What is a detailed explanation for why friction does zero work on an object rolling on a surface without slipping?

Question

I'm working through Taylor's Mechanics and he seems to assume this implicitly at a few points. Intuitively, I agree that this should be the case. For while at any time during the rolling there ought to be a force of friction acting on the point of the object in contact with the surface, this force never "acts through a distance" since the object just rolls away. Trying to think more rigorously, I've come up with the following:

Thinking only in two dimensions for brevity's sake (although I see no reason why this argument can't be extended to three), and taking an arbitrary point on the object along its surface, the force of friction acting on this point is zero everywhere except for the point in time when it is in contact with the surface on which it is rolling. Assuming the path of the point can be given a continuous parameterization $(g_1(t),g_2(t))$ (I think this is a safe assumption), and taking an arbitrary partition of this curve $a = t_0<t_1...<tn=b$ and $c_i \in [t_{i-1}, t_i]$, each term in the approximating sums $\sum_{i=1}^{n}F_1(g(c_i))\cdot (g_1(t_i)-g_1(t_{i-1}))=S_1$ and $\sum_{i=1}^{n}F_2(g(c_i))\cdot (g_2(t_i)-g_2(t_{i-1}))=S_2$, where $(F_1, F_2)$ is a vector field defined on $A \supset g([a,b])$ representing the friction, is zero except in the case that $c_i$ is the point of contact of the point on the object and the surface. However since $g_1, g_2$ are continuous, for a sufficiently fine mesh of the partition, even if there is a $c_i$ corresponding to the point of contact in these sums, this term will be sufficiently small so as to make $|S_j-0| \lt \epsilon$ true for any $\epsilon \gt 0$, $j \in \{1,2\}$. Since these are the approximating sums of the Stieltjes integrals $\int (F_1\circ g)dg_1$ and $\int (F_2\circ g)dg_2$, $\int (F_1\circ g)dg_1 = \int (F_2\circ g)dg_2 = 0$. Whence we get $W_{fric} = \int_{g}\left\langle F, dx\right\rangle = \int_{g}F_1dx + \int_{g}F_2dy = 0$.

I am fairly confident in this argument, however I realize, as is often the case in physics, that there may be complicating factors that make a narrow mathematical argument like this untrue in general. So is this reasoning correct? If not, what is an argument for friction doing zero work on an object that is rolling without slipping? Or, is this not true at all?

Answer 1

Friction can arise through various interaction mechanisms at the interface: electrostatic attraction, mechanical interlocking of surface features, mechanical deformation and fracture, etc.

The friction mechanism being described as contributing zero in the case of perfect rolling is one specifically modeled as a force acting perpendicular to the interface, which acts on the two objects in contact if the surfaces slide against each other. This is just one possible friction mechanism.

As you note, no mutual sliding occurs, and so the force never acts through a distance; consequently, no work is performed. You can parameterize the path in a more precise way if you like, and express the process with more sophisticated mathematics, but your simpler statement that the relevant displacement is zero seems convincing enough when concluding that this particular mechanism does no work.

Again, though, other dissipative mechanisms exist, such as microscopic topography and reentrant structures deforming, interlocking, and fracturing even if the surfaces approach each other only perpendicularly and never undergo mutual sliding. Thus, the above argument doesn't imply that friction never does work in real-life cases of rolling without slipping.

I'm not sure this adds much that isn't already in your question, but perhaps you'll find it useful.