M2 June 2011 Q4

Figure 1: lamina ABCDE made of rectangle ABDE (AB = 4a, AE = 2a) and triangle BCD, with F the midpoint of BD and FC = a — Figure 1

Figure 1 shows a uniform lamina \(ABCDE\) such that \(ABDE\) is a rectangle, \(BC = CD\), \(AB = 4a\) and \(AE = 2a\). The point \(F\) is the midpoint of \(BD\) and \(FC = a\).

(a) Find, in terms of \(a\), the distance of the centre of mass of the lamina from \(AE\). (4)

The lamina is freely suspended from \(A\) and hangs in equilibrium.

(b) Find the angle between \(AB\) and the downward vertical. (3)

Lamina ABCDE with centre of mass G on the line of symmetry; the line AG makes angle phi with AB — Mark scheme (a)

Scheme	Marks
\(\begin{array}{lccc} & ABDE & BCD & \text{Lamina} \\ \text{Mass ratio} & 8a^2\rho & a^2\rho & 9a^2\rho \\ & 8 & 1 & 9 \end{array}\)	B1
\(\begin{array}{lccc} \text{Dist of C of M from AE} & 2a & 4\frac{1}{3}a & \bar{x} \end{array}\)	B1
\(8 \times 2a + 1 \times \dfrac{13}{3}a = 9\bar{x}\)	M1
\(\bar{x} = \dfrac{61}{27}a\quad (2.26a)\)	A1
	(4)

Scheme	Marks
\(\tan\phi = \dfrac{a}{\frac{61}{27}a} = \dfrac{27}{61}\)	M1 A1 ft
\(\phi = 23.87\ldots = 24^\circ\quad\) (accept 23.9), 0.417 radians	A1
	(3)
	(7 marks)