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Here is the formula I am trying to derive: $$\tau = \omega\times I \omega + I\dot{\omega}$$ where $\tau$ is the torque. $\dot{\omega}$ is the angular acceleration. $I$ is the moment of inertia. $\omega$ is the angular velocity.

This is what I tried. I know that the angular momentum: $$L = I\omega $$ So $$\tau = \frac{dL}{dt} = \frac{d(I\omega)}{dt} = \frac{dI}{dt}\omega + \frac{d\omega}{dt}I $$
$$\tau = \frac{dI}{dt}\omega + \dot{\omega}I $$

It is close, but I can not go further. I only know that $I = m r^2$, and $\frac{dI}{dt}$ does not give me the result I wanted. I must have missed something.

Di Wang
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  • Unfortunately no, here $I$ is a matrix and is not equal to $mr^2$. Are you familiar with the formula $I = RI_{\rm{body}}R^T$? –  Dec 05 '18 at 02:40
  • No, looks like a matrix decomposition. Yeah, that makes sense now, I think I can look into that. – Di Wang Dec 05 '18 at 17:33
  • See https://en.wikipedia.org/wiki/Rigid_body_dynamics#Rotation_in_three_dimensions – cgiovanardi Dec 05 '18 at 19:32

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