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Homework Help: Moment Of Inertia

  1. Jan 6, 2010 #1
    1. The problem statement, all variables and given/known data
    When we put together 4 rod in case they will construct a square and when we hang it from the corner. How can we find its total moment of inertia?

    Length of rods: L
    Mass of rods: M



    2. Relevant equations
    Paralel Axis Theorem: I = I + Md^2


    3. The attempt at a solution
    As my book said answer is 10ML^2/3. Can you please explain it how can i calculate it? Thanks a lot.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 6, 2010 #2

    rock.freak667

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    Find the moment of inertia of one rod about its centre. Find the distance to the centre of the square, then use the parallel axis theorem.
     
  4. Jan 6, 2010 #3
    Does it make sense firstly finding all of four rods moment of inertia at its centre then taking it and finding the new moment of inertia secondly at the corner of square???
    Actually i cannot understand precisely, can you please be more clear?
     
  5. Jan 6, 2010 #4

    rock.freak667

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    Sorry, I didn't see the hanging from the corner part.

    I=(1/12)ML2.

    You need to find the distances of the centres of each rod to the corner and then use the parallel axis theorem. Then just simply add them up.
     
  6. Jan 6, 2010 #5
    But again i cannot get the result that is given by book. So do i have to take perpendicular distance to the corner for the each rod? I mean for example; for top two rods it is easy:
    I = (1/12)ML^2 + M(L/2)^2 but when it comes to bottom rods, it is being complicated, unfortunately. Again what distance i will use?
    L or squareroot of M(5/4)L^2 ???
    Still i could not get the right answer despite the deal with two cases.
    If you have any suggestion, i will appreciate it. Thanks.
     
  7. Jan 6, 2010 #6

    Doc Al

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    Md^2 = M(5/4)L^2
     
  8. Jan 6, 2010 #7

    vela

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    This approach would work. The moment of inertia about the center of the square would work out to be [tex]4/3 ML^2[/tex]. Then you can apply the parallel-axis theorem to the square as a whole. You just have to be a little careful with this step because the mass of the square is 4M, not just M.
     
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