Rod released from horizontal position

Given: $m = 6$ kg, $L = 1.5$ m, hinge at A, released from rest in the horizontal position.

   A (hinge) o==========o B
             |-- G --|
   W = mg  v  (at L/2)

At the instant of release the rod is at rest, so $\omega = 0$ and the centripetal term $\omega^2 (L/2)$ vanishes. The mass centre G has only a tangential (vertical) acceleration.

Moment of inertia about A

$$I_A = \frac{mL^2}{3} = \frac{6(1.5)^2}{3} = 4.50\ \text{kg·m}^2$$

Angular acceleration

Moments about the hinge (only the weight has a moment; the hinge reaction passes through A):

$$mg\,\frac{L}{2} = I_A\alpha \;\Rightarrow\; \alpha = \frac{44.14}{4.50} = 9.81\ \text{rad/s}^2\ \left(=\frac{3g}{2L}\right)$$

Acceleration of G

$$a_G = \alpha\,\frac{L}{2} = 9.81 \times 0.75 = 7.357\ \text{m/s}^2\ \text{(downward)}$$

Hinge reaction

Horizontal: $A_x = 0$ (there is no horizontal acceleration or applied horizontal force).

Vertical (downward positive): $mg - A_y = m\,a_G$

$$A_y = 6(9.81 - 7.357) = 14.72\ \text{N (upward)}$$

This is one quarter of the weight ($mg/4 = 58.86/4 = 14.72$ N), a known result for a uniform rod released from the horizontal position.

Answer: α = 9.81 rad/s²; a_G = 7.36 m/s² downward; hinge reaction = 14.7 N upward (A_x = 0).