Centre of Mass — Continuous Body

From discrete to continuous

For a discrete system: $\vec{r}_{cm} = \frac{\sum m_i \vec{r}_i}{M}$

For a continuous body, the sum becomes an integral. Instead of point masses $m_i$, we have infinitesimal mass elements $dm$ at position $\vec{r}$:

$\vec{r}_{cm} = \frac{\int \vec{r} \, dm}{M} = \frac{\int \vec{r} \, dm}{\int dm}$

In components:

$x_{cm} = \frac{\int x \, dm}{M}, \quad y_{cm} = \frac{\int y \, dm}{M}, \quad z_{cm} = \frac{\int z \, dm}{M}$

How to express $dm$

The key step is expressing $dm$ in terms of a coordinate variable and the density.

Linear mass density $\lambda$ (mass per unit length, for rods, wires, arcs):

$dm = \lambda \, dl$

Surface mass density $\sigma$ (mass per unit area, for laminas, shells):

$dm = \sigma \, dA$

Volume mass density $\rho$ (mass per unit volume, for solid bodies):

$dm = \rho \, dV$

For uniform bodies, $\lambda$, $\sigma$, or $\rho$ are constant and can be taken outside the integral.

General method

1. Choose a coordinate system with origin at a convenient point (often a symmetry point)
2. Choose an appropriate mass element $dm$ (strip, shell, disc, etc.)
3. Express $dm$ in terms of the coordinate variable
4. Set up and evaluate the integral
5. Divide by total mass $M$

Example: Uniform rod of length $L$

Take the left end as origin. Linear mass density $\lambda = M/L$ (uniform).

$dm = \lambda \, dx = \frac{M}{L} dx$

$x_{cm} = \frac{1}{M}\int_0^L x \cdot \frac{M}{L} dx = \frac{1}{L}\int_0^L x \, dx = \frac{1}{L} \cdot \frac{L^2}{2} = \frac{L}{2}$

CM is at the midpoint — expected by symmetry.

Example: Uniform triangular lamina

For a triangle with base $b$ and height $h$, base along the $x$-axis:

At height $y$, the width of the strip is $b\left(1 - \frac{y}{h}\right)$ (by similar triangles).

$dm = \sigma \cdot b\left(1 - \frac{y}{h}\right) dy$

$y_{cm} = \frac{\int_0^h y \cdot \sigma b\left(1-\frac{y}{h}\right) dy}{\sigma \cdot \frac{1}{2}bh}$

Numerator: $\sigma b \int_0^h \left(y - \frac{y^2}{h}\right) dy = \sigma b \left[\frac{h^2}{2} - \frac{h^2}{3}\right] = \frac{\sigma b h^2}{6}$

$y_{cm} = \frac{\sigma b h^2/6}{\sigma b h/2} = \frac{h}{3}$

CM is at $h/3$ from the base — consistent with the triangle result.

Symmetry shortcut

For bodies with a line or plane of symmetry, the CM lies on that symmetry element. This can eliminate one or more integrals entirely.

A semicircular disc has a line of symmetry along the $y$-axis → $x_{cm} = 0$. Only $y_{cm}$ needs to be computed.