Two beads each of mass \(m\) are threaded on to a smooth straight rod one end of which is freely hinged to a fixed point. They are connected by an elastic string of natural length \(l\) and modulus \(\lambda\). The rod is set in uniform rotation in a horizontal plane with angular velocity \(\omega\). Show that, if \(2\lambda< ml\omega^2\), the string must in general ultimately break.
Show that \[ \sum_{r=0}^n r(r+1)\dots(r+k-1) = \frac{1}{k+1}n(n+1)\dots(n+k). \] Deduce that, if \(a_{r+1} = a_r/(1+ra_r)\), then \[ \sum_{r=0}^n \frac{1}{a_ra_{r+2}} = \frac{n+1}{a_0} + \frac{n+1}{3a_0^2}(n^2+2n+3) + \frac{1}{20}(n-1)n(n+1)(n+2)(n+3). \]
The sequence \(a_1, a_2, a_3, \dots\) is defined by means of the relations \[ a_1=3, \quad a_{p+1} = \frac{a_p^2+5}{2a_p} \quad (p>0). \] Prove that \[ 0 < a_{p+1} - \sqrt{5} \le \frac{(3-\sqrt{5})^{2^p}}{(2\sqrt{5})^{2^p-1}} < 6 \times \left(\frac{2}{11}\right)^{2^p}. \] Hence show that \(a_p \to \sqrt{5}\) as \(p\to\infty\). Use this process to calculate \(\sqrt{5}\) correct to four places of decimals.
A regimental dinner is attended by \(n\) officers who leave their caps in an ante-room before going in to dine. At the conclusion of the dinner there is a certain amount of confusion in the ante-room with the result that each officer emerges wearing a cap which is not his own. If \(p_r\) is the chance of this occurring for a dinner of \(r\) officers show that \[ p_n + \frac{p_{n-1}}{1!} + \frac{p_{n-2}}{2!} + \dots + \frac{p_0}{n!} = 1, \] where \(p_0=1\). By forming the series \(\displaystyle\sum_{n=0}^\infty p_n x^n\) and multiplying by \(e^x\), or otherwise, deduce that \[ p_n = 1 - \frac{1}{1!} + \frac{1}{2!} - \dots + (-1)^n \frac{1}{n!}. \]
\(ABCD\) is a cyclic quadrilateral whose diagonals \(AC, BD\) meet in \(X\). \(E\) and \(F\) are the feet of the perpendiculars from \(X\) to \(AD\) and \(BC\) respectively, and \(G\) is the mid-point of \(AB\). Points \(P\) and \(Q\) are taken on \(AD, BC\) respectively such that \(EP=AE, FQ=BF\). Show that the triangles \(AXQ\) and \(PXB\) are congruent, and hence prove that \(GE=GF\).
A man can walk at the rate of 100 yd. a minute, which is \(n\) times faster than he can swim. He stands at one corner of a rectangular pond 80 yd. long and 60 yd. wide. To get to the opposite corner he may walk round the edge, swim straight across or walk part of the way along the longer side and then swim the rest. If he is to make the trip in the least time, how should he proceed and how long does he take if (i) \(n=\frac{5}{3}\), (ii) \(n=\frac{4}{3}\), (iii) \(n=\frac{5}{4}\)?
A uniform heavy rod of weight \(6w\) and length \(3a\) is freely hinged at one end and kept horizontal by a support distant \(2a\) from the hinge. A weight \(w\) is suspended from the rod midway between the hinge and the support, and a weight \(w\) is suspended from the free end of the rod. Calculate the bending moment at all points of the rod and illustrate your result by a sketch. Show that the bending moment has a stationary value at a point distant less than \(a\) from the hinge, but attains its greatest numerical value at the support.
The ends \(A, B\) of a heavy uniform rod of weight \(w\) and length \(2a\) are attached by two light inextensible strings each of length \(b\) to two points \(C, D\) at the same level a distance \(2c\) apart where \(a+c>b>\sqrt{(a^2+c^2)}\). The rod is now moved in such a way that it always remains horizontal and so that the mid-point of \(AB\) remains vertically below the mid-point of \(CD\). Find the potential energy of the rod as a function of the angle between \(AB\) and \(CD\), if both strings remain taut and do not cross. Also find the couple necessary to keep \(AB\) perpendicular to \(CD\).
The point of suspension of a simple pendulum with a bob of mass \(m\) is made to move in a horizontal straight line with constant acceleration \(f\); if \(3f^2=g^2\), and if the string is initially at rest and vertical with the bob vertically below the point of suspension, find the greatest angle the string makes with the vertical in the motion. Also show that the maximum tension in the string is \(2(\sqrt{3}-1)mg\).
A particle of mass \(m\) is suspended from a fixed point \(O\) by a light elastic string of natural length \(l\). The particle hangs in equilibrium under gravity, and the length of the string is \(l+a\); an upward vertical impulse is then applied to the particle and it first comes to rest at \(O\). Show that the magnitude of the impulse is \(m[g(a+2l)]^{\frac{1}{2}}\), and find the time the particle takes to reach \(O\).