Lecture 1. Rotating reference systems
Consider the system \(S’\) rotating at constant angular velocity \(\vec{\omega}\) counterclockwise with respect to the reference system \(S\). At any instant the vector position of the point mass \(m\) is \(\vec{r}=\vec{r}^{\ \prime}\). at time \(t\) the position vectors are \(\vec{r}(t)\) and \(\vec{r}^{\ \prime}(t)\) and coincide. at time \(t+\Delta t\) the vector \(\vec{r}(t)\) is still the same, while the vector in \(S’\) has undergo a rotation and is now \(\vec{r}^{\ \prime}(t)=\vec{r}(t)+\vec{\omega}\times\vec{r}(t)\Delta t\). the current position of the mass at \(t+\Delta t\) is now \(\vec{r}(t+\Delta t)=\vec{r}^{\ \prime}(t+\Delta t)\). Therefore
\begin{equation}
\vec{r}(t+\Delta t)-\vec{r}(t)=\vec{r}^{\ \prime}(t+\Delta t)-\left[\vec{r}^{\ \prime}(t)-\vec{\omega}\times\vec{r}(t)\Delta t\right]
\end{equation}
so that the velocity vectors are related by
\begin{equation}
\vec{v}(t)=\vec{v}^{\ \prime}(t)+\vec{\omega}\times\vec{r}^{\ \prime}(t)\ .
\end{equation}
We learn from the preceding equation the rule for differentiation with respect to time:
\begin{equation}
\left.\frac{d}{dt}\right|_{in}=\left.\frac{d}{dt}\right|_{non in}+\vec{\omega}\times
\end{equation}
To get the acceleration we apply the rule of differentiation to the velocity:
\begin{equation}
\left.\frac{d}{dt}\right|_{in}\vec{v}=\left.\frac{d}{dt}\right|_{non in}\left[\vec{v}^{\ \prime}(t)+\vec{\omega}\times\vec{r}^{\ \prime}(t)\right]+\vec{\omega}\times\left[\vec{v}^{\ \prime}(t)+\vec{\omega}\times\vec{r}^{\ \prime}(t)\right]
\end{equation}
which is equivalent to
\begin{equation}
\vec{a}=\vec{a}^{\ \prime}(t)+2\vec{\omega}\times\vec{v}^{\ \prime}(t)+\vec{\omega}\times\left[\vec{\omega}\times\vec{r}^{\ \prime}(t)\right]
\end{equation}