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

Answer 1

Answer: (C)

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