A photographer on Marine Drive in Mumbai leaves the shutter of her camera open for a few seconds at night. A cyclist rides past with a small reflector fixed to the rim of the front wheel. The photograph shows a chain of bright arches along the road. Each arch rises smoothly, and each one comes down to the road in a sharp point.
The curve a rim point draws
Follow one point P on the rim of a wheel that rolls without slipping. Lesson 4 gave its speed: \(2v\sin(\phi/2)\), where \(\phi\) is the angle the wheel has turned since P last touched the road. When P touches the road it is the contact point, and its speed is zero. Halfway through the turn it is at the top, moving at \(2v\).
The valve cap on the rim of a rolling bicycle wheel comes down to the road. What does it do there?
The path in the figure above is a cycloid. It is a chain of arches, one for each turn of the wheel. Each arch is \(2R\) high, because P reaches the top of the wheel halfway through the turn. Each arch is \(2\pi R\) long along the road, because the wheel moves forward by one circumference in one turn.
Where the arches meet the road, the curve comes to a sharp point, a cusp. The dashed line in the figure joins the contact point to P, and the velocity of P is always at right angles to it. As P comes down to the road, that dashed line shrinks and turns until it is almost level. So P arrives moving straight down, stops for an instant, and leaves moving straight up.
The shape follows from translation plus rotation. Put the origin where P touches the road, and let the wheel roll to the right.
\[ x_C = R\phi, \quad y_C = R \]
Why. after turning through \(\phi\), the wheel has laid an arc \(R\phi\) on the road, so its centre C has moved forward by \(R\phi\). It stays at height \(R\).
\[ x_P - x_C = -R\sin\phi, \quad y_P - y_C = -R\cos\phi \]
Why. P started straight below C. The wheel turns clockwise, so after an angle \(\phi\) the point P is \(\phi\) round from straight down, towards the back.
\[ x = R(\phi - \sin\phi), \quad y = R(1 - \cos\phi) \]
Why. add the position of the centre to the position of P relative to it.
\[ v_P = 2v\sin(\phi/2) \]
Read it. differentiate with \(d\phi/dt = \omega\) and \(\omega R = v\): \(v_x = v(1 - \cos\phi)\) and \(v_y = v\sin\phi\), and \(v_x^2 + v_y^2 = 2v^2(1 - \cos\phi) = 4v^2\sin^2(\phi/2)\).
Check the equations. At \(\phi = 0\) they give \((0, 0)\): P is on the road at the origin. At \(\phi = \pi\) they give \((\pi R, 2R)\): the top of the arch, halfway along it. At \(\phi = 2\pi\) they give \((2\pi R, 0)\): the next cusp. The speed from step 4 is the rule of lesson 4 again, found this time by calculus.
The path of a point on the rim of a rolling wheel meets the road in a sharp point. Explain why it is not a smooth curve there.
When the point touches the road it is the contact point of a rolling wheel, so its velocity is zero at that instant. Just before, it moves almost straight down, and just after, almost straight up. The direction of motion reverses where the speed passes through zero. So the path has a sharp corner there, the cusp.
Path against displacement
After one full turn, P is back on the road, \(2\pi R\) further along. Its displacement over the turn is \(2\pi R\), straight along the road, exactly the distance the centre moved. But P did not travel in a straight line. It went up and over an arch. So the length of its path must be larger than \(2\pi R\).
The length of the path is the area under the speed-time graph, as Motion in One Dimension showed. The speed of P is \(2v\sin(\omega t/2)\), and one turn takes \(T = 2\pi/\omega\).
\[ s = \int_0^{T} 2v\sin\dfrac{\omega t}{2}\,dt \]
Why. the distance covered is the integral of the speed over the time of one turn.
\[ s = 2v\left[-\dfrac{2}{\omega}\cos\dfrac{\omega t}{2}\right]_0^{T} \]
Why. the integral of \(\sin kt\) is \(-\cos(kt)/k\), here with \(k = \omega/2\).
\[ s = \dfrac{4v}{\omega}\,(1 - \cos\pi) = \dfrac{8v}{\omega} \]
Why. at \(t = T\) the angle \(\omega T/2\) is \(\pi\), and \(\cos\pi = -1\).
\[ s = 8R \]
Read it. the rolling condition gives \(v/\omega = R\).
So one arch of a cycloid is \(8R\) long, and it spans \(2\pi R \approx 6.28R\) of road. The path is longer than the displacement by a factor of \(8/2\pi \approx 1.27\). Fig 5.1 shows why: the rim point is slower than the centre near the road and faster near the top, and the fast part wins.
The bicycle wheel of lesson 3 has radius \(0.35\,\text{m}\) and rolls at \(7\,\text{m/s}\). For one turn, find (a) the path length and the displacement of the valve cap on its rim. Find also (b) the time of the turn and (c) the average speed and average velocity of the cap.
(a) The path is \(8R = 8 \times 0.35 = 2.8\,\text{m}\). The displacement is \(2\pi R \approx 2.20\,\text{m}\), along the road.
(b) One turn takes \(T = 2\pi R/v = 2.20/7 \approx 0.314\,\text{s}\).
(c) The average speed is \(2.8/0.314 \approx 8.9\,\text{m/s}\), which is \(4v/\pi\). The average velocity is \(2.20/0.314 = 7\,\text{m/s}\) along the road, the speed of the cycle.
A wheel of radius \(0.5\,\text{m}\) rolls without slipping. How long is the path traced by a point on its rim during one full turn, in metres?
A point on the rim of a rolling wheel of radius \(R\) starts on the road. How long is its path when it first reaches the top of the wheel?
A student cycles \(10\,\text{km}\) to college without the tyres slipping. Estimate how far the valve cap on the rim of the front wheel travels, in km.
The cycloid fascinated the mathematicians of the seventeenth century. Galileo named it, and Roberval and Torricelli found the area under one arch. In 1658 the young English astronomer Christopher Wren found its length. He proved that one arch is exactly four times the diameter of the wheel, \(8R\). Wren later left mathematics for architecture and designed St Paul's Cathedral in London.
Points inside the rim
Now paint a mark on a spoke, at a distance \(r\) from the axle, with \(r\) less than \(R\). The same derivation applies with \(r\) in place of the \(R\) of step 2. The centre still moves forward by \(R\phi\), but the mark circles it at the smaller radius \(r\).
\[ x = R\phi - r\sin\phi, \qquad y = R - r\cos\phi \]
The lowest point of this path is at height \(R - r\), above the road. So the mark never touches the road and never becomes the contact point.
Its speed follows from the distance to the contact point, as always. At its lowest, the mark is \(R - r\) above the contact point, so it moves at \(\omega(R - r) = v(1 - r/R)\), forward. At its highest it is \(R + r\) from the contact and moves at \(v(1 + r/R)\). The speed never falls to zero, so the path has no cusps. Such a smooth wave is a curtate cycloid. At \(r = 0\), the axle itself, the wave flattens into a straight line.
A train wheel has a flange, a lip that reaches below the top of the rail to keep the wheel on the track. A point on the edge of the flange is at a distance \(r\) greater than \(R\), the radius at which the wheel rolls. At its lowest it is below the contact point, and it moves at \(v(1 - r/R)\), which is negative. So some part of a moving train is always moving backward. The path of a flange point has a small loop at each turn.
A mark is painted on a spoke of a rolling wheel, halfway between the axle and the rim. Which statement about its path is correct?
Return to the photograph on Marine Drive. The reflector sits on the rim, so it traces a cycloid: a chain of arches \(2R\) high, each \(2\pi R\) long along the road. Each arch meets the road in a cusp, where the reflector stops for an instant. Along each arch the reflector covers \(8R\) of path. A reflector fixed to a spoke, nearer the axle, would draw a smooth wave instead.
- Taking the path of a rim point in one turn to be \(2\pi R\).\(2\pi R\) is its displacement. Its path over one arch is \(8R\), about 27 % longer.
- Drawing the cycloid as a smooth curve at the road.A rim point stops at the road, so the curve meets it in a sharp cusp, arriving straight down and leaving straight up.
- Giving a point inside the rim cusps.Only points that touch the road stop. A point inside the rim traces a smooth wave and never stops.