The center of mass of a system of particles is the mass-weighted average position of the system.
For particles in a plane, the center of mass has both an x-coordinate and a y-coordinate.
1. Vector definition
Each particle has a position vector
\[
\vec r_i=x_i\hat{\imath}+y_i\hat{\jmath}.
\]
For point masses \(m_1,m_2,\ldots,m_n\), the center-of-mass vector is
\[
\vec r_{\mathrm{cm}}
=
\frac{\sum_i m_i\vec r_i}{\sum_i m_i}.
\]
This means the whole system can often be treated as if its total mass were concentrated at
\(\vec r_{\mathrm{cm}}\) when studying translational motion.
2. Coordinate formulas
Splitting the vector equation into coordinates gives
\[
x_{\mathrm{cm}}
=
\frac{\sum_i m_i x_i}{\sum_i m_i},
\qquad
y_{\mathrm{cm}}
=
\frac{\sum_i m_i y_i}{\sum_i m_i}.
\]
The denominator is the same for both coordinates:
\[
M=\sum_i m_i.
\]
The numerators are the x-moment and y-moment sums:
\[
S_x=\sum_i m_i x_i,
\qquad
S_y=\sum_i m_i y_i.
\]
3. Why the center of mass is a weighted average
A simple average would treat every point equally. The center of mass does not.
A heavier particle contributes more strongly because its coordinate is multiplied by its mass:
\[
\text{x-weighted term}=m_i x_i,
\qquad
\text{y-weighted term}=m_i y_i.
\]
Therefore, the center of mass lies closer to heavier particles.
4. Relation to the 1D formula
If all particles lie on the x-axis, then \(y_i=0\) for every particle. The y-coordinate of the center of mass is
\[
y_{\mathrm{cm}}=0,
\]
and the x-coordinate reduces to the 1D formula:
\[
x_{\mathrm{cm}}=\frac{\sum_i m_i x_i}{\sum_i m_i}.
\]
5. Negative coordinates
Coordinates can be positive, zero, or negative. A negative coordinate simply means the particle is located on the
negative side of the chosen origin.
\[
x_i<0 \Rightarrow m_i x_i<0,
\qquad
y_i<0 \Rightarrow m_i y_i<0.
\]
The formulas work the same way in all quadrants.
6. Solving for a missing mass
Suppose one unknown mass \(m_x\) is placed at known coordinates \((x_x,y_x)\), and a target center of mass is known.
Let
\[
M=\sum_{\mathrm{known}}m_i,
\qquad
S_x=\sum_{\mathrm{known}}m_i x_i,
\qquad
S_y=\sum_{\mathrm{known}}m_i y_i.
\]
Then
\[
x_{\mathrm{cm}}=\frac{S_x+m_xx_x}{M+m_x},
\qquad
y_{\mathrm{cm}}=\frac{S_y+m_xy_x}{M+m_x}.
\]
From the x-coordinate equation,
\[
m_x=\frac{S_x-x_{\mathrm{cm}}M}{x_{\mathrm{cm}}-x_x}.
\]
The y-coordinate gives a consistency check. A physically meaningful missing mass must be positive and must satisfy
both coordinate equations.
7. Solving one missing coordinate
If a particle has known mass \(m_x\) and known y-coordinate, but unknown x-coordinate, then
\[
x_x=\frac{x_{\mathrm{cm}}(M+m_x)-S_x}{m_x}.
\]
Similarly, if the y-coordinate is unknown,
\[
y_x=\frac{y_{\mathrm{cm}}(M+m_x)-S_y}{m_x}.
\]
8. Worked example
Consider three point masses:
\[
m_1=2\ \mathrm{kg},\quad (x_1,y_1)=(0,0)\ \mathrm{m},
\]
\[
m_2=3\ \mathrm{kg},\quad (x_2,y_2)=(4,0)\ \mathrm{m},
\]
\[
m_3=1\ \mathrm{kg},\quad (x_3,y_3)=(0,6)\ \mathrm{m}.
\]
Total mass:
\[
\sum m_i=2+3+1=6\ \mathrm{kg}.
\]
x-moment:
\[
\sum m_i x_i=(2)(0)+(3)(4)+(1)(0)=12\ \mathrm{kg\,m}.
\]
y-moment:
\[
\sum m_i y_i=(2)(0)+(3)(0)+(1)(6)=6\ \mathrm{kg\,m}.
\]
Therefore,
\[
x_{\mathrm{cm}}=\frac{12}{6}=2.0\ \mathrm{m},
\qquad
y_{\mathrm{cm}}=\frac{6}{6}=1.0\ \mathrm{m}.
\]
So the center of mass is
\[
\boxed{(x_{\mathrm{cm}},y_{\mathrm{cm}})=(2.0,\ 1.0)\ \mathrm{m}}.
\]
Formula summary
| Concept |
Formula |
Use |
| Total mass |
\(\sum_i m_i\) |
Add all point masses. |
| x-moment sum |
\(\sum_i m_i x_i\) |
Mass-weighted sum of x-coordinates. |
| y-moment sum |
\(\sum_i m_i y_i\) |
Mass-weighted sum of y-coordinates. |
| x-coordinate of COM |
\(x_{\mathrm{cm}}=\frac{\sum_i m_i x_i}{\sum_i m_i}\) |
Find horizontal coordinate of the center of mass. |
| y-coordinate of COM |
\(y_{\mathrm{cm}}=\frac{\sum_i m_i y_i}{\sum_i m_i}\) |
Find vertical coordinate of the center of mass. |
| Vector form |
\(\vec r_{\mathrm{cm}}=\frac{\sum_i m_i\vec r_i}{\sum_i m_i}\) |
Compact formula for 2D or 3D systems. |
Key idea: compute the center of mass one coordinate at a time. The same total mass divides both moment sums.
