# Angular velocity of rigidly rotating orbit in 3D

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 ($$\omega$$).

Problem

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 $$\vec\omega$$?

• 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 $\overrightarrow {\omega }\times \overrightarrow {r}=\tilde \omega \cdot r$ – Eli Feb 18 at 16:23
• 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. – usumdelphini Feb 19 at 10:13
• It would be good if you could state clearly the meaning of each of the variables in your equations, such as $\vec{n}$ and $r_0$. – Dlamini Feb 27 at 8:31
• @Dlamini I have already, in the text. – usumdelphini Feb 27 at 12:48