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:

relation of angles and bases

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 $\overrightarrow{e}_{3}(t)$:

projections of basis vectors

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?


closed as off-topic by ACuriousMind Dec 19 '17 at 15:12

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Hi and welcome to physics.SE! Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. $\endgroup$ – ACuriousMind Dec 19 '17 at 15:12
  • $\begingroup$ What's wrong is that you assumed $\alpha$ is negligibly small. In that limit, you'll get $[-\dot\alpha, \dot\theta \sin\alpha, \dot\theta \cos\alpha] \to [-\dot\alpha, 0, \dot\theta]$. $\endgroup$ – David Hammen Dec 19 '17 at 16:17
  • $\begingroup$ @ACuriousMind Hi and thank you for the Correction. Should I rather repost my changed question or should I edit it here? $\endgroup$ – B. Preiss Dec 19 '17 at 18:35
  • $\begingroup$ Please edit it here - once it is edited by you, it will automatically be placed in a review queue where users can decide whether it is now on-topic. $\endgroup$ – ACuriousMind Dec 19 '17 at 18:38
  • $\begingroup$ so, I posted my question here link as a follow-up. Hope it is better that way! $\endgroup$ – B. Preiss Dec 19 '17 at 23:58