M3 June 2012 Q6
6.
Figure 3 shows a uniform equilateral triangular lamina \(PRT\) with sides of length \(2a\).
(a) Using calculus, prove that the centre of mass of \(PRT\) is at a distance \(\dfrac{2\sqrt{3}}{3}a\) from \(R\). (6)
The circular sector \(PQU\), of radius \(a\) and centre \(P\), and the circular sector \(TUS\), of radius \(a\) and centre \(T\), are removed from \(PRT\) to form the uniform lamina \(QRSU\) shown in Figure 4.
(b) Show that the distance of the centre of mass of \(QRSU\) from \(U\) is \(\dfrac{2a}{3\sqrt{3} - \pi}\) (6)
| Scheme | Marks |
|---|---|
| Mass of lamina \(= \rho\dfrac{1}{2} \times 2a \times \sqrt{3}a = \sqrt{3}\rho a^2\) | B1 |
| \(\displaystyle\sum \rho x \times \frac{2x}{\sqrt{3}} \times \delta x = \rho\int_0^{\sqrt{3}a}\frac{2x^2}{\sqrt{3}}\,\mathrm{d}x\) | M1 |
| \(= \rho\left[\dfrac{2x^3}{3\sqrt{3}}\right]_0^{\sqrt{3}a}\) | A1 |
| \(= \rho\dfrac{2 \times 3\sqrt{3}a^3}{3\sqrt{3}} = 2\rho a^3\) | A1 |
| Distance from vertex \(= \dfrac{2\rho a^3}{\sqrt{3}\rho a^2} = \dfrac{2}{3}a\sqrt{3}\) ** | M1A1 |
| (6) |
| Scheme | Marks |
|---|---|
| Area of each sector \(= \dfrac{1}{6}\pi a^2\) | B1 |
| Using sector formula, \(d = h\sin\alpha = \dfrac{2a\sin\alpha}{3\alpha}\sin\alpha = \dfrac{a}{3\dfrac{\pi}{6}} \times \dfrac{1}{2} = \dfrac{a}{\pi}\) | B2,1,0 |
| Taking moments: \(\left(\sqrt{3}a^2 - 2 \times \dfrac{\pi a^2}{6}\right)D = \sqrt{3}a^2 \times \dfrac{\sqrt{3}a}{3} - 2 \times \dfrac{\pi a^2}{6} \times \dfrac{a}{\pi}\) | M1A1 |
| \(D = \dfrac{\dfrac{2a^3}{3}}{\left(\sqrt{3} - \dfrac{\pi}{3}\right)a^2} = \dfrac{2a}{3\sqrt{3} - \pi}\) ** | A1 |
| (6) | |
| (12 marks) |