Problems

Find the limit of \[\left(\frac{\beta x^{\beta-1}}{x^\beta - a^\beta} - \frac{1}{x-a}\right)\] as \(x \to a\).

The equation \[\sin x = \lambda x\] (where \(\lambda > 0\), \(x > 0\)) has a finite number \(N\) of non-zero solutions \(x_i\), \(i = 1, \ldots, N\), where \(N\) depends on \(\lambda\), provided \(\lambda < 1\).

1. Prove that there are no non-zero solutions for \(\lambda > 1\), while for \(\lambda = 1 - \varepsilon^2\) (\(\varepsilon \ll 1\)) there is a non-zero solution approximately equal to \(\varepsilon\sqrt{6}\).
2. Suppose now that all the non-zero roots are distinct, and that \(x_1 < x_2 < \ldots < x_N\). By drawing an appropriate graph explain why \(N\) is always odd, and why \[(2n - 2)\pi < x_{2n-1} < (2n - 1)\pi \text{ for } n = 1, \ldots, \frac{N+1}{2}\] \[2n\pi < x_{2n} < (2n + \tfrac{1}{2})\pi \text{ for } n = 1, \ldots, \frac{N-1}{2}.\]

Hence derive an approximate value for \(N\) as a function of \(\lambda\) when \(\lambda\) is very small.

A uniform circular cylinder of mass \(m\) and radius \(a\) moves under the action of a horizontal force \(P\) on a rough horizontal table. The force \(P\) is directed through the centre of mass of the cylinder. The table moves with acceleration \(f\) in the same direction as the force \(P\). Both \(P\) and \(f\) are perpendicular to the axis of the cylinder. The coefficient of friction between the cylinder and the table is \(\mu\). Initially the cylinder and the table are at rest, and there is no slipping in the subsequent motion. When the cylinder has moved a distance \(x\) perpendicular to its axis, it has rotated through an angle \(\theta\) about its axis. Show that \[\ddot{x} + a\ddot{\theta} = f\] and find \(\ddot{\theta}\). Show that \[\mu \geq \tfrac{1}{3}|mf - P|/mg\] where \(g\) is the acceleration due to gravity.

A small bead can slide on the spoke of a wheel of radius \(b\) that is constrained to rotate about its axle with angular velocity \(\omega\). Initially the bead is at rest relative to the wheel, at a distance \(d > 0\) from the centre. Show that if gravity can be ignored, the bead will always slide to the rim of the wheel, whatever the coefficients \(\mu_1\) and \(\mu_2\) of static and sliding friction. When its distance from the centre of the axle is \(a\), the bead has an outward velocity of magnitude \([\sqrt{(1 + \mu_1^2)} + \mu_2]a\omega\). Show that the bead hits the rim of the wheel at a time \[t = \frac{\ln(b/a)}{[\sqrt{(1 + \mu_1^2)} - \mu_2]\ \omega}\] later. [You may assume the spoke to lie in the radial direction.]

A light rod of length \(a\) rests horizontally with its ends on equally rough fixed planes inclined at angles \(\alpha\) and \(\beta\) to the horizontal. The vertical plane through the rod is perpendicular to the line of intersection of the rough planes. The coefficient of friction between the ends of the rod and the planes on which they lie is \(\tan \lambda\), and \(\alpha < \lambda < \pi/2 - \alpha\), \(\beta < \lambda < \pi/2 - \beta\). Show that the length of that section of the rod on which a weight can be placed without disturbing the equilibrium is \[\frac{a \sin 2\lambda \cos(\alpha - \beta)}{\sin(\alpha + \beta)}.\]

An harmonious population with ample space and food is liable to grow at a rate proportional to its size. However, disunity induces mortal combat, so that in practice the ratio \(x_n\) of the number in any generation to a fixed number \(k\) satisfies \[x_{n+1} = \alpha x_n(1 - x_n)\] where \(\alpha\) is a positive constant. It is known that under certain circumstances the solution to this equation is of the form \begin{align*} x_n = x_{n+2} = x_{n+4} = \ldots = p,\\ x_{n+1} = x_{n+3} = x_{n+5} = \ldots = q. \end{align*} Show that aside from the trivial solution \(p = q = 0\), the relation \[\alpha^2(1-p)(1-q) = 1\] can then be satisfied, together with either \(p = q\) or \[\alpha(p+q-1) = 1.\] Hence, or otherwise, establish the ranges of \(\alpha\) for which (a) the population can oscillate with a period of 2 generations and (b) a non-zero steady state exists.