Problems

A small ring of mass \(m\) is free to slide without friction on a hoop of radius \(a\). The hoop is fixed in a vertical plane. The ring is connected by a light elastic string of natural length \(a\) to the highest point of the hoop. The ring is initially at rest at the lowest point of the hoop and is then slightly displaced. In the subsequent motion the angle of the string to the downward vertical is \(\phi\). Given that the ring first comes to rest just as the string becomes slack, find an expression for the modulus of elasticity of the string in terms of \(m\) and \(g\). Show that, throughout the motion, the magnitude \(R\) of the reaction between the ring and the hoop is given by \[ R = ( 12\cos^2\phi -15\cos\phi +5) mg \] and that \(R\) is non-zero throughout the motion.

Two identical particles \(P\) and \(Q\), each of mass \(m\), are attached to the ends of a diameter of a light thin circular hoop of radius \(a\). The hoop rolls without slipping along a straight line on a horizontal table with the plane of the hoop vertical. Initially, \(P\) is in contact with the table. At time \(t\), the hoop has rotated through an angle \(\theta\). Write down the position at time \(t\) of \(P\), relative to its starting point, in cartesian coordinates, and determine its speed in terms of \(a\), \(\theta\) and \(\dot\theta\). Show that the total kinetic energy of the two particles is \(2ma^2\dot\theta^2\). Given that the only external forces on the system are gravity and the vertical reaction of the table on the hoop, show that the hoop rolls with constant speed.

Show Solution

---

Solution:

+ - ↺

We can see that the position of \(O\) is \(\begin{pmatrix} a \theta \\ a \end{pmatrix}\) since the hoop is not slipping. \(P\)'s position relative to \(O\) is \(\begin{pmatrix} -a\sin\theta\\a(1-\cos \theta) \end{pmatrix}\), therefore the position of \(P\) is \(\begin{pmatrix} a(\theta-\sin\theta) \\ a(1-\cos \theta) \end{pmatrix}\). We can now calculate \(\mathbf{v}_P = a \begin{pmatrix} (\dot{\theta}-\dot{\theta}\cos\theta) \\ \dot{\theta}\sin \theta \end{pmatrix} = a \dot{\theta} \begin{pmatrix} (1-\cos\theta) \\ \sin \theta \end{pmatrix}\) We can also see that \begin{align*} && |\mathbf{v}_P|^2 &= a^2\dot{\theta}^2 \l \l 1 - \cos \theta \r^2 + \sin^2 \theta \r \\ && &= a^2\dot{\theta}^2 ( 2 - 2\cos \theta) \\ && &= 2a^2\dot{\theta}^2 ( 1 - \cos \theta) \\ && &= a^2\dot{\theta}^2 4 \sin^2 \frac{\theta}{2} \\ \Rightarrow |\mathbf{v}_P| &= 2a \dot{\theta} \left | \sin \frac{\theta}2 \right | \end{align*} Not that the position of \(Q\) is \(\begin{pmatrix} a(\theta+\sin\theta) \\ a(1+\cos \theta) \end{pmatrix}\) Therefore \begin{align*} && |\mathbf{v}_Q|^2 &= a^2\dot{\theta}^2 \l \l 1 + \sin \theta \r^2 + \l 1 + \cos \theta \r^2 \r \\ && &= a^2\dot{\theta}^2 \l \l 1 + \sin \theta \r^2 + \cos^2 \theta \r \\ && &= 2a^2\dot{\theta}^2 \l 1 + \cos \theta \r \\ \end{align*} Therefore \[ \text{K.E.} = \frac12m|\mathbf{v}_P|^2 + |\mathbf{v}_Q|^2 = \frac12m2a^2 \dot{\theta}^2 (1 - \cos \theta + 1-\cos \theta) = 2ma^2 \dot{\theta}^2\] Since there are no external forces acting conservation of energy tells us that kinetic energy is constant, ie \(4ma^2 \dot{\theta}\ddot{\theta} = 0 \Rightarrow \ddot{\theta} = 0\), ie the hoop is rolling with constant speed.

A circular hoop of radius \(a\) is free to rotate about a fixed horizontal axis passing through a point \(P\) on its circumference. The plane of the hoop is perpendicular to this axis. The hoop hangs in equilibrium with its centre, \(O\), vertically below \(P\). The point \(A\) on the hoop is vertically below \(O\), so that \(POA\) is a diameter of the hoop. A mouse \(M\) runs at constant speed \(u\) round the rough inner surface of the lower part of the hoop. Show that the mouse can choose its speed so that the hoop remains in equilibrium with diameter \(POA\) vertical. Describe what happens to the hoop when the mouse passes the point at which angle \(AOM = 2 \arctan \mu\,\), where \(\mu\) is the coefficient of friction between mouse and hoop.

Calculate the moment of inertia of a uniform thin circular hoop of mass \(m\) and radius \(a\) about an axis perpendicular to the plane of the hoop through a point on its circumference. The hoop, which is rough, rolls with speed \(v\) on a rough horizontal table straight towards the edge and rolls over the edge without initially losing contact with the edge. Show that the hoop will lose contact with the edge when it has rotated about the edge of the table through an angle \(\theta\), where \[ \cos\theta = \frac 12 +\frac {v^2}{2ag}. \] %Give the corresponding result for a smooth hoop and table.