Answer
$\omega = 500~rad/s$
Work Step by Step
We can find the angular speed:
$\int_{0}^{3.0}~r~F(t)~dt = I~\omega$
$r~\int_{0}^{3.0}~(0.50t+0.30t^2)~dt = I~\omega$
$(r)~(0.25t^2+0.10t^3)\Big \vert_{0}^{3.0} = I~\omega$
$\omega = \frac{(0.10~m)~[(0.25)(3.0)^2+(0.10)(3.0)^3]}{1.0\times 10^{-3}~kg~m^2}$
$\omega = 500~rad/s$
