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In a spherical coordinates system ($r$, $\theta$, $\phi$ ), assuming an angular rotation $\omega_z$ around the z-axis, the tangential velocity of a point can be expressed as:

$$V_x = -\omega_z R \sin\theta \sin\phi $$ $$ V_y = \omega_z R \sin\theta \cos\phi$$

What happens if I have a rotation $\omega_x$ around the $x$-axis? What are the equation for the $V_y$ and $V_z$ velocity components of the point?

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  • $\begingroup$ Frankly, by far the easiest expressions come out when/if you redefine your coordinate azimuth and ascension angles to be defined w.r.t. x instead of z. In that case, you merely have (x,y,z)↦(y,z,x). $\endgroup$ Oct 18, 2021 at 19:08

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The equation is always

$$ \boxed{\; \boldsymbol{v} = \boldsymbol{\omega} \times \boldsymbol{r} \; } $$

where $\boldsymbol{\omega} = \pmatrix{\omega_x \\ \omega_y \\ \omega_z } $ and $\boldsymbol{r} = \pmatrix{ r \cos \phi \sin \theta \\ r \sin \phi \sin \theta \\ r \cos \theta} $

In your first case $\omega_z \neq 0$ while others are 0, and in the second case $\omega_x \neq 0$.

$$\boldsymbol{v} = \begin{bmatrix} 0 & r\cos \theta & -r \sin \phi \sin \theta \\ -r \cos \theta & 0 & r \cos \phi \sin \theta \\ r \sin \phi \sin \theta & -r \cos \phi \sin \theta & 0 \end{bmatrix} \pmatrix {\omega_x \\ \omega_y \\ \omega_z } $$

If rotation is about the x-axis, then use the first column of the matrix above and multiply it with $\omega_x$.

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