Question

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

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

Answer 1

The correct option is C $\frac{7u^2}{10g}$

at no slipping condition for sphere,

$u=\omega r$

$\Rightarrow \omega =\frac{u}{r}$

From energy conservation principle,

Initial energy= Final energy

$\Rightarrow$Linear kinetic energy+rotational energy=final potential energy

$\Rightarrow$ $\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh$

$\Rightarrow \frac{1}{2}m{u}^2+\frac{1}{2}\times \frac{2}{5}m{r}^2{\left(\frac{u}{r}\right)}^2=mgh$

$\Rightarrow \frac{1}{2}m{u}^2+\frac{1}{5}m{u}^2=mgh$

$\Rightarrow \frac{7}{10}m{u}^2=mgh$

$\Rightarrow h=\frac{7u^2}{10g}$