Problems

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

One end of a thin uniform inextensible, but perfectly flexible, string of length \(l\) and uniform mass per unit length is held at a point on a smooth table a distance \(d(< l)\) away from a small vertical hole in the surface of the table. The string passes through the hole so that a length \(l-d\) of the string hangs vertically. The string is released from rest. Assuming that the height of the table is greater than \(l\), find the time taken for the end of the string to reach the top of the hole.

A librarian wishes to pick up a row of identical books from a shelf, by pressing her hands on the outer covers of the two outermost books and lifting the whole row together. The covers of the books are all in parallel vertical planes, and the weight of each book is \(W\). With each arm, the librarian can exert a maximum force of \(P\) in the vertical direction, and, independently, a maximum force of \(Q\) in the horizontal direction. The coefficient of friction between each pair of books and also between each hand and a book is \(\mu.\) Derive an expression for the maximum number of books that can be picked up without slipping, using this method. {[}You may assume that the books are thin enough for the rotational effect of the couple on each book to be ignored.{]}

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Solution:

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The force acting vertically on each of the outer books must be (by symmetry) \(\frac{nW}{2}\). The force acting horizontally on the outer books (and between each book in the horizontal direction) will be the same (we might as well say \(Q\) since increasing this force doesn't make any task less achievable. Looking at an end book, it will have force \(\frac{nW}{2}\) acting on one side, but it this force needs to not slip, ie \(\frac{nW}{2} \leq \mu Q\) \begin{align*} && \frac{nW}{2} &\leq \mu Q \\ \Rightarrow && n &\leq \frac{2\mu Q}{W} \\ && \frac{nW}{2} & \leq P \\ && n & \leq \frac{2P}{W} \\ \Rightarrow && n &\leq \frac2{W}\min \left (P, \mu Q \right) \end{align*}

Two particles of mass \(M\) and \(m\) \((M>m)\) are attached to the ends of a light rod of length \(2l.\) The rod is fixed at its midpoint to a point \(O\) on a horizontal axle so that the rod can swing freely about \(O\) in a vertical plane normal to the axle. The axle rotates about a vertical axis through \(O\) at a constant angular speed \(\omega\) such that the rod makes a constant angle \(\alpha\) \((0<\alpha<\frac{1}{2}\pi)\) with the vertical. Show that \[ \omega^{2}=\left(\frac{M-m}{M+m}\right)\frac{g}{l\cos\alpha}. \] Show also that the force of reaction of the rod on the axle is inclined at an angle \[ \tan^{-1}\left[\left(\frac{M-m}{M+m}\right)^{2}\tan\alpha\right] \] with the downward vertical.

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Solution:

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The accelerations of \(M\) and \(m\) are \(l \sin \alpha \omega^2\) and \(-l \sin \alpha \omega^2\) so the forces \(R_M\) and \(R_m\) are \(M\binom{l \sin \alpha \omega^2}{g}, \,m \binom{-l \sin \alpha \omega^2}{g}\). Since the axle is rotating freely, the moment about \(O\) for the rod must be \(O\). The moment for \(M\) will be \(M\binom{l \sin \alpha\omega^2}{g} \cdot \binom{-l\cos \alpha}{l \sin \alpha} = lM\sin\alpha (g - l \cos \alpha\omega^2)\). The moment for \(m\) will be \(m \binom{-l \sin \alpha\omega^2}{g} \cdot \binom{-l\cos \alpha\omega^2}{l \sin \alpha} = lm \sin \alpha(g+l \cos \alpha\omega^2)\) Therefore \begin{align*} && lM\sin\alpha (g - l \cos \alpha\omega^2) &= lm \sin \alpha(g+l \cos \alpha\omega^2) \\ && M(g - l \cos \alpha \omega^2) &= m(g + l \cos \alpha \omega^2 ) \\ \Rightarrow && g(M-m) &= l \cos \alpha (M+m) \omega^2 \\ \Rightarrow && \omega^2 &= \left (\frac{M-m}{M+m} \right) \frac{g}{l \cos \alpha} \end{align*} as required. The total force on the rod is \(\mathbf{0}\) so the reaction force must be \(M\binom{l \sin \alpha \omega^2}{g}+ \,m \binom{-l \sin \alpha \omega^2}{g} = \binom{l \sin \alpha \omega^2 (M-m)}{(M+m)g}\) Therefore the angle this makes with downward vertical is: \begin{align*} \theta &= \tan^{-1} \left ( \frac{l \sin \alpha \omega^2 (M-m)}{(M+m)g} \right) \\ &= \tan^{-1} \left ( \frac{l \sin \alpha (M-m)}{(M+m)g} \omega^2\right) \\ &= \tan^{-1} \left ( \frac{l \sin \alpha (M-m)}{(M+m)g} \left (\frac{M-m}{M+m} \right) \frac{g}{l \cos \alpha}\right) \\ &= \tan^{-1}\left[\left(\frac{M-m}{M+m}\right)^{2}\tan\alpha\right] \end{align*} as required.

An examination consists of several papers, which are marked independently. The mark given for each paper can be an integer from \(0\) to \(m\) inclusive, and the total mark for the examination is the sum of the marks on the individual papers. In order to make the examination completely fair, the examiners decide to allocate the mark for each paper at random, so that the probability that any given candidate will be allocated \(k\) marks \((0\leqslant k\leqslant m)\) for a given paper is \((m+1)^{-1}\). If there are just two papers, show that the probability that a given candidate will receive a total of \(n\) marks is \[ \frac{2m-n+1}{\left(m+1\right)^{2}} \] for \(m< n\leqslant2m\), and find the corresponding result for \(0\leqslant n\leqslant m\). If the examination consists of three papers, show that the probability that a given candidate will receive a total of \(n\) marks is \[ \frac{6mn-4m^{2}-2n^{2}+3m+2}{2\left(m+1\right)^{2}} \] in the case \(m< n\leqslant2m\). Find the corresponding result for \(0\leqslant n\leqslant m\), and deduce the result for \(2m< n\leqslant3m\).

Find the probability that the quadratic equation \[ X^{2}+2BX+1=0 \] has real roots when \(B\) is normally distributed with zero mean and unit variance. Given that the two roots \(X_{1}\) and \(X_{2}\) are real, find:

1. the probability that both \(X_{1}\) and \(X_{2}\) are greater than \(\frac{1}{5}\);
2. the expected value of \(\left|X_{1}+X_{2}\right|\);