A particle is moving in a circular path of radius a under the action of an attractive potential $\mathrm{U}=-\frac{\mathrm{k}}{2 \mathrm{r}^{2}}$. Its total energy is:

(A) $-\frac{3}{2} \frac{\mathrm{k}}{\mathrm{a}^{2}}$

(B) $-\frac{\mathrm{k}}{4 \mathrm{a}^{2}}$

(C) $\frac{\mathrm{k}}{2 \mathrm{a}^{2}}$

JEE Main · 2018 · 08 Apr 2018 · Offline · Work, Power & Energy

A particle is moving in a circular path of radius a under the action of an attractive potential

U = - \frac{k}{2 r ^{2}}

. Its total energy is:

Answer: (D)

The force comes from the potential through

F = - \frac{d U}{d r}

; with

U = - \frac{k}{2 r ^{2}}

this is an attractive force of magnitude

\frac{k}{r ^{3}}

. For the circular orbit of radius

a

this supplies the centripetal requirement,

\frac{m v ^{2}}{a} = \frac{k}{a ^{3}}

, so

m v^{2} = \frac{k}{a ^{2}}

and the kinetic energy is

K = \frac{1}{2} m v^{2} = \frac{k}{2 a ^{2}}

. Adding the potential energy at that radius,

U = - \frac{k}{2 a ^{2}}

, the total energy is

E = K + U = \frac{k}{2 a ^{2}} - \frac{k}{2 a ^{2}} = 0

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option