A particle is moving in a circle of radius $50 \mathrm{~cm}$ in such a way that at any instant the normal and tangential components of it's acceleration are equal. If its speed at $\mathrm{t}=0$ is $4 \mathrm{~m} / \mathrm{s}$, the time taken to complete the first revolution will be $\frac{1}{\alpha}\left[1-e^{-2 \pi}\right] \mathrm{s}$, where $\alpha=$ _________.

<p>$$\begin{aligned} & \left|\vec{a}_c\right|=\left|\vec{a}_t\right| \\ & \frac{v^2}{r}=\frac{d v}{d t} \\ & \Rightarrow \int_\limits4^v \frac{d v}{v^2}=\int_\limits0^t \frac{d t}{r} \\ & \Rightarrow\left[\frac{-1}{v}\right]_4^v=\frac{t}{r} \\ & \Rightarrow \frac{-1}{v}+\frac{1}{4}=2 t \end{aligned}$$</p> <p>$$\begin{aligned} & \Rightarrow \mathrm{v}=\frac{4}{1-8 \mathrm{t}}=\frac{\mathrm{ds}}{\mathrm{dt}} \\ & 4 \int_0^{\mathrm{t}} \frac{\mathrm{dt}}{1-8 \mathrm{t}}=\int_0^{\mathrm{s}} \mathrm{ds} \\ & (\mathrm{r}=0.5 \mathrm{~m} \\ & \mathrm{s}=2 \pi \mathrm{r}=\pi) \\ & 4 \times \frac{[\ell \mathrm{n}(1-8 \mathrm{t})]_0^{\mathrm{t}}}{-8}=\pi \\ & \ell \mathrm{n}(1-8 \mathrm{t})=-2 \pi \\ & 1-8 \mathrm{t}=\mathrm{e}^{-2 \pi} \\ & \mathrm{t}=\left(1-\mathrm{e}^{-2 \pi}\right) \frac{1}{8} \mathrm{~s} \end{aligned}$$</p> <p>So, $\alpha=8$</p>