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1995 Paper 3 Q9
D: 1700.0 B: 1470.3
moments of inertia circular disc non-uniform density rotational dynamics stable equilibrium energy conservation optimisation compound pendulum

A thin circular disc of mass \(m\), radius \(r\) and with its centre of mass at its centre \(C\) can rotate freely in a vertical plane about a fixed horizontal axis through a point \(O\) of its circumference. A particle \(P\), also of mass \(m,\) is attached to the circumference of the disc so that the angle \(OCP\) is \(2\alpha,\) where \(\alpha\leqslant\pi/2\).

  1. In the position of stable equilibrium \(OC\) makes an angle \(\beta\) with the vertical. Prove that \[ \tan\beta=\frac{\sin2\alpha}{2-\cos2\alpha}. \]
  2. The density of the disc at a point distant \(x\) from \(C\) is \(\rho x/r.\) Show that its moment of inertia about the horizontal axis through \(O\) is \(8mr^{2}/5\).
  3. The mid-point of \(CP\) is \(Q\). The disc is held at rest with \(OQ\) horizontal and \(C\) lower than \(P\) and it is then released. Show that the speed \(v\) with which \(C\) is moving when \(P\) passes vertically below \(O\) is given by \[ v^{2}=\frac{15gr\sin\alpha}{2(2+5\sin^{2}\alpha)}. \] Find the maximum value of \(v^{2}\) as \(\alpha\) is varied.

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1993 Paper 3 Q13
D: 1700.0 B: 1484.0
moments of inertia energy conservation variable mass chain circular disc rotational dynamics tension Newton's second law

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A uniform circular disc with radius \(a\), mass \(4m\) and centre \(O\) is freely mounted on a fixed horizontal axis which is perpendicular to its plane and passes through \(O\). A uniform heavy chain \(PS\) of length \((4+\pi)a\), mass \((4+\pi)m\) and negligible thickness is hung over the rim of the disc as shown in the diagram: \(Q\) and \(R\) are the points of the chain at the same level as \(O\). The contact between the chain and the rim of the disc is sufficiently rough to prevent slipping. Initially, the system is at rest with \(PQ=RS =2a\). A particle of mass \(m\) is attached to the chain at \(P\) and the system is released. By considering the energy of the system, show that when \(P\) has descended a distance \(x\), its speed \(v\) is given by $$ (\pi+7)av^2 = 2g(x^2+ax). $$ By considering the part \(PQ\) of the chain as a body of variable mass, show that when \(S\) reaches \(R\) the tension in the chain at \(Q\) is $$ {5\pi -2 \over \pi +7} mg. $$

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1992 Paper 3 Q13
D: 1700.0 B: 1500.0
work-energy elastic string Hooke's law circular disc moment of inertia rotational dynamics energy conservation reaction force rigid body modulus of elasticity

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A uniform circular disc of radius \(2b,\) mass \(m\) and centre \(O\) is free to turn about a fixed horizontal axis through \(O\) perpendicular to the plane of the disc. A light elastic string of modulus \(kmg\), where \(k>4/\pi,\) has one end attached to a fixed point \(A\) and the other end to the rim of the disc at \(P\). The string is in contact with the rim of the disc along the arc \(PC,\) and \(OC\) is horizontal. The natural length of the string and the length of the line \(AC\) are each \(\pi b\) and \(AC\) is vertical. A particle \(Q\) of mass \(m\) is attached to the rim of the disc and \(\angle POQ=90^{\circ}\) as shown in the diagram. The system is released from rest with \(OP\) vertical and \(P\) below \(O\). Show that \(P\) reaches \(C\) and that then the upward vertical component of the reaction on the axis is \(mg(10-\pi k)/3\).

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