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 it's accelerations are equal. If it's speed at $t=0$ is $v_0$. The time taken to complete the first revolution is

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

Answer 1

The correct option is C $\frac{R}{v_0}(1-e^{-2\pi })$

Let at time t, the normal component of acceleration, $a_r$ and tangential component $a_t$ are equal.
So, $a_t=a_r....\left(1\right)$

We know that, $a_t=\frac{dv}{dt}$ and $a_r=\frac{v^2}{R}$

From equation $\left(1\right)$, we have
$\frac{dv}{dt}=\frac{v^2}{R}$

Rearranging and integrating the velocity from $v_0$ to $v$ and time from $0$ to $t$,

$\Rightarrow {\int }_{v_0}^v{v}^{-2}dv={\int }_0^t\frac{1}{R}dt$

$\Rightarrow {[-\frac{1}{v}]}_{v_0}^v=\frac{1}{R}[t{]}_0^t$

$\Rightarrow [\frac{-1}{v}-(-\frac{1}{v_0})]=\frac{1}{R}[t-0]$

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

$\Rightarrow \frac{1}{v}=\frac{R-t{v}_0}{v_{0}R}$

$\Rightarrow v=\frac{v_{0}R}{R-t{v}_0}$

$($As, $v=\frac{dx}{dt})$

$\Rightarrow \frac{dx}{dt}=\frac{v_{0}R}{R-t{v}_0}$

$\Rightarrow {\int }_0^{2\pi R}dx={\int }_0^t\frac{v_{0}R}{R-t{v}_0}dt$

$($In one revolution, distance travelled $=2\pi R)$

$\Rightarrow 2\pi R=v_{0}R{\left[\frac{ln(R-t{v}_0)}{-v_0}\right]}_0^t$

$\Rightarrow 2\pi =-1\left[ln\right(R-t{v}_0)-ln(R-0\times v_0\left)\right]$

$\Rightarrow 2\pi =lnR-ln(R-t{v}_0)$

$\Rightarrow 2\pi =ln\frac{R}{R-t{v}_0}$

$\Rightarrow e^{2\pi }=\frac{R}{R-t{v}_0}$

$\Rightarrow R-t{v}_0=R{e}^{-2\pi }$

$\Rightarrow t{v}_0=R(1-e^{-2\pi })$

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

Hence, option $\left(c\right)$ is correct answer.