M5 June 2009 Q4

Figure 1: right-angled triangle OAB with OA = a horizontal, AB = 2a vertical and the right angle at A — Figure 1

A uniform lamina of mass \(M\) is in the shape of a right-angled triangle \(OAB\). The angle \(OAB\) is 90°, \(OA = a\) and \(AB = 2a\), as shown in Figure 1.

(a) Prove, using integration, that the moment of inertia of the lamina \(OAB\) about the edge \(OA\) is \(\tfrac{2}{3}Ma^2\).

(You may assume without proof that the moment of inertia of a uniform rod of mass \(m\) and length \(2l\) about an axis through one end and perpendicular to the rod is \(\tfrac{4}{3}ml^2\).)

(6)

The lamina \(OAB\) is free to rotate about a fixed smooth horizontal axis along the edge \(OA\) and hangs at rest with \(B\) vertically below \(A\). The lamina is then given a horizontal impulse of magnitude \(J\). The impulse is applied to the lamina at the point \(B\), in a direction which is perpendicular to the plane of the lamina. Given that the lamina first comes to instantaneous rest after rotating through an angle of 120°,

(b) find an expression for \(J\), in terms of \(M\), \(a\) and \(g\). (7)

Mark scheme (a)

Scheme	Marks
\(\delta m = \dfrac{2Mx\delta x}{a^2}\)	M1 A1
\(\delta I = \dfrac{1}{3}\dfrac{2Mx\delta x}{a^2}(2x)^2\)	M1 A1
\(I = \displaystyle\int_0^a \dfrac{8Mx^3\,\mathrm{d}x}{3a^2}\) \(= \dfrac{8M}{3a^2}\left[\dfrac{x^4}{4}\right]_0^a\)	DM1
\(= \dfrac{2}{3}Ma^2\) *	A1
	(6)

Mark scheme (b)

Scheme	Marks
\(J.2a = \dfrac{2}{3}Ma^2\omega\)	M1 A1
\(\dfrac{1}{2}\dfrac{2}{3}Ma^2\omega^2 = Mg\dfrac{2a}{3}(1 + \cos 60^\circ)\)	M1 A2
solving for \(J\)	DM1
\(J = M\sqrt{\dfrac{ag}{3}}\)	A1
	(7)
	(13 marks)