Inertia Tensor as a Transformation

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uliuli
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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?

Thanks!
 
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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!