Problems

One end of a thin heavy uniform inextensible perfectly flexible rope of length \(2L\) and mass \(2M\) is attached to a fixed point \(P\). A particle of mass \(m\) is attached to the other end. Initially, the particle is held at \(P\) and the rope hangs vertically in a loop below \(P\). The particle is then released so that it and a section of the rope (of decreasing length) fall vertically as shown in the diagram.

A comet in deep space picks up mass as it travels through a large stationary dust cloud. It is subject to a gravitational force of magnitude \(M\!f\) acting in the direction of its motion. When it entered the cloud, the comet had mass \(M\) and speed \(V\). After a time \(t\), it has travelled a distance \(x\) through the cloud, its mass is \(M(1+bx)\), where~\(b\) is a positive constant, and its speed is \(v\).

1. In the case when \(f=0\), write down an equation relating \(V\), \(x\), \(v\) and \(b\). Hence find an expression for \(x\) in terms of \(b\), \(V\) and \(t\).
2. In the case when \(f\) is a non-zero constant, use Newton's second law in the form \[ \text{force} = \text{rate of change of momentum} \] to show that \[ v = \frac{ft+V}{1+bx}\,. \] Hence find an expression for \(x\) in terms of \(b\), \(V\), \(f\) and \(t\). Show that it is possible, if \(b\), \(V\) and \(f\) are suitably chosen, for the comet to move with constant speed. Show also that, if the comet does not move with constant speed, its speed tends to a constant as \(t\to\infty\).

A chain of mass \(m\) and length \(l\) is composed of \(n\) small smooth links. It is suspended vertically over a horizontal table with its end just touching the table, and released so that it collapses inelastically onto the table. Calculate the change in momentum of the \((k+1)\)th link from the bottom of the chain as it falls onto the table. Write down an expression for the total impulse sustained by the table in this way from the whole chain. By approximating the sum by an integral, show that this total impulse is approximately \[ {\textstyle \frac23} m \surd(2gl) \] when \(n\) is large.

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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 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 \[ \frac{\mathrm{d}\omega}{\mathrm{d}t}=\frac{3\alpha\beta au}{2[Ma^{2}+2\alpha l(3a^{2}+l^{2})]}. \]

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Solution:

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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