The Center-of-Mass Theorem

Derivation

The Center-of-Mass Theorem

D-026 Home PU-101 Threads force · matter Depends on Momentum Conservation from Newton's Third Law

Statement

For a system of \(N\) particles with total mass \(M=\sum_i m_i\), the mass-weighted mean position \(\vec{R}=\frac{1}{M}\sum_i m_i \vec{r}_i\) obeys \(M\,\ddot{\vec{R}}=\vec{F}^{\,\text{ext}}\), where \(\vec{F}^{\,\text{ext}}\) is the vector sum of all forces exerted on the system by bodies outside it. The internal forces, however complicated, contribute nothing to the motion of \(\vec{R}\).

Why it matters

This theorem is what licenses the entire "point-particle" idealisation used throughout mechanics. When we say a thrown hammer, a planet, or a diver "follows a parabola", we are silently invoking it: one privileged point of any extended, spinning, deforming body traces the trajectory a single mass \(M\) would, no matter how violently the parts move relative to one another.

It also cleanly separates the two halves of rigid-body dynamics. Translation of the whole is governed by \(\vec{F}^{\,\text{ext}}\) acting at \(\vec{R}\); everything else — rotation, vibration, internal energy — is decoupled from the centre-of-mass drift. Momentum conservation for an isolated system is the special case \(\vec{F}^{\,\text{ext}}=\vec{0}\).

Assumptions

Newtonian point particles.If speeds approach \(c\), \(\vec{p}=m\vec{v}\) fails and one must use \(\vec{p}=\gamma m\vec{v}\); the simple \(M\ddot{\vec{R}}\) no longer equals the external force because relativistic mass–energy shifts the effective centre.

Constant particle masses.If the \(m_i\) change (ablation, rockets, condensation), \(\frac{d}{dt}(m_i\vec{v}_i)\neq m_i\vec{a}_i\) and extra "thrust" terms appear; the bare theorem must be augmented.

Internal forces obey the strong third law (equal, opposite, and collinear).If \(\vec{f}_{ij}=-\vec{f}_{ji}\) fails — as for the magnetic force between two moving charges — the internal forces no longer sum to zero and the centre of mass can accelerate with no external agent, unless field momentum is included.

A single inertial frame.In a rotating or accelerating frame, fictitious forces act on every \(m_i\) and enter \(\vec{F}^{\,\text{ext}}\) as \(M\)-proportional pseudo-forces; the clean split still holds but "external force" must be reinterpreted.

Derivation

\[ \vec{R}\;\equiv\;\frac{1}{M}\sum_{i=1}^{N} m_i\,\vec{r}_i,\qquad M\equiv\sum_{i=1}^{N} m_i \]

Definition of the centre of mass as the mass-weighted mean of the position vectors. A

\[ M\,\vec{R}=\sum_{i=1}^{N} m_i\,\vec{r}_i \]

Clear the denominator; \(M\) is constant, so this is an identity valid at all times. A

\[ M\,\dot{\vec{R}}=\sum_{i=1}^{N} m_i\,\dot{\vec{r}}_i=\sum_{i=1}^{N}\vec{p}_i=\vec{P} \]

Differentiate once in time; \(m_i\) constant lets \(\tfrac{d}{dt}(m_i\vec{r}_i)=m_i\dot{\vec{r}}_i\). The total momentum \(\vec{P}\) is exactly \(M\dot{\vec{R}}\). A

\[ M\,\ddot{\vec{R}}=\dot{\vec{P}}=\sum_{i=1}^{N} m_i\,\ddot{\vec{r}}_i \]

Differentiate again; constancy of \(M\) and \(m_i\) keeps them outside the derivative. A

\[ m_i\,\ddot{\vec{r}}_i=\vec{F}_i=\vec{F}_i^{\,\text{ext}}+\sum_{j\neq i}\vec{f}_{ij} \]

Newton's second law for particle \(i\); split the total force into the external part and the sum of internal forces \(\vec{f}_{ij}\) from every other particle \(j\). A

\[ M\,\ddot{\vec{R}}=\sum_{i}\vec{F}_i^{\,\text{ext}}+\sum_{i}\sum_{j\neq i}\vec{f}_{ij} \]

Substitute Step 5 into Step 4 and separate the two double sums. The first sum defines \(\vec{F}^{\,\text{ext}}\). B

\[ \sum_{i}\sum_{j\neq i}\vec{f}_{ij}=\sum_{i

Pair each ordered term \((i,j)\) with its partner \((j,i)\); Newton's third law gives \(\vec{f}_{ij}=-\vec{f}_{ji}\), so every pair cancels. This is where momentum-conservation-from-the-third-law enters. C

\[ M\,\ddot{\vec{R}}=\vec{F}^{\,\text{ext}},\qquad \vec{F}^{\,\text{ext}}\equiv\sum_{i}\vec{F}_i^{\,\text{ext}} \]

Drop the vanished internal sum. The centre of mass moves as a single particle of mass \(M\) driven by the net external force. A

Result

\[ M\,\ddot{\vec{R}}=\vec{F}^{\,\text{ext}} \]

Reading. The mass-weighted average position of any system, closed or not, accelerates exactly as a point of mass \(M\) subjected to the resultant of the external forces alone. Internal forces — springs, collisions, explosions, gravity between parts — can never shift the centre of mass on their own; they only redistribute matter around it.

Units check. \([M][\ddot{\vec{R}}]=\text{kg}\cdot\text{m s}^{-2}=\text{kg m s}^{-2}=\text{N}\), which matches the dimensions of a force \(\vec{F}^{\,\text{ext}}\). Both sides are vectors, so the equation holds component by component.

Limiting cases

\(\vec{F}^{\,\text{ext}}=\vec{0}\): \(\dot{\vec{R}}=\) const, i.e. \(\vec{P}\) conserved — the isolated-system momentum law is recovered.
Single particle (\(N=1\)): \(\vec{R}=\vec{r}_1\) and the theorem collapses to \(m_1\ddot{\vec{r}}_1=\vec{F}_1\), ordinary Newton II.
Rigid body in free fall: \(\vec{F}^{\,\text{ext}}=M\vec{g}\), so \(\ddot{\vec{R}}=\vec{g}\) independent of spin or shape — the parabola of projectile motion.
Symmetric explosion of a stationary body: fragments fly apart yet \(\vec{R}\) stays put, since only internal forces acted.

Breaks when

Open/variable-mass systems. For a rocket or a raindrop growing by accretion the \(m_i\) are not constant, so Step 3 fails; \(M\ddot{\vec{R}}=\vec{F}^{\,\text{ext}}\) must be replaced by a momentum equation carrying an \(\dot{m}\,\vec{u}\) thrust term.
Third-law violations / field momentum. Two moving charges exert non-collinear, unequal magnetic forces; the internal sum in Step 7 no longer vanishes and \(\vec{R}\) of the charges alone appears to accelerate. Consistency is restored only by counting the momentum stored in the electromagnetic field.
Relativistic speeds. With \(\vec{p}=\gamma m\vec{v}\), \(M\ddot{\vec{R}}\) is no longer the total force; the useful invariant becomes the centre of energy/momentum, not the Newtonian centre of mass.
Non-inertial frames without correction. In a rotating frame, Coriolis and centrifugal pseudo-forces act on the parts; ignoring them makes the bare theorem give the wrong \(\ddot{\vec{R}}\).

Failure modes

Counting internal forces. Adding a spring or collision force between two parts of the system into \(\vec{F}^{\,\text{ext}}\); these cancel and must be excluded.
Using the geometric centroid. Taking \(\vec{R}\) as the shape's centre rather than the mass-weighted mean — wrong whenever density is non-uniform.
Confusing \(\vec{R}\) with a real particle. Believing something material sits at \(\vec{R}\); for a ring or boomerang the centre of mass is in empty space.
Thinking an explosion moves the centre of mass. Expecting fragments to shift \(\vec{R}\); internal forces cannot, only external ones can.
Forgetting weight acts at \(\vec{R}\). Placing gravity's resultant somewhere other than the centre of mass in a uniform field, botching torque balance.
Dropping the mass weighting when combining subsystems. Averaging sub-centres of mass without weighting by their masses.

Discussion

The deep content of the theorem is a decoupling. Any \(N\)-body motion can be written as the drift of \(\vec{R}\) plus the motion of the parts relative to \(\vec{R}\). The drift obeys a clean, closed law depending only on external forces and total mass; all the internal complexity — rotation, elastic waves, chemical rearrangement — lives entirely in the relative coordinates and never leaks into \(\ddot{\vec{R}}\). This is why we may analyse a tumbling wrench's flight and its spin as two independent problems.

Notice how little was assumed. We never specified the nature of the internal forces — gravitational, electrostatic, contact, whatever — only that they come in third-law pairs. That single symmetry is enough to annihilate the entire internal double sum. The theorem is therefore an early, concrete instance of the profound link between a symmetry and a conservation-type law: pairwise reciprocity of forces guarantees that internal dynamics cannot move the mass centroid.

The theorem is also the bridge from particle mechanics to continuum and rigid-body mechanics. Replacing \(\sum_i m_i(\cdots)\) by \(\int \rho\,dV\,(\cdots)\) carries \(\vec{R}=\frac{1}{M}\int \vec{r}\,\rho\,dV\) and \(M\ddot{\vec{R}}=\vec{F}^{\,\text{ext}}\) over unchanged, because the internal stresses inside a continuous body are exactly the field version of third-law pairs and integrate to zero over the interior.

At the frontier the theorem quietly redefines itself. When third-law collinearity fails — as in electrodynamics — the resolution is not that momentum is lost but that the correct conserved bookkeeping includes the momentum of the fields, \(\vec{P}_{\text{tot}}=\sum_i \vec{p}_i+\int \epsilon_0(\vec{E}\times\vec{B})\,dV\). The "centre of mass" generalises to the centre of energy, and the theorem survives in the covariant statement that the total energy–momentum four-vector's spatial part moves inertially for a closed system. The Newtonian result is the low-speed shadow of this relativistic conservation law.

Common misconceptions. The centre of mass is not "where the mass is" — it is a weighted average that can lie outside the material entirely. And \(M\ddot{\vec{R}}=\vec{F}^{\,\text{ext}}\) says nothing about how the object rotates or deforms; a body can spin wildly, shatter, or pulse while its centre of mass glides along a perfectly smooth path.

Worked examples

Example 1 — Two skaters pulling on a rope.

\[ \vec{F}^{\,\text{ext}}\approx\vec{0}\;\Rightarrow\;\vec{R}=\text{const} \]

Frictionless ice, horizontal plane: the only horizontal forces are the internal rope tensions, which cancel. A

\[ X_{\text{cm}}=\frac{m_A x_A+m_B x_B}{m_A+m_B} \]

One-dimensional centre of mass along the line joining them. A

\[ m_A\,\Delta x_A=-\,m_B\,\Delta x_B \]

Since \(X_{\text{cm}}\) cannot move, the mass-weighted displacements must be equal and opposite. B

\[ m_A=60\,\text{kg},\;m_B=40\,\text{kg},\;\text{gap}=10\,\text{m};\quad \Delta x_A+\lvert\Delta x_B\rvert=10\,\text{m} \]

Insert numbers; they meet at the fixed centre of mass, so their travel distances sum to the initial separation. A

\[ 60\,\Delta x_A=40(10-\Delta x_A)\;\Rightarrow\;100\,\Delta x_A=400 \]

Solve the linear relation from Steps 3–4. A

\[ \Delta x_A=4.0\ \text{m},\qquad \lvert\Delta x_B\rvert=6.0\ \text{m} \]

Reading. The lighter skater covers more ground; each moves inversely to their mass, and they meet 4 m from the heavier one — exactly at the unmoved centre of mass.

Example 2 — Fragmenting shell at the top of its arc.

\[ \ddot{\vec{R}}=\vec{g}\quad\text{throughout, before and after the burst} \]

The only external force on the shell-plus-fragments is gravity; the explosion is internal, so \(\vec{R}\) keeps falling as an unbroken projectile. A

\[ x_{\text{cm}}(t)=x_{\text{apex}}+v_x\,t \]

Horizontally \(\vec{F}^{\,\text{ext}}\) has no component, so the centre of mass continues at the launch horizontal velocity. A

\[ M\,x_{\text{cm}}=m_1 x_1+m_2 x_2 \]

The centre of mass is the mass-weighted mean of the two fragment landing positions (equal masses \(m_1=m_2=M/2\)). A

\[ v_x=40\,\text{m s}^{-1},\;t_{\text{fall}}=3.0\,\text{s},\;x_{\text{apex}}=120\,\text{m};\quad x_{\text{cm}}=120+40(3.0) \]

A shell splits in two at apex; one fragment drops straight down (\(x_1=x_{\text{apex}}=120\,\text{m}\)). Compute where the centre of mass lands. A

\[ x_{\text{cm}}=240\,\text{m};\quad \tfrac{M}{2}(120)+\tfrac{M}{2}x_2=M(240)\;\Rightarrow\;x_2=360\,\text{m} \]

Impose Step 3 with the known centre-of-mass landing point to solve for the second fragment. B

\[ x_{\text{cm}}=240\ \text{m},\qquad x_2=360\ \text{m} \]

Reading. Although the shell blew apart, its centre of mass landed 240 m downrange exactly as an intact shell would. Knowing that anchors the second fragment at 360 m.

Problems

A 3.0 kg and a 5.0 kg mass sit at \(x=0\) and \(x=0.80\,\text{m}\). Find the centre of mass.
Solution
\(X_{\text{cm}}=\dfrac{(3.0)(0)+(5.0)(0.80)}{3.0+5.0}=\dfrac{4.0}{8.0}=0.50\,\text{m}\) from the origin, i.e. 0.30 m from the 5.0 kg mass.
A 70 kg person stands at one end of a 30 kg, 4.0 m canoe on frictionless water and walks to the other end. How far does the canoe move relative to the water?
Solution
\(\vec{F}^{\,\text{ext}}=0\) so \(X_{\text{cm}}\) is fixed. Let the canoe shift \(d\) opposite to the walk. The person moves \(4.0-d\) relative to water. Conservation of centre of mass: \(70(4.0-d)=30\,d\Rightarrow 280=100\,d\Rightarrow d=2.8\,\text{m}\). The canoe slides 2.8 m.
Three equal 2.0 kg masses sit at the corners of an equilateral triangle of side 1.0 m. Locate the centre of mass.
Solution
By symmetry \(\vec{R}\) is the centroid. Place corners at \((0,0)\), \((1,0)\), \((0.5,\,\tfrac{\sqrt3}{2})\). \(X_{\text{cm}}=\tfrac{0+1+0.5}{3}=0.50\,\text{m}\), \(Y_{\text{cm}}=\tfrac{0+0+0.866}{3}=0.29\,\text{m}\). The centre of mass is at \((0.50,\,0.29)\,\text{m}\).
A 2.0 kg cart at rest on a frictionless track holds a compressed spring that launches a 0.50 kg block at \(6.0\,\text{m s}^{-1}\). Find the cart's recoil speed and verify the centre of mass stays at rest.
Solution
Internal spring force only, so \(\vec{P}=0\) throughout. \(0=(0.50)(6.0)+(2.0)v\Rightarrow v=-1.5\,\text{m s}^{-1}\); the cart recoils at 1.5 m/s. Check: \(M\dot{X}_{\text{cm}}=\sum p_i=(0.50)(6.0)+(2.0)(-1.5)=3.0-3.0=0\), so \(\dot{X}_{\text{cm}}=0\). The centre of mass remains at rest.
A 1200 kg fireworks shell is launched and, at the top of its trajectory 90 m downrange with horizontal velocity \(25\,\text{m s}^{-1}\), splits into a 400 kg piece and an 800 kg piece. The 400 kg piece is found 30 m behind the apex point (upstream). If both land after the same 4.0 s fall, where does the 800 kg piece land horizontally? Ignore air resistance.
Solution
External force is gravity only, so the centre of mass continues: \(x_{\text{cm}}=90+25(4.0)=190\,\text{m}\). The 400 kg piece lands at \(x_1=90-30=60\,\text{m}\). Centre-of-mass condition: \(1200(190)=400(60)+800\,x_2\Rightarrow 228000=24000+800\,x_2\Rightarrow x_2=\dfrac{204000}{800}=255\,\text{m}\). The 800 kg piece lands 255 m downrange.