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 Sikes^{1} 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 $KE=12mv2+12Icm\omega 2=710mv2$ assuming that $Icm=25mR2$ and $v=R\omega $ for a solid ball of radius *R* that rolls without slipping. Since the rolling friction force $F=\mu Rmg$, where *μ _{R}* is the coefficient of rolling friction, the authors concluded that

where *s* is the total distance traveled by the ball during its back and forth motion, *v _{i}* is the initial speed of the ball, and

*v*is the speed after traveling the distance

_{f}*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 $\Delta KE=\u2212Fs$. 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 $Fs=\mu RsN$. If it is assumed that both forces act at single points and *μ _{Rs}* is independent of rolling speed, then the acceleration of the ball is $a=dv/dt=\u2212\mu Rsg$ so $vf2=vi2\u22122\mu Rsgs$. Consequently,

For the decelerating ball, the torque due to the friction force acts in a direction to *increase* the rotational speed $\omega $. 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 $\tau =FsR\u2212ND=Icmd\omega /dt,\u2009Fs=\u2212mdv/dt$ and $v=R\omega $, it is easy to show that $\mu Rs=Fs/N=5D/(7R)$. 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 = $NDs/R=7Fss/5=7m(vi2\u2212vf2)/10$, consistent with the fact that the total kinetic energy of the ball is $7/10\u2009mv2$. Additional theoretical and experimental results concerning rolling friction are readily available.^{3–5}