Physics - Circular Motion Question with Solution | TestHub

Let velocity at the lowest point is "u" and velocity at the top most point is "v"

At highest point  $\frac{m{v}^2}{\text{R}}=mg\Rightarrow v=\sqrt{g\mathrm{R}}$

now $v^2=u^2-2g\left(2\mathrm{R}\right)\Rightarrow u^2=5g\mathrm{R}$

Minimum velocity required is $u=\sqrt{5\mathrm{g}\mathrm{R}}$