The center of mass is the mass-weighted average position of a system. It is the point that moves as if the whole
system’s mass were concentrated there and acted on by the net external force.
Center of mass position
For particles with masses \(m_i\) at positions \(\vec r_i\), the center of mass is:
Vector formula.
\[
\vec r_{cm}=\frac{\sum m_i\vec r_i}{\sum m_i}
\]
In one dimension:
\[
x_{cm}=\frac{\sum m_ix_i}{\sum m_i}
\]
In two dimensions:
\[
x_{cm}=\frac{\sum m_ix_i}{\sum m_i},
\qquad
y_{cm}=\frac{\sum m_iy_i}{\sum m_i}
\]
Center of mass velocity
The velocity of the center of mass is the mass-weighted average of particle velocities:
\[
\vec v_{cm}=\frac{\sum m_i\vec v_i}{\sum m_i}
\]
This connects directly to total momentum:
Total momentum relation.
\[
\vec P_{\mathrm{total}}=\sum m_i\vec v_i=M\vec v_{cm}
\]
where \(M=\sum m_i\) is the total mass.
Center of mass acceleration
The center of mass acceleration is:
\[
\vec a_{cm}=\frac{\sum m_i\vec a_i}{\sum m_i}
\]
Newton’s second law for the entire system is:
System form of Newton’s second law.
\[
\vec F_{\mathrm{net,ext}}=M\vec a_{cm}
\]
Internal forces between particles cancel in pairs, so they do not change the motion of the center of mass.
Motion with constant acceleration
If each particle has constant acceleration, its position and velocity at time \(t\) are:
\[
x_i(t)=x_{i0}+v_{xi0}t+\frac12a_{xi}t^2
\]
\[
v_{xi}(t)=v_{xi0}+a_{xi}t
\]
In 2D, use the same formulas separately for \(y\):
\[
y_i(t)=y_{i0}+v_{yi0}t+\frac12a_{yi}t^2,
\qquad
v_{yi}(t)=v_{yi0}+a_{yi}t
\]
Explosions and separating systems
In an explosion, the particles may fly apart, but if the net external force is zero, the center of mass keeps moving
at constant velocity. The internal explosion forces can change the relative motion of the fragments, but not the COM motion.
\[
\vec F_{\mathrm{net,ext}}=0
\quad\Rightarrow\quad
\vec a_{cm}=0
\quad\Rightarrow\quad
\vec v_{cm}=\mathrm{constant}
\]
Kinetic energy split
The total kinetic energy can be split into kinetic energy of center-of-mass motion and kinetic energy of motion relative
to the center of mass.
\[
K_{\mathrm{total}}=K_{cm}+K_{\mathrm{internal}}
\]
\[
K_{cm}=\frac12M v_{cm}^2
\]
The internal kinetic energy describes motion relative to the center of mass, such as fragments flying apart in an explosion.
Worked example
Suppose there are two masses: \(m_1=3\ \mathrm{kg}\) at \(x_1=0\ \mathrm{m}\), and
\(m_2=5\ \mathrm{kg}\) at \(x_2=4\ \mathrm{m}\).
Step 1: total mass.
\[
M=3+5=8\ \mathrm{kg}
\]
Step 2: center of mass.
\[
\begin{aligned}
x_{cm}
&=
\frac{m_1x_1+m_2x_2}{m_1+m_2}\\
&=
\frac{(3)(0)+(5)(4)}{3+5}\\
&=
\frac{20}{8}\\
&=
2.5\ \mathrm{m}
\end{aligned}
\]
The COM is closer to the heavier \(5\ \mathrm{kg}\) mass, so it lies at \(x=2.5\ \mathrm{m}\), not halfway between
the two particles.
Formula summary
| Concept |
Formula |
Meaning |
| Total mass |
\(M=\sum m_i\) |
Mass of the whole particle system |
| COM position |
\(\vec r_{cm}=\frac{\sum m_i\vec r_i}{M}\) |
Mass-weighted average position |
| COM velocity |
\(\vec v_{cm}=\frac{\sum m_i\vec v_i}{M}\) |
Mass-weighted average velocity |
| Total momentum |
\(\vec P=M\vec v_{cm}\) |
The system momentum equals total mass times COM velocity |
| COM acceleration |
\(\vec a_{cm}=\frac{\sum m_i\vec a_i}{M}\) |
Mass-weighted average acceleration |
| External-force relation |
\(\vec F_{\mathrm{net,ext}}=M\vec a_{cm}\) |
The COM responds only to net external force |
| Constant-acceleration position |
\(x_i(t)=x_{i0}+v_{xi0}t+\frac12a_{xi}t^2\) |
Particle position at time \(t\) |
| Energy split |
\(K_{\mathrm{total}}=K_{cm}+K_{\mathrm{internal}}\) |
Separates whole-system motion from relative motion |
The center of mass is not always located on an object or particle. It is a weighted point that describes the overall translational motion of the system.
