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

C

Explanation

`U = -K/(2r^(2))`
Force can be calculated from potential energy as follows :
`F = (dU)/(dr) = - K/2((-2)/(r^(3)))=-K/(r^(3))`
Negative sign indicates that it is central force which acts as centripetal force for the circular motion of particle . Hence we ca write the following equation using circular dynamics :
`K/(r^(3)) = (mv^(2))/r rArr mv^(2) = K/(r^(2))`
`rArr K.E =1/2 mv^(2) = K/(2r^(2))`
Total energy `T.E = P.E +K.E = 0 `