AcademyMomentum Systems
Academy
Center of Mass
Level 1 - Physics topic page in Momentum Systems.
Principle
The center of mass moves as if all external force acted on total mass there.
Notation
\(M\)
total mass
\(\mathrm{kg}\)
\(\vec{r}_{\mathrm{cm}}\)
center-of-mass position
\(\mathrm{m}\)
\(\vec{v}_{\mathrm{cm}}\)
center-of-mass velocity
\(\mathrm{m\,s^{-1}}\)
\(\vec{a}_{\mathrm{cm}}\)
center-of-mass acceleration
\(\mathrm{m\,s^{-2}}\)
\(\vec{P}\)
total momentum
\(\mathrm{kg\,m\,s^{-1}}\)
\(\vec{F}_{\mathrm{ext}}\)
net external force
\(\mathrm{N}\)
Method
Center of mass is a mass-weighted position, so differentiating it connects geometry to momentum.
Total mass
\[M=\sum_i m_i\]
Weighted position
\[\vec{r}_{\mathrm{cm}}=\frac{1}{M}\sum_i m_i\vec{r}_i\]
Differentiate once
\[\vec{v}_{\mathrm{cm}}=\frac{1}{M}\sum_i m_i\vec{v}_i=\frac{\vec{P}}{M}\]
Differentiate momentum
\[\sum\vec{F}_{\mathrm{ext}}=\frac{d\vec{P}}{dt}=M\vec{a}_{\mathrm{cm}}\]
The diagram places the center of mass closer to the heavier object on the line.
That placement is a weighted average, not the geometric midpoint.
Rules
These are the compact center-of-mass definitions and dynamics.
Total mass
\[M=\sum_i m_i\]
CM position
\[\vec{r}_{\mathrm{cm}}=\frac{1}{M}\sum_i m_i\vec{r}_i\]
CM momentum
\[\vec{P}=M\vec{v}_{\mathrm{cm}}\]
CM dynamics
\[\sum\vec{F}_{\mathrm{ext}}=M\vec{a}_{\mathrm{cm}}\]
Examples
Question
Masses
\[1.0\,\mathrm{kg}\]
at \[x=0\]
and \[3.0\,\mathrm{kg}\]
at \[x=2.0\,\mathrm{m}\]
lie on a line. Find \[x_{\mathrm{cm}}\]
Answer
\[x_{\mathrm{cm}}=\frac{1.0(0)+3.0(2.0)}{4.0}=1.5\,\mathrm{m}\]
Checks
- The center of mass can lie outside the material.
- Internal motion cannot accelerate the center of mass.
- With no external force, center-of-mass velocity is constant.
- Heavier masses pull the center closer.