Centre of mass — definition for system of particles: Worked examples

This is a deep-dive companion to the parent note on Centre of Mass. There we derived the formula $r_{c m} = \frac{1}{M} \sum_{i} m_{i} r_{i}, M = \sum_{i} m_{i} .$ Here we do the opposite of proving: we stress-test it. We hunt down every kind of input the formula can meet — positive coordinates, negative coordinates, zeros, equal masses, a mass so big it swallows the others, a mass that vanishes, and finally a real-world word problem and an exam trap. If you can walk through all of these, no COM question can surprise you.

The scenario matrix

Every COM problem for point particles falls into one of these cells. The examples that follow are labelled with the cell they cover, and together they hit all of them.

Cell	What makes it tricky	Example
A. All-positive, 1-D	plain baseline, no sign issues	Ex 1
B. Mixed sign, 1-D	positions on both sides of origin — cancellation	Ex 2
C. Equal masses	weights cancel → COM is plain midpoint/centroid	Ex 3
D. Degenerate: a zero mass	one particle contributes nothing; formula must survive	Ex 4
E. 2-D, mixed sign	do $x$ and $y$ separately, signs in both	Ex 5
F. Limiting: one mass $\to$ huge	COM collapses onto the dominant mass	Ex 6
G. Real-world word problem	translate a scene into masses + positions	Ex 7
H. Exam twist: find the unknown	COM is given, solve for a mass or position	Ex 8

The one rule we will lean on in every cell:

Cell A — All-positive, 1-D (the baseline)

Worked example Ex 1: Two masses, both to the right

$m_{1} = 2$ kg at $x_{1} = 1$ m, $m_{2} = 6$ kg at $x_{2} = 9$ m. Find $x_{c m}$ .

Forecast: The $6$ kg mass is three times heavier and sits at $9$ m. Guess: COM should be much closer to $9$ than to $1$ . Somewhere around $7$ ? Write your number down.

Step 1 — Total mass. $M = 2 + 6 = 8$ kg. Why this step? It is the denominator; nothing is a position until we divide by it.

Step 2 — Build the numerator (the moment). $\sum m_{i} x_{i} = 2 (1) + 6 (9) = 2 + 54 = 56 kg\cdotm .$ Why this step? Each mass "casts $m_{i}$ votes" for its own position; we add all votes.

Step 3 — Divide. $x_{c m} = \frac{56}{8} = 7 m .$ Why this step? Division turns the kg·m moment back into a metre position.

Verify: $7$ m is between $1$ and $9$ ✔, and it leans toward the heavy mass at $9$ ✔ (distance $2$ from $9$ vs. distance $6$ from $1$ — ratio $2 : 6 = m_{1} : m_{2}$ ✔, the see-saw balance). Matches the forecast.

Figure — Centre of mass — definition for system of particles

Cell B — Mixed sign, 1-D (cancellation)

Here comes the first real trap: positions on both sides of the origin. The moment of a mass at negative $x$ is negative — it pulls the average left.

Cell C — Equal masses (weights cancel)

Worked example Ex 3: Three equal masses at the corners of a triangle

Three masses, each $m = 2$ kg, at $(0, 0)$ , $(6, 0)$ , $(3, 6)$ . Find the COM.

Forecast: Equal masses vote equally — so the mass drops out and the COM is the plain centroid (average of the corners). Guess before computing.

Step 1 — Note the cancellation. With every $m_{i} = m$ , $x_{c m} = \frac{m x _{1} + m x _{2} + m x _{3}}{3 m} = \frac{x _{1} + x _{2} + x _{3}}{3} .$ Why this step? The common factor $m$ appears in top and bottom and cancels — that is why equal masses give the ordinary average.

Step 2 — Average the $x$ 's. $x_{c m} = \frac{0 + 6 + 3}{3} = \frac{9}{3} = 3$ m.

Step 3 — Average the $y$ 's. $y_{c m} = \frac{0 + 0 + 6}{3} = \frac{6}{3} = 2$ m.

Verify: COM $= (3, 2)$ m. It sits inside the triangle ✔ (a mass average of the corners can never fall outside their convex hull). The centroid of a triangle is the average of its vertices — matches ✔.

Cell D — Degenerate input: a zero mass

What if one particle has mass $0$ ? The formula must not break; the zero-mass particle simply contributes nothing.

Worked example Ex 5: Four particles, all quadrants

$m_{1} = 1$ kg at $(2, 3)$ , $m_{2} = 2$ kg at $(- 4, 1)$ , $m_{3} = 1$ kg at $(0, - 5)$ , $m_{4} = 2$ kg at $(3, - 1)$ . Find the COM.

Forecast: The $x$ -votes: right ( $+ 2, + 3$ ) vs. left ( $- 4$ , weight $2$ ). Left has a heavy voter — expect $x_{c m}$ near zero or slightly negative. The $y$ -votes have a lone downward pull at $- 5$ — expect $y_{c m}$ below zero.

Step 1 — Total mass. $M = 1 + 2 + 1 + 2 = 6$ kg.

Step 2 — $x$ -moment (signs!). $\sum m_{i} x_{i} = 1 (2) + 2 (- 4) + 1 (0) + 2 (3) = 2 - 8 + 0 + 6 = 0 kg\cdotm .$ Why this step? The right and left votes exactly cancel — a genuine mixed-sign cancellation.

Step 3 — $x_{c m}$ . $x_{c m} = \frac{0}{6} = 0$ m. A COM sitting on the $y$ -axis.

Step 4 — $y$ -moment. $\sum m_{i} y_{i} = 1 (3) + 2 (1) + 1 (- 5) + 2 (- 1) = 3 + 2 - 5 - 2 = - 2 kg\cdotm .$

Step 5 — $y_{c m}$ . $y_{c m} = \frac{- 2}{6} = - \frac{1}{3} \approx - 0.333$ m.

Verify: COM $= (0, - 0.33)$ m. It landed on the $y$ -axis (as $x$ -votes cancelled) and slightly below the $x$ -axis (net downward pull) ✔ — both match the forecast. It sits in empty space, touching no particle — perfectly legal for a COM.

Cell F — Limiting case: one mass dominates

Cell G — Real-world word problem

Worked example Ex 7: Two people on a boat

A $60$ kg woman stands at the front and an $80$ kg man at the back of a light boat. The boat is $4$ m long; put the origin at the back, so the man is at $x = 0$ m and the woman at $x = 4$ m. (Treat the boat's own mass as negligible.) Where is the COM of the two people?

Forecast: The man ( $80$ kg) is heavier and at $0$ ; the woman ( $60$ kg) at $4$ . COM should lie past the midpoint ( $2$ m) toward the man, i.e. below $2$ m. Guess $\approx 1.7$ ?

Step 1 — Translate the scene. masses ${80, 60}$ kg at positions ${0, 4}$ m. Why this step? A word problem is just a table of $(m_{i}, x_{i})$ in disguise; the physics starts once it's a table.

Step 2 — Total mass. $M = 80 + 60 = 140$ kg.

Step 3 — Moment and divide. $x_{c m} = \frac{80 ( 0 ) + 60 ( 4 )}{140} = \frac{240}{140} = \frac{12}{7} \approx 1.714 m .$

Verify: $1.714$ m is between $0$ and $4$ ✔ and below the midpoint $2$ m, leaning toward the heavier man ✔ — matches the forecast. (Physical bonus: if they now swap places on a frictionless lake, this COM stays put — see Conservation of Linear Momentum — and the boat slides to compensate.)

Cell H — Exam twist: solve for the unknown

Here the COM is given and you must find a missing mass or position. Same formula, run backwards.

Worked example Ex 8: Find the missing mass

$m_{1} = 3$ kg sits at $x_{1} = 0$ ; an unknown mass $m_{2}$ sits at $x_{2} = 8$ m. The COM is measured at $x_{c m} = 6$ m. Find $m_{2}$ .

Forecast: COM at $6$ m is closer to the mass at $8$ than to the mass at $0$ . The heavier a mass, the closer COM leans to it — so $m_{2}$ must be the heavier one, bigger than $3$ kg. Guess $\approx 9$ ?

Step 1 — Write the COM equation with the unknown. $6 = \frac{3 ( 0 ) + m _{2} ( 8 )}{3 + m _{2}} .$ Why this step? We know everything except $m_{2}$ ; put it in and it becomes one equation in one unknown.

Step 2 — Clear the denominator. $6 (3 + m_{2}) = 8 m_{2} \Rightarrow 18 + 6 m_{2} = 8 m_{2} .$ Why this step? Multiplying out removes the fraction so we can isolate $m_{2}$ .

Step 3 — Solve. $18 = 2 m_{2} \Rightarrow m_{2} = 9 kg .$

Verify: Plug back: $x_{c m} = \frac{3 ( 0 ) + 9 ( 8 )}{3 + 9} = \frac{72}{12} = 6$ m ✔. And $m_{2} = 9 > 3$ kg — heavier, as forecast ✔. See-saw check: distance ratio $6 : 2$ (from each mass to COM) should equal inverse mass ratio $m_{2} : m_{1} = 9 : 3 = 3 : 1$ ; and $6 : 2 = 3 : 1$ ✔.

Recap of the matrix

Recall Which trap does each cell teach?

A — baseline, lean toward heavy mass. B — keep negative signs; COM can be negative. C — equal masses cancel → plain average. D — zero mass contributes nothing; $M = 0$ is undefined. E — do $x$ and $y$ separately; COM can be empty space. F — dominant mass swallows the COM in the limit. G — word problems are just $(m_{i}, x_{i})$ tables. H — COM given → solve backwards for the unknown.

Connections

Weighted Average and Moments — the see-saw / lever check used in Ex 1 and Ex 8 is the moment balance.
Newton's Second Law for a System of Particles — once you can locate COM, you track its motion.
Conservation of Linear Momentum — the boat problem (Ex 7) continues here.
Collisions — Elastic and Inelastic — these worked-example skills feed the COM-frame analysis.
Centre of Mass of Continuous Bodies — replace the sums here with integrals.