Problems

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 \[ \frac{mu^2}{\sqrt{b^2 - 2aut}}. \]

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Solution:

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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

A heavy particle lies on a smooth horizontal table, and is attached to one end of a light inextensible string of length \(L\). The other end of the string is attached to a point \(P\) on the circumference of the base of a vertical post which is fixed into the table. The base of the post is a circle of radius \(a\) with its centre at a point \(O\) on the table. Initially, at time \(t=0\), the string is taut and perpendicular to the line \(OP.\) The particle is then struck in such a way that the string starts winding round the post and remains taut. At a later time \(t\), a length \(a\theta(t)\ (< L)\) of the string is in contact with the post. Using cartesian axes with origin \(O\), find the position and velocity vectors of the particle at time \(t\) in terms of \(a,L,\theta\) and \(\dot{\theta},\) and hence show that the speed of the particle is \((L-a\theta)\dot{\theta}.\) If the initial speed of the particle is \(v\), show that the particle hits the post at a time \(L^{2}/(2av).\)

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Solution:

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As the string wraps around, the total length in contact will be \(a \theta\). The end contact point will be at \((a\cos \theta, a\sin \theta)\) and the string will be tangential to that. The tangent (unit) vector will be \(\binom{-\sin \theta}{\cos \theta}\), and so the particle will be at \(\binom{a\cos \theta - (L-a\theta) \sin \theta}{a \sin \theta + (L-a \theta) \cos \theta}\). The velocity will be: \begin{align*} \frac{\d}{\d t} \binom{a\cos \theta - (L-a\theta) \sin \theta}{a \sin \theta + (L-a \theta) \cos \theta} &= \binom{-a \sin \theta \cdot \dot{\theta} -(L-a \theta) \cos \theta \cdot \dot{\theta} + a \sin \theta \cdot \dot{\theta} }{a \cos \theta \cdot \dot{\theta} + (L-a \theta) \sin \theta \cdot \dot{\theta} - a \cos \theta \cdot \dot{\theta}} \\ &= \binom{-(L-a \theta) \cos \theta \cdot \dot{\theta} }{ (L-a \theta) \sin \theta \cdot \dot{\theta}} \\ \end{align*} Therefore the speed is \((L-a\theta) \dot{\theta}\). By conservation of energy, we must have that speed is constant, ie: \begin{align*} && (L - a \theta)\dot{\theta} &= v \\ \Rightarrow && \int_0^{L/a} (L - a \theta)\d \theta &= \int_0^T v \d t \\ \Rightarrow && vT &= \frac{L^2}{a} - a\frac{L^2}{2a^2} \\ &&&= \frac{L^2}{2a} \\ \Rightarrow && T &= \frac{L^2}{2av} \end{align*} as requried