Give an empty bicycle a push along a smooth, level road and let it go. It rolls on for a long way, and nobody has to hold the wheels to the road. Now ride the same bicycle onto loose gravel and pedal hard. The rear wheel spins and throws stones backward, and the bicycle hardly moves. On the road the ground held the wheel. On the gravel it could not. What does the ground do to a rolling wheel, and when?
The contact point is at rest
A body that rolls without slipping obeys \(v_{cm} = \omega R\). Here \(v_{cm}\) is the speed of the centre of mass, \(\omega\) the angular velocity and \(R\) the radius of the rolling edge. The chapter on Rolling Motion showed what this condition means. The point of the body that touches the ground is at rest at that instant. It does not slide over the ground.
So the friction at the contact of a rolling body is static friction, not kinetic. Recall from the chapter on Friction what static friction does. It has no fixed size. It takes whatever value is needed to stop sliding, up to the limit \(\mu_s N\). It also has no fixed direction; it points whichever way the need demands. Here \(\mu_s\) is the coefficient of static friction and \(N\) the normal force.
When a body rolls without slipping, the friction at the contact is static friction, \(|f| \le \mu_s N\). Its size and its direction are set by the motion, not by the direction in which the body moves.
Static friction does no work here
Work is a force times the displacement of the point where the force acts. The friction on a rolling body acts at the contact point, and that point is at rest while the force acts on it. The next instant a new point of the rim touches the ground, and it too is at rest. So the friction never acts through any displacement. It does no work on the body.
This fact carries a large reward. On a rough surface, a body that rolls without slipping loses no mechanical energy to friction. The ground can still exert a force and a torque on it, and can shift energy between translation and rotation. But it cannot turn any of it into heat, because nothing rubs. The next lesson uses this to find speeds down a slope from the height alone.
A hoop rolls without slipping down a rough slope. How much work does the friction on the hoop do?
When rolling needs no friction
Take a rigid wheel rolling at a steady speed on level ground. Its centre moves with a constant velocity, so the net force on it is zero. Its angular velocity is constant, so the net torque about its centre is zero. The weight acts at the centre and the normal force acts straight up through the centre. Neither has any torque about the centre, and together they balance. Nothing is left for friction to do, so the friction is zero.
So rolling itself does not need friction. A change in rolling does. Friction appears when the motion must change, or when some other force tries to make the contact slide. The four cases in the figure show the usual agents. In each case think of the contact point and ask which way it would slide if the ground were ice. Friction points the other way.
Look at the driven wheel in Fig 1.1. The engine turns the wheel, and the wheel tries to push the ground backward. The ground pushes the wheel forward. That forward friction is the only outside horizontal force on a car, so it is what accelerates the car. On loose gravel the ground cannot supply enough static friction. The contact slides, the wheel spins, and the bicycle of the opening hardly moves.
Why does a real wheel slow down on a level road if friction is zero? A real tyre and a real road both deform a little where they touch. The contact is a small patch, and the ground pushes a little harder on its front half than on its back half. So the normal force acts slightly ahead of the centre, and it has a small torque that opposes the turning. This is rolling resistance. It comes from deformation, not from friction, and every body in this chapter is taken as perfectly rigid.
A rigid wheel rolls without slipping at a steady speed along a level road. What is the friction on it?
A cyclist speeds up on a dry, level road. The chain drives the rear wheel, and the front wheel turns freely on its axle. Which statements are true?
One procedure for every case
No direction needs to be memorised, because the algebra reports it. Write three equations. The first is Newton's second law for the centre of mass, along the motion. The second is \(\tau_{cm} = I_{cm}\alpha\) about the centre of mass, from the chapter on Dynamics of Rotation. Here \(\tau_{cm}\) is the net torque about the centre, \(I_{cm}\) the moment of inertia about it and \(\alpha\) the angular acceleration. The third is the rolling condition in its accelerating form, \(a = \alpha R\), from the chapter on Rolling Motion.
Write the moment of inertia as \(I_{cm} = Mk^2\), with \(M\) the mass and \(k\) the radius of gyration. Then the shape of the body enters only through the ratio \(k^2/R^2\). It is \(1\) for a ring, \(\tfrac12\) for a disc or a solid cylinder, \(\tfrac23\) for a hollow sphere and \(\tfrac25\) for a solid sphere. This chapter calls it the shape number of the body.
Assume a direction for the friction \(f\). Write \(F_{net} = Ma\) for the centre of mass, \(\tau_{cm} = I_{cm}\alpha\) about the centre of mass, and \(a = \alpha R\). Solve. A negative \(f\) means the arrow points the other way. Last, check that \(|f| \le \mu_s N\); if it is not, the body cannot roll without slipping.
A hoop of mass \(2\,\text{kg}\) rolls without slipping on a rough level floor. A horizontal force of \(10\,\text{N}\) pulls it at its centre. Find its acceleration and the friction on it. Assume, on purpose, that friction points forward.
Take forward as positive and clockwise turning as positive, since a hoop moving forward turns clockwise. For a hoop \(I_{cm} = MR^2\), so \(k^2/R^2 = 1\).
Force: \(10 + f = 2a\). Torque about the centre: \(-fR = MR^2\alpha\). Rolling: \(a = \alpha R\), so the torque equation gives \(f = -Ma = -2a\).
Put \(f = -2a\) into the force equation: \(10 - 2a = 2a\), so \(a = 2.5\,\text{m/s}^2\) and \(f = -5\,\text{N}\).
The minus sign says the guess was wrong. The friction is \(5\,\text{N}\) backward, and that backward friction is what spins the hoop up. Rolling needs \(\mu_s \ge 5/20 = 0.25\).
The example is the figure in numbers. Half of the pull goes into speeding up the centre and half is used by the friction to spin the hoop up. The wrong guess cost nothing, because the sign corrected it. Only the last step needs care. If the floor gave \(\mu_s = 0.2\), the \(5\,\text{N}\) needed would exceed the \(4\,\text{N}\) available, and the hoop would slide as it turned.
A solid cylinder of mass \(4\,\text{kg}\) rolls without slipping on a rough level floor, pulled at its axle by a horizontal force of \(12\,\text{N}\). Find the size of the friction on it, in newtons.
A student assumes that friction on a rolling body points forward and finds \(f = -3\,\text{N}\). What should she conclude?
- Using \(f = \mu N\) for a rolling body.Rolling without slipping means static friction. Find \(f\) from the equations; \(\mu_s N\) is only its upper limit.
- Believing that friction always opposes the motion of the body.Static friction opposes the tendency of the contact to slide. On a driven wheel it points forward, along the motion.
- Believing that friction on a rolling body wastes energy.The contact point is at rest, so the friction does no work. Only sliding turns energy into heat.