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Gravitation and Archimedes principle as a spring?

  1. Jul 30, 2011 #1
    Hey all,

    I have seen that these two forces combined together have the potential energy of a spring with a constant 'K' as a combination of some sort of the two forces.
    I have no idea how am I to find this K.
    All I know
    Mgh + ρVgh gives the potential energy of a given system lets say.

    help?
     
  2. jcsd
  3. Jul 30, 2011 #2

    PeterO

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    Can you try to re-phrase what you are having difficulty with please?
     
  4. Jul 30, 2011 #3
    The problem was:
    A cylinder is placed inside water, and it is in equilibrium when half of the cylinder height is submerged. So I Know there is the archimedes principle which causes a force upwards and then there is gravity downwards. This problem was part of question somewhere, but in the explanation of the answer concerning the energies of the system when it is in motion, they somehow did not use mgh of either the ρVgh. They did a combination of those to act as a spring, which I did not get a clear answer on how to do that.
     
  5. Jul 30, 2011 #4

    tiny-tim

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    hi uriwolln! :smile:

    if the restoring force is proportional to the distance from the equilibrium position,

    the the equation is exactly the same as the equation for a spring :wink:
     
  6. Jul 30, 2011 #5
    :)
    Did not think of it this way.
    But then, what will be 'k' equal to?
    How do I work the algebra for it?
    and will the potential energy of it all will indeed be k(x^2)/2?
     
  7. Jul 30, 2011 #6

    tiny-tim

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    try it and see :wink:
     
  8. Jul 30, 2011 #7
    So lets see:
    kx=mg - ρVg ==> k = (mg - ρVg)/2 ?
    and then potential energy: k(x^2)/2?
     
  9. Jul 30, 2011 #8

    tiny-tim

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    but V depends on x :redface:
     
  10. Jul 30, 2011 #9
    :)
    true.... forgot that for a sec..... but is this the equation I should go for?
    K being a function of x? Then the whole potential energy is going wrong, because K is not constant, or does it? ( I think K should be constant to use k(x^2)/2)
     
  11. Jul 30, 2011 #10

    tiny-tim

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    erm :redface: … first, find the equation

    then analyse it! :wink:
     
  12. Jul 30, 2011 #11
    Here goes:

    Kx = (pir2H)g - ρ(pir2(H-x)g/2)
    K = (pir2Hg/2x) + ρ(pir2g)/2
     
  13. Jul 30, 2011 #12

    tiny-tim

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    no, you mean mx'' = …

    carry on from there :wink:
     
  14. Jul 30, 2011 #13
    mx?
    why am I meaning mx? m as in mass? and x as displacement?
    Then where does K come in?

    (I am not being disrespectful I hope, I really liked your comment, which made me laugh, because I really do not know why should I use mx)
     
  15. Jul 30, 2011 #14

    tiny-tim

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    no, mx''

    (ie md2x/dt2)
     
  16. Jul 30, 2011 #15
    :)
    ρcpir2Hx'' = ρcpir2Hg - ρl(pir2(H/2 - x)
    x'' + x(ρlg/ρcH) = g(1-ρl/2ρc)
    Which x then I know is oscillating.
    But then where does K come in?
     
  17. Jul 30, 2011 #16

    tiny-tim

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    you have x'' = -Ax + B

    if it was just x'' = -Ax, what would k be? :wink:
     
  18. Aug 1, 2011 #17
    then K should be:
    k = (ρlg/ρcH)
    So potential energy is as well 0.5Kx2?
     
  19. Aug 1, 2011 #18

    tiny-tim

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    yes :smile:, but I'm not sure whether you're just guessing :wink:

    how did you get rid of the B?
    maybe … what's x?
     
  20. Aug 1, 2011 #19
    :)
    actually I did guess.
    Well, I mean, I figured because it was the coefficient of x, then it would be reasonable to assume that, that is K.
    X would be the answer for the differential equation I got previous.
     
  21. Aug 1, 2011 #20

    tiny-tim

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    :biggrin:
    ok, sensible guess for k :smile:, but it doesn't tell you how to combine it with the distance (in kx or 1/2 kx2) :redface:

    when you have x'' = -Ax + B,

    rewrite that as (x - B/A)'' - -A(x - B/A) …

    now you have it in the same form as the standard spring equation :wink:
     
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