- #1

- 58

- 0

Just looking at the scaler case, angular momentum is [itex]L = I\omega[/itex]

Then, according to the 2nd law, [itex]\tau = \frac{d}{dt}(L) = \dot{I}\omega + I\alpha[/itex]

Is this correct? I understand the problem with doing this in the linear case (i.e. it does not respect Galilean invariance), but does the rotational case hold? I'm not sure it does...I mean, in the absence of torque ([itex]\tau = 0[/itex]), the equation says that [itex]\alpha \neq 0[/itex] if [itex]\dot{I} \neq 0[/itex] which doesn't quite make sense. Does the [itex]\dot{I}\omega[/itex] term just get lumped into [itex]\tau[/itex]? (like the linear case)