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2006 Paper 3 Q9
D: 1700.0 B: 1484.0
work-energy potential energy equilibrium bead on wire inextensible string constraint equations differentiation smooth ring conservation of energy

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.

\psset{xunit=0.8cm,yunit=0.8cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.5pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(-3.93,-2.62)(3.97,4.92) \psplot[plotpoints=200]{-3.2}{3.2}{4-0.6*x^(2)} \rput{0}(0,0){\psellipse(0,0)(0.61,0.16)} \psline(0,0)(1.73,2.19) \psline[linestyle=dashed,dash=1pt 1pt](-5,0)(5,0) \rput[tl](0.46,0.58){\(\theta\)} \rput[tl](0.06,-0.27){\(O\)} \rput[tl](0.15,-1.35){\(P\)} \psline[linestyle=dashed,dash=1pt 1pt](1.73,2.19)(1.73,0) \rput[tl](1.85,1){\(y\)} \rput[tl](0.77,1.7){\(r\)} \rput[tl](1.84,2.68){\(B\)} \rput[tl](0.06,4.47){\(H\)} \rput[tl](-0.36,2.3){\(h\)} \psline[linestyle=dashed,dash=1pt 1pt](0,7)(0,0) \psline(0,0)(0,-2) \begin{scriptsize} \psdots[dotsize=5pt 0,dotstyle=*](1.73,2.19) \psdots[dotsize=5pt 0,dotstyle=*](0,-2) \psdots[dotstyle=*,linecolor=darkgray](0,0) \end{scriptsize} \end{pspicture*}
The shape of the wire is such that the system can be in static equilibrium for all positions of the bead. By considering potential energy, show that the equation of the wire is \(y+r =2h\). The bead is initially at \(H\). It is then projected along the wire with initial speed \(V\). Show that, in the subsequent motion, \[ \dot \theta = -\frac {h \dot r }{r \sqrt{rh -h^2}}\, \] where \(\theta\) is given by \(\theta = \arcsin(y/r)\). Hence show that the speed of the particle \(P\) is \(V \Big(\dfrac{r-h}{2r-h}\Big)^{\!\frac12}\,\). \noindent[{\it Note that \(\arcsin \theta\) is another notation for \(\sin^{-1}\theta\).}]

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1997 Paper 1 Q10
D: 1516.0 B: 1484.0
circular motion vertical conical pendulum smooth ring string tension trigonometric identity resolving forces approximation

The point \(A\) is vertically above the point \(B\). A light inextensible string, with a smooth ring \(P\) of mass \(m\) threaded onto it, has its ends attached at \(A\) and \(B\). The plane \(APB\) rotates about \(AB\) with constant angular velocity \(\omega\) so that \(P\) describes a horizontal circle of radius \(r\) and the string is taut. The angle \(BAP\) has value \(\theta\) and the angle \(ABP\) has value \(\phi\). Show that \[\tan\frac{\phi-\theta}{2}=\frac{g}{r\omega^{2}}.\] Find the tension in the string in terms of \(m\), \(g\), \(r\), \(\omega\) and \(\sin\frac{1}{2}(\theta+\phi)\). Deduce from your results that if \(r\omega^2\) is small compared with \(g\), then the tension is approximately \(\frac{mg}{2}\)

Solution: None \begin{multicols}{2}

TikZ diagram
\columnbreak \begin{align*} N2(\uparrow): && T \cos \theta - T \cos \phi - mg &= 0 \\ N2(\rightarrow): && T \sin \theta + T \sin \phi &= m r \omega^2 \\ \\ && T \cos \theta - T \cos \phi &= mg \tag{\(*\)}\\ && T \sin \theta + T \sin \phi &= m r \omega^2 \tag{{\(**\)}} \end{align*} \end{multicols} Dividing \((*)\) by \((**)\) we obtain: \begin{align*} \frac{g}{r\omega^2} &= \frac{\cos \theta - \cos \phi}{\sin \theta + \sin \phi} \\ &= \frac{2 \sin \left ( \frac{\theta + \phi}2 \right )\sin \left (\frac{\phi - \theta}2 \right )}{2 \sin \left ( \frac{\theta + \phi}2 \right )\cos \left (\frac{\phi - \theta}2 \right )} \\ &= \tan \left ( \frac{\phi - \theta}2 \right ) \end{align*} as required. Squaring and adding \((*)\) and \((**)\) we obtain: \begin{align*} && m^2(g^2 + r^2 \omega^4) &= T^2(2 + \sin \theta \sin \phi - \cos \theta \cos \phi) \\ && &= T^2(2 - 2\cos (\theta + \phi)) \\ && &= T^2(2 - 2(1 - 2 \sin^2 \left ( \frac{\theta + \phi}2 \right ) )) \\ && &= T^2(4 \sin^2 \left ( \frac{\theta + \phi}2 \right )) \\ \Rightarrow && T &= \frac{m\sqrt{g^2 + r^2 \omega^4}}{2 \sin \left ( \frac{\theta + \phi}2 \right )} \\ \Rightarrow && T &= \frac{mg\sqrt{1 + \frac{r^2 \omega^4}{g^2}}}{2 \sin \left ( \frac{\theta + \phi}2 \right )} \end{align*} If \(r \omega^2 \ll g\) then \(\tan \l \frac{\phi - \theta}2 \r\) is very large, so \(\phi - \theta \approx \pi\) and so \(\phi + \theta \approx \pi\). We can then say that \[ T \approx \frac{mg}{2}\]

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