Given data

• Radius of circle: $r=50\text{ cm}=0.5\text{ m}$
• Initial speed: $v_0=4\text{ m/s}$
• At every instant, normal acceleration equals tangential acceleration.

Write the two accelerations

• Normal (centripetal) acceleration: $a_n=\frac{v^2}{r}$
• Tangential acceleration: $a_t=\frac{dv}{dt}$

Given $a_n=a_t$, so $\frac{dv}{dt}=\frac{v^2}{r}$

Relate speed with angular displacement Since motion is on a circle, $v=r\frac{d\theta}{dt}$ Hence, $\frac{d\theta}{dt}=\frac{v}{r}$

Now, $\frac{dv}{dt}=\frac{dv}{d\theta}\cdot\frac{d\theta}{dt}$ so $\frac{dv}{d\theta}\cdot\frac{v}{r}=\frac{v^2}{r}$

Multiplying by $\frac{r}{v}$: $\frac{dv}{d\theta}=v$

Solve for $v(\theta)$ $\frac{dv}{v}=d\theta$ Integrating, $\ln v=\theta+C$ $v=Ce^{\theta}$

At $\theta=0$, $v=4$, so $C=4$. Therefore, $v=4e^{\theta}$

Find time for first revolution One revolution means $\theta:0\to 2\pi$.

Using $dt=\frac{r\,d\theta}{v}$ we get

Compare with the given form Given, $t=\frac{1}{\alpha}\left(1-e^{-2\pi}\right)$ Comparing, $\frac{1}{\alpha}=\frac{1}{8}$ so $\alpha=8$

Final answer $\boxed{8}$