Question

A particle is moving in a circular path of radius '$a$' under the action of an attractive potential, $U=-\frac{k}{2r^2}$ (where $r$ is the radial distance). Its total energy is

• A. Zero
• B. $-\frac{3}{2}\frac{k}{a^2}$
• C. $-\frac{k}{4a^2}$
• D. $\frac{k}{2a^2}$

Answer 1

The correct option is A Zero

Given, $U=\frac{-k}{2r^2}$
We know that,
$F_x=\frac{-dU}{dx}$ where $\overrightarrow{x}$ & $\overrightarrow{F}$ are in the same direction.
But here, radius vector $\overrightarrow{r}$ and centripetal force $\overrightarrow{F_r}$ are in opposite directions.

$\therefore F_r=\frac{dU}{dr}=-k\frac{(-2)}{2r^3}=\frac{k}{r^3}$
$\Rightarrow \frac{m{v}^2}{r}=\frac{k}{r^3}\Rightarrow m{v}^2=\frac{k}{r^2}$
$\Rightarrow K.E.=\frac{1}{2}m{v}^2=\frac{k}{2r^2}$

$\therefore$ Total energy $=K.E+P.E=\frac{k}{2r^2}-\frac{k}{2r^2}=0$