Orbital Angular Momenta - Conversion to a Different Origin

[MisterX]

Suppose we have some object with angular momenta $\mathbf{L}_1$ about rotation center $O_1$ and we have another object with angular momentum $\mathbf{L}^\prime_2$ about rotation center $O_2$. $O_2$ has some velocity $\mathbf{v}_2$ relative to $O_1$. Then we wonder what is the angular momentum of the second object with respect to $O_1$ so that we may express the total angular momentum of the system. Is there anything to say in general about this?

What about the case where the second object is a sphere, and it is rotating on its axis parallel to $\mathbf{r} \times \mathbf{v}_2$, where $\mathbf{r}$ is a vector from $O_1$ to $O_2$ ? For example consider a rotating sphere in a circular orbit, where the sphere's axis of rotation is parallel to the plane of the orbit. What is the total angular momentum about the center of the circular orbit?

 

[MisterX]

I don't know why this didn't occur to me until I took a shower.

$\mathbf{x} = \mathbf{x}^\prime + \mathbf{O}_2$

[itex][/itex]

$\mathbf{L}_2 = \sum m_i \mathbf{x}_i \times \dot{\mathbf{x}}_i = \sum m_i (\mathbf{x}_i^\prime + \mathbf{O}_2)\times \dot{\mathbf{x}}_i$

$= \sum m_i (\mathbf{x}_i^\prime + \mathbf{O}_2)\times \left(\dot{\mathbf{x}}^\prime_i + \dot{\mathbf{O}}_2\right)$

$=\mathbf{L}_2^\prime + \sum m_i \mathbf{O}_2 \times \dot{\mathbf{O}}_2 + \sum m_i \mathbf{x}_i^\prime \times \dot{\mathbf{O}}_2 + \sum m_i \mathbf{O}_2 \times \dot{\mathbf{x}}^\prime_i$

So at least the first two terms have a nice form.

$=\mathbf{L}_2^\prime + M_{total} \mathbf{O}_2 \times \dot{\mathbf{O}}_2 + \sum m_i \mathbf{x}_i^\prime \times \dot{\mathbf{O}}_2 + \sum m_i \mathbf{O}_2 \times \dot{\mathbf{x}}^\prime_i$