Centres of mass of particles and laminas

What you’ll learn

• Calculate mass-weighted coordinates.
• Handle composite laminas, holes and suspension.

Understand the idea

The centre of mass is the point at which a body's weight can be considered to act in a uniform gravitational field. For discrete particles, each coordinate is a mass-weighted mean. Symmetry can determine a coordinate immediately and should be used before doing arithmetic.

Split a composite uniform lamina into familiar shapes with known centroids. With constant thickness and density, mass is proportional to area; represent a removed piece by a negative area. When a body hangs freely in equilibrium from one point, its centre of mass lies vertically below the suspension point.

Key formulae

$$\bar x=\frac{\sum m_ix_i}{\sum m_i},\qquad\bar y=\frac{\sum m_iy_i}{\sum m_i}$$

$$\bar x=\frac{\sum A_ix_i}{\sum A_i}\quad\text{(uniform lamina)}$$

$$(\bar x,\bar y)_{\text{triangle}}=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$$

Worked example

Particles of masses $2\,\mathrm{kg}$ and $3\,\mathrm{kg}$ are at $(0,1)$ and $(5,6)$ metres. Find their centre of mass.

1. The total mass is $5\,\mathrm{kg}$.
2. The x-coordinate is $(2(0)+3(5))/5=3$.
3. The y-coordinate is $(2(1)+3(6))/5=4$.

Answer: $(3,4)\,\mathrm m$.