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Inertia Tensor as a Transformation

  1. Jul 8, 2011 #1
    Hi everyone,

    I was thinking about the relationship between angular velocity and angular momentum for a rigid body: [itex]I \omega = L[/itex]. In particular, I'm trying to gain a little bit of intuition as to what transformations [itex]I[/itex] can perform on [itex]\omega[/itex].

    Let's use a reference frame at the center of mass of the body. We can rewrite the inertia tensor as: [itex]I = R I_0 R^T[/itex], where [itex]I_0[/itex] is the diagonalized inertia tensor and [itex]R[/itex] rotates the coordinate system from a principle-axes-aligned frame to the current frame. When I now apply [itex]I[/itex] to [itex]\omega[/itex], I am equivalently rotating [itex]\omega[/itex] by some arbitrary rotation, scaling each component by a positive (and in general, different for each component) number, and rotating by the inverse of the first rotation. In general, what does this concatenation of transformations give me?

  2. jcsd
  3. Jul 9, 2011 #2
    To reply to my own thread :):

    The more I think about this, the less I think one can say about the family of transforms that [itex]I[/itex] can perform. I can 'design' [itex]I_0[/itex] to be arbitrarily close to a projection onto a cartesian axis, I can make [itex]R[/itex] any rotation I want... there is a lot of freedom.

    About the only concrete restriction on the transform I've come up with is that [itex]\omega[/itex] and [itex]L[/itex] have a positive inner product: [itex]\omega^T L = \omega^T R I_0 R^T \omega = (R^T \omega)^T I_0 (R^T \omega) = \tilde{\omega}^T I_0 \tilde{\omega} > 0 [/itex] due to the positive definiteness of [itex]I_0[/itex].

    Ohh well!
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