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Rotational Inertia of a Cube

  1. Jan 1, 2007 #1
    Does anyone know what the rotational inertia of a cube of uniform density is when it is rotated about an edge? Any help is appreciated!
     
  2. jcsd
  3. Jan 1, 2007 #2

    radou

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    I suggest you start with the definition of rotational inertia of a rigid body of uniform density.
     
  4. Jan 1, 2007 #3
    I=integral(r^2,m)

    but then what do I do?
    I'd use the parallel axis theorem but I don't know how to find the rotational inertia for a cube about it's center of mass.
     
    Last edited: Jan 1, 2007
  5. Jan 1, 2007 #4
    You need to be able to derive the equation; you can't look it up (in your textbook, etc.)?
     
  6. Jan 1, 2007 #5
    [tex]
    I = \int r^2 \,dm = \rho \int r^2 \,dV
    [/tex]
    where [itex]\rho[/itex] is the density of the cube (assumed to be uniform).

    You can then convert r into a Cartesian equivalent, then split dV into dx, dy, dz and do a triple integral.
     
  7. Jan 1, 2007 #6
    I think you should consider the cube from above a plane with greater mass--that is-- 1/12(m)(2L^2). Then parallel axis it.
     
  8. Jan 2, 2007 #7
    So would it be [tex]
    I = \iiint_{0}^{L} x^2+y^2+z^2 \,dxdydz
    [/tex] where L is the length of a side?
     
    Last edited: Jan 2, 2007
  9. Jan 2, 2007 #8
    Also, Ja4Coltrane, I don't understand what you mean. Could you please elaborate?
     
  10. Jan 2, 2007 #9

    AlephZero

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    r is the distance from the axis of rotation, not from the centre. If you are rotating about the z axis, then r^2 = x^2 + y^2.

    Re Ja4Coltane's post, possibly he means the inertia of the cube about one edge is essentially the same as the inertia of a square about one corner. The thickness contributes to the mass, but it doesn't affect the geometric part of the formula.
     
  11. Jan 2, 2007 #10
    So is [tex] I = \rho \iiint_{0}^{L} x^2+y^2+z^2 \,dxdydz [/tex] the rotational inertial of a cube rotated about a corner?
     
    Last edited: Jan 2, 2007
  12. Jan 2, 2007 #11

    AlephZero

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    What do you mean by "the rotational inertial about the origin"? I know what rotation inertia about a line is, and I know what the rotation inertia tensor about a point is (and it's got 6 independent components, not just one).

    The rotation inertia about the edge defined by (x=0, y=0) is [tex] I = \iiint_{0}^{L} x^2+y^2 \,dxdydz [/tex] That's very similar to the inertia of a square about one corner.
     
  13. Jan 2, 2007 #12

    radou

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    You can use the rotational inertia of a thin plate around the axis perpendicular to the plane of the plate to derive the moment of inertia of a cube around an edge.

    Edit: this link may be of some use: http://hypertextbook.com/physics/mechanics/rotational-inertia/.
     
    Last edited: Jan 2, 2007
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