24 problems found
Two light elastic springs each have natural length \(a\). One end of each spring is attached to a particle \(P\) of weight \(W\). The other ends of the springs are attached to the end-points, \(B\) and \(C\), of a fixed horizontal bar \(BC\) of length \(2a\). The moduli of elasticity of the springs \(PB\) and \(PC\) are \(s_1 W\) and \(s_2 W\) respectively; these values are such that the particle \(P\) hangs in equilibrium with angle \(BPC\) equal to \(90^\circ\).
A particle \(P\) of mass \(m\) moves freely and without friction on a wire circle of radius \(a\), whose axis is horizontal. The highest point of the circle is \(H\), the lowest point of the circle is \(L\) and angle \(PHL = \theta\). A light spring of modulus of elasticity \(\lambda\) is attached to \(P\) and to \(H\). The natural length of the spring is \(l\), which is less than the diameter of the circle.
A light elastic spring \(AB\), of natural length \(a\) and modulus of elasticity \(kmg\), hangs vertically with one end \(A\) attached to a fixed point. A particle of mass \(m\) is attached to the other end \(B\). The particle is held at rest so that \(AB > a\) and is released. Find the equation of motion of the particle and deduce that the particle oscillates vertically. If the period of oscillation is \(\dfrac{2\pi}{\Omega}\), show that \(kg = a\Omega^2\). Suppose instead that the particle, still attached to \(B\), lies on a horizontal platform which performs simple harmonic motion vertically with amplitude \(b\) and period \(\dfrac{2\pi}{\omega}\). At the lowest point of its oscillation, the platform is a distance \(h\) below \(A\). Let \(x\) be the distance of the particle above the lowest point of the oscillation of the platform. When the particle is in contact with the platform, show that the upward force on the particle from the platform is \[ mg + m\Omega^2(a + x - h) + m\omega^2(b - x). \] Given that \(\omega < \Omega\), show that, if the particle remains in contact with the platform throughout its motion, \[ h \leqslant a\left(1 + \frac{1}{k}\right) + \frac{\omega^2 b}{\Omega^2}. \] Find the corresponding inequality if \(\omega > \Omega\). Hence show that, if the particle remains in contact with the platform throughout its motion, it is necessary that \[ h \leqslant a\left(1 + \frac{1}{k}\right) + b, \] whatever the value of \(\omega\).
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 \(\sqrt{2ag/3\,}\,\).
Solution: \begin{align*} \text{N2}(\downarrow): && mg -T &= m\ddot{y} \\ \text{N2}(\rightarrow): && T &= 2m\ddot{x} \\ \Rightarrow && g &= \ddot{y}+2\ddot{x} \\ \Rightarrow && \tfrac12gt^2 &= y + 2x \end{align*} At time \(T = \sqrt{6a/g}\), we have \(y + 2x = 3a\), note also that \(\dot{y}+2\dot{x} = gt\) \begin{array}{ccc} & \text{KE} & \text{GPE} & \text{EPE} \\ \text{Initial} & 0 & 0 & 0 \\ \text{Final} & \frac12m\dot{y}^2 + \frac12(2m)\dot{x}^2 & -mgy & \frac{\lambda (y-x)^2}{2a} \end{array} Also note when we head over the table, \(x = a\) and \(y = a\) \begin{align*} \text{COE}: && 0 &= \frac12m(gT-2\dot{x})^2+m\dot{x}^2-mga+\frac{\lambda(0)^2}{2a} \\ \Rightarrow && 0 &= (gT-2\dot{x})^2+2\dot{x}^2-2ga \\ &&&= (\sqrt{6ag}-2\dot{x})^2+2\dot{x}^2-2ga \\ &&&= 6\dot{x}^2-4\sqrt{6ag}+4ag \\ \Rightarrow &&&= (\sqrt{6}\dot{x} - 2\sqrt{ag})^2 \\ \Rightarrow && \dot{x} &= \sqrt{2ag/3} \end{align*} as required.
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 \[ 2\pi \sqrt{ \frac{4mla}{3 (4a-\sqrt3 \, l)\lambda } \; }\,. \]
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 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 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.
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 particle is attached to a point \(P\) of an unstretched light uniform spring \(AB\) of modulus of elasticity \(\lambda\) in such a way that \(AP\) has length \(a\) and \(PB\) has length \(b\). The ends \(A\) and \(B\) of the spring are now fixed to points in a vertical line a distance \(l\) apart, The particle oscillates along this line. Show that the motion is simple harmonic. Show also that the period is the same whatever the value of \(l\) and whichever end of the string is uppermost.
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 \[ v^{2}=(k+2)gl. \]
Solution: \begin{align*} && \text{Energy at the top} &= mgh \\ && \text{Energy at the bottom} &= \frac12\frac{\lambda (h-l)^2}{l} \\ \Rightarrow && mgh & = \frac{\frac{mg}{k}(h-l)^2}{2l} \\ \Rightarrow && 2hkl &= (h-l)^2 \\ \Rightarrow && 0 &= h^2-2lh-2hlk+l^2 \\ &&0&= h^2-2hl(k+1)+l^2 \\ \Rightarrow && \frac{h}{l} &= \frac{2(k+1)\pm \sqrt{4(k+1)^2-4}}{2} \\ &&&= (k+1) \pm \sqrt{k^2+2k} \\ \Rightarrow && h &= l \left ( (k+1) \pm \sqrt{k^2+2k} \right) \end{align*} Since the negative root is less than \(1\), she would have not fully extended the cord. Therefore \(h = l \left ( (k+1) + \sqrt{k^2+2k} \right)\) Her maximum speed will be when her acceleration is \(0\), ie \(g = \text{force from cord}\) ie \(mg = \frac{\lambda x}{l}\) or \(x = \frac{mgl}{\lambda} = \frac{mglk}{mg} = kl\). At this point by conservation of energy we will have \begin{align*} && mgh &= mg(h-l-x) + \frac12 m v^2+\frac{1}{2} \frac{mgx^2}{kl} \\ \Rightarrow && mg\left ( l + kl \right) &= \frac12 m v^2 + \frac12 \frac{mgl^2k^2}{kl} \\ \Rightarrow && 2g\left ( l + kl \right) &= v^2 + glk \\ \Rightarrow && v^2 &= gl(2+k) \end{align*}
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 \[ u_{2}^{2}=e^{4}u_{0}^{2}+\frac{\lambda}{lm}h^{2}(1-e^{4}). \]
A small lamp of mass \(m\) is at the end \(A\) of a light rod \(AB\) of length \(2a\) attached at \(B\) to a vertical wall in such a way that the rod can rotate freely about \(B\) in a vertical plane perpendicular to the wall. A spring \(CD\) of natural length \(a\) and modulus of elasticity \(\lambda\) is joined to the rod at its mid-point \(C\) and to the wall at a point \(D\) a distance \(a\) vertically above \(B\). The arrangement is sketched below. \noindent
\(ABCD\) is a horizontal line with \(AB=CD=a\) and \(BC=6a\). There are fixed smooth pegs at \(B\) and \(C\). A uniform string of natural length \(2a\) and modulus of elasticity \(kmg\) is stretched from \(A\) to \(D\), passing over the pegs at \(B\) and \(C\). A particle of mass \(m\) is attached to the midpoint \(P\) of the string. When the system is in equilibrium, \(P\) is a distance \(a/4\) below \(BC\). Evaluate \(k\). The particle is pulled down to a point \(Q\), which is at a distance \(pa\) below the mid-point of \(BC\), and is released from rest. \(P\) rises to a point \(R\), which is at a distance \(3a\) above \(BC\). Show that \(2p^2-p-17=0\). Show also that the tension in the strings is less when the particle is at \(R\) than when the particle is at \(Q\).