A soaring bird of weight \(mg\) experiences a lift force \(L\) perpendicular to the velocity of the air relative to the bird, and a drag force \(D\) parallel to it. This relative velocity has magnitude \(U\) and makes an angle \(\theta\) with the horizontal. \(L\) and \(D\) are given by \begin{align} L = \alpha U^2, \quad D = \beta U^2 + \gamma/U^2 \end{align} where \(\beta\) and \(\gamma\) are constants and \(\alpha\) can be varied by the bird up to a maximum value \(\alpha_{\text{max}}\). Show that the bird can glide with constant relative velocity \(U\) only if \(U\) satisfies \begin{align} \left(\frac{mg\cos\theta}{\alpha_{\text{max}}}\right)^{\frac{1}{2}} \leq U < \left(\frac{mg\sin\theta}{\beta}\right)^{\frac{1}{2}} \end{align} Find the value of \(U\) at which \(D\) is a minimum. The bird glides stably if a small increase in \(U\), with \(\theta\) fixed, leads to a resultant force in the direction of \(-\mathbf{U}\), and a small decrease to a resultant force in the opposite direction. Show that stable gliding is possible only if \begin{align} \left(\frac{mg\sin\theta}{2\beta}\right)^{\frac{1}{2}} < U \end{align}
Consider a simple pendulum of length \(l\) and angular displacement \(\theta\) which is not assumed to be small. Show that \begin{align} \frac{1}{2}l\left(\frac{d\theta}{dt}\right)^2 = g(\cos\theta - \cos\gamma) \end{align} where \(\gamma\) is the maximum value of \(\theta\). Show also that the period \(P\) is given by \begin{align} P = 2\sqrt{\frac{l}{g}}\int_0^{\gamma}\left(\sin^2(\gamma/2) - \sin^2(\theta/2)\right)^{-\frac{1}{2}}d\theta \end{align} By using the substitution \(\sin(\theta/2) = \sin(\gamma/2)\sin\alpha\), or otherwise, show that for small values of \(\gamma\) the period is approximately \begin{align} 2\pi\sqrt{\frac{l}{g}}\left(1+\frac{\gamma^2}{16}\right) \end{align}
In the theory of relativity the following relations hold for a particle: \begin{align} E = mc^2, \quad m = m_0\left(1-\frac{v^2}{c^2}\right)^{-\frac{1}{2}}, \\ p = mv \text{ and } F = \frac{dp}{dt} \end{align} [Here \(E\) is energy, \(m\) is mass which varies with speed, \(m_0\) is the constant rest mass, \(v\) is velocity, \(p\) is momentum, \(F\) is force and \(c\) is the constant speed of light.] Show that \begin{align} Fv = c^2\frac{dm}{dt} \end{align} and \begin{align} p^2c^2 = E^2 - E_0^2 \end{align} where \(E_0 = m_0c^2\), the rest energy. If \(T = E - E_0\), show that for small values of \(v/c\), \(T\) is approximately \(\frac{1}{2}mv^2\), whereas for values of \(v/c\) close to but smaller than 1, \(T\) is approximately \(pc\).
A light frictionless pulley is supported by a mounting of mass \(m\), which is attached to the ceiling of a room by an elastic string with force constant \(k\). A light inextensible string has one end attached to the floor of the room. It passes over the pulley and carries a load of mass \(M\) at its other end. The whole system rests in equilibrium with the straight sections of both strings being vertical. Find the extension of the elastic string. The load is now pulled vertically downwards through a distance \(a\) and then released. If neither string becomes slack in the subsequent motion, show that \(a\) must be less than \((m+4M)g/k\) and find the period of oscillation of the system. [The tension in the string is the product of the force constant and the extension.]
A simple gun consists of a smooth tube \(AB\) of length \(l\) whose end \(A\) is mounted at a fixed point on level ground. The tube is clamped at an angle \(\theta\) with the horizontal. Particles are projected up the tube, leaving \(A\) with a speed \(v\) and returning to ground level at a distance \(R\) from \(A\). If \(L = v^2/g\) and \(l/L\) is small compared with \(\sin\theta\), show that \(R\) is given approximately by \begin{align} L\sin 2\theta + 2l\cos\theta\cos 2\theta \end{align}
A bob of mass \(m\) is attached to a light string. The free end of the string is passed from below through a small smooth hole in a fixed horizontal metal plate, and is then fastened so that a length \(l\) of string hangs below the plate. The bob is set oscillating as a simple pendulum, so that the angle \(\theta\) between the string and the vertical at time \(t\) has the form \(\theta = \alpha\cos(\omega t)\), where \(\alpha\), \(\omega\) are constants and \(\alpha\) is small. Write down the value of \(\omega\) in terms of \(l\) and the acceleration of gravity \(g\). Neglecting powers of \(\alpha\) greater than \(\alpha^2\), evaluate as functions of time \begin{align} \text{(i)} &\text{ the kinetic energy } K \text{ of the bob,}\\ \text{(ii)} &\text{ the potential energy } V \text{ of the bob relative to the level of the plate,}\\ \text{(iii)} &\text{ the tension } T \text{ in the string.} \end{align} Hence show that the total energy \(E = K + V\) and the time-averaged value \(T_{\text{av}}\) of \(T\) are given by \begin{align} E = -mgl\left(1-\frac{1}{2}\alpha^2\right), \quad T_{\text{av}} = mg\left(1+\frac{1}4\alpha^2\right) \end{align} The string is now pulled up very slowly and steadily through the hole, so that \(l\) and \(\alpha\) both change slowly with time and \(dl/dt\) is constant. Use energy considerations to obtain a relation between \(T_{\text{av}}\) and the rate of change of \(E\), and deduce that \(l^3\alpha^4\) remains constant as \(l\) and \(\alpha\) change.