Question

A particle moves in a plane under the action of a force which is always perpendicular to the particle's velocity and depends on a distance to a certain point on the plane as $\frac{1}{r^n}$, where $n$ is a constant. At what value of $n$ will the motion of the particle along the circle be steady?

Answer 1

This is not central force problem unless the path is a circle about the said point. Rather here $F_t$ (tangential force) vanishes. Thus equation of motion becomes,
$v_t=v_0=$ constant
and, $\frac{m{v}_0^2}{r}=\frac{A}{r^2}$ for $r=r_0$
We can consider the latter equation as the equilibrium under two forces. When the motion is perturbed, we write $r=r_0+x$ and the net force acting on the particle is,
$-\frac{A}{{(r_0+x)}^n}+\frac{m{v}_0^2}{r_0+x}=\frac{-A}{r_0^n}(1-\frac{nx}{r_0})+\frac{m{v}_0^2}{r_0}(1-\frac{x}{r_0})=\frac{m{v}_0^2}{r_0^2}(1-n)x$
This is opposite to the displacement $x$, if $n<1$. ($\frac{m{v}_0^2}{r}$ is an outward directed centrifugal force while $\frac{-A}{r^n}$ is the inward directed external force).