A student in Pune fits a small cycle computer to the handlebar of her bicycle. A magnet on one spoke passes a sensor on the fork once every turn of the front wheel. The computer only counts turns and measures the time between them. Yet it shows her speed and the distance to college correctly. It has never seen the road.
What rolling means
Lesson 1 found that a wheel may slide on the road while it turns. Its lowest point moved at \(v_{cm} - \omega R\), which was not zero. A tyre that grips the road does something special. The point of the tyre touching the road does not slide over the road at all. This is rolling without slipping, and on a fixed road it means one thing: the contact point is at rest.
The contact point of a wheel turning in the rolling sense has velocity \(v_{cm} - \omega R\), forward. Set it to zero, and the centre speed and the rim speed must be equal.
\[ v_{\text{contact}} = v_{cm} - \omega R = 0 \quad\Longrightarrow\quad v_{cm} = \omega R \]
This one equation is the rolling condition. It ties the two motions of the wheel together. Once you know how fast the centre moves, you know how fast the wheel turns, and the other way round.
A body rolls without slipping on a surface when its point of contact has zero velocity relative to that surface. On a fixed surface this gives \(v_{cm} = \omega R\) for a round body of radius \(R\).
There is a second way to see the same result. As a wheel rolls, each piece of its rim touches the road once and is laid down on it. No piece slides, so the length of rim laid down equals the length of road covered. When the wheel turns through an angle \(\theta\), it lays down an arc \(R\theta\), and the centre moves forward by the same \(R\theta\).
Divide \(x = R\theta\) by time and the rolling condition returns: \(v_{cm} = R\,d\theta/dt = \omega R\). One full turn lays the whole rim, \(2\pi R\), on the road, as Fig 3.1 shows at its end. So a wheel that rolls \(n\) turns covers \(n \times 2\pi R\) of road.
From now on, for a body rolling on a fixed surface, we write \(v\) for \(v_{cm}\) whenever the meaning is clear.
A bicycle wheel of radius \(0.35\,\text{m}\) rolls without slipping while the bicycle moves at \(7\,\text{m/s}\). Find (a) the angular velocity of the wheel and (b) the number of turns each second. Find also (c) the distance covered in one minute, by two routes.
(a) The rolling condition gives \(\omega = v/R = 7/0.35 = 20\,\text{rad/s}\).
(b) One turn is \(2\pi\) rad, so the wheel makes \(20/2\pi \approx 3.18\) turns every second.
(c) By kinematics, the distance is \(vt = 7 \times 60 = 420\,\text{m}\).
By counting turns, the wheel turns through \(20 \times 60 = 1200\,\text{rad}\), about \(191\) turns. Each turn lays \(2\pi \times 0.35 \approx 2.20\,\text{m}\) of rim on the road, and \(191 \times 2.20 \approx 420\,\text{m}\). The two routes agree because the tyre does not slide.
A car moves at \(54\,\text{km/h}\) on tyres of radius \(0.3\,\text{m}\) that roll without slipping. What is the angular velocity of each wheel, in rad/s?
A car travels \(1000\,\text{km}\) on a highway trip. Its tyres have a radius of about \(0.3\,\text{m}\) and roll without slipping. Estimate how many turns each tyre makes, in lakh.
When the speed changes
Suppose the wheel keeps rolling while it gains speed. Then \(v = \omega R\) holds at every instant, not just at one. Two quantities that are equal at every instant change at equal rates. Differentiate both sides with respect to time, keeping \(R\) fixed.
\[ \frac{dv}{dt} = R\,\frac{d\omega}{dt} \quad\Longrightarrow\quad a_{cm} = \alpha R \]
Here \(a_{cm}\) is the acceleration of the centre and \(\alpha\) is the angular acceleration of the wheel.
The condition matters. The relation \(a_{cm} = \alpha R\) is not true of every wheel. It holds only over a stretch of time in which rolling continues. A wheel that rolls at one instant and skids the next obeys \(v = \omega R\) at that instant, but not \(a_{cm} = \alpha R\).
A bus leaves a stop in Chennai from rest with a steady acceleration of \(1.5\,\text{m/s}^2\). Its wheels have radius \(0.5\,\text{m}\) and roll without slipping. Find the angular acceleration of a wheel, its angular velocity after \(4\,\text{s}\), and the number of turns it makes in those \(4\,\text{s}\).
Rolling continues throughout, so \(a_{cm} = \alpha R\) holds for the whole \(4\,\text{s}\).
\(\alpha = a_{cm}/R\), then \(\omega = \alpha t\) and \(\theta = \tfrac12 \alpha t^2\), since the wheel starts from rest.
\(\alpha = 1.5/0.5 = 3\,\text{rad/s}^2\). After \(4\,\text{s}\), \(\omega = 3 \times 4 = 12\,\text{rad/s}\). The angle is \(\tfrac12 \times 3 \times 16 = 24\,\text{rad}\), which is \(24/2\pi \approx 3.8\) turns.
The wheel accelerates at \(3\,\text{rad/s}^2\), turns at \(12\,\text{rad/s}\) after \(4\,\text{s}\), and makes about \(3.8\) turns. The bus moves \(\tfrac12 \times 1.5 \times 16 = 12\,\text{m}\), which equals \(R\theta = 0.5 \times 24\).
When does the relation \(a_{cm} = \alpha R\) hold for a wheel?
Two ways to lose grip
The contact point can slide in either direction, and each case has a name. In a forward skid, the centre moves faster than the rim turns, \(v_{cm} > \omega R\). The contact point slides forward over the road. A car whose brakes lock the wheels does this: \(\omega = 0\) while the car still moves.
In a wheel spin, the rim turns faster than the centre moves, \(\omega R > v_{cm}\). The contact point slides backward over the road. A tractor stuck in a muddy field does this: its big wheels turn fast, throw mud backward, and the tractor hardly moves. In both cases the contact slides, and neither \(v_{cm} = \omega R\) nor \(a_{cm} = \alpha R\) may be used.
The slider in the figure above tells the whole story in one line. The velocity of the contact point is \(4 - 0.5\,\omega\). It is positive in a skid, negative in a spin and zero only at \(\omega = 8\,\text{rad/s}\), where \(v = \omega R\). Watch the axis I as well. It sits \(v/\omega\) below the centre, below the road in a skid and inside the wheel in a spin. Only in rolling does it sit exactly at the contact point.
A car brakes so hard that its wheels lock and stop turning, while the car still slides forward. During the slide, which statement is correct?
In which of these does the contact point slide backward over the ground?
A locked wheel skids, and a skidding tyre cannot steer the vehicle: it slides in whatever direction it was moving. An anti-lock braking system senses a wheel that is about to stop turning and eases the brake many times a second. The tyre stays close to rolling, with its contact point nearly at rest, and the rider can still steer. India made such brakes compulsory on new motorcycles above 125 cc from 2019.
Rolling on something that moves
The rolling condition is really a statement about relative velocity. No sliding means that the contact point of the body and the surface under it move together. On a fixed road the surface is at rest, so the contact point is at rest. On a moving plank, a conveyor belt or another roller, the contact point must move with that surface instead.
Take the forward direction as positive, and count \(\omega\) as positive in the rolling sense. The contact point at the bottom of a body moves at \(v_{cm} - \omega R\). If the surface under it moves at \(v_s\), rolling without slipping means
\[ v_{cm} - \omega R = v_s. \]
Put \(v_s = 0\) and the fixed-road condition returns. The same idea works at a contact on the top of a roller. There the contact point moves at \(v_{cm} + \omega R\), and it must match the body resting on top.
A cylinder of radius \(0.2\,\text{m}\) rolls without slipping on a plank. The plank is pulled forward at \(1.0\,\text{m/s}\) over a smooth floor, and the centre of the cylinder moves forward at \(0.4\,\text{m/s}\). Find the angular velocity of the cylinder and the sense in which it turns.
Forward is positive, and \(\omega\) is positive in the rolling sense (clockwise, for motion to the right). The surface under the cylinder moves at \(v_s = +1.0\,\text{m/s}\).
\(v_{cm} - \omega R = v_s\), so \(0.4 - 0.2\,\omega = 1.0\).
\(0.2\,\omega = -0.6\), so \(\omega = -3\,\text{rad/s}\).
The cylinder turns at \(3\,\text{rad/s}\) anticlockwise. Relative to the plank, its centre moves backward at \(0.6\,\text{m/s}\), and it rolls backward along the plank, exactly as a rolling cylinder should.
A heavy stone moved on logs uses both kinds of contact at once. Each log rolls on the ground, so its bottom point is at rest and \(v_{log} = \omega r\). The stone rests on the tops of the logs and does not slide on them, so it moves with the top points, at \(v_{log} + \omega r = 2v_{log}\). The stone always moves twice as fast as the logs under it.
That is why workers who move a heavy machine on steel pipes keep carrying pipes forward. Every pipe falls behind the load at the speed of the pipes themselves. In Fig 3.2 the stone gains \(1.2\,\text{m}\) on the logs in \(2.4\,\text{s}\), and the rear log is left behind. It must be lifted and placed again at the front.
A plank rests on a long cylinder, which rests on the ground. Nothing slides. The plank moves forward at \(1.2\,\text{m/s}\). How fast does the centre of the cylinder move, in m/s?
Return to the cycle computer at the start. The tyre rolls without slipping, so every turn lays exactly one circumference of rim on the road. The computer multiplies the count of turns by the circumference, \(2\pi R\), entered once by the owner. On a wet road where the tyre skids, or on sand where it spins, the count and the road disagree, and the reading goes wrong.
- Using \(v_{cm} = \omega R\) for any wheel that touches the ground.It holds only when the contact does not slide. Check first for a skid or a spin.
- Setting the contact point at rest on a moving surface.The contact point moves with the surface. Write \(v_{cm} - \omega R = v_s\) with signs.
- Using \(a_{cm} = \alpha R\) from a single rolling instant.The relation needs rolling to continue over a stretch of time, because it is the rate of change of \(v_{cm} = \omega R\).