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

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$z e r o$

$- \frac{3}{2} \frac{k}{a^{2}}$

$- \frac{k}{4 a^{2}}$

$\frac{k}{2 a^{2}}$

answer is D.

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Detailed Solution

$\begin{array}{l} F = - \frac{d U}{d r} = - \frac{d (- K / 2 r^{2})}{d r} = \frac{K}{2} \frac{d r^{- 2}}{d r} (o n d i f f e r e n t i a t i n g w i t h r e s p e c t t o r) \\ F = \frac{K}{2} (- 2 r^{- 3}) = - \frac{K}{r^{3}} \end{array}$

Since it is performing circular motion FORCE F= $\frac{m v^{2}}{r} = \frac{K}{r^{3}}$

$\begin{array}{l} m v^{2} = \frac{K}{r^{2}} \Rightarrow K . E = \frac{1}{2} m v^{2} = \frac{K}{2 r^{2}} \\ g i v e n P E = U = \frac{K}{2 r^{2}} \\ T o t a l E n e r g y = K E + P E = - \frac{K}{2 r^{2}} + \frac{K}{2 r^{2}} = z e r o \end{array}$