Question

If a particle of mass m is moving in a horizontal circle of radius $r$ with a centripetal force $\left(-\frac{K}{r^2}\right)$ , total energy is

• A. $-\frac{K}{2r}$
• B. $-\frac{K}{r}$
• C. $-\frac{2K}{r}$
• D. $-\frac{4K}{r}$

Answer 1

Answer: (A)

Given, centripetal force, $F=-\frac{K}{r^2}$
Also, $\frac{K}{r^2}=\frac{m{v}^2}{r}$ or $m{v}^2=\frac{K}{r}$
$\therefore$ Kinetic energy, $K.E.=\frac{1}{2}m{v}^2=\frac{K}{2r}$
For central forces, $P.E.=-2K.E.=-\frac{K}{r}$
Total energy, $T.E.=P.E.+K.E.$
$=\frac{-K}{r}+\frac{K}{2r}=-\frac{K}{2r}$