M2 January 2010 Q7
7. [The centre of mass of a semi-circular lamina of radius \(r\) is \(\dfrac{4r}{3\pi}\) from the centre]
A template \(T\) consists of a uniform plane lamina \(PQROS\), as shown in Figure 3. The lamina is bounded by two semicircles, with diameters \(SO\) and \(OR\), and by the sides \(SP\), \(PQ\) and \(QR\) of the rectangle \(PQRS\). The point \(O\) is the mid-point of \(SR\), \(PQ = 12\) cm and \(QR = 2x\) cm.
(a) Show that the centre of mass of \(T\) is a distance \(\dfrac{4\left|2x^2 - 3\right|}{8x + 3\pi}\) cm from \(SR\). (7)
The template \(T\) is freely suspended from the point \(P\) and hangs in equilibrium.
Given that \(x = 2\) and that \(\theta\) is the angle that \(PQ\) makes with the horizontal,
(b) show that \(\tan\theta = \dfrac{48 + 9\pi}{22 + 6\pi}\). (4)
| Scheme | Marks |
|---|---|
| \(\begin{array}{ccc} \text{Rectangle} & \text{Semicircles} & \text{Template, } T \\[4pt] 24x & 4.5\pi \qquad 4.5\pi & 24x + 9\pi \\[4pt] x & \dfrac{4 \times 3}{3\pi} \qquad \dfrac{4 \times 3}{3\pi} & \bar{x} \end{array}\) | B2 B2 |
| \(24x^2 - 4.5\pi \times \left(\dfrac{4 \times 3}{3\pi}\right) - 4.5\pi \times \left(\dfrac{4 \times 3}{3\pi}\right) = (24x + 9\pi)\bar{x}\) | M1 A1 |
| distance \(= \left|\bar{x}\right| = \dfrac{4\left|2x^2 - 3\right|}{(8x + 3\pi)}\ \ **\) | A1 |
| (7) |
| Scheme | Marks |
|---|---|
| When \(x = 2,\quad \left|\bar{x}\right| = \dfrac{20}{16 + 3\pi}\) | B1 |
| \(\tan\theta = \dfrac{6}{4 - \left|\bar{x}\right|} = \dfrac{6}{4 - \dfrac{20}{16 + 3\pi}}\) | M1 A1 |
| \(= \dfrac{48 + 9\pi}{22 + 6\pi}\). | A1 |
| (4) | |
| (11 marks) |