My attempt:

I thought of breaking up the problem into two cases and the combining them at the end.

case1: Pretend no rotation:

With no rotation the spool has forces Tension acting on it. \(\displaystyle T = Ma\)

The mass attached to the string has forces Tension and gravity. solved for \(\displaystyle T' = mg - ma\)

Since the acceleration for both we can get to \(\displaystyle a = (mg)/(M+m)\)

So, we can get a final velocity of \(\displaystyle v = (2*(mg)/(M+m)*)^1/2\) where I started with \(\displaystyle vf^2 = vi^2+2al\), l being the displacement of the spool on the table.

Case2: Pretend no translation:

With no translation, I believe then the Tension and Torque are equal to eachother. Then we can get \(\displaystyle alpha = (torque/I)\). and we can get \(\displaystyle theta = L/(pi*d)\), (note: used 'L' instead of lowercase 'l' to see clear difference between the I, moment of inertia.).

What I end up using is the angular kinematics to get \(\displaystyle omegaf= (2*(torque/I)*(L/p(i*d)))^1/2\)

So this is my work...am I on the rigth track or completely wrong? And how can I relate these two to get a uniform equation?