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Moment of inertia of a cube along the diagonal.

  1. Feb 4, 2014 #1
    1. The problem statement, all variables and given/known data
    Calculate the moment of inertia of a cube which rotates along an axis along its diagonal.

    2. Relevant equations
    Moment of inertia definition: [itex]I = \int \rho (\vec{r}) \vec{r} ^2 dV[/itex]
    Angular velocity vector; [itex]\vec{\omega}=\omega (\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})[/itex]


    3. The attempt at a solution
    My biggest problem is on finding the position vector for the volume element. My try at it goes like this:
    [itex]
    \vec{r} = \vec{p}- \vec{p}·\vec{\omega} \frac{\vec{\omega}}{\omega}
    [/itex]
    Where [itex]\vec{p}[/itex] is the point of interest. With this we calculate the difference between the point's vector and a vector with the direction of [itex]\vec{\omega}[/itex] and magnitude equal to that of the component of [itex]\vec{p}[/itex] along [itex]\vec{\omega}[/itex]. This should be the shortest vector from the axis of rotation to the point. Am I right?

    I think, however, that there is an easier way to solve this. Isn't there any shortcut similar to the Steiner Theorem, only for rotates axes?

    Thank you.
     
  2. jcsd
  3. Feb 4, 2014 #2

    SteamKing

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    The moment of inertia values for a body form a tensor. Instead of trying to calculate the moment of inertia directly using the diagonal as an axis, calculate the moments using an axis aligned with the edges of the cube passing through the centroid and rotate them to the diagonal.

    This discussion is relevant:

    http://www.eng.auburn.edu/~marghitu/MECH2110/staticsC4.pdf [Broken]

    See Eq. 4.8 on page 5
     
    Last edited by a moderator: May 6, 2017
  4. Feb 4, 2014 #3
    The section you refer to talks about translation of axis. Isn't this problem more about a rotation of axes? Anyway, thanks for the link. I used the results on section 4.2 for a general axis and I obtained this:
    [itex] I = \frac{\rho}{\sqrt{3}}(-\frac{1}{2} ab^2c^2+\frac{1}{2} a^2bc^2-\frac{1}{2} a^2b^2c + \frac{2}{3}b^3)[/itex]
    Does it make any sense? It doesn't for me. Why is the horizontal dimension given appear on its own? Shouldn't it be symmetrical?
     
    Last edited by a moderator: May 6, 2017
  5. Feb 4, 2014 #4

    SteamKing

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    Eq. 4.8 (and Sec. 4.2) come before Sec. 4.3, Translation of Coordinate Axes.

    As to the validity of your calculations, I don't know. Why don't you show the full work up, starting with your calculation of the inertia tensor before rotation?
     
  6. Feb 5, 2014 #5
    Ah, right, I misread you answer.

    My calculations look like this:
    [itex]
    I = \int dV| \vec{u} x \vec{r}|^2 = \int dV \left | \begin{array}{ccc}
    \vec{i} & \vec{j} & \vec{k} \\
    \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\
    x & y & z \\ \end{array} \right | = \int dV \frac{1}{\sqrt{3}} (z - y, -z + x, y - x)^2 =
    [/itex]
    [itex]
    \int dV \frac{1}{\sqrt{3}} (z^2 -2zy + y^2 -z^2 + 2zx -x^2 + y^2 - 2yx + x^2) = \frac{1}{\sqrt{3}} \int dV ( -2zy + y^2 + 2zx + y^2 - 2yx) =[/itex]

    [itex] \frac{1}{\sqrt{3}} ( -\frac{ab^2c^2}{2} + \frac{a^2bc^2}{2} - -\frac{a^2b^2c}{2}+ \frac{2}{3}b^3 )
    [/itex]
     
    Last edited: Feb 5, 2014
  7. Feb 5, 2014 #6

    SteamKing

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    It's not clear what the length of the edge of the cube is.

    You also have 2 i's in your integral determinant.

    If the edge length of the cube is say 'a', then the inertia tensor of the cube of uniform density about its centroidal axes will have entries only on the main diagonal (the products of inertia are identically zero because of symmetry). These values on the main diagonal will be

    I[itex]_{xx}[/itex] = I[itex]_{yy}[/itex] = I[itex]_{zz}[/itex] = [itex]\frac{m}{12}[/itex]*(a[itex]^{2}[/itex]+a[itex]^{2}[/itex]) = [itex]\frac{m}{6}[/itex]*a[itex]^{2}[/itex]

    m = ρ a[itex]^{3}[/itex], where ρ is the density

    and off diagonal

    I[itex]_{xy}[/itex] = I[itex]_{yz}[/itex] = I[itex]_{xz}[/itex] = 0

    You can apply Eq. 4.8 to this tensor without evaluating the integral.
     
  8. Feb 5, 2014 #7

    AlephZero

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    Something went wrong in post #5, because the last term in your answer has the different dimensions (length^3) compared with the others (length^5).

    And if this is a cube, why are there three different dimensions a b and c?

    The really neat way to do this is not Eq 4.8, but look further down at sections 4.4 and 4.5. For a cube, the "ellipsoid of inertia" is not a general ellipsoid. The three axes are the sane length, so it is a sphere. That means the inertia is the same about any line through the centroid. The inertia tensor about any set of axes through the centroid is a diagonal matrix with the three equal terms on the diagonal.
     
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