Question

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. The total energy of the particle is

• A. $ -\frac{k}{r} $
• B. $ -\frac{k}{2r} $
• C. $ \frac{k}{2r} $
• D. $ \frac{2k}{r} $

Answer 1

Answer: (B)

Since the particle is moving in horizontal circle, centripetal force,
$ F = \frac{mv^{2}}{r} = \frac{k}{r^{2}} $ ;
$ mv^{2} = \frac{k}{r} \ldots\left(i\right) $
Kinetic energy of the particle,
$ K = \frac{1}{2}mv^{2} = \frac{k}{2r}$ (Using $ \left(i\right) $ )
As $ F = \frac{-dU}{dr} $
$ \therefore $ Potential energy,
$ U = -\int\limits_{\infty}^{r} F \,dr = -\int\limits_{\infty}^{r}\left(\frac{-k}{r^2}\right)dr $
$ = k \int\limits^{r}_{\infty } r^{-2}dr = \frac{-k}{r} $ .
$ \therefore $ Total energy $ = K + U = \frac{k}{2r}-\frac{k}{r} = \frac{-k}{2r} $ .