Centre of mass — definition for system of particles: Question bank

Recall the one formula everything below leans on:

Two pictures to keep in your head

Before the traps, two geometric words appear repeatedly. Meet them once with a picture so they never trip you.

Figure — Centre of mass — definition for system of particles

The second picture is the simplest possible case — two masses on a line — which almost every trap secretly reduces to:

One more property the traps rely on — and that beginners rarely notice:

True or false — justify

Each claim is stated the way a student might believe it. Decide true/false, then give the reason.

The centre of mass is always located at one of the particles.

False — it is an average position, so it usually sits in empty space between the particles (e.g. midway between two equal masses, where no mass exists).

Adding more particles at the current COM position shifts the COM.

False — a particle placed exactly at the existing

r_{c m}

contributes the same position as the average, so the weighted average (and hence the COM) does not move; only

M

grows.

If you double every mass in the system, the COM stays in the same place.

True — doubling every

m_{i}

multiplies numerator and denominator by

2

, which cancels; the COM depends on mass ratios, not absolute masses.

The COM must lie inside the bounding box (smallest upright rectangle) that contains all the particles.

True — a weighted average of coordinates can never exceed the largest or fall below the smallest coordinate on any axis, so it stays within those four edges (see the yellow box above).

Two systems with the same total mass always have the same COM.

False — the COM depends on where each mass sits, not just on the total; two systems can share

M

yet have wildly different mass distributions and COMs.

Sliding a particle purely along the

x

-axis changes

x_{c m}

but leaves

y_{c m}

untouched.

True — each axis is an independent weighted average, and

y_{c m}

uses only the

y_{i}

values; moving a particle without changing its

y_{i}

cannot change

y_{c m}

The COM of a system can lie outside the convex hull (the rubber-band shape) of the particles.

False — a weighted average with positive weights (masses) is always a convex combination, so it lies within the rubber-band shape, never poking outside it.

If all masses are equal, the COM is the plain geometric average of the positions.

True — equal weights cancel in numerator and denominator, leaving

r_{c m} = \frac{1}{n} \sum r_{i}

, the ordinary centroid.

The COM's location changes if you move your coordinate origin.

False — the COM is a physical point fixed among the masses; shifting the origin by

a

shifts its coordinates by

- a

but leaves the actual point exactly where it was (see the two-origin figure).

Spot the error

Each item is a wrong worked statement. Name the mistake and correct the reasoning.

x_{c m} = \sum m_{i} x_{i}

, so the COM of

2

kg at

0

and

3

kg at

5

m is

15

m."

The error is forgetting to divide by

M

\sum m_{i} x_{i} = 15

has units kg·m (a moment), not metres; dividing by

M = 5

kg gives the true

x_{c m} = 3

"COM of a

1

kg mass at

0

and a

9

kg mass at

10

m is the midpoint,

5

m."

The error is ignoring mass weighting. The heavier

9

kg mass drags the average toward itself, giving

x_{c m} = \frac{0 + 90}{10} = 9

m, far from the midpoint.

"To find

y_{c m}

, plug the

x

-positions into the

y

formula."

The error mixes axes.

y_{c m}

uses only the

y_{i}

coordinates; each axis is a separate scalar average, so

x

-values must never appear in the

y

calculation.

"The COM velocity is

v_{c m} = \sum m_{i} v_{i}

Missing the division by

M

again. Total momentum is

P = \sum m_{i} v_{i}

, but

v_{c m} = P / M

; the momentum itself is not the velocity.

"Since particle 3 is heaviest, the COM sits exactly on particle 3."

The error is treating "leans toward heavy" as "sits on heavy." The COM only shifts toward the heaviest mass; it reaches a particle only if all the other masses are zero.

"A massless particle at

x = 100

m pulls the COM toward

100

m."

The error assumes every particle influences the average. A particle with

m_{i} = 0

contributes

0 \cdot x_{i} = 0

to the numerator and

0

M

, so it has no effect on the COM at all.

"The COM of an L-shaped set of masses is the corner where the arms meet."

The error assumes the COM lies at a visually "central" particle. The COM is a mass-weighted average of the actual positions and generally sits off any single particle, somewhere inside the L's spread.

Why questions

These probe the reasoning behind the formula, not the arithmetic.

Why do we divide by the total mass

M

instead of by the number of particles

n

Dividing by

M

makes each mass contribute in proportion to how heavy it is (a weighted average); dividing by

n

would treat a

100

kg and a

1

kg mass as equally important. Concretely,

2

kg at

0

and

3

kg at

5

m gives

\frac{15}{5} = 3

m (mass-weighted), not

\frac{5}{2} = 2.5

m (plain average) — the extra kilogram earns the heavy mass more pull.

Why is the COM defined through momentum rather than, say, energy?

Because Newton's laws act on momentum (

F = d P / d t

), so anchoring the COM to

P = M v_{c m}

is precisely what lets the whole system obey

F_{e x t} = M a_{c m}

like one particle — see Newton's Second Law for a System of Particles. Energy is a scalar and carries no direction, so it could never pin down where the representative point sits.

Why does the COM lean toward the heavier mass?

Because each position is weighted by its mass in the sum; a large

m_{i}

multiplies its

x_{i}

heavily, dragging the weighted average toward that particle. Picture the see-saw figure: to balance, the pivot must sit close to the big kid so their smaller lever arm times bigger weight equals the small kid's larger arm times smaller weight.

Why can we treat

x

y

, and

z

separately when the definition is a vector equation?

A vector equation holds component-by-component, and the mass weighting

m_{i}

is the same number in each component. So the one vector average splits cleanly into three independent scalar averages —

x_{c m}

never needs any

y

-value, and vice versa.

Why does the COM velocity stay constant when there is no external force?

Because

F_{e x t} = M a_{c m}

, so zero external force means zero COM acceleration, i.e. constant

v_{c m}

— this is Conservation of Linear Momentum seen from the COM. Internal forces come in action–reaction pairs that cancel in the sum, so they can never budge the COM.

Why is analysing a collision easier in the COM frame?

In the COM frame the total momentum is zero, so the two objects always approach and separate symmetrically, collapsing many variables into one — the backbone of Collisions — Elastic and Inelastic.

Why does replacing

\sum

with

\int

give the COM of a solid body?

Chop the body into

n

tiny chunks, each a "particle" of mass

Δ m_{k}

at position

r_{k}

; the discrete formula gives

\frac{1}{M} \sum_{k} r_{k} Δ m_{k}

. Now shrink the chunks so

n \to \infty

and

Δ m_{k} \to 0

: that limit of a sum of tiny pieces is exactly the definition of the integral

\frac{1}{M} \int r d m

— see Centre of Mass of Continuous Bodies.

Why is the COM formula identical to the "moment" idea from a see-saw?

Both are weighted averages: balancing torques

\sum m_{i} g (x_{i} - x_{c m}) = 0

about the COM rearranges to

x_{c m} = \frac{\sum m _{i} x _{i}}{M}

, which is why Weighted Average and Moments shares the maths. The COM is literally the balance point where all the mass-moments cancel.

Edge cases

Boundary and degenerate scenarios the formula must still survive.

Where is the COM of a single particle?

Exactly at that particle — with one mass,

r_{c m} = \frac{m r}{m} = r

, so the average of one thing is that thing.

What is the COM of a uniform ring of particles?

The geometric centre of the ring, which is empty space — a clean proof that the COM need not coincide with any mass.

If one particle's mass grows without bound while others stay fixed, where does the COM go?

It approaches that particle's position, because in the limit

m_{1} \to \infty

the term

m_{1} r_{1}

dominates both numerator and denominator, so

r_{c m} \to r_{1}

Can the COM of a two-particle system fall outside the segment joining them?

No — with positive masses it is a convex combination

r_{c m} = \frac{m _{1} r _{1} + m _{2} r _{2}}{m _{1} + m _{2}}

, which always lands between the two points, never beyond.

What happens to the COM if all masses are equal and symmetrically placed about a point?

The COM sits exactly at that symmetry point, since equal weights make it the centroid and symmetry balances every displacement.

If two particles occupy the same position, what is the COM?

That shared position — both terms point to the same place, so the weighted average is that place regardless of the mass split.

Does giving a system zero total mass make sense in this formula?

No —

M = 0

makes the definition divide by zero; physically every real system has positive total mass, so this case is excluded, not answered.

Recall One-line summary of the trap families

The traps come in a few flavours: (1) dropping the mass weighting $m_{i}$ ; (2) forgetting to divide by $M$ ; (3) mixing up axes; (4) assuming the COM must sit on a particle or at a "visually central" spot; and (5) thinking the origin or absolute mass scale matters. Guard all five and the COM never surprises you.

Connections

Parent: COM definition — the formulas these traps test.
Newton's Second Law for a System of Particles — why momentum anchors the definition.
Conservation of Linear Momentum — the "no external force" edge cases.
Collisions — Elastic and Inelastic — where the COM frame pays off.
Centre of Mass of Continuous Bodies — the $\sum \to \int$ limit case.
Weighted Average and Moments — the see-saw analogy behind the "why" items.