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  • A particle is moving in a circular path of radius a under the action of an attractive potential U=  . Its total energy is :

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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 :

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@Shilpa

U=-\frac{K}{2r^{2}}

\therefore F=\frac{-dU}{dr}=\frac{-K}{r^{3}}

\Rightarrow \frac{K}{r^{3}}=\frac{mv^{2}}{r} or  mv^{2}=\frac{K}{r^{2}}

K.E.=\frac{1}{2}mv^{2}=\frac{K}{2r^{2}}

Total energy = K.E.+P.E.=0

 

 

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avinash.dongre

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