A particle of mass `m` is moving in a horizontal circle of radius r, under a centripetal force equal to `- (K//r^(2))`, where K is constant. What is the total energy of the particle ?

Explanation

As the particle is moving in a circle
`(mv^(2))/(r)=(K)/(r^(2))["for circular motion"|F|=(mv^(2))/(r)]`
`KE = (1)/(2)mv^(2)=(K)/(2r) " " ...........(1)`
Now as `F = - (dU//dr).`
`thereforeU=-underset(oo)overset(r)int Fdr=-underset(oo)overset(r)int-(k)/(r^(2))dr`
`U = - (K)/(r) " " .........(2)`
So Total energy E= U + K
`=-(K)/(r)+(K)/(2r)From (1) &(2) i.e., E=(K)/(2r)`