= Solution
Assume the particle <masses> $m_i$ are fixed and their total <mass> $M=\sum_i m_i$ is positive. The <centre of mass> and the relative <position> vectors are
$$
\boxed{\mathbf R=\frac1M\sum_i m_i\mathbf x_i,\qquad\mathbf y_i=\mathbf x_i-\mathbf R.}
$$
Their weighted sum is zero, so <differentiation> gives $\sum_i m_i\dot{\mathbf y}_i=0$. Expanding the <kinetic energy> with $\dot{\mathbf x}_i=\dot{\mathbf R}+\dot{\mathbf y}_i$ makes the mixed term vanish:
$$
T=\frac12\sum_i m_i\bigl(|\dot{\mathbf R}|^2+2\dot{\mathbf R}\cdot\dot{\mathbf y}_i+|\dot{\mathbf y}_i|^2\bigr)
=\boxed{\frac12M|\dot{\mathbf R}|^2+\frac12\sum_i m_i|\dot{\mathbf y}_i|^2}.
$$
This <kinetic-energy decomposition about the centre of mass> separates translation from internal motion without assuming any particular <force> law.
For the <rigid body>, its relative <velocity> is $\dot{\mathbf y}_i=\omega\mathbf n\times\mathbf y_i$. The <cross product> identity for a <unit vector> gives $|\mathbf n\times\mathbf y_i|^2=|\mathbf y_i|^2-(\mathbf n\cdot\mathbf y_i)^2$, the squared perpendicular distance to the rotation axis. Hence the <moment of inertia> of the particle is $I_i=m_i[|\mathbf y_i|^2-(\mathbf n\cdot\mathbf y_i)^2]$, and summing the <rotational kinetic energy> yields
$$
\boxed{T=\frac12M|\dot{\mathbf R}|^2+\frac12I\omega^2,\qquad I=\sum_i I_i.}
$$
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