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).
