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Bead sliding down a smooth cord.

  1. Dec 21, 2012 #1
    Hello,
    1. The problem statement, all variables and given/known data
    I am asked to show that the time it would take a bead to slide down a smooth cord, positioned at an angle beta wrt the vertical axis, is independent on that angle (between the cord and the
    axis). The bead starts its slide from rest.

    2. Relevant equations



    3. The attempt at a solution
    I was initially not sure whether I should write down force equations or use energy conservation, so I have tried both (should I have used impulse, insead?).
    In any case, for the forces acting on the bead sliding down I got:
    mgsin(beta) = N ; mgcos(beta) - T = md2x/dt2
    And from conservation of energy I got:
    mgy = 1/2*mv2

    My problem is that I am really not sure any of these equations are flawless as they stand, AND they all seem to be dependent on the angle beta.
    Could someone please advise?
     
  2. jcsd
  3. Dec 21, 2012 #2

    lewando

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    Looks okay.
    Where did this come from? Is "T" tension? If so, not relevant.
    True. May not need it, though.

    Since time is important, you'll need an expression containing this. Use one of the standard kinematic equations. Try to come up with a general expression for time as a function of β. If you can do that then time is dependant on β. If your general expression for time is only dependant on m,g, and L, then you have shown otherwise.
     
  4. Dec 21, 2012 #3
    Why would T, tension, be irrelevant here? Doesn't "smooth" merely imply without friction?
    Moreover, supposing my choise of coordinate system is such where my x axis is parallel to the slope, then will the bead not be accelerating only along the x axis (hence, a_y = 0)?
    In any case, x = 1/2*a_x*t^2, where x should be L*cos(beta) with L denoting the distance the bead has passed along the cord. Hence, t = sqrt(2Lcos(beta)/a_x) where a_x = gcos(beta) (from the equation in my previous post). Thus, t = sqrt(2gL).
    Is it correct?
     
  5. Dec 21, 2012 #4

    lewando

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    Comments in blue
     
  6. Dec 21, 2012 #5
    Are you suggesting that t should have been 2L/[gcos(beta)], yielding t infinite for cos pi/2? Please remember that I am to show that the time of the slide from the top of the cord of slope tan(beta) is independent on beta.
     
  7. Dec 21, 2012 #6

    lewando

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    if you consider two cases, β = 90 and β = 0, you will get two different times, yes?
     
  8. Dec 21, 2012 #7
    It should, yes. But again, the time has to be independent on beta as that is how the question is formulated: "show that the time of the slide from A to B is not dependent on the angle beta." Just to make sure, you are stating that it should depend on the angle, right?
     
  9. Dec 21, 2012 #8

    lewando

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    Yes, that is what I am asserting. And you should be comfortable asserting that too, based on the 0/90 cases.

    So I am reviewing the original question and the assumption I am making is that the cord is in tension such that it is essentially rigid and stays that way (no letting go of the bottom end upon release of the bead). I can't think of any more gotchas.
     
  10. Dec 21, 2012 #9
    Would the expression for the time remain the same?
     
  11. Dec 21, 2012 #10

    lewando

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    I think you were trying to ask if t = sqrt(2L/(gcos(β))) was the result. I think it is.
     
  12. Dec 21, 2012 #11
    I have attached the question's diagram. Hopefully it would assist in somewhat clarifying the set up. The bead is sliding, from rest, from A to B, down the cord, as shown.
    I doubt I'd be asked to prove the independence of the time on the angle beta, were it not the case.
    I'd be happy to share your thoughts on this, as I do deem it rather perplexing.
     

    Attached Files:

  13. Dec 21, 2012 #12
    I mean, having looked at the diagram, can you now come up with an explanation why t = sqrt(2L/(gcos(β))) might not be the right answer?
     
  14. Dec 21, 2012 #13

    lewando

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    It appears that L is not constant. It also appears that we are talking about a chord not a cord. Thanks for sharing the diagram, it will be a big help moving forward.
     
  15. Dec 21, 2012 #14
    I never wrote L was constant. I actually explicitly stated that it was the distance traversed by the bead along the chord.
    Should it then indeed be sqrt(2L/g)?
     
  16. Dec 21, 2012 #15

    lewando

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    What is L in this expression then? It appears to be a constant. What it really is is the length of chord AB as a function of β.
     
  17. Dec 21, 2012 #16
    I am a bit confused myself. Are you claiming that t is dependent on the angle beta still?
     
  18. Dec 21, 2012 #17

    lewando

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    I am going to hold off on making any new claims for now :smile:. That was my claim based on incomplete information and making the assumption that L was constant. Now that all the cards are on the table, I am observing (rather than claiming) that the lenghth of the chord is a function of β, as you can observe as well. To move forward, can you come up with this relationship?
     
  19. Dec 21, 2012 #18
    I believe it should be L*cos(beta). I am simply wondering how you would account for the fact that the question specifically asks to demonstrate that t is NOT dependent on beta (assuming you are still convinced that it is). To put it more clearly - I am not suggesting you are indubitably wrong, simply that there seems to be a contradiction, wouldn't you reckon?
     
  20. Dec 21, 2012 #19

    lewando

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    I am not convinced it is. in my original understanding of the question, it was. Questions can be badly worded every now and then, so I was not alarmed at that possibility. Let's just see where this goes. Another thing to consider... I originally discounted the value of the energy approach, I now think that will come back into play.
     
  21. Dec 21, 2012 #20

    lewando

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    So after some scribbling, I see that t is not dependent on β. Presently, I cannot communicate it elegantly (though I'm pretty sure that's not my job :wink:). I can tell you the general approach I used was:
    1. Equate the PE lost in descending a vertical distance, h (from point A), to the KE at B.
    2. Find Vy at point B.
    3. Use that result in the kinematic equation h = 1/2(Vy)t, solve for t. Trouble is, you are left with a cosβ term and an h term (both dependent on β). For specific angles, 0, 30, and 45, you get specific h's (at fractions of the constant diameter) and these reduce to an expression for time with no β and no h. Unfortunately, I am unable to state this in a more general way at this time.
     
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