Problems

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.

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Solution:

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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.

1. A wheel consists of a thin light circular rim attached by light spokes of length \(a\) to a small hub of mass \(m\). The wheel rolls without slipping on a rough horizontal table directly towards a straight edge of the table. The plane of the wheel is vertical throughout the motion. The speed of the wheel is \(u\), where \(u^2
2. Two particles, each of mass \(m/2\), are attached to a light circular hoop of radius \(a\), at the ends of a diameter. The hoop rolls without slipping on a rough horizontal table directly towards a straight edge of the table. The plane of the hoop is vertical throughout the motion. When the centre of the hoop is vertically above the edge of the table it has speed \(u\), where \(u^2

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.