Question

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. $\frac{3 u^{2}}{4 g}$
• B. $\frac{5 u^{2}}{2 g}$
• C. $\frac{7 u^{2}}{10 g}$
• D. $\frac{11 u^{2}}{9 g}$

Answer 1

Answer: (C)

The rolling sphere has rotational as well as translational kinetic energy.
$\therefore $ Kinetic energy $=\frac{1}{2} m u^2+\frac{1}{2}\left(\frac{2}{5} m r^2\right) \omega^2$
$=\frac{1}{2}mu^{2}+\frac{1}{2}\left(\frac{2}{5} m r^{2}\right)\left(\omega \right)^{2}$
$=\frac{1}{2}mu^{2}+\frac{1}{5}mu^{2}=\frac{7}{10}mu^{2} \, \, \, \left(\right.\because u=r\omega \left.\right)$
From conservation of energy
$ie$ , $mgh=\frac{7}{10}mu^{2}$
or $h=\frac{7 u^{2}}{10 g}$