Q. The centripetal acceleration of a particle varies inversely with the square of the radius $r$of the circular path .The KE of this particle varies directly as
A) $r$
B) $r^{2}$
C) $r^{- 2}$
D) $r^{- 1}$
Explanation
(d) Centripetal acceleration
$\frac{v^{2}}{r} \propto r^{- 2}$
Or $\frac{v^{2}}{r} = \frac{k}{r^{2}}$
Or $v^{2} = \frac{k}{r}$
$\therefore\ \ KE = \frac{1}{2}mv^{2}$
$= \frac{1}{2}\ \frac{mk}{r}\ $
Or $KE \propto r^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.