Systems of Particles and the Center of Mass

Throw a wrench spinning across a room and every point on it traces a wild, looping curve — except one. A single special point sails through the air on a clean, textbook parabola, exactly as if the wrench were a featureless pebble. Strap a camera to a firework and the story repeats at a grander scale: the shell bursts into a hundred screaming fragments, yet the average of all that chaos — each fragment weighted by its mass — keeps gliding along the same parabola the unexploded shell was following. The explosion, for all its violence, cannot budge it. That point is the center of mass, and the reason for its serenity is one of the deepest bookkeeping facts in mechanics. Every force the fragments exert on each other comes paired with an equal and opposite partner, courtesy of Newton's third law. Sum over the whole system and the internal mayhem cancels identically, leaving $M\ddot{\mathbf{R}} = \mathbf{F}_{\mathrm{ext}}$ : the center of mass moves as a single particle of the total mass, driven by the external forces alone. Euler carved this out as the first law of mechanics for bodies — it is what lets us treat a planet, a diver, or a 10²⁸-particle wrench as "a mass $M$ at position $\mathbf{R}$ " and get away with it. This article builds the center of mass from its definition, proves why internal forces can never move it, splits kinetic energy and angular momentum into clean CM-plus-internal parts, and puts numbers on it — from the Earth–Moon barycenter hiding inside the Earth to the walk-on-a-boat problem that momentum conservation solves in two lines.

Key takeaways

The center of mass is the mass-weighted average position $\mathbf{R} = \frac{1}{M}\sum_i m_i\mathbf{r}_i$ (sum → integral $\frac1M\int\mathbf{r}\,\mathrm{d}m$ for continuous bodies), with $M$ the total mass.
Its momentum is the total momentum: $\mathbf{P} = M\dot{\mathbf{R}}$ — the CM is the unique point whose motion carries the system's entire momentum budget.
Internal forces cancel in Newton's-third-law pairs, so $M\ddot{\mathbf{R}} = \mathbf{F}_{\mathrm{ext}}$ : no explosion, collision, or internal engine can move the center of mass.
If $\mathbf{F}_{\mathrm{ext}} = 0$ the CM moves in a straight line at constant velocity forever — momentum conservation in geometric form.
Positions relative to the CM satisfy $\sum_i m_i\mathbf{r}_i' = 0$ , which kills every cross term: kinetic energy and angular momentum split exactly into a CM part plus an internal part (König's theorems).

The core insight

Internal forces cancel pairwise (Newton's third law), so $M\ddot{\mathbf{R}}=\mathbf{F}_{\text{ext}}$ — the center of mass moves like a single particle of mass $M$ driven by the net external force, no matter how chaotic the internal motion.

Where this lives today

Exoplanet hunting is center-of-mass physics as an industry. A star with an unseen planet must orbit their common barycenter, so its spectrum swings periodically blue and red; the radial-velocity method has logged thousands of worlds by measuring stellar wobbles down to below a metre per second — walking pace — and ESA's Gaia mission is now delivering the astrometric version, watching stars trace tiny barycentric ellipses on the sky, with its 2026 data release expected to yield thousands of giant planets from wobble alone. The Earth–Moon barycenter from Worked Example 1 is the same calculation pointed at home; even the Sun orbits the solar system's barycenter, which Jupiter regularly drags outside the solar surface. The other frontier is balance. A humanoid robot is a many-body system whose controller must keep track of one non-material point: the whole-machine center of mass. Modern walking controllers plan CM trajectories and foot placements dozens of times per second so that the CM's projection stays dynamically supported — the same constraint a high-jumper exploits in reverse, arching so the body clears the bar while the CM, which must follow the ballistic parabola set at takeoff (equation 4), passes underneath it. From Fosbury's flop to warehouse robots catching their balance in 2026, the physics is one line: internal actuation is free to rearrange the parts, but only external forces steer the average.

The physical picture

Start with the firework. Before the burst, the shell arcs along a parabola set by gravity. The explosion is an internal affair: chemical energy hurls fragment against fragment, but every push is answered by an equal and opposite push on the pusher. Add up the forces over all fragments and the sum is exactly what it was before the explosion — just gravity. So the mass-weighted average position keeps flying the original parabola while the fragments spread symmetrically around it, a ghost trajectory only mathematics can see.

Figure

The shell explodes but its center of mass keeps flying the original parabola

Why is the average weighted by mass rather than, say, taken over the geometry? Because the point is engineered to carry the momentum. The total momentum is $\sum m_i\dot{\mathbf{r}}_i$ , and if we want one representative point $\mathbf{R}$ whose motion reproduces it as $M\dot{\mathbf{R}}$ , the mass-weighted average is the unique choice. The center of mass is not a geometric nicety — it is where the momentum lives. The best mental model for the third-law cancellation: people shoving each other inside a drifting boat. However hard they push, wrestle, or leap about inside, the boat-plus-passengers center of mass keeps drifting exactly as before, because every shove is paid back in kind within the system. Only something external — the water, the wind, an oar — can move the CM. Walk toward the bow and the boat obligingly slides backward beneath you by just enough to keep the average fixed; we will put numbers on that below.

Figure

The lever rule — the center of mass sits closer to the heavier body, dividing the separation in inverse mass ratio

Defining and computing the center of mass

For $N$ point masses the definition is the mass-weighted average position:

\mathbf{R} = \frac{1}{M}\sum_{i=1}^{N} m_i\,\mathbf{r}_i, \qquad M = \sum_{i=1}^{N} m_i \qquad (1)

where: $\mathbf{r}_i$ — position of particle $i$ . Two masses illustrate everything: the CM lies on the line joining them, dividing it in inverse proportion to the masses — four times closer to a body four times heavier. It need not lie inside any material at all: the CM of a ring is its empty center, and the CM of the Earth–Moon pair floats in rock 1,700 km beneath your feet. A solid body is the same sum taken to the continuum limit, chopping the body into mass elements $\mathrm{d}m = \rho\,\mathrm{d}V$ :

\mathbf{R} = \frac{1}{M}\int \mathbf{r}\,\mathrm{d}m = \frac{1}{M}\int \mathbf{r}\,\rho(\mathbf{r})\,\mathrm{d}V \qquad (2)

where: $\rho$ — the mass density. In practice symmetry does most of the work: the CM of any uniform symmetric body sits on every symmetry axis and plane, so a uniform rod's is at its middle and a sphere's at its center. When integration is genuinely needed it is usually one coordinate: a uniform solid hemisphere of radius $a$ , sliced into disks, gives $Z = \frac{1}{M}\int_0^a z\,\rho\pi(a^2 - z^2)\,\mathrm{d}z = 3a/8$ above the flat face — a classic worth having seen once. Composite bodies decompose: treat each simple piece as a point mass at its own CM and apply (1), including negative masses for holes. Differentiate (1) once and the payoff appears — the momentum identity:

\mathbf{P} = \sum_i m_i\dot{\mathbf{r}}_i = M\dot{\mathbf{R}} = M\mathbf{V} \qquad (3)

where: $\mathbf{V}$ — the CM velocity. The system's total momentum is exactly the momentum of one fictitious particle of mass $M$ riding at the CM. This identity is why momentum conservation and center-of-mass motion are the same subject in two vocabularies: $\mathbf{P}$ constant means the CM coasts uniformly.

Derivation AEuler's first law: the center of mass obeys Newton's second law

We are deriving the master result $M\ddot{\mathbf{R}} = \mathbf{F}_{\mathrm{ext}}$ , equation (4) — the statement that the center of mass of any system, however complicated its internal machinery, accelerates as a single point particle under the net external force alone. The route is to write Newton's second law for every particle separately, keeping internal and external forces in separate ledgers, and then sum over the whole system; the entire derivation turns on one cancellation, so the step to watch is the pairing of internal forces — it is Newton's third law, and nothing else, that erases the internal ledger. This is also where the result's limits live: exotic internal forces that violated the third law would leak through the cancellation.

Step 1. Newton's second law for particle $i$ , splitting the force into an external part and the internal forces from every other particle $j$ :

m_i\ddot{\mathbf{r}}_i = \mathbf{F}_i^{\mathrm{ext}} + \sum_{j\neq i}\mathbf{f}_{ij}

Why the split: the coming cancellation applies only to forces exerted by system members on each other, so the bookkeeping must keep them separate from the start.

Step 2. Sum over all particles:

\sum_i m_i\ddot{\mathbf{r}}_i = \sum_i \mathbf{F}_i^{\mathrm{ext}} + \sum_i\sum_{j\neq i}\mathbf{f}_{ij}

Step 3. Kill the internal double sum. Every term $\mathbf{f}_{ij}$ appears alongside its partner $\mathbf{f}_{ji}$ , and Newton's third law says $\mathbf{f}_{ji} = -\mathbf{f}_{ij}$ :

\sum_i\sum_{j\neq i}\mathbf{f}_{ij} = \sum_{\mathrm{pairs}}\left(\mathbf{f}_{ij} + \mathbf{f}_{ji}\right) = 0

Why this is the crux: the cancellation is pairwise and exact, independent of the nature, strength, or complexity of the internal forces — collisions, springs, explosions, gravity between the parts, all erased in one line.

Step 4. Recognize the left side. Differentiating $M\mathbf{R} = \sum_i m_i\mathbf{r}_i$ twice gives $M\ddot{\mathbf{R}} = \sum_i m_i\ddot{\mathbf{r}}_i$ , so

M\ddot{\mathbf{R}} = \mathbf{F}_{\mathrm{ext}} \qquad (4)

where: $\mathbf{F}_{\mathrm{ext}} = \sum_i\mathbf{F}_i^{\mathrm{ext}}$ — the net external force. Corollary: if $\mathbf{F}_{\mathrm{ext}} = 0$ , then $\mathbf{P} = M\mathbf{V}$ is constant — the conservation of linear momentum, now visibly a theorem about the third law. Why the projectile case works out: fragments under gravity alone contribute $\sum m_i g = Mg$ to the net external force, so equation (4) reduces to exactly the same equation of motion the intact shell obeyed — the CM continues the undisturbed parabola.

Splitting motion into CM plus internal

Once the CM is known, refer every particle to it: $\mathbf{r}_i = \mathbf{R} + \mathbf{r}_i'$ . The primed coordinates obey a constraint that looks innocent and does enormous work:

\sum_i m_i\,\mathbf{r}_i' = \sum_i m_i\mathbf{r}_i - M\mathbf{R} = 0 \qquad (5)

where: the vanishing is the CM definition read backwards — relative to the center of mass, the mass-weighted positions balance to zero, and so (differentiating) do the mass-weighted velocities. This zero is the assassin of cross terms: any expression bilinear in "CM motion" and "internal motion" dies on it.

Go deeper · Derivation B · 1 minKoenig's theorem: the kinetic energy splits with no cross term

Derivation BKoenig's theorem: the kinetic energy splits with no cross term

The goal is the exact decomposition $T = \tfrac12 MV^2 + T_{\mathrm{internal}}$ , equation (6) — bulk translational energy plus energy of motion about the CM, with nothing left over. The strategy is plain substitution: write every velocity as CM velocity plus relative velocity and expand the square; the derivation is three lines, and its entire content is the vanishing of the cross term via the constraint (5). Watch that step — it is the same cancellation that will split angular momentum, and (in the collisions articles) the same one that decides how much energy a collision can dissipate. If you remember one algebraic move from many-body mechanics, make it this one.

Step 1. Substitute $\dot{\mathbf{r}}_i = \mathbf{V} + \dot{\mathbf{r}}_i'$ into the definition $T = \tfrac12\sum_i m_i \dot{\mathbf{r}}_i\cdot\dot{\mathbf{r}}_i$ and expand the dot product:

T = \tfrac12\sum_i m_i V^2 \;+\; \mathbf{V}\cdot\sum_i m_i\dot{\mathbf{r}}_i' \;+\; \tfrac12\sum_i m_i v_i'^2

Step 2. Evaluate the middle term. By the constraint (5), $\sum_i m_i\mathbf{r}_i' = 0$ at all times, so its time derivative $\sum_i m_i\dot{\mathbf{r}}_i'$ vanishes identically. Why this is not an approximation: the constraint is an identity built into the definition of $\mathbf{R}$ , so the cross term is exactly zero for any motion whatsoever — tumbling, exploding, or rigid.

Step 3. Read off the split:

T = \tfrac12 M V^2 + \tfrac12\sum_i m_i v_i'^2 \qquad (6)

where: the first term is the kinetic energy of "the system as a particle"; the second — call it $T_{\mathrm{internal}}$ — contains rotation, vibration, and relative motion about the CM. The same substitution in $\mathbf{L} = \sum_i \mathbf{r}_i\times m_i\dot{\mathbf{r}}_i$ kills its cross terms too, yielding the angular-momentum split of equation (7). Why this recovers the rolling-body formula: for a rigid body the only internal motion is pure rotation, so $T_{\mathrm{internal}} = \tfrac12 I_{\mathrm{cm}}\omega^2$ and equation (6) becomes $T = \tfrac12MV^2 + \tfrac12 I_{\mathrm{cm}}\omega^2$ — the standard result for any rolling or spinning rigid body.

The angular-momentum version of the split, quoted for completeness because rigid-body dynamics is built on it:

\mathbf{L} = \mathbf{R}\times M\mathbf{V} \;+\; \sum_i \mathbf{r}_i'\times m_i\dot{\mathbf{r}}_i' \qquad (7)

where: the first term is the orbital angular momentum (the CM treated as a point mass about the origin) and the second the spin (angular momentum about the CM). A thrown, tumbling wrench is exactly this: orbital part on the parabola, spin part doing the tumbling, each evolving by its own clean law. Finally, for two bodies the internal motion collapses to a single coordinate. With $\mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2$ the relative separation, the internal dynamics obeys

\mu\,\ddot{\mathbf{r}} = \mathbf{f}(\mathbf{r}), \qquad \mu = \frac{m_1 m_2}{m_1 + m_2} \qquad (8)

where: $\mu$ — the reduced mass, and $\mathbf{f}$ the internal force of one body on the other. Two interacting bodies = one free CM + one effective particle of mass $\mu$ . This is the doorway to the two-body problem, planetary orbits, and the collision articles downstream.

Which formula answers which question

you want	use	notes
locate the CM of point masses	eq. (1)	inverse-lever rule for two bodies
locate the CM of a solid body	eq. (2)	symmetry first, integrate one slice variable if needed
the system's total momentum	$\mathbf{P} = M\mathbf{V}$ , eq. (3)	identity, not approximation
how the CM moves	$M\ddot{\mathbf{R}} = \mathbf{F}_{\mathrm{ext}}$ , eq. (4)	internal forces cannot appear
split energy or angular momentum	eqs. (6)–(7)	cross terms die by eq. (5)
two bodies interacting	eq. (8)	reduced mass, relative coordinate

Limiting cases

Case	Result	Meaning
$\mathbf{F}_{\mathrm{ext}} = 0$	CM coasts in a straight line at constant $\mathbf{V}$	momentum conservation as geometry; true through any internal violence
one particle ( $N=1$ )	$\mathbf{R} = \mathbf{r}_1$ , eq. (4) is Newton's second law	the many-body law contains the one-body law
two equal masses	CM at the midpoint; $\mu = m/2$	symmetric split of separation and of internal inertia
$m_2 \gg m_1$	CM sits essentially at body 2; $\mu \to m_1$	why "the Sun doesn't move" is a good first approximation
rigid body, pure rotation about CM	$T = \tfrac12 I_{\mathrm{cm}}\omega^2$ , $\mathbf{L}_{\mathrm{spin}} = I_{\mathrm{cm}}\boldsymbol{\omega}$	König's split specialized to rigid constraints
uniform gravity	center of gravity = center of mass	the two centers separate only in non-uniform fields (tides)

Common trapInternal forces moved the system, so they must be able to move the center of mass

No — and the confusion usually comes from watching a part instead of the whole. A person walking toward the bow of a boat certainly moves; so does the boat, sliding backward beneath them; but the mass-weighted average of person-plus-boat has not budged, equation (4) with zero external horizontal force. Two sibling errors ride with this one. First, "the rocket pushes on its own exhaust, so it can't accelerate" — the rocket body accelerates forward precisely because the exhaust accelerates backward; the combined CM of rocket plus expelled gas is what coasts undisturbed. Second, treating the CM as a material point: nothing needs to be at the CM (a ring's is in its hole, a high-jumper's passes under the bar). The CM is a bookkeeping point — the unique one whose motion carries the total momentum — not a particle you could paint red.

Worked examples

1 · The Earth–Moon barycenter (confidence builder)

The Moon ( $m_M = 7.35\times10^{22}$ kg) orbits the Earth ( $m_E = 5.97\times10^{24}$ kg) at a mean separation $d = 384{,}400$ km. Locate the center of mass of the pair. Earth's radius is 6,371 km.

Step 1. Put the Earth's center at the origin, the Moon on the $x$ -axis. Equation (1) with two bodies: $X = \dfrac{m_M\,d}{m_E + m_M}$ .

Step 2. Total mass: $m_E + m_M = 5.97\times10^{24} + 0.0735\times10^{24} = 6.04\times10^{24}$ kg. Mass fraction: $m_M/(m_E+m_M) = 0.0735/6.04 = 0.01216$ .

Step 3. $X = 0.01216 \times 384{,}400\ \mathrm{km} = \boxed{4{,}670\ \mathrm{km}}$ from Earth's center — about 1,700 km below Earth's surface.

Step 4. Read the physics: the Moon does not orbit "the Earth"; both bodies orbit this buried barycenter every 27.3 days. The Earth's monthly wobble around it — a circle of radius 4,670 km — is exactly the kind of reflex motion that lets astronomers detect unseen companions of distant stars: the companion may be invisible, but the mass-weighted average must sit still (or coast), so the star's wobble betrays the planet.

2 · Walking on a boat (exam level)

A 60 kg person stands at the stern of a 240 kg, 5.0 m boat floating at rest on calm water (neglect water resistance). She walks to the bow. How far does the boat move, and how far does the person move relative to the shore?

Step 1. Identify the law. No external horizontal force acts (gravity and buoyancy are vertical), so the system CM stays fixed: $m_p\,\Delta x_p + m_b\,\Delta x_b = 0$ , the differentiated form of equation (4).

Step 2. Relate the two displacements. Relative to the boat she walks the full length $L = 5.0$ m toward the bow, so her shore-frame displacement is $\Delta x_p = L + \Delta x_b$ (the boat's displacement $\Delta x_b$ will come out negative — backward).

Step 3. Substitute into the CM condition: $m_p(L + \Delta x_b) + m_b\,\Delta x_b = 0$ , so $\Delta x_b = -\dfrac{m_p\,L}{m_p + m_b} = -\dfrac{60 \times 5.0}{300} = \boxed{-1.0\ \mathrm{m}}$ . The boat slides 1.0 m sternward.

Step 4. Her displacement over the ground: $\Delta x_p = 5.0 - 1.0 = \boxed{+4.0\ \mathrm{m}}$ .

Step 5. Verify the CM stayed put: $60 \times (+4.0) + 240 \times (-1.0) = 240 - 240 = 0$ . ✓ Notice the general structure: the boat's recoil is the person's share of the total mass, $m_p/(m_p+m_b) = 1/5$ of the walk length — the same lever arithmetic as Example 1, now in motion.

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · St. Petersburg and Berlin Academies

Euler is the most prolific mathematician who ever lived — his collected works pass 80 volumes — and classical mechanics as we actually use it is largely his construction. Newton's laws, as stated in the Principia, applied to point masses; it was Euler, in his 1750 memoir "Discovery of a new principle of mechanics," who proclaimed $M\ddot{\mathbf{R}} = \mathbf{F}_{\mathrm{ext}}$ as an independent axiom valid for any body — solid, fluid, or fragmenting — making the center of mass the licensed representative of every extended object. (Its rotational twin, the torque law, is now called Euler's second law, and his rigid-body equations still fly every spacecraft.) He computed through the loss of sight in both eyes — "now I will have fewer distractions" — dictating to assistants at a pace that increased after blindness. When he died in 1783, the Marquis de Condorcet's eulogy supplied the line every mathematician knows: "he ceased to calculate and to live."

History

1687

— Newton's Principia proves (Corollary IV) that the common center of gravity of interacting bodies "is either at rest or moves uniformly in a right line."

1750

— Euler elevates

M\ddot{\mathbf{R}} = \mathbf{F}_{\mathrm{ext}}

to a first principle valid for all bodies, point-like or extended — Euler's first law.

1751

— Johann Samuel König publishes the decomposition of kinetic energy into CM plus internal parts, König's theorem.

1846

— Le Verrier predicts Neptune from Uranus's wobbles — anomalous motion of a body betraying an unseen partner sharing its center of mass.

1995

— Mayor and Queloz detect 51 Pegasi b from its star's 50-metre-per-second reflex wobble about the star–planet barycenter: the first exoplanet around a Sun-like star, found by center-of-mass bookkeeping.

Going further

Texts. Morin, Introduction to Classical Mechanics, ch. 5 (Momentum) and ch. 8 (Angular Momentum) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §1.2 · Landau & Lifshitz, Mechanics, 3rd ed., §8 (Centre of mass) · Taylor, Classical Mechanics, ch. 3. No empirical constants are used; the hemisphere result $Z=3a/8$ is exact. The worked-example numbers — Earth–Moon barycenter at 4,670 km using $m_M/(m_E+m_M) = 0.01216$ ; boat recoil 1.0 m and walker displacement 4.0 m for 60 kg/240 kg/5.0 m — are arithmetic consequences of the boxed formulas.

Builds on: Newton's Second Law · Newton's Third Law · Vectors in Kinematics and Coordinate Systems · Velocity and Acceleration. Leads to: Relative Motion and Galilean Transformations · Newton's First Law and Inertial Frames · Projectile Motion.

Is the center of mass the same thing as the center of gravity?

In every laboratory situation, yes — but they are conceptually different points. The center of mass is the mass-weighted average position, equation (1), defined with no reference to any force. The center of gravity is the point where the net gravitational torque vanishes, and it coincides with the CM exactly when gravity is uniform across the body. For a skyscraper the difference is nanometers; for a body large enough to feel gravity's variation — the Moon in Earth's field, a satellite with a long gravity-gradient boom — the two separate, and the offset produces real torques (that is how gravity-gradient stabilization orients satellites). Default assumption in mechanics problems: uniform $g$ , so the terms are interchangeable.

Why is the average weighted by mass and not just the geometric middle of the object?

Because the point is built to do a job: carry the system's momentum. The total momentum is $\sum m_i\dot{\mathbf{r}}_i$ , and if a single point $\mathbf{R}$ is to represent it as $M\dot{\mathbf{R}}$ — equation (3) — then $\mathbf{R}$ must be $\frac1M\sum m_i\mathbf{r}_i$ ; no other weighting works. The geometric center (centroid) coincides with the CM only for uniform density. Load one end of a rod with lead and the centroid stays at the middle while the CM shifts toward the lead — and it is the CM, not the centroid, that flies the clean parabola when you throw it.

If the CM of exploding fireworks keeps flying the same parabola, why do real fireworks not seem to work that way?

Two honest caveats hide in "internal forces cancel." First, air resistance is an external force, and after the burst the fragments present far more total surface area than the intact shell did — so the real CM decelerates faster after the explosion; the perfect-parabola statement is exact only in vacuum. Second, once fragments start landing, the ground exerts new external forces and the airborne-CM bookkeeping ends. Within those limits the theorem is ironclad: high-speed footage of shells bursting in near-still air shows the luminous cloud's centroid continuing smoothly through the burst. The theorem's power is its indifference to the explosion itself — chemistry, violence and all, it is invisible to equation (4).

The proof leans entirely on Newton's third law. Do any real forces violate it, and what happens to the theorem then?

Contact forces, gravity, and electrostatics obey the third law and the theorem is safe in all of everyday mechanics. The textbook exception is magnetism: the forces two moving charges exert on each other are not exactly equal and opposite, so the mechanical momentum of the charges alone is not conserved. The resolution — one of the deepest bookkeeping lessons in physics — is that the electromagnetic field carries momentum too; add it, and the total is conserved again with a suitably generalized center-of-energy theorem. For this article's purposes: whenever the interactions are third-law pairs (every mechanics problem you will meet at this tier), $M\ddot{\mathbf{R}} = \mathbf{F}_{\mathrm{ext}}$ holds without exception.

When I jump, does the Earth really recoil?

Yes — the theorem demands it. While you crouch and push, you and the Earth are one system with no relevant external force along the line of the jump, so the combined CM stays put: the Earth moves opposite to you by the mass ratio $m_{\mathrm{you}}/m_\oplus \approx 70/(6\times10^{24}) \approx 10^{-23}$ . For a one-metre jump the planet's displacement is about $10^{-23}$ m — twelve orders of magnitude smaller than a proton. The bookkeeping is exact and the number is absurd, which is the point: conservation laws do not care about detectability. The identical logic at detectable scale is the boat of Worked Example 2 and the star–planet wobble of Example 1.

How does a high-jumper's body pass over the bar if the center of mass goes under it?

Nothing needs to be at the CM — it is an average, not a particle. In the Fosbury flop the jumper arches backward over the bar so that, at every instant, most body mass hangs below the arch: head and legs droop on either side while the hips clear the bar. The mass-weighted average of that draped shape can sit outside the body, under the bar, even as every body part passes over it in sequence. Since takeoff fixes the CM's parabola — equation (4) allows no mid-air steering — the flop converts a fixed CM height budget into a higher cleared bar. The same "CM outside the body" fact is trivial for a ring or a boomerang; athletes just monetized it.

For the continuous integral, how do I actually find the CM of an odd-shaped object in practice?

Three tools, in order. Symmetry: the CM lies on every symmetry plane and axis, which pins it completely for rods, disks, spheres, and rectangles. Decomposition: split the body into simple pieces, replace each by a point mass at its own CM, and apply the discrete formula (1) — holes enter as negative masses, which handles brackets and plates with cutouts. Integration, only for genuinely smooth asymmetry: slice along one coordinate so each slice's CM is known, reducing (2) to a single integral — the solid hemisphere sliced into disks gives $Z = 3a/8$ in three lines. And for a real irregular object, physics beats calculus: hang it from two different points; the CM lies on each plumb line through the pivot, so it sits at their intersection.