Answer
$\alpha = 9.72~rad/s^2$
Work Step by Step
We can write a general expression for the torque about the pivot:
$\tau = r\times F = r~F~sin~\theta$
$r$ is the displacement vector from the pivot to the point where the force is applied
$F$ is the force vector
$\theta$ is the angle between these two vectors
We can find the magnitude of the angular acceleration $\alpha$:
$\tau = I~\alpha$
$R~F_1~sin~90^{\circ}-R~F_2~sin~90^{\circ}-r~F_3~sin~90^{\circ}+R~F_4~sin~0^{\circ} = \frac{1}{2}MR^2~\alpha$
$\alpha = \frac{(2)[R~F_1-R~F_2-r~F_3]}{MR^2}$
$\alpha = \frac{(2)[(0.12~m)~(6.0~N)-(0.12~m)~(4.0~N)-(0.050~m)~(2.0~N)]}{(2.0~kg)(0.12~m)^2}$
$\alpha = 9.72~rad/s^2$
