The center of mass of a system of particles is the average position of the system’s mass distribution.
For point masses on a straight line, the center of mass is a weighted average of positions.
1. Center of mass formula in one dimension
For point masses \(m_1,m_2,\ldots,m_n\) located at positions \(x_1,x_2,\ldots,x_n\), the center of mass is
\[
x_{\mathrm{cm}}=\frac{\sum_i m_i x_i}{\sum_i m_i}.
\]
Written out explicitly,
\[
x_{\mathrm{cm}}
=
\frac{m_1x_1+m_2x_2+\cdots+m_nx_n}
{m_1+m_2+\cdots+m_n}.
\]
The numerator is called the total moment about the origin. The denominator is the total mass.
2. Why it is a weighted average
A simple average of positions would treat every point equally. The center of mass does not do that.
A heavier particle contributes more strongly because its position is multiplied by its mass.
\[
\text{weighted term}=m_i x_i.
\]
This is why the center of mass lies closer to heavier point masses.
3. Balance interpretation
The center of mass is the point where the total mass could be imagined to balance. At the center of mass,
the weighted deviations cancel:
\[
\sum_i m_i(x_i-x_{\mathrm{cm}})=0.
\]
Expanding this equation gives
\[
\sum_i m_i x_i-x_{\mathrm{cm}}\sum_i m_i=0,
\]
so
\[
x_{\mathrm{cm}}=\frac{\sum_i m_i x_i}{\sum_i m_i}.
\]
4. Negative positions
Positions may be negative if the origin is chosen somewhere between or to the right of some masses.
The formula still works:
\[
x_i<0
\quad\Rightarrow\quad
m_i x_i<0.
\]
Negative weighted terms pull the center of mass toward the negative side of the axis.
5. Solving for a missing mass
Sometimes the desired center of mass is known and one mass is unknown. Suppose the unknown mass is \(m_x\)
at known position \(x_x\). Let
\[
M=\sum_{\mathrm{known}} m_i,
\qquad
S=\sum_{\mathrm{known}} m_i x_i.
\]
Then
\[
x_{\mathrm{cm}}=\frac{S+m_xx_x}{M+m_x}.
\]
Solving for the unknown mass gives
\[
m_x=\frac{S-x_{\mathrm{cm}}M}{x_{\mathrm{cm}}-x_x}.
\]
A physically meaningful missing mass must be positive.
6. Solving for a missing position
If the unknown point has known mass \(m_x\), but unknown position \(x_x\), then
\[
x_{\mathrm{cm}}=\frac{S+m_xx_x}{M+m_x}.
\]
Solving for \(x_x\),
\[
x_x=\frac{x_{\mathrm{cm}}(M+m_x)-S}{m_x}.
\]
7. Worked example
Three point masses are placed on the x-axis:
\[
m_1=2\ \mathrm{kg},\quad x_1=0\ \mathrm{m},
\]
\[
m_2=3\ \mathrm{kg},\quad x_2=4\ \mathrm{m},
\]
\[
m_3=5\ \mathrm{kg},\quad x_3=10\ \mathrm{m}.
\]
First compute the total mass:
\[
\sum m_i=2+3+5=10\ \mathrm{kg}.
\]
Then compute the weighted position sum:
\[
\sum m_i x_i=(2)(0)+(3)(4)+(5)(10)=0+12+50=62\ \mathrm{kg\,m}.
\]
Divide by the total mass:
\[
x_{\mathrm{cm}}=\frac{62}{10}=6.2\ \mathrm{m}.
\]
So the center of mass is
\[
\boxed{x_{\mathrm{cm}}=6.2\ \mathrm{m}}.
\]
This point is closer to the \(5\ \mathrm{kg}\) mass at \(10\ \mathrm{m}\) than a simple unweighted average would be.
Formula summary
| Concept |
Formula |
Use |
| Total mass |
\(\sum_i m_i\) |
Add all point masses. |
| Total moment |
\(\sum_i m_i x_i\) |
Add each mass multiplied by its position. |
| Center of mass |
\(x_{\mathrm{cm}}=\frac{\sum_i m_i x_i}{\sum_i m_i}\) |
Find the mass-weighted average position. |
| Missing mass |
\(m_x=\frac{S-x_{\mathrm{cm}}M}{x_{\mathrm{cm}}-x_x}\) |
Find an unknown mass from a target center of mass. |
| Missing position |
\(x_x=\frac{x_{\mathrm{cm}}(M+m_x)-S}{m_x}\) |
Find where a known mass must be placed. |
| Balance condition |
\(\sum_i m_i(x_i-x_{\mathrm{cm}})=0\) |
Shows why the center of mass is the balance point. |
Key idea: the center of mass is not usually the middle of the positions. It is the middle of the mass distribution.
