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Moment of Inertia about the line x=y

  1. Nov 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Let [tex]\Omega[/tex] represent a two dimensional material region who density is given by [tex]\rho[/tex](x,y). Establish an integral formula for the moment of inertia of the material region about the line y=x.


    2. Relevant equations
    The moment of inertia about the x axis: [tex]\int[/tex][tex]\int[/tex]x2[tex]\rho[/tex](x,y)dA



    3. The attempt at a solution
    I am attempting to transform the moment of inertia about the x axis to the line x=y, how would I go about doing this?

    Thanks for any help, Jim.
     
  2. jcsd
  3. Nov 28, 2009 #2

    Dick

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    The moment of inertia is the integral of r^2 rho(x,y) dA where r is the distance from the point (x,y) to the axis. That means the expression you gave is the moment of inertia about the y axis, not the x axis. Now you have to replace r in the integral with the distance from a point (x,y) to the line x=y. How many ways do you know to find the distance from a point to a line?
     
  4. Nov 28, 2009 #3
    Thanks for your help. You were right, I mistyped the formula for the moment about the x axis. Using points (x,y) and (y,x) I calculated r2=2x2-4xy+2y2 So the formula would be the integral of this multiplied by rho dA.
     
    Last edited: Nov 28, 2009
  5. Nov 29, 2009 #4

    Dick

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    That seems close, but aren't you off by a factor of four there? The distance from (1,0) to the line x=y is sqrt(2)/2. Isn't it? So that makes r^2 for (1,0) r^2=1/2.
     
  6. Nov 29, 2009 #5
    I see. I was using the wrong distance formula. So using the formula for the distance between a point (m,n) and a line Ax+By+c=0 the distance is r= |Am+Bn+C|/ sqrt(A2+B2)
    Using the point (x,y) and the line x-y=0 I calculated the distance r=|x-y|/sqrt(2)
    Does this sound correct?
     
  7. Nov 29, 2009 #6

    Dick

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    Yes, that sounds correct. So r^2=(x-y)^2/2.
     
  8. Nov 29, 2009 #7
    Yep, thanks so much Dick.
     
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