How Can Angular Velocity Be Expressed as a Function of Theta for a Falling Rod?

[moo5003]

Question:
"A uniform rod of mass m and length 2a stands vertically on a rough horizontal floor and is allowed to fall. Assuming that slipping has no occured, write the angular velocity of the rod as a function of the angle Theta the rod makes with the vertical."

WORK DONE:
Ic = (ml^2)/3
Torque = mgd
d = asin(theta)
Torque = angular accel * I

mgasin(theta) = ang accel * (ml^2)/3
l = 2a

gasin(theta) = ang accel * (m(2a)^2)/3
gsin(theta) = 4/3 * ang acell * a

Ang Accel = 3/(4a) * gsin(theta)

MAIN QUESTION: How do I substitute Angular Acceleration such that I can find an equation that solves angular velocity?

 

[Astronuc]

Angular acceleration \alpha\,=\,\dot\omega\,=\,\ddot\theta

Angular velocity \omega\,=\,\dot\theta

 

[moo5003]

Angular acceleration \alpha\,=\,\dot\omega\,=\,\ddot\theta
Angular velocity \omega\,=\,\dot\theta

I considered using calc to solve for this. But the integral of Angular Acceleration is a function of time. I'm unsure how I can incorporate that together and produce an equation as a function of Theta.

W = Integral [ 3/4 * 1/a * gsin(theta) ] dt

Would I just slap on a variable T and give that as my answer (There is no initial angular velocity right)? Or can I somehow subtitute T as a function of Theta?