Kinetic energy decomposition about the center of mass ($T=\frac{1}{2}M|\dot{R}{|}^2+\frac{1}{2}{\sum }_i{m}_i|{\dot{\rho }}_i{|}^2$)

Writing particle positions relative to the center of mass as ${\rho }_i=r_i-R$ makes ${\sum }_i{m}_i{\rho }_i=0$, so the cross term in the squared velocities vanishes. The total kinetic energy splits into center-of-mass translation and relative motion as displayed. For a rigid body, the latter is $\frac{1}{2}I{\omega }^2$, where $I$ is the moment of inertia about the instantaneous rotation axis through the center of mass.