Question

A particle is moving along a circular path of radius $R$ in such a way that at any instant magnitude of radial acceleration & tangential acceleration are equal. If at $t=0$ velocity of particle is $v_0$, then find the time period of first revolution of the particle .

Answer 1

At any moment let the speed be $v$
Therefore the normal component of acceleration is $\frac{v^2}{R}$
Tangential Acceleration:
$a=v^2/R$
$v\frac{dv}{ds}=\frac{v^2}{R}$
$\frac{dv}{v}=\frac{ds}{R}$
${}_{v_0}^v\frac{dv}{v}={}_0^s\frac{ds}{R}$
$\ln \left(\frac{v}{v_0}\right)=\frac{s}{R}$
$v=v_0{e}^{s/R}$ and $x=R\left[\ln \right(v)-\ln (v_0\left)\right]$
Therefore velocity at one revolution:
$v_{1rev}=v_0{e}^{2\pi R/R}=v_0{e}^{2\pi }$
Now using $ds/dt=v$
$\frac{ds}{dt}=v$
$ds=\frac{Rdv}{v}$
$\frac{Rdv}{v}=vdt$
${\int }_{v_0}^{v_{1}rev}\frac{Rdv}{v^2}={}_0^{t}dt$
$R[-\frac{1}{v_{1rev}}+\frac{1}{v_0}]=t$
$t=R[\frac{1}{v_0}-\frac{1}{v_0{e}^{2\pi }}]=\frac{R}{v_0}(1-e^{-2\pi })$