Study Companion: S19 — Composite Bodies and the Theorems of Pappus

S18 integrated to find the centroid of one shape. Most real parts are several shapes joined together: a plate with a hole, a bracket welded to a shaft, a hammer's steel head on a wooden handle. Once the centroid of each piece is known (from S18, or from a table), no further integration is needed: the pieces combine by the same principle of moments, each acting as one particle at its own centre. This lesson does that, extends it to shapes no formula fits, and then turns the centroid round to use it: the area and volume swept out by a revolving shape are its size times the distance its centroid travels, the theorems of Pappus.

Sources: Meriam, Kraige and Bolton, Engineering Mechanics: Statics, articles 5/4 ("Composite Bodies and Figures; Approximations") and 5/5 ("Theorems of Pappus"), with Sample Problems 5/6–5/10. Marghitu, Dupac and Madsen, Statics with MATLAB, section 3.6 ("Theorems of Guldinus-Pappus"), the decomposition method of section 3.3 and Examples 3.9–3.11. Harper, Solving Statics Problems in MATLAB, "Composite Bodies" (Sample Problem 5/8). Meriam's Sample Problem 5/6 is in inches, converted here: 1 in = 25.4\,\mathrm{mm}.

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 claw hammer: a 330\,\mathrm{mm} wooden handle of mass 0.15\,\mathrm{kg}, and a steel head of 0.45\,\mathrm{kg} whose centre is 320\,\mathrm{mm} from the end of the grip.

Lay the hammer across a finger. Where does it balance? The plausible guess is "near the middle of the handle, since the handle is most of the hammer's length". The answer is in Nugget 03.

Nugget 01: Parts With Known Centres Combine Like Particles

Claim. If a body divides into parts whose mass centres are known, each part can be treated as a particle of its own mass at its own centre. The mass centre of the whole is then the particle formula over the parts: \bar X = \sum m\bar x/\sum m. For uniform bodies the masses become lengths, areas or volumes.

The picture. Marghitu's Example 3.9: a region made of two rectangles.

Each rectangle's centroid is at its centre, C_1 and C_2. The whole region's centroid C lies on the line joining them, closer to the larger part.

Check with numbers. With a = 4\,\mathrm m, b = 2\,\mathrm m: part 1 is 4 \times 4, area 16\,\mathrm{m^2}, centre (2, 2)\,\mathrm m; part 2 is 4 \times 2, area 8\,\mathrm{m^2}, centre (6, 1)\,\mathrm m.

Marghitu integrates each part (A = 3ab, M_y = \frac52a^2b, M_x = \frac52ab^2) and gets C = (5a/6, 5b/6), the same point. C divides C_1C_2 in the ratio 8 : 16: one third of the way from C_1.

The idea. Meriam (5/4): when a body divides conveniently into parts whose mass centres are easily found, apply the principle of moments to each part as a finite element of the whole. For three parts, (m_1 + m_2 + m_3)\bar X = m_1\bar x_1 + m_2\bar x_2 + m_3\bar x_3, and in general

with L, A or V in place of m for composite lines, areas and volumes. Marghitu (3.3) states the same decomposition method in three steps: divide the body into simple shapes (particles, curves, surfaces, solids), locate each part's centre, and combine with x_C = \sum x_{Ci}\Delta_i/\sum\Delta_i, where \Delta_i is the part's length, area or volume (Eqs. 3.7–3.8). The centroids of common shapes come from tables (Meriam's Tables D/3–D/4, Marghitu's Fig. 3.6), which are S18's integrals done once.

The equation (mathematical identity: S17's particle formula applied to parts, valid because each part's moment equals its mass times its centre's coordinate):

Worked number. A shelf bracket made of two steel bars welded in an L: a 0.40\,\mathrm m bar along x (1.2\,\mathrm{kg}, centre (0.20, 0)) and a 0.25\,\mathrm m bar along y (0.75\,\mathrm{kg}, centre (0, 0.125)). \bar X = 1.2(0.20)/1.95 = 0.123\,\mathrm m, \bar Y = 0.75(0.125)/1.95 = 0.048\,\mathrm m.

Real example. The loading sheet of an aircraft or a ship is Eq. 5/7 as a table: each item's mass and its centre, summed into the craft's mass centre.

Plausible wrong answer. "Average the parts' centres." That gives (4, 1.5)\,\mathrm m for Example 3.9; the larger part must weigh more: (3.33, 1.67)\,\mathrm m.

Check. Two equal parts: where is the composite centre? (Midway between their centres.)

Narration spine. The two-rectangle region; each rectangle shrinks to a dot at its centre, sized by its area. A dashed line joins the dots and a balance point slides along it to one third of the way from the big dot.

Bridge. Real parts have holes. A hole is a part that is missing.

Nugget 02: A Hole Is a Part With Negative Size

Claim. A hole or cavity is entered as a part with negative mass, area or volume, at the hole's own centroid. The solid outline is added whole; the hole is subtracted. No integral is needed for either.

The picture. Meriam's Sample Problem 5/6, in millimetres, and its four parts.

Part 1 is a 304.8 \times 254\,\mathrm{mm} rectangle, part 2 a right triangle 152.4\,\mathrm{mm} wide added on its right, part 3 a semicircular notch of radius 76.2\,\mathrm{mm} cut from the base and part 4 a 50.8 \times 101.6\,\mathrm{mm} rectangular hole. Parts 3 and 4 are negative.

Check with numbers. Origin at the bottom left corner; centroids from Table D/3 (triangle at a third of its width from its vertical side, semicircle at 4r/3\pi):

Meriam's answers in inches, 7.50 and 5.08, are the same point.

The idea. Meriam (5/4): if a hole or cavity is taken as one of the parts of a composite body or figure, its mass (or area, or volume) is a negative quantity. Marghitu (3.3) says the same: holes are pieces of negative size. The reason is the principle of moments itself: the moment of the solid outline counts the material that would fill the hole, and subtracting the hole's moment removes it. A tabulation keeps signs and units straight.

The equation (mathematical identity: moments of the outline minus the hole):

Worked number. S17's L-plate as a square minus a square: \bar x = [0.16(0.2) - 0.09(0.25)]/(0.16 - 0.09) = 0.1357\,\mathrm m.

Real example (one you can see). Lightening holes in an aircraft rib or a racing car's chassis plate remove mass where it is not needed; a designer computes the shifted centre exactly this way, one negative row per hole.

Plausible wrong answer. "Subtract the hole's area but keep its moment out of the sum." The moment must be subtracted too, or \bar X is computed for an outline with the wrong area. Dropping parts 3 and 4's moments here gives \bar X = 18.68\times10^{6}/82\,490 = 226\,\mathrm{mm}, outside the true answer by 36\,\mathrm{mm}.

Check. A hole is drilled exactly at a plate's centroid. Does the centroid move? (No: the hole's moment about the centroid is zero.)

Narration spine. The plate fills in; a rectangle and a triangle slide in with plus signs, a semicircle and a small rectangle lift out with minus signs. A table writes itself row by row, the two totals divide, and C drops onto the plate.

Bridge. All the parts so far were one material. When they are not, each part must carry its own mass.

Nugget 03: Different Materials: Weight Each Part by Its Own Mass

Claim. When the parts are of different materials, Eq. 5/7 is used with masses, each part's mass computed from its own density, or mass per unit area for sheet. The mass centre then differs from the centroid of the shape.

The picture. Meriam's Sample Problem 5/8: a bracket of sheet metal on a thicker base, carrying a steel shaft.

The face is sheet metal of 25\,\mathrm{kg/m^2}, the base sheet of 40\,\mathrm{kg/m^2}, the shaft solid steel of 7830\,\mathrm{kg/m^3}. By symmetry \bar X = 0.

Check with numbers. Origin at the shaft's end on the face, y along the shaft, z up.

The semicircle's \bar z = 4r/3\pi = 4(50)/3\pi = 21.2\,\mathrm{mm} (SP 5/3); the triangle's centroid is a third of its 75\,\mathrm{mm} altitude above its base, \bar z = -(150 - 25 - 25) = -100\,\mathrm{mm} (SP 5/2).

(Carried without rounding the masses, \sum m = 2.643\,\mathrm{kg} and \bar Y = 53.2\,\mathrm{mm}; the book's rounded table gives 53.3.) The shaft is 56\,\% of the mass and drags G out along it.

The idea. Meriam (5/4) uses Eq. 5/7 with masses: each part's mass is easily calculated from its own material, the triangular cut-out is a negative mass, and the coordinates of the parts' centres come from earlier results or by inspection. With several materials, only masses (not areas or volumes) may be summed: a square metre of base plate weighs 1.6 times a square metre of face.

The equation (mathematical identity: Eq. 5/7 with each part's mass from its own material):

Worked number. The puzzle's hammer, x from the grip end: handle 0.15\,\mathrm{kg} at 0.165\,\mathrm m, head 0.45\,\mathrm{kg} at 0.320\,\mathrm m:

39\,\mathrm{mm} from the head's centre, not near the middle of the handle at 0.165\,\mathrm m: the head is three times the handle's mass.

Real example (one you can see). Balance a hammer on a finger: the balance point is a few centimetres from the head. That is why a hammer feels head-heavy and swings with authority, and why a carpenter's grip is at the far end.

Plausible wrong answer. "Use the areas of the bracket's parts, as in Nugget 02." Areas give the shape's centroid, which ignores that the shaft is solid steel and the base is heavier sheet.

Check. If the shaft were aluminium (2700\,\mathrm{kg/m^3}), would G move toward the face or away? (Toward the face: the shaft would weigh only 0.51\,\mathrm{kg}.)

Narration spine. The bracket explodes into its five parts, each labelled with its material and a mass bar: the shaft's bar towers over the rest. They reassemble and G slides along the shaft to 53\,\mathrm{mm}. Then the hammer: its head's mass bar is three times the handle's, and a finger slides under it to 39\,\mathrm{mm} from the head.

Bridge. Harper keeps the shaft's length as a variable, to see how the mass centre follows it.

Nugget 04: Let One Part Vary and Watch the Mass Centre Move

Claim. With one dimension left as a parameter, the composite formula becomes a function. Harper writes the bracket's mass centre as functions of the shaft length L and plots its path. The path also shows something the table cannot: the mass centre's \bar Y is least when it sits at the end of the shaft.

The picture. Harper's two plots: \bar Y and \bar Z against L, and \bar Z against \bar Y.

As the shaft grows from nothing to 200\,\mathrm{mm}, G rises toward the shaft's axis and moves out along it. On the right, the same results are drawn as the path of G in the bracket's plane of symmetry.

Check with numbers. Only part 5 changes: m_5 = 7830\pi(0.02)^2L/1000 = 0.00984L\,\mathrm{kg} (L in mm), at \bar y_5 = L/2, \bar z_5 = 0. The other four parts total 1.166\,\mathrm{kg}, \sum m\bar y = 30, \sum m\bar z = -120.73:

L = 0: (25.7, -103.5)\,\mathrm{mm}. L = 150\,\mathrm{mm}: (53.3, -45.7)\,\mathrm{mm}, Meriam's answer. L = 200\,\mathrm{mm}: (72.4, -38.5)\,\mathrm{mm}.

The idea. Harper (5.1) keeps Meriam's table and changes only the last row. For part (c), \bar Z against \bar Y, it notes that eliminating L is messy (two roots), and uses a parametric plot instead: both coordinates are computed from the common parameter L and plotted against each other. The first plot shows clearly how each coordinate depends on L; the second shows where G actually is in space but not the value of L at each point, which is why the ends are labelled.

An observation of ours, not Harper's. \bar Y dips before it rises: at first the short shaft adds mass close to the face, below the current \bar Y, and pulls G back. Setting d\bar Y/dL = 0 gives 0.00492L^2 + 1.166L - 30 = 0, L = 23.4\,\mathrm{mm}, and there \bar Y = 23.4\,\mathrm{mm} = L: the minimum is where G lies exactly at the end of the shaft. Adding a slice of shaft beyond G pushes G out; adding it short of G pulls G in.

The equation (mathematical identity: the composite formula with one part's mass and centre as functions of a parameter):

Worked number. L = 100\,\mathrm{mm}: m_5 = 0.984\,\mathrm{kg}, \sum m = 2.150\,\mathrm{kg}, \bar Y = (30 + 49.2)/2.150 = 36.84\,\mathrm{mm}, \bar Z = -120.73/2.150 = -56.15\,\mathrm{mm}.

Real example (one you can see). Filling a soft-drink can: the empty can's mass centre is at mid-height; as liquid is poured in, the centre first drops, then rises again, and it is lowest exactly when it reaches the liquid's surface: the same arithmetic as the shaft, with liquid added below or above G.

Plausible wrong answer. "A longer shaft always moves G outward." Not while the shaft ends inside \bar Y: below L = 23.4\,\mathrm{mm} the added steel pulls G back.

Check. For L = 50\,\mathrm{mm}, is \bar Y more or less than L? (Less: \bar Y = 42.3/1.658 = 25.5\,\mathrm{mm}, so the shaft's end is beyond G and lengthening it pushes G out.)

Narration spine. The shaft grows out of the face like a telescope. G first creeps backward, touches the shaft's end at 23.4\,\mathrm{mm}, then travels out with it, tracing the curve of the right-hand plot while the two left-hand curves draw themselves.

Bridge. Every part so far had a formula. When the outline is irregular, strips and a table do the work.

Nugget 05: When No Formula Fits the Shape, Sum Strips in a Table

Claim. An irregular area is divided into strips of measured average height; each strip is a part with an easy centroid, and Eq. 5/7 sums them. An irregular volume is reduced to an area: plot its cross-sectional area against x, and the centroid of the area under that curve is the volume's \bar x.

The picture. Meriam's Sample Problem 5/7: a body 1\,\mathrm m long whose cross-section varies along it.

Each strip of the plot, width \Delta x and height A, has area A\Delta x, which is the volume of the corresponding slice of the body. The area under the curve is the body's volume, and its centroid's x is the body's \bar X.

Check with numbers. Five sections of 0.2\,\mathrm m:

The idea. Meriam (5/4), An approximation method: when the boundary cannot be described by simple shapes or by a formula, divide the area into strips of width \Delta x and variable height h; each strip's area h\Delta x times its centroid's coordinates gives its moments, and a systematic table gives A, \sum A\bar x_c and \sum A\bar y_c. Narrower strips are more accurate; estimate each strip's average height; strips of equal width are convenient but not required, and any element shape that approximates the area well will do. Irregular volumes: plot the area A of the cross-sections normal to x against x (Fig. 5/15). A strip under the curve, A\Delta x, is the slice's volume \Delta V, so \bar x = \sum(A\Delta x)\bar x_c/\sum A\Delta x = \sum V\bar x_c/\sum V. Meriam's hint to SP 5/7: the body's shape in y and z does not affect \bar X.

The equation (approximation: the integral of S18 replaced by a sum of finite strips):

Worked number. The body's volume is 4.48\,\mathrm{m^3}, read off as the area under its A(x) curve, before any centroid is found.

Real example. A ship's displacement and centre of buoyancy are found exactly this way, from the areas of its hull sections at regular stations along its length (the naval architect's "Bonjean curves" and Simpson's rule).

Plausible wrong answer. "\bar X is where the cross-section is largest." The largest section is at about 0.6\,\mathrm m; the moments put \bar X at 0.533\,\mathrm m.

Check. Would doubling every cross-section's area change \bar X? (No: every V doubles, and the ratio is unchanged.)

Narration spine. A lumpy body lies along the x-axis; slices peel off it and stand up as bars whose heights are their areas, building the plot. A balance point slides under the bars and settles at 0.533\,\mathrm m.

Bridge. The centroid can also be used the other way round, to find areas and volumes of revolution with no integral at all.

Nugget 06: A Revolved Curve Sweeps Its Length Times the Path of Its Centroid

Claim. Revolve a plane curve of length L about an axis in its plane that does not cross it: the surface it sweeps has area 2\pi\bar yL, where \bar y is the distance from the axis to the curve's centroid. For a partial turn through \theta radians, \theta\bar yL.

The picture. Meriam's Fig. 5/16 (Marghitu's Fig. 3.4).

Each element dL, at distance y from the axis, sweeps a thin ring of circumference 2\pi y and slant width dL.

Check with numbers. A cylinder's side: a straight line of length h at distance r, parallel to the axis, has \bar y = r: A = 2\pi rh. A sphere: a semicircular arc of radius r, length \pi r, centroid 2r/\pi from its diameter (S18): A = 2\pi(2r/\pi)(\pi r) = 4\pi r^2. A frustum's slant side from radius 1\,\mathrm m to 2\,\mathrm m over a height of 2\,\mathrm m: length \sqrt{1^2 + 2^2} = \sqrt5\,\mathrm m, centroid at radius 1.5\,\mathrm m: A = 2\pi(1.5)\sqrt5 = 21.07\,\mathrm{m^2}.

The idea. Meriam (5/5): the ring swept by dL has area dA = 2\pi y\,dL; the total is A = 2\pi\int y\,dL, and since \bar yL = \int y\,dL,

The generated area equals the lateral area of a cylinder of length L and radius \bar y. Marghitu (3.6, Theorem 1) proves it the same way and adds two points: the axis must not intersect the generating curve, though it may touch it; and for a composite generating curve, A = 2\pi\sum L_i\bar y_i (Eq. 3.15), the first moment of the composite line about the axis. A closed can of radius 50\,\mathrm{mm} and height 120\,\mathrm{mm} is generated by three segments (two radii and the side): A = 2\pi[2(0.05)(0.025) + 0.12(0.05)] = 0.0534\,\mathrm{m^2}. Meriam: for a partial revolution through \theta (in radians), A = \theta\bar yL (5/8a). The theorems are attributed to Pappus of Alexandria (third century); Paul Guldin (1577–1643) claimed them, which is why Marghitu names both.

The equation (mathematical identity: Pappus's first theorem):

Worked number (Meriam's Sample Problem 5/9, area). A torus is swept by a circle of radius a (length 2\pi a, centroid at its centre) revolved at distance R: A = 2\pi R(2\pi a) = 4\pi^2Ra. Marghitu's Example 3.11, a = 1\,\mathrm m, R = 2\,\mathrm m: A = 78.957\,\mathrm{m^2}.

Real example (one you can see). The paint for a water tower, a cooling tower or a lampshade: their surfaces are revolved profiles, and the area is the profile's length times the path of its centroid.

Plausible wrong answer. "Use the path of the curve's farthest point." That over-counts; it is the centroid's path, 2\pi\bar y, that multiplies L.

Check. A semicircular arc revolved about the tangent at one end crosses nothing. Can Pappus be used? (Yes, it only touches the axis: Marghitu allows a curve that touches.)

Narration spine. A wavy curve above an axis. One element glows and spins about the axis, painting a thin ring; then the whole curve spins, painting a surface. The centroid's own circle draws in dashes, and a cylinder of that radius and length L appears beside it with the same area.

Bridge. The same argument with an area instead of a curve gives volumes.

Nugget 07: A Revolved Area Sweeps Its Area Times the Path of Its Centroid

Claim. Revolve a plane area A about an axis in its plane that does not cross it: the solid it sweeps has volume 2\pi\bar yA, or \theta\bar yA for a turn of \theta radians. Read backwards, a known volume gives the centroid.

The picture. Meriam's Sample Problem 5/10 and a torus.

The 60\,\mathrm{mm} right triangle's centroid is 50\,\mathrm{mm} from the z-axis; revolved through 180^\circ it travels half a circle. The torus's circle has its centroid at its centre, R from the axis.

Check with numbers. SP 5/10: \bar r = 30 + \frac13(60) = 50\,\mathrm{mm}, A = \frac12(60)(60) = 1800\,\mathrm{mm^2}, \theta = \pi:

in steel. The torus (SP 5/9): V = 2\pi R(\pi a^2) = 2\pi^2Ra^2; Marghitu's Example 3.11, a = 1\,\mathrm m, R = 2\,\mathrm m: V = 39.478\,\mathrm{m^3}, which Marghitu checks by washers, V = \pi\int_{-r}^{r}(f_2^2 - f_1^2)\,dx with f_{1,2} = b \mp \sqrt{r^2 - x^2}.

The idea. Meriam (5/5): the area dA at distance y sweeps a ring of volume dV = 2\pi y\,dA, so V = 2\pi\int y\,dA = 2\pi\bar yA (5/9): the generating area times the circumference of the circle its centroid describes; V = \theta\bar yA (5/9a) for a partial turn, with \theta in radians. Marghitu (3.6, Theorem 2, Eq. 3.16) proves it the same way: the volume is 2\pi times the first moment of the generating area about the axis, which must not cross the area (it may be tangent at the boundary). Meriam adds the inverse use: when the area or volume of revolution is known, dividing it by 2\pi times the length or area gives the centroid's distance from the axis.

Second view, reconciled. Marghitu's Example 3.10 finds the volume of a frustum (h = R = 2\,\mathrm m, r = 1\,\mathrm m) by discs: V = \pi\int_0^hf^2\,dx = \pi h(R^2 + Rr + r^2)/3 = 14.661\,\mathrm{m^3}. Pappus gets the same from the generating trapezoid: area h(R + r)/2 = 3\,\mathrm{m^2}, centroid (R^2 + Rr + r^2)/[3(R + r)] = 7/9\,\mathrm m from the axis, V = 2\pi(7/9)(3) = 14.661\,\mathrm{m^3}.

The equation (mathematical identity: Pappus's second theorem):

Worked number (inverse use). A sphere of radius r is a semicircular area \frac12\pi r^2 revolved about its diameter. Knowing V = \frac43\pi r^3: \bar y = \frac43\pi r^3/(2\pi\cdot\frac12\pi r^2) = 4r/3\pi, S18's semicircle centroid with no integral.

Real example (one you can see). A bicycle inner tube is a torus. For a 700\,\mathrm{mm} wheel with the tube's centre at R = 0.35\,\mathrm m and a 25\,\mathrm{mm} tube (a = 12.5\,\mathrm{mm}): V = 2\pi^2(0.35)(0.0125)^2 = 1.08\times10^{-3}\,\mathrm{m^3}, about a litre of air.

Plausible wrong answer. "Use \theta = 180 for a half turn." \theta must be in radians, \pi; in degrees the volume comes out 57.3 times too large.

Check. A rectangle b wide and h tall, one side on the axis, is revolved fully. Does Pappus give the cylinder's volume? (V = 2\pi(b/2)(bh) = \pi b^2h. Yes.)

Narration spine. The triangle stands beside the z-axis; its centroid lights and a dashed half circle draws from it as the triangle swings round, leaving a solid behind. Then a circle spins about a distant axis into a torus, its centroid's circle labelled 2\pi R.

Bridge. With centroids in hand, a distributed load on a beam can be replaced by one force, through the centroid of the load's own diagram.

Synthesis

The hammer balances 281\,\mathrm{mm} from the end of its grip, 39\,\mathrm{mm} from the head's centre, not in the middle of its handle. Two numbers and one division found it, because each part, once its centre is known, acts like a single particle of its own mass. Holes are parts with negative mass; different materials just give different masses; a part whose size changes makes the mass centre a function; and an outline with no formula becomes a table of strips.

Pappus turns the same first moment into a measuring tool. The area or volume a shape sweeps when it revolves is its length or area times the distance its centroid travels, and the theorem can be read either way: to find a volume from a centroid, or a centroid from a volume. A composite body is a weighted average of its parts' centres, holes negative; and a revolved shape sweeps its size times the path of its centroid.

• Central idea: known centres combine by Eq. 5/7, without integration.
• Key relationship: \bar X = \sum m\bar x/\sum m; A = \theta\bar yL, V = \theta\bar yA.
• Most important assumption: each part's own centre is correct (from a table or S18), and for Pappus the axis does not cross the generating figure.
• Open question: a load spread along a beam has a diagram whose area is the load and whose centroid is its line of action. How do beams carry such loads? (S20.)

Computed views

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

Python (numpy, scipy). S19_composite_bodies.py writes every composite table as one weighted mean, with holes as negative rows. It runs Harper's bracket over the shaft length L, finds the lowest \bar Y with a bounded search, and applies Pappus.

Part 1: composite tables
  hammer: X = 0.281 m
  L-plate as a square minus a square: x = 0.1357 m
  welded L bracket: (0.123, 0.048) m
  SP 5/6 plate: A = 82492 mm^2, C = [190.5 129.1] mm
  body from its A(x): V = 4.48 m^3, x = 0.533 m
Part 2: Harper's bracket, shaft length L
  L =   0 mm: (Y, Z) = (25.71, -103.46) mm
  L = 100 mm: (Y, Z) = (36.82, -56.13) mm
  L = 150 mm: (Y, Z) = (53.24, -45.68) mm
  L = 200 mm: (Y, Z) = (72.35, -38.51) mm
  lowest Y = 23.4 mm at L = 23.4 mm
Part 3: Pappus
  torus a = 1, R = 2: A = 2 pi R (2 pi a) = 78.957 m^2, V = 2 pi R (pi a^2) = 39.478 m^3
  sphere: y_bar = V / (2 pi A) = 0.4244 r (4 / 3 pi = 0.4244)
  tube R = 0.35 m, a = 12.5 mm: V = 1.08 litres

The bracket rows use the masses unrounded. That is why L = 100\,\mathrm{mm} gives (36.82, -56.13) against Nugget 04's (36.84, -56.15) from the table's three-figure masses. The plate's area also sums to 82\,492\,\mathrm{mm^2}, against the table's rounded 82\,490. The bounded search confirms the observation in Nugget 04: the lowest \bar Y is where \bar Y = L.

MATLAB (Octave-compatible). S19_composite_bodies.m does the same with matrix products and fminbnd.

SP 5/6 plate: A = 82492 mm^2, C = (190.5, 129.1) mm
hammer: X = 0.281 m
L =   0 mm: (Y, Z) = (25.71, -103.46) mm
L = 100 mm: (Y, Z) = (36.82, -56.13) mm
L = 150 mm: (Y, Z) = (53.24, -45.68) mm
L = 200 mm: (Y, Z) = (72.35, -38.51) mm
lowest Y = 23.4 mm at L = 23.4 mm
torus: A = 78.957 m^2, V = 39.478 m^3

Traps

• Averaging the parts' centres without weights. Each centre is weighted by its part's mass, length, area or volume. (From Nugget 01.)
• Adding a hole's area or forgetting its moment. A hole's area and moment are both negative. (From Nugget 02.)
• Mixing areas and masses. With more than one material, only masses may be summed. (From Nugget 03.)
• Assuming a larger part always pushes the centre outward. Material added short of G pulls it back. (From Nugget 04.)
• Using the largest cross-section as the centroid. Moments, not peaks, locate it. (From Nugget 05.)
• Using the outer radius in Pappus. The centroid's path multiplies the size. (From Nugget 06.)
• Revolving about an axis that crosses the figure. The two sides cancel and the theorem fails; split the figure at the axis. (From Nugget 06.)
• Putting \theta in degrees. Eqs. 5/8a and 5/9a need radians. (From Nugget 07.)

Self-check

1. Meriam's Sample Problem 5/6 in SI without the rectangular hole (part 4). Find the new centroid.
2. A 300 \times 200\,\mathrm{mm} plate has a 60\,\mathrm{mm} diameter hole whose centre is 220\,\mathrm{mm} from the left edge and 100\,\mathrm{mm} from the bottom. Find \bar X.
3. A uniform wire is a straight 2\,\mathrm m piece along the x-axis from (0, 0) to (2, 0)\,\mathrm m, closed by a semicircular arc of radius 1\,\mathrm m above it. Find its centroid.
4. Meriam's bracket (SP 5/8) with an aluminium shaft (2700\,\mathrm{kg/m^3}). Find \bar Y and \bar Z.
5. Harper's bracket with L = 100\,\mathrm{mm}: find \bar Y and \bar Z.
6. For Harper's bracket, find the shaft length that makes \bar Y least, and show that \bar Y then equals L.
7. A body 1\,\mathrm m long has cross-sectional areas 1.0, 2.0, 2.6, 3.0, 3.2 and 3.3\,\mathrm{m^2} at x = 0, 0.2, \ldots, 1.0\,\mathrm m. Using the average of each interval's end values, estimate its volume and \bar X.
8. Use Pappus to find the area of a sphere, and then, from the sphere's volume, the centroid of a semicircular area.
9. SP 5/10's triangle revolved through a full turn: find the volume and the mass in steel.
10. A bicycle inner tube: R = 0.35\,\mathrm m to the tube's centre, tube radius a = 12.5\,\mathrm{mm}. Find the volume of air and the area of the tube's surface.
11. Marghitu's Example 3.10: find the frustum's volume (h = R = 2\,\mathrm m, r = 1\,\mathrm m) by Pappus.
12. Find the lateral (slant) area of the same frustum by Pappus, and compare with \pi(R + r)s, s the slant length.

Fully worked solutions

1. Parts 1–3 only: A = 77\,419 + 19\,355 - 9121 = 87\,653\,\mathrm{mm^2}; \sum A\bar x = (11.80 + 6.88 - 1.39)\times10^{6} = 17.29\times10^{6}\,\mathrm{mm^3}, \sum A\bar y = (9.83 + 1.64 - 0.295)\times10^{6} = 11.18\times10^{6}\,\mathrm{mm^3}. \bar X = 197.3\,\mathrm{mm}, \bar Y = 127.5\,\mathrm{mm}: removing the hole's negative area moves C toward where the hole was.

2. Plate 60\,000\,\mathrm{mm^2} at 150\,\mathrm{mm}; hole \pi(30)^2 = 2827\,\mathrm{mm^2} at 220\,\mathrm{mm}, negative. \bar X = [60\,000(150) - 2827(220)]/(60\,000 - 2827) = 8\,378\,000/57\,173 = 146.5\,\mathrm{mm}.

3. Straight piece: L = 2\,\mathrm m, centre (1, 0). Arc: L = \pi\,\mathrm m, centre (1, 2/\pi) (S18, 2r/\pi above the diameter). \bar X = 1\,\mathrm m by symmetry; \bar Y = [2(0) + \pi(2/\pi)]/(2 + \pi) = 2/5.142 = 0.389\,\mathrm m.

4. m_5 = 2700\pi(0.02)^2(0.15) = 0.509\,\mathrm{kg}, \sum m = 1.167 + 0.509 = 1.676\,\mathrm{kg}. \bar Y = (30 + 0.509 \times 75)/1.676 = 40.7\,\mathrm{mm}; \bar Z = -120.73/1.676 = -72.0\,\mathrm{mm}. G moves toward the face and down.

5. m_5 = 0.984\,\mathrm{kg}, \sum m = 2.150\,\mathrm{kg}; \bar Y = (30 + 0.00492 \times 10\,000)/2.150 = 79.2/2.150 = 36.84\,\mathrm{mm}; \bar Z = -120.73/2.150 = -56.15\,\mathrm{mm}.

6. d\bar Y/dL = 0: 0.00984L(1.166 + 0.00984L) - 0.00984(30 + 0.00492L^2) = 0, which reduces to 0.00492L^2 + 1.166L - 30 = 0, L = [-1.166 + \sqrt{1.166^2 + 4(0.00492)(30)}]/(2 \times 0.00492) = 23.4\,\mathrm{mm}. Then \bar Y = (30 + 2.69)/(1.166 + 0.230) = 23.4\,\mathrm{mm} = L. (At the stationary point the condition 0.00984L\sum m = 0.00984\sum m\bar y is exactly L = \bar Y.)

7. Interval averages 1.5, 2.3, 2.8, 3.1, 3.25\,\mathrm{m^2}; volumes (\times0.2\,\mathrm m) 0.30, 0.46, 0.56, 0.62, 0.65\,\mathrm{m^3}, total 2.59\,\mathrm{m^3}. \sum V\bar x = 0.030 + 0.138 + 0.280 + 0.434 + 0.585 = 1.467\,\mathrm{m^4}. \bar X = 1.467/2.59 = 0.566\,\mathrm m.

8. Area: the semicircular arc, L = \pi r, \bar y = 2r/\pi: A = 2\pi(2r/\pi)(\pi r) = 4\pi r^2. Centroid: the semicircular area \frac12\pi r^2 sweeps V = \frac43\pi r^3, so \bar y = V/(2\pi A) = \frac43\pi r^3/(\pi^2r^2) = 4r/3\pi.

9. V = 2\pi(50)(1800) = 5.655\times10^{5}\,\mathrm{mm^3} = 5.655\times10^{-4}\,\mathrm{m^3}; m = 7830(5.655\times10^{-4}) = 4.43\,\mathrm{kg}, twice the half turn's 2.21\,\mathrm{kg}.

10. V = 2\pi^2Ra^2 = 2\pi^2(0.35)(0.0125)^2 = 1.079\times10^{-3}\,\mathrm{m^3} = 1.08\,\mathrm L; A = 4\pi^2Ra = 4\pi^2(0.35)(0.0125) = 0.173\,\mathrm{m^2}.

11. The generating trapezoid: parallel sides r = 1\,\mathrm m and R = 2\,\mathrm m (distances from the axis), width h = 2\,\mathrm m. Area h(R + r)/2 = 3\,\mathrm{m^2}; centroid distance from the axis (R^2 + Rr + r^2)/[3(R + r)] = 7/9\,\mathrm m. V = 2\pi(7/9)(3) = 14\pi/3 = 14.661\,\mathrm{m^3}, Marghitu's value by discs.

12. Slant side length s = \sqrt{2^2 + 1^2} = \sqrt5\,\mathrm m, its centroid at radius 1.5\,\mathrm m: A = 2\pi(1.5)\sqrt5 = 21.07\,\mathrm{m^2}. The formula \pi(R + r)s = 3\pi\sqrt5 = 21.07\,\mathrm{m^2}: the same, since 2\pi\cdot\frac{R + r}{2} = \pi(R + r).