# Tangential velocity - Spherical coordinates

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?

• 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). Oct 18, 2021 at 19:08

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