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How is the integral expression for contact time derived?

  1. Jan 14, 2016 #1
    Don't know if this is the correct place to post this, this is not an assignment question, but I am terribly stuck with this.

    While going through the derivation of contact time for a hertzian contact as given in problem 3 at the following link http://s17.postimg.org/t1kq6mlxr/Capture.png , I am not able to understand how the integral form for contact time has come into picture. I understand that the twice the integral of displacement over velocity from x=0 to x=x0 gives the total contact time. But can anyone please explain the in-between steps to get the same (how did the 1/sqrt(1-(x/x0)^5/2) come about? I understand that this is a very trivial problem but it will be a great help to understand the steps.
     
  2. jcsd
  3. Jan 14, 2016 #2

    Ray Vickson

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    You have an equation of the form
    [tex] \left( \frac{dx}{dt}\right)^2 + c\, x^{5/2} = K [/tex]
    so
    [tex] \frac{dx}{dt} = \sqrt{K -c\, x^{5/2}}[/tex]
    (assuming ##dx >0##), hence
    [tex] dt = \frac{dx}{\sqrt{K -c\, x^{5/2}}} [/tex]
     
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