A particle is moving in a circular path of radius -a- under the action of an attractive potential- U-k2r2 where r is the radial distance- Its total energy is

Given, U= k2r^2 We know that, F_x= dUdx where x amp; F are in the same direction. But here, radius vector r and centripetal force F_r are in opposite direct ...

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Standard XII

Physics

Conservative Force as Gradient of Potential

A particle is...

Question

A particle is moving in a circular path of radius ' $a$ ' under the action of an attractive potential, $U = - \frac{k}{2 r^{2}}$ (where $r$ is the radial distance). 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
Given, $U = \frac{- k}{2 r^{2}}$
We know that,
$F_{x} = \frac{- d U}{d x}$ where $\to x$ & $\to F$ are in the same direction.
But here, radius vector $\to r$ and centripetal force $_{r}$ are in opposite directions.

$∴ F_{r} = \frac{d U}{d r} = - k \frac{(- 2)}{2 r^{3}} = \frac{k}{r^{3}}$
$\Rightarrow \frac{m v^{2}}{r} = \frac{k}{r^{3}} \Rightarrow m v^{2} = \frac{k}{r^{2}}$
$\Rightarrow K . E . = \frac{1}{2} m v^{2} = \frac{k}{2 r^{2}}$

$∴$ Total energy $= K . E + P . E = \frac{k}{2 r^{2}} - \frac{k}{2 r^{2}} = 0$

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A particle is moving in a circular path of radius 'a' under the action of an attractive potential, U=−k2r2 (where r is the radial distance). Its total energy is

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