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Question

A particle of mass $$m$$ is moving in a circle of radius $$r$$ under a centripetal force equal to $$ \dfrac{-K}{r^2} \hat{r} $$ where $$K$$ is constant. What is the total energy of the particle ?

Solution
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Centripetal Area $$ = \dfrac{-K}{r^2} $$

$$ \dfrac{mv^2}{r} = \dfrac{K}{r^2} $$

$$ mv^{2} = \dfrac{K}{r} $$

Kinetic energy $$ = \dfrac{mv^2}{2} = \dfrac{K}{2r} $$

Potential energy at a distance $$r$$ is given as $$ U = \dfrac{-K}{r} $$

So total energy $$ = U + K $$

$$ = \dfrac{-K}{2R} $$

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