Why are Euler's equations of motion coupled? Physical explanation I have a problem with one of my study questions for an oral exam:

Euler’s equation of motion around the $z$ axis in two dimensions is $I_z\dot{\omega}_z = M_z$, whereas it in three dimensions is $I_z\dot{\omega}_z =-(I_y-I_x)\omega_x\omega_y+M_z$, assuming that the $xyz$ coordinate systems is aligned with the principal axis. Why does Euler’s equation of motion for axis $z$ contain the rotational velocities for axes $x$ and $y$?

How can one explain this physically? I mean I can derive Euler's equation of motion, but how can I illustrate that the angular velocities are changing in 3 dimensions?
 A: As explained on Wikipedia, the nice tensor form of the equations is
$$ \mathbf{I} \cdot \dot{\boldsymbol\omega} + \boldsymbol\omega \times \left( \mathbf{I} \cdot \boldsymbol\omega \right) = \mathbf{M} $$
This reduces to your equations if one diagonalizes the tensor of the moment of inertia $I$ and labels the diagonal entries etc.
The three components are mixed with each other because quantities like $\vec\omega$ and $\vec M$ are really associated with rotations in space and rotations around the axis $x,y,z$ don't commute with each other – unlike translations. Translations commute with each other which is why the 3 components in $\vec F=m\vec a$ don't mix with each other.
For example, take the Earth, rotate it by 90 degrees around the $x$ axis, then 90 degrees around $y$ axis, then you rotate back by 90 degrees but first around $x$ axis again, so that you aren't undoing the $y$ rotation immediately, but then you undo the $y$ rotation, too. You don't get back where you have been: instead, you end up rotating the Earth around the $z$ axis. We say that rotations form the group $SO(3)$ which is non-abelian, $gh\neq hg$. The moment of force wants to rotate the rigid body around an axis but because it was already rotating around another axis given by $\vec \omega$ and the rotations don't commute with each other, the effect of the moment of force also influences the "remaining third" component.
A natural way to write the vectors $\vec \omega, \vec M$ is actually an "antisymmetric tensor" – they're "pseudovectors", not actual vectors. At any rate, when you correctly derive the equations, you should reproduce what Euler got.
A: If you are familiar aircraft flight dynamics, then please remember an aerobatic maneuver so called "Immelmann turn". Let us take x,y and z axis for longitudinal, lateral and vertical coordinate axes fixed to the aircraft. Then, if we pull control a stick and let the aircraft turn in vertical plane around the y axis and turn the stick to the right or left and make a rolling motion around the x axis at the same time. When the both two angular motion accomplished 180 degree turn, then the aircraft flies in the opposite direction before this maneuver.  This means that the aircraft turned 180 degree around the z axis. This is my interpretation of the Euler equation of rigid body motion.
