What is the direction of rolling resistance?

Question

I would like to know the direction of rolling resistance between an object and a surface. From my research, rolling resistance causes a torque around the center of mass of the object, and since torque is a vector, it must have a direction. In case of a rolling wheel, it is clear: the rolling resistance must oppose the rolling motion. So if the rolling object has spin angular velocity $\vec{w}$, then the rolling resistance torque is in the $-\vec{w}$ direction. However, for general rolling shapes I am getting confused.

So consider a shape (could be a wheel, a coin, a ball, an egg etc) that is moving across a horizontal surface. I say moving, not rolling, because the object could be partly sliding and partly rolling. Suppose that the center of mass has velocity $\vec{v}$, and the object is rotating around its center of mass with spin angular velocity $\vec{w}$. The rolling resistance does creates a torque $\vec{\tau}_{rr}$ around the center of mass of the ball. This torque affects $\vec{w}$.

Now I have two contradicting ideas for the direction of the rolling resistance.

• On the one hand, I would expect it to oppose the rotation of the object, like in the wheel case. That means that $\vec{\tau}_{rr}$ is in the $-\vec{w}$ direction.
• On the other hand, I have understood so far that rolling resistance is caused by the distribution of vertical normal forces at the contact 'point' with the ground. Therefore, I would expect the induced torque $\vec{\tau}_{rr}$ to have only x and y components.

The two ideas are contradictory, because $\vec{w}$ might have a z-component.

The Wikipedia page on rolling resistance does not answer this question. It does say that rolling resistance 'opposes the motion'. But which motion is this? I could think of an 'imaginary' force $\vec{F}_{roll}$ at the contact point, which would correspond to the torque via $\vec{\tau}_{rr} = \vec{r}\times \vec{F}_{roll} $. Here $\vec{r}$ is the vector from center of mass to contact point. Then $\vec{F}_{roll}$ could oppose the motion $\vec{v}$, or it could oppose the instanteneous motion at the contact point $\vec{v} +\vec{\omega}\times \vec{r}$.

Which of my ideas is correct, and how can you derive this from physical arguments?

Answer 1

Am I wrong, but don’t you speak about the angular acceleration? This could change the way we have to consider the situation…

Answer 2

Your first argument, or rather, the first notion that the torque created due to rolling resistance opposes the relative angular velocity is correct. In your second idea, you start off by capturing the cause of rolling resistance correctly, but your inference that $\vec{\tau}_{rr}$ must not have a component whether opposing or supporting (we come to that later) the angular velocity seems incorrect.

Consider, without any loss of generality, a rubber tyre moving along in the positive x-axis with some angular velocity $\vec{w}$ in the $-\hat{z}$ direction. As you zoom in on the contact point, you will see that it is more of a patch than a point, and that the normal force is not acting at the geometric center of this patch. This follows from the fact that the part to the left of the center is recovering from the deformation and the rightward part is "leaning into" the ground. Thus, for this model, we could argue that a normal force $\vec{N}$ $ \hat{j}$ acts at a point, say, $d$ units right of the "true center". This point will thus have the lever arm vector $\vec{R}_{P} = d \hat{i} - R \hat{j}$, where $R$ is the radius of this tyre.

When you take the cross product of the lever arm vector and the normal force to find $\vec{\tau}_{rr}$, you will find that one of the terms in the resulting product vanishes and you are left only with $dN$ $\hat{z}$. To be fair, the magnitude of this torque is of little importance to us here, what is important is that it is precisely opposing the relative angular velocity of the tyre which was in the $-\hat{z}$ direction.

EDIT: I had previously omitted the reason as to why I think $\vec{\tau}_{rr}$ would even equal this cross product in the first place. The deformation of the wheel, like I mentioned, implies that the normal force does not directly point towards the axle, rather, it is offset by some small distance (here, $d$ units) rightwards, as opposed to an ideal case where it would point directly towards the axle. The torque that this normal force has is thus precisely the opposing torque that rolling resistance offers.