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,
$F=\frac{m{v}^2}{r}=\frac{k}{r^2}$ ;
$m{v}^2=\frac{k}{r}\dots \left(i\right)$
Kinetic energy of the particle,
$K=\frac{1}{2}m{v}^2=\frac{k}{2r}$ (Using $\left(i\right)$ )
As $F=\frac{-dU}{dr}$
$\therefore$ Potential energy,
$U=-\underset{\infty }{\overset{r}{\int }}Fdr=-\underset{\infty }{\overset{r}{\int }}\left(\frac{-k}{r^2}\right)dr$
$=k\underset{\infty }{\overset{r}{\int }}{r}^{-2}dr=\frac{-k}{r}$ .
$\therefore$ Total energy $=K+U=\frac{k}{2r}-\frac{k}{r}=\frac{-k}{2r}$ .