A bar subject to a rolling disk which is released on an inclined plane

[Telemachus]

Well, it's my second post about rigid body. I originally posted this on introductory physics, but as nobody answered this, or the previous topic, I've decided to post this here.

Homework Statement

I have this other exercise rigid in the plane, with which I am having problems.

The rod of mass m and length l, is released based on the vertical position of rest with the small roller end A resting on the slope. Determine the initial acceleration A.

(neglect friction and mass of the roller A)

The answer given by the book is a_A=\displaystyle\frac{g\sin \theta}{1-\frac{3}{4}\cos^2\theta}

Homework Equations

I try to raise the moment equation, and Newton. But not me, not that I'm doing wrong. For this consider that the bar rotates about its center of mass.
I_{cm}=\displaystyle\frac{mL^2}{12}
N-mg\cos\theta=0
mg\sin\theta=ma_{cm}
I_{cm}\alpha=\displaystyle\frac{L}{2}mg\sin\theta\cos\theta

The Attempt at a Solution

\alpha=\displaystyle\frac{6g\sin\theta\cos\theta}{L}
a_cm=g\sin\theta

Then: a_A=a_{cm}+\displaystyle\frac{L}{2}\alpha=g\sin\theta+3g\cos\theta\sin\theta

Greetings and thanks for posting.

 

[Telemachus]

I've found the answer on the internet. But I still having some doubts about how he makes the reasoning of it.

I don't realize how he makes the reasoning for this:
\displaystyle\sum_{}{M_A}=I\alpha+\displaystyle\sum_{}{}mad
0=\displaystyle\frac{ml^2}{12}\alpha\alpha+m\displaystyle\frac{l}{2}\alpha\displaystyle\frac{l}{2}-ma_A\displaystyle\frac{l}{2}\cos\theta

I realize about the trivial part: \displaystyle\sum_{}{M_A}=0 but I don't know how to get to the other part of the equality, the one concerning to I\alpha+\displaystyle\sum_{}{}mad

Can somebody help me a little bit please?