Locating the Centre of Mass

Every commercial flight you board has had this article's arithmetic done for it: loadmasters compute the aircraft's centre of gravity from passenger, fuel and cargo placement and certify it inside the manufacturer's envelope — a COM too far aft and the aircraft cannot recover from a stall; too far forward and it cannot flare to land. The Fosbury flop remains sport's cleanest exploitation: arching the body throws the COM outside it, under the bar, so the jumper "clears" a height the mass headquarters never visits — worth roughly 10–20 cm of free altitude over the straddle technique it replaced. Humanoid robots stand and walk by computing their COM hundreds of times a second and steering it over the support polygon of their feet — Boston Dynamics' parkour is centre-of-mass management at combat speed. And the Earth–Moon calculation of worked example 2 is now a discovery instrument: a star and its unseen planet orbit their common COM, and the star's tiny wobble about that buried point is how the radial-velocity method has weighed hundreds of exoplanets.

Why should a single point represent an extended, lumpy object? Because of what "average" does. Each particle of the bat pulls the average toward itself with a strength proportional to its mass — the massive blade tugs the average bladeward far harder than the light handle resists — and the resulting point is the system's mass headquarters. Two properties make it more than a definition. It is the balance point: supported there, gravity's turning effects from the two sides cancel exactly (derivation A makes this precise). And — the next article's headline — it is the point that moves simply: throw the bat spinning through the air, and while every other point corkscrews, the COM sails on a clean parabola.

Three habits of the point are worth installing before any formula. It hugs the heavy side: for two bodies it divides the separation in inverse mass ratio, so the Earth–Moon COM sits deep inside the Earth. It respects symmetry: a uniform rod, disc, sphere or ring balances at its geometric centre, no computation required — mass to the left mirrors mass to the right. It owes the material nothing: the ring's centre is empty air, an L-shaped plate's COM can sit in the notch, a boomerang's lies outside the wood entirely — and the Fosbury flopper exploits exactly this, draping her body around a COM that slips under the bar. "The COM is a point of the object" is a misconception; it is a point of the description. One more distinction, cheap now and worth marks later: the centre of gravity is the average position weighted by weight ($m_ig_i$) rather than mass — the point where gravity's net pull effectively acts. Wherever $g$ is the same for every particle (any lab, any exam problem not set on a megastructure), weights are proportional to masses and the two points coincide. They part company only when $g$ varies across the object — a skyscraper's CoG sits measurably below its COM because the lower floors feel marginally stronger gravity.

The definition, in coordinates (one line per axis):

where: $m_i, (x_i, y_i)$ — each particle's mass and coordinates, all measured from one origin of your choice. Read it as a lever statement: each mass "votes" for its own position with $m_i$ ballots, and the COM is the election result. The denominator is the total mass $M$ — dividing by the number of particles instead (a plain average) is the first classic error; it is correct only when all masses are equal. For two bodies, (1) collapses to the geometry worth memorizing:

where: $r_1, r_2$ — distances of $m_1, m_2$ from the COM along the line joining them. Inverse ratio: the COM sits closer to the heavier mass, dividing the separation $d$ into $r_1 = \frac{m_2}{m_1+m_2}d$ and $r_2 = \frac{m_1}{m_1+m_2}d$. Special case that anchors intuition: equal masses → the midpoint. For continuous uniform bodies, symmetry does the integral's work: the COM lies on every axis of symmetry, hence at the geometric centre of rods, rings, discs, rectangles, spheres and cylinders. Shapes with only partial symmetry carry standard results (derived by integration at UG level — quote them at this level, and know which is which):

For composite bodies — the exam's main course — use the theorem that makes (1) scale: each part may be replaced by its total mass sitting at its own COM, and the parts then averaged as if they were particles. An L-plate is two (or three) rectangles; a hammer is a rod plus a head; and a body with a hole is the full body plus the removed piece carrying negative mass — the trick that turns the JEE disc-with-hole from an integral into three lines (derivation B).

where: masses of uniform plates are proportional to their areas (volumes in 3D), so areas may stand in for masses throughout. The sign discipline: the hole's term enters with a minus in both numerator and denominator.

1 · Board level — three particles on a triangle (CBSE pattern, 3 marks)

Particles of mass $1.0$, $2.0$ and $3.0\ \mathrm{kg}$ sit at the corners $(0,0)$, $(1.0, 0)$ and $(0, 1.0)$ (metres). Locate the centre of mass.

Step 1. Total mass $M = 6.0$ kg. One axis at a time.

Step 2. $x_{\rm cm} = \dfrac{1(0) + 2(1.0) + 3(0)}{6} = \boxed{\dfrac{1}{3}\ \mathrm{m}}$.

Step 3. $y_{\rm cm} = \dfrac{1(0) + 2(0) + 3(1.0)}{6} = \boxed{\dfrac{1}{2}\ \mathrm{m}}$.

Step 4. Sanity: the point $(0.33, 0.50)$ leans toward the 3 kg corner (up) and the 2 kg corner (right), inside the triangle — and it is not the centroid $(1/3, 1/3)$, which only equal masses would give. State the comparison; it shows the examiner you know what the weighting does.

2 · Board level — the Earth–Moon headquarters (CBSE/ICSE pattern)

The Moon's mass is $1/81$ of the Earth's, and their centres are $3.84 \times 10^5\ \mathrm{km}$ apart. How far from the Earth's centre is the COM of the Earth–Moon system? (Earth's radius: $6371\ \mathrm{km}$.)

Step 1. Two-body split: Earth's distance from the COM is $r_E = \dfrac{m_M}{m_E + m_M}\,d = \dfrac{1}{82}\,d$.

Step 2. $r_E = \dfrac{3.84\times10^5}{82} = \boxed{4.7\times10^3\ \mathrm{km}}$ from the Earth's centre.

Step 3. Compare: $4700\ \mathrm{km} < 6371\ \mathrm{km}$ — the system's balance point lies about $1700$ km beneath the Earth's surface. The Moon does not orbit the Earth's centre; both bodies waltz around this buried point — the wobble that astronomers use to weigh distant planets.

3 · Competitive level — the disc with a hole (JEE-Main pattern, ~2.5 minutes)

From a uniform disc of radius $R$ centred at $O$, a circular hole of radius $R/2$ is punched out with its centre at $R/2$ from $O$. Locate the COM of the remaining plate.

Step 1. Areas stand for masses (uniform): full $\pi R^2$, hole $\pi R^2/4$ at $x = +R/2$.

Step 2. Negative-mass formula: $x_{\rm cm} = \dfrac{\pi R^2(0) - \frac{\pi R^2}{4}\cdot\frac{R}{2}}{\pi R^2 - \frac{\pi R^2}{4}} = \dfrac{-R/8}{3/4} = \boxed{-\dfrac{R}{6}}$ — a shift of $R/6$ on the side opposite the hole.

Step 3. Audits: direction away from the hole ✓ (mass was removed from $+x$); magnitude less than the plug's own offset ✓; hole → 0 kills the shift ✓. Timing note: ~2.5 min, of which two are sign discipline. Set the origin at the FULL body's centre (its COM term then vanishes) and let the hole carry both minus signs. Variants — square plate with circular bite, sphere with spherical cavity — run on the identical three lines with areas → volumes for solids; the cavity-sphere version is a JEE-Advanced regular.

4 · Competitive level — the L-plate (JEE-Main/NEET pattern, ~3 minutes)

A uniform L-shaped plate is made of three squares, each of side $a$: two along the x-axis and one stacked on the first (corners at the origin). Locate its COM.

Step 1. Decompose: three equal squares (equal areas ⇒ equal masses $m$), COMs at their centres: $A(\tfrac a2, \tfrac a2)$, $B(\tfrac{3a}2, \tfrac a2)$, $C(\tfrac a2, \tfrac{3a}2)$.

Step 2. Equal masses — a plain average is legal here: $x_{\rm cm} = \dfrac{\tfrac a2 + \tfrac{3a}2 + \tfrac a2}{3} = \boxed{\dfrac{5a}{6}}$, $\;y_{\rm cm} = \dfrac{\tfrac a2 + \tfrac a2 + \tfrac{3a}2}{3} = \boxed{\dfrac{5a}{6}}$.

Step 3. Read the result: $(5a/6, 5a/6)$ lies on the L's diagonal symmetry line ($y = x$ — the shape is mirror-symmetric about it) ✓, and just inside the material: the point sits in the corner square, a whisker from the inner notch. Tilt the L slightly and its balance point would hover over the void — the "COM outside the material" phenomenon one bite away.

Step 4. Cross-check by a different decomposition (the professional habit): a $2a \times a$ rectangle (mass $2m$, COM $(a, a/2)$) plus the top square ($m$ at $(\tfrac a2, \tfrac{3a}2)$): $x = \frac{2m(a) + m(a/2)}{3m} = \frac{5a}{6}$ ✓. Same point, any legal chopping. Timing note: ~3 min including the cross-check — which is worth its 30 seconds in a multi-mark question: two decompositions agreeing is proof against arithmetic slips. For unequal pieces, remember the weights: only equal masses permit Step 2's bare average.

Unweighted averages. Dividing $\sum m_ix_i$ by the particle count, or averaging part-COMs without their masses, is the founding error of the topic. Legal only when the masses are equal — and then say so.

One origin per problem. All positions in (1) must be measured from the same origin. Mixing "distance from A" with "distance from B" mid-formula produces confident nonsense. (The COM point itself is origin-independent — only its coordinates change — so choose the origin that kills the most terms.)

Inverse ratio inverted. The COM is closer to the heavier mass: $r_1/r_2 = m_2/m_1$. Writing the ratio the intuitive-but-wrong way round survives dimensional checks and dies on the equal-mass corner — audit with $m_1 = m_2 \Rightarrow$ midpoint.

Ring vs disc values swapped. Semicircular ring $2R/\pi \approx 0.64R$; semicircular disc $4R/3\pi \approx 0.42R$. The disc's is smaller (its mass crowds toward the flat side). Cone solid $h/4$ vs hollow $h/3$ swap the same way. One mnemonic: spread the mass outward (ring, shell), push the COM outward.

Negative-mass sign slips. The hole's mass subtracts in numerator and denominator, and its coordinate is the hole-centre's, not the edge's. Set the origin at the full body's COM and the formula becomes one clean fraction.

"COM must be inside the body." Rings, Ls, horseshoes, high-jumpers. The reliable bound is the convex hull, nothing tighter.

COM vs CoG on principle questions. They coincide for uniform $g$ — every numerical problem at this level — but definition questions ("distinguish…", an ISC favourite) want the weighting difference stated: mass-weighted vs weight-weighted, parting company when $g$ varies over the body.

Areas-for-masses without uniformity. The proportionality $m \propto A$ needs uniform plates. A problem that varies the density (two materials welded) is telling you to weight by actual masses — read for the word "uniform" before borrowing areas.

Texts. NCERT Physics Class 11, Ch. 7 (Systems of Particles — centre of mass sections) · HC Verma, Concepts of Physics Vol. 1, Ch. 9 · Archimedes, On the Equilibrium of Planes for the origins — and any JEE archive under "centre of mass of remainder".

Provenance. All results are standard. A verifier should recompute: the moment-cancellation identity in derivation A; the $-R/6$ shift in derivation B (and via formula (3)); Example 1's $(1/3, 1/2)$; Example 2's $3.84\times10^5/82 = 4.68\times10^3$ km $<$ 6371 km; Example 4's $(5a/6, 5a/6)$ by both decompositions; the standard-results table against NCERT/HC Verma.

Prerequisites. Scalars & Vectors · Components & Resolution of Vectors. Leads to. Motion of the Centre of Mass · Rigid-Body Equilibrium, CoG & Moments (rotation chapter) · Systems of Particles & Centre of Mass (UG deep dive).