Q. A particle is moving in a circle of radius $R$ in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $t = 0\ is\ v_{0}$, the time taken to complete the first revolution is
A) $\frac{R}{v_{0}}$
B) $\frac{R}{v_{0}}(1 - e^{- 2\pi})$
C) $\frac{R}{v_{0}}e^{- 2\pi}$
D) $\frac{2\pi R}{v_{0}}$
Explanation
(b) When a particle moves in a circular motion, it is acted upon by centripetal force directed towards the centre. Hence, centripetal acceleration is
$a_{N} = \frac{dv}{dt} = \frac{v^{2}}{R}$
$or\ \ \int_{0}^{t}{\frac{dt}{R} = \int_{v_{0}}^{v}\frac{dv}{v^{2}}}$
$or\ \ \ t = - R\left\lbrack \frac{1}{v} \right\rbrack_{v_{0}}^{v}$
$v = \frac{v_{0}R}{R - v_{0}t}$
$Also\ \ \ \ \frac{dr}{dt} = \frac{v_{0}R}{\left( R - v_{0}t \right)}$
$\int_{0}^{2\pi\ \ R}{dr = v_{0}R\ \int_{0}^{T}\frac{dt}{R - v_{0}t}}$
$\Longrightarrow \ \ \ T = \frac{R}{v_{0}}\left( 1 - {1 - e}^{- 2\pi} \right)\ $
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.