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Math Help - Moment of inertia.

  1. #1
    Super Member Showcase_22's Avatar
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    Moment of inertia.

    Find the moment of inertia with respect to the y axis of a thin sheet of constant density \delta=1 bounded by the curve y=\frac{sin^2(x)}{x^2} and the interval \pi \leq x \leq 2 \pi.
    I need to evaluate I_y=\int_{\pi}^{2 \pi} \int_0^{\frac{sin^2(x)}{x^2}} y^2 \ dy \ dx=\frac{1}{3} \int_{\pi}^{2 \pi} \frac{sin^6 (x)}{x^6} dx

    As far as I know, this can't be integrated. This suggests that I have done something wrong with my initial integral.

    Can anyone see what it is?
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  2. #2
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    maybe not the way you want it done ...

    I = \int r^2 \, dm

    r = x

    dm = \delta \cdot y \, dx = \frac{\sin^2{x}}{x^2} \, dx

    I = \int_{\pi}^{2\pi} x^2 \cdot \frac{\sin^2{x}}{x^2} \, dx

    I = \frac{1}{2} \int_{\pi}^{2\pi} 1 - \cos(2x) \, dx

    I = \frac{1}{2} \left[x - \frac{\sin(2x)}{2}\right]_{\pi}^{2\pi}<br />

    I = \frac{\pi}{2}
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  3. #3
    Senior Member Twig's Avatar
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    hey skeeter

    when you write  dm = \delta \cdot y \, dx , does this come from  m = \delta \cdot V \mbox{ thus, in differential form } dm = \delta \, dV ?
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  4. #4
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    actually, it's dm = \delta \cdot dA since the region is planar, but you've got the right idea.
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  5. #5
    Senior Member Twig's Avatar
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    hi

    Ok I see. Thanks!
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  6. #6
    Super Member Showcase_22's Avatar
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    Sorry for the last reply, I had an exam!

    <br /> <br />
I = \int r^2 \, dm<br />
    Is this an alternate definition for the moment of inertia?

    In my book it has:

    I_x=\int y^2 \delta (x,y) dA

    I_y=\int x^2 \delta (x,y) dA

    I_0=\int (x^2+y^2) \delta (x,y) dA=I_x+I_y

    So I can see that what you've done is the last one, but haven't you used polar co-ordinates (ie. x^2+y^2=r^2)?
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