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 its acceleration are equal. If its speed at \(t=0\) is \(4 \mathrm{~m} / \mathrm{s}\), the time taken to complete the first revolution will be \(\frac{1}{\alpha}\left[1-\mathrm{e}^{-2 \pi}\right] \mathrm{s}\), where \(\alpha=\) _______.

\( \frac{\mathrm{v}^2}{\mathrm{r}}=\frac{\mathrm{dv}}{\mathrm{dt}} \)

\( \Rightarrow \int_4^{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{v}^2}=\int_0^{\mathrm{t}} \frac{\mathrm{dt}}{\mathrm{r}} \)

\( \Rightarrow\left[\frac{-1}{\mathrm{v}}\right]_4^{\mathrm{v}}=\frac{\mathrm{t}}{\mathrm{r}} \)

\( \Rightarrow \frac{-1}{\mathrm{v}}+\frac{1}{4}=2 \mathrm{t}\)

\( \Rightarrow \mathrm{v}=\frac{4}{1-8 \mathrm{t}}=\frac{\mathrm{ds}}{\mathrm{dt}} \)

\( 4 \int_0^{\mathrm{t}} \frac{\mathrm{dt}}{1-8 \mathrm{t}}=\int_0^5 \mathrm{ds} \)

\( (\mathrm{r}=0.5 \mathrm{~m} \)

\( \mathrm{s}=2 \pi \mathrm{r}=\pi) \)

\( 4 \times \frac{[\ln (1-8 \mathrm{t})]_0^{\mathrm{t}}}{-8}=\pi \)

\( \ln (1-8 \mathrm{t})=-2 \pi \)

\( 1-8 \mathrm{t}=\mathrm{e}^{-2 \pi} \)

\( \mathrm{t}=\left(1-\mathrm{e}^{-2 \pi}\right) \frac{1}{8} \mathrm{~s}\)

So, \(\alpha=8\)