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I've had a look on the problem of a ball on a rotating circular track where the ball reaches a stable position at a fixed height that depends on the angular velocity and radius of the track only. For images and a video see https://www.leifiphysik.de/mechanik/kreisbewegung/versuche/kugel-rotierender-rinne (German language)

I do understand the derivation for the height $h=R-\frac{g}{\omega^2}$ and also understand that the derivation implicitly assumes $0\leq h\leq R$.

Taken together this suggests a lower bound for the angular velocity $\omega_\mathrm{min}=\sqrt{\frac{g}{R}}$ below which there is no stable motion. This value seems to correspond to the case where the gravitation force equals the centrifugal force at Radius $R$.

What I don't understand is the physical reason for such minimal angular speed and what it would have to do with rotation at radius $R$ (since in this case the radius of rotation would be closer to 0 than $R$). I imagine it has to do something with the forces, but don't really see how a minimum comes about.

One idea I had is that the shape of the track plays a role. Since it is rather flat in the center I would expect the normal force to be almost the same as the ball moves away from the center position, while the centrifugal force grows with $r^2$. However this is contrary to experiment, where the ball only starts to rise if the speed is sufficiently big.

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A stable equilibrium exists for all angular velocities. It is located at $r=0$ if $\omega^2 < g/R$, and at a finite $r$ if $\omega^2 > g/R$.

The motion of the ball in the rotating (non-inertial) frame can be described as motion in the potential field $$ U(r) = mg\,h(r) - \frac{1}{2} m \omega^2 r^2, $$ where $h(r) = R - \sqrt{R^2 - r^2}$. For $\omega = 0$, the potential profile is convex, so the equilibrium at $r=0$ is stable. The derivatives of $U(r)$ are $$ \frac{dU}{dr} = m g r \left( \frac{1}{R - h} - \frac{\omega^2}{g} \right), $$ $$ \frac{d^2U}{dr^2} = m g \left( \frac{R^2}{(R - h)^3} - \frac{\omega^2}{g} \right). $$ All equilibrium positions and their stability follow from these expressions. For $\omega^2 < g/R$, the minimum of $U(r)$ is at $r=0$ and stable. When $\omega^2 = g/R$, the curvature at $r=0$ vanishes, giving neutral stability. For $\omega^2 > g/R$, the point $r=0$ becomes unstable, and a new stable equilibrium appears at finite $r$, where $h = R - {g}/{\omega^2}$. The figure below illustrates the behaviour of $U(r)$.

enter image description here

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  • $\begingroup$ Thanks, that makes sense. Is there a way to understand why this happens, when $m\cdot \omega^2\cdot R = m\cdot g$ ? $\endgroup$ Commented Oct 5 at 14:24
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    $\begingroup$ @user1583209 Yes. Near the bottom, where $r \ll R$, the reaction force $\vec{F}_N$ is nearly vertical, so $F_N \simeq mg$. Its horizontal component is $F_{ZP} \simeq (r/R) F_N$. The equilibrium at $r=0$ becomes neutral when $F_{ZP} = m \omega^2 r$ for $r \to 0$, which gives $\omega^2 = g/R$. $\endgroup$ Commented Oct 5 at 15:13

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