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

zero

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

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

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

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Solution

The correct option is A zero
$U = - \frac{K}{2 r^{2}} = P . E$
$F = - \frac{d u}{d r} = - (- \frac{K}{2} (- \frac{2}{r^{3}})) = - \frac{K}{r^{3}}$
$\frac{K}{r^{3}} = \frac{m v^{2}}{r} \Rightarrow m v^{2} = \frac{K}{r^{2}}$
$K . E . = \frac{1}{2} m v^{2} = \frac{K}{2 r^{2}}$
$E = P . E . + K . E . = 0$

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