The coefficient of rolling friction is usually very small when a smooth ball rolls on a smooth surface, making it hard to measure. However, Minkin and Sikes1 have described a relatively simple method, where a ball is allowed to roll back and forth along a curved track. At any given time, the kinetic energy of the ball is given by $KE=12mv2+12Icmω2=710mv2$ assuming that $Icm=25mR2$ and $v=Rω$ for a solid ball of radius R that rolls without slipping. Since the rolling friction force $F=μRmg$, where μR is the coefficient of rolling friction, the authors concluded that

$μR=710(vi2−vf2)gs,$
(1)

where s is the total distance traveled by the ball during its back and forth motion, vi is the initial speed of the ball, and vf is the speed after traveling the distance s.

Equation (1) is based on a phenomenological definition of friction, where the frictional force equates directly to a decrease in kinetic energy through the relationship $ΔKE=−Fs$. However, students may find this definition confusing, since we also teach that when an object rolls without slipping, the only friction force (other than air resistance) is the static friction that acts on the point of contact. Therefore, this static friction might reasonably be called rolling friction.

Static rolling friction provides the force that decreases the translational speed of the ball, but it does no work. In a similar way, static rolling friction provides the force to accelerate a vehicle, but the work is done by the motor. If a ball rolls on a horizontal surface, then the normal force N = mg and the friction force $Fs=μRsN$. If it is assumed that both forces act at single points and μRs is independent of rolling speed, then the acceleration of the ball is $a=dv/dt=−μRsg$ so $vf2=vi2−2μRsgs$. Consequently,

$μRs=12(vi2−vf2)gs.$
(2)

For the decelerating ball, the torque due to the friction force acts in a direction to increase the rotational speed $ω$. The reason that v and ω both decrease with time is that the normal reaction force, N, acts at a point located a distance, D, ahead of the middle of the ball.2 The normal force is actually distributed over the whole contact region, but the net effect is equivalent to a single force N acting at a point ahead of the center of mass since the compression force at the front of the ball is larger than the expansion force at the rear of the ball. The opposing torque, ND, is larger than the torque due to friction, $FsR$. Since $τ=FsR−ND=Icmdω/dt, Fs=−mdv/dt$ and $v=Rω$, it is easy to show that $μRs=Fs/N=5D/(7R)$. The work done to decrease the total kinetic energy, while the ball and/or the surface compresses and expands, is given by torque × angular displacement = $NDs/R=7Fss/5=7m(vi2−vf2)/10$, consistent with the fact that the total kinetic energy of the ball is $7/10 mv2$. Additional theoretical and experimental results concerning rolling friction are readily available.3–5

1.
L.
Minkin
and
D.
Sikes
, “
Coefficient of rolling friction: Lab experiment
,”
Am. J. Phys.
86
,
77
78
(
2018
).
2.
R.
Cross
, “
Rolling to a stop down an inclined plane
,”
Eur. J. Phys.
36
,
065047
(
2015
).
3.
R.
Cross
, “
Coulomb's law for rolling friction
,”
Am. J. Phys.
84
,
221
230
(
2016
).
4.
K.
Maslova
,
V.
de Jesus
, and
D.
Sasaki
, “
Understanding the effect of rolling friction in the inclined track experiment
,”
Phys. Educ.
55
,
055010
(
2020
).
5.
C.
Hanish
and
M.
Ziese
, “
Rolling friction in a 3D printed stringless pendulum
,”
Eur. J. Phys.
42
,
045004
(
2021
).