Example 18 Given $\angle A O B=\theta\left(\theta\right.$ is a constant and $\left.0<\theta<\frac{\pi}{2}\right)$, moving points $P, Q$ are on rays $O A, O B$ respectively, such that the area of $\triangle O P Q$ is always 36. Let the centroid of $\triangle O P Q$ be $G$, and point $M$ is on ray $O G$ and satisfies $|O M|=$ $\frac{3}{2}|O G|$.
(1) Find the minimum value of $|O G|$.
(2) Find the trajectory of the moving point $M$.
Analysis: Since $O A, O B$ are symmetric about the bisector of $\angle A O B$, to simplify the calculation and maintain the symmetry of the trajectory equation, we establish a coordinate system with $O$ as the origin and the bisector of $\angle A O B$ as the $x$-axis.

Solution: Establish a Cartesian coordinate system with $O$ as the origin and the bisector of $\angle A O B$ as the $x$-axis. Then we can set $P\left(a \cos \frac{\theta}{2}, a \sin \frac{\theta}{2}\right), Q\left(b \cos \frac{\theta}{2}, -b \sin \frac{\theta}{2}\right)$, where $|O P|=a, |O Q|=b$. Thus, the coordinates of the centroid $G$ of $\triangle A O B$ are
$$
\begin{array}{l}
x_{G}=\frac{1}{3}\left(a \cos \frac{\theta}{2}+b \cos \frac{\theta}{2}+0\right)=\frac{1}{3}(a+b) \cos \frac{\theta}{2}, \\
y_{G}=\frac{1}{3}\left(a \sin \frac{\theta}{2}-b \sin \frac{\theta}{2}+0\right)=\frac{1}{3}(a-b) \sin \frac{\theta}{2}, \\
\begin{aligned}
|O G|^{2} & =x_{G}^{2}+y_{G}^{2}=\frac{1}{9}\left(a^{2}+b^{2}\right)+\frac{2}{9} a b\left(\cos ^{2} \frac{\theta}{2}-\sin ^{2} \frac{\theta}{2}\right) \\
& =\frac{1}{9}\left(a^{2}+b^{2}\right)+\frac{2}{9} a b \cos \theta \geqslant \frac{1}{9} \cdot 2 a b+\frac{2}{9} a b \cos \theta \\
& =\frac{4}{9} a b \cos ^{2} \frac{\theta}{2} .
\end{aligned}
\end{array}
$$

Given that $S_{\triangle P Q}=\frac{1}{2} a b \sin \theta=36$, i.e., $a b=\frac{72}{\sin \theta}$, we have
$$
|O G| \geqslant \sqrt{\frac{4}{9} \cdot \frac{72}{\sin \theta} \cdot \cos ^{2} \frac{\theta}{2}}=4 \sqrt{\cot \frac{\theta}{2}},
$$

with equality holding if and only if $a=b=\sqrt{\frac{72}{\sin \theta}}$. Therefore, the minimum value of $|O G|$ is $4 \sqrt{\cot \frac{\theta}{2}}$.
(2) Let the coordinates of $M$ be $M(x, y)$. Then, from $|O M|=\frac{3}{2}|O G|$, we get
$$
\begin{array}{l}
x=\frac{3}{2} x_{G}=\frac{1}{2}(a+b) \cos \frac{\theta}{2}>0, \\
y=\frac{3}{2} y_{G}=\frac{1}{2}(a-b) \sin \frac{\theta}{2} .
\end{array}
$$

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 $a b=\frac{72}{\sin \theta}$, we get $\frac{x^{2}}{\cos ^{2} \frac{\theta}{2}}-\frac{y^{2}}{\sin ^{2} \frac{\theta}{2}}=\frac{72}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$,
i.e., $\frac{x^{2}}{36 \cot \frac{\theta}{2}}-\frac{y^{2}}{36 \tan \frac{\theta}{2}}=1(x>0)$.
Hence, the trajectory of point $M$ is the right branch of the hyperbola $\frac{x^{2}}{36 \cot \frac{\theta}{2}}-\frac{y^{2}}{36 \tan \frac{\theta}{2}}=1$.