How to derive relation for time derivative in a rotating reference frame I am looking for an appropriate derivation of the $(\frac{d}{dt})_{\text{laboratory}} = (\frac{d}{dt})_{\text{rotating}} + \omega \times $ relationship that enables one to calculate all desired quantities in a rotating referece frame. Does anybody know a good way how to understand this transformation?
 A: The components of any vector function can be written any any desired basis.  In particular, let
\begin{align}
  \mathbf A_L(t) = (A^1_L(t) , A^2_L(t), A^3_L(t))
\end{align}
denote the components of a vector function as written in an orthonormal basis fixed in the laboratory, and let
\begin{align}
  \mathbf A_R(t) = (A^1_R(t), A^2_R(t), A^3_R(t))
\end{align}
denote the components of the same vector as written in a rotating orthonormal basis.  These components will be related by a time-dependent special orthogonal matrix (rotation);
\begin{align}
  \mathbf A_L(t) = R(t)\mathbf A_R(t)
\end{align}
In particular, note that taking time derivatives on both sides gives
\begin{align}
  \dot{\mathbf A}_L(t) 
  &=  R(t) \dot{\mathbf A}_R(t) + \dot R(t) \mathbf A_R(t) \\
  &= R(t) \dot{\mathbf A}_R(t) + \dot R(t) R(t)^T\mathbf A_L(t)
\end{align}
Since $R(t)$ is an orthogonal matrix, we have
\begin{align}
  R(t)R(t)^T = I
\end{align}
and taking derivatives of both sides, and using the fact that time derivatives and matrix transposes commute, we find that
\begin{align}
  \dot R(t) R(t)^T = -(\dot R(t)R(t)^T)^T
\end{align}
in other words, $\dot R(t) R(t)^T$ is an antisymmetric matrix.  We lose no generality by therefore writing
\begin{align}
  \dot R(t) R(t)^T = \Omega(t)
\end{align}
where
\begin{align}
  \Omega(t) = \begin{pmatrix}
          0 & -\omega^3(t)  & \omega^2(t) \\
          \omega^3(t) & 0 & -\omega^1(t) \\
          -\omega^2(t) & \omega^1(t) & 0 \\
        \end{pmatrix}
\end{align}
for some vector of functions $\boldsymbol\omega=(\omega^1, \omega^2, \omega^3$).  Therefore, we have
\begin{align}
  \dot {\mathbf A}_L(t) = R(t)\dot{\mathbf A}_R(t) + \Omega(t)\mathbf A_L(t)
\end{align}
It is straightforward to explicitly show that multiplication by $\Omega(t)$ is equivalent to a cross product by $\boldsymbol\omega(t)$, so we can write
\begin{align}
  \Omega(t) \mathbf A_L(t) = \boldsymbol\omega(t)\times\mathbf A_L(t)
\end{align}
and we are therefore let to the expression
\begin{align}
  \boxed{\dot{\mathbf A}_L(t) = R(t)\dot{\mathbf A}_R(t) + \boldsymbol\omega(t)\times\mathbf A_L(t)}
\end{align}
If we make the identifications
\begin{align}
  \dot{\mathbf A}_L(t) &=\left(\frac{d\mathbf A}{dt}\right)_\mathrm{laboratory} \\
  R(t)\dot{\mathbf A}_R(t) &=\left(\frac{d\mathbf A}{dt}\right)_\mathrm{rotating}
\end{align}
then we see that this is equivalent to your formula.  In my opinion, the physicist notation in the formula you wrote down is extremely confusing, and I prefer to use the more descriptive notation in the boxed expression above; I find that it leads to less errors and is more conceptually clear.
A: I'll consider two-dimensional reference frames, and i will use galilean transformation, adapted to your particular problem. Here is how i imagine this: 

Here, $\vec{r}$ is the position vector of a point, as seen from the laboratory frame;  $\vec{r_0}$ is the position vector of the center of the other (moving) frame; and $\vec{r'}$ is the position of the point as seen from the primed frame. Note that $\vec{r'}$ will describe a circle, so we can write it as:
$$\vec{r'}=r'\cos(\omega t) \vec{i}+r'\sin(\omega t) \vec{j}$$
(assuming that the particle starts at an angle $0$ when $t=0$).
Now, clearly, we have $\vec{r}=\vec{r_0}+\vec{r'}$. Differentiating this with respect to time, we get:
$$ {\frac{d\vec{r}}{dt}}={\frac{d\vec{r_0}}{dt}}+{\frac{d\vec{r'}}{dt}}$$
which is the same as 
$$\vec{v}=\vec{v_0}+\vec{v'}$$
where $\vec{v}$ is the speed as seen from the laboratory frame, $\vec{v_0}$ is the speed of the moving frame, and $\vec{v'}$ is the speed in the moving frame.
But $\vec{v'}={\frac{d\vec{r'}}{dt}}=-\omega r'\sin(\omega t)\vec{i}+\omega r'\sin(\omega t)\vec{j}$, so that we finally get:
$$\vec{v}=\vec{v_0}+\omega r' [-\sin(\omega t)\vec{i}+\cos(\omega t)\vec{j}]$$
To find the accelerations, you can differentiate again this relation.
