Starting with 2D case, the intertial components of the position vector
\begin{align*} &\vec{R}=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}= \underbrace{\begin{bmatrix} \cos(\phi(t)) & -\sin(\phi(t)) & 0 \\ \sin(\phi(t)) & \cos(\phi(t)) & 0 \\ 0 & 0 &1 \\ \end{bmatrix}}_{\mathbf S}\,\begin{bmatrix} u_x \\ u_y \\ 0 \\ \end{bmatrix} \end{align*}
thus the velocity
\begin{align*} &\vec{v}=\frac{d}{dt}\,\vec{R}=\frac{d}{dt}\,\phi(t) \begin{bmatrix} -\sin(\phi(t)) & -\cos(\phi(t)) & 0 \\ \cos(\phi(t)) & -\sin(\phi(t)) & 0 \\ 0 & 0 &0 \\ \end{bmatrix}\,\begin{bmatrix} u_x \\ u_y \\ 0 \\ \end{bmatrix}=\vec{\omega}\times\vec{R}\quad,\text{where}\\ & \vec{\omega}= \begin{bmatrix} 0 \\ 0 \\ \frac{d}{dt}\,\phi(t) \\ \end{bmatrix} \end{align*}
3D case \begin{align*} & \vec{R}=\mathbf{S}\left[~\phi_x(t)~,\phi_y(t),~\phi_z(t)~\right]\,\vec{u} \quad\Rightarrow\\ &\vec v=\vec{\dot{R}}=\mathbf{\dot{S}}\,\vec{u}\quad ,\text{with}\quad \mathbf{\dot{S}}=\left[ \begin {array}{ccc} 0&-\omega_{{z}}&\omega_{{y}} \\ \omega_{{z}}&0&-\omega_{{x}}\\ -\omega_{{y}}&\omega_{{x}}&0\end {array} \right] \,\mathbf{S}\quad\Rightarrow\\ &\vec v=\vec{\omega}\times\vec{R} \end{align*}
- $\mathbf S~$ is the transformation matrix between body-system and inertial-system