Playground physics

With the transformation matrix $D = \left( \begin{array}{cc} \cos \omega t & \sin \omega t \\ - \sin \omega t & \cos \omega t \end{array}\right)$ we can derive the path of the ball in the rotational frame $\begin{aligned} & & \vec r\,' &= D \cdot (- r_0 + v_0 t) \vec e_y = (- r_0 + v_0 t) \left(\begin{array}{c} \sin \omega t \\ \cos \omega t \end{array} \right) \\ \Rightarrow & & \vec v\,' &= \frac{d \vec r\,'}{dt} = v_0\left(\begin{array}{c} \sin \omega t \\ \cos \omega t \end{array} \right) + (- r_0 + v_0 t) \omega \left(\begin{array}{c} \cos \omega t \\ -\sin \omega t \end{array} \right)\\ \Rightarrow & & |v'| &= \sqrt{v_0^2 + \omega^2(v_0 t - r_0)^2} = v_0 \sqrt{1 + \omega^2 (t - t_0/2)^2} \\ \Rightarrow & & \overline{v'} &= \frac{v_0}{t_0} \int_0^{t_0} \sqrt{1 + \omega^2 (t - t_0/2)^2} dt = \frac{2 v_0}{\pi} \int_{-\pi/4}^{\pi/4} \sqrt{1 + \tau^2} d \tau \\ & & & =\frac{v_0}{\pi}\left[ \tau \sqrt{1 + \tau^2} + \text{arcsinh}(\tau) \right]_{-\pi/4}^{\pi/4} \\ & & & = \frac{2}{\pi} \left[ \frac{\pi}{4} \sqrt{1 + \frac{\pi^2}{16}} + \text{arcsinh}\left(\frac{\pi}{4}\right) \right] v_0 \approx 1.095 \cdot v_0 \end{aligned}$

The average velocity is thus almost 10% higher than in the stationary reference system. The velocity value varies between the maximum value $v' \approx 1.27 v_0$ on the edge and the minimum value $v' = v_0$ in the center of the disk. The path is determined by the centrifugal force $\vec F_\text{cf}$ and the Coriolis force $\vec F_\text{C}$ given by $\begin{aligned} \vec F_\text{cf} &= m \vec \omega \times (\vec r\,' \times \vec \omega) \\ \vec F_\text{C} &= 2 m (\vec v\,' \times \vec \omega) \end{aligned}$ with the vector $\vec \omega = \omega \vec e_z$ of the angular velocity. The centrifugal force is directed radially outwards and brakes the ball on its way from $A$ to $O$ and accelerates it again from $O$ to $B$ . What is more decisive, however, is the effect of the Coriolis force, which only changes the velocity direction and deflects the ball to the right, so that the path curve corresponds virtually to a semicircle. Because of the ratio $| F_\text{C} / F_\text{cf} | > 3$ of the two forces, here the centrifugal force has only a minor role.