Is the angular acceleration the same when calculated from the center of gravity and from a non-rotating point?

Question

I got this problem in Greiner's Classical Mechanics. The problem's statement is here.

A bar of length $2l$ and mass $M$ is fixed at point A, so that it can rotate only in the vertical plane. The external force $\vec F$ acts on the center of gravity. Calculate the reaction force $\vec F_r$ at point A!

A sketch of the given solution is the following : First calculate the torque around A:$$\tau_A=F.l\\ \therefore \dot{\omega}=\frac {\tau_A} {I_A}=\frac{3F}{4Ml}$$ Now calculate the torque again but this time around $S$, the center of gravity:$$\tau_S=F_r.l \\ \therefore \dot{\omega}=\frac{\tau_S}{I_S}=\frac{3F_r}{Ml}$$ From these two equations of angular acceleration they conclude that :$$F_r=\frac 1 4 F$$
Now my questions are :

1. Why there is such a reaction force?
2. Why did they take the two expressions of the angular acceleration to be the same? (In the book they said that no matter what point we choose to calculate torque, these two expressions must be equal. I don't get it.)
3. Why did they neglect gravity?

Thank you.

Answer 1

(1) If there was no force from a horizontal axle at point A, the rod wold move down without rotation. (2) If a solid object males one revolution, every point in the object will have revolved about every other point. (3) Since the indicated force is pointed down, it may include gravity.

Answer 2

1. If the clamp was not there, the rod would fall straight down without rotating. So, to prevent that, the clamp must be exerting a force on the rod, which they have termed $F_r$
2. It is the property of rigid bodies that angular velocity and acceleration of every point in the rigid body about every other point is the same. If a point in the body rotates around the COM in $t$ time, then from the frame of reference of that point, the COM will also rotate around it in $t$ time.
3. Gravity is not mentioned, however, $F$ seems to be analogous to gravity in this case.