Answer
$\omega = 3.1~rad/s$
Work Step by Step
We can find the height $h$ the center of mass falls from the initial position to the horizontal position:
$\frac{h}{L/2} = sin~\theta$
$h = \frac{L~sin~\theta}{2}$
$h = \frac{(2.0~m)~sin~40^{\circ}}{2}$
$h = 0.6428~m$
We can use conservation of energy to find the angular speed:
$\frac{1}{2}~I~\omega^2 = mgh$
$(\frac{1}{2})~(\frac{1}{3}mL^2)~\omega^2 = mgh$
$\omega^2 = \frac{6gh}{L^2}$
$\omega = \frac{\sqrt{6gh}}{L}$
$\omega = \frac{\sqrt{(6)(9.8~m/s^2)(0.6428~m)}}{2.0~m}$
$\omega = 3.1~rad/s$
