Normal force on bar after letting go of it

Question

I want to understand the following question, both from a theoretical and practical standpoint. Given the following image:

Originally, the system is in equilibrium thanks to the force A. I am asked what the normal force the table exerts on the bar would be right after the force A stopped being applied. I thought it was a relatively simple question, as using newtons second law: $$N+A-mg=0$$ $$NL/2-AL/2=0$$ Which gives that both A and L should have a magnitude of mg/2. Thus I thought that right after A stopped being applied N wouldn't change, and remain $mg/2$ But the correct answer is $N=mg/4$, which is obtained through a complex process involving polar coordinates. Can this answer be obtained in an intuitive way? Why is my initial approach wrong?

Answer 1

When A becomes zero the system is no longer in equilibrium, it has acceleration.

Vertical force. N-mg=ma_c Clockwise Moment about C. NL/2 =I alpha

where I is moment of inertia about C.

The third equation is circular motion of C about A

-(L/2)alpha = a_c.

Answer 2

Your equations are correct until rod is in equilibrium. But, after force A is no longer applied, the rod will rotate in a circular path about B for a very short period.
$mg - N' = ma$ (a is the acceleration the center of gravity of rod)
$mgL/2 = mL²/3*α$ (Torque equation about B)
$α = 3g/2L$
$a = α*L/2$ (the acceleration of the center of gravity of rod is due to rotation of rod about B)
On solving, we get $N' = mg/4$