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Moment of Inertia of a Square?

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data

    See below for the diagram.

    Use this result to calculate the moment of inertia of the shape below about the axis shown by the dashed line. The object is a square with a side of length L, total mass of 4M, and a uniform density.

    2. Relevant equations



    3. The attempt at a solution

    I got

    I = (ML^2)/3

    My work is attached below. The reason why I ask this question is because I can't find the answer any were online for moment of inertia of a square about this particular axis and just wanted to make sure that my answer was correct. Thanks for any help you can provide if it's wrong.
     

    Attached Files:

  2. jcsd
  3. Oct 14, 2011 #2

    vela

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    Your work is fine.

    Wikipedia has a list of area moments of inertia.

    http://en.wikipedia.org/wiki/List_of_area_moments_of_inertia

    If you look at the first rectangular one, it's basically the result you want if you identify bh, the area of the rectangular, with the mass, and set h=L. Then you get
    [tex]I = \frac{bh^3}{12} = \frac{4M L^2}{12} = \frac{1}{3}ML^2[/tex]
     
  4. Nov 28, 2011 #3
    interesting, my professor marked my answer wrong =/
     
  5. Nov 29, 2011 #4
    Ok the square is hollow and it's just a wire frame in the shape of a square and is why I got the problem wrong.

    Taking this information into consideration...

    I took the two bottom sides (the sides in which the axis of rotation goes through them) and added them together knowing that the moment of inertia of a line through were the axis of rotation goes through the midpoint of the line is (ML^2)/12 and got
    (ML^2)/6

    For the two other sides of the square in which the axis of rotation does not directly intersect the side I simply did this

    I = integral r^2 dm
    rho = dm/dL
    dm = rho dL
    r = L/2
    I said this because the axis of rotation is this far away from the side and is constant
    I = rho * integral[0,L] (L/2)^2 dL
    = rho/4 * integral[0,L] L^2 dL
    = (rho*L^3)/12
    rho = M/L
    I = (M*L^3)/(12*L) = (ML^2)/12
    but I have to multiply this by two because of the other side
    I = (ML^2)/6

    So I got for the total moment of inertia
    (ML^2)/6 + (ML^2)/6 = (ML^2)/3

    taking into consideration that the total mass is 4M
    I = (4ML^2)/3

    Is this correct?
     
  6. Nov 30, 2011 #5
    B.U.M.P. - went onto second page
     
    Last edited: Nov 30, 2011
  7. Nov 30, 2011 #6

    Doc Al

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    Yep.

    So far, so good.

    I don't understand what you're trying to integrate here. All the mass of these pieces is a distance L/2 from the axis. (The fact that they are rods is irrelevant.) So what's the moment of inertia?

    Redo the second piece of this.

    The mass of each piece is already included in the sum. (Don't multiply by 4!)
     
  8. Dec 12, 2011 #7
    So I'm getting 2ML^2. Is this correct? IF the mass wasn't 4M and for example 6M would I just find the moment of inertia of the each side of the square and replace M with (6M)/4 or (3M)/2?
     
  9. Dec 12, 2011 #8

    Doc Al

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    No. Show how you got this.

    You've already calculated the moment of inertia of the horizontal sides. What did you get for each vertical side?
    Not sure what you're asking here. If the total mass were 6M, then the mass of each side would be 6M/4.
     
  10. Dec 12, 2011 #9
    Sorry I got numbers mixed up

    The moment of inertia of a vertical side I got
    I = integral r^2 dm, r = L/2, I = integral (L/2)^2 dm = L^2/4 * integral dm = (ML^2)/4
    this is for one side.

    I took the two bottom sides (the sides in which the axis of rotation goes through them) and added them together knowing that the moment of inertia of a line through were the axis of rotation goes through the midpoint of the line is (ML^2)/12 and got
    (ML^2)/6
    I come I ignore the fact that the mass of both the horizontal sides combined is M/2...

    Oh I think I get it now. I can ignore this fact because when I add them together I would get 1+1+1+M and just get 4M

    So the moment of inertia for just one vertical side is (ML^2)/4 and them combined is (ML^2)/2

    Adding the two moment of inertia together I now get
    I = (ML^2)/6 + (ML^2)/2 = (2ML^2)/3

    Is this correct? I'm not sure how the fact that the total mass is 4M plays into this.
     
  11. Dec 12, 2011 #10

    Doc Al

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    Good!
    That's what allowed you to deduce that each side had mass M.
     
  12. Dec 12, 2011 #11
    So if each side had mass (3M)/2 would I have for one vertical side

    I = (ML^2)/4

    would become

    I = (3ML^2)/(4*2) = (3 ML^2)/8?
     
  13. Dec 12, 2011 #12

    Doc Al

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    Yep.
     
  14. Dec 12, 2011 #13
    Ok thank you very much
     
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