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Calculating Moment of Inertia for Simple Function

  1. Nov 7, 2006 #1
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    A uniform plate of height H = 1.96 m is cut in the form of a parabolic section. The lower boundary of the plate is defined by: [tex]y=.35x^2[/tex]. The plate has a mass of 7.27 kg. Find the moment of inertia of the plate (in kgm2) about the y-axis.

    I have absolutely no clue where to start... I know [tex]I=\int(r^2)dm[/tex] but I'm not sure how I can relate this to this problem.
     
  2. jcsd
  3. Nov 7, 2006 #2

    OlderDan

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    You would do well to start by finding the area of the plate. There is a double integral involved, and I think most calculus books do this problem two ways. Either way, one of the integrals is usually done for you by the way the problem is set up in the first place. Either you divide the plate into vertical strips of varying length L(x)(which is equivalent to doing the y integral in your head) and then integrating over x, or you divide the plate into horizontal strips of varying length L(y) (which is equivalent to doing the x integral in your head) and then integrating over y. For this problem, you will want to use the vertical strip approach because when it comes time to do the integral for the moment of inertia, vertical strips have only one r that applies to the whole strip.
     
  4. Nov 7, 2006 #3
    I started off by finding the value of x when y=H, which is 2.366 and -2.366. Then we know Area = [tex]\int(x^2dx)[/tex] between those values of x which is 3.09 m^2. I know [tex]\frac{dm}{dA}=\rho=\frac{M}{A}[/tex]. Therefore, [tex]dm=\frac{M}{A}dA[/tex]. We are given that [tex]\frac{M}{A}=2.353[/tex]. This leaves us with [tex]I=\int 2.353x^2dA=\int 2.353x^2dxdy[/tex]. I know how to do multiple integrals. Is that what is required here. I doubt it since this class is just a basic Physics class with Calc 1 and 2 prerequisites, no Multivariable stuff.

    What do I do next?
     
    Last edited: Nov 7, 2006
  5. Nov 7, 2006 #4

    OlderDan

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    See annotations. Fix the area integral. Once you get that and find the density correctly you will need to do the same type of integral but with the integrand multiplied by x^2 as you have done.
     
  6. Nov 7, 2006 #5
    OK, I've got correct area and density (I think) [tex]A=6.184[/tex] [tex]\rho=1.1756[/tex]. Now I'm stuck on the next part. I know [tex] \int((\rho)x^2dA)[/tex] is the equation I want, right? But how do I go from here? Is this a double integral? I tried just [tex]\int((\rho)x^2dx)[/tex] but didn't get the right answer. Where do I include the y? I'm lost, help!:confused:
     
  7. Nov 7, 2006 #6

    OlderDan

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    Assuming you used dA = [H - y(x)]dx for the area integral, you now do everything the same except this time you integrate x^2dA = x^2[H-y(x)]dx. ρ is a constant, so it is just a scale factor multiplying the integral.
     
  8. Nov 7, 2006 #7
    Thank you so much, this problem had me stumped. I'm pretty sure I got it now.
     
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