A uniform rod of mass m and length l can rotate in a vertical plane...

Taking torque equation about hinge H,
$$\begin{gathered} {\tau }_{net}=I\alpha \\ \Rightarrow mgcos{37}^{\circ }\times \frac{l}{2}=\frac{m{l}^2}{3}\alpha \\ \Rightarrow \alpha =\frac{6g}{5l} \end{gathered}$$
[torque due to hinge reaction is zero]
Let the hinge force components be N1 and N2
$a_t=\frac{l}{2}\alpha =\frac{3g}{5}$
Force equation perpendicular to the rod,
$$\begin{gathered} mg\cos {37}^{\circ }-N_1=m{a}_t \\ \Rightarrow mg\times \frac{4}{5}-N_1=m\times \frac{3g}{5} \\ \Rightarrow N_1=\frac{mg}{5} \end{gathered}$$
Angular velocity (ω)=0 at the time of release and net force along the rod is zero since there is no acceleration towards the center,
$N_2=mg\sin {37}^{\circ }=\frac{3mg}{5}$
Hinge force (N) is
$N=\sqrt{N_1^2+N_2^2}=\sqrt{(\frac{mg}{5}{)}^2+(\frac{3mg}{5}{)}^2}=\frac{mg\sqrt{10}}{5}$