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Calculating Moment of Inertia of a cube

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the moment of Inertia of a cube (mass M, length L) around an axis going through one of the edges.

    2. Relevant equations

    I=Ʃmr2

    3. The attempt at a solution

    Well I imagined the cube is formed of infinitely many square plates. The width of the square plate is dL.
    ρ=M/L3
    dm=ρ*L2*dL

    so now im stuck with the integral what is r? is r = L√2/2 from pythagorean thm ?
     
  2. jcsd
  3. Dec 6, 2011 #2

    gneill

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    It would be better to deal with a linear density for the plates, rather than the volumetric density. Otherwise you'll need to be performing a triple integral in order to determine the volume of each plate as you move up the stack, and you'll lose all the benefits of approaching the problem as a stack of plates!

    Since the density of the cube is taken to be constant, you can safely make use of a linear density: if the height of the cube is L and the mass is M, then ρ = M/L is the linear density of the cube for the vertical direction. Thus, for a plate of thickness dz, the mass element becomes dm = (M/L)dz.
     
  4. Dec 7, 2011 #3
    Wow. But I still can not understand how we can consider a plate as a linear object? I mean, isnt that applicable to one dimensional rods ?
     
  5. Dec 7, 2011 #4
    Btw, your dm and the one i used are the same, arent they?
     
  6. Dec 7, 2011 #5

    gneill

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    We're modelling the cube as a linear stack of plates of thickness dz. Each plate has a mass dm. the sum of all the dm's is the mass of the cube, M.

    There is no need to calculate every dm as a volume density multiplied by L x L x dz. You can combine the ρv x L x L into a new "linear density" parameter ρ so that the mass of an individual plate is given by dm = ρ dz. It's a "linear" density insofar as it pertains to linear positions along the z-axis.

    They are, but mine don't require you to integrate over each plate's area as you go.
     
  7. Dec 7, 2011 #6
    Ok I will definitely think about it. Now we have ρ=M/L and thus dm=(M/L)dz

    SO I=∫dmr2
    I=∫M/Lr2dz

    Now I got confused how to integrate this one?
     
  8. Dec 7, 2011 #7

    gneill

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    What you want to integrate here is the moments of inertia of the plates. Now, you can either sum the moments about the centers of the plates to find the moment of inertia of the cube about a central axis and later apply the parallel axis theorem to move the axis to an edge, or you can apply the parallel axis theorem first to the moment of inertia of the individual plates. Your choice.

    First off, what's the formula for the moment of inertia of a square plate about its center?
     
  9. Dec 7, 2011 #8
    Well as far as I know, it is ML^2/6. But the problem is that I can not find it myself, I dont want to memorize them (the formulas for moment of Inertia).
     
  10. Dec 8, 2011 #9

    gneill

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    It's handy to memorize a few of them, and often the ones required will be given to you on an exam (unless one of the questions is to derive it for a particular shape).

    Finding the formula for the moment of inertia of a given shape is an exercise in geometry and integration. Do you want to do that first?
     
  11. Dec 8, 2011 #10
    Of course, that was my main and actual argument.
     
  12. Dec 8, 2011 #11

    gneill

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    I meant, do you want to solve for the moment of inertia of plate first?

    Okay, here's a drawing showing a flat rectangular plate comprising the area bounded by the coordinate axes and the blue line. Width is a, height is b. Assume the density is [itex]\rho[/itex], and that we want the moment of inertia about an axis passing through the origin. How would you set up the integration to sum all the [itex]r^2 dm[/itex]?

    attachment.php?attachmentid=41731&stc=1&d=1323376895.gif
     

    Attached Files:

  13. Dec 8, 2011 #12
    Ok, the only possible way (at least for me) is performing a triple integral. This integral, however is easy to solve, but a bit tedious.
    Make r^2 = x^2 + y^2. Integrate dm = p dV over the region ocupied by the cube (the limits of integration are 0 to L for all three axis). Remember that dV = dx dy dz and p is constant.
     
  14. Dec 9, 2011 #13
    To gneill, I would say density=ρ=M/ab
    so what şs dm equal to ?
    Is it dm=ρ*x*y
    r2=x2+y2
    r=sqrt(x2+y2)

    So now, ∫dm.r2, but dm is not the one above..

    To BetoG93, yes i found it (method to find it with triple Integral) on the net, but that lacked explanation unfortunately
     
  15. Dec 9, 2011 #14

    gneill

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    [itex] dm = \rho \; dx \; dy [/itex]

     
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