Centre of Mass of Solid Bodies

A real object isn’t a handful of dots. Treat it as continuous, slice it into infinitely many tiny pieces, and the sum becomes an integral.

From sums to integrals#

For a body with continuously spread matter, replace “particle i” with a tiny element of mass $dm$ at position (x, y, z):

The art is choosing the element: as the integration variable runs between its limits, the elements must cover the whole body exactly once, and every part of one element should have (almost) the same coordinate you are integrating.

(a) A uniform rod

Rod of mass M and length L along the x-axis from 0 to L. Mass per unit length is M/L, so an element between x and x + dx has $dm = (M/L)\,dx$:

The midpoint, as symmetry predicts.

(b) A uniform semicircular wire

Radius R, centre at the origin, ends on the x-axis. Choose the element between angles θ and θ + dθ: length $R\,d\theta$, mass $dm = \dfrac{M}{\pi R}R\,d\theta = \dfrac M\pi d\theta$, coordinates $(R\cos\theta, R\sin\theta)$. As θ goes from 0 to π the elements cover the wire.

(c) A uniform semicircular plate

No new integral over angles is needed. Slice the plate into thin semicircular strips of radius r and width dr. Each strip is a semicircular wire, whose centre of mass we just found: at height $2r/\pi$. Its mass is its area $\pi r\,dr$ times the mass per area $M/(\pi R^2/2)$, i.e. $dm = \dfrac{2M}{R^2}r\,dr$. Replace each strip by a point mass at its own centre of mass and add: