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.

Two particles \(A\) and \(B\) of masses \(m\) and \(2 m\), respectively, are connected by a light spring of natural length \(a\) and modulus of elasticity \(\lambda\). They are placed on a smooth horizontal table with \(AB\) perpendicular to the edge of the table, and \(A\) is held on the edge of the table. Initially the spring is at its natural length. Particle \(A\) is released. At a time \(t\) later, particle \(A\) has dropped a distance \(y\) and particle \( B\) has moved a distance \(x\) from its initial position (where \(x < a\)). Show that \( y + 2x= \frac12 gt^2\). The value of \(\lambda\) is such that particle \(B\) reaches the edge of the table at a time \(T\) given by \(T= \sqrt{6a/g\,}\,\). By considering the total energy of the system (without solving any differential equations), show that the speed of particle \(B\) at this time is \(\sqrt{2ag/3\,}\,\).

Three pegs \(P\), \(Q\) and \(R\) are fixed on a smooth horizontal table in such a way that they form the vertices of an equilateral triangle of side \(2a\). A particle \(X\) of mass \(m\) lies on the table. It is attached to the pegs by three springs, \(PX\), \(QX\) and \(RX\), each of modulus of elasticity \(\lambda\) and natural length \(l\), where \(l < \frac{ \ 2 }{\sqrt3}\, a\). Initially the particle is in equilibrium. Show that the extension in each spring is \(\frac{\ 2}{\sqrt3}\,a -l\,\). The particle is then pulled a small distance directly towards \(P\) and released. Show that the tension \(T\) in the spring \(RX\) is given by \[ T= \frac {\lambda} l \left( \sqrt{\frac {4a^2}3 + \frac{2ax}{\sqrt3} +x^2\; }\; -l\right) , \] where \(x\) is the displacement of \(X\) from its equilibrium position. Show further that the particle performs approximate simple harmonic motion with period \[ 2\pi \sqrt{ \frac{4mla}{3 (4a-\sqrt3 \, l)\lambda } \; }\,. \]

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 \[ \sqrt {\frac {32a}{kv}} \int_0^1 \frac 1 {\sqrt{1-u^4}} \, \d u \, . \]

Two particles \(X\) and \(Y\), of equal mass \(m\), lie on a smooth horizontal table and are connected by a light elastic spring of natural length \(a\) and modulus of elasticity \(\lambda\). Two more springs, identical to the first, connect \(X\) to a point \(P\) on the table and \(Y\) to a point \(Q\) on the table. The distance between \(P\) and \(Q\) is \(3a\). Initially, the particles are held so that \(XP=a\), \(YQ= \frac12 a\,\), and \(PXYQ\) is a straight line. The particles are then released. At time \(t\), the particle \(X\) is a distance \(a+x\) from \(P\) and the particle \(Y\) is a distance \(a+y\) from \(Q\). Show that \[ m \frac{\.d ^2 x}{\.d t^2} = -\frac\lambda a (2x+y) \] and find a similar expression involving \(\dfrac{\.d^2 y}{\.d t^2}\). Deduce that \[ x-y = A\cos \omega t +B \sin\omega t \] where \(A\) and \(B\) are constants to be determined and \(ma\omega^2=\lambda\). Find a similar expression for \(x+y\). Show that \(Y\) will never return to its initial position.

An equilateral triangle, comprising three light rods each of length \(\sqrt3a\), has a particle of mass \(m\) attached to each of its vertices. The triangle is suspended horizontally from a point vertically above its centre by three identical springs, so that the springs and rods form a tetrahedron. Each spring has natural length \(a\) and modulus of elasticity \(kmg\), and is light. Show that when the springs make an angle \(\theta\) with the horizontal the tension in each spring is \[ \frac{ kmg(1-\cos\theta)}{\cos\theta}\,. \] Given that the triangle is in equilibrium when \(\theta = \frac16 \pi\), show that \(k=4\sqrt3 +6\). The triangle is released from rest from the position at which \(\theta=\frac13\pi\). Show that when it passes through the equilibrium position its speed \(V\) satisfies \[ V^2 = \frac{4ag}3(6+\sqrt3)\,. \]

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 light spring is fixed at its lower end and its axis is vertical. When a certain particle \(P\) rests on the top of the spring, the compression is \(d\). When, instead, \(P\) is dropped onto the top of the spring from a height \(h\) above it, the compression at time \(t\) after \(P\) hits the top of the spring is \(x\). Obtain a second-order differential equation relating \(x\) and \(t\) for \(0\le t \le T\), where \(T\) is the time at which \(P\) first loses contact with the spring. Find the solution of this equation in the form \[ x= A + B\cos (\omega t) + C\sin(\omega t)\,, \] where the constants \(A\), \(B\), \(C\) and \(\omega\) are to be given in terms of \(d\), \(g\) and \(h\) as appropriate. Show that \[ T = \sqrt{d/g\;} \left (2 \pi - 2 \arctan \sqrt{2h/d\;}\;\right)\,. \]

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.

Two thin discs, each of radius \(r\) and mass \(m\), are held on a rough horizontal surface with their centres a distance \(6r\) apart. A thin light elastic band, of natural length \(2\pi r\) and modulus \(\dfrac{\pi mg}{12}\), is wrapped once round the discs, its straight sections being parallel. The contact between the elastic band and the discs is smooth. The coefficient of static friction between each disc and the horizontal surface is \(\mu\), and each disc experiences a force due to friction equal to \(\mu mg\) when it is sliding. The discs are released simultaneously. If the discs collide, they rebound and a half of their total kinetic energy is lost in the collision.

1. Show that the discs start sliding, but come to rest before colliding, if and only if \mbox{\(\frac23 <\mu <1\)}.
2. Show that, if the discs collide at least once, their total kinetic energy just before the first collision is \(\frac43 mgr(2-3\mu)\).
3. Show that if \(\frac 4 9 > \mu^2 >\frac{5}{27}\) the discs come to rest exactly once after the first collision.

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 \(\lambda\)\,.