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$\text{Rotational Energy} = \frac{1}{2} I \omega^2$. What $I$ should be used? $I$ as a inertia tensor matrix = stepRotation * inverse moment of inertia * inverse stepRotation; Or I as moment of inertia? Inertia tensor matrix gives changing energy due to rotation changes. So I don't believe its the right answer, but I'm not sure. Moment of inertia remains the same in simulation because it only represents the mass resistance to rotation in a fixed axis calculated by integration of massparticles * radius from object center.

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Moment of Inertia in this case. omega doesnt really have structure here and the tensor is overkill. –  madtowneast Dec 1 '11 at 16:30
    
Welcome to Physics.SE! MathJax is active on this site which allows you to use LaTeX math formatting. I've done your first equation to give you an idea of what is available to you, and there is a microscopic amount of help in the FAQ. –  dmckee Dec 1 '11 at 16:53
    
Only the tensor is correct. The reason you have changing energy is that you didn't update $\omega$--- it doesn't stay constant. Only the angular momentum stays constant. –  Ron Maimon Dec 1 '11 at 18:36
    
If I in the Rotational Energy(RE) is the same as I in Angular Momentum = I * w; might resolve this issue. The problem was that I had a fluctuating RE with a constant Angular Momentum. I will check this. –  Palax Dec 5 '11 at 16:35
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Um... what? The moment of inertia is a tensor. So $K = \frac{1}{2}\omega^T I\omega$ is always correct (for a rigid body).

The formula only reduces to the "scalar" case $K = \frac{1}{2}I\omega^2$ if the object is rotating around one of its principal axes.

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if you have a constant arbitrary axis Angular Momentum the Rotation Energy does not remain the same in time? –  Palax Dec 5 '11 at 16:44
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