Question

A sphere is rolling on a frictionless surface as shown in the figure with a translational velocity $v{\text{ms}}^{-1}$. If it is to climb the inclined surface, then $v$ should be

• A. $\ge \sqrt{\frac{10}{7}gh}$
• B. $\ge \sqrt{2gh}$
• C. $2gh$
• D. $\frac{10}{7}gh$

Answer 1

The correct option is A $\ge \sqrt{\frac{10}{7}gh}$

To climb the hill the kinetic energy should be greater than the increase in potential energy
$\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2\ge mgh$
or $\frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\omega }^2+\frac{1}{2}m{v}^2\ge mgh$--------(1)
Also for pure rolling, $v=\omega r$-----------(2)
Putting equation (2) in (1)
$\frac{7}{10}m{v}^2\ge mgh$
$v\ge \sqrt{\frac{10}{7}gh}$