Very well! So you got the correct result that the reaction force is mg/4 at the hinge upward.
You would get the same from the x,y components of acceleration. I like this method, as I usually get confused with the radial and tangential components.
The coordinates of the COM of a homogeneous rod with the ends at (x1,y1) and (x2,y2) are Xcom=(x1+x2)/2, Ycom=(y1+y2)/2. (x1,y1)=(0,0), and (x2,y2)=(x,y) here, so Xcom=x/2, Ycom=y/2.
You derived the x component of acceleration already, the same has to be done with the y component.
y=Lsin(θ) y'=Lcos(θ) θ',
y''= L(-sin(θ)θ'2+cos(θ)θ")=L(-sinθω2+cos(θ)ω')
When loosing contact, y"=L(-1/2 ω2+√3/2 ω')
From the condition that the x component of acceleration is zero, you have got that
ω'=-√3 ω2 and ω2=3g/4L, therefore y"=-3g/2,
and Ycom"=-3g/4.
ehild