Why is sliding friction more than rolling friction? [duplicate]

Question

Frictional force doesn't depend on area of contact. So why is co efficient of rolling friction less than coefficient of sliding friction?

Answer 1

So one useful way to view friction is to imagine that the surfaces of the materials we're talking about are loaded with a bunch of springs which, after they hit a certain displacement, "break", stealing a constant vibration energy of $\epsilon = k \lambda^2/2$ over some length scale $\lambda$ that they need to break. The exact functional form of this energy expression doesn't matter, I don't think.

Since you're going at a very large speed by atomic standards, you can just picture that there must be some average number-of-interactions $n$, at any given time, on the surface which doesn't depend much on velocity. It probably depends on lots of other things. For example, it depends certainly on the area of the surfaces: more area means more interactions, right? But it probably also depends on the pressure of the surfaces: more pressure means the surfaces are atomically closer together which probably proportionately increases their interactions. The same basically holds for static friction, but $n$ is going to be larger because once the surfaces are still next to each other (at atomic distances) you expect a lot more connections. But at the atomic scale things are still really "gappy" so a higher pressure probably proportionately increases the contact at atomic scales, and a higher area probably does the same.

So for static friction, you have to pay some energy $n \epsilon$ over some distance $\lambda$ in order to move the surfaces "at all", manifesting as a critical force $n \epsilon / \lambda$, with $n = k_s P A$ for the pressure $P$ between the surfaces and the area $A$ between them, and some constant $k_s$ not depending on either of those.

For dynamic friction, you have to pay a different amount of energy depending on how fast you go: you break $v / \lambda$ connections per second, losing an energy $n \epsilon$ for each of them. So your power loss is proportional to $k_d P A v / \lambda$. This is a power loss characteristic of a different constant force, $k_d P A / \lambda$. Now the punchline is that if P is an intersurface pressure and A is a surface area, $F_N = P A$ is a total intersurface force between those. So that's why the "normal force" comes into these expressions, and why $\mu_s = k_s/\lambda$ is so much larger than $\mu_d = k_d/\lambda$.

Now we come to your question: what about rolling friction? Well, rolling friction is a special subtype of static friction: the surfaces are not sliding next to each other, so the cost paid is not proportional to $v$ in the same way dynamic friction is. Instead of blazing past the surface sideways, the atom is lowered into contact, held there for a while, and raised until contact breaks.

The ideal wheel has a shape profile $y(x) = R - \sqrt{R^2 - x^2} = R(1 - \sqrt{1 - (x/R)^2})$. Now $1 - \sqrt{1 - u^2} \approx u^2/2 + u^4/8 + u^6/16 + \dots$, so let's just pretend that this is the parabola $y(x) = R (x/R)^2 / 2 = x^2/(2 R)$.

If we approximate the wheel's trajectory as a parabola $y = x^2 / (2 R)$, where $R$ is essentially a "radius of curvature" when the wheel's not perfectly round, then the wheel comes within $y < \lambda$ and then leaves $y < \lambda$ for the $x$ values $-\sqrt{2 R \lambda} < x < \sqrt{2 R \lambda}$, or the distance $\sqrt{8 R \lambda}$. As you can see, we've got a geometric mean between $\lambda$ and $8R$ and it's going to be a lot bigger, therefore, than $\lambda$ was.

This means that the dynamic friction cost is not paid with a frequency $v / \lambda$ but with a frequency $v / \sqrt{8 R \lambda}$. Since the denominator is so much larger, the frequency is so much lower. And that's some rough indication of why rolling friction is so much lower of a force: it uses a lot less energy because it's a form of static friction where the surfaces stay in contact for a much longer time. It's not even "much longer" by macroscopic standards: but it's a tremendous increase in the microscopic world.

Answer 2

'Rolling friction' isn't actually a frictional force in the same sense as sliding friction. The term 'rolling friction' is a little misleading, and is better named 'rolling resistance'. When something rolls, e.g. a wheel on an axis, a lot of forces go into resisting the rolling. These forces can include the friction of the bearings, the momentum of the tire, etc. We can describe the overall effects of this resistance to rolling as some coefficient times the normal force of the object in order to draw a connection to regular sliding friction.

Since the forces that resist rolling are usually pretty small (such as low friction on a greased axle), the so-called 'rolling friction' coefficient turns out to be pretty small as well.

This website has a good short discussion of rolling friction near the bottom of the page: http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html