The motion of a ball thrown in a rotating frame of reference

Question

Assume, there are 2 people staying next to each other on a rotating disk, s.t. there is a straight line from one to the other going through the center of the disk. One throws a ball to the other.

I want to discribe the motion of the ball relative to the person, who threw it:

Let $\vec{r_p}$ be the position of the person and $\vec{r_b}$ be the position of the ball, relative to the rotation axis.

Then, the position of the ball, relative to the person $\vec{r}^p_{b} = \vec{r_b} - \vec{r_p}$. Therefore, the relative velocity is $$ \frac{d}{dt}\vec{r}^p_{b} = \frac{d}{dt}\vec{r_b} - \frac{d}{dt}\vec{r_p} = \frac{d}{dt}\vec{r_b} - \vec{\omega} \times \vec{r_p} $$ and the acceleration $$ \frac{d^2}{dt^2}\vec{r}^p_{b} = \frac{d^2}{dt^2}\vec{r_b} - \frac{d}{dt}\vec{\omega} \times \vec{r_p} - \vec{\omega} \times (\vec{\omega} \times \vec{r}_p) $$

Now, the question is, why am I losing the coriolis term here (i.e. where and why did it go wrong)?

Answer 1

$\def \b {\mathbf}$ The equations

\begin{align*} & u= r_b- r_p\quad\text{vector $~u~$ rotate with }~\omega\,t\\ &0=\dot{r}_b+\omega\times\,r_b-\dot{r}_p-\omega\times\,r_p= \dot r'+\omega\times r'\\ &0= \ddot r'+\underbrace{\omega\times\dot r'+\omega\times \dot r'}_{2\,(\omega\times v')}+\omega\times(\omega\times r') \end{align*}

you obtain

\begin{align*} &\ddot{r}'=-2\,(\omega\times\,v')-\omega\times(\omega\times r') \end{align*}