homework and exercises - Angular Velocity expressed via Euler Angles

On the top of the fourth page from here, the author trivially derives the components of angular velocity, expressed via Euler angles of the system. I fail to understand how the components of angular velocity were derived. May you please enlighten me!

Answer

I assume you know about rotation matrices, and so for a sequence rotations about Z-X-Z with angles $\phi$, $\theta$ and $\psi$ repsectively you have

$$ \vec{\omega} = \dot{\phi} \hat{z} + T_1 \left( \dot{\theta} \hat{x} + T_2 \left( \dot{\psi} \hat{z} \right) \right) $$

The logic here is apply a local spin of $\dot{\phi}$, $\dot{\theta}$ and $\dot{\psi}$ on the local axes in the sequence.

1. Apply spin $\dot{\phi}$ about local Z and then rotate by $T_1$


2. Apply spin $\dot{\theta}$ about local X (rotated by $T_1$) and then rotate by $T_2$


3. Apply spin $\dot{\psi}$ about local Z (rotated by $T_1 T_2$).

Update

There is a way to formally derive the above using the identity $\dot T = \vec\omega \times T$ but it is rather involved for 3 degrees of freedom.

For two degrees of freedom it goes like this. With a rotation matrix $T= T_1 T_2$ (defined as above) the time derivative is

$$ \begin{aligned} \frac{{\rm d} T}{{\rm d} t} & = \dot{T}_1 T_2 + T_1 \dot{T}_2 \\ & = \left( (\dot{\psi} \hat{z}) \times T_1 \right) T_2 + T_1 \left( (\dot{\theta} \hat{x}) \times T_2 \right) \\ & = (\dot{\psi} \hat{z}) \times \left( T_1 T_2 \right) + (T_1 (\dot{\theta} \hat{x})) \times \left( T_1 T_2 \right) \\ & = \left(\dot{\psi} \hat{z} + T_1 (\dot{\theta} \hat{x}) \right) \times (T_1 T_2) \\ & = \left(\dot{\psi} \hat{z} + T_1 (\dot{\theta} \hat{x}) \right) \times T = \vec{\omega} \times T \end{aligned} \\ \vec{\omega} = \dot{\psi} \hat{z} + T_1 (\dot{\theta} \hat{x}) $$

using the distributed property $T (\vec{a} \times \vec{b} ) = (T \vec{a}) \times (T \vec{b} )$.