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I have never understood what's the meaning of the sentence "rolling without slipping". Let me explain.

I'll give an example. Yesterday my mechanics professor introduced some concepts of rotational dynamics. When he came to talk about spinning wheels he said something like:

"If the wheel is rolling without slipping, what's the velocity of the point at the base of the wheel?? It is... zero! Convince yourself that the velocity must be zero. Since if it wasn't zero, the wheel wouldn't be rolling without slipping. So the wheel is rolling without slipping if and only if the point at the base has velocity zero, i.e. if and only if the tangential speed equals the speed of the center of mass."

Well, what I really don't understand is this: is the "rolling without slipping" condition defined as "Point at the base has zero speed"? If not, what's the proper definition for that kind of motion?

Looking across the internet, I have found more or less the same ideas expressed in the quotation. Furthermore, if it was a definition, then it would be totally unnecessary to say "convince yourself" and improper to talk about necessary and sufficient conditions.

I'd like to point out that I'm not really confused about the mathematics behind this or with the meaning of the condition above. What puzzles me is why are those explanations always phrased as if the condition $v'=0$ (where $v'$ is the relative velocity beetween the point at base and the surface) is some necessary and sufficient condition to be "rolling without slipping". Seems to me that this is exactly the definition of "rolling without slipping" and not an "iff".

Any help is appreciated, thanks.

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7 Answers 7

up vote 9 down vote accepted

You can always decompose a motion like this into two parts: (1) rolling without slipping and (2) slipping without rolling.

What is slipping without rolling? It means the object moves uniformly in one direction along the surface, with no angular velocity about the object's own center of mass. For instance, a box that is pushed along the ground can easily slip without rolling.

Unfortunately, most people seem to assume that you can infer some physically important information from your own notion of what slipping is, without having to define it. I believe this is done to try to connect to intuition, but in the process, things get a lot more nebulous and ill-defined.

To me, it's easier to think about this in terms of the object's rotation--it was never obvious to me that the point in contact with the ground doesn't have velocity at the instant it touches. I prefer to think instead that an object that rolls without slipping travels 1 circumference along the ground for every for every full rotation it makes. And object that travels more than this distance (or that doesn't rotate at all) is slipping in some way.

Then, eventually, we can get to the notion that the point in contact during rolling cannot have nonzero velocity through whatever logical or physical arguments necessary.

But as is usual in physics, it's not really clear what definition should be considered "fundamental" with other results stemming from it. This emphasizes that physics is not built axiomatically.

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Thank you, this answer gives me a more precise idea about the matter. I understand that there's some degree of arbitrariness in the use of those terms and, for communicating, maybe this way is better. Your last sentence makes a good observation. If I may add something, I believe that a direct consequence of physics not being built axiomatically, is that it must be given more attention to the equivalences beetween definitions (and, ofcourse, one has to consider the existence of different definitions). However, I'm starting to go OT. Thank you for the answer. –  pppqqq May 14 '13 at 16:56
@Muphrid, see: mathandcode.com/img/diskrollnoslip.gif would you agree that this example shows that your bolded sentence isn't true on non-flat surfaces? It illustrates a circle radius $8/9$ rolling inside a circle radius $1$. The circle rolls an angle of $2\pi/8$ while its point of contact with the ground travels a distance $2\pi$ and the center of the circle travels a distance $2\pi/9$ (not $2\pi/8$). –  NeuroFuzzy Nov 26 at 19:59
@NeuroFuzzy Sure, if you define the overall travel of the wheel with respect to the center of the wheel. If instead you measure the overall distance traveled by whatever point is in contact with the ground, I think the bold statement still holds. Regardless, travel of a wheel along a curved surface is more involved, and I think it's good you point that out. –  Muphrid Nov 26 at 20:05
@Muphrid I included the overall distance travelled by the point of contact! I'm just looking at rolling without slipping on curved surfaces and many online solutions write $r d\theta=ds$, but I'm pretty sure (citing my example above) the correct version is $rd\theta=(1-kr)ds$, where $k$ is the curvature of the surface. Actually I derived that formula using the "a point in contact with the ground is momentarily stationary" condition. I'm mostly posting this for a sanity check. –  NeuroFuzzy Nov 26 at 20:24
(with "s" being the arc length movement of the point of contact with the curve) –  NeuroFuzzy Nov 26 at 20:25

If the wheel is rolling without slipping, what's the velocity of the point at the base of the wheel?? It is... zero! Convince yourself that the velocity must be zero. Since if it wasn't zero, the wheel wouldn't be rolling without slipping.

So far the explanation is correct. "No slipping" refers really to some non-zero interval of time, and to the state of the contact faces during this time. When there is no slipping the faces can exert greater tangential force on each other than in the state of sliding, and there is no loss of mechanical energy involved.

This can happen when two bodies are in physical contact for some non-zero time and the parts in contact have the same velocities during that interval.

It is not good to define no slipping just for an instant via the requirement that $v' = 0$ at that instant, because that can happen even if the bodies slide on each another at all other instants.

So the wheel is rolling without slipping if and only if the point at the base has velocity zero,

the word "so" is not very good here, and it should be added at the end that "contact point has velocity zero _all_the_time". Then it is OK.

It is interesting though that in practice there seems to be no case of perfect no-slip rolling. There is always some slipping and hence friction involved, even the train wheels on rails do slip a bit. The no-slip condition is thus a convenient approximation.

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Basically , it means at each instant the bottom most point has $0 $ velocity , it doesn't mean that the point has no acceleration . But at an instant it has $0$ velocity . And because of that at each instant $v_{cm}=\omega r$ for the bottom most point , and if this doesn't happen , then static friction acts to make it $0$ .

Its like suppose you are walking , your press your feet on the ground and road pushes you forward but your feet doesn't slide w.r.t. road in horizontal direction , however you can always lift it up . But road upto a certain extent resists motion in horizontal direction .

Also this'll help http://www.youtube.com/watch?v=9I1KSagocdE.

And don't worry , this is a normal thing to be confused upon . Almost everyone gets confused here .

Just remember velocity is $0$ at an instant , but acceleration is still there , meaning it can move at later instants of time . Like a block performing SHM at extreme position is at rest at an instant , but that doesn't mean it'll remain at rest , however in pure rolling it happens that at each instant there is point whose velocity gets $0$.

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Thank you for the link. I'm not really confused about accelerations, what puzzles me is why is this condition always phrased as if having the bottom most point with speed zero w.r.t. the surface is some necessary and sufficient condition to be "rolling without slipping". –  pppqqq May 14 '13 at 13:56
Because otherwise kinetic friction will act and you'll slip –  nonagon May 14 '13 at 15:12

The relative speed of the point of contact of the rolling body w.r.t. the surface on which it rolls is zero.

If the surface is at rest then the velocity of the point of contact of rolling body and surface is zero.


enter image description here

$$v_1 -\omega R=v_2$$

Also we can get the reltation in accelerations ..... Differentiate the above eq. $$a_1-\alpha R=a_2$$

Where $\alpha $ is the angular acceleration.

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Thank you, but what do you mean by "the surface is at rest"? If we are in the surface reference frame? –  pppqqq May 14 '13 at 13:34
@Kazz8: I mean $v_2=0$, and that angular acceleration is $\omega$ with direction as shown. –  ABC May 14 '13 at 13:36
I see. But it is 0 respect to what? To the ground? However, if by speed we mean relative speed beetween the surface and the wheel, this is exactly my definition: in this case there's no reason to say "rolling without slipping if and only if the point is at rest w.r.t the surface". –  pppqqq May 14 '13 at 13:52
@Kazz8 Yes.$v_2$ is velocity wrt ground.The words in ""__"" describe the rolling correctly.. –  ABC May 14 '13 at 14:16

The formal definition of rolling without slipping is the following:

Suppose you have two (rigid) bodies in mutual contact between each other. There are three different points at the point of contact: one of them (namely A) is a material point and belongs to the first body, the second one (namely B) to the other body, and the remaining one is the geometrical point. Of course, since the two bodies are in mutual contact, all these three points are at the same place at that moment.

Rolling without slipping ocurrs when the velocity of the material points (A and B) in contact are the same for any time.

Example: a disc (center O, point of contact P, radius a) rolling without slipping on a table.

Let x denote the coordinate of the center of mass. Using distribution of velocities for a rigid body and rolling without slipping, we have:

$$0 = \vec{v_P} = \vec{v_0} + \vec{w}\wedge(\vec{r_P}-\vec{r_O}) = \dot{x}\hat{i} + \vec{w}\wedge(-a\hat{j})$$

Since $$\vec{w} = -\dot{\theta}\hat{k}$$ (suppose angle is measured counterclockwise)

Then: $$ 0 = (\dot{x} -a\dot{\theta})\hat{i} \Rightarrow \dot{x} = a\dot{\theta}$$

This is what's usually find on introductory physics courses: the velocity of the center of mass is equal to w (angular velocity) times a (radius).

To sum up: in your example since the ground is -of course- at rest, then the velocity of the point at the base of the wheel is 0. Altough that's not the definition of rolling without slipping.

Physics works, and rigurous definitions exists. It's just a matter of finding them.

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Suppose one has a piece of thin string which is sitting on the ground in front of the wheel, wraps around the wheel once, and then continues along the ground behind the wheel. As the wheel rolls, different parts of the string will be wrapped around it. At any point where the string is in contact with the wheel, its velocity will match that of the wheel. At any point where the string is in contact with the ground, its velocity will match the ground. At the two points where the string is in contact with both the wheel and the ground, its velocity will match both, meaning corresponding parts of the wheel and the ground must have the same velocity.

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watch the following video for a great explanation: http://www.youtube.com/watch?v=xbXsSEtbkzU

and read this article for the interesting causes of rolling resistance/friction: //www.school-for-champions.com/science/friction_rolling.htm

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