M2 June 2012 Q4

Figure 2: circular disc centre O radius 4a with perpendicular diameters PQ and ST, circular hole radius 2a centred at R on OP, lamina shaded — Figure 2

A uniform circular disc has centre \(O\) and radius \(4a\). The lines \(PQ\) and \(ST\) are perpendicular diameters of the disc. A circular hole of radius \(2a\) is made in the disc, with the centre of the hole at the point \(R\) on \(OP\) where \(OR = 2a\), to form the lamina \(L\), shown shaded in Figure 2.

(a) Show that the distance of the centre of mass of \(L\) from \(P\) is \(\dfrac{14a}{3}\). (4)

The mass of \(L\) is \(m\) and a particle of mass \(km\) is now fixed to \(L\) at the point \(P\). The system is now suspended from the point \(S\) and hangs freely in equilibrium. The diameter \(ST\) makes an angle \(\alpha\) with the downward vertical through \(S\), where \(\tan\alpha = \dfrac{5}{6}\).

(b) Find the value of \(k\). (5)

Mark scheme (a)

Scheme

Marks

\(\pi(4a)^2\)	\(\pi(2a)^2\)	\((\pi(4a)^2 - \pi(2a)^2)\)
4	1	3
\(4a\)	\(2a\)	\(\bar{x}\)

B1 B1

\((4 \times 4a) - (1 \times 2a) = 3\bar{x}\)

\(\dfrac{14a}{3} = \bar{x}\ \ *\)

(4)

Notes

B1 Correct mass ratios

B1 Distance of c of m from \(P\) (or from a point on \(QP\)).

M1 Moments about axis through \(P\), or about a parallel axis then convert the answer to distance from \(P\). Condone a sign slip.

A1 Answer given – check working carefully. Must reach positive answer legitimately.

Diagram: S above line CP through O and G, with SO = 4a, CO = 2a/3, angles beta and alpha at S — Mark scheme (b)

Scheme	Marks
\(OG = 4a\tan\alpha = \dfrac{10a}{3}\ \left(\Rightarrow PG = \dfrac{2a}{3}\right)\)	M1 A1
\(M(P),\ (m + km)g.\dfrac{2a}{3}\cos\alpha = mg.\dfrac{14a}{3}\cos\alpha\) \(M(G):\ km \times \dfrac{2}{3}a = m \times \left(\dfrac{10}{3}a + \dfrac{2}{3}a\right) = 4ma\) \(M(O):\ m(1 + k) \times \dfrac{10}{3}a + m \times \dfrac{2}{3}a = km \times 4a\) \(M(C):\ \dfrac{12}{3}a \times (1 + k)m = \dfrac{14}{3}a \times km\) \(M(Q):\ \dfrac{22}{3}a \times m(1 + k) = \dfrac{10}{3}a \times m + 8a \times km\)	M1 A1
\(k = 6\)	A1
	(5)
	(9 marks)

\((k + 1)m \times PG = m \times \dfrac{14}{3}a\)	M1
\(PG = \dfrac{14a}{3(k + 1)}\)	A1
\(\tan\alpha = \dfrac{OG}{4a} = \dfrac{4a - \dfrac{14a}{3(k + 1)}}{4a}\ \left(= 1 - \dfrac{7}{6(k + 1)}\right)\)	M1
\(\dfrac{5}{6} = 1 - \dfrac{7}{6(k + 1)}\ \ \ \ k = 6\)	A1, A1


\(\tan(\angle CSO) = \tan\beta = \dfrac{\tfrac{2a}{3}}{4a} = \dfrac{1}{6}\)	M1
\(km.\sqrt{32}a.\sin(45 - \alpha) = m.\sqrt{16\tfrac{4}{9}}a.\sin(\alpha + \beta)\)	M1A1
\(k.\sqrt{32}.\left(\dfrac{6 - 5}{\sqrt{2}.\sqrt{61}}\right) = \dfrac{\sqrt{148}}{3}.\dfrac{6 \times 5 + 1 \times 6}{\sqrt{37}.\sqrt{61}}\)	A1
\(4k = \dfrac{2}{3} \times 36,\ \ k = 6\)	A1