Question

A body of mass $m$ is tied to one end of a string of length $l$ and revolves vertically in a circular path. At the lowest point of circle, what must be the $K.E.$ of the body so as to complete the circle

• A. $5\mspace{6mu} mgl$
• B. $4\mspace{6mu} mgl$
• C. $2.5\mspace{6mu} mgl$
• D. $2\mspace{6mu} mgl$

Answer 1

Answer: (C)

$2.5\mspace{6mu} mgl$

Answer 2

Minimum velocity at lowest point to complete vertical loop$= \sqrt{5gl}$

So minimum kinetic energy $= \frac{1}{2}m(v^{2})$ $= \frac{1}{2}m(\sqrt{5gl})^{2}$

$= \frac{5}{2}mgl = 2.5mgl$