A particle is moving in a circle of radius $50\ \text{cm}$ 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 $4\ \text{m s}^{-1}$, the time taken to complete the first revolution will be $\dfrac1\alpha\left[1 - e^{-2\pi}\right]\ \text{s}$, where $\alpha =$ ______.
Numerical value type. Enter your answer.
Answer: 8
$a_t = a_n$ gives $v\dfrac{dv}{ds} = \dfrac{v^2}{R}$, so $\dfrac{dv}{v} = \dfrac{ds}{R}$ and
$$v = v_0e^{s/R}$$
Time for one revolution ($s = 2\pi R$):
$$t = \int_0^{2\pi R}\frac{ds}{v_0e^{s/R}} = \frac{R}{v_0}\left(1 - e^{-2\pi}\right) = \frac{0.5}{4}\left(1 - e^{-2\pi}\right) = \frac18\left(1 - e^{-2\pi}\right)$$
So $\alpha = 8$.
Solution by Sreeraj P, M.Sc Physics