Feynman's wobbling plate Let us consider a symmetric top, i.e., a body whose mass distribution has axial symmetry (a cylinder, a disk, a cone, etc) free of any external torque. The Euler equations for this body are
$$I_1\dot\omega_1-(I_1-I_3)\omega_2\omega_3=0,$$
$$I_1\dot\omega_2-(I_3-I_1)\omega_1\omega_3=0,$$
$$I_3\dot\omega_3=0,$$
where $I_i$ are the inertia principal moments (moments of inertia along the principal axes) and due to the axial symmetry $I_1=I_2$. The components of the angular velocity $\vec \omega$ are given along the principal axes.
The third equation gives 
$$\omega_3=s,$$
where $s$ (spin) is constant. The first two equations can be combined into simple harmonic oscillator equations and the solution is
$$\omega_1=A\cos(\Omega t+\delta),\quad \omega_2=A\sin(\Omega t+\delta),\quad \Omega=\frac{I_3-I_1}{I_1}s.$$
In the body frame (principal axes) this means that the angular velocity vector has constant projection on the principal axis $\rho_3$ but its projection on the plane $\rho_1\rho_2$ rotates with angular velocity $\Omega$. This can be viewed as a precession of $\vec\omega$ around the symmetry axis of the body. The angular momentum is $\vec L=I\vec\omega$, so
$$\vec L=I_1A\cos(\Omega t+\delta)\hat\rho_1+I_2A\sin(\Omega t+\delta)\hat\rho_2+I_3s\hat\rho_3,$$
Hence it lies in the same plane as $\hat\rho_1$ and $\omega$ and shows the same precession as the latter.
In the inertial frame we see the symmetry axis and $\vec\omega$ are precessioning with frequency $\Omega$ around $\vec L$. To me it is this precession that is seen as a wobbling (see this at 5:26).  
If we consider a homogeneous disk, $I_3=2I_1$ so $\Omega=s$. However the classic result is a wobbling of frequency $2s$. The experimental demonstration can be viewed here at 0:50. This result can be obtained by writing the angular velocity components in terms of Euler angles and then solving for $\dot\alpha$. Looking at the Euler angles it seems in fact that the revolution of the line of nodes (denoted by N in the figure bellow) corresponds to the wobbling and both periods should be equal.

So my question is: Why does not the precession rate of the angular velocity vector give exactly the frequency of wobbling? In other words, how come the precession of $\vec\omega$ is different to the line of nodes rotation.
 A: 
Why does not the precession rate of the angular velocity vector give
  exactly the frequency of wobbling? In other words, how come the
  precession of $\vec\omega$ is different to the line of nodes rotation?

Short answer: Because the precession rate $\Omega$ corresponds to the precession of $\vec\omega$ in the body frame, not in the Earth frame. The variables in the Euler equations, $I_i$, $\omega_i$ and $\tau_i$, are in the body frame.
The angular velocity of the disk has two contributions, the component $\vec\omega'$ due to the spin and the component $\vec\omega''$ due to a rotation of the inclined disc about a vertical axis. The latter corresponds to the wobbling as viewed by someone in the Earth frame. 

As we can see from the figure, the resultant angular velocity $\vec\omega$ is always off the symmetry axis of the disk, which means it has a non-vanishing projection in the plane of the disk (dashed line). At same time, the axes fixed in the plane of the disk, $\rho_1$ and $\rho_2$ rotates with spin $s$ (viewed by someone in the Earth frame). Hence, someone in the disk frame will see the axes fixed and the projection of $\omega$ in this plane rotating with rate $s$. That is why the precession rate equals the spin.
On the other hand, the wobbling effect is due to the non-vanishing $\vec\omega''$. A $2\pi$ rotation about the vertical line corresponds to a complete oscillation (wobbling) therefore the frequency of the wobbling is actually equal to the magnitude of $\vec\omega''$.
