I have come across a lot of questions that say something like:

A ball rolls down a hill without slipping...

But I have done the maths and found that a ball would only 'not slip' if the friction is precisely:

$$mgI\sin\theta \over mR^2 +I$$ Where $I$ is the moment of inertia, and $\theta$ is the angle made to the horizontal by the plane. A friction of this value would never appear exactly which makes me thing that no-slipping is an idealist case. So here is my question:

Am I right in saying that no-slipping (i.e. when $\dot x =R \dot \theta$) is very uncommon and to have no-slipping on an incline plane you must have friction of exactly the value I have given?

  • $\begingroup$ Friction is weird. You first make an observation (a ball rolls without slipping) and then calculate what the friction force has to be to match your observation. $\endgroup$
    – gigacyan
    Commented Jun 1, 2015 at 8:40

2 Answers 2


I think you have misunderstood how friction works here. The friction you have written down is (typically) the minimum friction needed for the 'non slip' to occur.

Imagine a very faint slope. You will only need a small amount of friction to avoid slipping. As you make the slope steeper more friction is needed for the 'non slip'. Eventually the friction will not be enough and the ball will begin to slip (as well as roll).

So, no, non-slipping is not as idealized as you suggest.

  • $\begingroup$ Correct me if I am wrong but I think we may be taking about a different type of slipping. You (I think) are referring to the case were the ball (or a box say) would slide down the slope where when I say 'no slipping' I mean the case where for every full rotation of the ball its centre of mass has moved a translational distance of exactly $2\pi R$ $\endgroup$ Commented May 29, 2015 at 13:02
  • 1
    $\begingroup$ @Joseph I believe what Mikael is saying is that while you say the friction must be "precisely" a certain value, in actuality it must only be "at least" this value. For example, imagine you had a ramp with a coefficient of friction of exactly 1 (i.e. nothing could ever slip on it at all). What would happen with the ball placed on that slope? Is there any option other than that it will roll without slipping? $\endgroup$
    – Asher
    Commented May 29, 2015 at 13:17

When a tire rolls without slipping it does not leave skid marks.

A ball sitting on a level surface has a force mg downward. It does not move because the surface exerts an upward reaction force just strong enough to hold it still.

Friction is like this. When a ball rolls down a hill, the sum of the gravity and reaction forces is parallel to the surface in a downhill direction. The ball accelerates in that direction. Friction acts on the contact point on the bottom. Friction exerts a backward force just strong enough to keep the contant point still.

  • $\begingroup$ The converse of your first sentence is not true: a tire that moves without leaving skid marks may or may not be slipping. $\endgroup$
    – rob
    Commented May 29, 2015 at 15:42
  • $\begingroup$ @rob - True, but still... For a tire on a car it is largely true. If it slips it typically does leave a skid mark. It is true enough to illustrate the idea. $\endgroup$
    – mmesser314
    Commented May 30, 2015 at 1:26

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