Cycle to college in Mumbai during the monsoon without a mudguard, and you will reach college with a stripe of mud up your back. The mud is flung from the tyre, but not from every part equally. Lorries on Indian highways carry rubber mud flaps behind their wheels for the same reason. Which part of a rolling tyre throws mud the hardest?
Two equal parts
Lesson 1 gave the velocities of a wheel whose centre moves at \(v_{cm}\) while it turns at \(\omega\). Rolling adds one condition: \(\omega R = v\). So at every rim point the rotation part has exactly the same size as the translation part, \(v\). Only its direction changes round the rim. Adding two arrows of equal size gives a very tidy picture.
Take the chips of Fig 4.1 in order. With translation alone, every arrow is the same, \(v\) forward. With rotation alone, the centre is still, and each rim point has \(v\) along the rim. That is forward at the top, backward at the bottom, down at the front and up at the back. With both, the arrows add, point by point.
The top of a rolling wheel is its fastest point. It moves at twice the speed of the axle. The contact point is the only point at rest. The two side points, level with the axle, move faster than the axle, at \(\sqrt2\,v\), because their two equal parts meet at a right angle.
A motorbike moves at \(60\,\text{km/h}\) on a straight road, and its tyres roll without slipping. Relative to the road, the top of each tyre moves at
The contact point as the axis
Lesson 2 said that a point known to be at rest lies on the instantaneous axis. The contact point of a rolling wheel is exactly such a point. So at each instant the whole wheel turns about its contact point C. Every point moves at right angles to its line from C, with speed \(\omega\) times its distance from C.
This gives every speed in one step. The centre is \(R\) from C, so it moves at \(\omega R = v\). The top is \(2R\) from C, so it moves at \(2v\). A side point is \(R\sqrt2\) from C, so it moves at \(\sqrt2\,v\). These agree with the sums of Fig 4.1, without adding a single arrow.
For any rim point, the distance from C is a chord of the rim. Let the point be an angle \(\theta\) round the rim from C, measured at the centre. The derivation below finds its speed. One step is left for you.
\[ v_P = \omega \cdot CP \]
Why. C is at rest, so the wheel turns about C at this instant.
?
Your turn. the triangle formed by the centre, C and P has two sides \(R\) and the angle \(\theta\) between them. Which line is it?
\[ v_P = 2\omega R\sin(\theta/2) = 2v\sin(\theta/2) \]
Read it. put the chord into step 1 and use the rolling condition \(\omega R = v\).
Check the rule at three places. At the top, \(\theta = 180°\) and \(v_P = 2v\sin 90° = 2v\). At the side, \(\theta = 90°\) and \(v_P = 2v\sin 45° = \sqrt2\,v\). At \(\theta = 60°\), \(v_P = 2v\sin 30° = v\): that rim point moves exactly as fast as the centre.
The bicycle wheel of lesson 3 has radius \(0.35\,\text{m}\) and rolls at \(7\,\text{m/s}\). Find the speeds of the top and of a side point level with the axle. Find also the speeds of rim points \(60°\) and \(120°\) round from the contact.
Use \(v_P = 2v\sin(\theta/2)\) with \(v = 7\,\text{m/s}\) throughout.
Top, \(\theta = 180°\): \(2 \times 7 \times \sin 90° = 14\,\text{m/s}\).
Side, \(\theta = 90°\): \(2 \times 7 \times \sin 45° \approx 9.9\,\text{m/s}\).
At \(60°\): \(2 \times 7 \times \sin 30° = 7\,\text{m/s}\), the speed of the axle. At \(120°\): \(2 \times 7 \times \sin 60° \approx 12.1\,\text{m/s}\).
Check the side point about C: its chord is \(0.35\sqrt2 \approx 0.495\,\text{m}\), and \(\omega = 20\,\text{rad/s}\), so \(20 \times 0.495 \approx 9.9\,\text{m/s}\).
Equal distance, not equal height
Speed depends on the distance from C, not on the height above the road. The dashed circle in Fig 4.2 has radius \(R\) and is centred on C. It passes through the centre of the wheel and through the two rim points \(60°\) round from C. All three move at \(v\), although the rim points sit much lower than the centre.
The side points are at the same height as the centre, yet they move faster, at \(\sqrt2\,v\). They are farther from C than the centre is. Height is not what matters; distance from the contact point is.
The velocity of every rim point of a rolling wheel points straight at the top of the wheel. The line from C to the top T is a diameter. The angle CPT at any rim point P is an angle in a semicircle, so it is \(90°\). The velocity of P is at right angles to CP, so it lies along PT.
A car moves at \(20\,\text{m/s}\), and its tyres roll without slipping. What is the speed of a point on the rim of a tyre that is level with the axle, in m/s?
On a wheel rolling without slipping, which of these points move at the same speed as the centre?
Accelerations in steady rolling
Now let the wheel roll at a steady speed, so \(a_{cm} = 0\) and \(\alpha = 0\). Watch it from a frame that moves with the centre. In that frame the centre is still and every particle goes round a circle at a steady speed. A particle at distance \(r\) from the centre therefore has a centripetal acceleration \(\omega^2 r\), pointing at the centre.
That frame moves at a constant velocity, so it measures the same accelerations as the ground does. So on the ground too, every point of the wheel accelerates at \(\omega^2 r\) towards the centre, and the centre alone has no acceleration. The contact point is included. Its velocity is zero, but its acceleration is \(\omega^2 R = v^2/R\), pointing straight up at the centre.
That acceleration is large. For the bicycle wheel of lesson 3, \(\omega = 20\,\text{rad/s}\) and \(R = 0.35\,\text{m}\), so the contact point accelerates at \(20^2 \times 0.35 = 140\,\text{m/s}^2\), fourteen times \(g\). It grows as \(v^2\): at twice the speed it is four times as large.
Zero velocity does not mean zero acceleration. The contact point of a rolling wheel is at rest for one instant and accelerates hard at that instant. That acceleration is exactly what lifts it from the road a moment later. This is lesson 2's warning in its sharpest form.
A car moves at a steady \(15\,\text{m/s}\) on tyres of radius \(0.3\,\text{m}\) that roll without slipping. What is the magnitude of the acceleration of the point of a tyre touching the road, in m/s²?
Accelerations while the wheel gains speed
Now let the wheel gain speed while it rolls. The frame of the centre accelerates, so we add its acceleration to what that frame sees. Each point has three parts. The first is \(a_{cm}\), forward, the same at every point. The second is a tangential part \(\alpha r\), at right angles to the radius. The third is the centripetal part \(\omega^2 r\), towards the centre.
At the contact point, the tangential part \(\alpha R\) points backward, against \(a_{cm}\). Rolling makes \(a_{cm} = \alpha R\), so the two cancel exactly. Only \(\omega^2 R\), straight up, is left. The contact point accelerates the same way whether the wheel rolls steadily or gains speed.
At the top, the tangential part points forward, with \(a_{cm}\). The two add to \(2a_{cm}\), and the centripetal part \(\omega^2 R\) points down. At the front and the back the three parts mix, and each point needs its own vector sum.
The wheel of Fig 4.3 has radius \(0.25\,\text{m}\) and rolls at \(1\,\text{m/s}\) while it gains speed at \(1.5\,\text{m/s}^2\). Find the acceleration of the top point and of the contact point.
\(\omega = v/R = 4\,\text{rad/s}\), so \(\omega^2 R = 4\,\text{m/s}^2\). Rolling continues, so \(\alpha = a_{cm}/R = 6\,\text{rad/s}^2\) and \(\alpha R = 1.5\,\text{m/s}^2\).
Top: \(a_{cm} + \alpha R\) forward and \(\omega^2 R\) down. Contact: \(a_{cm} - \alpha R\) forward and \(\omega^2 R\) up.
Top: \(3\,\text{m/s}^2\) forward and \(4\,\text{m/s}^2\) down, so \(\sqrt{3^2 + 4^2} = 5\,\text{m/s}^2\). Contact: \(0\) forward and \(4\,\text{m/s}^2\) up.
The top accelerates at \(5\,\text{m/s}^2\), tilted forward and down. The contact accelerates at \(4\,\text{m/s}^2\) straight up, exactly as it would in steady rolling.
A wheel rolls without slipping and is gaining speed, so \(a_{cm}\) and \(\alpha\) are both non-zero. The acceleration of its contact point is
The contact point of a rolling wheel is at rest. Explain why it does not stay on the road.
Its velocity is zero only at that instant. Its acceleration is \(\omega^2 R\), pointing up towards the centre, so an instant later it has a small upward velocity and leaves the road. A new particle of the tyre becomes the contact point, and the old one rises along the rim.
Mud leaves the tread with the velocity of the point it leaves from. On the road that is up to \(2v\), from near the top of the tyre. In the frame of the vehicle itself the wheel only turns, so mud leaves the tread along the rim at speed \(v\). It flies backward from the bottom, straight up from the back and forward from the top. A flap behind a lorry wheel stops the spray thrown backward, and a mudguard over a cycle wheel stops the rest.
Return to the monsoon ride. On the road, the top of the tyre moves at \(2v\), twice the speed of the cycle. So mud leaving from near the top leaves fastest. Seen from the saddle, the rear wheel only turns, and mud from the back of the tyre flies straight up at \(v\). The rider moves along with the wheel, so that mud rises right behind the saddle and lands on the rider's back.
- Giving every point of a rolling wheel the speed of the axle.Only the centre moves at \(v\). The top moves at \(2v\) and the contact point is at rest.
- Judging speeds by height above the road.Speed is \(\omega\) times the distance from the contact point. Points level with the axle move faster than the axle.
- Setting the acceleration of the contact point to zero.Its velocity is zero, but it accelerates at \(\omega^2 R\) towards the centre, in steady and in accelerated rolling alike.