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Angular moment cube

  1. Dec 14, 2013 #1
    1. The problem statement, all variables and given/known data
    An homogeneous cube of mass M and side 2a spins around one diagonal of the faces with constant angular velocity w. Show that the size of the angular moment in relation to one of the fixed vertexes is [itex] \sqrt{\frac{43}{3}}Ma^2w [/itex]

    What I visualize here is a cube with one of its vertexes on the origin and then it spits in a way that the vertex on the origin is always keeps there.



    2. Relevant equations


    [itex]\vec{L}=\vec{L}_{cm;O} + \vec{L}_{rel cm} = I \vec{w} + I_{cm} \vec{w}[/itex]

    where I are inertia tensors




    3. The attempt at a solution

    I was able to calculate the inertia tensors:

    [itex] I =Ma^2 \begin{pmatrix} \frac{2}{3} & 0 & 0 \\ 0 & \frac{2}{3} & 0 \\0 & 0 & \frac{2}{3} \end{pmatrix} [/itex]

    [itex] I_{cm} =Ma^2 \begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\-1 & -1 & 2 \end{pmatrix} [/itex]

    but now I'm stuck on what the vector w should represent, I think it has something to do with euler angles.
     
  2. jcsd
  3. Dec 14, 2013 #2

    mfb

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    ##\omega## points along the axis of the rotation = the axis of fixed points.
     
  4. Dec 14, 2013 #3
    you mean like [itex]\vec{w} = (cos (\frac{pi}{4}) w ; 0 ; sen(\frac{pi}{4}) w) [/itex]? considering that the diagonal which upon the cube rotates is in the xOz plane.

    I tried that with no success.
     
  5. Dec 14, 2013 #4

    mfb

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    That's a complicated way to express that vector, but yes.

    What went wrong?

    I don't understand your Icm, and I don't think it is necessary to treat everything as tensors, but it should be possible.
     
  6. Dec 14, 2013 #5
    Sorry for the above post, somehow I managed to post a duplicate reply and now I cant delete it.

    Using the formulas for the first part of the angular momentum I got:

    [itex] L_{x}=\frac{2Ma^2}{3\sqrt{2}}
    L_{y}=0
    L_{z}=\frac{2Ma^2}{3\sqrt{2}}[/itex]

    for the second part of the equation:

    [itex] L_{x}=\frac{Ma^2}{\sqrt{2}}
    L_{y}=\frac{-2Ma^2}{\sqrt{2}}
    L_{z}=\frac{Ma^2}{\sqrt{2}} [/itex]

    adding:
    [itex]
    L_{x}=\frac{5Ma^2}{3\sqrt{2}}
    L_{y}=\frac{-2Ma^2}{\sqrt{2}}
    L_{z}=\frac{5Ma^2}{3\sqrt{2}}[/itex]

    then calculating the norm gives the result [itex] \frac{\sqrt{43}Ma^2w}{3} [/itex], well looks like this time I got it right, seems like I had transcribed the question wrong, the 3 goes outside the root.

    I just used the tensor formula but instead of integrating I assumed the whole mass was in the point of coordinates (a,a,a), for example:
    [itex] I_{11}=M[ (a^2 +a^2+a^2) - a^2] = 2Ma^2 [/itex]

    Honestly I didn't quite understand why we have to calculate the second component on the right side of the angular momentum equation since it only says "relatively to the origin", are both of the componets always necessary ? could you give me some insight on it? And perhaps tell me how you would solve the problem. Thanks!
     
    Last edited: Dec 14, 2013
  7. Dec 14, 2013 #6

    mfb

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    The moment of inertia of a cube for a rotation around its center of mass does not depend on the axis (as all 3 principal axes have the same moment of inertia), afterwards you just need the distance of the axis of rotation to the center of the cube and you can use the parallel axis theorem.

    I don't get that sqrt(43).
     
  8. Dec 14, 2013 #7
    Its just the result of the L vector norm :

    [itex]w\sqrt{(\frac{5Ma^2}{3\sqrt{2}})^2+ (\frac{-2Ma^2}{\sqrt{2}}) ^2 +(\frac{5Ma^2}{3\sqrt{2}})^2}=\frac{\sqrt{43}Ma^2w}{3}[/itex]
     
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