Question

A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity v m/s. 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 $\ge \sqrt{\frac{10}{7}gh}$

Applying law of conservation of energy for rotating body,
$\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh$
$\frac{1}{2}m{v}^2+\frac{1}{2}\frac{2}{5}m{r}^2\times \frac{v^2}{r^2}=mgh$
$\frac{v^2}{2}+\frac{2v^2}{10}=gh$
$\frac{5v^2+2v^2}{10}=gh\Rightarrow v^2=\frac{10}{7}gh$
$v\ge \sqrt{\frac{10}{7}gh}$