Question

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

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

Answer 1

Answer: (A)

$\ge \sqrt{\frac{10}{7}gh}$

Rotation and kinetic energy at base should be greater than the potential energy at top point of incline.

$\frac{1}{2}I{\omega }^2+\frac{1}{2}m{\nu }^2\ge mgh$
or
$\frac{1}{2}\left(\frac{2}{5}m{r}^2\right){\omega }^2+\frac{1}{2}m{\nu }^2\ge mgh$
or
$\nu \ge \sqrt{\frac{10}{7}gh}$