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Rising ring

  1. Jun 14, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-6-14_22-13-53.png
    upload_2017-6-14_22-13-23.png



    2. Relevant equations


    3. The attempt at a solution

    For ##0 \ge \theta \ge \pi/2##

    Forces on the ring,

    ##Mg + 2N\cos \theta = F\qquad 1##

    Forces on the beads ##mv^2/R = - N + mg \cos \theta##

    By conservation of energy when the bead has fallen through some angle ##\theta##,

    ##mg (2R) = \dfrac12 mv^2 + mg(R + R\cos \theta)##

    From this equation and above we get

    ##F = -4m\cos\theta + 6m\cos^2 \theta + M##

    The discriminent is ##\Delta = 16m^2 - 24 m M \ge 0 \implies m \ge \dfrac32M##

    The rise will start when the vertical force on the ring is zero,

    So ##\theta = \arccos\left( \dfrac{2m + \sqrt{4m^2 - 6Mm}}{6m}\right)##

    Is this correct ? why did we omit ##m = 3/2 M## ?
     
    Last edited: Jun 14, 2017
  2. jcsd
  3. Jun 14, 2017 #2

    mfb

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    2016 Award

    Staff: Mentor

    You consider different directions for the force on the ring and the force on the beads, that is a bit confusing.

    The formula for ##\theta## should have 4m in front of the square root.

    What do you mean by "omit m=3/2 M"? The ring doesn't have a net upwards force for small angles and it doesn't have it for large angles either. For ##m>\frac 3 2 M## there is a region in between where the ring starts to rise, for ##m \leq \frac 3 2 M## it does not rise.
     
  4. Jun 14, 2017 #3
    Why should I have 4m ? I canceled out 2 from top and bottom to get 2m from 4m.

    By the use of discriminent we got ##m \ge 3/2M##, which means ##m = 3/2M## is also the case but in the question it was given to prove that ring will only rise if ##m > 3/2 M##. So I was asking why not considering ##m = 3/2M## ?

    Can you give some justification for

     
  5. Jun 14, 2017 #4

    mfb

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    Ah sorry, got a factor 2 wrong.

    For m=3/2 M, the force on the ring gets zero at one point, but the ring doesn't rise because the force never points upwards in this idealized setup (in reality the thread would act as a spring, and the ring would go upwards slightly).
    The condition for a rising ring is F<0.
    That's what your formula for F shows. You can plot it.
     
  6. Jun 14, 2017 #5
    Should not it be ##F > 0## ? because else the ring will go down.

    Also if I put ##M = 0##, I get ##\theta \approx 48^\circ## which is less than ##\pi/2##. In upper semicircle the normal force is downward and weight of the ring is down, so every force is downward. Isn't it weird that massless ring will go up even though all the forces are down ?
     
  7. Jun 14, 2017 #6

    mfb

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    The way you defined your forces it has to be F<0. Without the beads, F=Mg which is clearly downwards.
    It changes from inward to outwards within the upper semicircle.
     
  8. Jun 14, 2017 #7
    No I want it upwards. Will ##-F = Mg + N\cos \theta## work?
     
  9. Jun 14, 2017 #8

    mfb

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    You can flip the signs of F and/or N, but the more you change them the more confusing it gets, especially if you don't start with a clear definition.
     
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