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Tensor of inertia - hollow cube.

  1. Mar 26, 2014 #1
    Hi,
    1. The problem statement, all variables and given/known data
    I have found the tensor of inertia of a rectangle of sides a and b and mass m, around its center, to be I11=ma2/12, I22=mb2/12, I33=(ma2 + mb2)/12. All other elements of that tensor are equal to zero. I would now like to use this result to determine the tensor of inertia of a hollow cube of side a around its center of mass.


    2. Relevant equations



    3. The attempt at a solution
    I realise I have to use the parallel axis theorem. I have hence tried the following:
    I11=ma2/12 + m(a/2)2, which yielded the wrong answer.
    I know that the correct equation is I11=I22=I33=ma2/12+ma2/12+ma2/6+4(ma2/12 + m(a/2)2)=5/3*ma2
    I simply do not understand why this is correct. Could anyone please explain why this is the correct way to calculate the desired tensor of inertia? Also, why would I be summing all the diagonal elements in my tensor for the rectangle?
     
  2. jcsd
  3. Mar 26, 2014 #2

    TSny

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    Doesn't this give the contribution of only one of the sides of the cube? What about the other sides?
     

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