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Moment of inertiaFind the moment of inertia of the plate

  1. Mar 27, 2005 #1
    A uniform plate of height H=1.60 m is cut in the form of a parabolic section. The lower boundary of the plate is defined by: y=1.75x^2. The plate has a mass of 4.94 kg. Find the moment of inertia of the plate (in kgm^2) about the y-axis.

    I integrated 1.75x^2 to get .583x^3. Then put 1.60 in for x and multiplied by the mass 4.94. I got 4.02 kgm^2 as my answer, which isn't right. Can someone help?
     
  2. jcsd
  3. Mar 28, 2005 #2

    dextercioby

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    I have good reasons to believe that

    [tex] I_{y}=\frac{M}{S}\int_{0}^{1.6} dy\int_{0}^{\sqrt{\frac{y}{1.75}}} x^{2} \ dx [/tex]

    The area of the plate is

    [tex] S=S_{rectangle}-\int_{0}^{l_{x}} \left(1.75x^{2}\right) \ dx [/tex]

    ,where [itex] l_{x} [/itex] is the "x" coordinate of the point obtained by projecting the point of coordinates
    [tex]\left(\sqrt{\frac{1.6}{1.75}},1.6\right) [/tex]
    onto the Ox axis...

    I hope u see which rectangle i'm talking about.

    Daniel.
     
    Last edited: Mar 28, 2005
  4. Mar 31, 2005 #3
    I'm sorry but I have no idea what you've done in this problem...
     
  5. Mar 31, 2005 #4

    dextercioby

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    U've got a plate whose contour is made up of 2 right lines and an arch of a parabola...U have to find the plates' area which will be needed to evaluate the moment of inertia.

    Make a drawing and see how u can compute that area...

    Daniel.
     
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