M2 January 2012 Q4
4.
The trapezium \(ABCD\) is a uniform lamina with \(AB = 4\) m and \(BC = CD = DA = 2\) m, as shown in Figure 1.
(a) Show that the centre of mass of the lamina is \(\dfrac{4\sqrt{3}}{9}\) m from \(AB\). (5)
The lamina is freely suspended from \(D\) and hangs in equilibrium.
(b) Find the angle between \(DC\) and the vertical through \(D\). (5)
| Scheme | Marks |
|---|---|
| For an appropriate division of the trapezium into standard shapes with: | |
| correct ratio of masses | B1 |
| correct distances of c.o.m. from AB | B1 |
| e.g three equilateral triangles of height \(\sqrt{3}\), mass \(m\) kg, | |
| com \(\dfrac{\sqrt{3}}{3}\) from bases of each | |
| \(3md = \left(m \times \dfrac{2}{3} \times \sqrt{3}\right) + \left(2 \times m \times \dfrac{1}{3}\sqrt{3}\right) = \dfrac{4\sqrt{3}}{3}m\), | M1 A1 |
| \(d = \dfrac{4\sqrt{3}}{9}\quad\) AG | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| Horizontal distance of c of m from D = 1m | B1 |
| Vertical distance \(\quad \sqrt{3} - \dfrac{4\sqrt{3}}{9} = \dfrac{5\sqrt{3}}{9}\ (0.962\ldots)\) | B1 |
| \(\tan^{-1}\dfrac{0.962\ldots}{1}\) | M1 A1ft |
| Angle \(= 43.9^\circ\) | A1 |
| (5) | |
| (10 marks) |