M5 June 2014 (R) Q8

Figure 1: circular disc of radius 2a, centre O, with four circular holes of radius a/2 centred at P, Q, R and S — Figure 1

A uniform circular disc of radius \(2a\) has centre \(O\). The points \(P\), \(Q\), \(R\) and \(S\) on the disc are the vertices of a square with centre \(O\) and \(OP = a\). Four circular holes, each of radius \(\dfrac{a}{2}\), and with centres \(P\), \(Q\), \(R\) and \(S\), are drilled in the disc to produce the lamina \(L\), shown shaded in Figure 1. The mass of \(L\) is \(M\).

(a) Show that the moment of inertia of \(L\) about an axis through \(O\), and perpendicular to the plane of \(L\), is \(\dfrac{55Ma^2}{24}\) (8)

The lamina \(L\) is free to rotate in a vertical plane about a fixed smooth horizontal axis which is perpendicular to \(L\) and which passes through a point \(A\) on the circumference of \(L\). At time \(t\), \(AO\) makes an angle \(\theta\) with the downward vertical through \(A\).

(b) Show that \(\dfrac{\mathrm{d}^2\theta}{\mathrm{d}t^2} = -\dfrac{48g}{151a}\sin\theta\) (4)

The magnitude of the component, in a direction perpendicular to \(AO\), of the force exerted on \(L\) by the axis is \(X\).

(d) Find \(X\) in terms of \(M\), \(g\) and \(\theta\). (4)

Mark scheme (a)

Scheme	Marks
\(\rho = \dfrac{M}{\pi(2a)^2 - 4\pi\left(\frac{1}{2}a\right)^2} = \dfrac{M}{3\pi a^2}\)	M1
\(I\) of big disc \(= \dfrac{1}{2}\pi(2a)^2\dfrac{M}{3\pi a^2}(2a)^2 = \dfrac{8Ma^2}{3}\)	M1 A1
\(I\) of each small disc \(= \dfrac{1}{2}m^\prime\left(\tfrac{1}{2}a\right)^2 + m^\prime a^2 = \dfrac{9m^\prime a^2}{8}\)	M1 A1
\(= \dfrac{9a^2}{8}\,\dfrac{M}{3\pi a^2}\,\pi\left(\tfrac{1}{2}a\right)^2 = \dfrac{3Ma^2}{32}\)	M1
\(I\) of \(L = \dfrac{8Ma^2}{3} - 4\left(\dfrac{3Ma^2}{32}\right) = \dfrac{55Ma^2}{24}\) **	M1 A1
	(8)

Mark scheme (b)

Scheme	Marks
\(I_A = \dfrac{55Ma^2}{24} + M(2a)^2 = \dfrac{151Ma^2}{24}\)	M1 A1
\(Mg2a\sin\theta = -\dfrac{151Ma^2}{24}\ddot{\theta}\)	M1
\(-\dfrac{48g}{151a}\sin\theta = \ddot{\theta}\) PRINTED ANSWER	A1
	(4)

Scheme	Marks
For small \(\theta,\ \sin\theta \approx \theta\), \(-\dfrac{48g}{151a}\theta = \ddot{\theta}\) SHM	M1
\(T = 2\pi\sqrt{\dfrac{151a}{48g}}\)	A1
	(2)

Mark scheme (d)

Scheme	Marks
\(X - Mg\sin\theta = M \cdot 2a\ddot{\theta}\)	M1 A1
\(X = Mg\sin\theta + M \cdot 2a \cdot \left(-\dfrac{48g}{151a}\sin\theta\right)\)	M1
\(= \dfrac{55Mg\sin\theta}{151}\)	A1
	(4)
	(18 marks)