A straight uniform rod has mass \(m\). Its ends \(P_1\) and \(P_2\) are attached to small light rings that are constrained to move on a rough rigid circular wire with centre \(O\) fixed in a vertical plane, and the angle \(P_1OP_2\) is a right angle. The rod rests with \(P_1\) lower than \(P_2\), and with both ends lower than \(O\). The coefficient of friction between each of the rings and the wire is \(\mu\). Given that the rod is in limiting equilibrium (i.e. \ on the point of slipping at both ends), show that \[ \tan \alpha = \frac{1-2\mu -\mu^2}{1+2\mu -\mu^2}\,, \] where \(\alpha\) is the angle between \(P_1O\) and the vertical (\(0<\alpha<45^\circ\)). Let \(\theta\) be the acute angle between the rod and the horizontal. Show that \(\theta =2\lambda\), where \(\lambda \) is defined by \(\tan \lambda= \mu\) and \(0<\lambda<22.5^\circ\).
In this question, you may use without proof the results: \[ \sum_{r=1}^n r = \tfrac12 n(n+1) \qquad\text{and}\qquad \sum_{r=1}^n r^2 = \tfrac1 6 n(n+1)(2n+1)\,. \] The independent random variables \(X_1\) and \(X_2\) each take values \(1\), \(2\), \(\ldots\), \(N\), each value being equally likely. The random variable \(X\) is defined by \[ X= \begin{cases} X_1 & \text { if } X_1\ge X_2\\ X_2 & \text { if } X_2\ge X_1\;. \end{cases} \]
Solution: \begin{align*} \P(X = r) &= \P(X_1 = r, X_2 \leq r) + \P(X_2 = r, X_1 < r) \\ &= \P(X_1 = r) \P(X_2 \leq r) + \P(X_2 = r)\P( X_1 < r) \\ &= \frac{1}{N} \frac{r}{N} + \frac{1}{N} \frac{r-1}{N} \\ &= \frac{2r-1}{N^2} \end{align*} \begin{align*} \E(X) &= \sum_{r=1}^N r \P(X = r) \\ &= \sum_{r=1}^N \frac{2r^2 - r}{N^2} \\ &= \frac{1}{N^2} \l \frac{N(N+1)(2N+1)}{3} - \frac{N(N+1)}{2} \r \\ &= \frac{N+1}{N} \l \frac{4N-1}{6} \r \end{align*} When \(N = 100\), this is equal to \(\frac{101 \cdot 399}{6 \cdot 100} = \frac{101 \cdot 133}{200} = 67.165\) \begin{align*} &&\frac12 &\leq \P(X \leq m) \\ &&&=\sum_{r=1}^m \P(X=r) \\ &&&=\sum_{r=1}^m \frac{2r-1}{N^2} \\ &&&= \frac{1}{N^2} \l m(m+1) - m \r \\ &&&= \frac{m^2}{N^2} \\ \Rightarrow && m^2 &\geq \frac{N^2}{2} \\ \Rightarrow && m &\geq \frac{N}{\sqrt{2}} \\ \Rightarrow && m &= \left \lceil \frac{N}{\sqrt{2}} \right \rceil \end{align*} When \(N = 100\), \(100/\sqrt{2} = \sqrt{2}50\). \(\sqrt{2} > 1.4 \Rightarrow 50\sqrt{2} > 70\) \(\sqrt{2} < 1.42 \Rightarrow 50 \sqrt{2} < 71\), therefore \(\displaystyle \left \lceil \frac{100}{\sqrt{2}} \right \rceil = 71\) \begin{align*} \lim_{N \to \infty} \frac{\frac{(N+1)(4N-1)}{6N}}{ \left \lceil\frac{N}{\sqrt{2}} \right \rceil} &= \lim_{N \to \infty} \frac{\sqrt{2}}{3}\l \frac{4N^2 +3N - 1}{2N^2} \r \tag{since the floor will be irrelevant}\\ &= \lim_{N \to \infty} \frac{\sqrt{2}}{3}\l 2 + \frac{3}{2N} - \frac{1}{N^2} \r \\ &= \lim_{N \to \infty} \frac{2\sqrt{2}}{3} \end{align*}
Three married couples sit down at a round table at which there are six chairs. All of the possible seating arrangements of the six people are equally likely.