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:

(1) $-\frac {k}{4a^2}$
(2) $\frac {k}{2a^2}$
(3) Zero
(4) $-\frac {3}{2} \frac {k}{a^2}$

[Based on JEE Main 2018]

Solution

Let the mass of the particle be m,

$P.E.=mU=-\frac {1}{2} \frac {mk}{r^2}$

$T.M.E.=\frac {1}{2} mv^2 +\left(-\frac {1}{2} \frac {mk}{r^2}\right)$

Where v is the speed of the particle.

$T.M.E.=\frac {1}{2} m \left(v^2 – \frac {k}{r^2}\right)$

$F=- \frac {d}{dr} \left(-\frac {1}{2} \frac {mk}{r^2}\right)=-\frac {mk}{r^3}$

Since F provides centripetal force,

$\left| { – \frac{{mk}}{{{r^3}}}} \right| = \frac{{m{v^2}}}{r},\therefore \frac{k}{{{r^2}}} = {v^2}$

Hence, T.M.E. = 0

Hence, Option (3).

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