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Gyroscope, non symmetrical body

  1. Sep 4, 2014 #1
    Hello, first of all I know this post for help is not the clearest. I am looking for quick help but not necessarily the answer. I know I can solve the math of the problem but I am having trouble analyzing and setting up what I have. So I am asking for help with the approach.

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

    Problem number 8 on this link: http://home.iitk.ac.in/~mohite/Assignment_05_AE688.pdf

    2. Relevant equations

    Inertia of slender rod is (1/12)ml2

    3. The attempt at a solution

    So here are the problems I have run into and would like to discuss:

    1) do I consider xyz a non rotating reference frame?
    I found angular momentum with respect to the xyz

    Hx = ωsin2(phi)
    Hy = -ωsin(phi)cos(phi)
    Hz = phi

    2) is Ω = (ωi + pk)?

    3) The hint to the problem is to apply the general moment equation:

    Mx = H(dot) + HyΩz + HzΩy

    I get as far as assuming all terms are zero but HyΩz.

    Which then gives me an answer: ωp(1/12)ml2sin(phi)cos(phi)

    Thank you for the help.
    Last edited: Sep 4, 2014
  2. jcsd
  3. Sep 4, 2014 #2


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    Homework Helper
    Gold Member

    Hello, murstr. Welcome to PF!

    I think the xyz axes shown in the figure rotate about the axis of the yoke (the x-axis). So, the rod always lies in the xy plane of these axes. The x axis is therefore not rotating relative to the inertial frame of the laboratory, while the y and z axes rotate with the yoke.

    Note that the expressions on the right do not have the correct dimensions for angular momentum. (Also, I don't see how you got these expressions.)

    This is the correct expression for the angular velocity of the rod. However, the angular velocity of the yoke frame (i.e., the xyz frame in the figure) relative to the lab frame is just ##\vec{Ω} = ω \hat{i }## (see below for why this might be important.)

    Should there be a subscript on the first term on the right? Are the signs of all the terms correct?

    This appears to be one of http://en.wikipedia.org/wiki/Euler's_equations_(rigid_body_dynamics)]Euler's[/PLAIN] [Broken] equations of motion . If so, note that the Euler equations refer to principal axes that rotate with the rod. So, the x, y, z subscripts in this equation are not referring to the xyz axes shown in the figure.

    [EDIT] After thinking some more, I think you can use your (corrected) equation for Mx in the xyz coordinate system shown in the figure (without going to the principal axis frame of the rod). But you will need to work with components of the moment of inertia tensor for the rod in this frame. And Ωz and Ωy would then be components of the angular velocity of the xyz coordinate system relative to the lab frame. As noted above, these two components are zero. If you are careful, you can get the correct answer for the problem fairly quickly this way.

    The advantage of going to the principal axis frame of the rod (Euler's equations) is that the moment of inertia tensor is simplified in this frame. But, overall, I found the calculation to take about the same amount of work using either the xyz frame or using the principal axis frame.
    Last edited by a moderator: May 6, 2017
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