Angular momentum: rotating rod and a ball

[Karol]

Homework Statement

A rod of length l swivels around an axis, denoted O in the drawing, located on the floor.
The initial position is β=45°.
A ball B is left to fall at the same time the rod is left to rotate.
Who will arrive, first, to the floor: the ball or the edge A of the rod

Homework Equations

T(torque)=Iα
Moment of inertia of the rod around it's edge:
$$I_o=\frac{1}{3}ml^2$$

The Attempt at a Solution

The torque at different angles α:
$$\frac{l}{2}mg\cdot \cos \beta=\frac{1}{3}ml^2 \cdot \alpha$$
And the geometric relation:
$$l \cdot \alpha=a_A$$
Where a_A is the acceleration of the edge of the rod.
$$\rightarrow a_A=\frac{3lg\cdot \cos \beta}{2}$$
I integrate a_A in the range of β=45° to β=0° and get a number for the velocity of the edge of the rod A at the floor, but i cannot integrate further this velocity in order to make it equal to the length of the arc and extract the time.

 

[TSny]

$$\rightarrow a_A=\frac{3lg\cdot \cos \beta}{2}$$

Note that the units don't match on the left and right sides.

Maybe you won't need to find the value of the time to answer the question. If we race such that your acceleration is always greater than mine, who will get to the finish line first?