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Moment of Inertia about axis through body diagonal of a Cuboid

  1. Apr 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider a cuboid of lengths a, b and c along the x, y and z axes respectively, centred at the origin.

    yxeK5yE.png


    The task is to show that the moment of inertia of the cuboid of mass M and mass density ρ about an axis along the body diagonal, from (-a/2, -b/2, -c/2) to (a/2, b/2, c/2), assumes the form:

    ##I = \frac{M}{6}\frac{(ab)^2+(ac)^2+(bc)^2}{a^2+b^2+c^2}##


    2. Relevant equations

    This is supposed to be done by making use of previously determined results. These are, that the moments of inertia assume the form:

    ##Θ_{xx}=\frac{M}{12}(b^2+c^2)##
    ##Θ_{yy}=\frac{M}{12}(a^2+c^2)##
    ##Θ_{zz}=\frac{M}{12}(a^2+b^2)##

    and also, that all the products of inertia vanish.

    N.B. I proved these results by integrating between -a/2 and a/2, -b/2 and b/2, -c/2 and c/2, which is different to what I'm doing below - I don't know if this is correct!

    3. The attempt at a solution

    I attempted to calculate the principal moments as follows.

    ##I_{11}=ρ\int_v y^2+z^2 dv##
    ##I_{11}=ρ\int_v y^2+z^2 dxdydz##
    ##I_{11}=ρ\int_0^adx\int_0^c\int_0^b(y^2+z^2)dydz##

    I'm not sure that these limits are correct. I'm integrating from 0 to a, 0 to b, etc because I'm starting at one corner and moving 'a' in the x direction, 'b' in the y direction, etc. Nevertheless this comes out as:

    ##I_{11}=\frac{M}{3}(b^2+c^2)##


    I then attempted to calculate the products of inertia as follows.

    ##I_{12}=-ρ\int_v xydv##
    ##I_{12}=-ρ\int_v xydxdydz##
    ##I_{12}=-ρ\int_0^c dz \int_0^b\int_0^a xydxdy##

    Again I'm not sure if these limits are correct, but this comes out as:

    ##I_{12}=-\frac{M}{4}ab##

    Now, the other principal moments and products of inertia will be different than these. I could calculate all 9 terms in the moment of inertia tensor in a similar way, but I cannot see how this will bring me closer to obtaining a simple solution of the form:

    ##I = \frac{M}{6}\frac{(ab)^2+(ac)^2+(bc)^2}{a^2+b^2+c^2}##

    Any help as to how to approach this problem would be appreciated.
     
  2. jcsd
  3. Apr 12, 2014 #2

    SteamKing

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    For the cuboid, centered at the origin, the principal moments of inertia should be identical to the
    Θ[itex]_{xx}[/itex], Θ[itex]_{yy}[/itex], Θ[itex]_{zz}[/itex] you were given. The cross-products of inertia must all be zero, since the cuboid has axes of symmetry coincident with all three coordinate planes, so I can't account for your calculating a non-zero result.

    Up to now, all you have done is confirm the given moments of inertia, and thus are no closer to solving the original problem. You should think about how the moments of inertia transform when the coordinate axes are rotated from their original orientation. For the second moment of area, this relationship can be easily derived. You should investigate the case of how 3-D integrals are transformed when the coordinate axes are rotated.
     
  4. Apr 12, 2014 #3
    I thought that the products of inertia only vanish if the object rotates about its principal axes, which in the case of a cuboid will be the x, y and z axes (am I right?), and since the body diagonal is not a principal axis, then the products of inertia will be non-zero.

    I have been thinking about the rotation of the coordinate axes. Is the following picture accurate?

    1zwlzj6.png

    In this representation, the ##e_z## axis is rotated towards the body diagonal by an angle which will be equal to ##tan^{-1}\frac{\sqrt{a^2+b^2}}{c}##.

    Two things come to mind when considering the transformation. Under O(3) passive rotations, the moment of inertia tensor transforms as follows:

    ##O(3): I → I'=OIO^T##

    Secondly, perhaps the determinant of the Jacobian matrix for the coordinate transformation would need to be considered.

    Are either of these approaches the correct way to proceed?
     
  5. Apr 12, 2014 #4

    SteamKing

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    It wasn't clear in the OP if you were trying to calculate the products of inertia about the coordinate axes or about the diagonal. The products of inertia for the cuboid about the coordinate axes are all zero due to symmetry.

    For the case of finding MOI about an arbitrary axis, the following articles may be helpful:

    http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node67.html

    http://en.wikipedia.org/wiki/Principal_axis_(mechanics)#Principal_axes

    This article should be especially helpful:

    http://www.eng.auburn.edu/~marghitu/MECH2110/C_4.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Apr 12, 2014 #5
    The key equation is Equation 4.8 in the last article you linked. The symmetry of this particular problem can be exploited, in that we do not need to consider "direction cosines" because the axis along the body diagonal is in the direction of the unit vector: ##\frac{a}{\sqrt{a^2+b^2+c^2}}\hat{i} + \frac{b}{\sqrt{a^2+b^2+c^2}}\hat{j} + \frac{c}{\sqrt{a^2+b^2+c^2}}\hat{k}##
    where a, b, and c are simply the lengths of the cuboid in the x,y and z directions respectively.

    Now examining Equation 4.8, the last three terms can be ignored if the products of inertia, about the body diagonal axis, are zero. How does one know if they are zero? Bearing in mind that the body diagonal is neither a principal axis or an axis of symmetry.

    Nevertheless if the products of inertia are taken to zero, then Equation 4.8 can then be written as:

    ##I_Δ=I_{xx}\frac{a^2}{a^2+b^2+c^2} + I_{yy}\frac{b^2}{a^2+b^2+c^2} + I_{zz}\frac{c^2}{a^2+b^2+c^2}##

    Hence:

    ##I_Δ=\frac{M}{12}(b^2+c^2)\frac{a^2}{a^2+b^2+c^2} + \frac{M}{12}(a^2+c^2)\frac{b^2}{a^2+b^2+c^2} + \frac{M}{12}(a^2+b^2)\frac{c^2}{a^2+b^2+c^2}##

    And through simple algebraic rearrangements:

    ##I_Δ=\frac{M}{6}\frac{(ab)^2+(ac)^2+(bc)^2}{a^2+b^2+c^2}##

    as required.

    So it seems the products of inertia were indeed zero. How do you determine if they are zero? I thought they were only zero if the object rotates about one of its principal axes, (is this even right?) but the body diagonal is not a principal axis. It is not clear to me whether the fact they are zero is to be determined by calculation or whether it is an obvious result that exploits the symmetries of the problem.
     
    Last edited by a moderator: May 6, 2017
  7. Apr 12, 2014 #6

    TSny

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    On the right hand side of 4.8, the components of the inertia tensor are with respect to the x, y, z coordinate axes of figure 4.3. Aren't these the principal axes in your calculation?
     
    Last edited: Apr 12, 2014
  8. Apr 12, 2014 #7
    Yes. The exam question previous to this asked me to verify that the products of inertia, with respect to the principal axes, were all zero, which I did. So, this explains why the last three terms in Equation 4.8 are zero, and so I am satisfied as to how to arrive at the final answer.

    Even so, I still have questions about products of inertia in general:

    1. Is it even correct to say, that if an object rotates about one of its principal axes then the products of inertia, about these axes, will be zero?

    2. If the object rotates about an axis which is not a principal axis, then will the products of inertia about this axis be non-zero?

    3. In addition, why do the principal axes of a cuboid lie along the x, y and z axes? Do the principal axes of objects necessarily have to coincide with the geometrical lines of symmetry of the object?
     
    Last edited: Apr 12, 2014
  9. Apr 12, 2014 #8

    TSny

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    Yes. Products of inertia are zero relative to principal axes. I think this is the definition of principal axes.

    One or two might be zero, but not all three. See example 4.4 on page 22 of the last reference given by SteamKing

    If an object has a symmetry axis and the mass has uniform density, then you can expect the symmetry axis to correspond to a principal axis. Even if an object has no symmetry axes, there will still be three mutually perpendicular principal axes.
     
    Last edited: Apr 12, 2014
  10. Apr 12, 2014 #9
    I see, thanks for the clarifications.
     
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