Centres of mass of plane figures and equilibrium of a laminaEdexcel A-Level Further Maths: Revision notes

For a uniform lamina the mass is proportional to the area, so the centre of mass is the geometric centre of the shape (the centroid). If the figure has an axis of symmetry, the centre of mass lies on it. If it has two axes, it is where they cross. Results from the formulae book may be quoted without proof:

• rectangle or parallelogram: at the centre;
• triangle: $\frac23$ of the way along each median from the vertex, so $\frac13$ of the height from each side;
• semicircular lamina of radius $r$: $\frac{4r}{3\pi}$ from the diameter;
• sector with half-angle $\alpha$ (radians): $\frac{2r\sin\alpha}{3\alpha}$ from the centre. Example: a semicircle of radius 5 cm has its centre of mass $\frac{20}{3\pi}=2.12$ cm from its diameter.

A composite figure is made of simple shapes added together, or with a shape removed. Mass is proportional to area, so use areas as weights. Take moments about two convenient perpendicular axes: $A\bar x=\sum a_ix_i,\qquad A\bar y=\sum a_iy_i.$ For a hole, use a negative area. Draw a table of area, $x$, $y$, $ax$, $ay$ for each part. Example: a square of side 12 cm with a 4 cm square removed from a corner at $(8,0)$ to $(12,4)$. Then $\bar x=\frac{144(6)-16(10)}{128}=5.5$ and $\bar y=\frac{144(6)-16(2)}{128}=6.5$. Example 2: a $10\times6$ rectangle with a semicircle of radius 5 attached to a 10 cm side. Distance from the opposite side: $\bar d=\frac{60(3)+12.5\pi(8.12)}{60+12.5\pi}=5.03$ cm.

If the parts are made of different materials, mass is no longer proportional to area, so use each part's mass as the weight in the moments equation: $M\bar x=\sum m_ix_i.$ Find the mass of each part from its density, or from the information given. Each uniform part still has its own centre of mass at its centroid. Example: two uniform strips of masses 2 kg and 3 kg with centres at 3 cm and 8 cm from one edge give $\bar x=\frac{2(3)+3(8)}{5}=6$ cm.

A framework is made of rods or wire. Each straight rod is uniform, so its mass is proportional to its length and its centre of mass is at its midpoint. Treat each rod as a particle of mass proportional to its length at its midpoint. Example: an isosceles triangle with $AB=AC=13$, $BC=10$, and height 12. Moments about $BC$: $36\bar y=13(6)+13(6)+10(0)$, so $\bar y=\frac{13}{3}=4.33$ cm. A particle added to the framework is included with its own mass, at its own position.

A lamina in equilibrium under coplanar forces has zero resultant force and zero total moment about any point. When it is freely suspended from a point $P$, the only forces are the weight (at the centre of mass $G$) and the force at $P$. For zero moment about $P$, the line $PG$ must be vertical, so $G$ is directly below $P$. To find the angle an edge makes with the vertical, locate $G$ relative to $P$ and use $\tan\theta=\frac{\text{horizontal offset}}{\text{vertical offset}}$. Example: the square with a corner removed, hung from $O$, has $G=(5.5,\,6.5)$, so $OC$ makes $\tan^{-1}\frac{5.5}{6.5}=40.2^\circ$ with the vertical. With a loaded framework, find the new centre of mass of the whole system first, then use the same method.