Centre of mass of laminas and composite figuresEdexcel International A Level Further Maths: Revision notes

The centre of mass of a body is the single point through which its weight can be taken to act. A lamina is a flat body so thin that its thickness is ignored, and it is uniform if its mass is spread evenly, so the mass of any part is proportional to its area. For a uniform lamina the centre of mass is the same point as the centroid of the shape, so areas can be used in place of masses in every calculation. If a lamina has an axis of symmetry, its centre of mass lies on that axis. With two axes of symmetry (rectangle, circle), it is where they cross.

The formulae booklet gives the results below and you may quote them without proof (integration is not required). For a uniform lamina:

• Rectangle or circle (disc): at the geometric centre.
• Triangle: at the intersection of the medians, one third of the way up from each side. Its coordinates are the mean of the three vertices: $\bar{x}=\frac{x_1+x_2+x_3}{3}$, $\bar{y}=\frac{y_1+y_2+y_3}{3}$.
• Semicircle of radius $r$: on the axis of symmetry, $\frac{4r}{3\pi}$ from the diameter.
• Sector of angle $2\alpha$ (radians), radius $r$: on the axis of symmetry, $\frac{2r\sin\alpha}{3\alpha}$ from the centre. Do not mix these up with the results for a wire or arc: a semicircular arc has its centre of mass at $\frac{2r}{\pi}$ from the diameter.

To find the centre of mass of a lamina made of simple parts, treat each part as a particle at its own centre of mass with mass proportional to its area, then use $\left(\sum m_i\right)\bar{x}=\sum m_ix_i,\qquad\left(\sum m_i\right)\bar{y}=\sum m_iy_i.$ This says the moment of the whole about an axis equals the sum of the moments of the parts. Choose axes through a corner or along an axis of symmetry to keep the numbers small. Find $\bar{x}$ and $\bar{y}$ separately; if the shape is symmetric about a line you only need the coordinate perpendicular to it.

If a shape has a piece removed, treat the removed piece as a negative mass: whole $=$ remainder $+$ removed piece, so $M_{\text{whole}}\,\bar{x}_{\text{whole}}=M_{\text{rem}}\,\bar{x}_{\text{rem}}+M_{\text{hole}}\,\bar{x}_{\text{hole}}.$ Example: a disc of radius $12$ cm with a hole of radius $4$ cm centred $6$ cm from its centre $O$. Areas (in units of $\pi$) are $144$, $16$ and $128$. Moments about $O$: $0=128\bar{x}+16(6)$, so $\bar{x}=-0.75$ cm: the centre of mass is $0.75$ cm from $O$ on the side away from the hole.

L-shape: $R_1$ is $0\le x\le6$, $0\le y\le2$ and $R_2$ is $0\le x\le2$, $2\le y\le8$. Both areas are $12$ with centres $(3,1)$ and $(1,5)$. Then $24\bar{x}=12(3)+12(1)$ gives $\bar{x}=2$, and $24\bar{y}=12(1)+12(5)$ gives $\bar{y}=3$. Method: (1) split the shape; (2) write each area and centre of mass; (3) take moments about two axes; (4) answer in the units and form asked for. Use ratios of areas where possible, because common factors such as $\pi$ cancel. Quote booklet results rather than deriving them.