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Calculating an inertia tensor

  1. Nov 25, 2003 #1
    I'm trying to write a quick and dirty computer simulation for 3D rotational dynamics. This is the first time I've ever seen the sort of physics needed for doing this, so please bear with me.

    I've decided that I don't need a huge amount of accuracy in the final simulation, so I was hoping to represent any rotating object as an aggregation of several smaller, simpler objects. For example, a rotating ice cream cone I could fudge as a cone and a sphere. With this method, I was hoping to get around a lot of slow and complex mathematics.

    Here was my idea: If I know the inertia tensors for simple objects like spheres and rods around their primary axes, I can just shift my moments and sum them up into a final tensor, eliminating the need to transform coordinates and find eigenvalues.

    For example: if I had an object that was composed of a solid sphere and a solid circular cylinder, with the center of mass of the sphere 5 units to the left of the object, and the CoM of the cylinder 2 units down from the CoM of the object, I could say that

    Ixx = 2/5mr^2 + 25m
    All other moments stay as they were; the products of inertia still remain zero.

    Izz = 1/2mr^2 + 4m
    All other moments stay as they were; the products of inertia still remain zero.

    Then I could take these two inertia tensors and add them together to get the final tensor of my aggregate object.

    Will this work? Can I do the parallel axis shifts like that in 3D?

    Thanks in advance!
  2. jcsd
  3. Nov 29, 2003 #2


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    Mmmm... fudge swirl ice cream....

    Sorry about that: what you're saying sounds good. If you've got parallel axes, you can use the parallel axis theorem to recalculate inertia.
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