Centre of mass of particles and composite bodiesAQA A-Level Further Maths: Revision notes

The centre of mass of a body is the point through which its weight can be taken to act. For a body made of particles it is the weighted average of their positions. If a body is uniform, its mass is spread evenly, so the mass of each part is proportional to its length (rod), area (lamina) or volume (solid) as appropriate. For a lamina, a flat body of negligible thickness and uniform density, mass is proportional to area. Because the weights act at the same relative points, we use masses (or areas) in the sums, never the weights themselves with $g$ unless needed.

For particles of masses $m_1,m_2,\ldots$ at $(x_1,y_1),(x_2,y_2),\ldots$: $\bar{x}=\frac{\sum m_ix_i}{\sum m_i},\qquad \bar{y}=\frac{\sum m_iy_i}{\sum m_i}.$ In vector form, $\bar{\mathbf{r}}=\frac{\sum m_i\mathbf{r}_i}{\sum m_i}$. Each $m_ix_i$ is the moment of that particle about the $y$-axis. Example: $2$ kg at $(1,3)$, $3$ kg at $(-2,1)$ and $5$ kg at $(3,-1)$. The total mass is $10$, $\sum mx=2-6+15=11$ and $\sum my=6+3-5=4$, so the centre of mass is $(1.1,\,0.4)$.

For a uniform body, the centre of mass lies on any axis of symmetry. Standard results: the midpoint of a uniform rod; the centre of a rectangle or other parallelogram (where the diagonals meet); the centre of a circular disc; and for a triangle, the point where the medians meet, which is $\frac13$ of the way up from any side (so $\frac23$ of the way along a median from a vertex). Use these to place each part of a composite body before taking moments. The centres of mass of a semicircular lamina, a sector and similar shapes are in the formula booklet.

A composite body is joined from simple parts. Split it into rectangles, triangles or discs, find each part's mass (area for a uniform lamina) and centre of mass, then take moments about two convenient axes: $\left(\sum m\right)\bar{x}=\sum m_ix_i,\qquad \left(\sum m\right)\bar{y}=\sum m_iy_i.$ Example: a T-shape is a bar $10\times2$ (area $20$, centre $(5,5)$) on top of a stem $2\times4$ (area $8$, centre $(5,2)$). By symmetry $\bar{x}=5$ and $28\bar{y}=20(5)+8(2)=116$, so $\bar{y}=4.14$.

If a piece is cut out (a hole), treat it as a part with negative mass: moments of the whole shape equal moments of the remaining body plus moments of the removed piece. For a square of side $10$ with a circular hole of radius $2$ centred at $(7,5)$: $100(5)=4\pi(7)+(100-4\pi)\bar{x},$ so $\bar{x}=4.71$, with $\bar{y}=5$ by symmetry. To add a particle or rod to a body, include it in the sums with its own mass and position; if a mass is given only for the lamina, use it as the lamina's mass in the same units as the particle.