Today I did more Angular Momentum theory and exercises from HRK.
The most interesting thing I came across was this fact:
Consider an object of mass $M$ both translating and rotating, with angular velocity $\omega$ and center of mass velocity $v_\text{cm}$. At the instant the center of mass is at position $\vec{r_{\text{cm}}}$ with respect to the origin, the angular momentum of this object relative to the origin is $$L=L' + \vec{r_{\text{cm}}} \times M \vec{v_{\text{cm}}},$$ where $L'$ is the angular momentum of the object relative to its center of mass.
I was able to prove this on my own! (with HRK dividing the problem up into two parts, of course). At some point, I'll have to start thinking of these on my own, but I think for now just proving it part-by-part gives me enough challenge. Once I start recognizing the repeatedly used proving ideas, I'll be able to prove it from scratch on my own.
Proof:
Consider points on the object, with an arbitrary point of mass $m_i$ at $r_i$ and momentum $p_i$. Let that point have position $r_{i}'$ and momentum $p_{i}'$ with respect to the CM. The angular momentum about the origin is $$L=\sum (r_i \times p_i)$$
Note that $$p_i = m_i(v_\text{cm} + v_i') = m_iv_i' + m_i v_\text{cm} = p_i' + m_i v_\text{cm},$$ so
$$L = \sum (r_i \times p_i) = \sum (r_\text{cm} + r_i') \times (m_i v_\text{cm} + p_i')$$
$$=\sum (r_{\text{cm}} \times m_i v_{\text{cm}}) + \sum(r_{\text{cm}} \times p_i') + \sum(r_i' + m_i v_{\text{cm}}) + \sum(r_i' \times p_i') \qquad (\star)$$
Note that $$\sum(r_{\text{cm}} \times p_i') = r_{\text{cm}} \sum p_i' = r_{\text{cm}} \sum (p_i - m_i v_{\text{cm}}) = r_{\text{cm}}\left(\sum p_i - Mv_{\text{cm}}\right) = r_{\text{cm}}(0) = 0,$$ and $$\sum r_i' + m_i v_{\text{cm}} = v_{\text{cm}} \sum m_i r_i' = v_{\text{cm}}(0) = 0.$$
Hence, the middle two terms of $(\star)$ are $0$, which implies $$L=\sum (r_{\text{cm}} \times m_i v_{\text{cm}}) + \sum(r_i' \times p_i')$$
$$=\left(r_{\text{cm}} v _{\text{cm}} \sum m_i \right) + (L')$$
$$=L' + \vec{r_{\text{cm}}} \times M \vec{v_{\text{cm}}},$$ as desired. $\square$
This intuitively makes sense because $L'$ is the part of the momentum that has to do with the rotation around the CM, while $M \vec{v_{\text{cm}}}$ is the part that has to do with the translation of the object.
The best example for understanding this is to consider a ball that is rolling without slipping.
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