Answer
$$
\omega_s=3.63\left(10^3\right) \mathrm{rad} / \mathrm{s}
$$
Work Step by Step
$$
\begin{aligned}
& \omega_p=0.5 \mathrm{rad} / \mathrm{s} \\
& \Sigma M_x=-I \dot{\phi}^2 \sin \theta \cos \theta+I_z \dot{\phi} \sin \theta(\dot{\phi} \cos \theta+\dot{\psi}) \\
& 0.090(9.81)(0.06) \sin 45^{\circ}=-0.090(0.035)^2(0.5)^2(0.7071)^2 \\
& +0.090(0.018)^2(0.5)(0.7071)[0.5(0.7071)+\dot{\psi}] \\
& \omega_s=\psi=3.63\left(10^3\right) \mathrm{rad} / \mathrm{s} \\
&
\end{aligned}
$$
