 Solution

Let $M_{\mathrm{tot}}={\sum }_i{m}_i$ and introduce positions relative to the centre of mass, ${\mathbf{\rho }}_i={\mathbf{r}}_i-\mathbf{R}$. The definition gives ${\sum }_i{m}_i{\mathbf{\rho }}_i=0$, and differentiation gives ${\sum }_i{m}_i{\dot{\mathbf{\rho }}}_i=0$. Expand the total kinetic energy:The cross term vanishes. This proves the kinetic energy decomposition about the center of mass:The decomposition itself holds without rigidity.

For a rigid body rotating about its centre of mass, ${\dot{\mathbf{\rho }}}_i=\mathbf{\omega }\times {\mathbf{\rho }}_i$. With unit axis vector $\mathbf{n}=\mathbf{\omega }/|\mathbf{\omega }|$, the moment of inertia about that axis iswhere $d_i$ is the perpendicular distance from the axis. Therefore