A particle is moving in a circular path of radius a under the action of an attractive potential `U=-k/(2r^2)`. Its total energy is

Options

A. zero
B. `-3/2 k/(a^2)`
C. `-k/(4a^2)`
D. `k/(2a^2)`

Answer

A

Explanation

`F=-(dU)/(dr)=-k/(r^3)`
This force provides the required centripetal force for circular motion of the particle.
`therefore (mv^2)/r=k/(r^3)`
`or, mv^2=k/(r^2) or, 1/2 mv^2 =k/(2r^2)`
So, kinetic energy `=k/(2r^2)`
Total energy of the particle =kinetic energy + potential energy
`=k/(2r^2) -k/(2r^2)=0`