Problems

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 \(\ddot x = -g\tan\alpha\,\).

Particles \(P\) and \(Q\) have masses \(3m\) and \(4m\), respectively. They lie on the outer curved surface of a~smooth circular cylinder of radius~\(a\) which is fixed with its axis horizontal. They are connected by a light inextensible string of length \(\frac12 \pi a\), which passes over the surface of the cylinder. The particles and the string all lie in a vertical plane perpendicular to the axis of the cylinder, and the axis intersects this plane at \(O\). Initially, the particles are in equilibrium. Equilibrium is slightly disturbed and \(Q\) begins to move downwards. Show that while the two particles are still in contact with the cylinder the angle \(\theta\) between \(OQ\) and the vertical satisfies \[ 7a\dot\theta^2 +8g \cos\theta + 6 g\sin\theta = 10g\,. \]

1. Given that \(Q\) loses contact with the cylinder first, show that it does so when~\(\theta=\beta\), where \(\beta\) satisfies \[ 15\cos\beta +6\sin\beta =10. \]
2. Show also that while \(P\) and \(Q\) are still in contact with the cylinder the tension in the string is $\frac {12}7 mg(\sin\theta +\cos\theta)\,$.

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 \(0<\alpha<\pi\,\).

In the figure, \(W_{1}\) and \(W_{2}\) are wheels, both of radius \(r\). Their centres \(C_{1}\) and \(C_{2}\) are fixed at the same height, a distance \(d\) apart, and each wheel is free to rotate, without friction, about its centre. Both wheels are in the same vertical plane. Particles of mass \(m\) are suspended from \(W_{1}\) and \(W_{2}\) as shown, by light inextensible strings would round the wheels. A light elastic string of natural length \(d\) and modulus elasticity \(\lambda\) is fixed to the rims of the wheels at the points \(P_{1}\) and \(P_{2}.\) The lines joining \(C_{1}\) to \(P_{1}\) and \(C_{2}\) to \(P_{2}\) both make an angle \(\theta\) with the vertical. The system is in equilibrium. \noindent

• sep}{3mm}
• \(\bf (i)\) no equilibrium positions,
• \(\bf (ii)\) just one equilibrium position,
• \(\bf (iii)\) exactly two equilibrium positions,
• \(\bf (iv)\) more than two equilibrium positions?