Question

A hollow sphere rolls without slipping on the horizontal surface such that its translational velocity is $v$. Find that the maximum height attained by it on an inclined surface as shown in the figure.

• A. $\frac{5v^2}{6g}$
• B. $\frac{5v^2}{g}$
• C. $\frac{v^2}{6g}$
• D. $\frac{5v}{6g}$

Answer 1

The correct option is A $\frac{5v^2}{6g}$

In case of pure rolling, total mechanical energy remains conserved.
$\Rightarrow (v=r\omega ),v=v_{\text{CM}}$: Condition for pure rolling.
$\therefore$Applying the mechanical energy conservation between positions $\left(1\right)$ and $\left(2\right)$.


$K{E}_1+P{E}_1=K{E}_2+P{E}_2$
$(\frac{1}{2}m{v}_{\text{CM}}^2+\frac{1}{2}{I}_{\text{CM}}{\omega }^2)+0=0+0+mgh$
considering datum for $PE=0$ at the position $\left(1\right)$
$\Rightarrow \frac{1}{2}m{v}^2+(\frac{1}{2}\times \frac{2}{3}m{r}^2{\left(\frac{v}{r}\right)}^2)=mgh$
$\Rightarrow \frac{5}{6}m{v}^2=mgh$
$\therefore h=\frac{5v^2}{6g}$
Hence $h$ is the maximum height attained when it stops finally on the inclined surface i.e $v=0,\omega =0$