Problems

A plane circular road is bounded by two concentric circles with centres at point~\(O\). The inner circle has radius \(R\) and the outer circle has radius \(R + w\). The points \(A\) and \(B\) lie on the outer circle, as shown in the diagram, with \(\angle AOB = 2\alpha\), \(\tfrac{1}{3}\pi \leqslant \alpha \leqslant \tfrac{1}{2}\pi\) and \(0 < w < R\).

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1. Show that I cannot cycle from \(A\) to \(B\) in a straight line, while remaining on the road.
2. I take a path from \(A\) to \(B\) that is an arc of a circle. This circle is tangent to the inner edge of the road, and has radius \(R + d\) (where \(d > w\)) and centre~\(O'\). My path is represented by the dashed arc in the above diagram. Let \(\angle AO'B = 2\theta\).
   1. Use the cosine rule to find \(d\) in terms of \(w\), \(R\) and \(\cos\alpha\).
   2. Find also an expression for \(\sin(\alpha - \theta)\) in terms of \(R\), \(d\) and \(\sin\alpha\).
   You are now given that \(\dfrac{w}{R}\) is much less than \(1\).
3. Show that \(\dfrac{d}{R}\) and \(\alpha - \theta\) are also both much less than \(1\).
4. My friend cycles from \(A\) to \(B\) along the outer edge of the road. Let my path be shorter than my friend's path by distance~\(S\). Show that \[ S = 2(R+d)(\alpha - \theta) + 2\alpha(w - d). \] Hence show that \(S\) is approximately a fraction \[ \frac{\sin\alpha - \alpha\cos\alpha}{\alpha(1 - \cos\alpha)} \cdot \frac{w}{R} \] of the length of my friend's path.

Derive a formula for the position of the centre of mass of a uniform circular arc of radius \(r\) which subtends an angle \(2\theta\) at the centre.

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Show Solution

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Solution:

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Splitting the arc up into strips of width \(\delta \theta\), then we must have \begin{align*} && \sum r\cos \theta (r \delta \theta) &= \bar{x}\sum (r \delta \theta) \\ \lim_{\delta \theta \to 0}: && \int_{-\theta}^{\theta} r^2 \cos \theta \d \theta &= \bar{x}2 \theta r \\ \Rightarrow && 2r^2 \sin \theta &= \bar{x} 2 \theta r \\ \Rightarrow && \bar{x} &= \frac{r\sin \theta}{\theta} \end{align*}

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The centre of mass will lie on the line of symmetry. It also must lie at the center of the base of the semi-circle (see diagram). Using a coordinate frame where that point is the origin we must have \begin{align*} && 0 &= -2r \cdot 2h - 4h \cdot h + \pi r \frac{r}{\frac{\pi}{2}} \\ &&&= -4rh-4h^2+2r^2\\ \Rightarrow && 0 &= r^2-2rh-h^2 \\ \Rightarrow && \frac{r}{h} &= 1 \pm \sqrt{3} \\ \Rightarrow && r &= (1+\sqrt{3})h \\ && h & = \frac12 (\sqrt{3}-1) r \end{align*}