Problems

The lower end of a rigid uniform rod of mass \(m\) and length \(a\) rests at point \(M\) on rough horizontal ground. Each of two elastic strings, of natural length \(\ell\) and modulus of elasticity \(\lambda\), is attached at one end to the top of the rod. Their lower ends are attached to points \(A\) and \(B\) on the ground, which are a distance \(2a\) apart. \(M\) is the midpoint of \(AB\). \(P\) is the point at the top of the rod and lies in the vertical plane through \(AMB\). Suppose that the rod is in equilibrium with angle \(PMB = 2\theta\), where \(\theta < 45°\) and \(\theta\) is such that both strings are in tension.

1. Show that angle \(APB\) is a right angle. Show that the force exerted on the rod by the elastic strings can be written as the sum of
   • a force of magnitude \(\frac{2a\lambda}{\ell}\) parallel to the rod
   • and a force of magnitude \(\sqrt{2}\lambda\) acting along the bisector of angle \(APB\).
2. By taking moments about point \(M\), or otherwise, show that \(\cos\theta + \sin\theta = \frac{2\lambda}{mg}\). Deduce that it is necessary that \(\frac{1}{2}mg < \lambda < \frac{1}{2}\sqrt{2}mg\).
3. \(N\) and \(F\) are the magnitudes of the normal and frictional forces, respectively, exerted on the rod by the ground at \(M\). Show, by taking moments about an appropriate point, or otherwise, that \[N - F\tan 2\theta = \frac{1}{2}mg.\]

Show Solution

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Solution:

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1. Notice that \(AM = MB = MP\) in particular \(P\) lies on a semi-circle of radius \(a\) and therefore by Thales' theorem \(\angle APB = 90^{\circ}\). Notice that by angles in a triangle and the angles adding to \(90^{\circ}\), \(\angle APM = \theta\). Therefore, \begin{align*} && |PB| &= 2a \sin \theta \\ && |PA| &= 2a \cos \theta \\ && T_A &= \frac{\lambda}{l} \left (2a \cos \theta -l \right) \\ && T_B &= \frac{\lambda}{l} \left (2a \sin\theta -l \right) \\ \end{align*} Since \(T_A\) and \(T_B\) are perpendicular, we can consider the forces as having vector \(\frac{\lambda}{l}\binom{2a\cos \theta-l}{2a\sin \theta - l}\) in this coordinate system, ie the sum of a vector \(\frac{2\lambda a}{l}\binom{\cos \theta}{\sin \theta}\) and \(\displaystyle -\sqrt{2}\lambda \binom{\frac1{\sqrt{2}}}{\frac1{\sqrt{2}}}\) which are unit vectors parallel to the rod and along the bisector of \(APB\) respectively.
2. \begin{align*} \overset{\curvearrowright}{M}: && 0 &= \frac{a}{2} \cdot mg \cos 2 \theta - a\cdot \sqrt{2}\lambda \cos (90-(45-\theta))\\ \Rightarrow && \cos 2 \theta &= \frac{\lambda}{mg} 2 \sqrt{2} \cos (45 + \theta) \\ \Rightarrow && \cos^2 \theta - \sin^2 \theta &= \frac{2\lambda}{mg} (\cos \theta - \sin \theta) \\ \underbrace{\Rightarrow}_{\theta < 45^{\circ}} && \cos \theta + \sin \theta &= \frac{2\lambda}{mg} \end{align*} Over \((0, 45^{\circ})\), \(\cos \theta + \sin \theta\) ranges from \(1\) to \(\sqrt{2}\), therefore \(1 < \frac{2 \lambda}{mg} < \sqrt{2} \Rightarrow \frac12 mg < \lambda < \frac12 \sqrt{2} mg\) as required.
3. \begin{align*} \overset{\curvearrowright}{P}: && 0 &=- \frac{a}{2} \cdot \left ( mg \cos 2\theta \right) - a \cdot F \sin 2 \theta + a \cdot N \cos 2 \theta \\ \Rightarrow && \frac12 mg &= N - F \tan 2 \theta \end{align*} as required.

A uniform elastic string lies on a smooth horizontal table. One end of the string is attached to a fixed peg, and the other end is pulled at constant speed \(u\). At time \(t=0\), the string is taut and its length is \(a\). Obtain an expression for the speed, at time \(t\), of the point on the string which is a distance \(x\) from the peg at time \(t\). An ant walks along the string starting at \(t=0\) at the peg. The ant walks at constant speed \(v\) along the string (so that its speed relative to the peg is the sum of \(v\) and the speed of the point on the string beneath the ant). At time \(t\), the ant is a distance \(x\) from the peg. Write down a first order differential equation for \(x\), and verify that \[ \frac{\d }{\d t} \left( \frac x {a+ut}\right) = \frac v {a+ut} \,. \] Show that the time \(T\) taken for the ant to reach the end of the string is given by \[uT = a(\e^k-1)\,,\] where \(k=u/v\). On reaching the end of the string, the ant turns round and walks back to the peg. Find in terms of \(T\) and \(k\) the time taken for the journey back.

A particle \(P\) of mass \(m\) moves on a smooth fixed straight horizontal rail and is attached to a fixed peg \(Q\) by a light elastic string of natural length \(a\) and modulus \(\lambda\). The peg \(Q\) is a distance \(a\) from the rail. Initially \(P\) is at rest with \(PQ=a\). An impulse imparts to \(P\) a speed \(v\) along the rail. Let \(x\) be the displacement at time \(t\) of \(P\) from its initial position. Obtain the equation \[ \dot x^2 = v^2 - k^2 \left( \sqrt{x^2+a^2} -a\right)^{\!2} \] where \( k^2 = \lambda/(ma)\), \(k>0\) and the dot denotes differentiation with respect to \(t\). Find, in terms of \(k\), \(a\) and \(v\), the greatest value, \(x_0\), attained by \(x\). Find also the acceleration of \(P\) at \(x=x_0\). Obtain, in the form of an integral, an expression for the period of the motion. Show that in the case \(v\ll ka\) (that is, \(v\) is much less than \(ka\)), this is approximately \[ \sqrt {\frac {32a}{kv}} \int_0^1 \frac 1 {\sqrt{1-u^4}} \, \d u \, . \]

A small ring of mass \(m\) is free to slide without friction on a hoop of radius \(a\). The hoop is fixed in a vertical plane. The ring is connected by a light elastic string of natural length \(a\) to the highest point of the hoop. The ring is initially at rest at the lowest point of the hoop and is then slightly displaced. In the subsequent motion the angle of the string to the downward vertical is \(\phi\). Given that the ring first comes to rest just as the string becomes slack, find an expression for the modulus of elasticity of the string in terms of \(m\) and \(g\). Show that, throughout the motion, the magnitude \(R\) of the reaction between the ring and the hoop is given by \[ R = ( 12\cos^2\phi -15\cos\phi +5) mg \] and that \(R\) is non-zero throughout the motion.

Particles \(P\) and \(Q\), each of mass \(m\), lie initially at rest a distance \(a\) apart on a smooth horizontal plane. They are connected by a light elastic string of natural length \(a\) and modulus of elasticity \(\frac12 m a \omega^2\), where \(\omega\) is a constant. Then \(P\) receives an impulse which gives it a velocity \(u\) directly away from \(Q\). Show that when the string next returns to length \(a\), the particles have travelled a distance \(\frac12 \pi u/\omega\,\), and find the speed of each particle. Find also the total time between the impulse and the subsequent collision of the particles.

A small bead \(B\), of mass \(m\), slides without friction on a fixed horizontal ring of radius \(a\). The centre of the ring is at \(O\). The bead is attached by a light elastic string to a fixed point \(P\) in the plane of the ring such that \(OP = b\), where \(b > a\). The natural length of the elastic string is \(c\), where \(c < b - a\), and its modulus of elasticity is \(\lambda\). Show that the equation of motion of the bead is \[ ma\ddot \phi = -\lambda\left( \frac{a\sin\phi}{c\sin\theta}-1\right)\sin(\theta+\phi) \,, \] where \(\theta=\angle BPO\) and \(\phi=\angle BOP\). Given that \(\theta\) and \(\phi\) are small, show that $a(\theta+\phi)\approx b\theta$. Hence find the period of small oscillations about the equilibrium position \(\theta=\phi =0\).

A long string consists of \(n\) short light strings joined together, each of natural length \(\ell\) and modulus of elasticity \(\lambda\). It hangs vertically at rest, suspended from one end. Each of the short strings has a particle of mass \(m\) attached to its lower end. The short strings are numbered \(1\) to \(n\), the \(n\)th short string being at the top. By considering the tension in the \(r\)th short string, determine the length of the long string. Find also the elastic energy stored in the long string. A uniform heavy rope of mass \(M\) and natural length \(L_0\) has modulus of elasticity \(\lambda\). The rope hangs vertically at rest, suspended from one end. Show that the length, \(L\), of the rope is given by \[ L=L_0\biggl(1+ \frac{Mg}{2\lambda}\biggr), \] and find an expression in terms of \(L\), \(L_0\) and \(\lambda\) for the elastic energy stored in the rope.

A particle \(P\) of mass \(m\) is attached to points \(A\) and \(B\), where \(A\) is a distance \(9a\) vertically above \(B\), by elastic strings, each of which has modulus of elasticity \(6mg\). The string \(AP\) has natural length \(6a\) and the string \(BP\) has natural length \(2a\). Let \(x\) be the distance \(AP\). The system is released from rest with \(P\) on the vertical line \(AB\) and \(x = 6a\). Show that the acceleration \(\ddot{x}\) of \(P\) is \(\ds{4g \over a}(7a - x)\) for \(6a < x < 7a\) and \(\ds{g \over a}(7a - x)\) for \(7a < x < 9a\,\). Find the time taken for the particle to reach \(B\).

A particle \(P\) of mass \(m\) is constrained to move on a vertical circle of smooth wire with centre~\(O\) and of radius \(a\). \(L\) is the lowest point of the circle and \(H\) the highest and \(\angle LOP = \theta\,\). The particle is attached to \(H\) by an elastic string of natural length \(a\) and modulus of elasticity~\(\alpha mg\,\), where \(\alpha > 1\,\). Show that, if \(\alpha>2\,\), there is an equilibrium position with \(0<\theta<\pi\,\). Given that \(\alpha =2+\sqrt 2\,\), and that \(\displaystyle \theta = \tfrac{1}{2}\pi + \phi\,\), show that \[ \ddot{\phi} \approx -\frac{g (\sqrt2+1)}{2a }\, \phi \] when \(\phi\) is small. For this value of \(\alpha\), explain briefly what happens to the particle if it is given a small displacement when \( \theta = \frac{1}{2}\pi\).

A smooth cylinder with circular cross-section of radius \(a\) is held with its axis horizontal. A~light elastic band of unstretched length \(2\pi a\) and modulus of elasticity \(\lambda\) is wrapped round the circumference of the cylinder, so that it forms a circle in a plane perpendicular to the axis of the cylinder. A particle of mass \(m\) is then attached to the rubber band at its lowest point and released from rest.

1. Given that the particle falls to a distance \(2a\) below the below the axis of the cylinder, but no further, show that \[ \lambda = \frac{9\pi m g}{(3\sqrt3-\pi)^2} \;. \]
2. Given instead that the particle reaches its maximum speed at a distance \(2a\) below the axis of the cylinder, find a similar expression for \(\lambda\)\,.

The string \(AP\) has a natural length of \(1\!\cdot5\!\) metres and modulus of elasticity equal to \(5g\) newtons. The end \(A\) is attached to the ceiling of a room of height \(2\!\cdot\!5\) metres and a particle of mass \mbox{\(0\!\cdot\!5\) kg} is attached to the end \(P\). The end \(P\) is released from rest at a point \(0\!\cdot\!5\) metres above the floor and vertically below \(A\). Show that the string becomes slack, but that \(P\) does not reach the ceiling. Show also that while the string is in tension, \(P\) executes simple harmonic motion, and that the time in seconds that elapses from the instant when \(P\) is released to the instant when \(P\) first returns to its original position is $$ \left(\frac8{3g}\right)^{\!\frac12}+ \left(\frac3 {5g}\right)^{\!\frac12} {\Big(\pi - \arccos (3/7)\Big)}. $$ \noindent [Note that \(\arccos x\) is another notation for \(\cos^{-1} x\).]

A bungee-jumper of mass \(m\) is attached by means of a light rope of natural length \(l\) and modulus of elasticity \(mg/k,\) where \(k\) is a constant, to a bridge over a ravine. She jumps from the bridge and falls vertically towards the ground. If she only just avoids hitting the ground, show that the height \(h\) of the bridge above the floor of the ravine satisfies \[ h^{2}-2hl(k+1)+l^{2}=0, \] and hence find \(h.\) Show that the maximum speed \(v\) which she attains during her fall satisfies \[ v^{2}=(k+2)gl. \]

A particle hangs in equilibrium from the ceiling of a stationary lift, to which it is attached by an elastic string of natural length \(l\) extended to a length \(l+a\). The lift now descends with constant acceleration \(f\) such that \(0 < f < g/2\). Show that the extension \(y\) of the string from its equilibrium length satisfies the differential equation $$ {{\rm d}^2 y \over {\rm d} t^2} +{g \over a}\, y = g-f. $$ Hence show that the string never becomes slack and the amplitude of the oscillation of the particle is \(af/g\). After a time \(T\) the lift stops accelerating and moves with constant velocity. Show that the string never becomes slack and the amplitude of the oscillation is now \[\frac{2af}{g}|\sin {\textstyle \frac{1}{2}}\omega T|,\] where \(\omega^{2}=g/a\).

A smooth circular wire of radius \(a\) is held fixed in a vertical plane with light elastic strings of natural length \(a\) and modulus \(\lambda\) attached to the upper and lower extremities, \(A\) and \(C\) respectively, of the vertical diameter. The other ends of the two strings are attached to a small ring \(B\) which is free to slide on the wire. Show that, while both strings remain taut, the equation for the motion of the ring is $$2ma \ddot\theta=\lambda(\cos\theta-\sin\theta)-mg\sin\theta,$$ where \(\theta\) is the angle \( \angle{CAB}\). Initially the system is at rest in equilibrium with \(\sin\theta=\frac{3}{5}\). Deduce that \(5\lambda=24mg\). The ring is now displaced slightly. Show that, in the ensuing motion, it will oscillate with period approximately $$10\pi\sqrt{a\over91g}\,.$$

A small ball of mass \(m\) is suspended in equilibrium by a light elastic string of natural length \(l\) and modulus of elasticity \(\lambda.\) Show that the total length of the string in equilibrium is \(l(1+mg/\lambda).\) If the ball is now projected downwards from the equilibrium position with speed \(u_{0},\) show that the speed \(v\) of the ball at distance \(x\) below the equilibrium position is given by \[ v^{2}+\frac{\lambda}{lm}x^{2}=u_{0}^{2}. \] At distance \(h\), where \(\lambda h^{2} < lmu_{0}^{2},\) below the equilibrium position is a horizontal surface on which the ball bounces with a coefficient of restitution \(e\). Show that after one bounce the velocity \(u_{1}\) at \(x=0\) is given by \[ u_{1}^{2}=e^{2}u_{0}^{2}+\frac{\lambda}{lm}h^{2}(1-e^{2}), \] and that after the second bounce the velocity \(u_{2}\) at \(x=0\) is given by \[ u_{2}^{2}=e^{4}u_{0}^{2}+\frac{\lambda}{lm}h^{2}(1-e^{4}). \]