The total angular momentum of a system of particles relative to the origin $O$ of an inertial reference frame is given by $\overrightarrow{\mathbf{L}}=\Sigma\left(\overrightarrow{\mathbf{r}}_{i} \times \overrightarrow{\mathbf{p}}_{i}\right)$, where $\overrightarrow{\mathbf{r}}_{i}$ and $\overrightarrow{\mathbf{p}}_{i}$ are measured with respect
to $O$. $(a)$ Use the relations $\overrightarrow{\mathbf{r}}_{i}=\overrightarrow{\mathbf{r}}_{\mathrm{cm}}+\overrightarrow{\mathbf{r}}_{i}^{\prime}$ and $\overrightarrow{\mathbf{p}}_{i}=$
$m_{i} \overrightarrow{\mathbf{v}}_{\mathrm{cm}}+\overrightarrow{\mathbf{p}}_{i}^{\prime}$ to express $\overrightarrow{\mathbf{L}}$ in terms of the positions $\overrightarrow{\mathbf{r}}_{i}^{\prime}$ and momenta $\overrightarrow{\mathbf{p}}_{i}^{\prime}$ relative to the center of mass $C$; see Fig. $10-22$.
(b) Use the definition of center of mass and the definition of angular momentum $\overrightarrow{\mathbf{L}}^{\prime}$ with respect to the center of mass to obtain $\overrightarrow{\mathbf{L}}=\overrightarrow{\mathbf{L}}^{\prime}+\overrightarrow{\mathbf{r}}_{\mathrm{cm}} \times M \overrightarrow{\mathbf{v}}_{\mathrm{cm}} \cdot(c)$ Show how this result can
be interpreted as regarding the total angular momentum to be the sum of spin angular momentum (angular momentum relative to the center of mass) and orbital angular momentum (angular momentum of the motion of the center of mass $C$ with respect to $O$ if all the system's mass were concentrated at $C)$.