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No. The generic formula is ##T = \frac 1 2 \vec L \cdot \vec \omega = \frac 1 2 (I\vec \omega)\cdot \vec \omega##. This reduces to your simpler form only in the special case where the inertia tensor is diagonal.CAF123 said:Does this mean that relative to any other frame, ##T' = \frac{1}{2}(I_1'w_1'^2 + I_2'w_2'^2 + I_3'w_3'^2)##
One reason it's confusing is because rotations in three dimensional space don't commute. I didn't attain a true understanding of rigid body motion until I learned about Lie groups.One grad student I was talking to said that the topic of rigid body motion was overall more difficult to comprehend than some of the research he was working on. I guess that is not very encouraging, but it puts your point into perspective.