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2008 Paper 1 Q9
D: 1500.0 B: 1484.0
circular motion rolling without slipping hoop kinetic energy parametric cycloid energy conservation constant speed

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.

Solution:

TikZ diagram
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.

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1998 Paper 2 Q6
D: 1600.0 B: 1500.0
parametric equations cycloid tangent-normal curve sketching gradient differentiation parametric differentiation

Two curves are given parametrically by \[ x_{1}=(\theta+\sin\theta),\qquad y_{1}=(1+\cos\theta),\tag{1} \]and \[ x_{2}=(\theta-\sin\theta),\qquad y_{1}=-(1+\cos\theta),\tag{2} \] Find the gradients of the tangents to the curves at the points where \(\theta= \pi/2\) and \(\theta=3\pi/2\). Sketch, using the same axes, the curves for \(0\le\theta \le 2\pi\). Find the equation of the normal to the curve (1) at the point with parameter \(\theta\). Show that this normal is a tangent to the curve (2).

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1997 Paper 3 Q5
D: 1700.0 B: 1484.0
cycloid arc length integration parametric circle ratio trigonometric

Find the ratio, over one revolution, of the distance moved by a wheel rolling on a flat surface to the distance traced out by a point on its circumference.

Solution: The point on the circumference will have position \((a\cos t, a \sin t )\) relative to the circumference where \(t \in [0, 2\pi]\). the wheel will travel \(2\pi a\), therefore the position is \((a\cos t + at, a \sin t )\). The total distance travelled can be computed using the arc length: \begin{align*} && s &= \int_0^{2\pi} \sqrt{\left ( \frac{\d y}{\d t} \right)^2 +\left ( \frac{\d x}{\d t} \right)^2} \d t \\ &&&= \int_0^{2\pi} \sqrt{(a - a\sin t)^2 +(a \cos t)^2 } \d t \\ &&&= a \int_0^{2\pi} \sqrt{2 - 2 \sin t } \d t \\ &&&= \sqrt{2}a \int_0^{2 \pi} \sqrt{1 - \sin t} \d t \\ &&&= \sqrt{2}a \int_0^{2 \pi} \frac{|\cos t|}{\sqrt{1 + \sin t}} \d t \\ &&&= 2\sqrt{2} a \int_{-\pi/2}^{\pi/2} \frac{\cos t}{\sqrt{1+\sin t}} \d t \\ &&&= 2\sqrt{2} a \left [ 2\sqrt{1+\sin t} \right]_{-\pi/2}^{\pi/2} \\ &&& = 2\sqrt{2} a 2\sqrt{2} \\ &&&= 8a \end{align*} Therefore the ratio is \(\frac{4}{\pi}\)

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