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2021 Paper 3 Q10
D: 1500.0 B: 1500.0
circular motion inextensible string wrapping cartesian coordinates energy conservation tension involute

The origin \(O\) of coordinates lies on a smooth horizontal table and the \(x\)- and \(y\)-axes lie in the plane of the table. A cylinder of radius \(a\) is fixed to the table with its axis perpendicular to the \(x\)--\(y\) plane and passing through \(O\), and with its lower circular end lying on the table. One end, \(P\), of a light inextensible string \(PQ\) of length \(b\) is attached to the bottom edge of the cylinder at \((a, 0)\). The other end, \(Q\), is attached to a particle of mass \(m\), which rests on the table. Initially \(PQ\) is straight and perpendicular to the radius of the cylinder at \(P\), so that \(Q\) is at \((a, b)\). The particle is then given a horizontal impulse parallel to the \(x\)-axis so that the string immediately begins to wrap around the cylinder. At time \(t\), the part of the string that is still straight has rotated through an angle \(\theta\), where \(a\theta < b\).

  1. Obtain the Cartesian coordinates of the particle at this time. Find also an expression for the speed of the particle in terms of \(\theta\), \(\dot{\theta}\), \(a\) and \(b\).
  2. Show that \[ \dot{\theta}(b - a\theta) = u, \] where \(u\) is the initial speed of the particle.
  3. Show further that the tension in the string at time \(t\) is \[ \frac{mu^2}{\sqrt{b^2 - 2aut}}. \]

Solution:

TikZ diagram
  1. The line to the circle is tangent, and the point it meets the circle is \((a \cos \theta, a \sin \theta)\) and it will be a distance \(b - a \theta\) away, therefore it is at \((a \cos \theta - (b-a \theta) \sin \theta, a \sin \theta + (b-a \theta) \cos \theta)\)
  2. The velocity will be \(\displaystyle \binom{-a \dot{\theta}\sin \theta-b \dot{\theta}\cos \theta + a \dot{\theta} \sin \theta + a \theta \dot{\theta} \cos \theta}{ a \dot{\theta} \cos \theta - b \dot{\theta} \sin \theta -a \dot{\theta} \cos \theta + a \theta \dot{\theta} \sin \theta}= \binom{-b \dot{\theta}\cos \theta + a \theta \dot{\theta} \cos \theta}{ - b \dot{\theta} \sin \theta + a \theta \dot{\theta} \sin \theta}\) Therefore the speed will be \(\dot{\theta}(b-a\theta)\)
  3. Conservation of energy and the fact that the tension is perpendicular to the velocity means no work is being done on the particle and hence it's speed is unchanged. So \(u = \dot{\theta}(b-a\theta)\).
  4. Note that the acceleration is \begin{align*} && \mathbf{a} &= \frac{\d}{\d t} \left (-\dot{\theta}(b-a\theta) \binom{\cos \theta}{\sin \theta} \right) \\ &&&=-u \dot{\theta}\binom{-\sin \theta}{\cos \theta} \\ \Rightarrow && T &= ma \\ &&&= \frac{mu^2}{b - a \theta} \end{align*} It would be valuable to have \(\theta\) in terms of \(t\), so we want to solve \begin{align*} &&\frac{\d \theta}{\d t} (b-a\theta) &= u \\ \Rightarrow && b \theta - a\frac{\theta^2}{2} + C &= ut \\ t = 0, \theta = 0: && C &= 0 \\ \Rightarrow && b\theta - \frac{a}{2} \theta^2 &= ut \\ \Rightarrow && \theta &= \frac{b \pm \sqrt{b^2-2aut}}{a} \end{align*} At \(t\) increases, \(\theta\) increases so \(a\theta = b -\sqrt{b^2-2aut}\) or \(b-a \theta = \sqrt{b^2-2aut}\) and the result follows

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2011 Paper 2 Q8
D: 1600.0 B: 1484.7
centre of mass involute parametric curve area integral curve sketching circle integration by parts optimisation

The end \(A\) of an inextensible string \(AB\) of length \(\pi\) is attached to a point on the circumference of a fixed circle of unit radius and centre \(O\). Initially the string is straight and tangent to the circle. The string is then wrapped round the circle until the end \(B\) comes into contact with the circle. The string remains taut during the motion, so that a section of the string is in contact with the circumference and the remaining section is straight. Taking \(O\) to be the origin of cartesian coordinates with \(A\) at \((-1,0)\) and \(B\) initially at \((-1, \pi)\), show that the curve described by \(B\) is given parametrically by \[ x= \cos t + t\sin t\,, \ \ \ \ \ \ y= \sin t - t\cos t\,, \] where \(t\) is the angle shown in the diagram.

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Find the value, \(t_0\), of \(t\) for which \(x\) takes its maximum value on the curve, and sketch the curve. Use the area integral $\displaystyle \int y \frac{\d x}{\d t} \, \d t\,$ to find the area between the curve and the \(x\) axis for~\hbox{\(\pi \ge t \ge t_0\)}. Find the area swept out by the string (that is, the area between the curve described by \(B\) and the semicircle shown in the diagram).

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1992 Paper 3 Q14
D: 1700.0 B: 1500.0
circular motion involute parametric acceleration components friction differential equation string particle on table unwinding

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A horizontal circular disc of radius \(a\) and centre \(O\) lies on a horizontal table and is fixed to it so that it cannot rotate. A light inextensible string of negligible thickness is wrapped round the disc and attached at its free end to a particle \(P\) of mass \(m\). When the string is all in contact with the disc, \(P\) is at \(A\). The string is unwound so that the part not in contact with the disc is taut and parallel to \(OA\). \(P\) is then at \(B\). The particle is projected along the table from \(B\) with speed \(V\) perpendicular to and away from \(OA\). In the general position, the string is tangential to the disc at \(Q\) and \(\angle AOQ=\theta.\) Show that, in the general position, the \(x\)-coordinate of \(P\) with respect to the axes shown in the figure is \(a\cos\theta+a\theta\sin\theta,\) and find \(y\)-coordinate of \(P\). Hence, or otherwise, show that the acceleration of \(P\) has components \(a\theta\dot{\theta}^{2}\) and \(a\dot{\theta}^{2}+a\theta\ddot{\theta}\) along and perpendicular to \(PQ,\) respectively. The friction force between \(P\) and the table is \(2\lambda mv^{2}/a,\) where \(v\) is the speed of \(P\) and \(\lambda\) is a constant. Show that \[ \frac{\ddot{\theta}}{\dot{\theta}}=-\left(\frac{1}{\theta}+2\lambda\theta\right)\dot{\theta} \] and find \(\dot{\theta}\) in terms of \(\theta,\lambda\) and \(a\). Find also the tension in the string when \(\theta=\pi.\)

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