---
title: Center-of-Mass Systems
module: Dynamics
moduleNumber: 2
lessonNumber: 5
order: 205
summary: >
  A firework bursts into a dozen fragments, yet one point keeps gliding along the
  original parabola as though nothing had happened. That point is the centre of mass,
  and following it collapses a many-body tangle into a single equation of motion. We
  define $\vec R=\frac1M\sum_i m_i\vec r_i$ and its continuous form, show that internal
  forces cancel so that only external ones move it, $M\vec A_{\rm cm}=\sum\vec F_{\rm ext}$,
  and put the result to work on recoil, collisions viewed from the centre-of-mass
  frame, and rocket propulsion, where mass leaving the boundary carries momentum with it.
topics: [Dynamics]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 5 — Applications of Newton's Laws; §§5-4–5-5"
---

## Center of mass and momentum

Particles of total mass $M=\sum_i m_i$ have centre-of-mass position
mass-weighted average

$$
\vec R=\frac{1}{M}\sum_i m_i\vec r_i.
$$

The average is a vector; each coordinate is weighted separately, and negative
coordinates need no special treatment. A larger mass pulls the centre toward its
position. The centre need not lie in material: for a ring, a frame, or two separated
bodies it can sit in empty space.

$$
% caption: The centre of mass of two particles lies on the line joining them, closer to
% the heavier one, where the mass-weighted distances balance: m1 times d1 equals m2 times d2.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (0.3,0) -- (6.1,0);
  \draw[fill=black!12, draw=black, thick] (1.2,0.5) circle (0.26);
  \draw[fill=acc!12, draw=acc, thick] (5.1,0.5) circle (0.55);
  \draw[fill=acc, draw=acc] (3.9,0.5) circle (2.5pt);
  \draw[black, dashed] (3.9,0.5) -- (3.9,0);
  \node[above] at (1.2,0.85) {$m_1$};
  \node[above] at (5.1,1.15) {$m_2$};
  \node[below] at (3.9,0) {CM};
\end{tikzpicture}
$$

> **Worked example.** A $2.0\ \mathrm{kg}$ cart sits at $x=0.40\ \mathrm m$ and a
> $6.0\ \mathrm{kg}$ cart at $x=1.60\ \mathrm m$ on a common axis. Their centre of
> mass is the mass-weighted mean position,
> $$
> X=\frac{(2.0\ \mathrm{kg})(0.40\ \mathrm m)+(6.0\ \mathrm{kg})(1.60\ \mathrm m)}
> {2.0\ \mathrm{kg}+6.0\ \mathrm{kg}}
> =\frac{10.4\ \mathrm{kg\,m}}{8.0\ \mathrm{kg}}=1.30\ \mathrm m,
> $$
> nearer the heavier cart. The signed first moments about $X$ cancel as a check:
> $(2.0)(0.40-1.30)+(6.0)(1.60-1.30)=-1.8+1.8=0\ \mathrm{kg\,m}$.

Shifting the coordinate origin moves every listed position and the centre-of-mass
coordinate by the same amount; the physical point does not move. This translation
check catches a coordinate mismatch when one position was measured from a different
fixture.

**Continuous distributions.**

An extended body replaces the discrete sum with an integral over mass elements:

$$
\vec R=\frac{1}{M}\int \vec r\d m.
$$

The element $\d m$ matches the distribution:
- a uniform wire uses $\d m=\lambda\d l$ (linear density $\lambda$),
- a thin sheet uses $\d m=\sigma\d A$ (surface density $\sigma$),
- a solid body uses $\d m=\rho\d V$ (volume density $\rho$).

Symmetry can fix components before any integration: a uniform disk has its centre at
its geometric centre. The integration limits must span the occupied material,
including holes and attached pieces, which a density formula alone does not fix. A
composite body is handled by summing separate integrals or discrete component masses
in one coordinate system.

**The centre-of-mass momentum theorem.**

Differentiating the definition twice gives the equation of motion for the whole
system,

$$
M\ddot{\vec R}=\sum \vec F_{\rm ext}.
$$

Internal action-reaction pairs cancel in the total, so only external forces appear.
Internal forces still redistribute momentum among the pieces, deform the system, or
set it rotating, but they cannot change the total momentum $\vec P=M\dot{\vec R}$,
whose rate of change is the net external force.

$$
% caption: Internal action-reaction forces between two enclosed particles cancel in the
% system momentum balance; only a force crossing the dashed boundary can move the centre of mass.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[dashed, thick] (0.4,-0.3) rectangle (4.6,1.9);
  \node[below right] at (0.55,1.9) {system};
  \draw[fill=black!12, draw=black, thick] (1.50,0.55) circle (0.30);
  \draw[thick] (3.30,0.55) circle (0.34);
  \draw[->, black, thick] (1.85,0.70) -- (2.92,0.70);
  \draw[->, black, thick] (2.92,0.40) -- (1.85,0.40);
  \draw[->, acc, very thick] (3.42,0.92) -- (5.15,2.45);
  \node[acc, right] at (5.15,2.45) {external};
\end{tikzpicture}
$$

The theorem holds for rigid, deformable, and separated-particle systems alike. At
constant total mass the centre of mass moves like a single particle of mass $M$
under $\sum\vec F_{\rm ext}$; the individual equations are still needed for internal
stresses and relative motion. Fragments can fly apart in every direction while their
centre of mass stays on the trajectory set by external gravity. Neglect an external
force only when its impulse over the interval is small against the momentum change
being analysed.

## Coordinates, decomposition, and extended bodies

Split each particle position into the whole-system position and a displacement
relative to the centre of mass:

$$
\vec r_i=\vec R_{\rm cm}+\vec r_i'.
$$

Taking the centre of mass as the primed origin makes the mass-weighted primed
positions vanish,

$$
\sum_i m_i\vec r_i'=0,
$$

so a large primed displacement on one side demands a compensating mass-weighted
displacement elsewhere. Translating every particle shifts $\vec R_{\rm cm}$ and
leaves all relative coordinates unchanged.

$$
% caption: Each particle position splits into the centre-of-mass position and a relative
% coordinate measured from it; the mass-weighted relative coordinates sum to zero.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[fill=acc, draw=acc] (3.0,0.6) circle (2.5pt);
  \node[below] at (3.0,0.45) {CM};
  \draw[fill=black!12, draw=black, thick] (1.3,0.6) circle (0.3);
  \draw[thick] (5.0,0.6) circle (0.42);
  \draw[->, black, thick] (3.0,0.6) -- (1.65,0.6) node[midway, above] {$d_1$};
  \draw[->, thick] (3.0,0.6) -- (4.55,0.6) node[midway, above] {$d_2$};
\end{tikzpicture}
$$

Differentiating gives the matching velocity split
$\vec v_i=\vec V_{\rm cm}+\vec v_i'$. Substituting it into the total kinetic energy
kills the cross term, since the mass-weighted primed velocities also sum to zero:

$$
K=\frac12M V_{\rm cm}^2+\sum_i\frac12m_i{v_i'}^2.
$$

The first term is the translational kinetic energy of the whole system; the second
is the kinetic energy of internal motion relative to the centre of mass. External
forces change the first by accelerating the centre of mass; internal forces move
energy within the second through deformation, rotation, or relative acceleration but
cancel from the centre-of-mass momentum.

> **Worked example.** Two particles of mass $m$ and $3m$ are separated by
> $1.20\ \mathrm m$. Measuring from the lighter particle, the centre of mass is
> $$
> x_{\rm cm}=\frac{m(0)+3m(1.20\ \mathrm m)}{4m}=0.90\ \mathrm m,
> $$
> so it lies $0.90\ \mathrm m$ from the lighter particle and $0.30\ \mathrm m$ from
> the heavier one. If the pair translates right at
> $V_{\rm cm}=2.0\ \mathrm{m\,s^{-1}}$, the total momentum is
> $P=(4m)V_{\rm cm}=(4m)(2.0\ \mathrm{m\,s^{-1}})$, independent of any internal
> motion. The relative kinetic energy in the centre-of-mass frame must be added
> separately; system momentum alone does not fix it.

**Symmetry and continuous bodies.**

Density can vary with position, so a geometric centroid is the centre of mass only
when the relevant density is uniform. Symmetry removes components before any
integration: a mirror plane holds the centre of mass in that plane, two perpendicular
mirror planes fix it at their intersection, and rotational symmetry places it on the
axis. The mass distribution must share the symmetry — a uniform disk with an
off-centre hole no longer has its centre at the geometric centre.

$$
% caption: A uniform plate symmetric about two axes has its centre of mass at their
% intersection; symmetry fixes both coordinates before any integration.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[thick] (0.6,0.5) rectangle (5.0,3.0);
  \draw[black, dashed] (2.8,0.2) -- (2.8,3.3);
  \draw[black, dashed] (0.3,1.75) -- (5.3,1.75);
  \draw[fill=acc, draw=acc] (2.8,1.75) circle (2.5pt);
  \node[acc, below right] at (2.9,1.7) {CM};
\end{tikzpicture}
$$

> **Worked example.** A rod runs from $x=0$ to $x=L$ with linear density
> $\lambda(x)=\lambda_0(1+x/L)$, denser toward the far end. Its mass is
> $$
> M=\int_0^L \lambda_0\Bigl(1+\frac{x}{L}\Bigr)\d x
> =\lambda_0\Bigl(L+\frac{L}{2}\Bigr)=\frac{3}{2}\lambda_0 L,
> $$
> and its first moment about the origin is
> $$
> \int_0^L x\,\lambda_0\Bigl(1+\frac{x}{L}\Bigr)\d x
> =\lambda_0\Bigl(\frac{L^2}{2}+\frac{L^2}{3}\Bigr)=\frac{5}{6}\lambda_0 L^2.
> $$
> The centre of mass is the first moment over the mass,
> $$
> x_{\rm cm}=\frac{\tfrac{5}{6}\lambda_0 L^2}{\tfrac{3}{2}\lambda_0 L}
> =\frac{5}{9}L\approx 0.556\,L,
> $$
> past the midpoint toward the denser end. The scale $\lambda_0$ cancels, so only the
> gradient sets the position: the result holds even when the total mass is unknown.

A centre of mass can also be measured without mapping the density. Suspend a rigid
planar body from one point and let it settle; gravity acts through the centre of
mass, so the vertical through the pivot contains it. A second suspension gives a
second vertical, and their intersection locates the centre in the body plane. Two
known supports give an alternative through the balance of total weight and its
moment.

$$
% caption: Suspending a body from two different points and marking each plumb line locates
% the centre of mass at the intersection of the two verticals.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black, thick] (0.5,3.3) -- (5.5,3.3);
  \draw[thick]
    (1.7,0.7) .. controls (0.9,1.7) and (1.6,2.9) .. (2.8,2.95)
    .. controls (4.3,3.0) and (4.5,1.5) .. (3.7,0.8)
    .. controls (3.0,0.35) and (2.2,0.35) .. (1.7,0.7);
  \draw[black, dashed] (2.5,2.9) -- (2.98,0.55);
  \draw[black, dashed] (4.15,2.7) -- (1.95,1.35);
  \draw[fill=acc, draw=acc] (2.78,1.6) circle (2.5pt);
  \node[acc, right] at (2.9,1.72) {CM};
\end{tikzpicture}
$$

Geometry dominates the uncertainty as much as mass does: a poorly measured support
separation enlarges the moment arm, and suspension lines meeting at a shallow angle
amplify their intersection error. Coatings, fasteners, voids, and trapped moisture
also shift the physical centre away from an idealised drawing.

## Recoil, collisions, and variable mass

Momentum conservation applies to a chosen system over a stated interval. When the net
external impulse is zero the total momentum is fixed and the centre-of-mass velocity
is constant, even though internal forces during an ejection or explosion can be large.
Those forces act in equal and opposite pairs inside the boundary, so individual pieces
change momentum while the system total does not. Treating ejected material as having
zero momentum breaks a recoil balance.

The boundary must contain every part of the internal interaction. Over a firearm's
recoil interval, firearm and projectile together form one system and their contact
force is internal; for the firearm alone that same force is external. Enlarging a
system can simplify the balance but requires the momentum of each added object, so the
boundary follows the physical question and interval, not algebraic convenience.

$$
% caption: Recoil redistributes momentum inside the boundary: the two fragments leave with
% equal and opposite momenta, so the total stays zero and the centre of mass does not move.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[dashed, thick] (0.4,-0.5) rectangle (5.6,1.6);
  \draw[fill=black!12, draw=black, thick] (2.2,0.55) circle (0.3);
  \draw[fill=black!12, draw=black, thick] (3.8,0.55) circle (0.42);
  \draw[->, black, very thick] (1.9,0.55) -- (0.7,0.55) node[left] {$p_1$};
  \draw[->, black, very thick] (4.2,0.55) -- (5.3,0.55) node[right] {$p_2$};
  \draw[fill=acc, draw=acc] (3.05,0.55) circle (2pt);
  \node[below] at (3.05,0.2) {CM};
\end{tikzpicture}
$$

> **Worked example.** A $12\ \mathrm{kg}$ cart at rest ejects a $2.0\ \mathrm{kg}$
> package rightward at $6.0\ \mathrm{m\,s^{-1}}$ relative to the floor. With negligible
> external horizontal impulse the total horizontal momentum stays zero, so the
> remaining $10\ \mathrm{kg}$ cart moves left at
> $$
> v_{\rm cart}=-\frac{(2.0\ \mathrm{kg})(6.0\ \mathrm{m\,s^{-1}})}{10\ \mathrm{kg}}
> =-1.2\ \mathrm{m\,s^{-1}}.
> $$
> The two momenta are $+12$ and $-12\ \mathrm{kg\,m\,s^{-1}}$, equal and opposite.
> Equal momenta do not mean equal kinetic energy: with $K=p^2/2m$ the lighter package
> carries $K_{\rm pkg}=36\ \mathrm J$ against the cart's $K_{\rm cart}=7.2\ \mathrm J$.
> That energy comes from the internal mechanism, not from any external force.

External gravity changes the centre-of-mass trajectory without invalidating the
internal recoil balance. A shell exploding in flight has fragments whose combined
centre of mass follows the projectile trajectory that the intact shell would follow
under gravity alone, provided drag and boundary leakage are negligible. Horizontally
the centre moves at constant speed when horizontal external impulse is negligible;
vertically it has the same gravitational acceleration as the total mass before the
explosion.

$$
% caption: A shell bursts in flight, yet its fragments' centre of mass stays on the
% original parabola set by gravity, because the burst forces are internal to the system.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (6.3,0) node[right] {x};
  \draw[acc, very thick, domain=0.3:5.6, samples=90] plot (\x,{-0.16*(\x-0.3)*(\x-5.6)});
  \draw[fill=acc, draw=acc] (2.95,1.12) circle (2pt);
  \draw[->, black, thick] (2.95,1.12) -- (4.5,1.9);
  \draw[->, black, thick] (2.95,1.12) -- (2.5,2.4);
  \draw[->, black, thick] (2.95,1.12) -- (3.7,0.35);
  \node[acc, left] at (2.85,1.35) {CM};
\end{tikzpicture}
$$

Boundary leakage and external impulse limit the approximation. A long ejection
interval allows drag or support forces to matter. A rocket requires a changing-mass
treatment because exhaust crosses the boundary. List forces crossing the boundary and
estimate their impulse before applying momentum conservation.

**Center-of-mass frames and collision analysis.**

The laboratory frame records the velocities measured by a fixed observer. The
centre-of-mass frame moves at $\vec V_{\rm cm}=\vec P/M$. Subtracting this
constant vector from every laboratory velocity gives
$\vec v_i'=\vec v_i-\vec V_{\rm cm}$. In that frame the mass-weighted
velocity sum is zero: $\sum_i m_i\vec v_i'=0$. The transformation changes neither
relative velocities nor the physical collision; it separates collective translation
from the motion available for internal redistribution.

In centre-of-mass coordinates, the kinetic energy is

$$
K=\frac12M V_{\rm cm}^2+K',
\qquad
K'=\sum_i\frac12m_i{v_i'}^2.
$$

The first term is fixed by total momentum for an isolated system. The second is the
kinetic energy in relative motion. Internal collision forces can convert $K'$ into
deformation, heat, sound, or other relative motion without changing the
centre-of-mass translation. For two bodies, $K'=\tfrac12\mu u^2$, where
$\mu=m_1m_2/(m_1+m_2)$ is reduced mass and
$u=|\vec v_1-\vec v_2|$ is relative speed.

$$
% caption: A perfectly inelastic collision removes the relative kinetic energy to internal
% energy while leaving the centre-of-mass translational kinetic energy unchanged.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (0,1.5) -- (3.8,1.5);
  \draw[fill=black!10, draw=black, thick] (0.4,1.5) rectangle (1.1,2.1);
  \draw[->, acc, thick] (1.15,1.8) -- (1.95,1.8) node[right] {v};
  \draw[fill=black!10, draw=black, thick] (2.6,1.5) rectangle (3.5,2.2);
  \node[black, left] at (0,1.8) {before};
  \draw[black] (0,-0.3) -- (3.8,-0.3);
  \draw[fill=black!10, draw=black, thick] (1.5,-0.3) rectangle (2.2,0.3);
  \draw[fill=black!10, draw=black, thick] (2.2,-0.3) rectangle (3.1,0.4);
  \draw[->, acc, thick] (3.15,0.05) -- (3.8,0.05) node[right] {$V$};
  \node[black, left] at (0,0.05) {after};
\end{tikzpicture}
$$

> **Worked example.** A $1.0\ \mathrm{kg}$ cart at $4.0\ \mathrm{m\,s^{-1}}$ strikes
> and sticks to a stationary $3.0\ \mathrm{kg}$ cart. Momentum sets the
> centre-of-mass speed,
> $$
> V_{\rm cm}=\frac{(1.0)(4.0)+(3.0)(0)}{1.0+3.0}=1.0\ \mathrm{m\,s^{-1}},
> $$
> which is the common final speed. In the CM frame the initial velocities are $+3.0$
> and $-1.0\ \mathrm{m\,s^{-1}}$, with cancelling momenta, and the relative kinetic
> energy is
> $$
> K'=\tfrac12\mu u^2=\tfrac12\frac{(1.0)(3.0)}{4.0}(4.0)^2=6.0\ \mathrm J,
> $$
> all lost to internal energy in the perfectly inelastic collision. The remaining
> laboratory kinetic energy $\tfrac12(4.0)(1.0)^2=2.0\ \mathrm J$ is the CM
> translation, and $6.0+2.0=8.0\ \mathrm J$ matches the initial
> $\tfrac12(1.0)(4.0)^2$.

**Variable mass, exhaust, and control volumes.**

If the rocket alone is the system, it is not a closed fixed-mass body: exhaust crosses
the boundary carrying momentum. The centre-of-mass theorem still applies exactly to
the closed system of rocket plus all its exhaust, but a practical calculation uses a
control volume and adds the momentum flux carried by the material leaving the region.

A rocket of instantaneous mass $m$ and velocity $v$ ejects exhaust backward at speed
$u$ relative to itself, giving the exhaust inertial velocity $v-u$. With external
force $F_{\rm ext}$, the short-interval momentum balance is

$$
m\d v=-u\d m+F_{\rm ext}\d t,
$$

with $\d m<0$ for the ejecting rocket, so the flux term $-u\d m$ points forward. It is
not an internal force on the rocket but the momentum of mass leaving the boundary.
Integrating the ideal case — constant $u$, negligible external force — gives the
Tsiolkovsky mass-ratio result

$$
\Delta v=u\ln\frac{m_i}{m_f}.
$$

$$
% caption: A rocket's control boundary has exhaust crossing it. That momentum flux must be
% counted, or the boundary enlarged to a closed system, when the rocket alone is analysed.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[dashed, thick] (2.0,-0.2) rectangle (5.8,1.4);
  \draw[thick]
    (3.4,0.35) -- (5.0,0.35) -- (5.5,0.6) -- (5.0,0.85) -- (3.4,0.85) -- cycle;
  \draw[->, very thick] (5.55,0.6) -- (6.6,0.6) node[right] {rocket};
  \draw[->, acc, very thick] (3.3,0.6) -- (1.3,0.6) node[left] {exhaust};
  \node[below right] at (2.0,1.4) {boundary};
\end{tikzpicture}
$$

A closed explosion differs from a rocket because all fragments remain within the
selected system after the event. Their internal forces redistribute momentum, but
no continuing mass flux crosses the boundary. A rocket plus its previously expelled
exhaust can also be made closed in principle, although that system grows with time
and is inconvenient for calculation. The control-volume formulation keeps a compact
boundary by accounting explicitly for what leaves it.

> **Worked example.** A $1000\ \mathrm{kg}$ rocket ejects $5.0\ \mathrm{kg}$ of
> exhaust at $u=300\ \mathrm{m\,s^{-1}}$ relative to itself over a short interval, with
> negligible external impulse. The small expelled fraction lets the initial mass stand
> in for $m$, so the forward speed gain is
> $$
> \Delta v\approx\frac{u\,|\Delta m|}{m}
> =\frac{(300\ \mathrm{m\,s^{-1}})(5.0\ \mathrm{kg})}{1000\ \mathrm{kg}}
> =1.5\ \mathrm{m\,s^{-1}}.
> $$
> The exhaust takes $\Delta p_{\rm ex}\approx(5.0\ \mathrm{kg})(-300\ \mathrm{m\,s^{-1}})
> =-1500\ \mathrm{kg\,m\,s^{-1}}$ in the rocket's initial frame, and the rocket takes
> $+1500\ \mathrm{kg\,m\,s^{-1}}$, again $1.5\ \mathrm{m\,s^{-1}}$. The exact
> Tsiolkovsky form $\Delta v=(300)\ln(1000/995)=1.5\ \mathrm{m\,s^{-1}}$ agrees at this
> sig-fig level; the logarithm is needed only over a substantial mass change.

External gravity and drag enter through $F_{\rm ext}$ and reduce the realised
velocity change. The boundary and the frame must be stated: relative exhaust speed
is defined in the rocket frame, whereas momentum balance is evaluated in one
inertial frame. Mixing those frames is the most common source of an incorrect rocket
equation.

## Rotation and measurement

A system's motion separates into translation of its centre of mass and motion about
that centre. For angular momentum about a fixed inertial origin O,
$\vec L_O=\vec R_{\rm cm}\times M\vec V_{\rm cm}+\vec L_{\rm cm}$.
The first term is orbital angular momentum of the whole mass treated at the CM. The
second is angular momentum of particles measured relative to the CM. This is valid
for any particle system; rigidity is not required for the decomposition.

Kinetic energy has the matching split
$K=\tfrac12M V_{\rm cm}^2+K_{\rm cm}$. For a rigid body rotating about its CM,
the relative term is $K_{\rm cm}=\tfrac12I_{\rm cm}\omega^2$. A deforming particle
system can have relative kinetic energy that cannot be represented by one moment of
inertia and one angular velocity. The centre-of-mass decomposition is therefore more
general than a rigid-body formula.

Torque about the CM isolates the internal rotational balance. External torque about
the CM changes $\vec L_{\rm cm}$, while net external force changes CM momentum.
About an arbitrary moving origin, correction terms may appear because that origin
has its own motion. The CM avoids them because the primed mass first moment is zero.
The CM is therefore the natural origin for separating rotation from translation.

$$
% caption: A moving extended body carries centre-of-mass translation plus rotation about
% the centre; its total kinetic energy adds the translational and rotational contributions.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[thick] (2.2,0.8) ellipse (1.1 and 0.5);
  \draw[fill=acc, draw=acc] (2.2,0.8) circle (2pt);
  \draw[->, acc, very thick] (2.2,0.8) -- (4.2,0.8) node[right] {$V$};
  \draw[->, black, thick] (1.55,1.75) arc[start angle=140, end angle=-140, radius=0.5];
  \node[black, above] at (1.55,1.9) {spin};
\end{tikzpicture}
$$

> **Worked example.** A $2.0\ \mathrm{kg}$ uniform rod of length $1.20\ \mathrm m$
> translates at $3.0\ \mathrm{m\,s^{-1}}$ while spinning about its centre at
> $\omega=4.0\ \mathrm{rad\,s^{-1}}$. The translational part is
> $$
> \tfrac12MV_{\rm cm}^2=\tfrac12(2.0)(3.0)^2=9.0\ \mathrm J,
> $$
> and with $I_{\rm cm}=ML^2/12=(2.0)(1.20)^2/12=0.240\ \mathrm{kg\,m^2}$ the rotational
> part is
> $$
> \tfrac12 I_{\rm cm}\omega^2=\tfrac12(0.240)(4.0)^2=1.92\ \mathrm J.
> $$
> The total kinetic energy is $9.0+1.92=10.9\ \mathrm J$, not the value from one
> endpoint speed. Translation and rotation are separate contributions in a moving
> extended body.

**Centre-of-mass measurement in extended systems.**

Extended systems are often measured rather than integrated from a density map. A
balance method places the object on two supports at known positions. Static force
balance gives the total weight, and moment balance about either support gives the
centre-of-mass coordinate along the support line. This method assumes the supports
are level, reactions are vertical, and the object is at rest. Frictional horizontal
forces do not alter the vertical moment balance when their lines of action have no
relevant lever arm, but tilted supports or compliant surfaces require a fuller force
model.

Suspension provides a geometric alternative. A rigid planar object hung from a point
settles until its centre of mass lies vertically below that point. A plumb line
through the suspension point therefore contains the centre of mass. Repeating from a
second point gives two lines whose intersection locates the centre in the body plane.
The method assumes the object is rigid enough that its mass distribution does not
shift between suspensions and that air currents and pivot friction are small enough
to permit true equilibrium.

Known test masses sharpen a balance measurement: a small calibrated mass at a measured
location shifts the combined centre by a predictable amount, and comparing the observed
reaction change against that prediction checks the support spacing, scale calibration,
and sign convention. Several placements support a linear fit rather than one difference.
Pendular measurement instead reads the period of small oscillations about a pivot,
which fixes the pivot-to-centre distance together with the moment of inertia about the
pivot; it suits bodies that cannot rest on supports but needs a small-angle model and a
measured time base.

**Data reduction.**

Every centre-of-mass calculation starts from one declared origin, and every position
enters as a signed coordinate from it. Mixing a distance from a support with a distance
from an end face gives a plausible number with a wrong first moment. Combine the
mass-weighted positions from the measured values, not from rounded intermediate
averages. A mass uncertainty enters both the total mass and the first moment; a
position uncertainty matters most for a component with large mass or a long lever arm,
and correlated errors — a ruler zero, a scale calibration — must not be treated as
independent noise. A small mass can still dominate the uncertainty when it sits far
from the origin.

| Input class | Contribution to first moment | Independent check |
| --- | --- | --- |
| Component mass | $m_i x_i$ | repeated weighing or calibration mass |
| Component position | $m_i\,\delta x_i$ sensitivity | measurement from a second reference mark |
| Common ruler offset | correlated shift of several $x_i$ | origin or datum check |
| Support reaction | moment-balance residual | load-cell calibration and spacing |

## Models, validation, and reporting

A rigid-body model fixes the distances among all mass elements, so the centre of mass
is fixed in body coordinates and relative motion is a rotation about it. A deformable
body changes shape, redistributes density, and stores strain energy: the centre-of-mass
theorem still holds for its particles, but one rigid-body inertia no longer describes
the internal motion. In a uniform gravitational field the total weight sets the
centre-of-mass acceleration without resolving forces on each element; in a measurable
gradient, different parts feel different forces, and replacing the field by its central
value is an approximation whose error grows with body size. Mass transfer changes the
boundary itself — a rocket, a leaking tank, or a cart gathering sand is not a fixed set
of particles — so its momentum balance needs a flux term or a larger closed system.

A centre-of-mass result must carry its frame: external forces enter directly in an
inertial frame, while a rotating or accelerating frame needs its apparent-force terms.
Validation compares the modelled centre or momentum balance against independent support
reactions, suspension lines, or tracked particle data, and a disagreement is traced to
calibration, geometry, frame choice, or missing mass before it is read as new physics.
The balance record names which quantity the model predicts and which measurement tests
it.

| System choice | Governing statement | Measurement check |
| --- | --- | --- |
| Fixed collection of particles | $M\vec a_{\rm cm}=\sum\vec F_{\rm ext}$ | support reactions or tracked particle data |
| Closed recoil or explosion system | total momentum conservation | vector sum before and after the event |
| Open mass-flow system | momentum balance with flux | included material and crossing velocity |
| Extended body in a gradient | net force plus possible torque | load, strain, or relative-displacement residual |

**Reporting conventions.**

Report a centre-of-mass coordinate as a signed value from a named origin (ordered
components in space), with axes and units, and with no more digits than the geometry
and masses justify. For a system-momentum or changing-mass result, state the boundary,
the inertial frame, the time interval, the included masses with any crossing material,
and which external impulses are kept. These are what let another reader reproduce the
claim, and what separate a closed-explosion balance from a rocket control-volume
balance.

## Force gradients and numerical models

The centre-of-mass theorem holds for any collection of particles in an inertial frame
once all external forces on them are included; it needs no rigid body, uniform field,
or uniform density, and internal forces cancel even when they produce large
deformation, stress, or heat. It describes translation of the whole mass, not the
motion of every part.

An external field gradient changes the force from one part of an extended body to
another — in gravity, the near side of a long body is attracted slightly more strongly
than the far side. The sum of those forces still sets the centre-of-mass acceleration,
while their variation across the body creates internal stresses and torques. Replacing
the field by its central value is valid only when the field changes little across the
object.

Tidal effects provide a familiar example. Two freely falling particles separated
radially in a gravitational field can accelerate by slightly different amounts. Their
centre of mass follows the net external force divided by total mass, but their
separation changes because the field is nonuniform. A self-gravitating body,
satellite, or long tether can therefore deform or experience differential tension
without violating the centre-of-mass theorem. The theorem predicts the bulk
trajectory; it does not assert that every element follows that trajectory.

A force gradient can also produce rotation: if the external force elements have a net
moment about the centre of mass, the body gains angular momentum while its centre still
follows the total force. A spatially varying field can carry a small net force with a
large torque, or the reverse, so centre-of-mass acceleration alone does not fix the
rotational response. The same extended-body data then need two reductions, one for
translation and one for rotation or differential motion.

| Observed quantity | Aggregate relation | Missing information |
| --- | --- | --- |
| Centre-of-mass acceleration | $M\vec a_{\rm cm}=\sum\vec F_{\rm ext}$ | distribution of force over the body |
| Rotation | $\boldsymbol\tau_{\rm cm}=\sum\vec r'_a\times\vec F_a$ | moment arms and force directions |
| Differential motion | field variation across the body | separation, stiffness, and time scale |

A laboratory gravity field is usually uniform across a small apparatus, but support
reactions, air flow, thermal gradients, or stray electromagnetic fields need not be. A
residual that grows with object size or separation is evidence of a gradient effect;
one within coordinate and mass uncertainty supports the uniform-field model. The
gradient force is external for the body under study, while the stresses transmitting it
are internal, and enlarging the system to include the field source makes that pair
internal at the cost of new momentum carriers.

**Numerical implementation.**

A numerical centre-of-mass calculation is clearest with masses and signed coordinates
in structured arrays, one row per particle, so the same data feed the first-moment and
momentum checks. Convert units before forming the sums — mixing grams with kilograms
corrupts a weighted result with no obvious error. Symmetry gives a free check: a
symmetric pair of equal masses must leave the corresponding coordinate fixed, and the
mass-weighted relative coordinates must sum to zero once the computed centre is
subtracted. Keep the raw, unrounded inputs beside the converted values and the origin
definition, so a formatted table never becomes the only record of the data.

## System communication and consistency

Every system model needs an explicit boundary and frame. Name the objects inside the
boundary and the material that can cross it during the interval — a cart and its
payload, a projectile and its fragments, a rocket and its exhaust each have a different
momentum balance — because the boundary fixes which forces are internal and whether a
mass-flux term is needed. Positions, velocities, momenta, and impulses are all
frame-dependent even though the physical outcome is not; mixing a velocity from a
moving frame with an external force from the laboratory frame produces a spurious
momentum imbalance.

List the external impulses rather than assuming them away, and compare each against the
momentum change over a stated observation window: a large force can be negligible in a
millisecond collision, while a modest force can dominate a long interaction. The window
also fixes which material crosses the boundary, so it belongs beside every momentum or
centre-of-mass result, together with the validity conditions — rigidity, uniform
density, constant total mass, uniform field — under which the result is exact for the
model but only approximate for the apparatus.

**Final consistency checks.**

After a centre-of-mass calculation, form the relative coordinates
$\vec r_i'=\vec r_i-\vec R_{\rm cm}$. Their mass-weighted sum must vanish
within the stated numerical precision:

$$
\sum_i m_i\vec r_i'=\vec 0.
$$

A system-momentum calculation compares the measured change in total momentum against
the external impulse over the same interval, using one boundary, basis, and set of time
stamps. A disagreement points to an omitted impulse, a mass crossing the boundary, a
frame mismatch, or plain measurement uncertainty — listed before the algebra is
touched. Dimensional checks stay active: a mass-weighted position over total mass has
units of length, while total momentum and external impulse share units of mass times
velocity. A centre outside the span of the positive mass locations signals a sign
error.

## Variable mass and measurement design

A rocket needs a different boundary from a closed explosion. With the rocket alone as
the system, exhaust crosses the boundary and carries momentum away, so the balance
carries an outward flux term; combining rocket and exhaust into one material system
makes their mutual forces internal. For rocket mass $m$, inertial velocity $v$, and
exhaust speed $u$ backward relative to the rocket, the short-interval balance is again

$$
m\d v=-u\d m+F_{\rm ext}\d t,
$$

with $\d m<0$ making the flux term forward. The Tsiolkovsky relation needs constant
$u$ and negligible external impulse; gravity, drag, or varying exhaust speed require
extra terms or numerical integration.

A closed explosion has no continuing flux across the boundary: fragments take large
opposite momenta while the centre of mass follows the net external force. A missing
fragment, gas plume, or unaccounted impulse then shows up as an apparent momentum
imbalance, so the boundary inventory and frame accompany any recoil calculation.

**Centre-of-mass reference frames and measurement design.**

Subtracting the system velocity from every laboratory velocity gives the CM-frame
velocities, whose mass-weighted momentum sum is zero, so internal collision impulses
redistribute momentum without changing it; adding the CM velocity back returns the
laboratory description. A measurement plan separates internal contact from boundary
forces — for two carts, their mutual contact is internal while track friction, a
tether, or a wall is external — and sets a collision window that opens before contact
and closes after separation.

Video tracking adds position, time-base, and perspective uncertainty: a calibrated
length scale, frame timestamps, and a stated camera geometry are needed before finite
differences become velocities. Differentiation amplifies frame-to-frame noise, so a
fitted trajectory beats a single displacement difference, and a fixed reference object
exposes camera drift.

A quick sign check: two masses with laboratory momenta $+6.0$ and
$-2.0\ \mathrm{kg\,m\,s^{-1}}$ have total $+4.0\ \mathrm{kg\,m\,s^{-1}}$; subtracting
the CM velocity must send their mass-weighted CM momenta to zero, and adding it back
must rebuild the laboratory values.

> **Worked example.** A cart pair shows an initial-to-final momentum difference of
> $0.18\ \mathrm{kg\,m\,s^{-1}}$, while a force sensor records a support impulse of
> $0.15\pm0.05\ \mathrm{N\,s}$ over the same window. Because
> $$
> |0.18-0.15|=0.03\ \mathrm{kg\,m\,s^{-1}}<0.05\ \mathrm{N\,s},
> $$
> the momentum change is accounted for by the measured external impulse: the data
> support the external-impulse balance, not conservation for an isolated system. A
> residual still larger than the combined uncertainty after checking frame, boundary,
> and calibration points to an omitted force or unmeasured mass flux — record it by
> component, since a vertical discrepancy suggests support or gravity impulse and a
> horizontal one a different cause.