M4 June 2013 (R) Q6

EdexcelOld spec16 marksHooke's Law Work-Energy Principle

Figure 2: rod AB hinged at A making angle 2 theta with the upward vertical AE, AE = 3l, AD = 3l, string from E to D, particle km at B — Figure 2

A uniform rod \(AB\) has mass \(4m\) and length \(4l\). The rod can turn freely in a vertical plane about a fixed smooth horizontal axis through \(A\). A particle of mass \(km\), where \(k < 7\), is attached to the rod at \(B\). One end of a light elastic string, of natural length \(l\) and modulus of elasticity \(4mg\), is attached to the point \(D\) of the rod, where \(AD = 3l\). The other end of the string is attached to a fixed point \(E\) which is vertically above \(A\), where \(AE = 3l\), as shown in Figure 2. The angle between the rod and the upward vertical is \(2\theta\), where \(\arcsin\left(\dfrac{1}{6}\right) < \theta \leqslant \dfrac{\pi}{2}\).

(a) Show that, while the string is stretched, the potential energy of the system is \[8mgl\{(7 - k)\sin^2\theta - 3\sin\theta\} + \text{constant}\] (6)

There is a position of equilibrium with \(\theta \leqslant \dfrac{\pi}{6}\).

(b) Show that \(k \leqslant 4\) (5)

Given that \(k = 4\),

Rod AB with AE = 3l, string ED, weight 4mg at 2l from A, D at 3l, weight at B — Mark scheme (a)

Scheme	Marks
Length of string \(= 2 \times 3l\sin\theta\)	B1
Extension \(= 6l\sin\theta - l\)
E.P.E. \(= \dfrac{4mg}{2l}(6l\sin\theta - l)^2\)
G.P.E. of rod \(= 4mg \times 2l\cos 2\theta\)
G.P.E. of mass at \(B\) \(= kmg \times 4l\cos 2\theta\)
\(V = \dfrac{4mg}{2l}(6l\sin\theta - l)^2 + 8mgl\cos 2\theta + 4kmgl\cos 2\theta +\) const	M1 A2
\(V = \dfrac{4mg}{2l}(6l\sin\theta - l)^2 + 8mgl\left(1 - 2\sin^2\theta\right) + 4kmgl\cos 2\theta +\) const	M1
\(= 2mgl\left(36\sin^2\theta - 12\sin\theta - 8\sin^2\theta - 4k\sin^2\theta\right) +\) const
\(= 8mgl\left((7 - k)\sin^2\theta - 3\sin\theta\right) +\) constant	A1
	(6)

Mark scheme (b)

Scheme	Marks
\(\dfrac{\mathrm{d}V}{\mathrm{d}\theta} = 8mgl\left(2(7 - k)\sin\theta\cos\theta - 3\cos\theta\right)\)	M1
\(\dfrac{\mathrm{d}V}{\mathrm{d}\theta} = 0 \qquad \left(2(7 - k)\sin\theta - 3\right)\cos\theta = 0\)	M1
\(\sin\theta = \dfrac{3}{2(7 - k)} \qquad (\text{or } \cos\theta = 0, \text{ need not be seen})\)	A1
\(\theta \leqslant \frac{\pi}{6} \quad \Rightarrow \dfrac{3}{2(7 - k)} \leqslant \frac{1}{2}\)	M1
\(3 \leqslant 7 - k \quad k \leqslant 4 \qquad *\)	A1
	(5)

Notes

M1 Differentiate

M1 Set derivative = 0

M1 Use of \(\sin\theta \leqslant \dfrac{1}{2}\)

Mark scheme (c)

Scheme	Marks
\(k = 4\ \Rightarrow \theta = \dfrac{\pi}{6}\)	B1
\(\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = 8mgl\left[6\cos^2\theta - (6\sin\theta - 3)\sin\theta\right]\)	M1
\(= 8mgl\left[6 \times \left(\dfrac{\sqrt{3}}{2}\right)^2 - 6 \times \left(\dfrac{1}{2}\right)^2 + 3 \times \dfrac{1}{2}\right]\)	A1
\(\dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} > 0\)	M1
\(V\) is min. \(\quad \therefore\) stable equilibrium	A1
	(5)
	(16 marks)

Notes

M1 Second derivative (\(8mgl\) or \(24mgl\) not needed) [or differentiate \(8mgl(3\sin 2\theta - 3\cos\theta)\)]

A1 Numerical unsimplified

M1 by numerical evaluation or justification from trig terms \((36mgl)\)

A1 CSO