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Engineeing Dynamics - Inertia Dyad of Half Cylinder

  1. Nov 6, 2012 #1
    I'm getting desperate. The professor has assigned a project and has not clearly explained how to derive the answers. I'm doing the best I can but his TA's and recommended tutors for the class are always incapable of reproducing the answers either. It's a junior level dynamics class, but he's actually turned the class into a machine dynamics class taught in senior or masters level. The work needs to be done in mathematica, but all I need is help with theory since that's the part we don't learn in lecture.

    Anyways, here is the question...

    1. The problem statement, all variables and given/known data

    Develop the mass center expression and the inertia matrix and inertia dyadic for a half cylinder, then let the inner diameter approach the outer diameter to develop the same for a thin half shell. Use Mathematica for your work.

    2. Relevant equations

    Ii,i = ∫m(rj2+rk2)dm

    3. The attempt at a solution

    First is the Inertia-matrix of the half cylinder. I'm not sure how to derive all the terms, but I did my best.

    {y2 + z2, -x y, -x z}
    {-x y, x2 + z2, -y z}
    {-x z, -y z, x2 + y2}

    Then I write my position vector:

    BrP = x b[1] + y b[2] + z b[3]

    Where B is a point in the body frame, P is the endpoint under evaluation, and r is the vector r. b[1,2,3] are unit vectors in the body frame.

    Now is where I think I went wrong, if not before. The Volume is:

    V = ∫0W0H-L/2L/2rdxdydz

    Then mass is:

    m = ∫∫∫ρrdxdydz

    So center of mass is:

    C = (1/m)∫∫∫ρr(x,y,z)rdxdydz

    Then I get lost even more...

    S.O.S
     
  2. jcsd
  3. Nov 7, 2012 #2

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    If x,y,z are cartesian coordinates, dV = dxdydz, not r dxdydz
     
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