Answer
$F_2 = 0.140~N$
Work Step by Step
We can find the angular acceleration:
$\omega = \omega_0+\alpha~t$
$\alpha = \frac{\omega - \omega_0}{t}$
$\alpha = \frac{250~rad/s - 0}{1.25~s}$
$\alpha = 200~rad/s^2$
We can write a general expression for the torque about the center:
$\tau = r\times F = r~F~sin~\theta$
$r$ 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 $F_2$:
$\tau = I~\alpha$
$R~F_2~sin~90^{\circ}-R~F_1~sin~90^{\circ} = \frac{1}{2}MR^2~\alpha$
$F_2-F_1 = \frac{1}{2}MR~\alpha$
$F_2 = \frac{1}{2}MR~\alpha+F_1$
$F_2 = \frac{1}{2}(0.0200~kg)(0.0200~m)~(200~rad/s^2)+(0.100~N)$
$F_2 = 0.140~N$
