Q. The kinetic energy $k$ of a particle moving along a circle of radius $R$ depends upon the distance $s$ as $k = as^{2}$. The force acting on the particle is
A) $2a\frac{s^{2}}{R}$
B) $2as\left\lbrack 1 + \frac{s^{2}}{R^{2}} \right\rbrack^{1/2}$
C) $2as$
D) $2a$
Explanation
(b) Here $k = \frac{1}{2}mv^{2} = as^{2}$
$\therefore\ mv^{2} = 2as^{2}$
Differentiating w.r.t. time $t$
$2mv\frac{dv}{dt} = 4as\frac{ds}{dt} = 4asv,m\frac{dv}{dt} = 2as$
This is the tangential force, $F_{t} = 2as$
Centripetal force $F_{c} = \frac{mv^{2}}{R} = \frac{2as^{2}}{R}$
$\therefore$ Force acting on the particle
$F = \sqrt{F_{t}^{2} + F_{c}^{2}} = \sqrt{(2as)^{2} + \left( \frac{2as}{R} \right)^{2}} = 2as\sqrt{1 + s^{2}/R^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.