Centre of Mass: Uniform Laminae and Composite Bodies

A lamina is a thin flat plate — a two-dimensional rigid body of negligible thickness. This lesson extends the discrete centre-of-mass formula to such continuous bodies, building the standard results for the rectangle, triangle, semicircle, quarter-circle and sector, and then assembling them into composite bodies by addition and into shapes-with-holes by subtraction. We finish with the general integral method for non-standard regions and the classic "suspended lamina" equilibrium problem. This is one of the most exam-frequent topics in the Mechanics option, and a beautiful application of integration from the pure papers.

1 Where this sits in AQA 7367

This is Paper 3, the Mechanics option (7367/3M), following directly from discrete centre of mass. It draws heavily on integration (definite integrals, areas, the substitution method) from the compulsory-pure material, applied to a physical problem. The standard-results-and-composite work is mostly AO1; the subtraction (hole) problems and the suspended-lamina equilibrium are AO2/AO3, where you reason about negative areas and turning effects. Expect "find the centre of mass", "hence find the angle to the vertical", and occasional "show that" integral derivations.

2 Core theory — the integral definition and standard results

For a uniform lamina (constant surface density $\rho$ ), an element of area $\mathrm dA$ has mass $\mathrm dm = \rho\,\mathrm dA$ . The discrete sums become integrals:

\bar{x} = \frac{\int x\,\mathrm dm}{\int \mathrm dm} = \frac{\int x\,\mathrm dA}{\int \mathrm dA}, \qquad \bar{y} = \frac{\int y\,\mathrm dA}{\int \mathrm dA},

the density $\rho$ cancelling because it is constant. So for a uniform lamina the centre of mass is purely geometric — it is the centroid, the point that depends only on the shape, not the material. Two consequences are immediate and powerful:

Any axis of symmetry contains the centre of mass. A rectangle, circle or any figure with a line of symmetry has its centroid on that line; two perpendicular symmetry lines pin it exactly. Always exploit symmetry first — it can halve the work.
The result is the area-weighted average of sub-shapes (next section), which is why composite bodies reduce to a discrete-particle calculation with "areas" in place of "masses".

The standard results you must know (each derivable by the integral above, and several derived below):

Shape	Centre of mass
Rectangle $a \times b$	$(a/2,\ b/2)$ from a corner (the centre)
Triangle, vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$	$\left(\dfrac{x_1+x_2+x_3}{3},\ \dfrac{y_1+y_2+y_3}{3}\right)$ — the centroid, $\tfrac13$ up each median
Circle radius $r$	at the centre
Semicircle radius $r$	$\dfrac{4r}{3\pi}$ from the diameter
Quarter circle radius $r$	$\dfrac{4r}{3\pi}$ from each straight edge
Sector radius $r$ , half-angle $\alpha$	$\dfrac{2r\sin\alpha}{3\alpha}$ from the centre along the axis of symmetry

(Sanity check the sector result: as $\alpha\to 0$ it tends to $\tfrac23 r$ , the centroid of a thin triangle; at $\alpha = \tfrac{\pi}{2}$ it gives $\dfrac{2r\cdot1}{3\cdot\pi/2} = \dfrac{4r}{3\pi}$ , the semicircle — both limits agree, a reassuring check.) Building the habit of checking a formula against its limiting cases is one of the most valuable transferable skills in the whole option: it catches a misremembered constant in seconds, and examiners reward the brief sanity-check sentence that demonstrates it. Note too that every distance in the table is measured from a named reference (a corner, the diameter, the straight edges, the centre) — quoting a standard result without saying from where is a frequent and avoidable loss of marks.

Derivation: the semicircle

Take a semicircle of radius $r$ , flat side (diameter) along the $x$ -axis, centred at the origin. By symmetry $\bar x = 0$ . For $\bar y$ , slice into thin horizontal strips at height $y$ : each has width $2\sqrt{r^2 - y^2}$ , thickness $\mathrm dy$ , so area $\mathrm dA = 2\sqrt{r^2-y^2}\,\mathrm dy$ . Then

\bar y = \frac{\displaystyle\int_0^r y\cdot 2\sqrt{r^2 - y^2}\,\mathrm dy}{\tfrac12\pi r^2}.

For the numerator substitute $u = r^2 - y^2$ , $\mathrm du = -2y\,\mathrm dy$ :

\int_0^r 2y\sqrt{r^2-y^2}\,\mathrm dy = \int_{r^2}^{0} -\sqrt{u}\,\mathrm du = \int_0^{r^2}\sqrt{u}\,\mathrm du = \left[\tfrac23 u^{3/2}\right]_0^{r^2} = \tfrac23 r^3.

Hence

\bar y = \frac{\tfrac23 r^3}{\tfrac12\pi r^2} = \frac{4r}{3\pi} \approx 0.424\,r.

So the centroid sits a little under halfway up the radius — reassuringly inside the shape.

Derivation: the triangle

Take a triangle with its base of length $b$ on the $x$ -axis from $(0,0)$ to $(b,0)$ and apex at height $h$ . A horizontal strip at height $y$ has length proportional to its distance from the apex: by similar triangles its length is $b\bigl(1 - \tfrac{y}{h}\bigr)$ , so $\mathrm dA = b\bigl(1-\tfrac yh\bigr)\mathrm dy$ . The total area is $\tfrac12 bh$ , and

\bar y = \frac{\int_0^h y\,b\bigl(1 - \tfrac yh\bigr)\mathrm dy}{\tfrac12 bh} = \frac{b\left[\tfrac{y^2}{2} - \tfrac{y^3}{3h}\right]_0^h}{\tfrac12 bh} = \frac{b\bigl(\tfrac{h^2}{2} - \tfrac{h^2}{3}\bigr)}{\tfrac12 bh} = \frac{\tfrac{1}{6}bh^2}{\tfrac12 bh} = \frac{h}{3}.

So the centroid is at height $h/3$ above the base — i.e. $\tfrac13$ of the way up each median. Generalising to arbitrary vertices, the centroid is the simple average $\left(\tfrac{x_1+x_2+x_3}{3}, \tfrac{y_1+y_2+y_3}{3}\right)$ , exactly as in the table. This is one result you can also just write down in the exam, but seeing the integral confirms why the "averaging the vertices" shortcut is legitimate.

3 Worked examples with mark scheme

Example 1 — composite by addition (L-shaped lamina)

An L-shaped uniform lamina is a $6\,\text{cm}\times2\,\text{cm}$ rectangle (bottom, from $(0,0)$ to $(6,2)$ ) with a $2\,\text{cm}\times4\,\text{cm}$ rectangle on top of its left end (from $(0,2)$ to $(2,6)$ ). Find the centre of mass.

Split into two rectangles; for uniform laminae, mass $\propto$ area: (M1)

Bottom: $A_1 = 12\,\text{cm}^2$ , centroid $(3,1)$ .
Left: $A_2 = 8\,\text{cm}^2$ , centroid $(1,4)$ .

\bar x = \frac{A_1\bar x_1 + A_2\bar x_2}{A_1+A_2} = \frac{12(3)+8(1)}{20} = \frac{44}{20} = 2.2\,\text{cm}, \quad (\textbf{M1 A1})

\bar y = \frac{12(1)+8(4)}{20} = \frac{44}{20} = 2.2\,\text{cm}. \quad (\textbf{A1})

Centre of mass $(2.2, 2.2)$ . (M1 for splitting into known shapes; M1 for the area-weighted formula; A1 each coordinate. The symmetry of this particular L about the line $y=x$ explains why $\bar x = \bar y$ — a free check.)

Example 2 — composite by subtraction (disc with a circular hole)

A uniform disc of radius $6\,\text{cm}$ has a circular hole of radius $2\,\text{cm}$ cut out, the hole's centre $3\,\text{cm}$ from the disc's centre. Find the centre of mass of what remains.

Take the disc centre as origin, hole centred at $x = 3$ . Treat the hole as negative area: (M1)

Full disc: $A_1 = \pi(6)^2 = 36\pi$ , $\bar x_1 = 0$ .
Hole: $A_2 = \pi(2)^2 = 4\pi$ , $\bar x_2 = 3$ .

\bar x = \frac{A_1\bar x_1 - A_2\bar x_2}{A_1 - A_2} = \frac{36\pi(0) - 4\pi(3)}{36\pi - 4\pi} = \frac{-12\pi}{32\pi} = -0.375\,\text{cm}. \quad (\textbf{M1 A1})

By symmetry (the diameter through both centres is an axis of symmetry) $\bar y = 0$ . The centre of mass is $0.375\,\text{cm}$ from the disc's centre, away from the hole — sensible, since removing mass on the $+x$ side shifts the balance to $-x$ . (M1 for treating the hole as negative; M1 for the subtraction formula; A1 $-0.375$ . The $\pi$ cancels — leave it in until it does.)

Example 3 — rectangle with a rectangular hole

A $10\,\text{cm}\times8\,\text{cm}$ uniform lamina (origin at its bottom-left corner) has a $4\,\text{cm}\times3\,\text{cm}$ rectangle removed from the top-right, occupying $(6,5)$ to $(10,8)$ . Find the centre of mass.

Full rectangle: $A_1 = 80$ , centroid $(5,4)$ .
Hole: $A_2 = 12$ , centroid $(8, 6.5)$ . (M1)

\bar x = \frac{80(5) - 12(8)}{80 - 12} = \frac{400 - 96}{68} = \frac{304}{68} \approx 4.47\,\text{cm}, \quad (\textbf{M1 A1})

\bar y = \frac{80(4) - 12(6.5)}{68} = \frac{320 - 78}{68} = \frac{242}{68} \approx 3.56\,\text{cm}. \quad (\textbf{A1})

Centre of mass $\approx(4.47, 3.56)$ — shifted down-and-left from the centre $(5,4)$ , away from the removed corner. (M1 hole centroid; M1 subtraction; A1 each coordinate.)

Example 4 — rectangle plus a semicircular cap

A uniform lamina is a rectangle $8\,\text{cm}$ wide and $5\,\text{cm}$ tall (from $(0,0)$ to $(8,5)$ ) with a semicircle of radius $4\,\text{cm}$ added on top (flat side along the top edge $y=5$ , bulging upward). Find the height of the centre of mass above the base.

Combine by addition; by symmetry $\bar x = 4$ . (M1)

Rectangle: $A_1 = 40$ , centroid height $\bar y_1 = 2.5$ .
Semicircle: $A_2 = \tfrac12\pi(4)^2 = 8\pi \approx 25.13$ . Its centroid is $\dfrac{4r}{3\pi} = \dfrac{16}{3\pi}\approx1.698$ above its flat side, i.e. at height $\bar y_2 = 5 + 1.698 = 6.698$ . (M1)

\bar y = \frac{A_1\bar y_1 + A_2\bar y_2}{A_1 + A_2} = \frac{40(2.5) + 25.13(6.698)}{40 + 25.13} = \frac{100 + 168.3}{65.13} = \frac{268.3}{65.13} \approx 4.12\,\text{cm}. \quad (\textbf{M1 A1})

The centre of mass is about $4.12\,\text{cm}$ above the base — pulled up from the rectangle's $2.5\,\text{cm}$ by the added cap, but still below the rectangle's top because the rectangle carries more mass. (M1 for addition; M1 for shifting the semicircle's centroid to absolute height; M1 the formula; A1 the value. The commonest slip is forgetting the " $+5$ " that places the semicircle centroid above the join.)

4 Specimen-style exam question

(specimen-style — not from any past paper)

A uniform lamina consists of a square $OABC$ of side $4\,\text{cm}$ with $O$ at the origin, $A(4,0)$ , $B(4,4)$ , $C(0,4)$ , with a semicircular piece of radius $2\,\text{cm}$ (centre the midpoint of $OA$ ) removed from the bottom edge. (a) Find the centre of mass of the lamina. (b) The lamina is suspended freely from $C$ . Find the angle that $CO$ makes with the vertical.

Model solution. Use subtraction, taking the square minus the semicircle. The semicircle has its flat side on $OA$ (the $x$ -axis), centre $(2,0)$ , and bulges into the square (upwards), so its centroid is at $\bigl(2,\ \tfrac{4(2)}{3\pi}\bigr) = \bigl(2,\ \tfrac{8}{3\pi}\bigr)$ , with $\tfrac{8}{3\pi}\approx0.849$ .

Square: $A_1 = 16$ , centroid $(2,2)$ .
Semicircle (removed): $A_2 = \tfrac12\pi(2)^2 = 2\pi \approx 6.283$ , centroid $(2, 0.849)$ .

By symmetry $\bar x = 2$ . For $\bar y$ :

\bar y = \frac{A_1(2) - A_2(0.849)}{A_1 - A_2} = \frac{16(2) - 6.283(0.849)}{16 - 6.283} = \frac{32 - 5.334}{9.717} = \frac{26.666}{9.717} \approx 2.74\,\text{cm}.

So the centre of mass is at $(2, 2.74)$ . (a) Removing area near the bottom has pushed the centroid up from $y=2$ , as expected.

(b) Suspended from $C(0,4)$ , the line $C\bar G$ hangs vertical, where $\bar G = (2, 2.74)$ . The displacement from $C$ to $\bar G$ is $(2 - 0,\ 2.74 - 4) = (2, -1.26)$ . The edge $CO$ points from $C(0,4)$ to $O(0,0)$ , i.e. straight down, direction $(0,-1)$ . The angle $\theta$ between $CO$ and the (vertical) line $C\bar G$ satisfies

\tan\theta = \frac{\text{horizontal component of } C\bar G}{\text{vertical component}} = \frac{2}{1.26} \approx 1.587 \ \Rightarrow\ \theta \approx 57.8^\circ.

So $CO$ makes about $57.8^\circ$ with the vertical. The key steps are: find $\bar G$ by subtraction; then make $C\bar G$ vertical and measure $CO$ against it.

Example 5 — suspension of the L-shaped lamina

The L-shaped lamina of Example 1 (centre of mass $(2.2, 2.2)$ ) is freely suspended from the corner $(6,2)$ . Find the angle the long base (the edge from $(0,0)$ to $(6,0)$ ) makes with the vertical when it hangs in equilibrium.

The line from the pivot $(6,2)$ to the centre of mass $(2.2, 2.2)$ hangs vertical. Its horizontal and vertical components are

\Delta x = 6 - 2.2 = 3.8, \qquad \Delta y = 2.2 - 2 = 0.2.

Centre of Mass: Uniform Laminae and Composite Bodies

Centre of Mass: Uniform Laminae and Composite Bodies

1 Where this sits in AQA 7367

2 Core theory — the integral definition and standard results

Derivation: the semicircle

Derivation: the triangle

3 Worked examples with mark scheme

Example 1 — composite by addition (L-shaped lamina)

Example 2 — composite by subtraction (disc with a circular hole)

Example 3 — rectangle with a rectangular hole

Example 4 — rectangle plus a semicircular cap

4 Specimen-style exam question

Example 5 — suspension of the L-shaped lamina

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