= Solution
Set $s=\sin\theta$. The maximum height becomes the concave <quadratic function>
$$
y_{\max}=rs+\frac{r^2\omega^2}{2g}(1-s^2),\qquad \frac{dy_{\max}}{ds}=r-\frac{r^2\omega^2}{g}s.
$$
Under the stated high-<angular frequency> assumption, $r\omega^2>g$, its interior maximum is
$$
\boxed{\sin\theta_* =\frac g{r\omega^2},\qquad (x_*,y_*)=\left(-r\sqrt{1-\frac{g^2}{r^2\omega^4}},\frac g{\omega^2}\right).}
$$
Thus the optimum release lies on the upper-left quarter of the wheel for the printed clockwise rotation. The largest attained height is
$$
\boxed{Y(\omega)=\frac{r^2\omega^2}{2g}+\frac g{2\omega^2}.}
$$
If extended to $r\omega^2\leq g$, the maximum over release positions is the limiting height $r$ at the top of the rim. The interior solution is the relevant branch for reaching a ceiling above the wheel.
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