I tried to apply the procedure found in this link - derive angular velocity from infinitesimal rotation matrix by small angle approximation - to calculate $\omega$ from the rotation matrix $\mathbf{R}^{T}$. The specific situation, where I tried to do this can be found in the picture below:
Now in my example the basis of the inertial and moving coordinate system may be given by the vectors $\overrightarrow{e}_{x}$, $\overrightarrow{e}_{y}$, $\overrightarrow{e}_{z}$ and $\overrightarrow{e}_{1}(t)$, $\overrightarrow{e}_{2}(t)$, $\overrightarrow{e}_{3}(t)$ respectively.
Here are the projected views (projected along $\overrightarrow{e}_{x}$ and $\overrightarrow{e}_{z}$) of the basis-vectors, where $\alpha$ ($\dot{\alpha}=\Omega$) shall be the angle enclosed by $\overrightarrow{e}_{z}$ and $\overrightarrow{e}_{3}(t)$:
In the views each vector is noted with its projected length.
The coordinate vectors of any point shall be linked by the Rotation Matrix $\mathbf{R'}$:
$\underbrace{[\overrightarrow{u}]_{i}}_{\textrm{coordinate vector in inertial frame }}=\mathbf{R'}\cdot\underbrace{[\overrightarrow{u}]_{r}}_{\textrm{coordinate vector in rotating frame }}$
I think $\mathbf{R'}$ would be:
$\mathbf{R'}=\left[\begin{array}{ccc} \cos(\theta) & -\sin(\theta) & 0\\ \sin(\theta)\cdot\cos(\alpha) & \cos(\theta)\cdot\cos(\alpha) & \sin(\alpha)\\ -\sin(\theta)\cdot\sin(\alpha) & -\cos(\theta)\cdot\sin(\alpha) & \cos(\alpha) \end{array}\right]$.
According to the example I linked above, I can approximate $\mathbf{R'}$ as:
$\mathbf{R'}\simeq\left[\begin{array}{ccc} 1 & -\delta \theta & 0\\ \delta \theta & 1 & \delta \alpha\\ 0 & -\delta \alpha & 1 \end{array}\right]=\left[\begin{array}{ccc} 0 & -\delta \theta & 0\\ \delta \theta & 0 & \delta \alpha\\ 0 & -\delta \alpha & 0 \end{array}\right]+\boldsymbol{I}_{3x3}=S(\omega)+\boldsymbol{I}_{3x3}=\delta\overrightarrow{E}\times+\boldsymbol{I}_{3x3}$
where $S(\delta \theta,\delta \alpha)$ is a Skewsymmetric Matrix and $\delta \overrightarrow{E}=\left[\begin{array}{c} -\delta \alpha\\ 0\\ \delta \theta \end{array}\right]$.
Analogous to the example it follows:
$\frac{\delta}{\delta t}[\overrightarrow{u}]_{i}=\frac{\delta}{\delta t}E\times[\overrightarrow{u}]_{r}=\underbrace{\left[\overrightarrow{\omega}_{r}\right]_{i}}_{\textrm{angular velocity of rotating frame expressed in inertial frame}}\times[\overrightarrow{u}]_{r}$.
That means, that $\left[\overrightarrow{\omega}_{r}\right]_{i}=\left[\begin{array}{c} -\dot{\alpha}\\ 0\\ \dot{\theta} \end{array}\right]$.
That seems to be wrong, because by inspection I get:
$\left[\overrightarrow{\omega}_{r}\right]_{i}=\left[\begin{array}{c} -\dot{\alpha}\\ \dot{\theta\cdot\sin(\alpha)}\\ \dot{\theta\cdot\cos(\alpha)} \end{array}\right]\neq\left[\begin{array}{c} -\dot{\alpha}\\ 0\\ \dot{\theta} \end{array}\right]$.
Here is my question: What is wrong?
