M4 June 2011 Q7

EdexcelOld spec14 marksHooke's Law Work-Energy Principle

Figure 3: framework ABC with AB = 2a and BC = a at right angles at B, hinged at A; string from B through ring R, 2a from A on the same level, angle ARB theta — Figure 3

Figure 3 shows a framework \(ABC\), consisting of two uniform rods rigidly joined together at \(B\) so that \(\angle ABC = 90^\circ\). The rod \(AB\) has length \(2a\) and mass \(4m\), and the rod \(BC\) has length \(a\) and mass \(2m\). The framework is smoothly hinged at \(A\) to a fixed point, so that the framework can rotate in a fixed vertical plane. One end of a light elastic string, of natural length \(2a\) and modulus of elasticity \(3mg\), is attached to \(A\). The string passes through a small smooth ring \(R\) fixed at a distance \(2a\) from \(A\), on the same horizontal level as \(A\) and in the same vertical plane as the framework. The other end of the string is attached to \(B\).

The angle \(ARB\) is \(\theta\), where \(0 \lt \theta \lt \dfrac{\pi}{2}\).

(a) Show that the potential energy \(V\) of the system is given by\[V = 8amg\sin 2\theta + 5amg\cos 2\theta + \text{constant}\] (7)

(b) Find the value of \(\theta\) for which the system is in equilibrium. (4)

Diagram: R, A and B with angles theta at R and 2 theta at A; weights 4mg at the midpoint of AB and 2mg at the midpoint of BC — Mark scheme (a)

Scheme	Marks

\(\mathrm{BR} = 2 \times 2a\cos\theta = 4a\cos\theta\)	B1
\(\mathrm{EPE} = 3mg\dfrac{(4a\cos\theta)^2}{2 \times 2a}\)	M1
\(= 12mga\cos^2\theta = 6mga + 6mga\cos 2\theta\)	A1
GPE: taking AR as the level of zero GPE, GPE = GPE of AB + GPE of BC	M1+M1
\(= 4mg \times a\sin 2\theta + 2mg\left(2a\sin 2\theta - a/2\cos 2\theta\right)\)	A1
\(= 8mga\sin 2\theta - mga\cos 2\theta\)
\(\Rightarrow\) Total \(V = 8mga\sin 2\theta + 5mga\cos 2\theta + \text{constant}\), as required. **	A1
	(7)

Mark scheme (b)

Scheme	Marks
\(\dfrac{dV}{d\theta} = 16mga\cos 2\theta - 10mga\sin 2\theta\)	M1 A1
\(\dfrac{dV}{d\theta} = 0 \Rightarrow 10\sin 2\theta = 16\cos 2\theta\)	M1
\(\Rightarrow \tan 2\theta = \dfrac{8}{5} \Rightarrow \theta = 0.51\) radians \((29.0^\circ)\)	A1
	(4)

Notes

Alternative

Or: \(8mga\sin 2\theta + 5mga\cos 2\theta = \sqrt{89}mga\cos(2\theta - \alpha),\ \ \tan\alpha = \dfrac{8}{5}\)	M1 A1
t. pts when \(2\theta - \alpha = n\pi \Rightarrow \theta = 0.51\) rads.	M1 A1

Mark scheme (c)

Scheme	Marks
\(\dfrac{d^2V}{d\theta^2} = -32mga\sin 2\theta - 20mga\cos 2\theta\)	M1
\(\theta = 0.51 \Rightarrow \dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} \lt 0\), equilibrium is unstable. cso	M1 A1
	(3)
	(14 marks)

Notes

Alternative

Or: \(\ 2\theta - \alpha = 0 \Rightarrow \cos(2\theta - \alpha) = 1\)
Max value \(\Rightarrow\) equilibrium is unstable