= Solution
Let $M_{\rm tot}=\sum_i m_i$ and introduce positions relative to the <centre of mass>, $\boldsymbol\rho_i=\boldsymbol r_i-\boldsymbol R$. The definition gives $\sum_i m_i\boldsymbol\rho_i=0$, and differentiation gives $\sum_i m_i\dot{\boldsymbol\rho}_i=0$. Expand the total <kinetic energy>:
$$
\begin{aligned}
T&=\frac12\sum_i m_i|\dot{\boldsymbol R}+\dot{\boldsymbol\rho}_i|^2\\
&=\frac12M_{\rm tot}|\dot{\boldsymbol R}|^2
+\dot{\boldsymbol R}\cdot\sum_i m_i\dot{\boldsymbol\rho}_i
+\frac12\sum_i m_i|\dot{\boldsymbol\rho}_i|^2.
\end{aligned}
$$
The cross term vanishes. This proves the <kinetic energy decomposition about the center of mass>:
$$
\boxed{T=T_1+T_2,\qquad
T_1=\frac12M_{\rm tot}|\dot{\boldsymbol R}|^2,\quad
T_2=\frac12\sum_i m_i|\dot{\boldsymbol\rho}_i|^2.}
$$
The decomposition itself holds without rigidity.
For a <rigid body> rotating about its <centre of mass>, $\dot{\boldsymbol\rho}_i=\boldsymbol\omega\times\boldsymbol\rho_i$. With unit axis vector $\boldsymbol n=\boldsymbol\omega/|\boldsymbol\omega|$, the <moment of inertia> about that axis is
$$
I=\sum_i m_i\left[|\boldsymbol\rho_i|^2-(\boldsymbol n\cdot\boldsymbol\rho_i)^2\right]
=\sum_i m_i d_i^2,
$$
where $d_i$ is the perpendicular distance from the axis. Therefore
$$
\boxed{T_2=\frac12 I\omega^2,\qquad \omega=|\boldsymbol\omega|.}
$$
Back to article page
