A uniform billiard ball lies at rest on a horizontal table, the coefficient of friction between the ball and the table being \(\mu\). It is set in motion by an impulse applied at a point \(P\) of the ball. The diameter through \(P\) is initially horizontal. The direction of the impulse lies in the vertical plane through this diameter, and is inclined at an angle \(\alpha\) below the horizontal. Show that the ball will initially slip, rather than roll, provided \[2\cot\alpha > 7\mu - 5.\] In this case, show that if the initial velocity of the ball is \(v_0\), then the ball eventually rolls with uniform velocity \[\frac{5}{7}\left(\frac{\cot\alpha - 1}{\cot\alpha - \mu}\right)v_0.\]
A circular hoop of mass \(m\) is pivoted so as to be able to rotate freely in a horizontal plane about a point \(O\) on its circumference. A small, smooth ring of mass \(km\) is free to slide on the hoop. Initially the ring lies at \(P\), the opposite end of the diameter through \(O\), and the system is at rest. It is then set in motion by equal and opposite impulses \(I\) applied at \(P\) to the ring and the hoop. By using the principle of conservation of angular momentum, or otherwise, show that when the ring reaches \(O\), the hoop has rotated through an angle \[\frac{1}{2}\pi\{1 - (1 + 2k)^{-\frac{1}{2}}\}.\]
A heavy, uniform circular cylinder of radius \(r\) lies on a rough horizontal plane with its axis horizontal. A heavy, uniform rod \(AB\) of length \(l\) lies in the vertical plane which bisects the axis of the cylinder at right angles. Its end \(A\) rests on the plane and point \(C\), distinct from \(B\), is in contact with the cylinder. The coefficient of friction is the same value, \(\mu\), at each of the three points of contact between the rod, cylinder and plane. The rod makes an angle \(\alpha\) with the horizontal, and the friction is limiting at Show, by a geometrical method or otherwise, that for fixed values of \(r\), \(l\) and \(\alpha\), there exists a value of \(\mu\) such that this situation is a possible equilibrium state of the system provided \[r\cot(\frac{1}{2}\alpha) \leq l\cos\alpha \leq 2r\cot(\frac{1}{2}\alpha).\]
The earth may be assumed to be a homogeneous sphere and then the gravitational acceleration within it may be shown to be directed towards and to vary directly as the distance from the centre. A straight tunnel connects two points on the surface of the earth which subtend an angle \((\pi - 2\alpha)\) at the centre. A small particle is placed at one end of the tunnel. The limiting coefficient of static friction between the particle and the tunnel is \(\mu_s\), and the coefficient of dynamic friction is \(\mu_d\), where \[\mu_d < \mu_s < 1.\] Describe the subsequent motion of the particle and show that, if the particle moves initially and does not reach the half-way point in the tunnel, then \[\frac{1}{2}\cot\alpha < \mu_d < \mu_s < \cot\alpha.\]
A tumbler which has square cross-section of side \(2a\) and height \(Ka\) is closed at one end and this end rests on a rough horizontal table. The tumbler is filled to a height \(ka\) with liquid of uniform density. Assuming that no sliding takes place and that the weight of the tumbler is negligible compared with that of the liquid, show that when the table is tilted slowly through an angle \(\theta\) about an axis parallel to one face of the tumbler then, provided \[\tan\theta \leq k \leq K - \tan\theta,\] the tumbler will topple when \(\theta\) is given by \[\tan^3\theta + (3k^2 + 2)\tan\theta - 6k = 0.\] If it is required to tilt the table through an angle \(\tan^{-1}\frac{1}{2}\) without spillage, determine what height of tumbler is required and how full it can be.
Two players play a dice game on a board marked with squares numbered 0 to 13. Each player has a counter that is initially on square 0 and they take turns to throw a six-sided die. A player's counter is not moved until he throws a six, when it moves to square 6. Thereafter, if it is on square \(m\) and he throws an \(n\), it advances to square \(m + n\) if \(m + n \leq 13\), and 'rebounds' to square \(26 - (m + n)\) if \(m + n > 13\). The winner is the player whose counter first reaches square 13. Find (i) the probability that the first player to throw is the first to move his counter; (ii) the probability that the loser's counter never leaves square 0.