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Fast Particle in a Bowl

  1. Jul 2, 2005 #1
    I'm trying to solve a problem from the book of qualifier problems by Cahn, but I don't see how he got his solution.

    The problem is about a particle constrained to move on a smooth spherical surface with radius R. The particle starts at the equator of the sphere with an angular velocity of [tex] \omega [/tex], and the particle is fast in the sense that [tex]\omega ^2 R >> g[/tex].

    The problem is to show that the depth z below the level of the equator is

    [tex]z \approx \frac{2g}{\omega ^2} \sin ^2 \frac{\omega t}{2}[/tex].

    I found that the energy of the particle can be used to write:

    [tex] \frac{1}{2} [ \frac{(1+z^2)\dot{z}^2}{R^2-z^2} + \frac{\omega ^2 R^4}{R^2-z^2}]-gz= \frac{R^2 \omega ^2}{2}[/tex]

    using cylindrical coordinates.

    But the solutions claim instead that

    [tex]\frac{1}{2} [ \frac{R^2 \dot{z}^2}{R^2-z^2} + \frac{\omega ^2 R^4}{R^2-z^2}]-gz= \frac{R^2 \omega ^2}{2}[/tex].

    I don't see why I'm getting a different answer, or how these could be the same.

    Also the solutions claim that the condition [tex]\omega ^2 R >> g[/tex] leads to [tex]z<<R[/tex]. I can see intuitively why that would be true, but I'm not sure how to show it formally.

    Any help would be greatly appreciated.

  2. jcsd
  3. Jul 2, 2005 #2
    I assume that the particle moves outside the spherical surface, now draw the free body diagram. Normal reaction due to surface of the metal , mg downwards, friction (if it exists) between particle and surface.Now take the components . write the expression for centripedal force. The radius can be found out by using the 'z' coordinate.

  4. Jul 3, 2005 #3
    I don't think it's necessary to deal with the forces. Since we know the angular velocity of the particle at the equator of the sphere, we know its initial energy which we can equate to a general expression for the energy of the particle.

    [tex]\frac{m}{2}[ \dot{\rho}^2 + \rho ^2 \dot{\phi}^2 +\dot{z}^2]-mgz=\frac{mR^2 \omega ^2}{2}[/tex]

    So here z=0 is the equator, and the left hand side is the energy in general and the right hand side is the energy at the equator.

    Then we can use the constraint that the particle is stuck on the sphere to express the energy just in terms of z and [tex]\dot{z}[/tex].

    [tex] \rho^2=R^2-z^2[/tex]

    [tex] \dot{\rho}=\frac{-z \dot{z}}{\sqrt{R^2-z^2}}[/tex]

    And to get rid of [tex]\dot{\phi}[/tex]:

    [tex]\dot{\phi}=\frac{\omega R^2}{\rho^2}[/tex]

    That way we only have to solve one first order diff-eq. Now substituting the expression for [tex]\dot{\rho}[/tex] and [tex]\dot{\phi}[/tex] gives me:

    [tex] \frac{1}{2} [ \frac{(1+z^2)\dot{z}^2}{R^2-z^2} + \frac{\omega ^2 R^4}{R^2-z^2}]-gz= \frac{R^2 \omega ^2}{2}[/tex]

    My problem is that this expression doesn't match what the solutions show. Instead they get:

    [tex]\frac{1}{2} [ \frac{R^2 \dot{z}^2}{R^2-z^2} + \frac{\omega ^2 R^4}{R^2-z^2}]-gz= \frac{R^2 \omega ^2}{2}[/tex]
  5. Jul 3, 2005 #4
    Never mind. I made a really stupid mistake in simplifying after making the substitutions. I get the same result as the solutions now. And to get z<<R you have to set up an effective potential and set it equal to zero.
  6. Jul 3, 2005 #5

    George Jones

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    We all make these silly mistakes, but there's an easy technique that, surprisingly often, makes thes mistakes glaringly obvious - dimensional analysis. That [itex](1 + z^2)[/itex] seems to have a problem with units almost jumps off the page, er, computer screen.

    Once, this helped me in a pure math course - real analysis! The prof often would devise "interesting" questions for the test, i.e., test questions not based on assignment questions or examples from the lectures. I was pondering one such question on the theory of integration and getting nowhere, so I decided to see if giving plausible physical dimensions to each of the quantities in the question would help. This seemed to indicate that the question was impossible, so I told the prof that I thought there was a mistake in the question. After working on the question for a bit, he a agreed, and gave the class a corrected version.

    He asked me "How did you find this out?" When I replied "The units didn't match up," he just gave me a puzzled look. He was a pure mathematician, with no background in any of the physical sciences.

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