How can the motion of point A be determined in this system?

[ehild]

The answers become clear when we write the equations for the motion of the CM and point A in general. In this case, the string encloses the angle φ with the vertical while the beam encloses angle θ. Let be the length of the beam 2L, and r the length of the string.
x_A=rsin(φ), y_A=rcos(φ)
$\ddot x_A=−rsin(φ){\dot φ}^2+rcos(φ)\ddot φ$, $\ddot y_A=−rcos(φ){\dot φ}^2−rsin(φ)\ddot φ$
At t=0+, φ=0, $\dot φ=0$, so $\ddot x_A=r\ddot φ$, $\ddot y_A=0$, point A accelerates in horizontal direction.

$X_{CM}=L\sin(θ)+rsin(φ)$, $Y_{CM}=L\cos(θ)+r\cos(φ)$
$\ddot X_{CM}=-L\sin(θ){\dot θ}^2+L\cos(θ)\ddotθ-r\sin(φ){\dot φ}^2+r\cos(φ)\ddotφ$
$\ddot Y_{CM}=-L\cos(θ){\dot θ}^2-L\sin(θ)\ddot θ-r\cos(φ){\dot φ}^2-r\sin(φ)\ddotφ$
At t=0, $\ddot X_{CM}=L\cos(θ)\ddot θ+r\ddot φ$, $\ddot Y_{CM}=-L\cos(θ)\ddot θ$

For the translation of the CM:
$m\ddot X_{CM}=-T\sin (φ)$
$m\ddot Y_{CM}=-T\cos (φ)+mg$
The torque equation with respect to the CM:
$TLsin(θ-φ)=-\frac{1}{3}mL^2 \ddot θ$.
At t=0:
$\ddot X_{CM}=0$, $\ddot Y_{CM}=-T+mg$
$TLsin(θ)=-\frac{1}{3}mL^2 \ddot θ$.

T can be eliminated, $\ddot θ$and $\ddot φ$ determined for t=0.