A uniform thin rod of mass m and length L is hinged at its lower end and released from rest from the vertical position. Considering the instant when…

Energy conservation: the centre of mass falls by $L/2$, so $mgL/2=\tfrac12\left(\tfrac13mL^2\right)\omega^2$ and $\omega=\sqrt{3g/L}$; A is true. The torque about the hinge is $mgL/2$ when horizontal, so $\alpha=\dfrac{mgL/2}{mL^2/3}=\dfrac{3g}{2L}$. The centre of mass has tangential (vertical) acceleration $\alpha L/2=3g/4$ and centripetal (horizontal) acceleration $\omega^2L/2=3g/2$. Vertically $mg-N_y=m(3g/4)$, so $N_y=mg/4$; B is true. Horizontally $N_x=m\omega^2L/2=3mg/2$, not $mg/2$; C is false. The free-end speed is $\omega L=\sqrt{3gL}$; D is true. Thus A, B and D are correct. Note that the hinge force is not simply $mg$: at the horizontal position the large centripetal requirement makes the horizontal component the larger one.