Question

If a particle of mass 'm' is moving in a horizontal circle of radius 'r' with a centripetal force $\propto (-1/r^2)$, the total energy is

Answer 1

Mass ${}^{\prime }{m}^{\prime }$ revolves in a circle of radius ${}^{\prime }{r}^{\prime }$ experiences centripetal force :

$F_e\alpha -\frac{1}{r^2}$

$\Rightarrow F_e=-\frac{A}{r^2}$ ($A$ : some arbitrary constant )

If potential energy of system is $u$, the force can be written :-

$F_e=-\frac{du}{dr}$

$\Rightarrow \frac{du}{dr}=\frac{A}{r^2}\Rightarrow \int du=\int \frac{A}{r^2}dr$ [integrating]

$\Rightarrow u=\frac{-A}{r}\to \left(1\right)$

Also, kinetic energy $k$, is related to centripetal force as :

$K=\frac{|Fe|r}{2}=\frac{A}{2r}\to \left(2\right)[ask>0]$

from $\left(1\right)$ and $\left(2\right)$, total energy of system $\left(E\right)$ :

$E=u+k=\frac{-A}{r}+\frac{A}{2r}$

$\Rightarrow E=\frac{-A}{2r}$