Centre of Mass of a System of Particles and of a Rigid Body

The two-particle formula of Section 5.2 extends directly to any number $N$ of point masses

$m_1, m_2, \ldots, m_N$ at positions $\vec r_1, \vec r_2, \ldots, \vec r_N$:

where $M = \sum_i m_i$ is the total mass of the system. In component form, this is simply

$x_{cm} = \sum_i m_i x_i / M$, and likewise for $y_{cm}$ and $z_{cm}$.

A rigid body can be thought of as the limiting case of an enormous number of such point masses -- one

for every infinitesimally small mass element $dm$ making up the body -- packed so closely together that the

sum above becomes an integral over the whole body:

Carrying out this integral for a body of uniform density and a simple, symmetric shape always gives a

result that could equally well have been guessed from the body's own symmetry: the centre of mass of

any uniform body possessing a centre, a line, or a plane of symmetry always lies exactly on that centre,

line, or plane. This is why, without ever working through the integral explicitly, the centre of mass of a

uniform rod is at its midpoint, of a uniform ring or disc is at its geometric centre, of a uniform sphere

is at its centre, and of a uniform triangular lamina is at its centroid -- the accompanying table collects

these standard results for quick reference, exactly as the syllabus's phrase "examples of simple

geometrical bodies" intends.

One case is worth singling out, because it is the clearest illustration that the centre of mass is a

purely mathematical, mass-weighted point and not necessarily a physical part of the body at all: for a

uniform circular ring, whose material lies only along the circular boundary with nothing filling the

interior, the centre of mass still comes out at the ring's geometric centre -- a point that lies in the …