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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.

enter image description here

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.

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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.

enter image description here

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 $\vec\omega''$.

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