Answer
$a = \frac{2gb^2}{2b^2+R^2}$
$\alpha =\frac{2gb}{2b^2+R^2}$
$T = \frac{2mgR^2}{2b^2+R^2}$
Work Step by Step
We can set up a torque equation for the yo-yo.
$\tau = I\alpha$
$T~b = (\frac{1}{2}mR^2+\frac{1}{2}mR^2)(\frac{a}{b})$
$T = \frac{mR^2a}{b^2}$
We can use this expression in the force equation for the yo-yo.
$\sum F = (2m)~a$
$2mg - T = 2ma$
$2mg = 2ma+ \frac{mR^2a}{b^2}$
$2gb^2 = 2ab^2+ R^2a$
$a = \frac{2gb^2}{2b^2+R^2}$
We can use the acceleration $a$ to find the angular acceleration.
$\alpha = \frac{a}{b}$
$\alpha =\frac{2gb}{2b^2+R^2}$
We can find the tension in the string.
$T = \frac{mR^2a}{b^2}$
$T = (\frac{mR^2}{b^2})(\frac{2gb^2}{2b^2+R^2})$
$T = \frac{2mgR^2}{2b^2+R^2}$
