Centre of Mass: Uniform Laminae and Composite Bodies

This lesson extends centre-of-mass techniques to uniform laminae (flat shapes) and composite bodies made by combining or subtracting standard shapes.

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1 Standard Results for Uniform Laminae

For a uniform lamina (constant density throughout), the centre of mass is at the geometric centre (centroid).

Shape	Centre of mass
Rectangle $a \times b$a×b	$(a/2, b/2)$(a/2,b/2) from a corner
Triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$(x1​,y1​),(x2​,y2​),(x3​,y3​)	$\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$(3x1​+x2​+x3​​,3y1​+y2​+y3​​)
Circle radius $r$r	At the centre
Semicircle radius $r$r	$\frac{4r}{3\pi}$3π4r​ from the diameter
Quarter circle radius $r$r	$\frac{4r}{3\pi}$3π4r​ from each straight edge
Circular sector angle $2\alpha$2α, radius $r$r	$\frac{2r\sin\alpha}{3\alpha}$3α2rsinα​ from the centre along the axis of symmetry

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Derivation: Semicircle

Consider a semicircle of radius $r$r with the diameter along the $x$x-axis, centred at the origin. By symmetry, $\bar{x} = 0$xˉ=0.

For $\bar{y}$yˉ​, divide the semicircle into thin horizontal strips at height $y$y, width $2\sqrt{r^2 - y^2}$2r2−y2​, thickness $\mathrm{d}y$dy:

$$\bar{y} = \frac{\int_0^r y \cdot 2\sqrt{r^2 - y^2}\,\mathrm{d}y}{\frac{1}{2}\pi r^2}$$yˉ​=21​πr2∫0r​y⋅2r2−y2​dy​

Let $u = r^2 - y^2$u=r2−y2, $\mathrm{d}u = -2y\,\mathrm{d}y$du=−2ydy:

Numerator = $\int_{r^2}^{0} (-\sqrt{u})\,\mathrm{d}u = \int_0^{r^2} \sqrt{u}\,\mathrm{d}u = \left[\frac{2}{3}u^{3/2}\right]_0^{r^2} = \frac{2r^3}{3}$∫r20​(−u​)du=∫0r2​u​du=[32​u3/2]0r2​=32r3​

$\bar{y} = \frac{2r^3/3}{\pi r^2/2} = \frac{4r}{3\pi}$yˉ​=πr2/22r3/3​=3π4r​

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2 Composite Bodies — Addition Method

A composite lamina is made by joining two or more simple shapes. Use the principle:

$$\bar{x} = \frac{m_1 \bar{x}_1 + m_2 \bar{x}_2}{m_1 + m_2}$$xˉ=m1​+m2​m1​xˉ1​+m2​xˉ2​​

For uniform laminae of the same density $\rho$ρ, mass is proportional to area, so:

$$\bar{x} = \frac{A_1 \bar{x}_1 + A_2 \bar{x}_2}{A_1 + A_2}$$xˉ=A1​+A2​A1​xˉ1​+A2​xˉ2​​

Example 1 — L-Shaped Lamina

An L-shaped uniform lamina is formed by a 6 cm $\times$× 2 cm rectangle (bottom) and a 2 cm $\times$× 4 cm rectangle (left, on top of the bottom rectangle). Find the centre of mass.

Diagram: The bottom rectangle spans $(0,0)$(0,0) to $(6,2)$(6,2). The left rectangle spans $(0,2)$(0,2) to $(2,6)$(2,6).

Solution

Bottom rectangle: area $A_1 = 12$A1​=12 cm$^2$2, centre $(3, 1)$(3,1).

Left rectangle: area $A_2 = 8$A2​=8 cm$^2$2, centre $(1, 4)$(1,4).

$\bar{x} = \frac{12(3) + 8(1)}{20} = \frac{36 + 8}{20} = \frac{44}{20} = 2.2$xˉ=2012(3)+8(1)​=2036+8​=2044​=2.2 cm.

$\bar{y} = \frac{12(1) + 8(4)}{20} = \frac{12 + 32}{20} = \frac{44}{20} = 2.2$yˉ​=2012(1)+8(4)​=2012+32​=2044​=2.2 cm.

Centre of mass: $(2.2, 2.2)$(2.2,2.2).

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3 Composite Bodies — Subtraction Method

A shape with a hole is treated as the complete shape minus the removed part:

$$\bar{x} = \frac{A_{\text{whole}} \bar{x}_{\text{whole}} - A_{\text{hole}} \bar{x}_{\text{hole}}}{A_{\text{whole}} - A_{\text{hole}}}$$xˉ=Awhole​−Ahole​Awhole​xˉwhole​−Ahole​xˉhole​​

Example 2 — Disc with a Circular Hole

A uniform circular disc of radius 6 cm has a circular hole of radius 2 cm cut from it. The centre of the hole is 3 cm from the centre of the disc. Find the centre of mass of the remaining shape.

Solution

Take the centre of the full disc as the origin, and let the hole be centred at $x = 3$x=3.

Full disc: $A_1 = \pi(6)^2 = 36\pi$A1​=π(6)2=36π, $\bar{x}_1 = 0$xˉ1​=0.

Hole: $A_2 = \pi(2)^2 = 4\pi$A2​=π(2)2=4π, $\bar{x}_2 = 3$xˉ2​=3.