Consider a circle in 3-dimensional space. On this circular orbit, a rigid bead moves, thus changing its angle $\phi$ with a reference radius on the circle.

The intrinsic angular velocity is given by $\Omega=R\dot\phi$. The whole orbit rigidly rotates and translates with respect with the laboratory frame of reference.

The rigid motion is characterised by rotational velocity $\vec\omega$ and translational velocity $V=\dot{ \vec{r}}_0$, where $\vec{r}_0$ is the position of the centre of the circle and $\vec\omega$ can be written in terms of the Euler angles $\theta_1,\theta_2,\theta_3$ which identify the orientation of the circular orbit.

The position and velocity of the rigid bead can thus be written in the lab frame using the unit tangent $\vec t$ and normal $\vec n$ vectors, as follows

$$\vec r=\vec{r}_0+R\ \vec{n}\left(\theta_1,\theta_2,\theta_3\right)$$ $$\vec v=\dot{ \vec{r}}_0+\frac{\partial \vec r}{\partial{\theta_i}}\dot \theta_i+\Omega\ \vec t \left(\theta_1,\theta_2,\theta_3\right)$$

Let's assume that I know $\vec{v}$. If this was a 2-dimensional problem and the circle was costrained to rotate on the $x,y$ plane (we can reduce the notation to that case by imposing $\theta_2=\theta_3=0$), we could easily find $\omega=\dot\theta_1$ by using the fact that

$$(\vec v-\dot{ \vec{r}}_0)\cdot \vec n=\vec n\cdot\frac{\partial \vec n}{\partial{\theta_1}}\omega$$ but since $\vec n(\theta_1,\theta_2,\theta_3)$ is a unit vector and is just the rotated of $\vec n^\prime$, we have

$$\frac{\partial \vec n}{\partial{\theta_1}}\cdot\vec n=\frac{\partial \left[R(\theta_1)\cdot \vec{n}\right]}{\partial{\theta_1}}=-\sin\theta_1$$

where $R(\theta_1)$ is the 2-dimensional rotation matrix. So from $\vec v-\dot{ \vec{r}}_0$ I can derive $\omega$, which is (as expected) independent of $\dot{\phi}$. One relationship (projection along perpendicular to $\vec t$) and one unknown ($\omega$).


In 3 dimensions, I can use the same strategy and project the velocity vector onto $\vec n$ and $\vec \Omega=\vec n \times \vec t$. This gives me two relationships, but I have three unknowns, namely $\omega_x, \omega_y, \omega_z$.

Therefore, it does not seem possible to write $\omega$ as independent from $\Omega$.

And yet, this does not seem to me geometrically valid, because the two should be independent. How can I find a third relationship to find all the three components of $\vec\omega$?

  • $\begingroup$ For a given rotation matrix between body fixed coordinate system an intertial system $R=R\left( \theta _{1},\theta _{2},\theta _{3}\right) $ you get $\overrightarrow {\omega}$ from this equation $\dfrac {dR}{dt}=R\cdot \tilde \omega $ with $\overrightarrow {\omega }\times \overrightarrow {r}=\tilde \omega \cdot r$ $\endgroup$ – Eli Feb 18 at 16:23
  • $\begingroup$ That's clear, but I don't know $dR/dt$. That's what I want to find, either $\dot{\theta_i}$ (which is essentially $dR/dt$) or $\vec\omega$. My last calculation above is just a way of passing from $\dot{\theta_i}$ to $\vec\omega$, but I still need to find 3 geometrical relationships that have either $\omega_i$ or $\theta_i$ as unknown. At the moment I have only two. $\endgroup$ – usumdelphini Feb 19 at 10:13
  • $\begingroup$ It would be good if you could state clearly the meaning of each of the variables in your equations, such as $\vec{n}$ and $r_0$. $\endgroup$ – Dlamini Feb 27 at 8:31
  • $\begingroup$ @Dlamini I have already, in the text. $\endgroup$ – usumdelphini Feb 27 at 12:48

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