Lagrangian of a precessing wheel: gyroscope in Lagrangian mechanics? Let's consider this classical example:

I would write the Lagrangian as the one of the spherical pendulum plus the rotational energy of the wheel, but I fail to see how the precession motion would arise from that. I feel like the two components should be somehow coupled. I would greatly appreciate some guidance.

Here is my attempt which I believe is wrong, as I fail to see how that would create the precession motion. Taking the wheel as an ideal ring with radius $R$ and mass $m$ and indicating with $r$ the arm length:
$\displaystyle L=L_\text{pend} + L_\text{rot}$
where:
$\displaystyle L_\text{pend}=\frac{1}{2}
mr^2\left(
  \dot{\theta}^2+\sin^2\theta\ \dot{\phi}^2
\right)
+ mgr\cos\theta$
and
$\displaystyle L_\text{rot} = \frac{1}{2}I\dot\alpha^2 = \frac{1}{2}m R^2\dot\alpha^2 $

 A: 
I) choose  coordinate system $(x,y,z)$
II) choose the generalized coordinates $\varphi\,,\psi$
Gyroscope 
III) create the rotation matrix $R$ via the generalized coordinates and in your case also  time depended  $\Omega\,\tau$ . where $\Omega$ is the wheel rotation about the y axis and $\tau$ is the time
$$R=S_x(\varphi)\,S_z(\psi)\,Sy(\Omega\tau)$$
(I choose this sequence of the rotation matrix to avoid singularity ,this is important for numerical simulation )
Where:
$$S_x=\left[ \begin {array}{ccc} 1&0&0\\0&\cos \left( 
\varphi  \right) &-\sin \left( \varphi  \right) \\ 0
&\sin \left( \varphi  \right) &\cos \left( \varphi  \right) 
\end {array} \right] 
$$
$$S_z=\left[ \begin {array}{ccc} \cos \left( \psi \right) &-\sin \left( 
\psi \right) &0\\\sin \left( \psi \right) &\cos
 \left( \psi \right) &0\\ 0&0&1\end {array} \right] 
$$
$$S_y=\left[ \begin {array}{ccc} \cos \left( \Omega\,\tau \right) &0&\sin
 \left( \Omega\,\tau \right) \\ 0&1&0
\\ -\sin \left( \Omega\,\tau \right) &0&\cos \left( 
\Omega\,\tau \right) \end {array} \right] 
$$
IV) obtain the angular velocity vector $\vec{\omega}$ out of the rotation matrix $R$
$$\vec{\omega}=J_R\,\vec{{\dot{q}}}+\vec{\omega}_\tau$$
where
$$J_R= \left[ \begin {array}{cc} \cos \left( \psi \right) \cos \left( \Omega
\,\tau \right) &-\sin \left( \Omega\,\tau \right) 
\\ -\sin \left( \psi \right) &0\\ 
\cos \left( \psi \right) \sin \left( \Omega\,\tau \right) &\cos
 \left( \Omega\,\tau \right) \end {array} \right] 
$$
$$\vec{{\dot{q}}}= \left[ \begin {array}{c} \dot{\varphi} \\ {\it \dot{\psi}}
\end {array} \right] 
$$
$$\vec{\omega}_\tau=\left[ \begin {array}{c} 0\\ \Omega
\\ 0\end {array} \right] 
$$ 
V) Kinetic Energy 
$$T_g=\frac{1}{2}\vec{\omega}^T\,J\,\vec{\omega}$$
where J is the inertia tensor of the wheel with $J_z=J_x$
$$J=\left[ \begin {array}{ccc} J_{{x}}&0&0\\ 0&J_{{y}}&0
\\0&0&J_{{x}}\end {array} \right] 
$$
VI) generalized torques 
the torque about the x axis   is $\tau_\varphi=-m\,g\,a$
thus the generalized torques are:
$$\begin{bmatrix}
  \tau_x \\
  \tau_z \\
\end{bmatrix}=J_R^T\,\begin{bmatrix}
  \tau_\varphi \\
  0 \\
0\\
\end{bmatrix}=  \left[ \begin {array}{c} -\cos \left( \psi \right) \cos \left( \Omega
\,\tau \right) a\,m\,g\\ \sin \left( \Omega\,\tau
 \right) a\,m\,g\end {array} \right] 
$$
To get the generalized torques in the equations of motion,you have to extend the kinetic energy $T_g$ to
$$T_g\mapsto T_g+\tau_x\,\varphi+\tau_z\psi$$
Thus the Lagrange $L$ is:
$$L=T_g$$ 
I admit that this is quite elaborate calculations,so use symbolic manipulated program  like Maple to obtain the results.
