Centre of Mass: Discrete Particles

This lesson covers finding the centre of mass of a system of discrete particles. The centre of mass (also called the centre of gravity in a uniform gravitational field) is the single point at which the entire mass of the system can be considered to act.

---

1 Definition

For $n$n particles with masses $m_1, m_2, \ldots, m_n$m1​,m2​,…,mn​ at positions $(x_1, y_1), (x_2, y_2), \ldots, (x_n, y_n)$(x1​,y1​),(x2​,y2​),…,(xn​,yn​):

$$\boxed{\bar{x} = \frac{\sum m_i x_i}{\sum m_i}, \quad \bar{y} = \frac{\sum m_i y_i}{\sum m_i}}$$xˉ=∑mi​∑mi​xi​​,yˉ​=∑mi​∑mi​yi​​​

In vector form:

$$\bar{\mathbf{r}} = \frac{\sum m_i \mathbf{r}_i}{\sum m_i} = \frac{\sum m_i \mathbf{r}_i}{M}$$rˉ=∑mi​∑mi​ri​​=M∑mi​ri​​

where $M = \sum m_i$M=∑mi​ is the total mass.

---

2 Physical Interpretation

• If the system is supported at the centre of mass, it balances.
• The system moves as if all external forces act at the centre of mass.
• In a uniform gravitational field, the centre of mass coincides with the centre of gravity.

---

3 Worked Examples

Example 1 — Two Particles on a Line

Masses 3 kg at $x = 2$x=2 and 5 kg at $x = 6$x=6. Find $\bar{x}$xˉ.

Solution

$\bar{x} = \frac{3(2) + 5(6)}{3 + 5} = \frac{6 + 30}{8} = \frac{36}{8} = 4.5$xˉ=3+53(2)+5(6)​=86+30​=836​=4.5

The centre of mass is at $x = 4.5$x=4.5, closer to the heavier mass.

---

Example 2 — Three Particles in 2D

Masses: 2 kg at $(1, 3)$(1,3), 4 kg at $(5, 1)$(5,1), 6 kg at $(3, 5)$(3,5).

Solution

$M = 2 + 4 + 6 = 12$M=2+4+6=12 kg.

$\bar{x} = \frac{2(1) + 4(5) + 6(3)}{12} = \frac{2 + 20 + 18}{12} = \frac{40}{12} = \frac{10}{3} \approx 3.33$xˉ=122(1)+4(5)+6(3)​=122+20+18​=1240​=310​≈3.33

$\bar{y} = \frac{2(3) + 4(1) + 6(5)}{12} = \frac{6 + 4 + 30}{12} = \frac{40}{12} = \frac{10}{3} \approx 3.33$yˉ​=122(3)+4(1)+6(5)​=126+4+30​=1240​=310​≈3.33

Centre of mass: $(10/3, 10/3)$(10/3,10/3).

---

Example 3 — Particles on Vertices of a Triangle

Three particles of masses 1 kg, 2 kg, and 3 kg are placed at the vertices $(0, 0)$(0,0), $(6, 0)$(6,0), and $(3, 4)$(3,4) respectively.

Solution

$M = 6$M=6 kg.

$\bar{x} = \frac{1(0) + 2(6) + 3(3)}{6} = \frac{0 + 12 + 9}{6} = \frac{21}{6} = 3.5$xˉ=61(0)+2(6)+3(3)​=60+12+9​=621​=3.5

$\bar{y} = \frac{1(0) + 2(0) + 3(4)}{6} = \frac{12}{6} = 2$yˉ​=61(0)+2(0)+3(4)​=612​=2

Centre of mass: $(3.5, 2)$(3.5,2).

---

4 Centre of Mass on a Rod

For particles attached to a rod (or along a line), the problem reduces to one dimension.

Example 4

A light rod AB has length 2 m. Masses of 4 kg and 6 kg are attached at A and B respectively. Find the distance of the centre of mass from A.

Solution