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Force diagram for ring/bead problem

  1. Nov 17, 2012 #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. 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
    F = ma
    Kinetic/potential energy


    3. The attempt at a solution
    jtkZN.png

    The ring is supposed to rise when the Normal force overcomes the gravitational force on the ring, but how is the Normal force supposed to point directly upward? There is no component of the Normal force that goes up.
     
  2. jcsd
  3. Nov 17, 2012 #2

    lewando

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    I assume the beads are constrained to travel along the ring and either miss each other or collide in a perfectly elastic manner at the bottom.
     
  4. Nov 17, 2012 #3
    Yes, but for our problem we don't care what happens after the ring rises (assuming the ring rises before the beads collide, in which case here the ring does indeed rise before the rings reach the bottom).
     
  5. Nov 17, 2012 #4

    lewando

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    I cannot visualize how the ring can rise when the beads are falling. I can, when the beads are rising, though.
     
  6. Nov 17, 2012 #5
    Right, I thought it was unintuitive as well. But apparently the mathematics of the physics work out, which is what I am trying to do with the force diagram, but I can't seem to get it
     
  7. Nov 17, 2012 #6
    The kinetic energy of the beads depends on their angular position. Find it, and find the corresponding velocity.

    Then, since the motion if circular you should be able to find out the centripetal acceleration of the beads at any given angle. From that, you should be able to figure out the forces the beads are exerting on the hoop; then find their vertical components.
     
  8. Nov 17, 2012 #7
    I've already found the KE/velocity: KE = mg*(R - Rcos(x)) and I can get the velocity from there. And I know the net force on the bead is N - mgcos(x) which is = m(v^2/R). But I don't know how to make any force diagram that will make the ring go vertically upwards.
     
  9. Nov 17, 2012 #8
    You have two forces exerted by a bead: one related to its curved motion, which acts radially outward; another is due to its weight, acting vertically down. You can find the vertical component of the radial force; the horizontal components of the two beads' radial forces will cancel each other.
     
  10. Nov 17, 2012 #9
    This is what I don't understand. If I decompose the Normal force and the gravitational force based on the axes I set up in my picture, there is a net force pointing southeast. So what cancels out here?
     
  11. Nov 17, 2012 #10

    tiny-tim

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    Hi PhizKid! :smile:
    I don't understand. :redface:

    The normal force on/from the beads is at an angle, and always has a component in the vertical direction (except at the ±90° positions).

    (but you don't need the forces …

    how can the center of mass accelerate faster than g? :wink:)
     
  12. Nov 17, 2012 #11
    But then you would need to make a new set of axes:

    PKYQv.png

    But I don't think this would be of any help?
     
  13. Nov 17, 2012 #12

    tiny-tim

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    what's new about vertical and horizontal? :confused:

    even the ancient greeks knew about them! :rolleyes:

    get on with it! :smile:
     
  14. Nov 17, 2012 #13
    I don't know how to solve this without using forces. But the center of mass isn't moving but it would move vertically upwards if the Normal force pointing upwards was greater than the force of gravity
     
  15. Nov 17, 2012 #14
    You have a radial force, at some angle from the vertical. You can surely project the force to the vertical and horizontal axes.
     
  16. Nov 17, 2012 #15
    The radial force is the same thing as the normal force, or is it something different?
     
  17. Nov 17, 2012 #16
    Not exactly. The centripetal acceleration is due to two forces: the reaction force (normal) and gravity. The radial component of this combined force is what I call the radial force (even though its direction is the same as that of the normal force).
     
  18. Nov 17, 2012 #17
    The only force to decompose is the gravitational force, in which one component is parallel to the normal force radially, and the other component is tangential to the ring. I don't understand where this is going
     
  19. Nov 17, 2012 #18
    Again, get the radial force first. That is easy to do, because all you need is the velocity and the radius, which you have. Then get the vertical component of this force, and subtract the force of gravity from it; this gives you the vertical component of the normal force, but mind the signs.
     
  20. Nov 17, 2012 #19

    TSny

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    I think you're trying to see conceptually how either bead can exert a normal force on the wire that has a vertical component that is upward. Near the top of the loop, the wire exerts a normal force on the bead that is radially outward, as you have drawn in your first drawing. So, the bead exerts a normal force on the wire that is radially inward (and therefore has a vertical component downward.) But think about what happens as the bead continues farther along the loop. You might be familiar with the problem of something sliding down a frictionless spherical surface (see attachment). At some point the object leaves the surface! What prevents the bead from leaving the wire loop?
     

    Attached Files:

  21. Nov 18, 2012 #20
    I'm not sure what "radial force" means. I thought the radial force was N - mgcos(x) = m(v^2/R)? What other force is there?

    The object leaves the surface because the Normal force becomes 0 since it loses contact. In this problem, the bead can't because it's attached to the ring.

    So do I have to think about the forces on the beads and the forces on the ring separately?
     
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