Q.Using a funnel and a marble or a ball bearing try to work out the situation in the above question (the ball inside a rotating funnel). Try to realize that as the marble goes towards the brim, its linear speed increases but its angular speed decreases. When nearing the base, it is the other way.

Roll a marble inside a funnel and watch its circles at different heights. …

For a ball circling at radius r on a funnel wall of semi-vertical angle θ, the steady-circle condition gives $v^2 = rg\tan\theta$ and hence $\omega^2 = v^2/r^2 = (g/r)\tan\theta$. So as the marble spirals OUT towards the brim, r grows: its linear speed $v=\sqrt{rg\tan\theta}$ grows with $\sqrt{r}$, while its angular speed $\omega=\sqrt{g\tan\theta/r}$ falls as $1/\sqrt{r}$. Nearing the base, r shrinks and the roles reverse: the marble's circles get slower in v but visibly FASTER in revo …

Verify with timing: count revolutions per second at a high circle vs a low circle — the ratio of frequencies should follow $n \propto r^{-1/2}$ …

Expecting v and ω to rise and fall together — on a funnel they are linked by $v=r\omega$ with $v\propto\sqrt{r}$, so ω must go as $1/\sqrt{r}$: larger circles …