# Transforming Moment of Inertia/Angular Momenta in Lab Frame for Rigid Bodies

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?

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.