Answer
$$
\dot{\phi}=2.28 \mathrm{rad} / \mathrm{s}^2
$$
Work Step by Step
Here, $\quad \dot{\psi}=\omega_t=15 \mathrm{rad} / \mathrm{s}, \quad I=m k_y^2=0.25\left(0.13^2\right)=0.004225 \mathrm{~kg} \cdot \mathrm{m}^2$ $I_z=m k_z^2=0.25\left(0.042^2\right)=0.000441 \mathrm{~kg} \cdot \mathrm{m}^2$.
$$
\begin{aligned}
\dot{\psi}=\frac{I-I_z}{I_z} \dot{\phi} \cos \theta ; \quad 15 & =\left(\frac{0.004225-0.000441}{0.000441}\right) \dot{\phi} \cos 40^{\circ} \\
\dot{\phi} & =2.282 \mathrm{rad} / \mathrm{s}^2=2.28 \mathrm{rad} / \mathrm{s}^2
\end{aligned}
$$
