Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

10 Answers 10

up vote 15 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.

share|cite|improve this answer
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: 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 '14 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 '14 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 '14 at 20:24
(with "s" being the arc length movement of the point of contact with the curve) – NeuroFuzzy Nov 26 '14 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.

share|cite|improve this answer

The above answers are all good but i want to give another example which really helped me with understanding what does it mean that the point of contact has a velocity of zero.

Think of the 'spinning circular object' not as a ball, but instead think of it as a star polygon with an infinite amount of edges, for the sake of the example a very large number will suffice - 9 edges

enter image description here

At any given moment only one of the edges is touching the ground. Think of the star's movement - if it's not slipping, the point that is touching the ground is not moving, it pushes against the ground, 'fighting' the force of friction.

Another nice example is the human cartwheel, but it has 2 points touching the ground at the same time, which is why i like it less...

enter image description here

Now this is why wheels are good for transporting things, as a result of the contact point not moving only the static friction force is working on the wheel, and static friction maintains conservation of energy.

I hope this gives another point of view to anyone googling the subject and running into this question...

share|cite|improve this answer

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

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 $0$.

share|cite|improve this answer
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 – user23503 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.

share|cite|improve this answer
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.

share|cite|improve this answer

A physical scenario will help you visualise some of the other answers, particularly Muphrid's and nonagon's.

Imagine an aeroplane coming in to land, as a wheel of its landing gear touches the ground. If the tyre isn't spinning before contact, the point of contact on the tyre is moving at the same speed as the aeroplane, approximately $70{\rm m\,s^{-1}}$, and so the rubber is dragged along the ground at this speed, and there's a huge skid with heaps of smoke. Naturally, a huge torque on the wheel results and its angular velocity swiftly rises until there is no more skidding.

In practice, however, many planes' wheels are spun by motors so that there is less skidding.

Now imagine this scenario with different wheel angular speeds at contact. It's also helpful to think from the plane's stationary frame of reference. The wheel's angular speed makes the bottom of the wheel move at a speed $\omega\,r$ backwards relative to the plane. The ground is moving backwards at some speed relative to the plane, and to get the best life out of your tyres, you want the backwards motion of the point of contact to exactly match the backwards motion of the ground. If the wheel's initial angular speed is too slow, the ground is moving backwards relative to the point of contact, and there is a skid that tends to increase the wheel's angular speed. However, suppose we spin the wheel really fast, so that the point of contact is moving backwards relative to the plane faster than the ground just before contact. There is then a skid in the opposite direction, which tends to decrease the wheel's angular speed. When your initial spin angular velocity is such that the point of contact's motion relative to the aeroplane is the same as that of the ground, you get no skid.

share|cite|improve this answer

To understand rolling without slipping, first consider the case of rolling only case about the center of mass.In this case, a point on the top rim will have a speed $v=\omega\cdot R$ and a velocity $v=-\omega\cdot r$, at bottom of the rim, as observed by you. However in case of rolling without slipping, We observe the velocity of center of mass is $v$ (only translational) and velocity at the top rim is translational and rotational is equal $v+\omega R$ and velocity at bottom, at contact, is $\omega R-v$ (translational and rotational).

Remember, $v$ here is the velocity of center of mass. Since $v=\omega\cdot R$, we conclude the velocity at contact is zero and at top is $2\cdot v$.

share|cite|improve this answer
I mean, first consider the case of only spinning – Dr. A Feb 19 '15 at 11:11
You can click the little "edit" button to edit your answer - that would make things clearer (comments are not permanent here). Also you might want to expand on what you mean by "case of only spinning". – Neuneck Feb 19 '15 at 11:16

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.

share|cite|improve this answer

watch the following video for a great explanation:

and read this article for the interesting causes of rolling resistance/friction: //

share|cite|improve this answer

protected by Qmechanic Feb 19 '15 at 11:21

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.