Mathematics/Lessons/Moments/Centres of Mass
Mechanics · Moments · Year 2

Centres of Mass

Every rigid body has a single point where its whole weight can be treated as acting. For a uniform rod that's just the midpoint — but for a non-uniform one, you have to find it using moments.

11 min read 2 worked examples

Uniform vs non-uniform bodies

So far, every rod or beam you've met has been uniform — its mass is spread evenly along its length, so its centre of mass sits exactly at the midpoint. That's why the weight of a uniform rod always acts "at the middle" in earlier moment equations.

A non-uniform rod doesn't have this symmetry: it might be thicker at one end, or made of different materials along its length. Its centre of mass can lie anywhere, and you're usually given enough information — such as the reactions at two supports — to calculate exactly where.

Centre of mass

The centre of mass of a rigid body is the single point at which its entire weight can be considered to act. For a uniform rod or beam, this is the midpoint of its length. For a non-uniform body, its position must be found from the given information — typically by taking moments.

Using a known centre of mass

If you already know where the centre of mass of a non-uniform body is, you treat the problem exactly like any other rigid-body equilibrium question: the weight acts as a single downward force at that point, and you resolve and take moments as usual.

Worked example 1

A non-uniform plank pivoted at its midpoint
A non-uniform plank $AB$ has mass $22\ \text{kg}$ and length $5\ \text{m}$. It is pivoted at $M$, the midpoint of $AB$. The centre of mass of the plank is at $C$, where $AC = 2.1\ \text{m}$. A child of mass $16\ \text{kg}$ sits at $A$. A second child of mass $26\ \text{kg}$ sits at a point $x$ metres from $A$. Modelling the plank as a rod and the children as particles, find the value of $x$ for which the plank is horizontal.
MethodTake moments about the pivot M
1
$M$ is the midpoint of $AB$, so $M$ is $2.5\ \text{m}$ from $A$. The plank's weight acts at $C$, which is $2.5-2.1=0.4\ \text{m}$ from $M$, on the $A$ side.
Locate every force relative to the pivot before writing the moment equation.
2
The child at $A$ is $2.5\ \text{m}$ from $M$, also on the $A$ side. Both the plank's weight and this child produce a moment in the same rotational sense about $M$.
Forces on the same side of the pivot turn the plank the same way.
3
Taking moments about $M$, with the second child assumed to sit on the $B$ side of the pivot at distance $(x-2.5)$ from $M$:
$26g(x-2.5) = 22g(0.4) + 16g(2.5)$
Taking moments about $M$ eliminates the unknown reaction there entirely.
4
$26(x-2.5) = 8.8+40 = 48.8 \Rightarrow x-2.5 = 1.877\ldots$
Divide by $g$ first, then by $26$.
5
$x = 4.38\ \text{m}$ (3 s.f.).
This lies between $M$ and $B$, so the assumption in step 3 is consistent.
▪ solved
A horizontal plank AB of length 5 metres resting on a pivot at M, its midpoint. The centre of mass C is marked 2.1 metres from A, with a downward arrow there representing the plank's weight of 22g newtons. A downward arrow at A represents a child of mass 16 kilograms. A second downward arrow, at an unmarked point x metres from A on the far side of the pivot, represents a child of mass 26 kilograms.
The plank pivots at its midpoint M. The centre of mass C, and both children, act as downward forces at their respective distances from A.

Finding an unknown centre of mass

More often, the centre of mass itself is the unknown. If you're given the reactions at two supports — or a relationship between them — you can take moments about one support to find where the weight must act for the body to balance.

Worked example 2

Finding the centre of mass from two reactions
A non-uniform rod $AB$ is $4\ \text{m}$ long and has weight $30\ \text{N}$. It rests horizontally on supports at $C$ and $D$, where $AC = 1.2\ \text{m}$ and $AD = 3.2\ \text{m}$. The magnitude of the reaction at $D$ is twice the magnitude of the reaction at $C$. Find the distance of the centre of mass of the rod from $A$.
MethodResolve to find both reactions, then take moments about A
1
Let the reaction at $C$ be $R$, so the reaction at $D$ is $2R$.
Introduce a single unknown for the ratio given.
2
Resolving vertically: $R+2R=30 \Rightarrow 3R=30 \Rightarrow R=10$.
The rod is in equilibrium, so total upward reaction equals total downward weight.
3
So the reaction at $C$ is $10\ \text{N}$ and the reaction at $D$ is $20\ \text{N}$.
Substitute back.
4
Let the centre of mass act at a point $x\ \text{m}$ from $A$. Taking moments about $A$: $30x = 10(1.2) + 20(3.2)$
Taking moments about $A$ brings in both known reactions and the one unknown, $x$.
5
$30x = 12+64 = 76 \Rightarrow x = \dfrac{76}{30} = 2.53\ \text{m}$ (3 s.f.).
Solve for $x$.
▪ solved
A quick sanity check

The reaction at $D$ is twice the reaction at $C$, so more of the rod's weight is supported near $D$ — you'd expect the centre of mass to sit closer to $D$ than to $C$. Here $x=2.53$ is indeed closer to $D$ (at $3.2$) than to $C$ (at $1.2$), which confirms the answer is physically sensible.

A horizontal rod AB of length 4 metres resting on two supports at C and D. C is 1.2 metres from A, with an upward arrow labelled R newtons. D is 3.2 metres from A, with an upward arrow labelled 2R newtons. A downward arrow at an unmarked point x metres from A, between C and D, represents the rod's weight of 30 newtons acting at its centre of mass.
The rod rests on two supports; the ratio of the reactions at C and D pins down exactly where the centre of mass, and hence the weight, must act.
Common mistake

Assuming the centre of mass is at the midpoint just because a question doesn't draw attention to it. Always check whether a rod or beam is described as uniform or non-uniform — the word "uniform" is doing real mathematical work. If a body is non-uniform and no position is given for its centre of mass, you almost always need to find it first, usually via moments, before the rest of the problem can be solved.

Check yourself

A non-uniform rod $AB$ is $3\ \text{m}$ long and has weight $12\ \text{N}$. It rests horizontally on supports at $C$ and $D$, where $AC=0.5\ \text{m}$ and $AD=2.5\ \text{m}$. The reaction at $C$ is three times the reaction at $D$. Find the distance of the centre of mass of the rod from $A$.

Let the reaction at $D$ be $R$, so the reaction at $C$ is $3R$. Resolving vertically: $3R+R=12 \Rightarrow R=3\ \text{N}$, so $R_C=9\ \text{N}$ and $R_D=3\ \text{N}$.

Taking moments about $A$: $12x = 9(0.5)+3(2.5) = 4.5+7.5=12 \Rightarrow x=1\ \text{m}$ from $A$.

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