M5 June 2018 Q7

7. A pendulum consists of a uniform circular disc, of radius \(a\) and mass \(4m\), whose centre is fixed to the end \(B\) of a uniform rod \(AB\). The rod has mass \(3m\) and length \(4l\), where \(2l \gt a\). The rod lies in the same plane as the disc. The pendulum is free to rotate about a fixed smooth horizontal axis \(L\) which passes through \(A\) and is perpendicular to the plane of the disc. The moment of inertia of the pendulum about \(L\) is \(2m(a^2 + 40l^2)\).

(a) Find the approximate period of small oscillations of the pendulum about its position of stable equilibrium. (5)

The pendulum is held with \(B\) vertically above \(A\) and is then slightly displaced from rest. In the subsequent motion the midpoint of \(AB\) strikes a small peg, which is fixed at the same horizontal level as \(A\), and the pendulum rebounds upwards. Immediately before it strikes the peg, the angular speed of the pendulum is \(\omega\).

(b) Show that \(\omega^2 = \dfrac{22gl}{(a^2 + 40l^2)}\) (3)

Immediately after it strikes the peg, the angular speed of the pendulum is \(\dfrac{1}{2}\omega\).

(c) Find, in terms of \(m\), \(g\), \(a\) and \(l\), the magnitude of the impulse exerted on the peg by the pendulum. (4)

(d) Show that the size of the angle turned through by the pendulum, between it hitting the peg and it next coming to rest, is \(\arcsin\dfrac{1}{4}\). (4)

Mark scheme (a)

Scheme	Marks
\(3mg\cdot 2l\sin\theta + 4mg\cdot 4l\sin\theta = -2m(a^2 + 40l^2)\ddot{\theta}\)	M1 A1
For small \(\theta\), \(\sin\theta \simeq \theta\)
\(-\dfrac{11gl}{(a^2 + 40l^2)}\theta = \ddot{\theta}\)	M1
SHM; \(T = 2\pi\sqrt{\dfrac{(a^2 + 40l^2)}{11gl}}\)	M1 A1
	(5)

Notes

First M1 for equation of motion about \(A\) with usual rules

First A1 for a correct equation

Second M1 for using small angle approx. and putting into appropriate form (with − sign)

Third M1 for SHM and use of correct formula for period

Second A1 for correct answer

Mark scheme (b)

Scheme	Marks
\(4mg\cdot 4l + 3mg\cdot 2l = \dfrac{1}{2}2m(a^2 + 40l^2)\omega^2\)	M1 A1
\(\omega^2 = \dfrac{22gl}{(a^2 + 40l^2)}\) GIVEN ANSWER	A1
	(3)

Notes

M1 for conservation of energy equation with usual rules

First A1 for a correct equation

Second A1 for correct GIVEN ANSWER

Mark scheme (c)

Scheme	Marks
\(J\cdot 2l = 2m(a^2 + 40l^2)\left(\dfrac{1}{2}\omega - -\omega\right)\)	M1 A2
\(J = \dfrac{3m}{2}\sqrt{\dfrac{22g(a^2 + 40l^2)}{l}}\)	A1
	(4)

Notes

M1 for impulse-momentum equation with usual rules (\(I\) and \(\omega\) do not need to be substituted)

First A1 for a correct equation without \(I\) and \(\omega\) substituted

Second A1 for a correct equation with \(I\) and \(\omega\) substituted

Third A1 for correct answer (must be positive)

Mark scheme (d)

Scheme	Marks
\(\dfrac{1}{2}2m(a^2 + 40l^2)\left(\dfrac{1}{2}\omega\right)^2 = 3mg\cdot 2l\sin\theta + 4mg\cdot 4l\sin\theta\)	M1 A1
\(\dfrac{1}{8}2m(a^2 + 40l^2)\dfrac{22gl}{(a^2 + 40l^2)} = 22mgl\sin\theta\)	DM1
\(\sin\theta = \dfrac{1}{4} \Rightarrow \theta = \sin^{-1}\dfrac{1}{4}\) GIVEN ANSWER	A1
	(4)
	(16 marks)

Notes

First M1 for conservation of energy equation with usual rules

First A1 for a correct equation without \(\omega\) substituted

Second DM1, dependent, for substituting for \(\omega\)

Second A1 for correctly obtaining GIVEN ANSWER