Study Companion: S17 — Centre of Mass and Centre of Gravity

Every force in Parts 1–4 acted at a point. Weight was one arrow at one place, a pin pushed at its centre, a tyre at its contact point. None of that is literally true: weight acts on every gram of a body, and every contact force is spread over an area, however small. This lesson starts Part 5, distributed forces, with the most common one, gravity. It asks where the single arrow that replaces all those tiny weights must be drawn, and it finds that point, the centre of gravity, by the principle of moments: the moment of the sum equals the sum of the moments.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, articles 5/1 ("Introduction") and 5/2 ("Center of Mass"), with the Chapter Review (5/10). Marghitu, Dupac and Madsen, Statics with MATLAB, sections 3.1 ("First Moment"), 3.2 ("Center of Mass of a Set of Particles"), 3.3 ("Center of Mass of a Body") and 3.5 ("Center of Gravity"). Neither book works an example on these sections, so the examples here are ours, built on their equations. Marghitu calls the centre of mass C; this lesson uses Meriam's G throughout. Axes: x to the right, y up, z toward the reader.

How to use this companion

Each nugget is one idea. It opens with a claim, then a picture shows why it holds, a number checks it, and an equation states it, labelled with the kind of statement it is. A worked example with units, a real use, the most common wrong answer and a check question follow. A bridge says why the next idea is needed, and the Narration spine summarises the video beat. Traps and a self-check with full solutions close the lesson.

The puzzle

A 1500\,\mathrm{kg} car is rolled onto four wheel scales, the kind used to set up a race car. The two front scales together read 870\,\mathrm{kg}, the two rear ones 630\,\mathrm{kg}. The axles are 2.70\,\mathrm m apart.

Where along the car is its centre of gravity, and can the scales also tell how high it is? The plausible guess is "about halfway between the axles, since the weight is spread along the whole car". The answers are in Nugget 03.

Nugget 01: Every Real Force Is Spread Out, and Concentrated Is a Model

Claim. No force acts at a mathematical point. A force may be treated as concentrated when the region it is spread over is small compared with the body's other dimensions and only its effect on the body as a whole is wanted. Otherwise it is a distributed force, described by its intensity: per unit length, per unit area or per unit volume.

The picture. The three kinds of distribution.

A cable carrying its own weight, or a deck hung from it, is loaded along a line: the intensity w is in newtons per metre. Water on the face of a dam pushes on an area: the intensity is a pressure, in pascals. Gravity pulls on every element of a body's volume: a body force, of intensity \rho g in newtons per cubic metre.

Check with numbers. A 1500\,\mathrm{kg} car on four tyres inflated to 220\,\mathrm{kPa}: each tyre flattens until its contact patch carries a quarter of the weight at roughly the inflation pressure, so each patch is about

A = \frac{1500(9.81)/4}{220\times10^{3}} = \frac{3679\,\mathrm N}{220\times10^{3}\,\mathrm{Pa}} = 0.01672\,\mathrm{m^2},

a square about 129\,\mathrm{mm} on a side. Against the 2.70\,\mathrm m wheelbase that is 4.8\,\%. For the forces on the car as a whole, each patch may be replaced by one force at its centre; for the stresses in the tyre's rubber it may not.

The idea. Meriam (5/1): "concentrated" forces do not exist in the exact sense. The pavement pushes on a tyre over its whole contact area, a hardened steel ball touches its race over a tiny but finite area, and a truss pin bears on its hole over an area. When the dimension b of that area is negligible against the other pertinent dimensions, the distributed contact forces are replaced by their resultant R for the external effects on the body. Their internal effects, the stresses near the contact, need the actual distribution and the material's properties: that is mechanics of materials, not statics. When the region is not small, the distribution is summed by integration, which needs the intensity at every point. Three categories:

Line distribution: intensity w, force per unit length (N/m), as in the load on a suspension cable.
Area distribution: force per unit area, called pressure for fluids and stress inside solids, in N/m², the pascal (Pa). The pascal is small, so fluid pressures are usually in kPa and stresses in MPa. (Meriam's conversion for US readers, 6895\,\mathrm{Pa} = 1\,\mathrm{lb/in^2}, is 6894.76\,\mathrm{Pa} to more figures.)
Volume distribution: a body force, the commonest being gravity, of intensity \rho g, the specific weight, in (\mathrm{kg/m^3})(\mathrm{m/s^2}) = \mathrm{N/m^3}.

The Chapter Review (5/10) gives the general recipe for every problem in this part: multiply the intensity by the element of length, area or volume it is expressed per, integrate to get the resultant, then place the resultant by the principle of moments.

The equation (definition of the three intensities):

dF = w\,dL, \qquad dF = p\,dA, \qquad dW = \rho g\,dV

Worked number. Steel, \rho = 7850\,\mathrm{kg/m^3}: \rho g = 7850(9.81) = 77.0\times10^{3}\,\mathrm{N/m^3} = 77.0\,\mathrm{kN/m^3}. Water: 9.81\,\mathrm{kN/m^3}. A power line of 1.2\,\mathrm{kg} per metre carries w = 1.2(9.81) = 11.77\,\mathrm{N/m} of its own weight, 3.53\,\mathrm{kN} over a 300\,\mathrm m span.

Real example (one you can see). Look at a car tyre from the side: the flat of the contact patch is visible, and a soft tyre's patch is visibly longer. Halve the pressure and the patch doubles in area for the same load.

Plausible wrong answer. "A force at a point is exact; distributed loads are an approximation." It is the other way round: the point force is the model, justified when the patch is small.

Check. A 1\,\mathrm m long beam rests on a 0.6\,\mathrm m wide pad. Is the pad's push a concentrated force? (Not for anything inside the beam near the pad, and only roughly for the beam as a whole: 0.6\,\mathrm m is not small against 1\,\mathrm m.)

Narration spine. A car tyre squashes onto the road and a coloured patch spreads under it; the patch's arrows merge into one arrow at its centre. Then three panels fill in turn: arrows along a cable labelled N/m, arrows on a dam growing with depth labelled Pa, arrows on every speck of a body labelled N/m³.

Bridge. Gravity is the body force that every structure carries. Its many small arrows must be replaced by one, through a point that can be found by experiment.

Nugget 02: Hang a Body From Two Points and the Cords Cross at G

Claim. Suspend a body from a point and it settles with its weight's line of action straight below the cord. Do it from two different points: the two lines cross at the centre of gravity G. Every further suspension line passes through the same point, provided the gravity field is uniform and parallel.

The picture. An L-shaped plate, both legs 0.4\,\mathrm m long and 0.1\,\mathrm m wide, hung from the end of each leg.

Hanging freely, the plate is a two-force body: the cord's pull and the resultant weight must be collinear (S11). So each plumb line, marked on the plate, contains G. The two lines, redrawn on the plate in (c), cross at a point inside the corner of the L, outside the material.

Check with numbers. By symmetry about the plate's diagonal, G lies on the line x = y. The two legs, as rectangles of area 0.04\,\mathrm{m^2} (centre (0.2, 0.05) m) and 0.03\,\mathrm{m^2} (centre (0.05, 0.25) m, the upright above the shared corner):

\bar x = \frac{0.04(0.2) + 0.03(0.05)}{0.07} = \frac{0.0095}{0.07} = 0.1357\,\mathrm m = \bar y.

Both coordinates exceed 0.1\,\mathrm m, the width of the legs, so G is in the empty corner. Hung from the top of the upright, the plumb line makes \tan^{-1}[0.1357/(0.4 - 0.1357)] = 27.2^\circ with the upright's edge.

The idea. Meriam (5/2): hang a body from A; it is in equilibrium under the cord's tension and the resultant W of the gravitational forces on all its particles, so W is collinear with the cord. Mark that line, hang it from B and C, mark again: for all practical purposes the lines are concurrent at a single point, the centre of gravity G. An exact analysis disagrees slightly. The gravity forces on different particles point toward the Earth's centre, so they are not quite parallel, and particles at different heights feel slightly different gravity. The lines would then not quite meet, and no unique centre of gravity would exist. For bodies small compared with the Earth this does not matter: we assume a uniform and parallel gravity field, and that assumption makes G unique (Nugget 06 measures how small the error is).

The equation (physical law: equilibrium of a freely hanging body, S11's two-force body):

\mathbf T + \mathbf W = \mathbf 0 \text{ and collinear} \;\Rightarrow\; G \text{ lies on every suspension line}

Worked number. For the L-plate, the suspension from the top of the upright gives one line, from the end of the other leg a second; by symmetry they meet on x = y at (0.1357, 0.1357)\,\mathrm m, 136\,\mathrm{mm} from each outer edge.

Real example (one you can see). A picture hung from a single nail settles with the nail straight above its centre of gravity; a crane hook is placed above a load's G before lifting, or the load swings until it is.

Plausible wrong answer. "The centre of gravity is a point of the body." For the L-plate it is in the air at the inside of the corner; a ring's is at its empty centre.

Check. A plate is hung from a corner and the plumb line passes through the opposite corner. What does that say about G? (It lies on that diagonal.)

Narration spine. The L-plate hangs from one corner and swings to rest; a dashed plumb line is drawn on it. It is rehung from the other leg, swings, another line. The plate rights itself, the two lines cross in the empty corner, and a dot labelled G appears there.

Bridge. Suspension finds G for a body in your hand. For a body on paper, a calculation is needed, and it comes from moments.

Nugget 03: The Moment of the Sum Equals the Sum of the Moments

Claim. The resultant weight W of a body must have the same moment about any axis as all the elemental weights dW together. Equating the two places G: \bar xW = \int x\,dW, and likewise for \bar y and \bar z.

The picture. Left: a body's element of weight dW at x, and the resultant W at \bar x. Right: the puzzle's car, its weight and the two axle loads.

The elements' weights are parallel forces (S06). Their resultant is their sum, W = \int dW, and its line of action is where its moment equals the sum of the elements' moments. That line passes through G.

Check with numbers (the puzzle). The car is a body in equilibrium on two supports, so its weight's moment about the front axle equals the rear axle load's moment: W\bar x = R_r L with \bar x measured back from the front axle. Mass cancels from both sides with g:

\bar x = \frac{630}{1500}(2.70) = 1.134\,\mathrm m \text{ behind the front axle.}

Check with moments about G itself, where the weight has no arm: the front load gives 870(1.134) = 986.6\,\mathrm{kg\,m} one way and the rear load 630(2.70 - 1.134) = 630(1.566) = 986.6\,\mathrm{kg\,m} the other. They balance. The halfway guess, 1.35\,\mathrm m, is 0.216\,\mathrm m too far back: the engine is in front.

The idea. Meriam (5/2) applies the principle of moments (2/6) to the parallel system of gravitational forces. The moment about the y-axis of an elemental weight is x\,dW, the sum of these over the body is \int x\,dW, and that sum must equal W\bar x, the moment of the sum. Similar expressions give \bar y and \bar z:

\bar x = \frac{\int x\,dW}{W}, \qquad \bar y = \frac{\int y\,dW}{W}, \qquad \bar z = \frac{\int z\,dW}{W} \qquad (5/1a)

To see the third as a physical moment, reorient the body so the z-axis is horizontal. Meriam's emphasis: in each expression the numerator is the sum of the moments, and W times the coordinate is the moment of the sum. This principle recurs throughout mechanics. The Chapter Review (5/10) states it as the method for every distributed force: evaluate the sum of the moments of the increments about a convenient axis, equate it to the moment of the resultant, and solve for the resultant's moment arm.

The equation (mathematical identity: the principle of moments for the parallel gravity forces):

W\bar x = \int x\,dW \qquad \text{(moment of the sum = sum of the moments)}

Worked number (the height, by tilting). Raise the car's rear axle until the axle line slopes at 10^\circ. If G lay on the axle line, the rear reading would not change (both arms shrink by \cos10^\circ). It falls to 605.5\,\mathrm{kg}:

With G a height h above the axle line, tilting moves it forward horizontally by h\sin10^\circ. Moments about the front contact, which lies straight below the front axle:

R_r(L\cos\theta) = W(\bar x\cos\theta - h\sin\theta) \;\Rightarrow\; h = \frac{\bar x - L R_r/W}{\tan\theta} = \frac{1.134 - 2.70(605.5/1500)}{\tan10^\circ} = \frac{0.0441}{0.1763} = 0.250\,\mathrm m.

With wheels of 0.30\,\mathrm m radius, G is 0.55\,\mathrm m above the ground. Every number came from scale readings and one angle.

Real example (one you can see). This is how race teams and vehicle test labs find a car's centre of gravity: corner-weight scales give its position in plan, and one reading with an axle lifted gives its height.

Plausible wrong answer. "Halfway between the axles." Only if the axle loads are equal; here G is 1.134\,\mathrm m from the front, not 1.35\,\mathrm m.

Check. If both scales read the same, where is G, and what does a tilt test then tell you? (Midway; the tilt still gives its height.)

Narration spine. A body breaks into a cloud of small downward arrows. One arrow and its arm x light up, the arms sweep over the body summing, and a single blue arrow W slides along until its moment matches: it stops at \bar x. The car appears on its scales, its arrow W settling 1.134\,\mathrm m behind the front axle; then the rear rises 10^\circ, the rear reading ticks down to 605.5\,\mathrm{kg}, and G lifts off the axle line to 0.25\,\mathrm m.

Bridge. The weights dW are one choice of "strength" attached to each point. Marghitu starts more generally: any strengths at any points have a centroid.

Nugget 04: The First Moment Gives the Centroid of Any Weighted Set of Points

Claim. Attach a scalar strength s_i to each point P_i. Its first moment about O is the vector s_i\mathbf r_i. The centroid C is the one point about which the first moments of all the points add to zero; it does not depend on where O is. With masses as strengths, C is the centre of mass.

The picture. Marghitu's Fig. 3.1, redrawn.

From any origin O, each point has a position \mathbf r_i and a first moment s_i\mathbf r_i. Measured from C instead, the positions are \mathbf r_i - \mathbf r_C, and C is chosen so that \sum s_i(\mathbf r_i - \mathbf r_C) = \mathbf 0.

Check with numbers. Three particles: 2\,\mathrm{kg} at (0, 0, 0), 3\,\mathrm{kg} at (1, 0, 0) and 5\,\mathrm{kg} at (0, 2, 1)\,\mathrm m; total 10\,\mathrm{kg}.

\mathbf r_C = \frac{2(0, 0, 0) + 3(1, 0, 0) + 5(0, 2, 1)}{10} = \frac{(3, 10, 5)}{10} = (0.3, 1.0, 0.5)\,\mathrm m.

From the origin O' = (1, 1, 1)\,\mathrm m the positions become (-1, -1, -1), (0, -1, -1), (-1, 1, 0), and \mathbf r'_C = [2(-1, -1, -1) + 3(0, -1, -1) + 5(-1, 1, 0)]/10 = (-0.7, 0, -0.5)\,\mathrm m. Adding back (1, 1, 1) gives (0.3, 1.0, 0.5)\,\mathrm m: the same point.

The idea. Marghitu (3.1) defines the first moment of a point P_i of strength s_i about O as \mathbf M_i = s_i\mathbf r_i; the strengths are scalars of one dimension, mass being one example. The centroid C of the set is the point with respect to which the sum of the first moments is zero:

\sum_{i=1}^{n} s_i(\mathbf r_i - \mathbf r_C) = \sum s_i\mathbf r_i - \mathbf r_C\sum s_i = \mathbf 0 \;\Rightarrow\; \mathbf r_C = \frac{\sum s_i\mathbf r_i}{\sum s_i}.

It does not depend on the reference point: moving O by \mathbf a changes every \mathbf r_i and \mathbf r_C by the same \mathbf a. If \sum s_i = 0 the centroid is not defined. The centre of mass of a set of particles (3.2) is the centroid with s_i = m_i: \mathbf r_C = \sum m_i\mathbf r_i/M, with M the total mass (Eq. 3.1).

Second view, reconciled. Marghitu's condition is Meriam's principle of moments with g divided out. Taking the moment of each weight m_ig about any axis through C and requiring the total to vanish is the same as requiring \sum m_i(\mathbf r_i - \mathbf r_C) = \mathbf 0, because the weights are parallel. The two definitions pick the same point.

The equation (definition: centroid of points of given strength; centre of mass when s_i = m_i):

\mathbf r_C = \frac{\sum s_i\mathbf r_i}{\sum s_i}, \qquad \mathbf r_G = \frac{\sum m_i\mathbf r_i}{\sum m_i}

Worked number. Strengths need not be masses. Two equal and opposite strengths, +2 at A and -2 at B, sum to zero: no centroid exists, and their total first moment, 2(\mathbf r_A - \mathbf r_B), is the same about every origin, like a couple's moment (S05). Negative strengths are useful, though: Marghitu (3.3) treats a hole as a part of negative size.

Real example. A loaded aircraft's centre of gravity is computed exactly this way, as a sum of mass times position over fuel tanks, passengers, cargo and the empty airframe, and checked against the allowed envelope before every flight.

Plausible wrong answer. "Moving the origin moves the centroid." Every position shifts by the same vector, and so does \mathbf r_C: the point stays put.

Check. Two particles, 1\,\mathrm{kg} at x = 0 and 3\,\mathrm{kg} at x = 4\,\mathrm m. Where is C? (x = 3\,\mathrm m, nearer the heavier one.)

Narration spine. Three dots of different sizes with arrows from O. A blue point C slides around; dashed arrows from C to each dot, scaled by its strength, rebalance until they sum to nothing, and C freezes. O jumps elsewhere, all the arrows from O redraw, and C does not move.

Bridge. A real body is not a few particles but a continuum. The sums become integrals.

Nugget 05: For a Continuous Body, Sum the Mass Elements

Claim. For a body, replace each particle by an element of mass dm and each sum by an integral: \bar{\mathbf r} = \int\mathbf r\,dm/m. When the density varies, write dm = \rho\,dV and keep \rho inside the integrals; the denser part pulls G toward itself.

The picture. A rod whose density grows linearly from \rho_0 at one end to 2\rho_0 at the other.

A slice of length dx at x has mass dm = \rho(x)\,dx, with \rho here mass per unit length. More of the mass sits toward the dense end, so G moves past the midpoint.

Check with numbers. With \rho = \rho_0(1 + x/L):

\bar x = \frac{\int_0^L x\rho_0(1 + x/L)\,dx}{\int_0^L \rho_0(1 + x/L)\,dx} = \frac{\frac{L^2}{2} + \frac{L^2}{3}}{L + \frac L2} = \frac{\frac56L^2}{\frac32L} = \frac{5L}{9} = 0.556L.

For L = 2\,\mathrm m and \rho_0 = 3\,\mathrm{kg/m}: mass 3(2 + 1) = 9\,\mathrm{kg}, \bar x = 1.111\,\mathrm m, against 1\,\mathrm m for a uniform rod.

The idea. Meriam (5/2): with W = mg and dW = g\,dm, Eq. 5/1a becomes

\bar x = \frac{\int x\,dm}{m}, \quad \bar y = \frac{\int y\,dm}{m}, \quad \bar z = \frac{\int z\,dm}{m} \quad (5/1b), \qquad \bar{\mathbf r} = \frac{\int\mathbf r\,dm}{m} \quad (5/2)

the components of one vector equation, with \mathbf r = x\mathbf i + y\mathbf j + z\mathbf k locating the element and \bar{\mathbf r} locating G. The density \rho is mass per unit volume, so dm = \rho\,dV; when \rho varies with position it stays inside both integrals:

\bar x = \frac{\int x\rho\,dV}{\int\rho\,dV}, \quad \bar y = \frac{\int y\rho\,dV}{\int\rho\,dV}, \quad \bar z = \frac{\int z\rho\,dV}{\int\rho\,dV} \qquad (5/3)

Marghitu (3.3) writes the same (Eqs. 3.2–3.3) and notes that the centre of mass is the point about which the body's total mass moment is zero. For a uniform body \rho cancels (Eq. 3.4), leaving pure geometry; for a uniform curve (\rho_l = m/L) and a uniform surface (\rho_s = m/A) the integrals run over length and area (Eqs. 3.5–3.6), the subject of S18. Marghitu 3.3 also gives the decomposition method for composite bodies: split the body into simple parts, locate each part's centre, and combine them with the particle formula, \bar x = \sum\bar x_i\Delta_i/\sum\Delta_i, with a hole entered as a part of negative size (Eqs. 3.7–3.8). The L-plate of Nugget 02 is an example: a 0.4\,\mathrm m square minus a 0.3\,\mathrm m square, \bar x = [0.16(0.2) - 0.09(0.25)]/(0.16 - 0.09) = 0.1357\,\mathrm m. S19 develops it.

The equation (definition of the mass centre of a continuum):

\bar{\mathbf r} = \frac{\int\mathbf r\,dm}{\int dm} = \frac{\int\mathbf r\rho\,dV}{\int\rho\,dV}

Worked number. Meriam's Problem 5/32, a triangular plate of base b and height h whose thickness grows linearly from t_0 at the base to 2t_0 at the apex. At height y its width is b(1 - y/h) and its thickness t_0(1 + y/h), so dm \propto (1 - y/h)(1 + y/h)\,dy:

\bar y = \frac{\int_0^h y(1 - y^2/h^2)\,dy}{\int_0^h (1 - y^2/h^2)\,dy} = \frac{h^2/4}{2h/3} = \frac{3h}{8},

higher than the uniform triangle's h/3, toward the thick apex.

Real example. A baseball bat, a golf club or a tapered mast has its centre of mass toward its heavy end. Bat makers shift it deliberately ("end-loaded" bats) by varying the wall thickness, which is a variable density along the length.

Plausible wrong answer. "Find the geometric centre, then weight it by the average density." The average density gives the mass, not the position; \rho must weight each element's moment.

Check. If \rho is the same everywhere, what happens to Eq. 5/3? (\rho cancels and G is the centroid of the shape.)

Narration spine. The rod's colour deepens from one end to the other. A thin slice slides along it, its mass bar growing with the colour; a running balance point drifts from L/2 toward the dark end and stops at 5L/9.

Bridge. Equations 5/1b, 5/2 and 5/3 no longer contain g. They define a point that would exist even without gravity.

Nugget 06: Centre of Mass and Centre of Gravity Agree Only in a Uniform Field

Claim. The centre of mass depends only on how mass is distributed; the centre of gravity depends on the gravity field as well. They coincide when the field is uniform and parallel, which near the Earth's surface is true to far better than engineering accuracy. Away from gravity, a body still has a centre of mass but no centre of gravity.

The picture. A tall tower in the real field, which weakens with height.

Each kilogram near the top weighs slightly less than one at the base, so the weights are not proportional to the masses, and the point where their resultant acts sits slightly below the centre of mass.

Check with numbers. A uniform 828\,\mathrm m tower, with g falling as g_0R^2/(R + z)^2 (R = 6.371\times10^{6}\,\mathrm m): at the top, g is 0.99974\,g_0. With g inside the integrals,

\bar z_{\text{gravity}} = \frac{\int_0^H z\,g(z)\,dz}{\int_0^H g(z)\,dz} \approx \frac H2 - \frac{H^2}{6R} = 414\,\mathrm m - 0.0179\,\mathrm m.

The centre of gravity is 17.9\,\mathrm{mm} below the centre of mass on an 828\,\mathrm m tower. Over a 100\,\mathrm m long body the directions of gravity differ by 100/R rad, 0.0009^\circ.

The idea. Meriam (5/2): Eqs. 5/1b, 5/2 and 5/3 contain no g, so they define a unique point that depends solely on the distribution of mass, the centre of mass. It coincides with the centre of gravity as long as the field is uniform and parallel. It is meaningless to speak of a centre of gravity for a body removed from the Earth's field, but the body still has its centre of mass, and it is the centre of mass that governs a body's dynamic response to unbalanced forces (dynamics). Meriam therefore refers to the centre of mass from here on. Marghitu (3.5) defines the centre of gravity as the point that locates the resultant weight, about which the moments of the particles' weights sum to zero, and writes it with g inside the integrals:

\bar x = \frac{\int x\rho g\,dV}{\int\rho g\,dV}, \quad \text{and likewise } \bar y, \bar z \qquad (3.13)

If g is constant throughout the body it cancels and the centre of gravity is the centre of mass; Marghitu quotes g = 9.81\,\mathrm{m/s^2}.

The equation (approximation: the uniform-field assumption):

g = \text{const} \;\Rightarrow\; \frac{\int\mathbf r\rho g\,dV}{\int\rho g\,dV} = \frac{\int\mathbf r\rho\,dV}{\int\rho\,dV}, \quad \text{error} \approx \frac{H^2}{6R} \text{ for a vertical body of height } H

Worked number. For a 50\,\mathrm m crane the gap is 50^2/(6 \times 6.371\times10^{6}) = 0.07\,\mathrm{mm}, far below any tolerance of manufacture.

Real example. On the International Space Station objects float; they have no weight to speak of and so no meaningful centre of gravity, yet a thrown tool still spins about its centre of mass.

Plausible wrong answer. "Centre of mass and centre of gravity are two names for one point." Only in a uniform field; in principle they differ, and in orbit only one exists.

Check. Does a satellite in orbit have a centre of mass? A centre of gravity? (Yes; a centre of gravity in the textbook sense is not meaningful there.)

Narration spine. A tall tower; arrows of weight on its slices, slightly shorter toward the top. Two dots, blue for mass and orange for gravity, start together at mid-height; a magnifier zooms in, and the orange dot sits a hair below: 17.9\,\mathrm{mm} on 828\,\mathrm m.

Bridge. The integrals are easiest when the axes are chosen well, and many can be skipped altogether.

Nugget 07: Symmetry and the Right Axes Do Half the Work

Claim. In a homogeneous body, G lies on every plane or line of symmetry, because symmetric elements pair off and their moments cancel. Choose axes along the symmetry, and coordinates that make the boundaries simple, and most of the integration disappears.

The picture. Meriam's Fig. 5/5.

A right circular cone is symmetric about its axis: G is on the axis, and one coordinate remains. A half cone has one plane of symmetry: two coordinates remain. A half ring has two planes of symmetry, and G lies on their intersection, the line AB: one coordinate remains.

Check with numbers. The L-plate of Nugget 02 is symmetric about the line y = x, so \bar x = \bar y, and only one of them was calculated: 0.1357\,\mathrm m. The three particles of Nugget 04 have no symmetry, and all three coordinates had to be computed.

The idea. Meriam (5/2): the calculation is simplified by an intelligent choice of reference axes, placed so that the equations of the boundaries are as simple as possible; polar coordinates suit circular boundaries. Whenever a homogeneous body has a line or plane of symmetry, a coordinate axis or plane should coincide with it. The centre of mass always lies on it, since the moments of symmetrically placed elements cancel: the body can be regarded as made of pairs of such elements. When symmetry exists, it is the easiest way to locate G.

The equation (geometric relationship: paired elements on either side of a symmetry plane x = 0):

\int x\,dm = \sum_{\text{pairs}} (x - x)\,dm = 0 \;\Rightarrow\; \bar x = 0

Worked number. A homogeneous cylinder 240\,\mathrm{mm} long and 120\,\mathrm{mm} in radius has three planes of symmetry: G is at its geometric centre, 120\,\mathrm{mm} from each end, with no integral at all. A quarter of it, cut lengthwise (Meriam's Problem 5/4), keeps the plane of symmetry across the middle of its length and the plane that bisects its quarter: the coordinate along the axis is still 120\,\mathrm{mm} from each end, and the other two need one integral between them, since G lies on the bisecting plane (S18 gives 4r/3\pi = 4(120)/3\pi = 50.9\,\mathrm{mm} from each flat face).

Real example (one you can see). A wheel is balanced by adding small weights until its mass centre lies on the axle, restoring the symmetry the tyre's irregularities broke; an unbalanced wheel shakes the steering at speed.

Plausible wrong answer. "A half ring's G is somewhere on the ring." It lies on the line AB, in the empty space between the ring and the cut.

Check. Where is the centre of mass of a homogeneous hemisphere, before any integral? (On its axis of symmetry; one coordinate remains.)

Narration spine. A cone spins about its axis and the axis lights; G slides along it. A half cone shows its cut face as a glowing plane, G free in that plane. A half ring shows two glowing planes; they intersect in a line AB, and G settles on it in the empty middle.

Bridge. Gravity is now one arrow through G, wherever G is.

Synthesis

The car's centre of gravity is 1.134\,\mathrm m behind its front axle, not halfway, and 0.55\,\mathrm m above the ground. Both came from scale readings and moments: the weight's moment about an axle must equal the sum of the moments of every gram's weight, and the scales measure that moment. That one statement, the moment of the sum equals the sum of the moments, is the centre of gravity.

Every distributed force in this part is handled the same way. Multiply an intensity by its element, integrate for the resultant, then integrate moments to place it. For gravity the intensity is \rho g; in a uniform field g drops out and the point depends only on mass, the centre of mass. Marghitu reaches the same point from the other side, as the centroid of points weighted by mass, about which the first moments vanish. A distributed force is replaced by one resultant through the point where its moment equals the sum of the moments of its parts; for weight in a uniform field, that point is the centre of mass.

Central idea: the resultant of distributed gravity acts through G, located by the principle of moments.
Key relationship: \bar{\mathbf r} = \int\mathbf r\,dm/m (Eq. 5/2), with dm = \rho\,dV; for particles \mathbf r_G = \sum m_i\mathbf r_i/\sum m_i.
Most important assumption: a uniform, parallel gravity field, without which the centre of gravity is not unique (error \approx H^2/6R).
Open question: for a homogeneous body, \rho cancels and only shape is left. Where is the centroid of a semicircle, a cone or a curved wire? (S18.)

Computed views

The same calculations, written twice so you can work in whichever language you prefer. The MATLAB follows Marghitu's programs; the Python is a translation, and both print the same numbers.

Python (numpy, sympy). S17_centre_of_mass.py takes the mass-weighted mean of Marghitu's three particles from two origins. It integrates the rod and the tapered triangle symbolically, so the answers come out as 5L/9 and 3h/8. It also locates the puzzle's car in plan and in height.

Part 1: particles (Marghitu 3.1)
  r_C = [0.3 1.  0.5] m
  from O' = (1, 1, 1): [-0.7  0.  -0.5] m
Part 2: varying density, by integration
  rod rho0 (1 + x/L): x_bar = 5*L/9; L = 2, rho0 = 3: m = 9 kg, x_bar = 1.111 m
  P 5/32 tapered triangle: y_bar = 3*h/8
  quarter cylinder, r = 120 mm: 4r / (3 pi) = 50.9 mm
Part 3: the car on wheel scales, wheelbase 2.70 m
  G is 1.134 m behind the front axle: 986.6 = 986.6 kg m
  rear reads 605.5 kg at 10 deg: h = 0.250 m above the axle line
Part 4: weights and loads
  steel rho g = 77.0 kN/m^3; line 1.2 kg/m: 11.77 N/m, 3.53 kN over 300 m
  tyre patch at 220 kPa: 0.01672 m^2
  50 m crane: centre of gravity and centre of mass 0.07 mm apart

MATLAB (Octave-compatible). S17_centre_of_mass.m does the same with matrix products and the numerical integral. Marghitu uses the symbolic int, which needs the Symbolic Math Toolbox.

r_C = [0.3 1.0 0.5] m, from (1, 1, 1): [-0.7 0.0 -0.5] m
rod: m = 9.0 kg, x_bar = 1.111 m
P 5/32: y_bar = 0.3750 h (3/8 = 0.375)
car: G 1.134 m behind the front axle, h = 0.250 m

Traps

Treating a point force as exact. It is a model, valid for external effects when the contact region is small. (From Nugget 01.)
Mixing up the intensities. A line load is N/m, a pressure is Pa (N/m²), a body force is N/m³; integrate each over length, area or volume. (From Nugget 01.)
Assuming G is inside the body. The L-plate's G is in its empty corner. (From Nugget 02.)
Placing G halfway between supports. It is where the support loads' moments balance. (From Nugget 03.)
Forgetting that tilting reveals height. A level body's scale readings give only the position in plan. (From Nugget 03.)
Defining a centroid of strengths that sum to zero. It does not exist. (From Nugget 04.)
Dropping a varying density from the numerator. \rho stays inside both integrals of Eq. 5/3. (From Nugget 05.)
Calling centre of gravity and centre of mass identical by definition. They coincide only in a uniform, parallel field. (From Nugget 06.)
Integrating what symmetry gives free. Put an axis on every symmetry line first. (From Nugget 07.)

Self-check

A 12\,\mathrm t lorry stands on six tyres at 600\,\mathrm{kPa}. Estimate each tyre's contact area, and say whether its contact force may be treated as concentrated for the lorry's axle loads.
Find the specific weight of aluminium (\rho = 2700\,\mathrm{kg/m^3}), and convert a tyre pressure of 30\,\mathrm{lb/in^2} to kPa.
Particles of 4\,\mathrm{kg} at (1, 0, 0), 2\,\mathrm{kg} at (0, 3, 0) and 6\,\mathrm{kg} at (0, 0, 2)\,\mathrm m. Find their centre of mass.
Recompute question 3 from the origin O' = (1, 1, 1)\,\mathrm m and show it is the same point.
Strengths +2 at A and -2 at B. Where is their centroid? What is their total first moment?
A 1600\,\mathrm{kg} car's front scales read 900\,\mathrm{kg} and its rear scales 700\,\mathrm{kg}; the wheelbase is 2.60\,\mathrm m. Locate G in plan.
The car of question 6 has its rear axle raised until the axle line slopes at 12^\circ; the rear scales then read 663.4\,\mathrm{kg}. How high is G above the axle line?
A rod of length L has density \rho = \rho_0(1 + 2x/L). Find \bar x, and its value for L = 1.2\,\mathrm m.
Meriam's Problem 5/32: a triangular plate (base b, height h) whose thickness grows linearly from t_0 at the base to 2t_0 at the apex. Find \bar y above the base.
A plate hung from A(0, 0.5)\,\mathrm m settles with its plumb line through (0.3, 0)\,\mathrm m; hung from B(0.6, 0.5)\,\mathrm m, through (0.1, 0)\,\mathrm m. Locate G.
How far below its centre of mass is the centre of gravity of a uniform 300\,\mathrm m mast?
How many coordinates of G must be calculated for a homogeneous cone, a half cone and a half ring, and why?

Fully worked solutions

1. Each tyre carries 12\,000(9.81)/6 = 19\,620\,\mathrm N; area 19\,620/(600\times10^{3}) = 0.0327\,\mathrm{m^2}, about 181\,\mathrm{mm} square. A lorry's axles are metres apart, so for the axle loads, yes; for the tyre's own stresses, no.

2. \rho g = 2700(9.81) = 26\,487\,\mathrm{N/m^3} = 26.5\,\mathrm{kN/m^3}. 1\,\mathrm{lb/in^2} = 4.448\,\mathrm N/(0.0254\,\mathrm m)^2 = 6894.76\,\mathrm{Pa}, so 30\,\mathrm{lb/in^2} = 206.8\,\mathrm{kPa}.

3. M = 12\,\mathrm{kg}. \bar x = 4(1)/12 = 0.333\,\mathrm m, \bar y = 2(3)/12 = 0.5\,\mathrm m, \bar z = 6(2)/12 = 1.0\,\mathrm m: G = (0.333, 0.5, 1.0)\,\mathrm m.

4. New positions (0, -1, -1), (-1, 2, -1), (-1, -1, 1)\,\mathrm m. \mathbf r'_G = [4(0, -1, -1) + 2(-1, 2, -1) + 6(-1, -1, 1)]/12 = (-8, -6, 0)/12 = (-0.667, -0.5, 0)\,\mathrm m. Adding (1, 1, 1): (0.333, 0.5, 1.0)\,\mathrm m, the same point.

5. \sum s_i = 0, so there is no centroid. The total first moment 2\mathbf r_A - 2\mathbf r_B = 2(\mathbf r_A - \mathbf r_B) is the same about every origin.

6. Moments about the front axle: 1600\bar x = 700(2.60), \bar x = 1.1375\,\mathrm m behind the front axle. Check about G: 900(1.1375) = 1023.8 and 700(1.4625) = 1023.8\,\mathrm{kg\,m}.

7. h = [\bar x - L R_r/W]/\tan\theta = [1.1375 - 2.60(663.4/1600)]/\tan12^\circ = (1.1375 - 1.0780)/0.2126 = 0.280\,\mathrm m.

8. \bar x = \int_0^L x(1 + 2x/L)\,dx \big/ \int_0^L (1 + 2x/L)\,dx = (L^2/2 + 2L^2/3)/(2L) = 7L/12. For L = 1.2\,\mathrm m: 0.700\,\mathrm m.

9. At height y: width b(1 - y/h), thickness t_0(1 + y/h), so with uniform material dm = \rho bt_0(1 - y^2/h^2)\,dy. \bar y = \int_0^h y(1 - y^2/h^2)\,dy \big/ \int_0^h (1 - y^2/h^2)\,dy = (h^2/4)/(2h/3) = 3h/8.

10. Line 1 through (0, 0.5) and (0.3, 0): y = 0.5 - \frac53x. Line 2 through (0.6, 0.5) and (0.1, 0): y = x - 0.1. Equal: 0.5 - \frac53x = x - 0.1, \frac83x = 0.6, x = 0.225\,\mathrm m, y = 0.125\,\mathrm m. G = (0.225, 0.125)\,\mathrm m.

11. H^2/(6R) = 300^2/(6 \times 6.371\times10^{6}) = 2.35\times10^{-3}\,\mathrm m = 2.35\,\mathrm{mm}.

12. Cone: one, along the axis (every plane through the axis is a plane of symmetry). Half cone: two, in its single plane of symmetry. Half ring: one, along the line AB where its two planes of symmetry meet.