Centre of Mass: Discrete Particles

The centre of mass is the single point at which the whole mass of a system may be regarded as concentrated. It is the point that moves as if all external forces acted there, the point about which gravity exerts no net turning effect, and the balance point of the system. This lesson establishes the defining formula for a set of discrete particles, develops it in one, two and three dimensions, and shows the powerful "remove-a-particle" technique that sets up the lamina work in the next lesson. Mastering discrete centre of mass is the foundation for laminae, composite bodies, toppling, and the moment-of-inertia and angular-momentum topics later in the option.

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1 Where this sits in AQA 7367

This is Paper 3, the Mechanics option (7367/3M). Centre of mass is a named topic and a gateway: the discrete formula here generalises by integration to the laminae of the next lesson, and underpins toppling, equilibrium of rigid bodies, and the parallel-axis reasoning of moments of inertia. The work is largely AO1 (apply a standard technique) with AO2 reasoning when you justify which particle to remove or interpret a balance condition; the inverse "find the missing mass/position given the centre of mass" problems are the AO3 stretch.

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2 Core theory — the defining formula

For $n$n particles of masses $m_1, m_2, \ldots, m_n$m1​,m2​,…,mn​ at position vectors $\mathbf{r}_1, \ldots, \mathbf{r}_n$r1​,…,rn​, the centre of mass has position vector

$$\boxed{\ \bar{\mathbf{r}} = \frac{\sum_i m_i \mathbf{r}_i}{\sum_i m_i} = \frac{1}{M}\sum_i m_i \mathbf{r}_i\ }, \qquad M = \sum_i m_i .$$ rˉ=∑i​mi​∑i​mi​ri​​=M1​i∑​mi​ri​ ​,M=i∑​mi​.

Resolving into Cartesian components, this is a pair (or in 3D a triple) of independent scalar equations:

$$\bar{x} = \frac{\sum_i m_i x_i}{M}, \qquad \bar{y} = \frac{\sum_i m_i y_i}{M}, \qquad \bar{z} = \frac{\sum_i m_i z_i}{M}.$$xˉ=M∑i​mi​xi​​,yˉ​=M∑i​mi​yi​​,zˉ=M∑i​mi​zi​​.

The quantity $\sum m_i x_i$∑mi​xi​ is the first moment of mass about the $y$y-axis; dividing by the total mass gives the mass-weighted average position. Two facts follow immediately and are worth internalising:

• The centre of mass is a weighted average, so it always lies within the convex hull of the particles — never outside the smallest region containing them — and is pulled towards the heavier particles.
• It is origin-independent as a physical point. You may put the origin anywhere convenient; the coordinates of $\bar{\mathbf r}$rˉ change with the origin but the point itself does not. Choosing a clever origin (a corner, a heavy particle) simplifies the arithmetic.

Why this is the balance point

Taking moments about the centre of mass, $\sum m_i (\mathbf r_i - \bar{\mathbf r}) = \sum m_i \mathbf r_i - M\bar{\mathbf r} = M\bar{\mathbf r} - M\bar{\mathbf r} = \mathbf 0$∑mi​(ri​−rˉ)=∑mi​ri​−Mrˉ=Mrˉ−Mrˉ=0. In a uniform gravitational field each weight is $m_i g$mi​g, so the total moment of the weights about $\bar{\mathbf r}$rˉ vanishes — the system balances there, and this is why the centre of mass coincides with the centre of gravity in uniform gravity.

The two-particle case as a check

For just two particles $m_1$m1​ at $x_1$x1​ and $m_2$m2​ at $x_2$x2​, the formula gives $\bar x = \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$xˉ=m1​+m2​m1​x1​+m2​x2​​. Writing the distances from $\bar x$xˉ as $d_1 = \bar x - x_1$d1​=xˉ−x1​ and $d_2 = x_2 - \bar x$d2​=x2​−xˉ, a line of algebra shows $m_1 d_1 = m_2 d_2$m1​d1​=m2​d2​ — the principle of moments (a "see-saw" balance): the centre of mass divides the line joining two particles in the inverse ratio of their masses. So a $2\,\text{kg}$2kg and a $6\,\text{kg}$6kg particle balance about a point three times closer to the $6\,\text{kg}$6kg mass. Keeping this picture in mind is an excellent sanity check on any two-particle answer.

Combining frameworks

Several particles already grouped into a sub-system can be replaced by a single particle of their combined mass placed at their centre of mass — this is the associativity of the weighted average, and it is what makes composite-body problems (and the laminae of the next lesson) tractable. For instance, three particles forming the top of a structure can be collapsed to one equivalent particle, then combined with the rest; the answer is identical to treating all of them at once, but the arithmetic is staged and less error-prone.

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3 Worked examples with mark scheme

Example 1 — two particles on a line

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

$$\bar{x} = \frac{3(2) + 5(6)}{3+5} = \frac{6 + 30}{8} = \frac{36}{8} = 4.5. \quad (\textbf{M1 A1})$$xˉ=3+53(2)+5(6)​=86+30​=836​=4.5.(M1 A1)

The centre of mass is at $x = 4.5$x=4.5 — between the particles but closer to the heavier $5\,\text{kg}$5kg mass, as expected. (M1 for $\sum m_i x_i / \sum m_i$∑mi​xi​/∑mi​; A1 for $4.5$4.5.)

Example 2 — three particles in two dimensions

Masses $2\,\text{kg}$2kg at $(1,3)$(1,3), $4\,\text{kg}$4kg at $(5,1)$(5,1), $6\,\text{kg}$6kg at $(3,5)$(3,5).

$M = 2 + 4 + 6 = 12\,\text{kg}.$M=2+4+6=12kg. (M1)

$$\bar{x} = \frac{2(1) + 4(5) + 6(3)}{12} = \frac{2 + 20 + 18}{12} = \frac{40}{12} = \frac{10}{3} \approx 3.33, \quad (\textbf{M1 A1})$$xˉ=122(1)+4(5)+6(3)​=122+20+18​=1240​=310​≈3.33,(M1 A1) $$\bar{y} = \frac{2(3) + 4(1) + 6(5)}{12} = \frac{6 + 4 + 30}{12} = \frac{40}{12} = \frac{10}{3} \approx 3.33. \quad (\textbf{A1})$$yˉ​=122(3)+4(1)+6(5)​=126+4+30​=1240​=310​≈3.33.(A1)

Centre of mass $\left(\tfrac{10}{3}, \tfrac{10}{3}\right)$(310​,310​). (M1 total mass; M1 the moment sum; A1 $\bar x$xˉ; A1 $\bar y$yˉ​. Compute $\bar x$xˉ and $\bar y$yˉ​ independently — a tabulated layout prevents mixing the columns.)

Example 3 — choosing the origin (rod)

A light rod $AB$AB of length $2\,\text{m}$2m carries $4\,\text{kg}$4kg at $A$A and $6\,\text{kg}$6kg at $B$B. Find the distance of the centre of mass from $A$A.

Take $A$A as origin, $B$B at $x = 2$x=2. (M1)

$$\bar{x} = \frac{4(0) + 6(2)}{4 + 6} = \frac{12}{10} = 1.2\,\text{m from } A. \quad (\textbf{M1 A1})$$xˉ=4+64(0)+6(2)​=1012​=1.2m from A.(M1 A1)

(M1 for a sensible choice of origin; M1 for the formula; A1 $1.2\,\text{m}$1.2m. The rod is "light", so it contributes no mass — only the attached particles count.)

Example 4 — three dimensions

Masses $2\,\text{kg}$2kg at $(1,0,3)$(1,0,3) and $3\,\text{kg}$3kg at $(4,2,1)$(4,2,1). Find the centre of mass.

$M = 5\,\text{kg}$M=5kg, and the three components are computed independently:

$$\bar{x} = \frac{2(1)+3(4)}{5} = \frac{14}{5} = 2.8, \quad \bar{y} = \frac{2(0)+3(2)}{5} = \frac{6}{5} = 1.2, \quad \bar{z} = \frac{2(3)+3(1)}{5} = \frac{9}{5} = 1.8.$$xˉ=52(1)+3(4)​=514​=2.8,yˉ​=52(0)+3(2)​=56​=1.2,zˉ=52(3)+3(1)​=59​=1.8.

Centre of mass $(2.8, 1.2, 1.8)$(2.8,1.2,1.8). (M1 A1 A1 A1) (The third dimension is no harder — one extra column in the same table. M1 for the method; one A1 per coordinate.)

Example 5 — an inverse problem

Particles of mass $2\,\text{kg}$2kg, $5\,\text{kg}$5kg and $m\,\text{kg}$mkg lie at $x = 0$x=0, $x = 4$x=4 and $x = 10$x=10 on a light rod. The centre of mass is at $x = 5$x=5. Find $m$m.

Set up $\bar x = 5$xˉ=5 as an equation in $m$m: (M1)

$$\frac{2(0) + 5(4) + m(10)}{2 + 5 + m} = 5 \ \Longrightarrow\ 20 + 10m = 5(7 + m) = 35 + 5m. \quad (\textbf{M1})$$2+5+m2(0)+5(4)+m(10)​=5 ⟹ 20+10m=5(7+m)=35+5m.(M1) $$5m = 15 \ \Longrightarrow\ m = 3\,\text{kg}. \quad (\textbf{A1})$$5m=15 ⟹ m=3kg.(A1)

(M1 for forming the equation; M1 for clearing the fraction; A1 $m = 3$m=3. Inverse problems are won by writing "$\bar x =$xˉ= given value" as an equation and solving — never by trial and error.)

Example 6 — a framework that hangs in equilibrium

Three particles of mass $1\,\text{kg}$1kg, $1\,\text{kg}$1kg and $2\,\text{kg}$2kg are fixed at the vertices $A(0,0)$A(0,0), $B(4,0)$B(4,0), $C(0,3)$C(0,3) of a light right-angled triangular framework. The framework is freely suspended from $A$A. Find the centre of mass, and the angle $AB$AB makes with the vertical when it hangs in equilibrium.

First the centre of mass: $M = 1+1+2 = 4\,\text{kg}$M=1+1+2=4kg,

$$\bar x = \frac{1(0)+1(4)+2(0)}{4} = \frac{4}{4} = 1, \qquad \bar y = \frac{1(0)+1(0)+2(3)}{4} = \frac{6}{4} = 1.5. \quad (\textbf{M1 A1 A1})$$xˉ=41(0)+1(4)+2(0)​=44​=1,yˉ​=41(0)+1(0)+2(3)​=46​=1.5.(M1 A1 A1)

When a body is freely suspended from a point, it hangs so that its centre of mass lies vertically below that point. So the line $A\bar{G}$AGˉ, from $A(0,0)$A(0,0) to $\bar G(1, 1.5)$Gˉ(1,1.5), becomes vertical. The angle $AB$AB (originally along the $x$x-axis) makes with the vertical equals the angle between $AB$AB and $A\bar G$AGˉ. The direction $A\bar G$AGˉ makes with the horizontal $AB$AB is

$$\tan\alpha = \frac{1.5}{1} = 1.5 \ \Rightarrow\ \alpha = \arctan 1.5 \approx 56.3^\circ. \quad (\textbf{M1 A1})$$tanα=11.5​=1.5 ⇒ α=arctan1.5≈56.3∘.(M1 A1)

Since $A\bar G$AGˉ is vertical in equilibrium, $AB$AB makes $56.3^\circ$56.3∘ with that vertical. (A1) (M1 A1 A1 for the centre of mass; M1 for using "centre of mass vertically below the suspension point"; A1 for $\tan\alpha = \bar y/\bar x$tanα=yˉ​/xˉ; A1 for the angle. Suspension problems always reduce to: find $\bar G$Gˉ, then make the line from the pivot to $\bar G$Gˉ vertical.)

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4 Specimen-style exam question

(specimen-style — not from any past paper)

Five particles are placed at the corners and centre of a rectangle as follows: $A$A ($1\,\text{kg}$1kg) at $(0,0)$(0,0), $B$B ($3\,\text{kg}$3kg) at $(4,0)$(4,0), $C$C ($2\,\text{kg}$2kg) at $(4,3)$(4,3), $D$D ($4\,\text{kg}$4kg) at $(0,3)$(0,3), $E$E ($5\,\text{kg}$5kg) at $(2,1)$(2,1). (a) Find the centre of mass of the system. (b) Particle $E$E is now removed. Find the new centre of mass.

Model solution. Tabulate and sum. $M = 1+3+2+4+5 = 15\,\text{kg}.$M=1+3+2+4+5=15kg.

$$\bar{x} = \frac{1(0)+3(4)+2(4)+4(0)+5(2)}{15} = \frac{0+12+8+0+10}{15} = \frac{30}{15} = 2,$$xˉ=151(0)+3(4)+2(4)+4(0)+5(2)​=150+12+8+0+10​=1530​=2, $$\bar{y} = \frac{1(0)+3(0)+2(3)+4(3)+5(1)}{15} = \frac{0+0+6+12+5}{15} = \frac{23}{15} \approx 1.53.$$yˉ​=151(0)+3(0)+2(3)+4(3)+5(1)​=150+0+6+12+5​=1523​≈1.53.

So the centre of mass is $\left(2, \tfrac{23}{15}\right)$(2,1523​). (a)

(b) Removing $E$E (mass $5$5 at $(2,1)$(2,1)) uses $\bar x_{\text{new}} = \dfrac{M\bar x - m_k x_k}{M - m_k}$xˉnew​=M−mk​Mxˉ−mk​xk​​:

$$\bar{x}_{\text{new}} = \frac{15(2) - 5(2)}{15 - 5} = \frac{30 - 10}{10} = 2, \qquad \bar{y}_{\text{new}} = \frac{15\cdot\frac{23}{15} - 5(1)}{10} = \frac{23 - 5}{10} = 1.8.$$xˉnew​=15−515(2)−5(2)​=1030−10​=2,yˉ​new​=1015⋅1523​−5(1)​=1023−5​=1.8.

New centre of mass $(2, 1.8)$(2,1.8). Removing the central particle leaves $\bar x$xˉ unchanged (it sat on the vertical line of symmetry of the remaining masses) but raises $\bar y$yˉ​, since $E$E was below the others.

A note on the removal formula

The "remove-a-particle" relation is just the defining formula with a negative mass. If the full system has moment $M\bar x = \sum m_i x_i$Mxˉ=∑mi​xi​, then deleting the term $m_k x_k$mk​xk​ leaves moment $M\bar x - m_k x_k$Mxˉ−mk​xk​ and total mass $M - m_k$M−mk​, so

$$\bar x_{\text{new}} = \frac{M\bar x - m_k x_k}{M - m_k}.$$xˉnew​=M−mk​Mxˉ−mk​xk​​.

Thinking of the removed particle as carrying negative mass $-m_k$−mk​ is the conceptual key that unlocks the subtraction method for laminae with holes in the next lesson — a hole is simply a region of negative area (and hence negative mass) superposed on the full shape.

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5 Synoptic links

• Vectors (A-Level Maths / Further pure). $\bar{\mathbf r} = \frac1M\sum m_i\mathbf r_i$rˉ=M1​∑mi​ri​ is a weighted vector sum; the unweighted (equal-mass) case is the centroid $\frac1n\sum\mathbf r_i$n1​∑ri​, exactly the mean position used in vector geometry.
• Laminae (next lesson). Replacing $\sum m_i$∑mi​ by $\int \mathrm{d}m$∫dm generalises every formula here to continuous bodies; the discrete "remove-a-particle" trick becomes the "subtract-a-hole" trick.
• Moments and equilibrium. The first moment of mass $\sum m_i x_i$∑mi​xi​ is the same quantity you take moments of when analysing a balanced beam.
• Statistics. $\bar x = \sum f_i x_i / \sum f_i$xˉ=∑fi​xi​/∑fi​ is literally the formula for the mean of grouped data — frequencies play the role of masses.

Physical interpretation worth carrying into the exam