M4 June 2005 Q5

EdexcelOld spec12 marksHooke's Law Work-Energy Principle

5. A non-uniform rod \(BC\) has mass \(m\) and length \(3l\). The centre of mass of the rod is at distance \(l\) from \(B\). The rod can turn freely about a fixed smooth horizontal axis through \(B\). One end of a light elastic string, of natural length \(l\) and modulus of elasticity \(\dfrac{mg}{6}\), is attached to \(C\). The other end of the string is attached to a point \(P\) which is at a height \(3l\) vertically above \(B\).

(a) Show that, while the string is stretched, the potential energy of the system is \[mgl(\cos^2\theta - \cos\theta) + \text{constant},\] where \(\theta\) is the angle between the string and the downward vertical and \(-\dfrac{\pi}{2} \lt \theta \lt \dfrac{\pi}{2}\). (6)

(b) Find the values of \(\theta\) for which the system is in equilibrium with the string stretched. (6)

P a height 3l above B; rod BC with G at distance l from B and 2l from C; angle theta at P and at C; angle 2theta between BG and the downward vertical; horizontal through B marked as zero PE — Mark scheme (a)

Scheme	Marks
GPE \(= -mgl\cos 2\theta\)	B1
EPE \(= \dfrac{mg}{6}\,\dfrac{(6l\cos\theta - l)^2}{2l}\)	M1
\(= \dfrac{mg}{12l}\left(36l^2\cos^2\theta - 12l^2\cos\theta + l^2\right)\)	M1
\(= mgl\left(3\cos^2\theta - \cos\theta\right) + C\)
\(V = -mgl\left(2\cos^2\theta - 1\right) + mgl\left(3\cos^2\theta - \cos\theta\right) + c\)	M1 M1 use of \(\cos 2\theta = \ldots\)
\(= mgl\left(\cos^2\theta - \cos\theta\right) + c^{\prime}\ \ *\)	A1
	(6)

Mark scheme (b)

Scheme	Marks
\(\dfrac{\mathrm{d}V}{\mathrm{d}\theta} = mgl(-2\cos\theta\sin\theta + \sin\theta) = 0\)	M1 A1; M1
\(\sin\theta(-2\cos\theta + 1) = 0\)
\(\sin\theta = 0\) or \(\cos\theta = \tfrac{1}{2}\)	M1
\(\theta = 0\) or \(\theta = \pm\dfrac{\pi}{3}\)	A1 A1
	(6)
	(12 marks)

Notes