Question

A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to $\left(\frac{-k}{r^2}\right)$ where k is constant. What is the total energy of the particle?

• A. -$\frac{K}{2r}$
• B. -$\frac{K}{2r^2}$
• C. –$\frac{K^2}{2r}$
• D. -$\frac{K}{r}$

Answer 1

The correct option is A -$\frac{K}{2r}$

As centripetal force is $\frac{m{v}^2}{r}$ hence $\frac{m{v}^2}{r}$ = $\frac{k}{r^2}$

$\therefore$ K.E. of particle = mv${}^2$ = $\frac{K}{2r}$.

The force acting on a particle is conservative in nature hence we can find potential energy of the particle as

F = -$\frac{dv}{dt}$ $\Rightarrow$ dv = -Fdr

$\Rightarrow$ U = ${\int }_r^{\infty }$Fdr ${\int }_r^{\infty }$$\frac{K}{r^2}$ dr

U = -$\frac{K}{r}$

$\therefore$ Total energy of particle = K.E. + P.E. = -$\frac{K}{r}$ + $\frac{K}{2r}$= -$\frac{K}{2r}$.