Problems

A woman stands in a field at a distance of \(a\,\mathrm{m}\) from the straight bank of a river which flows with negligible speed. She sees her frightened child clinging to a tree stump standing in the river \(b\,\mathrm{m}\) downstream from where she stands and \(c\,\mathrm{m}\) from the bank. She runs at a speed of \(u\,\mathrm{ms}^{-1}\) and swims at \(v\,\mathrm{ms}^{-1}\) in straight lines. Find an equation to be satisfied by \(x,\) where \(x\,\mathrm{m}\) is the distance upstream from the stump at which she should enter the river if she is to reach the child in the shortest possible time. Suppose now that the river flows with speed \(v\) ms\(^{-1}\) and the stump remains fixed. Show that, in this case, \(x\) must satisfy the equation \[ 2vx^{2}(b-x)=u(x^{2}-c^{2})[a^{2}+(b-x)^{2}]^{\frac{1}{2}}. \] For this second case, draw sketches of the woman's path for the three possibilities \(b>c,\) \(b=c\) and \(b< c\).

Show Solution

---

Solution:

+ - ↺

The distance to where she enters the water is \(\sqrt{a^2+(b-x)^2}\) and the distance through the water is \(\sqrt{x^2+c^2}\). The total time will be \(\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{\sqrt{x^2+c^2}}{v}\). To minimise this, we can differentiate. \begin{align*} \frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} + \frac{x}{v \sqrt{x^2+c^2}} &= 0 \\ \Rightarrow && v(b-x)(x^2+c^2)^{\frac12} &= xu(a^2+(b-x)^2)^{\frac12} \end{align*} When she is in the water, she can will move with velocity \(\begin{pmatrix} v \cos \theta \\ v \sin \theta -v \end{pmatrix}\). She needs to travel a distance \(\begin{pmatrix} c \\ -x \end{pmatrix}\), so we must have that \begin{align*} && \frac{x}{c} &= \frac{1-\sin \theta}{\cos \theta} \\ \Rightarrow && \sec \theta - \tan \theta &= \frac{x}{c} \\ \Rightarrow && \sec \theta &= \tan \theta + \frac{x}{c} \\ \Rightarrow && \sec^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\ \Rightarrow && 1 + \tan^2 \theta &= \tan^2 \theta + 2 \tan \theta \frac{x}{c} + \frac{x^2}{c^2} \\ \Rightarrow && \tan \theta &=\frac{c^2-x^2}{2xc} \\ \Rightarrow && \sin \theta &= \frac{c^2-x^2}{c^2+x^2} \\ && \cos \theta &= \frac{2xc}{c^2+x^2} \\ \end{align*} (where we have taken the positive value for \(\cos \theta\) since we must be heading towards the child). Since \(v \cos \theta t = c\) the time taken to reach the child in the water is \(\frac{c}{v} \frac{c^2+x^2}{2xc} = \frac{c^2+x^2}{2xv}\). So the total time is: \(\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{c^2+x^2}{2xv}\). To minimise this, we can differentiate. \begin{align*} \frac{\d}{\d x}: && \frac{-(b-x)}{u\sqrt{a^2+(b-x)^2}} -\frac{c^2}{2vx^2} + \frac{x^2}{2vx^2}&= 0 \\ \Rightarrow && u(x^2-c^2)\sqrt{a^2+(b-x)^2}&= 2vx^2(b-x) \end{align*} as required. When \(b = c\), the shortest path will be running directly to the bank (there's no quicker way to get to the bank) then swimming directly out (and letting the current take you downstream exactly as far as you need)). Therefore the path will be:

+ - ↺

If \(b > c\) then she should run a little downstream first.

+ - ↺

and if \(c > b\) she should actually run a little upstream to take advantage of the current:

+ - ↺

A firework consists of a uniform rod of mass \(M\) and length \(2a\), pivoted smoothly at one end so that it can rotate in a fixed horizontal plane, and a rocket attached to the other end. The rocket is a uniform rod of mass \(m(t)\) and length \(2l(t)\), with \(m(t)=2\alpha l(t)\) and \(\alpha\) constant. It is attached to the rod by its front end and it lies at right angles to the rod in the rod's plane of rotation. The rocket burns fuel in such a way that \(\mathrm{d}m/\mathrm{d}t=-\alpha\beta,\) with \(\beta\) constant. The burnt fuel is ejected from the back of the rocket, with speed \(u\) and directly backwards relative to the rocket. Show that, until the fuel is exhausted, the firework's angular velocity \(\omega\) at time \(t\) satisfies \[ \frac{\mathrm{d}\omega}{\mathrm{d}t}=\frac{3\alpha\beta au}{2[Ma^{2}+2\alpha l(3a^{2}+l^{2})]}. \]

Show Solution

---

Solution:

+ - ↺

The rocket principle states that the thrust generated by the rocket is \(-\frac{\d m}{\d t}u = \alpha \beta u\) This force is acting at a distance \(2a\) from \(O\) and therefore is generating a torque of \(2a \alpha \beta u\) on the system. Let's also consider the moments of inertia about \(O\). The fixed rod will have moment of inertia \(\frac13 M (2a)^2 = \frac43 M a^2\). The rocket will have moment of inertia \(I_{G} + md^2 = \frac1{12}m(t)(2l(t))^2 + m(t) ((2a)^2 + l(t)^2)= \frac43 ml^2+ 4ma^2\). Since our final equation doesn't involve \(m\), lets replace all the \(m\) with \(2al\) to obtain a total \(\displaystyle I = \frac43 Ma^2 + \frac83 \alpha l^3 + 8\alpha la^2\). Since \(\tau\) is constant, we can note that \(I\omega = 2a \alpha \beta u t\) (by integrating) and so \begin{align*} && \dot{\omega} &= \frac{\d }{\d t} \left ( \frac{2a \alpha \beta u t}{ \frac43 Ma^2 + \frac83 \alpha l^3 + 8\alpha la^2} \right) \\ &&&= \frac{\d }{\d t} \left ( \frac{3a \alpha \beta u t}{ 2Ma^2 +4\alpha l^3 + 4 \cdot 3 \cdot \alpha la^2} \right) \\ &&&= \frac{\d }{\d t} \left ( \frac{3a \alpha \beta u t}{ 2[Ma^2 +2\alpha l(l^2 + 3 a^2)]} \right) \\ \end{align*} This is, close, but not quite what they are after since the denominator also has a dependency on \(t\) we wont get exactly what they've asked for

A uniform rod, of mass \(3m\) and length \(2a,\) is freely hinged at one end and held by the other end in a horizontal position. A rough particle, of mass \(m\), is placed on the rod at its mid-point. If the free end is then released, prove that, until the particle begins to slide on the rod, the inclination \(\theta\) of the rod to the horizontal satisfies the equation \[ 5a\dot{\theta}^{2}=8g\sin\theta. \] The coefficient of friction between the particle and the rod is \(\frac{1}{2}.\) Show that, when the particle begins to slide, \(\tan\theta=\frac{1}{26}.\)

Show Solution

---

Solution:

+ - ↺

While the particle is not sliding, we can consider the whole system. Considering the moment of inertia about the end, we have: \begin{align*} I &= \frac13 \cdot 3m \cdot (2a)^2 + m a^2 \\ &= 5ma^2 \end{align*} Taking the level of the pivot as the \(0\) GPE level, the initial energy is \(0\). The energy once it has rotated through an angle \(\theta\) is: \begin{align*} && 0 &= \text{rotational ke} + \text{gpe} \\ &&&= \frac12 I \dot{\theta}^2 - 4mg \sin \theta \\ &&&= \frac12 5am \dot{\theta}^2 -4mg \sin \theta \\ \Rightarrow && 5a\dot{\theta}^2 &= 8g \sin \theta \end{align*} as required. We also have \(5a \ddot{\theta} = 4g \cos \theta\) The acceleration towards the pivot required to maintain circular motion is \(m \frac{v^2}{r} = m a \dot{\theta}^2\). When we are on the point of sliding:

+ - ↺

\begin{align*} \text{N2}(\nearrow): && R - mg\cos \theta &= -ma \ddot{\theta} \\ \Rightarrow && R &= mg \cos \theta - ma \frac{4mg \cos \theta}{5a} \\ &&&= \frac15mg \cos \theta \end{align*} Therefore we must have: \begin{align*} \text{N2}(\nwarrow):&&\mu R - mg \sin \theta &= ma \dot{\theta}^2 \\ && \frac12 \cdot \frac 15 mg \cos \theta &= m \frac{13}5 g \sin \theta \\ \Rightarrow && \tan \theta &= \frac{1}{26} \end{align*}

It is given that the gravitational force between a disc, of radius \(a,\) thickness \(\delta x\) and uniform density \(\rho,\) and a particle of mass \(m\) at a distance \(b(\geqslant0)\) from the disc on its axis is \[ 2\pi mk\rho\delta x\left(1-\frac{b}{(a^{2}+b^{2})^{\frac{1}{2}}}\right), \] where \(k\) is a constant. Show that the gravitational force on a particle of mass \(m\) at the surface of a uniform sphere of mass \(M\) and radius \(r\) is \(kmM/r^{2}.\) Deduce that in a spherical cloud of particles of uniform density, which all attract one another gravitationally, the radius \(r\) and inward velocity \(v=-\dfrac{\d r}{\d t}\) of a particle at the surface satisfy the equation \[ v\frac{\mathrm{d}v}{\mathrm{d}r}=-\frac{kM}{r^{2}}, \] where \(M\) is the mass of the cloud. At time \(t=0\), the cloud is instantaneously at rest and has radius \(R\). Show that \(r=R\cos^{2}\alpha\) after a time \[ \left(\frac{R^{3}}{2kM}\right)^{\frac{1}{2}}(\alpha+\tfrac{1}{2}\sin2\alpha). \]

A patient arrives with blue thumbs at the doctor's surgery. With probability \(p\) the patient is suffering from Fenland fever and requires treatment costing \(\pounds 100.\) With probability \(1-p\) he is suffering from Steppe syndrome and will get better anyway. A test exists which infallibly gives positive results if the patient is suffering from Fenland fever but also has probability \(q\) of giving positive results if the patient is not. The test cost \(\pounds 10.\) The doctor decides to proceed as follows. She will give the test repeatedly until either the last test is negative, in which case she dismisses the patient with kind words, or she has given the test \(n\) times with positive results each time, in which case she gives the treatment. In the case \(n=0,\) she treats the patient at once. She wishes to minimise the expected cost \(\pounds E_{n}\) to the National Health Service.

1. Show that \[ E_{n+1}-E_{n}=10p-10(1-p)q^{n}(9-10q), \] and deduce that if \(p=10^{-4},q=10^{-2},\) she should choose \(n=3.\)
2. Show that if \(q\) is larger than some fixed value \(q_{0},\) to be determined explicitly, then whatever the value of \(p,\) she should choose \(n=0.\)

1. \(X_{1},X_{2},\ldots,X_{n}\) are independent identically distributed random variables drawn from a uniform distribution on \([0,1].\) The random variables \(A\) and \(B\) are defined by \[ A=\min(X_{1},\ldots,X_{n}),\qquad B=\max(X_{1},\ldots,X_{n}). \] For any fixed \(k\), such that \(0< k< \frac{1}{2},\) let \[ p_{n}=\mathrm{P}(A\leqslant k\mbox{ and }B\geqslant1-k). \] What happens to \(p_{n}\) as \(n\rightarrow\infty\)? Comment briefly on this result.
2. Lord Copper, the celebrated and imperious newspaper proprietor, has decided to run a lottery in which each of the \(4,000,000\) readers of his newspaper will have an equal probability \(p\) of winning \(\pounds 1,000,000\) and their changes of winning will be independent. He has fixed all the details leaving to you, his subordinate, only the task of choosing \(p\). If nobody wins \(\pounds 1,000,000\), you will be sacked, and if more than two readers win \(\pounds 1,000,000,\) you will also be sacked. Explaining your reasoning, show that however you choose \(p,\) you will have less than a 60\% change of keeping your job.