Question

A sphere of mass $m$ is rolling on a frictionless surface (as shown in the figure) with a translation velocity $v\text{m/s}$. The sphere will climb on the inclined plane if $v$ is

• A. $\ge \sqrt{\frac{5}{7}gh}$
• B. $\ge \sqrt{2gh}$
• C. $\ge \sqrt{\frac{10}{7}gh}$
• D. $\ge \sqrt{gh}$

Answer 1

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

The sphere will climb on the inclined plane when
Total energy of the sphere is $\ge mgh$
Since the sphere have translational as well as rotational motion.
Total energy of the sphere = Rotational kinetic energy+Translational kinetic energy
$\therefore TE=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2$
($\omega =$ angular speed of sphere)

$\therefore \frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2\ge mgh$

$\Rightarrow \frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\omega }^2+\frac{1}{2}m{v}^2\ge mgh$
For pure rolling, $v=r\omega$
$\Rightarrow \frac{1}{5}m{v}^2+\frac{1}{2}m{v}^2\ge mgh$$\Rightarrow v\ge \sqrt{\frac{10}{7}gh}$

Why this question? Tips:- A rolling body contains rotational as well as translational kinetic energy.