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Expression for the moment of inertia

  1. May 20, 2014 #1
    1. The problem statement, all variables and given/known data
    The thin, homogeneous bent rod has the mass m and the total length of 4b. It rotates with the angular speed of ω = ω0(24i + 12j - 6.0k) (only rotation)

    Determine the expression for the moment of inertia with consideration of the center of mass of the rod

    The figure:
    http://s716.photobucket.com/user/Pitoraq/media/Mek4_zps8205c021.png.html

    2. Relevant equations

    H = (Ixxωx-Ixyωy-Ixzωz)i+(-Iyzωx+Iyyωy-Iyzωz)j+(-Izxωx-Izyωy+Izzωz)k

    m = ρb


    3. The attempt at a solution¨


    I plotted the figure in yx-axis and could see that

    Iyz=Iyx=Iyz=Izy=Ixz=Izx=0 is symmetric

    But i'm not sure about Ixx, Iyy and Izz
     
  2. jcsd
  3. May 20, 2014 #2

    adjacent

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    Gold Member

    Can you post your graph?
     
  4. May 20, 2014 #3
  5. May 20, 2014 #4

    SteamKing

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    Staff Emeritus
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    Homework Helper

    You might have to, you know, calculate Ixx, Iyy, and Izz. It's a shocking suggestion, I'm sure.
     
  6. May 27, 2014 #5
    The centre of mass is easy: it is at the origin of y-z axis.

    There are three parts to this problem: the rotating center bit, and the two sides. By symmetry, the MOI of the left and right side are equal.

    Calculate the MOI for rotation in the ##i## and ##j## and ##k## axes. Use parallel axes theorem to calculate the MOI of the sides.
     
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