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Falling Dumbbell

  1. Aug 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine force vs angle, that acts on the vertical wall from the falling dumbbell. Dumbbell falls without initial velocity. The mass of each bead of dumbbell is m.


    2. Relevant equations

    The problem need to be solved using only Newton's equations by writing them for each of the material points.

    3. The attempt at a solution

    I do not know how I find the acceleration of the points.
     
  2. jcsd
  3. Aug 29, 2013 #2

    tiny-tim

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    hi sergiokapone! :wink:

    start by drawing two free-body diagrams, one for each ball

    (call the positions of the two balls x and y, and find how x'' and y'' depend on θ and on each other)

    show us what you get :smile:
     
  4. Aug 29, 2013 #3

    haruspex

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    What wall? Is there a diagram missing?
     
  5. Aug 29, 2013 #4

    tiny-tim

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    images?q=tbn:ANd9GcTFdxNgkkuj_0J-yUYtCWa_3_c3ZQ-ScKJi3XiwiXGITWJK9Y0l.png

    wot? no wall? :biggrin:
     
  6. Aug 30, 2013 #5
    fd05004b914e9dc5f9299f38eea4f289.jpg
     
  7. Aug 30, 2013 #6
    x=lcosα
    y=lsinα
    l - the length of the conjunction
    x2+y2=l2
     
  8. Aug 30, 2013 #7

    tiny-tim

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    good morning! :smile:
    (ahh, it's that way round!)

    ok, that's the full diagram

    now draw the two free-body diagrams, one for each ball, marking the forces clearly on each
     
  9. Aug 30, 2013 #8
    I solved the problem. But in addition to Newton's laws, I had to use the law of conservation of energy.

    In the projections on the direction along the dumbbell:
    mgsinα-R=mv2/l
    R - reaction from the rod.
    x=lcosα
    y=lsinα
    Differentiating, we find velocity
    v=l dα/dt

    The law of conservation of energy: mgl=ml2dα/dt2/2+mglsinα

    Using Newton's third law, force, that acts on the vertical wall F=Rcosα
    or, finally
    F=mg(3sinα-2)cosα

    How can you do without the law of conservation of energy?
     
    Last edited: Aug 30, 2013
  10. Aug 30, 2013 #9

    haruspex

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    So far so good. What about the forces on the falling end?
     
  11. Aug 30, 2013 #10

    haruspex

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    Love it! Would he be Chad, Kilroy or Foo to you?
     
  12. Aug 30, 2013 #11
    See my previous post, please, where I have already written, it is the reaction force from the connecting rod and the force of gravity.
     
  13. Aug 30, 2013 #12

    tiny-tim

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    divide the reaction at the wall into Rx and Ry, and call the interaction between the two balls T

    do ∑F = 0 for the wall ball, and ∑F = ma for the other ball

    show us what you get :smile:

    he was (i understand) prolific in and after the second world war here in britain, announcing shortages, but i don't think he usually had a name, or was known by one (though apparently technically he was mr chad) :wink:
     
  14. Aug 30, 2013 #13
    OK
    1st ball:
    Rx=Tcosα
    Ry=Tsinα

    2nd (falling) ball:
    Tcosα=max
    Tsinα-mg=may
     
    Last edited: Aug 30, 2013
  15. Aug 30, 2013 #14

    tiny-tim

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    ok, now write ax and ay in terms of α (or write cosα and sinα in terms of x and y), and solve :wink:
     
  16. Aug 30, 2013 #15
    ???

    cosα = x/l
    sinα = y/l

    Concerning what to solve?
    I do not understand.
    Moreover, a(α) depend on second derivative of α, which is unknown.
    ax=-..α lsinα-l.α2cosα

    (.. - second time deriv, . - first time deriv )
     
  17. Aug 30, 2013 #16

    tiny-tim

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    might be easier to do it the other way …

    mx'' = Tcosα = Tx/l which is a lot easier to solve :wink:

    (write ' and '' for deriviatives)
     
  18. Aug 30, 2013 #17
    Real exponent?
     
  19. Aug 30, 2013 #18

    tiny-tim

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  20. Aug 30, 2013 #19
    x=aexp(sqrt T/ml t)
     
  21. Aug 30, 2013 #20

    tiny-tim

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    oh, hold on, T isn't constant, so you'll have to eliminate T first, before trying to integrate …

    that should give you the same equation you'd get if you treated it as a single rigid-body and took moments of forces about the corner
     
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