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Moment of inertia of rectangle without dd int

  1. Aug 2, 2014 #1
    1. The problem statement, all variables and given/known data
    I found the moment of inertia of a rectangle, with the axis perpendicular and going through the center using double integration. But I also used what i thought to be an equivalent method, but it didn't work.


    2. Relevant equations
    [tex]dI=r^2dm[/tex]
    [tex]\int dI [/tex]





    3. The attempt at a solution
    I was able to solve the double integration of the problem. But before that I tried using the double integrals, I used this method [tex]4((\int^\frac{a}{2}_0\frac{b^2}{4}+x^2)dx + (\int^\frac{b^2}{2}_0\frac{a^2}{4}+y^2))dy[/tex]

    As you can imagine I toke the integral of the equation [tex] x^2+y^2[/tex] by one of the length like b/2 as constant and integrating over the width a, from a/2 to -a/2 and because its an even function it be taken from a/2 to 0 and you can take a 2 out. And lastly add another to for the symmetry, do the same for the same for the onther side.

    You dont get the correct answer which is [tex]M\frac{a^2+b^2}{12}[/tex] and I dont understand why?
     
    Last edited: Aug 2, 2014
  2. jcsd
  3. Aug 2, 2014 #2

    vela

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    That should be ##dI = r^2\,dm##. You can't have a lone differential just sitting there.
    The simple answer is you did stuff that you're not justified doing. I'll assume you're using ##dm = dx\,dy##. Based on symmetry, you can say that
    $$ \int (x^2+y^2)\,dm = 4\int_0^\frac a2 \int_0^\frac b2 (x^2+y^2)\,dy\,dx,$$ but how can you say that
    $$4\int_0^\frac a2 \int_0^\frac b2 (x^2+y^2)\,dy\,dx = 4\left[ \int^\frac a2_0 \left(\frac{b^2}{4}+x^2\right)\,dx + \int^\frac b2_0 \left(\frac{a^2}{4}+y^2\right)\,dy\right]?$$How do you justify setting ##y=b/2## in one integral? Why couldn't it be ##y=b/4##? (I'm not saying that's right, but it would make more sense to use the average value of ##y## rather than the maximum.) How can you justify setting that variable to a constant value at all?

    In math, you need to be able to justify each step. If you think you can and then get the wrong answer, only then you can ask why. At this point, to me, it just appears like you assumed certain things were true without having a good reason to. It's no surprise then that your calculation didn't work out.
     
  4. Aug 2, 2014 #3
    My calculation are base on the physical interpretation of the event, the length and width of the rectangle b and a, so by making the one of them constant, say b/2 and taking the integral from a/2 to zero of x^2+(b^2/4) and multiplying that by 4 I get the moment of inertia of 4 section of the rectangle and I do the same for the other 4. The place where I believe I went wrong was the density of the mass, I use M/ab, instead of the the two M/length. Then you would have r^2/r for the integral equation which turns into the equation sqrt(x^2 + (b^2/4)).
     
    Last edited: Aug 3, 2014
  5. Aug 3, 2014 #4

    vela

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    What justification do you have to set y=b/2 in that integral? Can you articulate that? It's not enough to simply say you based it on the "physical interpretation of the event," whatever that means.

     
  6. Aug 3, 2014 #5
    Now I get why you stress the b/2, by taking the b/2 as constant Im not considering the moment of inertia of the inner sections of the rectangle. Alright go it, thanks, I'll just use double integrals.
     
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