Rolling Motion

This page describes the motion of rolling objects, including the forces acting on them and the types of energy they have.

The Main Idea

Rolling motion is the turning motion of a round object such as a sphere or cylinder along a surface against which it is pressed. The rotation of a rolling object is the result of Friction between its edge and the surface, which applies a Torque. When rolling happens without slipping, this friction is Static Friction because the edge of the rolling object does not move relative to the surface. The point of contact between the object and the surface constantly changes but does not slide. For an object rolling at a constant speed, the magnitude of this static friction force is 0. However, for an object accelerated by an external force, such as a disk rolling down a ramp, the static friction force has a nonzero magnitude to prevent the disk from slipping. Without this friction force, a disk placed on a ramp would simply slide down without rotating. Sometimes, slipping does occur. This can happen if, for example, the object is accelerated along the surface so quickly that the maximum static friction force cannot cause a sufficient corresponding angular acceleration, or if the object was already moving with an arbitrary Angular Velocity when it came into contact with the surface. In these situations, a Kinetic Friction force acts between the edge of the object and the surface. This kinetic friction force attempts to reduce the relative motion between the object's edge and the surface by adjusting the angular velocity of the object. Eventually, if no other forces act on the object or the surface, the angular velocity will be such that the object no longer slips.

Consider an accelerating object that rolls without slipping. Since the object is accelerating, there is a nonzero static friction force acting on its edge. Does this static friction force do work on the system? The answer depends on whether the rolling object is modeled as a Point Particle System or a Real System. If the system is treated as a point particle, the static friction force is treated as though it acts on the particle's center of mass. It is therefore exerted over a distance and does work, affecting the object's total kinetic energy. If the system is treated as a real object, the static friction force is recognized as acting on the object's stationary edge, and therefore does no work. For the real system, the object's total kinetic energy would be the same even if the static friction force at its edge were absent. The only difference is that now some of its kinetic energy is rotational rather than translational. Point particles cannot have rotational kinetic energy, so the static friction force is treated as though it does work to account for the difference in total kinetic energy.

A Mathematical Model

Solving problems about rolling objects

The effects of kinetic friction on an object that slips while rolling are usually relatively easy to find becasue the [[Kinetic Friction] force acting on the object has a known, easy to calculate magnitude ([math]\displaystyle{ \mu_k * N }[/math]) and acts in a known direction (the direction opposing motion). The kinetic friction force causes a translational acceleration according to Newton's Second Law: the Momentum Principle as well as an angular acceleration according to The Angular Momentum Principle.

The effects of static friction on objects that accelerate by rolling without slipping can be harder to find because the magnitude of the friction force is often not explicitly given. Recall that the magnitude of a static friction force can be anywhere between 0 and a maximum value ([math]\displaystyle{ \mu_s * N }[/math]). For problems involving rolling objects, the magnitude of the static friction force usually does not reach its maximum value. In fact, you may not even be told [math]\displaystyle{ \mu_s }[/math], only that it is high enough to prevent slipping. To solve these types of problems, the correct approach is often to set up a system of equations:

1. [math]\displaystyle{ a = \frac{f_{ext} - f_s}{m} }[/math] by Newton's Second Law: the Momentum Principle
2. [math]\displaystyle{ \alpha = \frac{f_s r}{I} }[/math] by The Angular Momentum Principle
3. [math]\displaystyle{ a = \alpha r }[/math] by geometry, since slipping does not occur

Where [math]\displaystyle{ f_{ext} }[/math] is the external force applied to the center of the object causing it to accelerate, [math]\displaystyle{ f_s }[/math] is the static friction force, [math]\displaystyle{ a }[/math] is the translational acceleration of the object, [math]\displaystyle{ \alpha }[/math] is the angular acceleration of the object, [math]\displaystyle{ r }[/math] is the radius of the object, [math]\displaystyle{ m }[/math] is the mass of the object, and [math]\displaystyle{ I }[/math] is the moment of inertia of the object.

These equations, along with the information given, are usually enough to solve any problem related to the acceleration of an object that rolls without slipping.

A Computational Model

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Examples

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Simple

A uniform cylinder of mass [math]\displaystyle{ M }[/math] and radius [math]\displaystyle{ R }[/math] is placed on a ramp inclined with an angle [math]\displaystyle{ \theta }[/math] above the horizontal. It rolls down without slipping. What is the translational acceleration of the cylinder?

Solution:

Let us use the system of equations set out in the "Solving problems about rolling objects" section:

1. [math]\displaystyle{ a = \frac{f_{ext} - f_s}{m} }[/math]
2. [math]\displaystyle{ \alpha = \frac{f_s r}{I} }[/math]
3. [math]\displaystyle{ a = \alpha r }[/math]

For this problem, [math]\displaystyle{ m }[/math] and [math]\displaystyle{ r }[/math] are given to be [math]\displaystyle{ M }[/math] and [math]\displaystyle{ R }[/math] respectively. [math]\displaystyle{ f_{ext} }[/math] is [math]\displaystyle{ Mg\sin\theta }[/math] because that is the component of gravity directed parallel to the ramp and therefore not balanced by the normal force. [math]\displaystyle{ I }[/math] is [math]\displaystyle{ \frac{1}{2}MR^2 }[/math] because the object in question is a cylinder. [math]\displaystyle{ f_s }[/math], [math]\displaystyle{ a }[/math], and [math]\displaystyle{ \alpha }[/math] are the three unknowns.

Making these substitutions, the system of equations is now

1. [math]\displaystyle{ a = \frac{Mg\sin\theta - f_s}{M} }[/math]
2. [math]\displaystyle{ \alpha = \frac{f_s R}{\frac{1}{2}MR^2} = 2 \frac{f_s}{MR} }[/math]
3. [math]\displaystyle{ a = \alpha R }[/math]

Substituting equation 2 into equation 3 yields

[math]\displaystyle{ a = 2 \frac{f_s}{MR} R = 2 \frac{f_s R}{M} }[/math]

Middling

Consider the cylinder from the previous problem. The coefficient of static friction between the cylinder and the ramp is [math]\displaystyle{ \mu_s }[/math]. What is the steepest the ramp can be without the cylinder slipping?

Difficult

A golf player putts a .046kg golf ball with radius of .043m. It slides across the green, initially without rotating, with an initial speed of 6m/s. The coefficient of kinetic friction between the ball and the turf is .12. How long does it take for the ball to stop slipping? Assume that the golf ball is a perfect sphere that has its mass uniformly distributed throughout.

Solution:

The magnitude of the friction force is given by

[math]\displaystyle{ F_k = \mu_k * N = .12 * .046 * 9.8 = .054 }[/math]N

This friction force has two effects. Firstly, it reduces the ball's translational speed. Secondly, it applies a torque to the ball, causing its angular speed to increase. The ball stops slipping when the ball's angular speed reaches a value corresponding to its translational speed so that the bottom-most point on the ball does not move relative to the ground.

First, let us find the translational speed of the ball as a function of time:

[math]\displaystyle{ v_f = v_0 + at }[/math]

[math]\displaystyle{ v_0 = 6 }[/math]m/s, [math]\displaystyle{ a = \frac{f}{m} = - \frac{.054}{.046} }[/math]m/s²

[math]\displaystyle{ v(t) = 6 - 1.17t }[/math]

Next, let us find the angular speed of the ball as a function of time:

[math]\displaystyle{ \omega_f = \omega_0 + \alpha t }[/math]

[math]\displaystyle{ \omega_0 = 0 }[/math], [math]\displaystyle{ \alpha = \frac{\tau}{I} = \frac{.054 * .043}{\frac{2}{5} * .046 * .043^2} = 68.25 }[/math]s[math]\displaystyle{ ^{-2} }[/math]

[math]\displaystyle{ \omega (t) = 68.25t }[/math]

When the ball no longer slips, [math]\displaystyle{ v = r \omega }[/math].

[math]\displaystyle{ 6 - 1.17t = .043 * 68.25t }[/math]

[math]\displaystyle{ 6 = 4.10t }[/math]

[math]\displaystyle{ t = 1.46 }[/math]s.

After 1.46s, the ball rolls without slipping and no longer slows down (except due to air resistance).

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