Will a disc or cylinder (rigid body) executing pure rolling on a rough surface stop, neglecting air drag and other heat losses and rolling friction but not static and kinetic friction? If yes, due to which friction it will stop, static or kinetic and how? Assume surface has no rolling friction.

  • $\begingroup$ What do you mean by the term rough surface? Is it just rough enough so that you can say there will be a friction force acting? Does pure rolling mean no slipping - no relative movement between the disc and the surface at the point of contact? Given these ideal conditions there will be no frictional force acting and the disc will roll for ever. $\endgroup$
    – Farcher
    Mar 27, 2017 at 10:33
  • 21
    $\begingroup$ ...neglecting one type of friction but not the others strikes me as inherently inconsistent, since they all arise from the same cause, roughness of the surfaces: You're basically saying "this surface is so smooth it has no friction but it's rough enough to have friction", which makes no sense. $\endgroup$
    – ACuriousMind
    Mar 27, 2017 at 12:22
  • 1
    $\begingroup$ @Farcher guesswork, but I'd say rough should simply mean on some local scale non-flat, independent of friction. $\endgroup$ Mar 27, 2017 at 16:08
  • 2
    $\begingroup$ How can you neglect heat loss and still dissipate energy (assuming you don't imply that the energy is somehow lost by transmitting motion to the infinitely massive surface)? $\endgroup$
    – Hot Licks
    Mar 27, 2017 at 19:39
  • 1
    $\begingroup$ @Farcher Pure rolling is another way to say that there is no slipping (The OP Is from India and I am from India too; this terminology is used in our textbooks). A surface is said to be rough if it has friction and almost always, we mean a flat surface (and smooth surfaces are said to be frictionless). These terms appear in school exams as well as national competitive exams. These terms are ambiguous; however, one can make a reasonable guess. The question is unclear anyway. $\endgroup$
    – Yashas
    Mar 28, 2017 at 12:31

4 Answers 4


As Yashas Samaga said, it will not stop on a smooth, but frictional surface. It will stop however on an actual rough surface (as it does in reality – e.g. a steel marble rolling on a rough stone surface will come to a halt quite quickly, although drag / rolling friction is as low as on a smooth glass plate, where the marble would indeed roll very far).

The reason is that a rough surface can in general not be continually tangent to the rolling body. Instead, if the object has rolled over a peak, it will not smoothly traverse the following trough but slightly collide with the next peak. If there's no rolling friction, then the collision will (ideally) be perfectly elastic, i.e. the cylinder will bounce off. When it hits the surface again, the vertical kinetic energy will regenerally not be fully reclaimed to movement in the original direction. In fact, while it has still some velocity in that direction, it will statistically more likely clash with yet another opposing front of the profile, thus losing yet more momentum.

So, I reckon ideally this would eventually lead to a random-walk kind of motion. In reality, this doesn't happen because the collisions are scarcaly sufficiently elastic – actually a good amount or kinetic energy is lost right when the roller hits the next peak.


Assumptions made in this answer:

  1. By rough surface, you meant a flat surface which has friction.
  2. The cylinder/sphere/disc/etc. are ideal; they do not deform.

This is my reasonable guess; I am aware of the terminology used in Indian high school textbooks and exams (I am from India too) but you should still edit your question and make it clear.

When a perfect/ideal cylinder (or a sphere, disc, ring, etc) pure rolls, the velocity of the lowermost point is zero (condition for pure rolling). As the relative velocity between the surfaces at the point of contact is zero, there is no "kinetic" friction (if there is no external force, there will be zero static friction).

Therefore, the cylinder will continue to roll forever in your case.


The cylinder will continue to roll forever unless it is acted upon by an external unbalanced force. There are situations where you can accelerate the object while pure rolling. One situation where this happens is shown in the figure below:

enter image description here

Let $f$ be the frictional force.

Let $F$ be the external force ($\le f_{max} = \mu N$).

The condition for an object pure rolling is:

$$v_{com} = \omega R \tag{1}$$

The translational velocity of the lowermost point cancels out the rotational motion of the lowermost point completely.

Differentiating the equation $(1)$ with respect to time, you get:

$$a_{com} = \alpha R \tag{2}$$

The translation acceleration can be compensated by angular acceleration such that as the translation velocity increases (or decreases), the angular velocity also increases (or decreases) to ensure that condition $(1)$ is satisfied.

In this case, there is no kinetic friction as the contact surfaces are still at rest. However, static friction acts (had it not, there would be relative motion as $v$ would change without affecting the value of $\omega$ which would cause the $(1)$ to fail).

The net force ($F_{net}$) and torque ($\tau_{net}$) can be calculated as follows:

$$F_{net} = ma = F - f \tag{3}$$

$$\tau_{net} = I\alpha = -fR \tag{4}$$

You have three equations (equation $(2)$, $(3)$ and $(4)$) and three unknowns ($f$, $a$ and $\alpha$). You can solve for $a$ and $\alpha$. From these values, you can calculate the time it takes for the body to stop rolling.

  • 3
    $\begingroup$ Lol I meant more like changing "As the relative velocity between the surfaces at the point of contact is zero, there is no friction." to "As the relative velocity between the surfaces at the point of contact is zero, there is no kinetic friction.". I.E. the static friction is purely for anti-slip. It was more or less semantics, not clarity. Adding in the detailed analysis doesn't hurt though, just took a bit more time :P $\endgroup$
    – JMac
    Mar 27, 2017 at 11:33
  • 2
    $\begingroup$ @YashasSamaga I think your discussion about forces is irrelevant to the subject (because there is no such unbalanced force in the OP's problem). But most of all, I think your assertion that the roller won't stop without such a force is wrong: your argumentation only applies to surfaces smooth enough that you have a continually shared tangent. That is precisely what's not given on a properly rough surface. $\endgroup$ Mar 27, 2017 at 15:40
  • $\begingroup$ The OP asked for a specific case of cylinder or disc (I assumed them to be perfect). By rough surface, the OP meant that the surface has frictional forces; in other words, assume that there is friction. The unbalanced forces was an additional information because it is important. $\endgroup$
    – Yashas
    Mar 27, 2017 at 15:43
  • $\begingroup$ Hm, ok I suppose that's a possible interpretation (the question should have been phrased better). But I don't consider “with friction” a very sensible definition of rough – a rubber surface can be very smooth yet give plenty of friction, whereas a textured steel surface actually has little friction but bumps that you can cling to. $\endgroup$ Mar 27, 2017 at 15:53
  • 1
    $\begingroup$ @JasonC The OP said the object is pure rolling on a surface; isn't that an indication that the OP is talking about ideal cases? With the rough surfaces you have mentioned, it is impossible to pure roll. In the case of #2, the body flips and then goes back into pure rolling. My bonus part talks about such situations. By rough surface, the OP probably meant a surface which has friction. If you consider the type of rough surface you have mentioned, it is impossible for a body to pure roll. It wouldn't start pure rolling in the first place. So you can logically eliminate that case. $\endgroup$
    – Yashas
    Mar 28, 2017 at 12:11

If both the cylinder and the surface are perfectly rigid, then yes, it will roll forever, at least until it encounters a bump that its kinetic energy is not sufficient to overcome.

But if the surface can deform, not all of the energy used to deform it will be returned to the cylinder. Some of it will propagate away from the point of contact in the form of sound waves, never to be seen again. The cylinder will lose energy and slowly slow down. Eventually it will encounter a bump that its remaining energy can't carry it over, and it will come to a stop.

Similarly, if the cylinder itself can deform, it will be building up internal vibrations from the roughness as well. This will also subtract from the rolling energy.

  • 2
    $\begingroup$ Right, though you don't seem to be using the word elastic in the standard physics sense. $\endgroup$ Mar 27, 2017 at 16:01
  • $\begingroup$ @leftaroundabout: I am not a physicist, but I think I am, so you'll have to explain in more detail what you mean. Remember, we're not talking about particles here, and the contact forces are not necessarily normal to either surface. $\endgroup$
    – Dave Tweed
    Mar 27, 2017 at 16:15
  • 3
    $\begingroup$ In physics, elastic means something can redirect momentum without dissipating energy. This does not say anything about deformability (in fact, rigid bodies are often more elastic than deformable ones – steel is highly elastic, play-doh or even rubber is pretty inelastic to a physicist). $\endgroup$ Mar 27, 2017 at 16:19
  • 2
    $\begingroup$ @DaveTweed I think the confusion is that what you're describing (loss of kinetic energy to internal energy) is exactly what a physicist would call an inelastic effect. $\endgroup$
    – hobbs
    Mar 27, 2017 at 18:48
  • 2
    $\begingroup$ Terminology notwithstanding, this is the right answer so +1. I would use the word deformable instead of elastic. $\endgroup$ Mar 27, 2017 at 21:23

Imagine an ideal (incompressible and with an ideal flat surface) cylinder (radius R) lying on an infinite (to avoid momentum transfer) ideal flat and incompressible plane to avoid elastic energy transfer. Applying two equally big but opposite forces $\vec F$, perpendicular to the central axis of the cylinder, on opposite sides of the cylinder will produce a torque $\vec{\tau}=2\vec F$x$\vec R$ along the central axis of the cylinder. This causes the cylinder to spin without imparting a linear velocity to it.

Once it spins imagine that the cylinder and the plane it lies on getting rougher and rougher so at some point (in time) the cylinder starts to roll with only static friction. The rough irregular shaped surfaces can stay ideally flat for the cylinder starting to roll. If we zoom in on the point of contact there isn't one point of contact. There may be all sorts of distortions of the plane and the cylinder which can cause it to roll (despite the "flat roughness"). Once it rolls some of those distortions are such that they can be broken by the rolling cylinder (incompressibility and ideal flatness don't imply unbreak ability), which obviously takes energy away from it, so eventually, after first starting to move, the cylinder stops moving.

The cylinder loses also energy by emitting e.m. radiation (very very little, though), and gravitational radiation (very very very little) because it's rotating motion.

  • $\begingroup$ I can't understand what you are saying. The force which causes a torque can also cause translational acceleration. $\endgroup$
    – Yashas
    Mar 28, 2017 at 13:56
  • $\begingroup$ If both the surface and the cylinder ideal flat and you apply the torque on the surface of the cylinder (perpendicular to its length axis), no translational motion will occur. I forgot to write about the force being perpendicular to the length axis of the cylinder. Thanks. $\endgroup$ Mar 28, 2017 at 14:03
  • $\begingroup$ You should mention the source of the torque. If there is a torque, then there is a force. If there is only one force present, then there HAS to be a translation acceleration. $\endgroup$
    – Yashas
    Mar 28, 2017 at 14:04
  • $\begingroup$ I'm getting red in the face! Of course, you're right! I made an edit. $\endgroup$ Mar 28, 2017 at 14:11
  • $\begingroup$ "Applying two opposite torques τ⃗ =F⃗ xR⃗ on two opposite sides of the cylinder with the force perpendicular to the length axis of the cylinder will only make the cylinder spin and impart no linear velocity to the cylinder", what are the two opposite torques? This answer is poorly written. It is hard to understand what you are trying to say. At the outset, this doesn't seem to answer the OP's question. $\endgroup$
    – Yashas
    Mar 28, 2017 at 14:20

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