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Find centroid of catenary (y=cosh(x))

  1. Sep 1, 2014 #1
    1. The problem statement, all variables and given/known data
    A uniform chain hangs in the shape of the catenary y=cosh(x) between x=−1 and x=1

    Find [itex]\bar{y}[/itex]

    2. Relevant equations
    ∫[itex]\bar{y}[/itex][itex]\rho[/itex]ds=∫y[itex]\rho[/itex]ds


    3. The attempt at a solution
    I can define ds (some small segment of the arc) by [itex]\sqrt{1+sinh(x)^2}[/itex]dx.

    Also, y is given as y=cosh(x)

    If I take the integral above from -1 to 1 as:

    ∫cosh(x)*[itex]\sqrt{1+sinh(x)^2}[/itex]dx

    I get ~2.8

    The problem I have with this answer is that the maximum y of this catenary is y=cosh(1) which is below y=2. The minimum value of the catenary is y=cosh(0)=1 so the y centroid should be somewhere between those two points.

    What am I doing wrong? Thanks!
     
    Last edited: Sep 1, 2014
  2. jcsd
  3. Sep 1, 2014 #2

    BvU

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    What does your ##\int ds## yield ?
     
  4. Sep 1, 2014 #3
    Arc length. This turns out to be about 2.4 between x=-1 and x=1.

    edit: from my book: (The example they use is y=x^2 between 0 and 1.)

    xvZs6V9.jpg


    Am I supposed to go about calculating the integral and then setting it equal to the form with [itex]\bar{y}[/itex] in it to solve for [itex]\bar{y}[/itex]?
     
    Last edited: Sep 1, 2014
  5. Sep 1, 2014 #4

    Ray Vickson

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    Isn't that exactly what the last equation says?
     
  6. Sep 1, 2014 #5
    You have to solve for y_COG from the equation as written in your notes.
     
  7. Sep 1, 2014 #6

    LCKurtz

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    If you factor the constant ##\bar y## out of the second equation of 3.6 and solve for ##\bar y## you will have the formula you need. You have just calculated part of it.
     
  8. Sep 1, 2014 #7

    BvU

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    Correct. And the 2.8 you had before divided by the 2.4 arc length gives you ##\bar y## as per your own relevant equation in post #1. Done !
     
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