Physics (2018) | JEE 2018 | Discussion Page

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

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

B) 0

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

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

Option B

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

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