Comment on “A computer model of classical rolling friction” [Am. J. Phys. 87, 720–728 (2019)]

Zengel, Keith

doi:10.1119/10.0001633

In a recent paper,¹ Knop presented a particle level model that explains the energy losses due to rolling resistance in great detail, though he left unclear an important relationship between the normal and static friction forces that act on rolling objects. Knop suggests in his introduction that a forward shift in the normal force for a deformed rolling object will explain the slowed rotation of the object, but a static friction force must be included to explain the slowed linear velocity of the object. The reasoning^1,2 behind this interpretation is that the normal force would slow the rotation of the object while having no effect on its linear velocity, so a static friction force is required to slow the object's linear velocity, even though it would tend to cause the object to spin faster. In this comment, I will discuss the curious nature of the static friction force for rolling objects, show the equivalency of energy losses due to static friction and energy losses due to the forward shifted normal force, and argue that these are not two independent phenomena but rather two ways of expressing the same phenomenon of rolling resistance.

Figure 1 shows a simplified diagram of the forces acting on the deformed rolling object. Here, mg is the weight of the object, F_sf is the static friction force, and N is the normal force. Note that it is necessary to include the force of static friction if the rolling object is going to experience acceleration and roll instead of simply sliding, whether or not the object is deformed. Note also that the force of kinetic friction, which Knop showed to be negligible, has been omitted. The torque equation for an axis that runs through the center of mass of the rolling object is

\sum τ = I α = R F_{s f} - N D,

(1)

where α is the angular acceleration, I is the center of mass moment of inertia of the object, R is the radius of the object, and D is the forward shift in the normal force, which is always ahead of the center of mass along the direction of travel and constant with respect to θ.³ Here, we are using the approximation that the deformation of the object is small relative to the radius of the object, and therefore, the lever arm for the static friction torque is roughly R. This is consistent with experimental results,^3–5 which suggest

D / R \approx 1 0^{- 3}

for a steel ball rolling on an aluminum track, for example. Assuming that the object rolls without slipping, we can define the linear

\hat{x}

acceleration as

a = α R .

(2)

Fig. 1.

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A diagram of the forces acting on a deformed but otherwise symmetric body rolling on perfectly uniform incline.

We can write the moment of inertia as

I = β m R^{2},

(3)

where β is the dimensionless coefficient that quantifies the distribution of mass of the ball or cylinder.

From Fig. 1, we see that only the center of mass and static friction forces affect the x-component of the center of mass velocity. Knop found the static friction force by calculating the additional force required for each particle composing the wheel to experience zero net horizontal force while in contact with the surface, but we can follow previous work^6–8 to solve directly for its magnitude,

F_{s f} = \frac{β}{1 + β} m g sin θ + \frac{N D}{(1 + β) R} .

(4)

This simple result is often overlooked because physicists typically eliminate F_sf to solve for the linear acceleration of the body. The first term in Eq. is the reaction to the component of the gravitational force acting parallel to the plane. The second term is the force of rolling resistance.

This static friction force vanishes for θ = 0 and D = 0, which means that an object rolling on a perfectly level surface with no deformation would experience no static friction force and therefore no change in its center of mass velocity. We can also see that the forward shift of the normal force is actually required for the rolling resistance term in the static friction force. So, while Knop was correct to point out that the forward shift of the normal force causes a torque that slows the rotation of the ball, he ought also to have pointed out that this slowed rotation would mean that the object would push forward on the rolling surface, which would in turn cause a backward pointing component of the static friction force that would slow the translational velocity of the object.

Furthermore, if we define the coefficient of rolling resistance as^2,9

μ_{r r} = \frac{D}{(1 + β) R},

(5)

then it is obvious that the rolling resistance term in Eq. can be written in the traditional form of a friction force,

F_{s f} = \frac{β}{1 + β} m g sin θ + μ_{r r} N .

(6)

It is also worth noting that the force equation for the

\hat{x}

direction can be written as

m a = - [\frac{N D}{R} + β m a] + m g sin θ,

(7)

where the bracketed term is the static friction force. In this form, we see that there is an acceleration term in F_sf. This atypical result is worth noting because according to Newton's second law, the acceleration of an object of a given mass is determined by the net force acting on it, but here the acceleration also affects the static friction force. This means that any additional forces, whether resistance to rolling or a parallel component of the gravitational force, will appear with an additional multiplying factor of

1 / (1 + β)

when Eq. is solved for a traditional form of

\sum F = m a

⁠. Note that this means that the torque due to rolling resistance in Eq. , which can be written as

N D / (1 + β)

⁠, will always be smaller in magnitude than the torque due to the forward shifted normal force,

- N D

⁠, which is why the object will spin slower even though the rolling resistance torque seems as if it would cause the object to spin faster. Note also the interesting result that the static friction force will point uphill for a ball rolling downhill and for a ball rolling uphill.⁷

We can also investigate the origin of the first term in Eq. . One way to understand this term is to look at the torque about the center of mass axis, as shown in Eq. , and the torque about the midpoint of the small area of contact,

\sum τ_{a c} = (β m R^{2} + m R^{2}) α = Rmg sin θ - N D .

(8)

For the case of no deformation, the ratio of Eqs. and is

\frac{\sum τ}{\sum τ_{a c}} = \frac{β}{1 + β} = \frac{F_{s f}}{m g sin θ} .

(9)

This means that the

β / (1 + β)

coefficient results from two principles: the fact that the angular acceleration around the midpoint of contact is equal in magnitude to the angular acceleration around the center of mass and the fact that the moment of inertia is smaller for rotation around the center of mass than for rotation around the midpoint of contact. In the case of an undeformed rolling object, a force parallel to the incline applied at the center of mass of a rolling object will induce a static friction force in the opposite direction with a reduced magnitude. For the case of a deformed object, the result of this ratio is equivalent to Eq. .

This analysis explains why, as Knop points out, the “angular work” from the deformation torque corresponds exactly to the kinetic energy losses of the object. The angular work around the midpoint of contact due to the deformation torque can be written as

\int (m R^{2} + I) α d θ = - \int N D d θ,

(10)

which reduces to

\frac{1}{2} m R^{2} (1 + β) (ω_{f}^{2} - ω_{i}^{2}) = - N D Δ θ,

(11)

where ω_f and ω_i are the final and initial angular velocities around the center of mass of the object. The change in linear velocity due to the rolling resistance term of the static friction force is given by

\int m a_{c m} d x = - \int μ_{r r} N d x,

(12)

which reduces to

\frac{1}{2} m (v_{f}^{2} - v_{i}^{2}) = - \frac{N D}{(1 + β) R} Δ x,

(13)

where v_f and v_i are the final and initial center of mass velocities. Using the condition of rolling without slipping, we can say that

R ω = v

and

R Δ θ

=

Δ x

⁠, which means that Eqs. and are equivalent. It is worth noting that Eq. comes with some interpretational difficulties, as it seems to suggest that the static friction force does work even though there is no displacement in the region where the force acts on the object. The important detail is that the left-hand side of Eq. includes the center of mass acceleration, not the acceleration of the midpoint of contact. So, while static friction does no work on the portion of the object in contact with the surface, it does contribute to a loss of translational kinetic energy at the center of mass of the object. Some authors have argued^10,11 that Eq. should be referred to as “pseudowork,” though the term is not without controversy.^12–14 The relevant point here is that, while the normal force does not reduce the translational velocity of a rolling object (because it acts in a direction perpendicular to the velocity of the object), the forward shift in the normal force is related to the static friction force, which acts in the opposite direction of the center of mass velocity and therefore can be understood as the phenomenon ultimately responsible for reducing the center of mass velocity of the object.

The model presented here is consistent with Knop's results and several models of rolling resistance referenced by Knop.^4,9,15 Knop's results explain why the normal force is forward shifted and how the energy loss occurs, which are new and valuable insights, but they fail to convey the relationship shown here between the forward shift in the normal force and the static friction force. It is not incorrect to suggest that rolling resistance requires both a forward shifted normal force and a static friction force, but teachers should emphasize the atypical role of static friction in the physics of rolling.