Question

A particle is moving in a circular path of radius $a$ with constant speed under the action of an attractive conservative force. Potential energy of the particle is given by the relation $U=\frac{-K}{2r^2}$, where $r$ is the radial distance of the particle from the centre of the circular path. Its total energy will be:

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

Answer 1

The correct option is B $\text{Zero}$

Relation between potential energy and conservative force is given by
$F=-\frac{dU}{dr}$
where $r$ is the radial distance of the particle describing circular path.

Here $U=\frac{-K}{2r^2}$
$\Rightarrow F=\frac{-dU}{dr}=\frac{-K}{2}\times (-2r^{-3})=\frac{K}{r^3}$
Since it is performing circular motion, force $F$ must provide centripetal force.
$\Rightarrow F=m{a}_c=\frac{m{v}^2}{r}$
$\Rightarrow \frac{m{v}^2}{r}=\frac{K}{r^3}$
$\therefore m{v}^2=\frac{K}{r^2}$

Kinetic energy of particle is given by,
$K.E=\frac{1}{2}m{v}^2=\frac{K}{2r^2}$
Total energy of particle is given by:
$T.E=P.E.+K.E.$
$\therefore T.E=\frac{-K}{2r^2}+\frac{K}{2r^2}=0$