This is just a quick question about transforming rigid bodies. Suppose we have a problem where we can find the moment of inertia tensor of some rotating rigid body in the body's frame (take it as rotating in the Z-axis). The angular momentum in the rotating frame is then just $L = Iω$, where $ω$ is a vector pointing in the Z-axis.
Now suppose we want to analyze this in the lab frame. Here's my question: do I have to transform the moment of inertia to the lab using the similarity formula, so $I' = λ*I*λ^{-1}$, where $λ$ is the rotation matrix that coincides with the Z-axis, then multiply by the original $ω$ for the lab angular momentum so that $L' = I'*ω$ OR can I simply transform the original angular momentum using $L' = λL$ where $λ$ is the rotation matrix. Or will they both give me the same answer?
Best Answer
Both results are the same
Transform angular momentum
$$ L' = \lambda (I \omega) $$
where $\omega$ and $I$ are in body basis vectors
Transform mass moment of inertia
$$ L' = (\lambda I \lambda^{-1}) \omega' =(\lambda I \lambda^{-1}) (\lambda \omega) = \lambda (I \omega)$$
As you can see, both are the same.