When reading a paper about magnetic bearings, I have stumbled upon the following ansatz which I don't fully grasp (page 3):

For a general body, the Euler equations are given in a body fixed reference frame, because of the constant matrix of inertia. However, in the case of rotational symmetric rotors, the matrix of inertia is also constant in a non-body fixed frame. The Euler equations in a non-body fixed frame are given as: $$\begin{equation*}\mathbf\Theta\mathbf{\dot\omega}+\mathbf\Omega\times(\mathbf\Theta\mathbf\omega)=\mathbf M\end{equation*}$$ where $\mathbf\Omega$ is the angular velocity vector of the chosen coordinate system, $\mathbf\omega$ is the angular velocity vector of the body, $\mathbf\Theta$ is the moment of inertia matrix and $\mathbf M$ is the torque vector.

I know that the interia tensor is invariant under rotations around its symmetry axis, so I guess what they one wants to do is to enter the co-rotating frame of reference around that axis and "project out" the major degree of freedom around that axis. As far as I can tell, in such a frame the off-diagonal elements of the tensor still won't vanish and it should be still time-dependent. The following equations however only contain time-constant moments of inertia and no off-diagonal terms.

Additionally, as far as I know, Euler's equations deliver the angular velocities in the co-moving frame, but there is no transformation between rotating and non-rotating system, bypassing all the hassle with using Euler angles transformations. They directly use the co-moving angular velocities for the laboratory system geometry.

I have tried to reproduce this by considering the rotation as the product of 3 Euler angles $ R(t)=R_z(\omega t)R_{x'}(\alpha(t))R_{z''}(\beta(t)) $ with no success. The first factor can be taken out by symmetry and the other two are tackled with some small angle approximations, I think.

Q: How is this working? How can I arrive at equations as simple as

$$ M_x=\Theta_x\ddot\alpha + \Theta_z\Omega\beta\\ M_y = \Theta_y\ddot\beta -\Theta_z\Omega\alpha $$

for a z-axis-symmetric body, where $ \alpha,\beta $ are the laboratory-frame x/y angles and $ \Omega $ is the (constant) angular velocity around the z-axis?

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    $\begingroup$ Did you use the fact that $R (a\times b) = (R a) \times (R b)$? Also note that $\dot{R} = \Omega \times R$. I agree with out that the EOM on rotating frame would look different. $\endgroup$ Feb 26 '18 at 1:12
  • $\begingroup$ @ja72 Certainly I have tried alot. When considering only infinitesimal rotations, I end up with angular velocity matrices that are numerically identical in body-fixed and non-body-fixed frames. The two Euler angles involved however (x' and z'') are not identical to the ones in the paper (y, x), where one just tilts on the lab-fixed coordinate axes. Something must be about this approach, as I found somewhat similar equations in other papers as well, but no explanations. $\endgroup$
    – Jodocus
    Feb 26 '18 at 7:33

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