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Really lost.

  1. Oct 11, 2006 #1
    A hemispherical bowl of mass M rests on a table. The inside surface of the bowl is frictionless, while the coefficient of friction between the bottom of the bowl and the table is μ = 1. A particle of mass m (and negligible size) is released from rest at the top of the bowl and slides down into it. What is the largest value of m/M for which the bowl will never slide on the table?
     
  2. jcsd
  3. Oct 11, 2006 #2
    And my teacher gives me the hint that the double angle formulas will be helpful, but I have no idea where those would be helpful b/c I would assume the only key moment is when the particle is at the base of the bowl.
     
  4. Oct 11, 2006 #3

    berkeman

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    Wow, u=1? Are you sure that isn't a typo? Kind of a simple problem otherwise. Has to be a typo.
     
  5. Oct 11, 2006 #4
    Hmm, yes I didn't think u = 1 was very practical but I'm almost positive that it's not a typo, so then how it would it work?
     
  6. Oct 11, 2006 #5

    berkeman

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    Sorry, I've never seen u=1. That means pinned, IMO.
     
  7. Oct 11, 2006 #6
  8. Oct 11, 2006 #7

    George Jones

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    Assuming the ball doesn't move, find the acceleration of the particle, which, since the speed of the particle is changing, is not strictly centripetal. There are two forces acting the particle - gravity and the contact force of the bowl on the particle. Use Newton's second law to find the contact force.

    There are four forces acting on the bowl - gravity, static friction, the contact forcle of the particle on the bowl, and the contact force of the table on the bowl.

    No, it just means that in order to start the object sliding, the applied force must be greater than the normal force. For example, the coefficient of friction between the tires on drag racers and the track often is greater than one.
     
  9. Oct 12, 2006 #8
    I'm sorry, I'm still quite puzzled, is there anything else you can tell me about the problem that might aid me.
     
  10. Oct 12, 2006 #9

    George Jones

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    What are the tangential and normal components of the particle's acceleration?
     
  11. Oct 12, 2006 #10
    tangential acc = d|v|/dt

    normal acc = v^2/r = 2gh/r = Normal force of the bowl onto the ball ?
     
  12. Oct 12, 2006 #11

    George Jones

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    Yes.

    Now, write the x and y component equations of F = ma. I would introduce an angle [itex]\theta[/itex] that is measured from horizontal, so, initially, [itex]\theta = 0[/itex] for the particle.
     
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