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I'm doing a physics problem in which a marble spins around a spinning bowl and both have angular velocity $\omega$. It rotates with radius $r$ around the central axis and the hemispherical bowl has radius $R$. I've solved for the radius $r$ in terms of $\omega$:

$$r=\sqrt{R^2-\frac{g^2}{\omega^4}}$$

But I can't figure out what this means when $\omega$ is really small ($<\sqrt{\frac{g}{R}}$ ).

Some hypotheses:

  • Something to do with friction
  • It falls through the center of the bowl (if it were hollow)
  • We're missing something in this model

Are any of these right? Is this more complex than it seems?

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Has somebody got a spinning bowl, a marble, and an experimentalist mindset ready? ;) [More seriously, why does the marble move in this setup at all? I imagine a marble spinning in the center of a bowl, but that's obviously not what you mean.] –  ACuriousMind Aug 8 at 1:06
    
If you show your derivation, it would be easier to comment. It can't be friction-there isn't any. You may have a small angle approximation you are using. It seems like $r$ should go smoothly to $0$ with $\omega$ –  Ross Millikan Aug 8 at 1:14
    
When you spin the ball slow enough, eventually it won't get off of the bottom of the bowl. –  Thoth19 Aug 8 at 1:23
    
how is $r$ to be related with $R$, i.e $r > R$ or $r < R$? –  Nikos M. Aug 8 at 2:04

1 Answer 1

up vote 4 down vote accepted

I have reproduced your calculation. If $\theta$ is the angle from the central vertical axis of the hemisphere to the ball, the tangential downward force is $g \sin \theta = g\frac rR$ The tangential upward force due to rotation is $\omega ^2 r \cos \theta=\omega ^2 r \sqrt {1-\frac {r^2}{R^2}}$ When $\omega \lt \sqrt{\frac gR}$ the downward force is always greater and the ball will sit at the bottom of the bowl. When $\omega \gt \sqrt{\frac gR}$ you get the right answer.

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