A yacht sails North with speed \(V\) into a wind of speed \(W\) coming from \(\theta^\circ\) East of North. Relative to the yacht, what is the speed and direction of the wind? By correctly setting the sail, it is possible to obtain a wind force on the yacht proportional to the square of the relative speed of the wind (with a constant of proportionality \(C_L\)) in a direction perpendicular to the relative direction of the wind. The sideways component of this wind force is absorbed by a fin under the yacht and causes negligible sideways drift of the yacht. When the yacht is travelling at a constant velocity the forwards component of the wind force balances the drag of the water on the hull, which is proportional to the square of the speed of the yacht with a constant of proportionality \(C_D\). Show that in the steady motion \begin{align*} C_D V^2 = C_L(W^2 + 2WV\cos\theta + V^2)W\sin\theta. \end{align*} By sketching a graph of the right-hand side of the above equation as a function of \(V\), or otherwise, show that there is just one solution for \(V\). Find its approximate value when \(C_D \gg C_L\) and when \(C_D \ll C_L\).
A vibrating carbon dioxide molecule can be thought of as three particles constrained to move along a line, the outer two particles each of mass 16 units being joined to the central particle of mass 12 units by identical springs. If the displacements of the three particles from their equilibrium positions are \(x_1\), \(x_2\) and \(x_3\) (\(x_2\) referring to the central particle), write down the equation of motion for each particle. Show that these equations can be satisfied by two modes of vibration \begin{align*} \text{(I)} \quad &x_1 = \cos \omega t, \quad x_2 = 0, \quad x_3 = -\cos \omega t\\ \text{and} \quad \text{(II)} \quad &x_1 = \cos \Omega t, \quad x_2 = -A \cos \Omega t, \quad x_3 = \cos \Omega t \end{align*} with suitable choices of \(\omega\), \(\Omega\) and \(A\). Show that the ratio of the frequencies of the two modes is \(\sqrt{\frac{4}{3}}\).
Two perfectly elastic balls collide without loss of energy. Show that the relative speed of the balls is the same before and after the collision. A child has a collection of perfectly elastic balls of various sizes. He arranges three of them separated by tiny gaps in a vertical line, the lightest being at the top, at a height much greater than any diameter and drops them simultaneously on to a rigid floor. Show that if the ratio of the masses is 1 : 2 : 6 the lightest ball rises to a height of \(9h\), the other balls being at rest. Show further that with a different choice of sizes the upper ball can move to a height of almost \(49h\).
When a soap film is punctured, a circular hole grows rapidly under the action of surface tension. It is observed that the mass of the film from the hole is concentrated on the rim of the hole and is spread evenly around the rim. Let the soap film, before being punctured, have a thickness \(h\) and a density \(\rho\), and let the radius of the hole at time \(t\) be \(r(t)\). How much mass is there in that segment of the rim which subtends a small angle \(\delta\theta\) at the centre of the hole? Write down Newton's equation of motion for this small segment given that the surface tension gives rise to a net outwards force on the segment of \(2T \cdot r\delta\theta\). Thence show that \begin{align*} r^4\ddot{r} = \frac{2T}{\rho h}r^4 + \text{constant}, \end{align*} and conclude that when the hole is large it grows like \begin{align*} r = t\left(\frac{2T}{\rho h}\right)^{\frac{1}{2}} + \text{constant}. \end{align*}
In a painting process, small charged paint drops move in an oscillating electric field. As a drop of mass \(m\) moves in the \(x\)-direction through the air it experiences a frictional force \(-k\dot{x}\), as well as the oscillating electric force whose amplitude varies in space, \(F(x)\cos\omega t\). When the electric field is weak, the motion of the paint drop can be obtained by successive approximations. To a zeroth approximation there is no electric field and the drop does not move, i.e. \(x \simeq x_0\), a constant. At the next approximation the drop oscillates a small distance about \(x \simeq x_0\) in the electric field which can be evaluated at \(x_0\) in this approximation. Thus, writing \(x \simeq x_0 + x_1(t)\), the first correction is governed by \begin{align*} m\ddot{x}_1 + k\dot{x}_1 = E(x_0)\cos \omega t. \end{align*} Find the forced response (i.e. particular integral for) \(x_1\) which has a frequency \(\omega\). At the following approximation it is necessary to take account of the small difference between the electric field at \(x = x_0 + x_1\) and \(x = x_0\), i.e. \begin{align*} x_1 \left.\frac{dE}{dx}\right|_{x_0} \cos \omega t. \end{align*} Thus, writing \(x \simeq x_0 + x_1(t) + x_2(t)\), the second correction \(x_2\) is governed by \begin{align*} m\ddot{x}_2 + k\dot{x}_2 = x_1(t) \left.\frac{dE}{dx}\right|_{x_0} \cos \omega t. \end{align*} The forced response (i.e. particular integral for) \(x_2\) consists of an oscillation with frequency \(2\omega\) and a steady velocity. Find just the steady velocity. If the paint drifts with the small steady velocity you have just calculated, where does it go?
A garden water sprinkler consists of a straight arm of length \(2l\) pivoted at its centre. The arm rotates in a horizontal plane at a steady angular velocity \(\Omega\), with a rusty pivot exerting a constant couple \(G\) on the arm. Water enters through the central pivot and leaves horizontally through small nozzles at the ends of the arm set at an angle \(\theta\) to the direction of the arm. In unit time a mass \(Q\) of water is discharged with a kinetic energy \(\frac{1}{2}QU^2\) relative to the ground. By considering the angular momentum imparted to the water in unit time, show that the angular velocity of the discharged water is \(G/l^2Q\) at the nozzles. Show further that \(\Omega\) is given by \begin{align*} \Omega l = \left[U^2 - \frac{G^2}{l^2Q^2}\right]^{\frac{1}{2}} \tan \theta - \frac{G}{lQ}. \end{align*} What is the total power which must be supplied to the sprinkler?