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Center of mass and moment of inertia of catenary

  1. Dec 26, 2013 #1
    1. The problem statement, all variables and given/known data
    A homogeneous catenary ##z=acosh(x/a)##, ##y=0## and ##x\in \left [ -a,a \right ]## is given. Calculate the center of mass and moment of inertia


    2. Relevant equations



    3. The attempt at a solution

    I started with ##x=at##, for##t\in \left [ -1,1 \right ]##, therefore y remains ##y=0## and ##z=acosh(ta/a)=acosh(t)##.
    Btw, a is constant greater than zero, not a sign of arcus or any other inverse function.

    Ok, now ##r(t)=(at,0,acosh(t))## and ##\dot{r}(t)=(a,0,asinh(t))## and ##\left |\dot{r} \right |=acosh(t)##

    So the length of the catenary is ##l=\int_{-1}^{1}acosh(t)dt=2asinh(1)##

    The catenary is symmetrical on x axis, but point z is given as ##z=\frac{\int_{-1}^{1}a^2cosh^2(t)dt}{\int_{-1}^{1}acosh(t)dt}=\frac{a^2}{2asinh(1)}\int_{-1}^{1}cosh^2(t)dt=\frac{asinh(2)}{4sinh(1)}=\frac{a}{2}cosh(1)##

    I would just like to know if this is ok, before I start calculating moment of inertia?
     
  2. jcsd
  3. Dec 26, 2013 #2

    vela

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    Your method looks fine, but I'm getting a different result for the integral of cosh2 t.
     
  4. Dec 26, 2013 #3
    Hmmm, lets take a look..

    ##a^2\int_{-1}^{1}cosh^2(t)dt=\frac{a^2}{2}\int_{-1}^{1}(1+cosh(2t))dt=\frac{a^2}{2}(t+\frac{1}{2}sinh(2t))## for ##t## form ##0## to ##2\pi ##. (I don't know how to type bar with integrating borders in latex...

    ##\frac{a^2}{2}(t+\frac{1}{2}sinh(2t))=\frac{a^2}{2}(1+\frac{1}{2}sinh(2)-1-\frac{1}{2}sinh(-2))=\frac{a^2}{2}sinh(2)##

    OR is it not true that ##cosh^2(t)=\frac{1}{2}(1+cosh(2t)## or do I have some other problems? :/
     
  5. Dec 26, 2013 #4

    vela

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    You messed up the constant term. It works out to 1-(-1) = 2, not 1-1=0.
     
  6. Dec 27, 2013 #5
    True... Thank you vela!
     
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