6 problems found
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.
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 long, light, inextensible string passes through a small, smooth ring fixed at the point \(O\). One end of the string is attached to a particle \(P\) of mass \(m\) which hangs freely below \(O\). The other end is attached to a bead, \(B\), also of mass \(m\), which is threaded on a smooth rigid wire fixed in the same vertical plane as \(O\). The distance \(OB\) is \(r\), the distance \(OH\) is \(h\) and the height of the bead above the horizontal plane through~\(O\) is \(y\), as shown in the diagram.
A train starts from a station. The tractive force exerted by the engine is at first constant and equal to \(F\). However, after the speed attains the value \(u\), the engine works at constant rate \(P,\) where \(P=Fu.\) The mass of the engine and the train together is \(M.\) Forces opposing motion may be neglected. Show that the engine will attain a speed \(v\), with \(v\geqslant u,\) after a time \[ t=\frac{M}{2P}\left(u^{2}+v^{2}\right). \] Show also that it will have travelled a distance \[ \frac{M}{6P}(2v^{3}+u^{3}) \] in this time.
Solution: While the force is constant, the train is accelerating at \(\frac{F}{M}\), and since \(u = \frac{F}{M} t_1 \Rightarrow t_1 = \frac{Mu^2}{Fu} = \frac{Mu^2}{P}\). Once the train is being driven at a constant rate, we can observe that change in energy will be power times time, ie \(Pt_2 = \frac{1}{2}M(v^2 - u^2) \Rightarrow t_2 = \frac{M}{2P} ( v^2 - u^2)\). Therefore the total time will be \(t_1 + t_2 = \frac{M}{2P} ( u^2 + v^2)\). During the first period, the distance will be: \(s_1 = \frac12 \frac{F}{M} t_1^2 = \frac12 \frac{F}{M} \frac{M^2u^2}{F^2} = \frac{Mu^3}{2P}\) In the second period, \(P = Fu\) and so \(\text{Force} = \frac{P}{v} \Rightarrow M v \frac{\d v}{\d x} = \frac{P}{v} \Rightarrow M \l \frac{v^3}{3} - \frac{u^3}{3}\r = Ps_2\) and therefore total distance will be: \(\frac{M}{6P}(2v^{3}+u^{3})\)
\(\,\)
The edges \(OA,OB,OC\) of a rigid cube are taken as coordinate axes and \(O',A',B',C'\) are the vertices diagonally opposite \(O,A,B,C,\) respectively. The four forces acting on the cube are \[ \begin{pmatrix}\alpha\\ \beta\\ \gamma \end{pmatrix}\mbox{ at }O\ (0,0,0),\ \begin{pmatrix}\lambda\\ 0\\ 1 \end{pmatrix}\mbox{ at }O'\ (a,a,a),\ \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix}\mbox{ at }B\ (0,a,0),\ \mbox{ and }\begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix}\mbox{ at }B'\ (a,0,a). \] The moment of the system about \(O\) is zero: find \(\lambda,\mu\) and \(\nu\).
Solution: \begin{align*} &&\mathbf{M} &= \begin{pmatrix}\lambda \\ 0\\ 1 \end{pmatrix} \times \begin{pmatrix}a\\ a \\ a \end{pmatrix} + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} \times \begin{pmatrix} 0 \\ a \\ 0 \end{pmatrix} + \begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix} \times \begin{pmatrix} a \\ 0 \\ a \end{pmatrix} \\ &&&= \begin{pmatrix} -a \\ -a(\lambda -1) \\ \lambda a \end{pmatrix} + \begin{pmatrix} -2a \\ 0 \\ -a \end{pmatrix} + \begin{pmatrix} \mu a \\ -a(1-\nu) \\ -a \mu \end{pmatrix} \\ &&&=a \begin{pmatrix} \mu - 3 \\ \nu - \lambda \\ \lambda-1-\mu \end{pmatrix} \\ \Rightarrow && \mu &= 3, \lambda = 4, \nu = 4 \end{align*}