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Moment of inertia

  1. Apr 2, 2009 #1
    Consider a thin plate of constant density which occupies the region in the first quadrant inside the curve:

    [itex]x^2+4y^2=4[/itex]


    Find moment of inertia about line x=-3


    Attempt to solution:


    [itex]y=\frac{\sqrt{4-x^2}}{2}[/itex]

    [itex]I(x=-3)=\frac{1\rho}{2}\int_0^2(x+3)^2\sqrt{4-x^2}[/itex]


    [itex]\text{Is there any easier way of integrating this thing without having to expand} (x+3)^2 \text{and then multiply it with}\sqrt{4-x^2}[/itex]
     

    Attached Files:

    Last edited: Apr 2, 2009
  2. jcsd
  3. Apr 2, 2009 #2
    In your solution for y(x), where did the 1/2 come from?

    I don't see any options except to do it piece by piece.
     
  4. Apr 2, 2009 #3

    [tex]y=\sqrt{\frac{4-x^2}{4}}[/tex]
     
  5. Apr 2, 2009 #4
    There is a conflict in your posted information. On the figure, it agrees with what you have used
    y(x) = (1/2) * sqrt(4-x^2)
    I see that now.
    In your original post, you also wrote
    LaTeX Code: x^2+y^2=4
    which leads to
    y(x) = sqrt(4-x^2)
    That is why I asked the question.
     
  6. Apr 2, 2009 #5
    Thanks for pointing out that mistake :smile:
     
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