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I am working on a simulation of a point moving in an elliptical 3D trayectory as shown in the image ellipse .

I wish to calculate the angular velocity vector of the motion . In this case I can't supose a single axis of rotation for all the trayectory because the orientation of the axis is changing every instant of time . Is there any method for calculating the angular velocity vector given the position at every instant of time ?

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  • $\begingroup$ Generally the answer would be the cross product $\vec{\omega}=\vec{r}\times\vec{v}/|\vec{r}|^2$ where $\vec{r}$ is the position with respect to the origin (which may change with time) and $\vec{v}$ is the velocity. But it depends on precisely how you want to define the 'angle'. Perhaps you'd prefer to project the trajectory onto a plane first? $\endgroup$ – lemon Nov 21 '17 at 16:52
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It's not clear what you mean by angular velocity. Angular velocity is only defined relative to some origin, and even then you must specify the definition clearly.

The usual definition is this: first, pick an origin. Let the position of the particle with respect to the origin be $\vec{x}(t)$ and $\dot{\vec{x}}(t)=\vec{v}(t)$. We can then decompose the velocity vector into a part that is parallel to the line from the origin to the particle and a perpendicular part. If $\hat{x} = \vec{x}/|\vec{x}|$ then $\vec{v}_{\parallel} = (\vec{v}\cdot \hat{x})\hat{x}$ and $\vec{v}_\perp =\vec{v}-(\vec{v}\cdot \hat{x})\hat{x}$.

Then if we use the cross product to define $\vec{\omega} = (\hat{x}\times\vec{v})/|\vec{x}|$ we can always get the perpendicular part of the velocity back via $$\vec{\omega}\times\vec{x} = \vec{v}_\perp$$ where that equality follows from the definition of $\vec{\omega}$ and some cross product identities.

Note that the angular velocity thus gives you information only about the perpendicular part of the velocity. The parallel or radial velocity has to be tracked separately.

And, once again, there are other definitions of angular velocity that you might want to use, so think carefully about what you need.

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  • $\begingroup$ Yes, I forgot to precise that I have a defined origin. Actually I'm trying to implement an algorithm that uses the data of the components ωx , ωy and ωz of the angular velocity . One of my simulation parameters is the magnitude of the vector, so I only have to find the orientation and then I could calculate each component .In the case of a planar motion it is easy beacause I know that it is always perpendicular to the plane. But in this case is quite complicate, I guess I should calculate ω for each instant of time with the formula you wrote above. $\endgroup$ – user176105 Nov 22 '17 at 8:46

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