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Rising of a hoop given the falling of beads

  1. Mar 9, 2017 #1
    1. The problem statement, all variables and given/known data
    A ring of mass M hangs from a thread, and two beads of mass m slide on it without friction, as shown. The beads are released simultaneously from the top of the ring and slide down opposite sides. Show that the ring will start to rise if m > 3M/2, and find the angle at which this occurs.

    2. Relevant equations


    3. The attempt at a solution
    First of all, I don't understand the situation physically. How does the ring move up exactly? What is the force propelling it upward?
     
  2. jcsd
  3. Mar 9, 2017 #2

    ehild

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    The whole system consists of the two beads and the loop. Initially, the loop hangs from the thread motionless, and the beads starts to slide down on it, accelerating with respect to the loop. The CoM of the whole system has a downward acceleration a, and the force corresponding to that is gravity and the tension in the thread, opposing it. When the CoM accelerates with g, the force of tension becomes zero, and the whole system is in free fall. But the beads accelerate downward with respect to the loop, so it has to accelerate upward.
     
  4. Mar 9, 2017 #3
    If the force of tension becomes zero, then what is accelerating the ring upward from a physical standpoint?
     
  5. Mar 9, 2017 #4

    collinsmark

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    If it helps, first imagine a slightly different problem. Consider the situation of two cars rolling off a cylinder (where the cars are not actually attached to the cylinder). Eventually, at some angle, they would roll off the edge of the cylinder, and would no longer touch the cylinder. (Don't take this too far though. The angle in question is different for this situation; it's not the answer you are looking for.)

    But in this problem (not my "different" problem above), the ring is "holding on" to the beads -- not letting them roll off the edge. What forces are involved in keeping these beads attached to the ring?
     
  6. Mar 9, 2017 #5

    ehild

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    The whole system would accelerate downward with g. If the downward acceleration of the beads are more than g, the loop must accelerate upward, to maintain the overall acceleration.
    As for the forces, there are normal forces between the loop and the beads. Those normal forces can accelerate the loop upward.
     
  7. Mar 9, 2017 #6

    andrevdh

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    Gravity is an external force to the system, so it is not similar to someone say walking forwards in a small boat and pushing the boat backwards.
     
  8. Mar 9, 2017 #7

    ehild

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    The forces acting on a bead is gravity and N, normal force from the loop. The forces acting on the loop are its weight, the tension of the string T and N', the force from the bead to the loop, equal and opposite to N.

    upload_2017-3-9_10-36-33.png
     
    Last edited: Mar 9, 2017
  9. Mar 9, 2017 #8

    ehild

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    It is not gravity, that pushes backward. Imagine you are in free fall, and have a stone with you. You throw the stone downward. Will your acceleration change during the time of throw, when you exert a downward force on the stone, so increasing its downward acceleration?
     
  10. Mar 9, 2017 #9

    andrevdh

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    Yes, your acceleration downwards will decrease, but I would think the blue N would then be on the bead and the orange N' on the hoop when the beads pass 90o from the top? (was hoping Mr Davis will respond)
     
  11. Mar 9, 2017 #10

    ehild

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    No, the bead is in general position. And the normal force on the bead can be both inward and outward. At small angles, when the speed of the bead is small, and the radial component of gravity is also small, it should point inward.
    Edit: At angles when the radial component of gravity is smaller than the centripetal force, the normal force points inward.
     
    Last edited: Mar 9, 2017
  12. Mar 9, 2017 #11

    andrevdh

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    That is what I was trying to say. When the bead is in the top half of the loop the normal force on it is outwards, but in the bottom half it is inwards. So the radial component of gravity and its speed (radial acceleration) detemines the magnitude (and direction?) of the normal force?
     
  13. Mar 9, 2017 #12

    ehild

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    I was not clear in the previous post. The normal force on the bead together with the radial component of gravity has to give the centripetal force. In the bottom half, the normal force must act inward. But in the top half, it can act in both ways.
     
  14. Mar 9, 2017 #13

    andrevdh

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    Aaaa, starting to get it! So one can use energy conservation and that should solve the problem - got it. I was not taking the centripetal acceleration into account!
     
  15. Mar 9, 2017 #14

    ehild

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    Yes, energy conservation is useful to get the speed of the beads at a given angle. But you need to find the condition that the tension in the string becomes zero.
     
  16. Mar 10, 2017 #15

    andrevdh

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    Thank you, think I can do it now (I didn't engage the problem fully - should have been able to solve it on my own). Will leave it to the op if he/she wants to know more.
     
  17. Mar 10, 2017 #16

    ehild

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    If the ring moves up the string becomes slack, so the tension is zero.
    Find the vertical acceleration of the beads in terms of the angle θ. Write y, and take the second derivative.
    Use conservation of energy to get the derivatives of θ. Derive the expression for the acceleration of the center of mass of mass, and find the condition, that it is equal to g.


    upload_2017-3-10_21-44-10.png
     
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