Diagonal of a thin rectangular foil, inertial principal axis? I'd like to know if the diagonal of a thin foil is an inertial principal axis.
I know that if an axis isn't a symmetry axis then it isn't a principal axis. In the rectangle the diagonal isn't a symmetry axis, so it shouldn't be a principal axis. Is it correct?
So, if I consider that the angle between the rotation axis and x-axis is $\theta$ and $\vec{\omega}$ is on xy-plane, I can't understand why
$$\tau_z=\omega_x \omega_y(I_{xx}-I_{yy})$$
 A: The inertia matrix for a thin rectangular foil (laying along the xy plane) in body coordinates is
$$ I_{body} = \begin{vmatrix} \frac{m}{12} b^2 & 0 & 0 \\ 0 & \frac{m}{12} a^2 & 0 \\ 0& 0 & \frac{m}{12}(a^2+b^2) \end{vmatrix} $$
where $a$ and $b$ are the side dimensions. 
The inertia matrix in world coordinates, while rotated by angle $\theta$ about the z axis is defined by
$$ I = {\rm Rot}(\hat{k},\theta) I_{body} {\rm Rot}(\hat{k},-\theta) $$
$$ I = \frac{m}{12} \begin{vmatrix} a^2 + (b^2-a^2)\cos^2 \theta & (b^2-a^2) \sin \theta \cos \theta & 0 \\ (b^2-a^2) \sin \theta \cos \theta & b^2 + (a^2-b^2) \cos^2 \theta & 0 \\ 0 & 0 & a^2+b^2 \end{vmatrix} $$
To make this a principal axis, the off-diagonal terms need to be zero. $$(b^2-a^2) \sin \theta \cos \theta = 0$$
This is only true if $a=b$ or $\theta = n \frac{\pi}{2}$ where $n=0,1,2 \ldots $
Once you have your inertia matrix in world coordinates you calculate the net torque on the center of mass as
$$ \vec{\tau} = I \dot{\vec{\omega}}+ \vec{\omega} \times I \vec{\omega} $$
The term you are getting is a result of the rotating frame torque.
