Rotation

Translation and rotation together yield the following velocity profile, with $\theta$ as the angle with the positive horizontal (counter-clockwise angle convention):

$\large{v_x = v_0 + v_0 sin\theta \hspace{1cm} v_y = -v_0 cos\theta}$

The differential mass is:

$\large{dm = \frac{d\theta}{2\pi}m}$

The differential kinetic energy is:

$\large{dE = \frac{1}{2} dm \, v^2 = \frac{1}{2} \frac{d\theta}{2\pi}m (v_x^2 + v_y^2) = \frac{1}{2} \frac{d\theta}{2\pi}m (2v_0^2)(1 + sin\theta) = \frac{mv_0^2}{2\pi} (1 + sin\theta)d\theta}$

Integrating $dE$ from $\theta = \pi$ to $\theta = 2\pi$ gives:

$\large{E = \frac{mv_0^2}{2\pi} \int_\pi^{2\pi} (1 + sin\theta)d\theta = \frac{mv_0^2}{2\pi} (\pi -2) = mv_0^2(\frac{1}{2} - \frac{1}{\pi}) = mv_0^2(\frac{1}{\alpha} - \frac{\beta}{\pi}) \\ \alpha + \beta = 2 + 1 = \boxed{3} }$