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2005 Paper 3 Q11
D: 1700.0 B: 1500.0
circular motion rigid body energy equilibrium stability trigonometric identity angular velocity potential energy spindle vertical plane rotation

A horizontal spindle rotates freely in a fixed bearing. Three light rods are each attached by one end to the spindle so that they rotate in a vertical plane. A particle of mass \(m\) is fixed to the other end of each of the three rods. The rods have lengths \(a\), \(b\) and \(c\), with \(a > b > c\,\) and the angle between any pair of rods is \(\frac23 \pi\). The angle between the rod of length \(a\) and the vertical is \(\theta\), as shown in the diagram. \vspace*{-0.1in}

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Find an expression for the energy of the system and show that, if the system is in equilibrium, then \[ \tan \theta = -\frac{(b-c) \sqrt{3}}{2a-b-c}\;. \] Deduce that there are exactly two equilibrium positions and determine which of the two equilibrium positions is stable. Show that, for the system to make complete revolutions, it must pass through its position of stable equilibrium with an angular velocity of at least \[ \sqrt{\frac{4gR}{a^2+b^2+c^2}} \, , \] where \(2R^2 = (a-b)^2+(b-c)^2 +(c-a)^2 \;\).

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2001 Paper 3 Q11
D: 1700.0 B: 1500.0
moments of inertia rotational dynamics energy conservation cylinder hanging mass angular velocity trigonometric equation optimisation

A uniform cylinder of radius \(a\) rotates freely about its axis, which is fixed and horizontal. The moment of inertia of the cylinder about its axis is \(I\,\). A light string is wrapped around the cylinder and supports a mass \(m\) which hangs freely. A particle of mass \(M\) is fixed to the surface of the cylinder. The system is held at rest with the particle vertically below the axis of the cylinder, and then released. Find, in terms of \(I\), \(a\), \(M\), \(m\), \(g\) and \(\theta\), the angular velocity of the cylinder when it has rotated through angle \(\theta\,\). Show that the cylinder will rotate without coming to a halt if \(m/M>\sin\alpha\,\), where \(\alpha\) satisifes \(\alpha=\tan \frac12\alpha\) and \(0<\alpha<\pi\,\).

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1999 Paper 1 Q11
D: 1500.0 B: 1486.1
circular motion gravitation inverse square law equilateral triangle centre of mass angular velocity three-body problem symmetry

The force of attraction between two stars of masses \(m_{1}\) and \(m_{2}\) a distance \(r\) apart is \(\gamma m_{1}m_{2}/r^{2}\). The Starmakers of Kryton place three stars of equal mass \(m\) at the corners of an equilateral triangle of side \(a\). Show that it is possible for each star to revolve round the centre of mass of the system with angular velocity \((3\gamma m/a^{3})^{1/2}\). Find a corresponding result if the Starmakers place a fourth star, of mass \(\lambda m\), at the centre of mass of the system.

Solution: The net force on the planets will always be towards the centre of mass (by symmetry or similar arguments). Therefore it suffices to check whether we can find a speed where the planets follow uniform circular motion, ie \(F = mr \omega^2\). (But clearly this is possible, we just need to find the speed)

TikZ diagram
\begin{align*} && F &= m r \omega^2 \\ && 2\frac{\gamma m^2}{a^2} \cos 30^{\circ} &= m \frac{a}{\sqrt{3}} \omega^2 \\ \Rightarrow && \frac{\sqrt{3}\gamma m^2}{a^2} &= \frac{ma \omega^2}{\sqrt{3}} \\ \Rightarrow && \omega^2 &= \frac{3\gamma m}{a^3} \\ \Rightarrow && \omega &= \left ( \frac{3\gamma m}{a^3}\right)^{1/2} \end{align*}
TikZ diagram
In the second scenario, we are interested in when: \begin{align*} && F &= m r \omega^2 \\ && \underbrace{2\frac{\gamma m^2}{a^2} \cos 30^{\circ}}_{\text{to other symmetric planets}} + \underbrace{\frac{\gamma \lambda m^2}{a^2}}_{\text{central planet}} &= m \frac{a}{\sqrt{3}} \omega^2 \\ \Rightarrow && \frac{(\sqrt{3}+\lambda)\gamma m^2}{a^2} &= \frac{ma \omega^2}{\sqrt{3}} \\ \Rightarrow && \omega^2 &= \frac{(3+\sqrt{3}\lambda)\gamma m}{a^3} \\ \Rightarrow && \omega &= \left ( \frac{(3+\sqrt{3}\lambda)\gamma m}{a^3}\right)^{1/2} \end{align*}

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1996 Paper 2 Q10
D: 1600.0 B: 1516.0
circular motion projectile optimisation angular velocity greatest height vertical circle release point

The plot of `Rhode Island Red and the Henhouse of Doom' calls for the heroine to cling on to the circumference of a fairground wheel of radius \(a\) rotating with constant angular velocity \(\omega\) about its horizontal axis and then let go. Let \(\omega_{0}\) be the largest value of \(\omega\) for which it is not possible for her subsequent path to carry her higher than the top of the wheel. Find \(\omega_{0}\) in terms of \(a\) and \(g\). If \(\omega>\omega_{0}\) show that the greatest height above the top of the wheel to which she can rise is \[\frac{a}{2}\left(\frac{\omega}{\omega_{0}} -\frac{\omega_{0}}{\omega}\right)^{\!\!2}.\]

Solution:

TikZ diagram
\begin{align*} \uparrow: && v &= u + at \\ \Rightarrow && T &= \frac{a \omega \sin \theta}{g} \\ && s &= ut + \frac12 gt^2 \\ \Rightarrow && s &= a\omega \sin \theta \cdot \frac{a \omega \sin \theta}{g} - \frac12 g \left ( \frac{a \omega \sin \theta}{g} \right) ^2 \\ &&&= \frac1{2g} a^2 \omega^2 \sin^2 \theta \\ s < \text{distance to top}: && \frac1{2g} a^2 \omega^2 \sin^2 \theta &< a(1- \cos \theta) \\ \Rightarrow && \omega^2 &< \frac{2g}{a} \frac{1-\cos \theta}{\sin^2 \theta} \\ &&&= \frac{2g}{a} \frac{2 \sin^2 \tfrac12 \theta}{4 \sin^2 \tfrac12 \theta \cos^2 \tfrac12 \theta} \\ &&&= \frac{g}{a} \sec^2 \tfrac12 \theta \\ &&&\leq \frac{g}{a} \tag{since it holds for all \(\theta\) it holds for min \(\theta\)}\\ \Rightarrow && \omega_0 &= \sqrt{\frac{g}{a}} \\ \\ && \text{max height} &= \frac1{2g} a^2 \omega^2 \sin^2 \theta - a(1-\cos \theta) \\ &&&= \frac1{2g} a^2 \omega^2 (1-\cos^2 \theta) - a(1-\cos \theta) \\ &&&= \frac{a}{2} \left (- \frac{\omega^2}{\omega_0^2} \cos^2 \theta + 2 \cos \theta + \frac{\omega^2}{\omega_0^2}-2 \right) \\ &&&= \frac{a}{2} \left (-\left (\frac{\omega_0}{\omega}- \frac{\omega}{\omega_0} \cos \theta \right)^2 + \frac{\omega^2}{\omega_0^2}-2+\frac{\omega_0^2}{\omega^2} \right) \\ &&&= \frac{a}{2} \left ( \frac{\omega}{\omega_0} - \frac{\omega_0}{\omega} \right)^2 - \frac{a}{2} \left (\frac{\omega_0}{\omega}- \frac{\omega}{\omega_0} \cos \theta \right)^2 \end{align*} If \(\omega > \omega_0\) we can find a \(\theta\) such that the second bracket is \(0\), hence the maximium height is as desired.

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1994 Paper 1 Q10
D: 1516.0 B: 1484.0
circular motion conical pendulum elastic string Hooke's law inequality angular velocity horizontal circle tension

One end \(A\) of a light elastic string of natural length \(l\) and modulus of elasticity \(\lambda\) is fixed and a particle of mass \(m\) is attached to the other end \(B\). The particle moves in a horizontal circle with centre on the vertical through \(A\) with angular velocity \(\omega.\) If \(\theta\) is the angle \(AB\) makes with the downward vertical, find an expression for \(\cos\theta\) in terms of \(m,g,l,\lambda\) and \(\omega.\) Show that the motion described is possible only if \[ \frac{g\lambda}{l(\lambda+mg)}<\omega^{2}<\frac{\lambda}{ml}. \]

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1989 Paper 3 Q12
D: 1700.0 B: 1500.0
circular motion rotating frame angular velocity acceleration components differential equation rod constraint pivoted rod non-inertial frame

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 \(OA\)

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1988 Paper 2 Q14
D: 1600.0 B: 1488.9
circular motion conical motion rotating rod rigid body angular velocity reaction force vertical axis rotation equilibrium condition

Two particles of mass \(M\) and \(m\) \((M>m)\) are attached to the ends of a light rod of length \(2l.\) The rod is fixed at its midpoint to a point \(O\) on a horizontal axle so that the rod can swing freely about \(O\) in a vertical plane normal to the axle. The axle rotates about a vertical axis through \(O\) at a constant angular speed \(\omega\) such that the rod makes a constant angle \(\alpha\) \((0<\alpha<\frac{1}{2}\pi)\) with the vertical. Show that \[ \omega^{2}=\left(\frac{M-m}{M+m}\right)\frac{g}{l\cos\alpha}. \] Show also that the force of reaction of the rod on the axle is inclined at an angle \[ \tan^{-1}\left[\left(\frac{M-m}{M+m}\right)^{2}\tan\alpha\right] \] with the downward vertical.

Solution:

TikZ diagram
The accelerations of \(M\) and \(m\) are \(l \sin \alpha \omega^2\) and \(-l \sin \alpha \omega^2\) so the forces \(R_M\) and \(R_m\) are \(M\binom{l \sin \alpha \omega^2}{g}, \,m \binom{-l \sin \alpha \omega^2}{g}\). Since the axle is rotating freely, the moment about \(O\) for the rod must be \(O\). The moment for \(M\) will be \(M\binom{l \sin \alpha\omega^2}{g} \cdot \binom{-l\cos \alpha}{l \sin \alpha} = lM\sin\alpha (g - l \cos \alpha\omega^2)\). The moment for \(m\) will be \(m \binom{-l \sin \alpha\omega^2}{g} \cdot \binom{-l\cos \alpha\omega^2}{l \sin \alpha} = lm \sin \alpha(g+l \cos \alpha\omega^2)\) Therefore \begin{align*} && lM\sin\alpha (g - l \cos \alpha\omega^2) &= lm \sin \alpha(g+l \cos \alpha\omega^2) \\ && M(g - l \cos \alpha \omega^2) &= m(g + l \cos \alpha \omega^2 ) \\ \Rightarrow && g(M-m) &= l \cos \alpha (M+m) \omega^2 \\ \Rightarrow && \omega^2 &= \left (\frac{M-m}{M+m} \right) \frac{g}{l \cos \alpha} \end{align*} as required. The total force on the rod is \(\mathbf{0}\) so the reaction force must be \(M\binom{l \sin \alpha \omega^2}{g}+ \,m \binom{-l \sin \alpha \omega^2}{g} = \binom{l \sin \alpha \omega^2 (M-m)}{(M+m)g}\) Therefore the angle this makes with downward vertical is: \begin{align*} \theta &= \tan^{-1} \left ( \frac{l \sin \alpha \omega^2 (M-m)}{(M+m)g} \right) \\ &= \tan^{-1} \left ( \frac{l \sin \alpha (M-m)}{(M+m)g} \omega^2\right) \\ &= \tan^{-1} \left ( \frac{l \sin \alpha (M-m)}{(M+m)g} \left (\frac{M-m}{M+m} \right) \frac{g}{l \cos \alpha}\right) \\ &= \tan^{-1}\left[\left(\frac{M-m}{M+m}\right)^{2}\tan\alpha\right] \end{align*} as required.

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