Question

A body is revolving in a vertical circle of radius r such that the sum of its $K.E.$ and $P.E.$ is constant. If the speed of the body at the highest point is $\sqrt{2gr}$ then the speed of the body at the lowest point will be:

• A. $\sqrt{7gr}$
• B. $\sqrt{6gr}$
• C. $\sqrt{8gr}$
• D. $\sqrt{9gr}$

Answer 1

The correct option is B $\sqrt{6gr}$

Applying WET between A and B.

$\frac{m(\sqrt{2gR}{)}^2}{2}+mg\left(2R\right)=\frac{m}{2}{v}_0^2$

$mgr+2mgR=\frac{m}{2}{v}_0^2$

$v_0=\sqrt{6gr}$