Center of Mass (Two Particles)

Equivalent forms

\vec r_{cm} = \frac{\sum_i m_i \vec r_i}{\sum_i m_i}

An ancient idea: the single 'balance point' that lets a system of bodies be replaced by a single particle.

Symbol	Name	SI unit	Dimension	Range
$x_cm$	Center of mass positionoutput Position of the system COM	m	L	-100 – 100
$m1$	Mass 1 Mass of body 1	kg	M	0.1 – 100
$m2$	Mass 2 Mass of body 2	kg	M	0.1 – 100
$x1$	Position 1 Position of body 1	m	L	-10 – 10
$x2$	Position 2 Position of body 2	m	L	-10 – 10

Unit systems

Symbol	SI	CGS	Imperial
$x_cm$	m	cm	ft
$m1$	kg	g	slug
$m2$	kg	g	slug
$x1$	m	cm	ft
$x2$	m	cm	ft

Where it holds

Non-relativistic; for relativistic systems use center of energy. Generalizes to N particles and to integrals for extended bodies.

Dimensional analysis

x_cm \to [L]

(M\cdot L)/M \to [L]

Metres on both sides.

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Discovery

Archimedes · -250

Archimedes' Method of Mechanical Theorems treated centers of mass geometrically — the conceptual ancestor of modern barycentric mechanics.

Try this

Where does the Earth–Moon system pivot in space?

Find the center of mass of a system with m₁ at x₁ and m₂ at x₂ — for the Earth–Moon, where does the barycenter sit relative to Earth's center?

Research status: stable

Real-world applications

Spacecraft attitude control
Earth–Moon barycenter (used by satellites)
Athletic high-jump 'Fosbury flop' (COM stays under the bar)
Robotic manipulator dynamics

Common misconceptions

COM must lie inside the body — a ring's COM is at its empty center
COM equals geometric center — only for uniform-density symmetric bodies
External forces act 'at the COM' — they act where they're applied; only the resultant moves the COM

Experimental verification

Balance points of meter sticks, gyroscope behavior, and orbital observations (Earth–Moon barycenter measured by lunar laser ranging).

Derivation

Define x_cm so that the total momentum equals M_total

\cdot v_cm

: p_total

= \Sigma m_i v_i = M

_total (dx_cm/dt).

Integrating once gives

x_cm = (\Sigma m_i x_i) / M

_total.

Newton's second law then takes the same form for the COM as for a single particle.

Limiting cases

m_{1} = m_{2}

⟶

x_cm = (x_{1}+x_{2})/2

Equal masses → geometric midpoint.

m_{2}

≫

m_{1}

⟶

x_cm \to x_{2}

COM is dominated by the heavy body.

m_{1} \to 0

⟶

x_cm \to x_{2}

Massless particle contributes nothing.

What if…

What if we add a third particle?

$x_cm = (\Sigma m_i x_i)/(\Sigma m_i)$ — extends trivially to N particles.

What if the body is continuous?

Replace sum with integral: $x_cm = \int x dm / M$ .

What if I move

m_{2} to

the right?

x_cm shifts toward $m_{2}$ proportionally to $m_{2}/(m_{1}+m_{2})$ .

1

Two particles on a stick

Given ·

m1:: 2
m2:: 8
x1:: -3
x2:: 5

Find · x_cm

Steps

Numerator: $-6 + 40 = 34$
Denominator: $2 + 8 = 10$
$x_cm = 3.4\,\mathrm{m}$ — closer to the heavier mass

Result ·

x_cm = (2\cdot (-3) + 8\cdot 5)/10 = 34/10 = 3.4\,\mathrm{m}

2

Earth–Moon barycenter

Given ·

m1:: 5.972e+24
m2:: 7.342e+22
x1:: 0
x2:: 384400000

Find · x_cm

Steps

Numerator: $0 + 7.342e22 \cdot 3.844e_{8} \approx 2.823e31$
Denominator: 6.045e24
$x_cm \approx 4.67\times 10^{6} m$ — inside Earth (radius $6.37\times 10^{6} m)$ , so Earth 'wobbles' around it

Result ·

x_cm \approx 4.67\times 10^{6} m

from Earth's center

Linear Momentum→Newton's Second Law→Moment of Inertia (Point Mass)→