Centres of Mass and Stability (AQA A-Level Further Maths): Revision Notes

Worked Example 1: Lamina under a hyperbolic curve

Question: Find the centre of mass, $G$, of the lamina formed from the area under the curve $y = \frac{12}{x}$ for $2 \leq x \leq 6$.

Solution:

Let the area density be $\rho$ and let $G$ be at the point $(\bar{x}, \bar{y})$.

Step 1: Calculate the total mass $M$.

The total mass equals density times area:

$M = \rho \int_2^6 y \, dx = \rho \int_2^6 \frac{12}{x} \, dx = 12\rho \int_2^6 \frac{1}{x} \, dx = 12\rho[\ln x]_2^6 = 12\rho \ln 3$

Step 2: Take moments about the y-axis to find $\bar{x}$.

$12\rho \ln 3 \times \bar{x} = \rho \int_2^6 xy \, dx = \rho \int_2^6 x \cdot \frac{12}{x} \, dx = 12\rho \int_2^6 1 \, dx = 12\rho[x]_2^6 = 48\rho$

Therefore: $\bar{x} = \frac{48\rho}{12\rho \ln 3} = \frac{48}{12 \ln 3} = \frac{4}{\ln 3} = 3.64 \text{ to 3 sf}$

Step 3: Take moments about the x-axis to find $\bar{y}$.

$12\rho \ln 3 \times \bar{y} = \rho \int_2^6 \frac{1}{2}y^2 \, dx = \rho \int_2^6 \frac{1}{2}\left(\frac{12}{x}\right)^2 \, dx = 72\rho \int_2^6 \frac{1}{x^2} \, dx = 72\rho\left[-\frac{1}{x}\right]_2^6 = 24\rho$

Therefore: $\bar{y} = \frac{24\rho}{12\rho \ln 3} = \frac{24}{12 \ln 3} = \frac{2}{\ln 3} = 1.82 \text{ to 3 sf}$

Answer: The centre of mass, $G$, is at the point (3.64, 1.82).

Key observation: Notice how $\rho$ cancels out in both calculations. This always happens for uniform laminae, so you can sometimes work without explicitly including $\rho$.

Worked Example 2: Solid of revolution

Question: Find the centre of mass, $G$, of the solid of revolution formed by rotating the area under the curve $y = \frac{12}{x}$ for $2 \leq x \leq 6$ through $2\pi$ radians about the x-axis. Give an exact answer.

Solution:

Let the volume density be $\rho$ and let $G$ be at the point $(\bar{x}, \bar{y})$.

Step 1: Calculate the total mass $M$.

$M = \rho \int_2^6 \pi y^2 \, dx = \rho\pi \int_2^6 \left(\frac{12}{x}\right)^2 \, dx = 144\rho\pi \int_2^6 \frac{1}{x^2} \, dx = 144\rho\pi\left[-\frac{1}{x}\right]_2^6 = 48\rho\pi$

Step 2: Take moments about the y-axis to find $\bar{x}$.

$48\rho\pi \times \bar{x} = \rho \int_2^6 x \times \pi y^2 \, dx = \rho\pi \int_2^6 x \cdot \frac{144}{x^2} \, dx = 144\rho\pi \int_2^6 \frac{1}{x} \, dx = 144\rho\pi[\ln x]_2^6 = 144\rho\pi \ln 3$

Therefore: $\bar{x} = \frac{144\rho\pi \ln 3}{48\rho\pi} = \frac{144 \ln 3}{48} = 3\ln 3$

Step 3: Use symmetry to find $\bar{y}$.

By symmetry about the x-axis, $\bar{y} = 0$.

Answer: The solid's centre of mass, $G$, is at the point (3ln 3, 0).

Time-saving tip: Always state that $\bar{y} = 0$ by symmetry for solids of revolution around the x-axis. This saves calculation time in the exam.

Worked Example 3: Suspended semicircular lamina

Question: A uniform semicircular lamina of centre $O$ with diameter $PQ = 2a$ is suspended from $P$ and hangs freely. Calculate the angle $\theta$ that $PQ$ makes with the vertical.

Solution:

Step 1: Draw a diagram with $PG$ vertical.

This is essential because in equilibrium, the centre of mass $G$ must hang vertically below the suspension point $P$.

Step 2: Quote the standard formula for a semicircular lamina.

For a semicircle lamina: $OG = \frac{2a\sin\alpha}{3\alpha}$

where $2\alpha$ is the angle of the full arc (in radians).

For a semicircle, $2\alpha = \pi$ radians, so:

$OG = \frac{2a\sin(\pi/2)}{3(\pi/2)} = \frac{2a \times 1}{3\pi/2} = \frac{4a}{3\pi}$

Step 3: Calculate the angle.

In triangle $OPG$:

• $OP = a$ (radius)
• $OG = \frac{4a}{3\pi}$
• $PG$ is vertical

Therefore: $\tan\theta = \frac{OG}{OP} = \frac{4a/3\pi}{a} = \frac{4}{3\pi}$

$\theta = \tan^{-1}\left(\frac{4}{3\pi}\right) = 23.0°$

Answer: The angle of $PQ$ to the vertical is θ = 23.0°.

Important note: The formula for the centre of mass of standard shapes (semicircle, hemisphere, cone, etc.) is given in the formula booklet. Always draw a clear diagram showing the vertical line through the suspension point.