At a Diwali mela a child lets a yo-yo drop from her finger. It does not fall like a stone. It falls slowly, spinning faster and faster, as the string unwinds. The string is clearly pulling upward on the yo-yo the whole way down. Yet the yo-yo still falls. So the string pulls, but it pulls less than the weight. How much less, and why does the spinning slow the fall?
A roller pulled at its axle
A garden roller is pulled by a handle fixed to its axle. The pull acts at the centre, so it has no torque about the centre. On ice the roller would slide forward without turning. On grass the ground stops that sliding with a backward friction at the contact, and that friction supplies the torque that spins the roller up.
A garden roller is a solid cylinder of mass \(40\,\text{kg}\). A gardener pulls its axle horizontally with a force of \(90\,\text{N}\), and it rolls without slipping on level grass. Take \(g = 10\,\text{m/s}^2\). Find its acceleration, the friction on it, and the least coefficient of static friction the grass must provide.
Assume friction \(f\) backward. Force: \(90 - f = 40a\). Torque about the centre: \(fR = \tfrac12(40)R^2(a/R)\), so \(f = 20a\).
Then \(90 = 60a\), so \(a = 1.5\,\text{m/s}^2\) and \(f = 30\,\text{N}\). The sign is positive, so the friction does point backward.
The normal force is \(Mg = 400\,\text{N}\), so rolling needs \(\mu_s \ge 30/400 = 0.075\). Grass easily gives that.
In general \(a = F/[M(1 + k^2/R^2)]\) and \(f = F\,(k^2/R^2)/(1 + k^2/R^2)\): one third of the pull, for a disc.
Pulling higher or lower
Now let the pull act at a height \(y\) above the centre, as when a thread is wound on a spool. Take \(y\) negative for a pull below the centre. The pull now has a torque \(Fy\) about the centre, which turns the body forward. The friction \(f\), taken backward, adds a torque \(fR\) in the same sense. The three equations are \(F - f = Ma\), \(Fy + fR = Mk^2a/R\) and the rolling condition that was already used.
Move the slider from the bottom. With the thread at \(0.8R\) below the centre, the friction is \(10.4\,\text{N}\) backward and the spool barely creeps forward at \(0.4\,\text{m/s}^2\). At the axle, \(y = 0\), the friction is \(4\,\text{N}\) backward and \(a = 2\,\text{m/s}^2\). At \(y = R/2\) the friction is exactly zero and \(a = 3\,\text{m/s}^2\). At the top of the rim it is \(4\,\text{N}\) forward and \(a = 4\,\text{m/s}^2\).
The friction vanishes when \(y = k^2/R\). At that height the pull alone gives exactly the spin that rolling needs, so the ground has nothing to add. For a disc this is half way up; for a ring it is the very top of the rim. A pull above that height spins the body too fast for its speed. The ground must then push forward to hold the spin back to \(v/R\).
A thin ring rolls without slipping on a level floor. A horizontal string, wound round it, pulls it at the very top of its rim. What is the friction on the ring?
A wheel driven by a torque
Now remove the pull and turn the wheel with a torque \(\tau\) about its axle, as a motor or a chain does. Nothing else pushes along the road. So the friction is the only horizontal force, and \(f = Ma\) points forward. The torque equation is \(\tau - fR = Mk^2a/R\), because the forward friction at the bottom opposes the turning.
Check the driven chip of Fig 3.1. With \(\tau = 0.6\,\text{N m}\), \(a = 0.6/(2 \times 0.1 \times 1.5) = 2\,\text{m/s}^2\), and \(f = Ma = 4\,\text{N}\) forward. The wheel needs \(\mu_s \ge 4/20 = 0.2\). On ice with \(\mu_s = 0.1\), the road cannot supply \(4\,\text{N}\), so the wheel spins in place. A car on a wet road, or a scooter on gravel, fails in the same way.
This settles an old puzzle. The engine of a car is inside the car, so its forces are internal and cannot change the momentum of the whole car. The engine only turns the wheels. The road pushes the car forward, through the static friction on the driven wheels. The free wheels, as on the cycle of the first lesson, feel a small backward friction that spins them up.
A car speeds up on a level road. Which outside force on the car points along its acceleration?
Rolling on a plank that is pulled
The chapter on Rolling Motion showed that rolling on a moving surface means no sliding between the two contact points. They share one velocity, and so one acceleration. A solid cylinder lies on a plank, and the plank is pulled forward along a smooth floor. The plank drags the cylinder's bottom forward, so friction pushes the cylinder forward and holds the plank back. The cylinder spins backward as the plank runs ahead.
A uniform solid cylinder of mass \(1\,\text{kg}\) rests on a plank of mass \(2\,\text{kg}\), which lies on a smooth floor. The plank is pulled forward by a horizontal force of \(14\,\text{N}\), and the cylinder rolls on it without slipping. Find the accelerations of the cylinder and of the plank.
Let \(f\) be the friction on the cylinder, forward, and \(\alpha\) its angular acceleration in the sense that moves its bottom point forward. Let \(a_c\) and \(A\) be the accelerations of the cylinder's centre and of the plank.
Cylinder: \(f = 1 \times a_c\) and \(fR = \tfrac12(1)R^2\alpha\), so \(\alpha R = 2f\). Plank: \(14 - f = 2A\). No sliding: \(a_c + \alpha R = A\).
The contact condition gives \(f + 2f = A\), so \(A = 3f\). Then \(14 - f = 6f\), so \(f = 2\,\text{N}\).
The cylinder moves forward at \(a_c = 2\,\text{m/s}^2\) and the plank at \(A = 6\,\text{m/s}^2\), three times as fast. The cylinder drifts backward along the plank while it moves forward over the floor.
The yo-yo
A yo-yo is the slope problem turned on its end. The string plays the part of the friction. It is the only force with a torque about the centre. And the point where the string leaves the yo-yo is at rest for an instant, as the contact of a rolling body is. So the string unwinds at exactly \(\omega r\), where \(r\) is the radius on which it is wound.
Take down as positive. Then \(Mg - T = Ma\), and about the centre \(Tr = Mk^2\alpha\) with \(a = \alpha r\). So \(T = Mk^2a/r^2\), and the force equation gives the result below. It is the slope result with \(\sin\theta = 1\). For a disc wound on its rim, \(k^2/r^2 = \tfrac12\), so \(a = \tfrac23 g\) and \(T = \tfrac13 Mg\).
The tension is less than the weight, which is why the yo-yo falls. But the tension is not zero, which is why it falls more slowly than a stone. A disc of mass \(0.12\,\text{kg}\) wound on its rim falls at \(6.7\,\text{m/s}^2\) while its string pulls with only \(0.4\,\text{N}\), a third of its weight.
A yo-yo of weight \(Mg\) unwinds from its string. Compare the tension in the string with the weight.
A thin ring unwinds from a light string wound on its rim, with the upper end of the string held fixed. Take \(g = 10\,\text{m/s}^2\). Find its acceleration, in m/s².
A real yo-yo has its string wound on a thin axle of radius about \(0.5\,\text{cm}\), and its radius of gyration is about \(2.5\,\text{cm}\). Estimate its downward acceleration, in m/s².
Now the yo-yo of the Diwali mela is explained. Its string pulls upward, but with less than the weight, so the yo-yo falls. The pull acts off the centre, so it spins the yo-yo, and the spin takes most of the energy of the fall. A string wound on a thin axle makes \(k^2/r^2\) large, so a real yo-yo falls at only a few per cent of \(g\).
The radius in the yo-yo formula is the radius where the string is wound, not the outer radius. In the slope formula it is the radius that touches the slope. Always use the radius of the point that is momentarily at rest.
- Taking friction on a pulled roller to point forward because the roller moves forward.A pull at the axle has no torque about the centre; backward friction supplies the spin.
- Saying that an engine pushes a car forward.The engine turns the wheels. The road pushes the car, through static friction on the driven wheels.
- Using the outer radius of a yo-yo in \(a = g/(1 + k^2/r^2)\).Use the radius on which the string is wound, where the point is momentarily at rest.