# How to find the instantaneous angular velocity vector?

Let $(R_0): O_0 \vec x_0 \vec y_0 \vec z_0$ and $(R): O\vec x \vec y \vec z$ be two given orthonormal frames.

The unique vector $\Omega_{0,1} = \Omega_{0,1}(t)$, given by the following three relations, is called the instantaneous angular velocity vector of $(R)$ with respect to $(R_0)$.

$$\begin{cases} d\vec x/dt = \Omega_{0,1} \times \vec x \\ d \vec y/dt = \Omega_{0,1} \times \vec y \\ d \vec z /dt = \Omega_{0,1} \times \vec z \end{cases}$$

I would like to know how, in general, one finds $\Omega_{0,1}$.

Let us consider an example:

Suppose that $(R)$'s position relative to $(R_0)$ is given by the three Euler Angles $(\psi, \theta, \phi)$; that's, $(R)$ is obtained from $(R_0)$ by the sequence of rotations:

$$x_0 y_0 z_0 \to^{\psi} nuz_0 \to^{\theta} nvz \to^{\phi} x y z$$

How does one find $\Omega_{0,1}$ here?

• Are you studying rigid body motion? :) I think there should be some answer to this in the analytical mechanics books. Take a look into Goldstein's book. – H. R. Nov 11 '15 at 11:54
• @H.R. Yes I am :) As for the question, I figured out later how to find $\Omega$. Thank you for the recommendation. – user258700 Nov 12 '15 at 1:20
• You can add what you find as an answer to this question, I am really interested to know myself! :) Would you please write an answer? :) – H. R. Nov 12 '15 at 9:07
• @H.R. Of course. Please check the answer. – user258700 Nov 12 '15 at 13:08
• You're welcome :) – user258700 Nov 12 '15 at 15:46

If $(R)$ is not rotating, then $\Omega_{0,1} = \vec 0$ because $\vec x, \vec y, \vec z$ are not changing with respect to time (seen as vectors in $(R_0)$, the absolute reference).

If $(R)$ is rotating, then I'll do an explanatory example.

Let $(R_0)$ be a given orthonormal frame and let the plane $(Ox_0y_0)$ be set in clockwise rotation about $(Oz_0)$. Let $(R)$ be this rotating frame; $(R): Oxyz_0$. Let $\Psi = \Psi(t) = \angle(Ox_0, Ox) = \angle(Oy_0, y)$.

We have:

$$\begin{cases} \vec x = \cos \Psi \vec x_0 + \sin \Psi \vec y_0 \\ \vec y = -\sin \Psi \vec x_0 + \cos \Psi \vec y_0 \\ \vec z = \vec z_0 \end{cases}$$

Hence:

$$\begin{cases} d\vec x/dt = \frac{d \Psi}{dt} \vec y \\ d\vec y /dt = - \frac{d \Psi}{dt} \vec x \\ d\vec z /dt = \vec 0\end{cases}$$

Thus:

$$\Omega_{0,1} = \frac{d \Psi}{dt} \vec z_0$$

And this is in general the expression of the IAVV of a frame $(R)$ with respect to a frame $(R_0)$ from which it was obtained through a rotation about one axis.

In case we have rotations about more than one axis, or in any arbitrary case in general, we can throughout the above formula derive the $\Omega$ by considering intermediary frames. So, for instance, in the case of Euler Angles (as described in the question), we would obtain:

$$\Omega_{0,1} = \frac{d\psi}{dt} \vec z_0 + \frac{d \theta}{dt} \vec n + \frac{d \phi}{dt} \vec z_1$$

Where: $\vec n = \cos \psi \vec x_0 + \sin \psi \vec y_0 \\ \vec z_1 = ...$