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Moment of inertia of a cuboid

  1. Apr 18, 2005 #1
    How do I calculate this about the z-axis, if the cuboid length is b in the y-direction, a in the x-direction and c in the z -direction?

    In my notes, I have I = ∫ r^2 dm = ∫ (x^2 + y^2) dm

    dm = ρdV = ρdxdydz

    This is what I did:

    I = ∫∫∫ (x^2 + y^2)ρ dxdydz

    I = ρ∫dz∫dy∫dx (x^2 + y^2)

    I = ρ∫∫dy [(1/3)x^3 + xy^2] {0->a}

    I = ρ∫∫[(1/3)a^3 + ay^2] dy

    I = ρ∫dz [(1/3)ya^3 + (1/3)ay^3] {0->b}

    I = ρ∫dz [(1/3)ba^3 + (1/3)ab^3]

    I = ρ[(1/3)zba^3 + (1/3)zab^3] {0->c}

    I = ρ[(1/3)cba^3 + (1/3)cab^3]

    I = (1/3)ρabc(a^2 + b^2)

    and ρabc = ρV = M, so I = (1/3)M(a^2 + b^2)

    However, the answer has (1/12) instead of (1/3). Where does the other 1/4 come from??

  2. jcsd
  3. Apr 18, 2005 #2


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    The moment of inertia depends on the axis about which you want to rotate.
    Your calculation is correct for the MOI about an axis that goes through the edge of the cuboid.

    The MOI about an axis through its center and parallel with the z-axis is (1/12)M(a^2+b^2).
  4. Apr 18, 2005 #3


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    The limits of integration need a little change...

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