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When does this ring leave the surface?

  1. Jun 4, 2014 #1
    1. The problem statement, all variables and given/known data
    In the figure below two point mass m, over the ring with mass M and radius R, are released from the rest in the highest point and there is no friction in the system. What is the maximum amount of m/M for which the ring doesn't leave the surface?
    View attachment untitled1.bmp



    2. Relevant equations
    According to the figure below

    [itex] mg cos\theta-N=mv^2/R [/itex]

    From conservation of energy we get

    [itex] v^2 =2gR(1-cos\theta) [/itex]

    View attachment untitled.bmp

    3. The attempt at a solution


    In my attempt, I get N as
    [itex] N=mg(3cos\theta -2) [/itex]
    Then I thing when the ring is in the threshold of leaving we should have
    [itex] 2Ncos\theta =Mg [/itex]
    But I can nor get the correct answer!
    Could anyone please help me?
     
    Last edited: Jun 4, 2014
  2. jcsd
  3. Jun 4, 2014 #2

    TSny

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    Check your signs in this equation. Otherwise, I think your approach looks good!
     
  4. Jun 4, 2014 #3

    BiGyElLoWhAt

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    So what's supposed to happen? The "small" masses 'm' swing around and (without hitting each other and stopping) reach the top and at some ratio m/M the ring "jumps" off of the table? Am I interpreting the quesion correctly?
     
  5. Jun 4, 2014 #4

    TSny

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    hokani,

    You need to think about the direction of the normal force acting on the point mass m when the ring is about to leave the surface. Is it in the direction that you drew in your diagram?
     
  6. Jun 4, 2014 #5
    No, we are concerning the problem when two masses haven't yet hit or they move conversely in opposite directions after collision. I think The ring would be in threshold of jumping from a value of m/M on. Find That value.
     
  7. Jun 4, 2014 #6
    I think considering the direction this way and proceeding to solve the problem would give the correct direction of N. The minus value means it is in opposite direction.
     
  8. Jun 4, 2014 #7

    BiGyElLoWhAt

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    OK in that case I would probably look at forces. Are you familiar with apparent weight?

    http://hyperphysics.phy-astr.gsu.edu/hbase/elev.html

    This is a 'similar' situation, in the fact that you're looking for a particular case. Maybe this will jog some ideas.
     
  9. Jun 4, 2014 #8
    Right. I have corrected it in my main post. But I get minus value for m/M!
     
  10. Jun 4, 2014 #9

    TSny

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    I think you still have a sign mistake that's due to not having the correct direction for the normal force acting on the point mass m at the point where the ring leaves the surface.

    If the normal force on m is in the direction that you indicated, then what would be the direction of the normal force on the ring M? Would that tend to lift the ring off the surface?

    When the point mass m first begins to slide from the top, the normal force on m is in the direction that you indicated. But, think about what happens to the normal force as the point mass slides farther along the ring.
     
  11. Jun 4, 2014 #10

    AlephZero

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    No, the masses slide down not up. if the ring leaves the table, this happens before they reach the bottom of the ring and hit each other. In fact, it happens before they are half way down.
     
  12. Jun 4, 2014 #11

    AlephZero

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    To get the signs sorted out, I suggest you draw two free body diagrams: one for the ring, and one for a small mass.

    Hint: Newton's third law.
     
  13. Jun 5, 2014 #12
    The masses m start from rest at the top.
     
  14. Jun 5, 2014 #13
    I have obtained the value M=2/3m. But I am not confident about that.
     
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