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!
