr^{2} \right)\omega^{2} = \frac{1}{2}mu^{2} + \frac{mu^{2}}{5} = \frac{7}{10}mu^{2}$
Potential energy = kinetic energy
$\therefore mgh = \frac{7}{10}mu^{2} \Rightarrow h = \frac{7u^{2}}{10g}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.