The discussion revolves around deriving Euler's equation of motion for a rigid body, specifically the equation $$\dot{\vec{L}} + \vec{\omega} \times \vec{L} = \vec{G}$$. Participants explore the concepts of angular momentum, rotational frames, and the implications of different reference systems in the context of rigid body dynamics.
The discussion is ongoing, with participants providing insights into the transport theorem and its application to the problem. Some participants are questioning the assumptions made about the relationship between different observers and their coordinate systems, while others are exploring the implications of these relationships on the understanding of angular momentum.
There are indications of confusion regarding the definitions of reference and coordinate systems, as well as the treatment of angular momentum in different frames. Participants are also grappling with the implications of the instantaneous coincidence of frames and how that affects the interpretation of vectors and their derivatives.
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.