Euler equation and angular acceleration explanation

Question

I can't understand this math. Can anyone help me sort it out? It's related to my homework.

Answer 1

The first equation with $\dot\omega$ is simply re-arranging the original Euler equation and solving for $\dot\omega$. If you have sufficient symmetry in the rotating object, then the product-of-inertia terms (the off-diagonal terms in the inertia tensor) are zero, which is what this text says. That means you have only the moment-of-inertia terms: the ones on the main diagonal. It's relatively straight-forward to invert a diagonal matrix, assuming there are no zero entries on the main diagonal: simply do the multiplicative inverse of each diagonal entry. So far, then, we have $$\dot\omega=I^{-1}(\tau-\omega\times(I\omega))=I^{-1}\tau-I^{-1}(\omega\times(I\omega)).$$ Note that $\tau$ is a vector, so the first term in the last equation there is just writing out $I^{-1}\tau$ in components. As for the cross product, recall that it's defined as follows: $$\mathbf{x}\times\mathbf{y}=\det\left[\begin{matrix}\mathbf{i} &\mathbf{j} &\mathbf{k} \\ x_1 &x_2 &x_3 \\ y_1 &y_2 &y_3\end{matrix}\right]=\mathbf{i}(x_2y_3-x_3y_2)-\mathbf{j}(x_1y_3-x_3y_1)+\mathbf{k}(x_1y_2-x_2y_1).$$ So, building up the expression they have, we do \begin{align*} I&=\left[\begin{matrix}I_{xx} &0 &0\\ 0 &I_{yy} &0 \\ 0 &0 &I_{zz}\end{matrix}\right], \\ \omega&=\left[\begin{matrix}\omega_x\\ \omega_y \\ \omega_z\end{matrix}\right], \quad \text{so} \\ I\omega&=\left[\begin{matrix}I_{xx}\omega_x \\ I_{yy}\omega_y \\ I_{zz}\omega_z\end{matrix}\right],\\ \omega\times(I\omega)&=\det\left[\begin{matrix}\mathbf{i} &\mathbf{j} &\mathbf{k} \\ \omega_x &\omega_y &\omega_z \\ I_{xx}\omega_x &I_{yy}\omega_y &I_{zz}\omega_z \end{matrix}\right]\\ &=\left[\begin{matrix}I_{zz}\omega_y\omega_z-I_{yy}\omega_y\omega_z \\ -(I_{zz}\omega_x\omega_z-I_{xx}\omega_x\omega_z)\\ I_{yy}\omega_x\omega_y-I_{xx}\omega_x\omega_y\end{matrix}\right] \\ &=\left[\begin{matrix}(I_{zz}-I_{yy})\omega_y\omega_z \\ (I_{xx}-I_{zz})\omega_x\omega_z\\ (I_{yy}-I_{xx})\omega_x\omega_y\end{matrix}\right], \; \text{so that finally,} \\ I^{-1}(\omega\times(I\omega))&=\left[\begin{matrix}\left(\frac{I_{zz}-I_{yy}}{I_{xx}}\right)\omega_y\omega_z \\ \left(\frac{I_{xx}-I_{zz}}{I_{yy}}\right)\omega_x\omega_z\\ \left(\frac{I_{yy}-I_{xx}}{I_{zz}}\right)\omega_x\omega_y\end{matrix}\right]. \end{align*} I've double-checked this work, and it looks to me like the authors might be off by a minus sign. Can anyone else confirm?

Does that help?