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: