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 acceleration are equal. If its speed at $t=0$ is $v_0$, the time taken to complete the first two revolutions is:

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

Answer 1

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

Given that, tangential acceleration = Normal acceleration
$a_t=\frac{v^2}{R}$
$\Rightarrow a_t=\frac{dv}{dt}=\frac{v^2}{R}$ ......(1)
Integrating both sides,
${\int }_{v_0}^v\frac{dv}{v^2}={\int }_0^t\frac{dt}{R}$
$\Rightarrow \frac{1}{v_0}-\frac{1}{v}=\frac{t}{R}$
$\Rightarrow t=R(\frac{1}{v_0}-\frac{1}{v}).....\left(2\right)$
Now
$a_t=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=v\frac{dv}{ds}$
$\Rightarrow \frac{vdv}{ds}=\frac{v^2}{R}$
$\Rightarrow \frac{dv}{ds}=\frac{v}{R}$
$\Rightarrow {\int }_{v_0}^v\frac{dv}{v}={\int }_0^{4\pi R}\frac{ds}{R}$
($4\pi R$ because performing two complete revolutions)
$\Rightarrow \text{ln}\frac{v}{v_0}=\frac{4\pi R}{R}$
$\Rightarrow \text{ln}\frac{v}{v_0}=4\pi$
$\Rightarrow v=v_0{e}^{4\pi }$...........(3)
Substitute $v$ in equation (2)
$t=R(\frac{1}{v_0}-\frac{1}{v_0{e}^{4\pi }})$
$\Rightarrow t=R(\frac{1}{v_0}-\frac{e^{-4\pi }}{v_0})$
$\Rightarrow t=\frac{R}{v_0}(1-e^{-4\pi })$
Hence, option $\left(c\right)$ is correct.