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Physical intepretation of mathematical impossibility

  1. Sep 21, 2009 #1
    Hi! I would appreciate your thoughts on something. :smile:

    Let's say you have a ring with radius R rotating with angular velocity [tex]\omega[/tex] about a vertical axis. A little bead is threaded onto the ring, and the friction between the bead and the ring is negligible. The bead follows the ring's rotation, and will for a given [tex]\omega[/tex] place itself on a position on the ring which makes an angle [tex]\theta[/tex] with the vertical.

    By applying Newton's laws, one obtains the following equations for a given omega (N is the magnitude of the normal force from the ring on the bead):

    [tex] N cos(\theta) = mg [/tex]

    and

    [tex] N sin(\theta) = m \omega^2 (R sin(\theta)) [/tex].

    This gives a formula for the cosine of theta;

    [tex] cos(\theta)=\frac{g}{\omega^2 r}[/tex].

    So the problem is: what is the physical intepretation of what happens when omega gets so small that the right side of the equation exceeds 1?

    My thoughts are that, since N both has to balance the force of gravity and simultaneously create a centripetal acceleration, N obviously has to be larger than mg. When omega sinks below a certain value, that just isn't the case anymore. So if you were to have it at an angle theta at a high angular velocity and then gradually lower omega until you hit that lower bound, it will just slide to the bottom.

    Am I right?:redface:
     
  2. jcsd
  3. Sep 21, 2009 #2

    sophiecentaur

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    It is a funny one, isn't it?
    I think the fallacy / flaw / loophole is that N has to be greater or equal to mg. I think that keeps physical reality from imaginary roots to the trig equation. It must mean that there is a minimum omega, below which the ring stays at the bottom.
    I can't 'feel' that, though.
    For a 1m radius ring, omega would need to be greater than about root g!
     
  4. Sep 22, 2009 #3
    I think I see the problem. Look at how we solve the equations to get the formula for [tex]\theta[/tex] which you found. We take

    [tex] N \cos(\theta) = mg [/tex]

    and divide to get

    [tex] N = \frac{mg}{\cos(\theta)} [/tex]

    Then we substitute into the other equation

    [tex] N \sin(\theta) = m \omega^2 R \sin(\theta) [/tex]

    to get

    [tex] m g \frac{\sin(\theta)}{\cos(\theta)} = m \omega^2 R \sin(\theta) [/tex]

    We multiply each side by the quantity [tex] \frac{\cos(\theta)}{m \omega^2 R} [/tex] to get

    [tex] \frac{g}{\omega^2 R} \sin(\theta) = \sin(\theta) \cos(\theta) [/tex]

    At this point we divide by [tex] \sin(\theta) [/tex]. Now, this is only permissible when [tex] \sin(\theta) \neq 0 [/tex], so there are actually multiple solutions to this equation. There are the solutions which come from solving

    [tex] \sin(\theta) = 0 [/tex]

    and those which come from solving

    [tex] \cos(\theta) = \frac{g}{\omega^2 R} [/tex]

    The first equation always has the solutions [tex]\theta = 0[/tex] and [tex]\theta = \pi[/tex], whereas the second equation can only be solved for [tex] \omega \geq \sqrt{\frac{g}{R}} [/tex]. The second equation also only applies when we don't have [tex]\theta = 0[/tex] or [tex] \theta = \pi [/tex], as in those cases we would have divided by 0 to get it.

    Physically, this corresponds to saying that there is a minimum value of [tex]\omega[/tex], below which the only fixed points are those at the top and bottom of the ring.
     
  5. Sep 24, 2009 #4

    sophiecentaur

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    Looks good. I must say I had my doubts about canceling the sin thetas but I'm not too confident with what you can and what you can't do in that area. 'Back to first principles' involves going 'back' too many years!
     
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