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2015 Paper 2 Q10
D: 1600.0 B: 1516.0
Newton's laws connected particles tension trigonometry acceleration constraint motion particle on floor normal reaction leaving surface

A particle of mass \(m\) is pulled along the floor of a room in a straight line by a light string which is pulled at constant speed \(V\) through a hole in the ceiling. The floor is smooth and horizontal, and the height of the room is \(h\). Find, in terms of \(V\) and \(\theta\), the speed of the particle when the string makes an angle of \(\theta\) with the vertical (and the particle is still in contact with the floor). Find also the acceleration, in terms of \(V\), \(h\) and \(\theta\). Find the tension in the string and hence show that the particle will leave the floor when \[ \tan^4\theta = \frac{V^2}{gh}\,. \]

Solution:

TikZ diagram
The length of the string is \(h/\cos \theta\), and it is decreasing at a rate \(V\). The distance along the ground is decreasing at a rate of \(V/\sin \theta\). Note that \(-V = \frac{\d}{\d t} \left ( \frac{h}{\cos \theta} \right) = \frac{h} {\cos^2 \theta} \sin \theta \cdot \dot{\theta} \Rightarrow \dot{\theta} = -\frac{V\cos^2\theta}{h \sin \theta}\). Note that \(a = \frac{\d}{\d t} \left ( \frac{V}{\sin \theta} \right) = -\frac{V}{\sin^2 \theta} \cos \theta \cdot \dot{\theta} = \frac{V^2 \cos^3 \theta}{h\sin^3 \theta}\). Resolving horizontally we must have \(T \sin \theta = ma \Rightarrow T = \frac{V^2m \cos^3 \theta}{h \sin^4 \theta}\). Resolving vertically at the point where we are about to leave the ground, we must have \(T\cos \theta = mg \Rightarrow \frac{V^2m \cos^4 \theta}{h \sin^4 \theta} = mg \Rightarrow \tan^4 \theta = \frac{V^2}{gh}\)

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2014 Paper 2 Q11
D: 1600.0 B: 1504.7
mechanics Newton's laws string and ring constraint motion horizontal rail constant angle trigonometric equation inextensible string particle on string acceleration tension

A small smooth ring \(R\) of mass \(m\) is free to slide on a fixed smooth horizontal rail. A light inextensible string of length~\(L\) is attached to one end,~\(O\), of the rail. The string passes through the ring, and a particle~\(P\) of mass~\(km\) (where \(k>0\)) is attached to its other end; this part of the string hangs at an acute angle \(\alpha\) to the vertical and it is given that \(\alpha\) is constant in the motion. Let \(x\) be the distance between \(O\) and the ring. Taking the \(y\)-axis to be vertically upwards, write down the Cartesian coordinates of~\(P\) relative to~\(O\) in terms of \(x\), \(L\) and~\(\alpha\).

  1. By considering the vertical component of the equation of motion of \(P\), show that \[ km\ddot x \cos\alpha = T \cos\alpha - kmg\,, \] where \(T\) is the tension in the string. Obtain two similar equations relating to the horizontal components of the equations of motion of \(P\) and \(R\).
  2. Show that \(\dfrac {\sin\alpha}{(1-\sin\alpha)^2_{\vphantom|}} = k\), and deduce, by means of a sketch or otherwise, that motion with \(\alpha\) constant is possible for all values of~\(k\).
  3. Show that \(\ddot x = -g\tan\alpha\,\).

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2007 Paper 3 Q9
D: 1700.0 B: 1484.0
SHM energy conservation elastic spring bead on wire circular motion horizontal period of oscillation constraint motion

Two small beads, \(A\) and \(B\), each of mass \(m\), are threaded on a smooth horizontal circular hoop of radius \(a\) and centre \(O\). The angle \(\theta\) is the acute angle determined by \(2\theta = \angle AOB\). The beads are connected by a light straight spring. The energy stored in the spring is \[ mk^2 a^2(\theta - \alpha)^2, \] where \(k\) and \(\alpha\) are constants satisfying \(k>0\) and \(\frac \pi 4< \alpha<\frac\pi2\). The spring is held in compression with \(\theta =\beta\) and then released. Find the period of oscillations in the two cases that arise according to the value of \(\beta\) and state the value of \(\beta\) for which oscillations do not occur.

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1997 Paper 3 Q11
D: 1700.0 B: 1500.0
moments of inertia rigid body compound pendulum angular momentum constraint motion separation condition clapper and bell Lagrangian mechanics

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A heavy symmetrical bell and clapper can both swung freely in a vertical plane about a point \(O\) on a horizontal beam at the apex of the bell. The mass of the bell is \(M\) and its moment of inertia about the beam is \(Mk^{2}\). Its centre of mass, \(G\), is a distance \(h\) from \(O\). The clapper may be regarded as a small heavy ball on a light rod of length \(l\). Initially the bell is held with its axis vertical and its mouth above the beam. The clapper ball rests against the side of the bell, with the rod making an angle \(\beta\) with the axis. The bell is then released. Show that, at the moment when the clapper and bell separate, the clapper rod makes an angle \(\alpha\) with the upwards vertical, where \[ \cot\alpha=\cot\beta-\frac{k^{2}}{hl}\mathrm{cosec}\beta. \]

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1988 Paper 3 Q14
D: 1700.0 B: 1484.0
circular motion bead on wire normal reaction radius of curvature parabola energy conservation projectile motion constraint motion

A small heavy bead can slide smoothly in a vertical plane on a fixed wire with equation \[ y=x-\frac{x^{2}}{4a}, \] where the \(y\)-axis points vertically upwards and \(a\) is a positive constant. The bead is projected from the origin with initial speed \(V\) along the wire.

  1. Show that for a suitable value of \(V\), to be determined, a motion is possible throughout which the bead exerts no pressure on the wire.
  2. Show that \(\theta,\) the angle between the particle's velocity at time \(t\) and the \(x\)-axis, satisfies \[ \frac{4a^{2}\dot{\theta}^{2}}{\cos^{6}\theta}+2ga(1-\tan^{2}\theta)=V^{2}. \]

Solution:

  1. The condition that the bead exerts no pressure on the wire is equivalent to the condition that the wire exerts no force on the bead. (Newton's Third Law). This is equivalent to the bead being projected under gravity. Notice that the initial projection is at \(45^{\circ}\) since \(\frac{dy}{dx}|_{x=0} = 1\). The position of the particle (under gravity) at time \(t\) is \(x = \frac{1}{\sqrt{2}}Vt\) and \(y = \frac{1}{\sqrt{2}}Vt - \frac12 gt^2 = x - \frac{1}{2}g \frac{2x^2}{V^2} = x - \frac{g}{V^2}x^2\). Therefore they follow the same trajectory if \(\frac{g}{V^2} = \frac{1}{4a} \Leftrightarrow V = 2\sqrt{ag}\)
  2. First note that the wire does no work as it is perpendicular to the velocity, so it is fine to use conservation of momentum. If we take our \(0\) GPE level to be be \(x = 0\), then we notice the initial energy is \(\frac12mV^2\). Secondly, notice that \(\tan \theta = \frac{\d y}{\d x} = 1- \frac{x}{2a} \Rightarrow x = 2a - 2a \tan \theta\) \begin{align*} y &= 2a(1-\tan \theta) - \frac{4a^2(1-\tan \theta)^2}{4a}\\ &= (1-\tan \theta)(2a-a(1-\tan \theta)) \\ &= a(1-\tan \theta)(1+\tan \theta) \\ &= a(1-\tan^2 \theta) \end{align*} GPE \(mga(1-\tan^2 \theta)\). To calculate the kinetic energy, notice that \(\dot{x} = v \cos \theta \dot{\theta}\) and \(\dot{x} = -2a\sec^2 \theta\dot{\theta} \Rightarrow v = -\frac{2a\dot{\theta} }{\cos^{3} \theta}\). Therefore, energy at time \(t\) is: \begin{align*} && \frac12 m V^2 &= \frac12 m \l - \frac{2a\dot{\theta}}{\cos^3 \theta} \r^2 + mga(1-\tan^2 \theta) \\ \Rightarrow && V^2 &= \frac{4a^2\dot{\theta}^2}{\cos^6 \theta} + 2ag(1-\tan^2 \theta) \end{align*}

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