Rotating Coordinate System How to visualize  rotating frame of reference ?  
 A: Say point C is the origin of the moving frame. Point P is the location of the particle. The kinematics are such that the coordinate of P in the fixed frame are $\boldsymbol{r}_P$, but in the moving frame $\boldsymbol{p}$. The origin of the moving frame is $\boldsymbol{r}_C$ and so we have:
$$\boldsymbol{r}_P = \boldsymbol{r}_C + \boldsymbol{p} $$
Take the derivative of the above to derive the velocity of P.
$$ \frac{{\rm d} \boldsymbol{r}_P}{{\rm d}t} = \frac{{\rm d} \boldsymbol{r}_C}{{\rm d}t} + \frac{{\rm d} \boldsymbol{p}}{{\rm d}t}$$
$$ \boldsymbol{v}_P = \boldsymbol{v}_C + \boldsymbol{\omega} \times \boldsymbol{p}$$
Regardless of the motion of the moving frame (and Point C) the result is a vector field where the magnitude of the velocity depends on the perpendicular distance to a rotation axis, and the direction is always in the "hoop" orientation about this axis.
The instantaneous axis of rotation goes through a point A, whose location depends on the velocity of the moving frame
$$\boxed{ \boldsymbol{r}_A = \boldsymbol{r}_C + \frac{ \boldsymbol{\omega} \times \boldsymbol{v}_C}{\| \boldsymbol{\omega} \|^2} }$$
If $\boldsymbol{v}_C=0$ then $\boldsymbol{r}_A = \boldsymbol{r}_C$.


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*Proof:
The instantaneous axis of rotation has the property that the linear velocity at A is zero  $\boldsymbol{v}_C + \boldsymbol{\omega}\times ( \boldsymbol{r}_A - \boldsymbol{r}_C ) =0$ or $\boldsymbol{v}_C = \boldsymbol{\omega}\times ( \boldsymbol{r}_C - \boldsymbol{r}_A )$.
Take $\boldsymbol{\omega} \times \boldsymbol{v}_C$ and expanded it using the vector triple product identity
$$\require{cancel} \boldsymbol{\omega} \times \boldsymbol{v}_C = \boldsymbol{\omega} \times \left(\boldsymbol{\omega}\times ( \boldsymbol{r}_C - \boldsymbol{r}_A )\right) = \boldsymbol{\omega} \left(\cancel{ \boldsymbol{\omega} \cdot (\boldsymbol{r}_C-\boldsymbol{r}_A)} \right) - (\boldsymbol{r}_C-\boldsymbol{r}_A)\,( \boldsymbol{\omega} \cdot \boldsymbol{\omega}) $$
$$ \frac{ \boldsymbol{\omega} \times \boldsymbol{v}_C }{ \| \boldsymbol{\omega} \|^2 } = \boldsymbol{r}_A - \boldsymbol{r}_C $$
Use the fact that the vector from point A to point C is perpendicular to the rotation vector.
