Discussion Forum

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

Answer:

Explanation:

 Force = $-\frac{\text{d}U}{\text{d}r}$

$\Rightarrow   F=-\frac{\text{d}}{\text{d}r}(\frac{-k}{2r^{2}})=$ $-\frac{k}{r^{3}}$

As particle is on circular path, this force must be centripetal force.

$\Rightarrow    \mid F \mid=\frac{mv^{2}}{r}$

So, 

$\frac{k}{r^{3}}=\frac{mv^{2}}{r}\Rightarrow\frac{1}{2}mv^{2}=\frac{k}{2r^{2}}$

 Total energy of particle = KE +PE

  = $\frac{k}{2r^{2}}-\frac{k}{2r^{2}}=0$

Total energy =0