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2001 Paper 1 Q11
D: 1484.0 B: 1469.5
energy conservation elastic string Hooke's law circular motion cylinder optimisation maximum speed geometry modulus of elasticity

A smooth cylinder with circular cross-section of radius \(a\) is held with its axis horizontal. A~light elastic band of unstretched length \(2\pi a\) and modulus of elasticity \(\lambda\) is wrapped round the circumference of the cylinder, so that it forms a circle in a plane perpendicular to the axis of the cylinder. A particle of mass \(m\) is then attached to the rubber band at its lowest point and released from rest.

  1. Given that the particle falls to a distance \(2a\) below the below the axis of the cylinder, but no further, show that \[ \lambda = \frac{9\pi m g}{(3\sqrt3-\pi)^2} \;. \]
  2. Given instead that the particle reaches its maximum speed at a distance \(2a\) below the axis of the cylinder, find a similar expression for \(\lambda\)\,.

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1996 Paper 1 Q9
D: 1500.0 B: 1485.6
energy conservation elastic string Hooke's law bungee jumping quadratic equation modulus of elasticity maximum speed optimisation

A bungee-jumper of mass \(m\) is attached by means of a light rope of natural length \(l\) and modulus of elasticity \(mg/k,\) where \(k\) is a constant, to a bridge over a ravine. She jumps from the bridge and falls vertically towards the ground. If she only just avoids hitting the ground, show that the height \(h\) of the bridge above the floor of the ravine satisfies \[ h^{2}-2hl(k+1)+l^{2}=0, \] and hence find \(h.\) Show that the maximum speed \(v\) which she attains during her fall satisfies \[ v^{2}=(k+2)gl. \]

Solution: \begin{align*} && \text{Energy at the top} &= mgh \\ && \text{Energy at the bottom} &= \frac12\frac{\lambda (h-l)^2}{l} \\ \Rightarrow && mgh & = \frac{\frac{mg}{k}(h-l)^2}{2l} \\ \Rightarrow && 2hkl &= (h-l)^2 \\ \Rightarrow && 0 &= h^2-2lh-2hlk+l^2 \\ &&0&= h^2-2hl(k+1)+l^2 \\ \Rightarrow && \frac{h}{l} &= \frac{2(k+1)\pm \sqrt{4(k+1)^2-4}}{2} \\ &&&= (k+1) \pm \sqrt{k^2+2k} \\ \Rightarrow && h &= l \left ( (k+1) \pm \sqrt{k^2+2k} \right) \end{align*} Since the negative root is less than \(1\), she would have not fully extended the cord. Therefore \(h = l \left ( (k+1) + \sqrt{k^2+2k} \right)\) Her maximum speed will be when her acceleration is \(0\), ie \(g = \text{force from cord}\) ie \(mg = \frac{\lambda x}{l}\) or \(x = \frac{mgl}{\lambda} = \frac{mglk}{mg} = kl\). At this point by conservation of energy we will have \begin{align*} && mgh &= mg(h-l-x) + \frac12 m v^2+\frac{1}{2} \frac{mgx^2}{kl} \\ \Rightarrow && mg\left ( l + kl \right) &= \frac12 m v^2 + \frac12 \frac{mgl^2k^2}{kl} \\ \Rightarrow && 2g\left ( l + kl \right) &= v^2 + glk \\ \Rightarrow && v^2 &= gl(2+k) \end{align*}

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