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Hexagon problem (mechanics)

  1. Mar 15, 2009 #1

    StatusX

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    Here's a mechanics problem that seems pretty straightforward, but I can't get anywhere on it. You have a system of six rods arranged in a regular hexagon, connected by hinges at each vertex, giving some non-rigid structure. Now you apply a force to one of the rods in a direction perpendicular to it, towards the center of the hexagon. This will have the effect of both adding to the total momentum of the hexagon system, as well as squashing it. The question is, if after you apply the force, the rod you were touching gets a velocity v, what is the velocity of the opposing rod? Even a hint on how to approach this would be appreciated.
     
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  3. Mar 15, 2009 #2

    tiny-tim

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    Hi StatusX! :smile:

    Assuming the rod does not turn, how does the "front" half of the hexagon compare with the "back" half? :wink:
     
  4. Mar 15, 2009 #3

    StatusX

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    I'm not sure what you're asking. Its a 2D problem, and the rod you push (call it the bottom rod) doesn't turn, it just moves upwards. Then I'm asking for the upward speed of the top rod.
     
  5. Mar 15, 2009 #4

    tiny-tim

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    Yes, and so what hapopens to the other rods? :smile:
     
  6. Mar 15, 2009 #5

    StatusX

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    That's part of what you need to figure out. Presumably, the bottom rod exerts some torque on them, they push the rods above them, and those push the top rod. But I'm not sure how all the forces work out.
     
  7. Mar 15, 2009 #6

    tiny-tim

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    never mind the forces …

    what's the geometry? :wink:
     
  8. Mar 15, 2009 #7

    StatusX

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    Here's a picture. Assuming we apply F for a (short) time t so that the vertical velocity of the bottom rod is v, the question asks for the velocity of the top rod. To be more precise, let's assume (for now) the 6 vertices are point masses, and they're connected by massless rods. But I'm also interested in the case when the mass is uniformly distributed along the rods.
     

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  9. Mar 15, 2009 #8
    It looks like you can assume left-right symmetry, so you only have to analyze half the problem. Then start writing impulse and momentum relations, remembering that the reaction at each pin has two components.
     
  10. Mar 16, 2009 #9

    StatusX

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    Can you elaborate a little, maybe explain how this would work for the first joint? I know this is very simply, but I haven't done this stuff in a while and I'm getting stuck.
     
  11. Mar 16, 2009 #10

    tiny-tim

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    just woke up … :zzz:

    from which the geometry tells you that there must also be front-back symmetry :wink:

    so the velocities of the front and back rods must … ? :smile:
     
  12. Mar 16, 2009 #11
    I'm not quite sure what "front" and "back" refer to, but left and right refer to the two sides of the picture and the fact that they are just alike. What happens on the left must also happen on the right in exactly the opposite sense horizontally, and the same sense vertically. So if a point in the right side moves to the right and upward, the symmetric point on the left side moves to the left and upward. Thus you really only need to look at on side or the other.
     
  13. Mar 17, 2009 #12

    StatusX

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    Let me be more specific with what's confusing me. For this, let's focus on a simpler system: two horizontal rods or length L connected by a frictionless hinge:

    -------O-------

    Now I want to know what happens if I apply an impulse I to the left end of the left rod. If there was no second rod, it would just start moving upward at speed I/m and spinning clockwise at angular velocity 6I/mL. But how does the second rod affect this? I'm not even sure what forces are present at the hinge. Can someone clear this up?
     
  14. Mar 17, 2009 #13

    rcgldr

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    Maybe this will help. For a given force, the linear reactions (f = m x a, a = f/m), are independent of any angular reactions. The upwards acceleration of the center of mass of the 2 rod system is equal to the applied force divided by the mass of the 2 rod system.

    In the hexagon case, even during the period the hexagon is collapsing, the center of mass accelerates via the same rule (a = f/m).

    The rest of this seems pretty complicated, somewhat similar to statics and dynamics in engineering classes.
     
  15. Mar 17, 2009 #14

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    I know it's complicated, but the problem is I don't even know how one would do it in principle. That is, I don't care how to solve the equations, I just want to know what the equations are. Specifically, what forces act on the second rod?
     
  16. Mar 17, 2009 #15
    The velocities of the front and back rods are equal and opposite with respect to...
     
  17. Mar 17, 2009 #16

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    the center of mass? Even if that is true, I don't see how it helps, because I don't know the motion of the center of mass (I'm only given the velocity of the bottom rod, not the applied force).
     
  18. Mar 17, 2009 #17
    That's right, the center of mass. It helps to get all the symmetries of the problem lined-up. Maybe Tiny see's something I don't, but the problem, where the mass is at the hinge points, is easier. Otherwise, you may have to account for the angular inertia of each stick.
     
    Last edited: Mar 17, 2009
  19. Mar 17, 2009 #18

    tiny-tim

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    Hi everyone! :smile:

    I've been watching this thread with interest …

    I think that's the best we can do on the information given …

    there's no reason to believe the collision is elastic, so anther equation is needed.

    We could simplify the problem by transferring to a frame in which forces of F/2 are applied on each side, but we still seem to need another equation. :redface:
     
  20. Mar 17, 2009 #19

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    I don't think another equation is needed. Surely you can build such a hexagonal system and apply this force, and something will happen. What could be missing? And what collision are you talking about?
     
  21. Mar 17, 2009 #20

    tiny-tim

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    The collision between the bottom rod and the two it's connected to …

    if energy is lost (and it usually is), then don't we need to know how much?
     
  22. Mar 17, 2009 #21

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    Why would you call that a collision? The rods are always touching. In any case, assume everything is ideal: no friction in the hinges, or energy transferred as heat to the rods. Do you see some problem with making these assumptions?
     
  23. Mar 17, 2009 #22

    tiny-tim

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    So are the balls in "Newton's cradle" (that "executive toy").
    Not even any noise?

    Well, if energy is conserved, then it's just a geometry problem …

    go to that frame I mentioned earlier, so the centre of mass doesn't move, and apply conservation of energy :wink:
     
  24. Mar 17, 2009 #23

    StatusX

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    ???


    I don't know what that toy or noise has to do with anything, and I don't see a way to solve it with conservation of energy. Do you want to sketch what you have in mind?
     
  25. Mar 17, 2009 #24

    berkeman

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    Would it help to first look at the problem when the top rod is fixed? Calculate the motion and velocities of the side rods as the hexagon collapses. Then do it again, with only the 5 masses (the 5 rods) opposing the motion induced by the force on the bottom rod.... It looks solvable, just a number of equations (translation of 6 rods and rotation of 4).
     
  26. Mar 17, 2009 #25

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    But again, I don't see what equations I should be using. If I knew all the forces on each rod, then it would be easy. But I don't. For example, what are all the forces on the lower left rod? Does the upper left rod apply a force to it, and in what direction? If you could even explain the simpler two rod system I metioned earlier, it would really help me.
     
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