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2011 Paper 3 Q11
D: 1700.0 B: 1484.0
circular motion rigid body equilibrium couple energy conservation trigonometry disc strings geometry

A thin uniform circular disc of radius \(a\) and mass \(m\) is held in equilibrium in a horizontal plane a distance \(b\) below a horizontal ceiling, where \(b>2a\). It is held in this way by \(n\) light inextensible vertical strings, each of length \(b\); one end of each string is attached to the edge of the disc and the other end is attached to a point on the ceiling. The strings are equally spaced around the edge of the disc. One of the strings is attached to the point \(P\) on the disc which has coordinates \((a,0,-b)\) with respect to cartesian axes with origin on the ceiling directly above the centre of the disc. The disc is then rotated through an angle \(\theta\) (where \(\theta<\pi\)) about its vertical axis of symmetry and held at rest by a couple acting in the plane of the disc. Show that the string attached to~\(P\) now makes an angle \(\phi\) with the vertical, where \[ b\sin\phi = 2a \sin\tfrac12 \theta\,. \] Show further that the magnitude of the couple is \[ \frac {mga^2\sin\theta}{\sqrt{b^2-4a^2\sin^2 \frac12\theta \ } \ }\,. \] The disc is now released from rest. Show that its angular speed, \(\omega\), when the strings are vertical is given by \[ \frac{a^2\omega^2}{4g} = b-\sqrt{b^2 - 4a^2\sin^2 \tfrac12\theta \;}\,. \]

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1990 Paper 3 Q14
D: 1700.0 B: 1484.0
moments rigid body cube 3D vectors couple force system equilibrium line of action work-energy cross product

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 \(\nu\).

  1. Given that \(\alpha=\beta=\gamma=0,\) find the system consisting of a single force at \(B\) together with a couple which is equivalent to the given system.
  2. Given that \(\alpha=2,\beta=3\) and \(\gamma=2,\) find the equation of the locus about each point of which the moment of the system is zero. Find the number of units of work done on the cube when it moves (without rotation) a distance in the direction of this line under the action of the given forces only.

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*}

  1. To find the force we add all vectors: \begin{align*} \mathbf{F} &= \begin{pmatrix}\lambda \\ 0\\ 1 \end{pmatrix} + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} + \begin{pmatrix}1\\ \mu\\ \nu \end{pmatrix} \\ &= \begin{pmatrix}4\\ 0\\ 1 \end{pmatrix} + \begin{pmatrix}-1\\ 0\\ 2 \end{pmatrix} + \begin{pmatrix}1\\ 3 \\ 4 \end{pmatrix} \\ &= \begin{pmatrix} 4 \\ 3 \\ 7 \end{pmatrix} \end{align*} Since the moment about \(O\) is \(0\), we have the moment about \(B\) is: \begin{align*} \mathbf{M} &= \begin{pmatrix} 0 \\ a \\ 0 \end{pmatrix} \times \begin{pmatrix} 4 \\ 3 \\ 7\end{pmatrix} \\ &= \begin{pmatrix} 7a \\ 0 \\ -4a\end{pmatrix} \end{align*}
  2. \begin{align*} \mathbf{0} &= \mathbf{r} \times \begin{pmatrix} 4 + 2 \\ 3+3 \\ 7+2 \end{pmatrix} \\ &= \mathbf{r} \times \begin{pmatrix} 6 \\ 6 \\ 9 \end{pmatrix} \\ \end{align*} Therefore \(\mathbf{r} = t\begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix}\) (ie a line) \begin{align*} \text{Work done} &= \text{Force} \cdot \text{distance} \end{align*} Since they are parallel, it's just the magnitude of the force, which is \(3\sqrt{2^2+2^2+3^2} = 3\sqrt{17}\)

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