3.2 - Centres of mass of composite bodies | Cambridge A-Level Further Mathematics

An L-shaped plate need not balance at the centre of its enclosing rectangle. In this part of 3.2, you will replace simple pieces by particles at their centres of mass, then combine their moments. The same method handles holes, joined solids and different densities.

An equivalent particle system

Where would you support a light bar carrying a $2\,\mathrm{kg}$ mass at one end and a $3\,\mathrm{kg}$ mass at the other? Its balance point lies nearer the larger mass. An unweighted average of the positions would miss that difference.

Replace each rigidly joined piece by a particle of the same mass at its own centre of mass. Their combined centre of mass preserves the gravitational moment. For particles on a horizontal line, equating moments about the origin gives

(m_1+\cdots+m_n)g\bar x=m_1gx_1+\cdots+m_ngx_n.

The common factor $g$ cancels. Use the same mass weighting separately for each coordinate.

Mass-weighted position

For masses $m_i>0$ at positions $(x_i,y_i)$ ,

\bar x=\frac{\sum m_ix_i}{\sum m_i}, \qquad \bar y=\frac{\sum m_iy_i}{\sum m_i}.

All coordinates must refer to the same origin and axes. A further coordinate is treated in the same way for a three-dimensional body.

Not in the formula booklet: learn it.

The weights are masses, not distances. You can use weights instead if all pieces are in the same gravitational field, since the common $g$ again cancels.

Worked example 1: unequal masses on a line

Question. Particles of masses $2\,\mathrm{kg}$ and $3\,\mathrm{kg}$ are fixed at $(0,0)$ and $(5,0)$ , in metres, on a light rigid frame. Find their centre of mass.

Plan. Divide the total first moment by the total mass.

\begin{aligned} \bar x&=\frac{2(0)+3(5)}{2+3}=3 \\ \bar y&=0. \end{aligned}

Answer. $G=(3,0)\,\mathrm m$ .

Check. About $x=3$ , the two weight moments have equal magnitudes: $2g(3)=3g(2)$ . The balance point is nearer the $3\,\mathrm{kg}$ mass.

Move only the second particle upwards. Its horizontal contribution stays the same, but you now need a second coordinate calculation.

Worked example 2: a second coordinate

Question. The $3\,\mathrm{kg}$ particle is moved to $(5,4)\,\mathrm m$ , while the $2\,\mathrm{kg}$ particle remains at $(0,0)$ . Find the new centre of mass.

Plan. Apply the same masses to the horizontal and vertical coordinates.

\begin{aligned} \bar x&=\frac{2(0)+3(5)}5=3 \\ \bar y&=\frac{2(0)+3(4)}5=\frac{12}5. \end{aligned}

Answer. $G=(3,12/5)\,\mathrm m$ .

Check. The point is three-fifths of the way from $(0,0)$ to $(5,4)$ , so its distances from the two masses are in the inverse mass ratio $3:2$ .

For more than two pieces, the same sums work. A table of masses and coordinates prevents a contribution from being omitted.

Watch out

$(x_1+x_2)/2$ is the correct average only for equal masses. Do not average the positions of unequal pieces without their mass factors.

Use signed coordinates when a particle lies to the left of the origin.

Composite uniform laminae

A uniform lamina has a common mass per unit area, so its pieces' masses are proportional to their areas. You may therefore use area in place of mass. Split the shape into non-overlapping pieces with known centres, or subtract a missing piece from a complete shape.

For a hole, subtract both its area and its first moment. A “negative mass” in the table is a bookkeeping device for removed material; it is not a physical mass left in the body. The remaining total mass must be positive.

[DIAGRAM: asset_name: An L-shaped lamina as two rectangles; asset_slug: cam9231-lamina-com; caption: Each rectangle contributes its area at its own centre; description: An L-shaped lamina occupies a 6 by 2 centimetre bottom rectangle and a 2 by 2 centimetre upper-left rectangle. Axes start at the lower-left corner. The component centres are at (3,1) and (1,3), with areas 12 and 4 square centimetres. The combined centre G is marked at (2.5,1.5). The missing upper-right 4 by 2 rectangle is outlined faintly.]
Each rectangle contributes its area at its own centre

Worked example 3: adding two rectangles

Question. A uniform L-shaped lamina is the union of the rectangles $0\le x\le6,\ 0\le y\le2$ and $0\le x\le2,\ 2\le y\le4$ , with lengths in centimetres. Find its centre of mass.

Plan. Use the areas of the two non-overlapping rectangles as masses.

Piece	Area	Centre
Bottom rectangle	$12$	$(3,1)$
Upper-left rectangle	$4$	$(1,3)$

The common mass per unit area cancels, giving

\begin{aligned} \bar x&=\frac{12(3)+4(1)}{16}=\frac52 \\ \bar y&=\frac{12(1)+4(3)}{16}=\frac32. \end{aligned}

Answer. $G=(5/2,3/2)\,\mathrm{cm}$ .

Check. Start instead with a $6$ by $4$ rectangle and remove the upper-right $4$ by $2$ rectangle: $\bar x=[24(3)-8(4)]/16=5/2$ , and $\bar y=[24(2)-8(3)]/16=3/2$ .

Move the missing corner from the upper right to the upper left. This reflects the shape in its vertical midline; subtraction now provides a short direct solution.

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3.2 - Centres of mass of composite bodies

An equivalent particle system

Composite uniform laminae

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