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I often see the relation that $\vec v=\vec v_0+ \vec \omega \times \vec r$ in a turning reference frame, but where does it actually come from and how do I arrive at the acceleration being $$\vec a=\vec a_0+ 2\vec \omega \times\vec v+ \vec \omega \times(\vec \omega \times \vec r)+\dot{\vec \omega} \times \vec r\,\,\text{?}$$

Is there a simple method to see this? All approaches that I saw use some non-intuitive change of differential operators and so on ($\frac{d}{dt} \rightarrow \frac{d}{dt}+\vec \omega \times$) and so on.

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The acceleration of a fixed point on a moving frame is $\vec{a}_0 + \dot{ \vec{\omega}} \times \vec{r} + \vec{\omega} \times ( \vec{v}-\vec{v}_0 )$. Do you have a reference for your equation? –  ja72 Jun 4 '13 at 16:20
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I don't think you can do much better than getting your head around the identity $$\frac{d}{dt} \rightarrow \frac{d}{dt}+\vec \omega \times,$$ which holds when the former is applied to vectors. The essential point of the identity is that even if a vector is stationary in one reference frame, it will have some rotational motion in the rotating frame.

It may help to rephrase this in matrix language: for any vector $\vec u$, it reads $$\frac{d}{dt} \begin{pmatrix}u_x\\u_y\\u_z\end{pmatrix} \rightarrow \frac{d}{dt} \begin{pmatrix}u_x\\u_y\\u_z\end{pmatrix} + \begin{pmatrix}0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x\\ -\omega_y & \omega_x & 0\end{pmatrix} \begin{pmatrix}u_x\\u_y\\u_z\end{pmatrix} = \begin{pmatrix} \frac{du_x}{dt} +\omega_y u_z-\omega_z u_y\\ \frac{d u_y}{dt}+\omega_z u_x-\omega_x u_z\\ \frac{du_z}{dt} +\omega_x u_y-\omega_y u_x\end{pmatrix} .$$ Thus, the rate of change of each vector component gets added a linear multiple of the other components, as they "rotate into it".

(For example, if $\vec \omega=\omega \hat{e}_z$, then $$\frac{d}{dt} \begin{pmatrix}u_x\\u_y\\u_z\end{pmatrix} \rightarrow \begin{pmatrix} \frac{du_x}{dt} -\omega u_y\\ \frac{d u_y}{dt}+\omega u_x\\ \frac{du_z}{dt} \end{pmatrix} ,$$ so that $u_x$ and $u_y$ transform into ($\pm$) each other as the frames rotate and the $x$ and $y$ axes rotate into ($\pm$) each other.)

That's the intuition behind the identity. Operationally, it is the easiest to apply (just substitute for $\frac{d}{dt}$), and it gives an unambiguous way to connect rates of change of vector components from one frame to another. What's not to love?

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This equation comes from the derivative on a rotating frame (as you mentioned). It might be easier to understand if you look at the two components of $$\frac{\rm d}{{\rm d}t} \rightarrow \frac{\partial}{\partial t}+\vec \omega \times$$ as the change as seen on the moving frame, plus the change due to the motion of the frame.

It all starts from the position kinematics. Consider a point A with position vector

$$ \vec{r}_A = \vec{r}_O + \vec{r} $$

where $\vec{r}_O$ is the position vector of the coordinate frame origin and $r$ relative position vector of A w.r.t. O

The time derivative of $\vec{r}$ is $\frac{{\rm d} \vec{r}}{{\rm d}t} =\dot{ \vec{r} } + \vec{\omega} \times \vec{r} $, based on derivatives on rotating frames, allowing for the differentiation of the above as

$$ \vec{v}_A = \vec{v}_O + \dot{ \vec{r} } + \vec{\omega} \times \vec{r} $$

Note that

$$\frac{{\rm d}}{{\rm d}t} \dot{\vec{r}}=\ddot{ \vec{r} } + \vec{\omega} \times \dot{ \vec{r}} $$

and

$$\frac{{\rm d}}{{\rm d}t} (\vec{\omega} \times \vec{r}) = \dot{\vec{\omega}} \times \vec{r} + \vec{\omega}\times \frac{{\rm d} \vec{r}}{{\rm d}t} = \dot{\vec{\omega}} \times \vec{r} + \vec{\omega}\times (\dot{ \vec{r} } + \vec{\omega} \times \vec{r}) $$

so together

$$ \vec{a}_A = \vec{a}_O + \ddot{ \vec{r} } + \vec{\omega} \times \dot{ \vec{r}} + \dot{\vec{\omega}} \times \vec{r} + \vec{\omega}\times (\dot{ \vec{r} } + \vec{\omega} \times \vec{r}) $$ $$ \vec{a}_A = \vec{a}_O + \ddot{ \vec{r} } + 2\vec{\omega} \times \dot{ \vec{r}} + \dot{\vec{\omega}} \times \vec{r} + \vec{\omega}\times \vec{\omega} \times \vec{r} $$

If the point is fixed to the frame then $\dot{\vec{r}}=0$, $\ddot{\vec{r}}=0$ and

$$ \vec{a}_A = \vec{a}_O + \dot{\vec{\omega}} \times \vec{r} + \vec{\omega}\times \vec{\omega} \times \vec{r} $$

As far as the equation on your question, I am not sure if it applies here because the $\vec{v}$ used should be the relative velocity of the point $\dot{\vec{r}}$ and the relative acceleration ignored $\ddot{\vec{r}}=0$ to match with correct equation. See also here and here.

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