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Equation of motion with proportional drag

  1. Dec 19, 2012 #1
    1. The problem statement, all variables and given/known data

    the forces will be like this

    [tex]m⋅dv/dt=-m⋅g-k⋅v[/tex]

    I need to find the velocity equation and the place equation, thereby meaning the integral of the velocity equation. for the condition t,v=0,v_0



    3. The attempt at a solution
    i have found this
    [tex]v(t)=mg/k*(-1+e^(-k/m⋅t))+v_0⋅e^(-k/m⋅t)[/tex] or in picture form http://imgur.com/WjksG

    But I am not sure it's correct
     
  2. jcsd
  3. Dec 19, 2012 #2

    ehild

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    It looks correct, but v0=0 .

    Edit: I misread vo, so it is correct.

    ehild
     
    Last edited: Dec 19, 2012
  4. Dec 19, 2012 #3

    haruspex

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    Your answer looks right to me. Why do you doubt it?
     
  5. Dec 19, 2012 #4

    haruspex

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    I read the OP condition as initially (t,v) = (0, v0)
     
  6. Dec 19, 2012 #5

    ehild

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    Sorry, I misread it. Edit my post.

    ehild
     
  7. Dec 19, 2012 #6
    that is what i meant. I doubt the results because when i draw them in a graph they dosen't seem to be correct. I have drawn them i geogebra if you are familiar with that software.
    My values are:
    m: 0,145 kg
    k: 0,0032
    g: (gravity acceleration) 9,82
    v_0: 9,93 m/s

    this is what i have entered in geogebra: g(x) =(0.145*9.82/0.0032)*(-1 + ℯ^(0x / 0.15)) + 9.93ℯ^(0x / 0.15)
    if you want to try for yourself (should save you some time). These values are not meant for a proportional drag, but still it seems weird that the graph looks like one of a constant function
    http://imgur.com/FFElN (the red one)
     
  8. Dec 19, 2012 #7
    Furthermore when i integrate the velocity equation i get this http://imgur.com/IO6G6
    Again not sure if I am correct
     
  9. Dec 19, 2012 #8

    haruspex

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    How did -k/m end up as 0 / 0.15? Looks like rounding error, and the red line graph seems to be a consequence.
     
  10. Dec 19, 2012 #9
  11. Dec 19, 2012 #10
    this is the equation of place (if that is the correct term in english) as far as I am concerned the height is not suposed to be negative to positive time values???
    http://imgur.com/CxAYC
     
  12. Dec 19, 2012 #11

    haruspex

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    Your velocity time graph shows velocity +ve at time 0, so the distance time graph should show distance increasing at time 0.
     
  13. Dec 20, 2012 #12
    Indeed it should, however it dosen't. Is this a wrong integration of the velocity equation?
    http://imgur.com/A6BKT
    Based on the gaph it does not seem to be correct.
     
  14. Dec 20, 2012 #13
    Just found the correct place equation i think
    http://imgur.com/quiLu
    can you see a good way to shoten it?
     
  15. Dec 20, 2012 #14

    haruspex

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    Looks right. I'd write it as
    [itex]\frac{mg}{k}\left(-t+\left(\frac{m}{k}+\frac{v_0}{g}\right)\left(1-e^{-\frac{kt}{m}}\right)\right)[/itex]
     
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