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Using inertia tensor in conservation of angular momentum

  1. May 10, 2009 #1
    Hi all, this is my first post. I'm under the gun at the moment because we were just told yesterday to expect a similar problem to the one below on our final, yet we were not taught how to work such a problem in class. (Our teacher does not write the final hence the discrepancy.)

    I have a solution manual I am trying to learn how to do this problem with the help of, but it is not going very well. My book only vaguely touches on the inertia tensor so i'm very confused about what's going on in this problem. I've posted the two pages from the solution manual. I keep trying but I can't seem to duplicate the numbers the solution manual gets.

    1)The problem statement, all variables and equations:
    Page 1: http://i43.tinypic.com/2j4d761.jpg
    Page 2: http://i41.tinypic.com/w1mhg.jpg

    2) Attempt at solution
    The 'I1' calculation seems to be Izz for a thin plate from the back of my book, where Izz = 1/12*m(a2 + b2). Yet the manual labels that value I1 and seems to use it in place of where Ixx should be in the inertia tensor (Imat.) I don't know if this is an error or not.

    And also for Ioa, i'm using the formula Ixx*ux2 + Iyy*uy2 + Izz*uz2 where the u's are the unit vectors for ro/a. Since ux is zero, I should get: Ioa = Iyy*uy2 + Izz*uz2. Again, i'm unsure of which is Iyy and Izz from the solution manual due to the obscure labeling of I1/I2/I3, and i'm unsure of how to identify which is which from the picture. (that's something I never figured out since we are usually given the values)

    If anyone can offer some tips or suggestions on this problem, I would be very grateful. After spending about 4 hours staring at it, googling, etc. i'm really at my wits end.
     
    Last edited: May 10, 2009
  2. jcsd
  3. May 10, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Mark24! Welcome to PF! :smile:

    The 1 2 and 3 refer to x y and z, so the I1 is Ixx, which they get by adding Iyy and Izz (the parallel axis theorem).

    I don't understand why you're confused … 1,2,3 ~ x,y,z is fairly simple notation :confused:

    Anyway, look at the diagram :smile: …​

    Izz clearly depends on a, and it's through the c.o.m., so it's small, so it's 1/12

    while Iyy clearly depends on b, and it's not through the c.o.m., so it's larger, so it's 1/3 :wink:
     
  4. May 10, 2009 #3
    Thanks tiny-tim, how they derived the moments of inertia makes sense now after you explained it. Since the inertias seem to be correct but i'm still getting the wrong answer, I must be messing up the final equation - the one listed in the second picture as:
    Imat(0,0,w1)oa = Ioa*w2

    For Ioa, I get Ioa = Iyy*uy2 + Izz*uz2 = .192

    For Imat, I get a diagonal matrix with values: .4167, .2667, .15. w1 is 2 rad/s.

    If I do Imat*(0,0,w1) = Ioa*w2, I get w2 = 1.56 in the z direction and 0 in the y, which is wrong. Any idea what I am missing? The solution has an 'oa' at the end on the left hand side of the equation. oa is the unit vector, so I don't understand why it would be needed in the cons. of angular momentum equation.
     
    Last edited: May 10, 2009
  5. May 11, 2009 #4

    tiny-tim

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    Hi Mark24! :smile:

    (have an omega: ω :wink:)
    Imatrix converts the angular velocity vector (0,0,ω1) to the angular momentum vector, Imat*(0,0,ω1).

    The external forces are at O and A, so angular momentum is only conserved about the OA axis, so we need the OA-component of the angular momentum vector:

    Imat*(0,0,ω1) "dot" oa :wink:
     
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