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Moment of inertia of a rectangular prism

  1. Jan 19, 2010 #1
    I was trying to derive the formula for the moment of inertia of a rectangular prism at its centre of mass. And ended up with this:


    Due to the limitation of my knowledge in Calculus, I was stuck here. Can anyone help me solve these two integral:

    [tex]\int arccos(\frac{a^{2}}{2r^{2}}-1)rdr[/tex]

    [tex]\int (\frac{\pi}{2}-arccos\frac{b/2}{r}-arccos\frac{a/2}{r})rdr[/tex]
  2. jcsd
  3. Jan 20, 2010 #2
  4. Jan 21, 2010 #3
    Yeah, the cuboid. Do you know how to do it? My formula doesn't look like it is going to work.
  5. Jan 21, 2010 #4
    I = ∫-y/2+y/2-x/2+x/2ρ r2 dx dy = ∫-y/2+y/2-x/2+x/2ρ[x2+y2] dx dy =2∫ox/2ρyx2dx +2∫oy/2ρxy2dy

    Where ρ is the density. So ρ = mass/xy

    You can take it from here.

    Bob S
    Last edited: Jan 22, 2010
  6. Jan 21, 2010 #5
    could you explain how this come about?
  7. Jan 22, 2010 #6
    (1/2)*M((1/2)*a)^2+(1/4)*M*(b^4*arccos((2*a^2-b^2)/b^2)*sqrt(-a^2*(a^2-b^2)/b^4)+2*a^4-2*a^2*b^2)/(sqrt(-a^2*(a^2-b^2)/b^4)*b^2*ab)+4*M*(int(((1/2)*Pi-arccos((1/2)*b/r)-arccos((1/2)*a/r))*r, r = (1/2)*b .. sqrt(a^2+b^2)))/ab
  8. Jan 22, 2010 #7
    I = ∫-y/2+y/2-x/2+x/2ρ r2 dx dy = ∫-y/2+y/2-x/2+x/2ρ[x2+y2] dx dy =2∫ox/2ρyx2dx +2∫oy/2ρxy2dy

    Where ρ is the density. So ρ = mass/xy

    I =(2m/xy)[ x·y3/(3·8) + y·x3/(3·8)] =(m/12)(y2 + x2)

    If x = y, then I = m·x2/6

    Compare this to a solid disk of diameter d: I = md2/8

    Bob S
  9. Jan 22, 2010 #8
    Oh, ya. It's just the idea that we have to use both dx and dy, and integrate them one after the other, also that even though x and y are in different intervals we only let the intervals on the intergral sign worry about it. I just can't grasp it.
  10. Jan 22, 2010 #9
    Do the integral for a solid disk of diameter d.
    Bob S
  11. Jan 22, 2010 #10
    ok, i would start like this: [tex]\sum r^{2}\Delta m[/tex]
    = [tex]\sum r^{2}\rho A[/tex]
    = [tex]\sum r^{2}\frac{M}{\pi(d/2)^{2}}2\pi r\Delta r[/tex]
    = [tex]8\frac{M}{d^{2}}\sum r^{3}\Delta r[/tex]
    = [tex]8\frac{M}{d^{2}}\int^{d/2}_{0}r^{3}dr[/tex]
    = [tex]8\frac{M}{d^{2}}\frac{(d/2)^{4}}{4}[/tex]
    = [tex]\frac{Md^{2}}{8}[/tex]

    Is this how you want me to do it? Or some other way?
  12. Jan 22, 2010 #11
    Looks good. Here is a slightly different way:

    Use r = d/2 and m= pi·d2/4

    I = ∫o2piorρ·r2 r·dr dθ

    I = 2·pi ∫orρ·r2 r·dr

    I = 2·pi·ρ·r4/4 = pi·ρ·d4/32 = m·d2/8

    Bob S
  13. Jan 22, 2010 #12
    Thanks, it works as it came out to be. I guess if I practise it more, the idea will come clearer.
  14. Jan 22, 2010 #13
    wait, how did the dy become y and dx become x? This was the derivation of the rectangular prism formula.
  15. Jan 22, 2010 #14
    It's a property of definite integrals.Integrating from -y/2 to 0 gives the same result as integrating from 0 to y/2
    Last edited: Jan 22, 2010
  16. Jan 22, 2010 #15
    I = ∫-y/2+y/2-x/2+x/2ρ r2 dx dy = ∫-y/2+y/2-x/2+x/2ρ[x2+y2] dx dy
    Let's take just the first term:

    I = ∫-y/2+y/2-x/2+x/2ρ·x2 dx dy

    Integrating by y first, then by x:

    I = ∫-x/2+x/2 ρ·y·x2 dx = 2∫o+x/2 ρ·y·x2 dx = [2ρ·y/3][x/2]3=2ρ·y·x3/24 = ρ·y·x3/12

    Now using ρ = m/xy we get

    I = m·x2/12 (just the first term; second term is m·y2/12 )

    Bob S
  17. Jan 22, 2010 #16
    Cool. Thanks Bob!!!
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