Question

A particle is moving in a circle of radius $R$ in such a way that at any instant the normal and tangential component of its acceleration are equal. If its speed at $t=0$ is $v_0$. The time taken to complete the first two revolutions is

• A. $\frac{R}{v_0}(1-e^{-4\pi })$
• B. $\frac{R}{v_0}(1+e^{-4\pi })$
• C. $\frac{R}{v_0}(1-e^{-2\pi })$
• D. $\frac{R}{v_0}(1+e^{-2\pi })$

Answer 1

The correct option is A $\frac{R}{v_0}(1-e^{-4\pi })$

Let the speed at instant $t$ be $v$.

So, radial accelaration, $a_c=\frac{v^2}{R}$

From the given question,

Tangential acceleration , $a_t=a_c$

$\Rightarrow a_t=\frac{v^2}{R}$

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

Integrating the above from time, $t=0$ to time, $t=t$

$\Rightarrow {\int }_{v_0}^v\frac{dv}{v^2}={\int }_0^t\frac{dt}{R}$

$\Rightarrow \frac{1}{v_0}-\frac{1}{v}=\frac{t}{R}$

$\Rightarrow t=R(\frac{1}{v_0}-\frac{1}{v})....\left(1\right)$

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

Where, $s$ is the length of the arc covered by the particle as it moves in the circle.

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

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

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

As, at $t=0$, $s=0$ and for first two revolution, if $t=T$ is the time taken then, $s=4\pi R$.

$\Rightarrow {\int }_0^{4\pi R}\frac{ds}{R}={\int }_0^T\frac{dt}{(\frac{R}{v_0}-t)}$

$\Rightarrow \frac{\left[s\right]{|}_0^{4\pi R}}{R}={[-\ln (\frac{R}{v_0}-t)]}_0^T$

$\Rightarrow \ln \frac{R}{v_0}-\ln (\frac{R}{v_0}-T)=\frac{4\pi R}{R}$

$\Rightarrow \ln [1-\frac{T{v}_0}{R}]=-4\pi$

$\Rightarrow \frac{T{v}_0}{R}=1-e^{-4\pi }$

$\therefore T=\frac{R}{v_0}(1-e^{-4\pi })$

Hence, option \((a) is the correct answer.