Problems

A rough ring of radius \(a\) is fixed so that it lies in a plane inclined at an angle \(\alpha\) to the horizontal. A uniform heavy rod of length \(b(>a)\) has one end smoothly pivoted at the centre of the ring, so that the rod is free to move in any direction. It rests on the circumference of the ring, making an angle \(\theta\) with the radius to the highest point on the circumference. Find the relation between \(\alpha,\theta\) and the coefficient of friction, \(\mu,\) which must hold when the rod is in limiting equilibrium.

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Solution:

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It is important to define clear coordinate axes, so let the \(x\)-axis point up the line of greatest slope of the ring. The \(z\)-axis perpendicular to the ring, and the \(y\)-axis perpendicular to both of these. Our method is going to be to take moments about \(O\) to avoid worrying about the force at the pivot. There are \(3\) forces we need to worry about:

• The mass of the rod
• The reaction where it meets the ring
• The friction at the ring

In our coordinate frame, the reaction will act in the \(z\)-direction, \(\displaystyle \begin{pmatrix} 0 \\ 0 \\ R \end{pmatrix}\), the friction force will act in the \(x-y\) plane: \(\displaystyle \begin{pmatrix} \mu R \sin \theta \\ -\mu R \cos \theta \\ 0 \end{pmatrix}\). We don't know the mass, but we know it will be acting "vertically", so \(\cos \alpha\) of it will act in the \(z\)-axis and \(\sin \alpha\) will act in the \(y\)-axis, ie it will act parallel to \(\displaystyle \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix}\). When taking moments, we need to consider \(\mathbf{r}\) the direction of the rod. This will be \(\displaystyle \begin{pmatrix} \cos \theta \\ \sin \theta \\ 0 \end{pmatrix}\). The moment of the weight will all be parallel to \(\mathbf{r} \times \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix}\). Similarly the moments of the contact forces will be \(\mathbf{r} \times \begin{pmatrix} \mu R \sin \theta \\ -\mu R \cos \theta \\ R \end{pmatrix}\). Since these moments sum to \(\mathbf{0}\) as we are in equilibrium, these vectors must be parallel. Therefore it is sufficient to check the vector triple product, \begin{align*} && 0 &= \begin{pmatrix} \cos \theta \\ \sin \theta \\ 0 \end{pmatrix} \cdot \left ( \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix} \times \begin{pmatrix} \mu \sin \theta \\ -\mu \cos \theta \\ 1 \end{pmatrix} \right ) \\ &&&= \cos \theta (\mu \cos \theta \cos \alpha)-\sin \theta (\sin \alpha - \mu \sin \theta \cos \alpha) \\ &&&= \mu((\sin^2 \theta+\cos^2 \theta) \cos \alpha) -\sin \theta \sin \alpha \\ \Rightarrow && \mu &= \tan \alpha \sin \theta \end{align*}

A long, inextensible string passes through a small fixed ring. One end of the string is attached to a particle of mass \(m,\) which hangs freely. The other end is attached to a bead also of mass \(m\) which is threaded on a smooth rigid wire fixed in the same vertical plane as the ring. The curve of the wire is such that the system can be in static equilibrium for all positions of the bead. The shortest distance between the wire and the ring is \(d(>0).\) Using plane polar coordinates centred on the ring, find the equation of the curve. The bead is set in motion. Assuming that the string remains taut, show that the speed of the bead when it is a distance \(r\) from the ring is \[ \left(\frac{r}{2r-d}\right)^{\frac{1}{2}}v, \] where \(v\) is the speed of the bead when \(r=d.\)

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Solution:

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Assume the total length of the string is \(l\). Then the total energy of the system (when nothing is moving) for a given \(\theta\) is: \(mg(r-l) + mgr \sin \theta\) Since for a point in static equilibrium, the derivative of this must be \(0\), this must be constant. So: \(r\l \sin \theta + 1\r = C \Rightarrow r = \frac{C}{1+\sin \theta}\) \(r\) will be smallest when \(\sin \theta = 1\), ie in polar coordinates, the equation should be \(r = \frac{2d}{1+\sin \theta}\) Alternatively, by considering forces, the shape must be a parabola with the ring at the focus. Considering the bead, it will have speed of \(r \dot{\theta}\) tangentially, and \(-\dot{r}\). The other particle will have speed \(\dot{r}\). Differentiating wrt to \(t\) \begin{align*} && 0 &= \dot{r}(\sin \theta + 1) + r \dot{\theta} \cos \theta \\ \Rightarrow && \dot{\theta} &= \frac{-\dot{r}(1+\sin \theta)}{r \cos \theta} \\ &&&= \frac{-\dot{r} 2d}{r^2 \sqrt{1-\l \frac{2d}{r}-1\r^2}} \\ &&&= \frac{-2d\dot{r}}{r^2\sqrt{\frac{r^2-(2d-r)^2}{r^2}}} \\ &&&= \frac{-d\dot{r}}{r\sqrt{dr-d^2}} \end{align*} By conservation of energy (since GPE is constant throughout the system, KE must be constant): \begin{align*} && \frac12 m (r^2 \dot{\theta}^2+\dot{r}^2) +\frac12 m \dot{r}^2 &= \frac12mv^2 \\ \Rightarrow && v^2 &= r^2 \dot{\theta}^2 + 2\dot{r}^2 \\ &&&= r^2 \frac{d^2\dot{r}^2}{r^2(dr-d^2)} + 2\dot{r}^2 \\ &&&= \dot{r}^2 \l \frac{d }{r-d} + 2 \r \\ &&&= \dot{r}^2 \l \frac{2r-d}{r-d} \r \\ \Rightarrow && v &= \dot{r} \l \frac{2r-d}{r-d} \r^{\frac12} \\ \Rightarrow && \dot{r} &= \l \frac{r-d}{2r-d} \r^{\frac12} v \\ \Rightarrow && u^2 &= r^2 \dot{\theta}^2+\dot{r}^2\\ &&&= \dot{r}^2 \l \frac{d }{r-d} + 1 \r \\ &&&= \l \frac{r-d}{2r-d} \r \l \frac{r}{r-d} \r v^2 \\ &&&= \l \frac{d}{2r-d} \r v^2 \\ \Rightarrow && u &= \l \frac{d}{2r-d} \r^{\frac12} v \end{align*}

Ice snooker is played on a rectangular horizontal table, of length \(L\) and width \(B\), on which a small disc (the puck) slides without friction. The table is bounded by smooth vertical walls (the cushions) and the coefficient of restitution between the puck and any cushion is \(e\). If the puck is hit so that it bounces off two adjacent cushions, show that its final path (after two bounces) is parallel to its original path. The puck rests against the cushion at a point which divides the side of length \(L\) in the ratio \(z:1\). Show that it is possible, whatever \(z\), to hit the puck so that it bounces off the three other cushions in succession clockwise and returns to the spot at which it started. By considering these paths as \(z\) varies, explain briefly why there are two different ways in which, starting at any point away from the cushions, it is possible to perform a shot in which the puck bounces off all four cushions in succession clockwise and returns to its starting point.

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The puck sets off at some velocity \(\displaystyle \binom{u_x}{u_y}\), after the first bounce off the wall parallel to the \(y\)-axis, it has velocity \(\displaystyle \binom{-eu_x}{u_y}\). After it bounces off the wall parallel to the \(x\)-axis, it has velocity \(\displaystyle \binom{-eu_x}{-eu_y}\) which is clearly parallel to the original velocity.

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If the puck bounces off 3 walls and returns to the same point the shape formed must be a parallelogram. We need to hit the point on the opposite side which is in a ratio of \(1:z\), but this must be possible if we aim towards the side further away.

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For a fixed path, as \(z\) increases we generate more parallelograms which cross ours (on two of the legs) twice. As they move the full length it will cover the full leg of the parallogram. Similarly going the other way will cover the other leg of the parallelogram. Therefore from every point there are two circuits round the table

A thin uniform elastic band of mass \(m,\) length \(l\) and modulus of elasticity \(\lambda\) is pushed on to a smooth circular cone of vertex angle \(2\alpha,\) in such a way that all elements of the band are the same distance from the vertex. It is then released from rest. Let \(x(t)\) be the length of the band at time \(t\) after release, and let \(t_{0}\) be the time at which the band becomes slack. Assuming that a small element of the band which subtends an angle \(\delta\theta\) at the axis of the cone experiences a force, due to the tension \(T\) in the band, of magnitude \(T\delta\theta\) directed towards the axis, and ignoring the effects of gravity, show that \[ \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+\frac{4\pi^{2}\lambda}{ml}(x-l)\sin^{2}\alpha=0,\qquad(0< t< t_{0}). \] Find the value of \(t_{0}.\)

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\begin{align*} \text{N2}(\nwarrow): && T\delta \theta \sin \alpha &= -m\frac{\delta \theta}{2\pi} \ddot{d} \end{align*} Notice that \(r = d \sin \alpha\) and \(x = 2 \pi r\), so \(x = 2\pi d \sin \alpha\) and \(\ddot{x} = 2\pi \sin \alpha \ddot{d} \Rightarrow \ddot{d} = \ddot{x} \frac{1}{2 \pi \sin \alpha}\) Notice also that \(T = \frac{\lambda}{l}(x-l)\) so. \begin{align*} && \frac{m}{4 \pi^2 \sin\alpha} \ddot{x} &= -\frac{\lambda}{l}(x-l) \sin\alpha \\ \Rightarrow && \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+\frac{4\pi^{2}\lambda}{ml}(x-l)\sin^{2}\alpha&=0 \end{align*} The solution to the differential equation we have is: \begin{align*} && x(t) &= A \sin \left (\sqrt{\frac{4 \pi^2 \lambda}{ml}\sin^2 \alpha} \cdot t \right) + B \sin \left (\sqrt{\frac{4 \pi^2 \lambda}{ml}\sin^2 \alpha} \cdot t \right) + l \\ &&&= A \sin \left (2 \pi \sin \alpha\sqrt{\frac{ \lambda}{ml}} \cdot t \right) +B \sin \left (2 \pi \sin \alpha\sqrt{\frac{ \lambda}{ml}} \cdot t \right) + l\\ && \dot{x}(0) = 0 \\ \Rightarrow && B &= 0 \\ && x(t) &= (x(0)-l) \sin \left (2 \pi \sin \alpha\sqrt{\frac{ \lambda}{ml}} \cdot t \right) + l \\ && x(t_0) &= l \\ \Rightarrow && t_0 &= \frac{1}{4\sin \alpha} \sqrt{\frac{ml}{\lambda}} \end{align*}

A train of length \(l_{1}\) and a lorry of length \(l_{2}\) are heading for a level crossing at speeds \(u_{1}\) and \(u_{2}\) respectively. Initially the front of the train and the front of the lorry are at distances \(d_{1}\) and \(d_{2}\) from the crossing. Find conditions on \(u_{1}\) and \(u_{2}\) under which a collision will occur. On a diagram with \(u_{1}\) and \(u_{2}\) measured along the \(x\) and \(y\) axes respectively, shade in the region which represents collision. Hence show that if \(u_{1}\) and \(u_{2}\) are two independent random variables, both uniformly distributed on \((0,V)\), then the probability of a collision in the case when initially the back of the train is nearer to the crossing than the front of the lorry is \[ \frac{l_{1}l_{2}+l_{2}d_{1}+l_{1}d_{2}}{2d_{2}\left(l_{2}+d_{2}\right)}. \] Find the probability of a collision in each of the other two possible cases.

My two friends, who shall remain nameless, but whom I shall refer to as \(P\) and \(Q\), both told me this afternoon that there is a body in my fridge. I'm not sure what to make of this, because \(P\) tells the truth with a probability of only \(p\), while \(Q\) (independently) tells the truth with probability \(q\). I haven't looked in the fridge for some time, so if you had asked me this morning, I would have said that there was just as likely to be a body in it as not. Clearly, in view of what \(P\) and \(Q\) told me, I must revise this estimate. Explain carefully why my new estimate of the probability of there being a body in the fridge should be \[ \frac{pq}{1-p-q+2pq}. \] I have now been to look in the fridge, and there is indeed a body in it; perhaps more than one. It seems to me that only my enemy \(A\), or my enemy \(B\), or (with a bit of luck) both \(A\) and \(B\) could be in my fridge, and this morning I would have judged these three possibilities to be equally likely. But tonight I asked \(P\) and \(Q\) separately whether or not \(A\) was in the fridge, and they each said that he was. What should be my new estimate of the probability that both \(A\) and \(B\) are in my fridge? Of course, I tell the truth always.