Gyroscope precession and Euler equations I've been trying for so long to solve this problem, but the solution I have found isn't the one I expected. Basically, I have to solve Euler's equations for a gyroscope with a weigth at a distance d. Given that the moment of inertia for the axis 1 and 2 is the same ($I_{1} = I_{2} = I$):
$$
\dot{\omega}_{1}I - (I-I_{3}) \omega_{2}\omega_{3} = 0
$$
$$
\dot{\omega}_{2}I + (I-I_{3}) \omega_{3}\omega_{1} = mgd
$$
$$
\dot{\omega}_{3}I = 0
$$
It is trivial that $\omega_3$ is constant, so we have that:
$$
(\dot{\omega_2} + i\dot{\omega_1}) - \Omega i (\omega_2 + i \omega_1) = \alpha
$$
$$
\dot{u} - \Omega i u = \alpha
$$
where $\Omega = \frac{I-I_3}{I}\omega_3$ and $\alpha = mgd/I$. If $\vec{\omega}_0 = (0,0, \omega_3)$ is the initial angular velocity, the solution for this equation is:
$$
\omega_2 = \frac{\alpha}{\Omega} \sin(\Omega t)
$$
$$
\omega_1 = \frac{\alpha}{\Omega} (1-\cos (\Omega t))
$$
Now comes what I don't fully understand: I get that this velocities are relative  to the body-fixed axes and, to understand them, I have to find the Eulerian angular velocities (I mean, the velocities relative to the Eulerian angles). When I do so, I find that I have to express the equations like this:
$$
\omega_1 = \dot{\phi} \sin{\theta} \sin{\psi} + \dot{\theta} \cos{\psi}
$$
$$
\omega_2 = \dot{\phi} \sin{\theta} \cos{\psi} - \dot{\theta} \sin{\psi}
$$
$$
\omega_3 = \dot{\phi}\cos{\theta} + \dot{\psi}
$$
If I suppose that $\theta = \pi /2$ and $\dot{\theta} = 0$ (from what I have seen there is no nutation) I get that $\omega_3 = \dot{\psi}$, the result I expected, but when I apply the same conditions to the other two equations I get:
$$
\dot{\phi} = \sqrt{\omega_1^2+ \omega_2} = \frac{\alpha \sqrt{2}}{\Omega} \sqrt{1-\cos (\Omega t)}
$$
I'm pretty sure that that answer is wrong: from what I have seen, the gyroscopic precession velocity is constant and it never stops. I was wondering if you could help me to find where I'm wrong and explain it to me. Thank you so much!
 A: This MIT description is the best I've ever seen for gyroscopic derivations: Lect. 30. Lessons 29 and 31 are also important to the solutions to such problems.
A: first solve the equations $~\omega_1=\ldots\,,\omega_2=\ldots\,,\omega_3=\ldots~$ to obtain $~\dot\varphi\,,\dot\vartheta\,,\dot{\psi}$
$$\left[ \begin {array}{c} \dot\varphi \\ \dot\vartheta 
\\ \dot\psi\end {array} \right] =
 \left[ \begin {array}{c} {\frac {\cos \left( \psi \right) \omega_{{2}
}+\omega_{{1}}\sin \left( \psi \right) }{\sin \left( \vartheta 
 \right) }}\\ -\sin \left( \psi \right) \omega_{{2}}
+\omega_{{1}}\cos \left( \psi \right) \\ {\frac {
\omega_{{3}}\sin \left( \vartheta  \right) -\cos \left( \vartheta 
 \right) \cos \left( \psi \right) \omega_{{2}}-\cos \left( \vartheta 
 \right) \omega_{{1}}\sin \left( \psi \right) }{\sin \left( \vartheta 
 \right) }}\end {array} \right] 
$$
substitute $\vartheta=\pi/2~$ and the solution for $~\omega_1=\omega_1(t)\,,\omega_2=\omega_2(t)\,,\omega_3=\text{const.}~$ and integrate you get the solution $~,\varphi(t)\,,\vartheta(t)\,,\psi(t)$
