Vector of angular acceleration  $\beta =\frac{d\omega }{dt}=\frac{d(\omega \text{'}+{\omega }_0)}{dt}=\frac{d{\omega }_0}{dt}(as\omega \text{'}=cons\tan t)$
 If any vector a (say) rotates with angular velocity  $\omega$  keeping its value constant then  $da/dl=\omega \times a$. The vector ${\omega }_0$  keeping its magnitude constant rotates about the $OO\text{'}$ axis with the angular velocity $\omega \text{'}$ .  So $d{\omega }_2/dt=(\omega \text{'}\times {\omega }_0)$  Hence $\beta =\omega \text{'}\times {\omega }_0$  The magnitude of the vector  $\beta$is equal to $\beta =\omega \text{'}{\omega }_0$  (as $(\omega \text{'}\perp {\omega }_0)$  

So, $\beta =\frac{\upsilon }{R\cot \alpha }\left(\frac{\upsilon }{R}\right)=\frac{{\upsilon }^2}{R^2}\tan \alpha =2.3rad/s^2$ ,  on putting the values of $\upsilon ,R$ and $\alpha$