[![enter image description here][1]][1]

**2D case**

 the components of the position vector to  the point P, given in inertial system are

\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


  [1]: https://i.sstatic.net/glS1s.png