4 problems found
A train moves westwards on a straight horizontal track with constant acceleration \(a\), where \(a > 0\). Axes are chosen as follows: the origin is fixed in the train; the \(x\)-axis is in the direction of the track with the positive \(x\)-axis pointing to the East; and the positive \(y\)-axis points vertically upwards. A smooth wire is fixed in the train. It lies in the \(x\)--\(y\) plane and is bent in the shape given by \(ky = x^2\), where \(k\) is a positive constant. A small bead is threaded onto the wire. Initially, the bead is held at the origin. It is then released.
A child is playing with a toy cannon on the floor of a long railway carriage. The carriage is moving horizontally in a northerly direction with acceleration \(a\). The child points the cannon southward at an angle \(\theta\) to the horizontal and fires a toy shell which leaves the cannon at speed \(V\). Find, in terms of \(a\) and \(g\), the value of \(\tan 2\theta\) for which the cannon has maximum range (in the carriage). If \(a\) is small compared with \(g\), show that the value of \(\theta\) which gives the maximum range is approximately \[ \frac \pi 4 + \frac a {2g}, \] and show that the maximum range is approximately \(\displaystyle \frac {V^2} g + \frac {V^2a}{g^2}. \)
Solution: \begin{align*} && s_x &= V \cos \theta t + \frac12at^2 \\ && s_y &= V \sin \theta t - \frac12 gt^2 \\ \Rightarrow && T &= \frac{2V \sin \theta}g \\ \Rightarrow && s_{max} &= \frac{2V^2 \sin \theta \cos \theta}{g} + \frac12a \frac{4V^2 \sin^2 \theta}{g^2} \\ &&&= (g \sin 2 \theta+2a\sin^2 \theta)\frac{V^2}{g^2} \\ && \frac{\d s_{max}}{\d \theta} &= (2g \cos 2 \theta +4 a \cos \theta \sin \theta)\frac{V^2}{g^2} \\ &&&= (2g \cos 2\theta + 2a \sin2 \theta) \frac{V^2}{g^2} \\ \Rightarrow && \tan 2\theta &= -\frac{a}{g} \\ \Rightarrow && 2 \theta &\in (\frac{\pi}2, \pi) \\ \Rightarrow && \tan \left (\frac{\pi}{2} - 2 \theta\right) &=-\frac{a}{g} \\ \Rightarrow && \frac{\pi}{2} - 2 \theta&\approx -\frac{a}{g} \\ \Rightarrow && \theta &\approx \frac{\pi}{4} + \frac{a}{2g} \\ \\ && s_{max} & \approx \left (g \sin \left (\frac{\pi}{2} + \frac{a}{g} \right)+2a\sin^2 \left ( \frac{\pi}{4} + \frac{a}{2g}\right)\right)\frac{V^2}{g^2} \\ &&&\approx \left (g \cdot 1+2a\left( \frac{1}{\sqrt{2}}(\frac{a}{2g}+1)\right)^2\right)\frac{V^2}{g^2} \\ &&&\approx \left (g+a\left(1+\frac{a}{g}\right)^2\right)\frac{V^2}{g^2} \\ &&&\approx \left (g+a\right)\frac{V^2}{g^2} \\ &&&= \frac{V^2}{g} + \frac{V^2a}{g} \end{align*}
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 horizontal plane rotates with constant angular velocity \(\Omega\) about a fixed vertical axis through a fixed point \(O\) of the plane. The point \(A\) is fixed in the plane and \(OA=a.\) A particle \(P\) lies on the plane and is joined to \(A\) by a light rod of length \(b(>a)\) freely pivoted at \(A\). Initially \(OAP\) is a straight line and \(P\) is moving with speed \((a+2\sqrt{ab})\Omega\) perpendicular to \(OP\) in the same sense as \(\Omega.\) At time \(t,\) \(AP\) makes an angle \(\phi\) with \(OA\) produced. Obtain an expression for the component of the acceleration of \(P\) perpendicular to \(AP\) in terms of \(\dfrac{\mathrm{d}^{2}\phi}{\mathrm{d}t^{2}},\phi,a,b\) and \(\Omega.\) Hence find \(\dfrac{\mathrm{d}\phi}{\mathrm{d}t}\), in terms of \(\phi,a,b\) and \(\Omega,\) and show that \(P\) never crosses \(OA.\)
Solution: Set up coordinate axes so that at time \(t\) \(OA\) is the \(x\)-axis, and all rotations are counter-clockwise. Then if \(OA = \mathbf{a}\), \(AP = \mathbf{x}\) and \(OP = \mathbf{p}\) we have: \begin{align*} \mathbf{a} &= \binom{a}{0} \\ \dot{\mathbf{a}} &= \binom{0}{-a \Omega} \\ \ddot{\mathbf{a}} &= \binom{-a \Omega^2}{0} \\ \\ \mathbf{x} &= \binom{b \cos \phi }{b \sin \phi } \\ \dot{\mathbf{x}} &= b \dot{\phi} \binom{-\sin \phi}{\cos \phi} \\ \ddot{\mathbf{x}} &= \binom{-b \ddot{\phi} \sin \phi-b \dot{\phi}^2 \cos \phi }{b \ddot{\phi} \cos \phi-b \dot{\phi}^2 \sin \phi} \\ \\ \ddot{\mathbf{p}} &= \binom{-a \Omega^2 +-b \ddot{\phi} \sin \phi-b \dot{\phi}^2 \cos \phi }{b \ddot{\phi} \cos \phi-b \dot{\phi}^2 \sin \phi} \end{align*} We can take a dot product with \(\mathbf{n} = \binom{-\sin \phi}{\cos \phi}\) to obtain the component perpendicular to \(AP\), which is: \begin{align*} && \binom{-\sin \phi}{\cos \phi} \cdot \ddot{x} &= a \Omega^2 \sin \phi + b \ddot{\phi} \end{align*} Noticing that this component must be \(0\) (since the only force acting on \(P\) is the rod), this must be equal to zero. \begin{align*} && 0 &= a \Omega^2 \sin \phi + b \ddot{\phi} \\ \Rightarrow && 0 &= a \Omega^2 \dot{\phi} \sin \phi + b\dot{\phi} \ddot{\phi} \\ \Rightarrow && C &= -a \Omega^2 \cos \phi + \tfrac12 b \dot{\phi}^2 \end{align*} Noticing that the initial conditions are \(\phi = 0\) and \(\dot{\phi} = 2\sqrt{\frac{a}{b}} \Omega\), so \begin{align*} && C &= -a \Omega^2+ \tfrac12 b \left ( 2\sqrt{\frac{a}{b}} \Omega \right)^2 \\ &&&= -a \Omega^2 + 2a \Omega^2 \\ &&&= a \Omega^2\\ \Rightarrow && \dot{\phi} &=\sqrt{\frac{2}{b} \left ( a \Omega^2 + a \Omega^2 \cos \phi \right)} \\ &&&= \Omega \sqrt{\frac{2a}{b}} \sqrt{1+ \cos \phi} \\ &&& = \Omega \sqrt{\frac{2a}{b}}\sqrt{2} \cos \tfrac{\phi}{2} \\ \Rightarrow && \int \sec \tfrac{\phi}{2} \d \phi &= 2 \Omega \sqrt{\frac{a}{b}}t \\ \Rightarrow && \tfrac12 \ln | \sec \tfrac{\phi}{2}+\tan \tfrac{\phi}{2} | &= 2 \Omega \sqrt{\frac{a}{b}}t + C \\ t = 0, \phi = 0: && C = 0 \\ \Rightarrow && \sec \tfrac{\phi}{2}+\tan \tfrac{\phi}{2} &= e^{4 \Omega \sqrt{\frac{a}{b}}t} \end{align*} Since when \(t > 0\) this is positive and larger than \(1\) we cannot have \(\phi = 0\) and since it remains below infinite \(\phi\) cannot reach \(\pi\). Therefore it cannot cross \(OA\)