14. Let $\angle A O B=\theta\left(\theta\right.$ be a constant and $\left.0<\theta<\frac{\pi}{2}\right)$, and let moving points $P$ and $Q$ be on rays $O A$ and $O B$ respectively, such that the area of $\triangle P O Q$ is always 36. Let the centroid of $\triangle P O Q$ be $G$, and point $M$ be on ray $O G$ such that
$$
|O M|=\frac{3}{2}|O G| \text {. }
$$
(1) Find the minimum value of $|O G|$;
(2) Find the equation of the locus of the moving point $M$.
14. (1) Taking $O$ as the origin and the bisector of $\angle A O B$ as the $x$-axis, establish a rectangular coordinate system $x O y$. Let
$$
\begin{array}{l}
P\left(a \cos \frac{\theta}{2}, a \sin \frac{\theta}{2}\right), Q\left(b \cos \frac{\theta}{2},-b \sin \frac{\theta}{2}\right), \\
G\left(x_{G}, y_{C}\right).
\end{array}
$$

Then $x_{G}=\frac{1}{3}(a+b) \cos \frac{\theta}{2}$,
$$
y_{G}=\frac{1}{3}(a-b) \sin \frac{\theta}{2}.
$$

Thus, $|O G|^{2}=x_{G}^{2}+y_{G}^{2}$
$$
=\frac{1}{9}\left(a^{2}+b^{2}\right)+\frac{2}{9} a b\left(\cos ^{2} \frac{\theta}{2}-\sin ^{2} \frac{\theta}{2}\right)
$$
$=\frac{1}{9}\left(a^{2}+b^{2}\right)+\frac{2}{9} a b \cos \theta$
$\geqslant \frac{1}{9} \cdot 2 a b+\frac{2}{9} a b \cos \theta=\frac{4}{9} a b \cos ^{2} \frac{\theta}{2}$.
Also, $S_{\triangle O P Q}=\frac{1}{2} a b \sin \theta=36 \Rightarrow a b=\frac{72}{\sin \theta}$.
Thus, $|O G| \geqslant \sqrt{\frac{4}{9} \cdot \frac{72}{\sin \theta} \cdot \cos ^{2} \theta}=4 \sqrt{\cot \frac{\theta}{2}}$,
and the equality holds when $a=b=\sqrt{\frac{72}{\sin \theta}}$.
Therefore, $|O G|_{\min }=4 \sqrt{\cot \frac{\theta}{2}}$.
(2) Let $M(x, y)$.
Then, by $|O M|=\frac{3}{2}|O G|$, we have
$$
\begin{array}{l}
x=\frac{3}{2} x_{G}=\frac{1}{2}(a+b) \cos \frac{\theta}{2}>0, \\
y=\frac{3}{2} y_{C}=\frac{1}{2}(a-b) \sin \frac{\theta}{2}.
\end{array}
$$

Thus, $a=\frac{x}{\cos \frac{\theta}{2}}+\frac{y}{\sin \frac{\theta}{2}}$,
$$
b=\frac{x}{\cos \frac{\theta}{2}}-\frac{y}{\sin \frac{\theta}{2}}.
$$

Substituting $a b=\frac{72}{\sin \theta}$ and simplifying, we get
$$
\frac{x^{2}}{36 \cot \frac{\theta}{2}}-\frac{y^{2}}{36 \tan \frac{\theta}{2}}=1(x>0).
$$

This is the equation of the trajectory of the moving point $M$.