M5 June 2015 Q6

6. A pendulum is modelled as a uniform rod \(AB\), of mass \(3m\) and length \(2a\), which has a particle of mass \(2m\) attached at \(B\). The pendulum is free to rotate in a vertical plane about a fixed smooth horizontal axis \(L\) which passes through \(A\). The vertical plane is perpendicular to the axis \(L\).

(a) Find the period of small oscillations of the pendulum about its position of stable equilibrium. (8)

The pendulum is hanging at rest in a vertical position, with \(B\) below \(A\), when it is given a horizontal impulse of magnitude \(J\). The impulse acts at \(B\) in a vertical plane which is perpendicular to the axis \(L\).

Given that the pendulum turns through an angle of 60\(^\circ\) before first coming to instantaneous rest,

(b) find \(J\). (8)

Mark scheme (a)

Scheme	Marks
\(I_L = \tfrac{1}{3}3m(2a)^2 + 2m(2a)^2 = 12ma^2\)	M1 A1
\(M(L),\quad -3mga\sin\theta - 2mg \cdot 2a\sin\theta = 12ma^2\ddot{\theta}\)	M1 A2 ft
\(-\dfrac{7g\sin\theta}{12a} = \ddot{\theta}\)
For small \(\theta,\ \sin\theta \approx \theta,\quad -\dfrac{7g\theta}{12a} = \ddot{\theta}\) so SHM with \(\omega = \sqrt{\dfrac{7g}{12a}}\)	M1
so, \(\ T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{12a}{7g}}\)	M1 A1
	(8)

Notes

First M1 for finding MI

First A1 for \(12ma^2\)

Second M1 for moments about the axis

Second and Third A1 ft, on their \(I\), for correct equation (A1 for each side)

They may use CM of rod + particle

Third M1 for small angle approx. and comparison with standard SHM to give an \(\omega\) value (need – sign in their DE)

Fourth M1 for use of \(2\pi/\omega\)

Fourth A1 cao for any equivalent answer

Mark scheme (b)

Scheme	Marks
\(3mga(1 - \cos 60^\circ) + 2mg2a(1 - \cos 60^\circ) = \tfrac{1}{2}12ma^2\omega^2\)	M1 A1 A1ft
	A1 (3rd DM)
\(J \cdot 2a = 12ma^2\omega\)	M1 A1 A1ft
\(J = 6ma\sqrt{\dfrac{7g}{12a}} = 6m\sqrt{\dfrac{7ga}{12}} = m\sqrt{21ag}\)	A1
	(8)
	(16 marks)

Notes

First M1 for energy equation

First A1 ft for KE term

Second A1 for PE terms

Third A1 is now 3^rd DM mark, dependent on both previous M marks, for solving for \(J\) and precedes final A mark

Second M1 for imp-momentum equation

Fourth A1 for LHS on scheme

Fifth A1 ft for RHS on scheme

Sixth A1 cao for any equivalent answer