Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Superposition of Moments of Inertia about Different CGs

  1. Feb 20, 2016 #1
    Hi all,

    You can use superposition to add moments of inertia when they're calculated about the same center of gravity (cg), but let's say you calculate the moments of inertia of several elements of a system about one cg and then use the Parallel Axis Theorem to then reference the total moments for these elements back to the total system cg. Do you get the same answer?

    I had assumed yes, but then did some math that's making me not so sure.

    Most basic example:

    Let's say I have 3 mass elements in a 3 dimensional space. I am given the mass and cgs (relative to a reference frame origin) of each element, as well as the moment of inertia of each element about its own cg.

    I lump together two of the elements and treat this as a new element. I find the mass, cg_lumped, and moments of this new lumped element relative to cg_lumped. Next, I want to find the TOTAL moments of inertia of the lumped element plus the 3rd element relative to the total system cg - call it cg_total.

    I end up with extra terms using the "lumped method." I guess it's nonlinear... I may have messed up my math and there may be a cleaner way to do it, but I just want a proof one way or another of whether the answer would be the same in general. My math focuses on the Ixx (inertia about the X axis through the cg) only.
    Setup and math here

    Thank you!
    Alexa.
     
  2. jcsd
  3. Feb 20, 2016 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    I've always found using a tabular form for calculating the moment of inertia of a composite body the clearest and most efficient way of organizing this type of calculation. It's a shame more textbooks don't illustrate this method, IMO.

    Let's say that our composite body is composed of three masses, m1, m2, and m3. Each of these masses has a c.g. referenced to the origin [for example, the c.g. of m1 is (x1, y1, z1) ], and there are the moments of inertia for each mass referenced about each mass's c.g. To simplify the calculation, let's assume that I1 = Ixx for m1, etc.

    The tabular form is set up like this:

    Code (Text):

    Item   Mass  xcg  ycg  zcg   M*x    M*y    M*z   M*x^2   M*y^2   M*z^2   I
    -----------------------------------------------------------------------------------------------
    1      m1    x1   y1   z1  m1*x1  m1*y1  m1*z1 m1*x1^2 m1*y1^2 m1*z1^2  I1
    2      m2    x2   y2   z2  m2*x2  m2*y2  m2*z2 m2*x2*2 m2*y2^2 m2*z2^2  I2
    3      m3    x3   y3   z3  m3*x3  m3*y3  m3*z3 m3*x3^2 m3*y3^2 m3*z3^2  I3
    -----------------------------------------------------------------------------------------------
    Total   M     x'   y'   z'   M*x    M*y    M*z   M*x^2   M*y^2   M*z^2   IT

    M = m1 + m2 + m3

    Mx = m1*x1 + m2*x2 + m3*x3
    My = m1*y1 + m2*y2 + m3*y3
    Mz = m1*z1 + m2*z2 + m3*z3

    Mx^2 = m1*x1^2 + m2*x2^2 + m3*x3^2
    My^2 = m1*y1^2 + m2*y2^2 + m3*y3^2
    Mz^2 = m1*z1^2 + m2*z2^2 + m3*z3^2

    IT = I1 + I2 + I3

    x' = Mx / M
    y' = My / M
    z' = Mz / M

    To calculate IT' about the centroid (x', y', z'):

    IT' = IT + My^2 + Mz^2 - M * (y'^2 + z'^2)

     
    Here, it's understood that IT' represents Ixx'.

    The tabular form can be easily extended to handle Iyy and Izz, not to mention the products of inertia Ixy, Iyz, and Ixz.

    Tablular form calculations can also be easily programmed into a spreadsheet to make evaluating actual numerical values that much easier.
     
  4. Feb 20, 2016 #3
    Thanks for the reply!

    I should have mentioned that I have a MATLAB script calculating mass properties, and have also used a spreadsheet calculation to double check. However, using the two different methods I describe above I have not been able to get the answers to match. I resorted to the hand calcs to prove to myself that they should theoretically should match, but in fact I seem to have proven the opposite.
     
  5. Feb 20, 2016 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    I would recommend that you post your calculations, especially in the spreadsheet form. There may be errors lurking within t which you may not have uncovered yet.
     
  6. Feb 20, 2016 #5
    I would just like to know if it is theoretically possible to use the lumped method I described and attempted to prove valid mathematically. My question is really regarding the validity of my proof, not my actually mass properties calculations. For the sake of this question one could really assume I'm not actually doing any actual calculations but just wanted to understand if i could use the lumped approach in theory :). The actual numbers are not important.
     
  7. Feb 20, 2016 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You should get the same result regardless of how you lump the individual masses. I didn't look at all of your derivation because it gave me a headache. It would be much easier to follow one of your calculations, IMO.

    I suspect that you have made an error in applying the Parallel Axis Theorem between the different groupings of masses. To apply the PAT correctly, you must first calculate everything about a common reference and then move it to a centroidal location using the PAT. If you are using multiple groupings of masses, this can get tricky. The tabular form eliminates problem this since you just add another row for an additional mass.
     
  8. Feb 20, 2016 #7
    Haha it gave me a headache as well. I will try to get a sample calc and post it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook