Question

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

Answer 1

For circular motion, the radial acceleration, $a_r$ is given by $\frac{v^2}{R}$, and the tangential acceleration, $a_t$ is given by $\frac{dv}{dt}$

Since these accelerations are equal, so $\frac{v^2}{R}=\frac{dv}{dt}$

Hence,$\left(\frac{R}{v^2}\right)dv=dt$

Upon integration

$\int \frac{R}{v^2}dv=\int dt$

$\frac{R}{v}=t+C$

To evaluate integration constant C

Put $v=v_0$ at $t=0$

Therefore, $C=-\frac{R}{v_0}$

The relation between v and t is ,

$\frac{R}{v}=t-\frac{R}{v_0}$

$t=R\left[\frac{v-v_0}{v_0.v}\right]$

Now, $v=\frac{ds}{dt}$,

where, s id the length of the arc covered by the particle as it moves in the circle

Therefore, $t=\frac{R}{v_0}-\frac{R}{\frac{ds}{dt}}$

$t=\frac{R}{v_0}-R.\frac{dt}{ds}$

$tds=\frac{R}{v_0}-Rdt$

$ds(\frac{R}{v_0}-t)=Rdt$

$\frac{ds}{R}=\frac{dt}{\frac{R}{v_0}-t}$

Upon integrating

$\int \frac{ds}{R}=\int \frac{dt}{\frac{R}{v_0}-t}$

$\frac{s}{R}=-\ln (\frac{R}{v_0}-t)+C$

putting at t=0 and s=0

$C=\frac{\ln R}{v_0}$

Therefore, the relation takes the form

$\frac{s}{R}=\frac{\ln R}{v_0}-\ln (\frac{R}{v_0}-t)=\ln \left(\frac{R}{R-v_{0}t}\right)$

For complete revolution, putting $s=2\pi R,t=T$

$\frac{2\pi R}{R}=\ln \left(\frac{R}{R-v_{0}T}\right)$

$T=\frac{R}{v_0}(1-e^{-2\pi })$