When a rigid ball rolls on a plane with no other forces, it is known that no friction applies and the ball keeps rolling. I was asked if this $F_{\text{friction}}=0\,$ can be derived from the usual law of static friction:

Static friction applies in the opposite direction of the tangential part of the external force, with equal magnitude.

Of course, this must be applied to a small particle at the contact point of the ball and plane. The external force would be the force from the other particles in the rigid body, to maintain its shape. So, how could this be derived? I want a logically complete explanation.


Static friction opposes impending relative motion between two surfaces (if relative motion actually occurs then we have kinetic friction). The external force being referred to is a force that has a component parallel to the ground such that it tries to slide the surfaces past each other.

The force acting on your point in contact with the ground in order to keep the object rigid acts upwards, i.e. with no horizontal component$^*$. Therefore this force does not "activate" any static friction.

If a ball is rolling on an (ideal) flat surface, and there is no external force acting on the ball itself in the horizontal direction, then there is no static friction force. Hence no torques or horizontal forces at all, and the ball will roll along at a constant speed.

$^*$ Let's show why this is the case. The easiest thing to do is just to consider moving along with the ball at its constant speed. Then you will just observe a particle undergoing uniform circular motion, and hence the acceleration must be directed upwards.

However, let's say you are still not convinced. Maybe you are resistant to changing your inertial frame of reference, and you want to see the argument done as you watch the ball roll by. The path a point on the outside of the circle traces out is called a cycloid. The position vector as a function of time then for a point on the circle is given by $$\mathbf r=R(\omega t-\sin\omega t)\hat x+R(1-\cos\omega t)\hat y$$ where $R$ is the radius of the circle, $\omega$ is the angular frequency of the rotation, and we are looking at a point of the circle that starts touching the ground at the origin at $t=0$

Through simple derivatives, we can show that the velocity $\mathbf v$ and the accleration $\mathbf a$ are $$\mathbf v=R(\omega-\omega\cos\omega t)\hat x+R(\omega\sin\omega t)\hat y$$ $$\mathbf a=R(\omega^2\sin\omega t)\hat x+R(\omega^2\cos\omega t)\hat y$$

The particle hits the ground every time $R(1-\cos\omega t)=0$ i.e., $t=2\pi n/\omega$ for any non-negative integer $n$. Plugging this time into our acceleration vector, we get $$\mathbf a\left(t=\frac{2\pi n}{\omega}\right)=0\hat x+R\omega^2\hat y$$

Hence the acceleration only points upwards when the point is in contact with the ground.

  • $\begingroup$ Why is the force acting on the contact point upwards?? I was aware of the parallel part, I forgot to put it in my definition. $\endgroup$ – Quasi07 Jun 27 at 13:57
  • $\begingroup$ It is due to the centripetal acceleration as the particle moves in a circle $\endgroup$ – Aditya Garg Jun 27 at 13:58
  • $\begingroup$ @Quasi07 Aditya beat me to it. But that is correct, $\endgroup$ – Aaron Stevens Jun 27 at 13:59
  • $\begingroup$ Yes I had this thought, but I don't think it is logically correct. The thing I'm trying to show is that there is no static friction, hence no angular acceleration. If there was angular acceleration, the acceleration wouldn't be pointing up. Is there something I'm missing? $\endgroup$ – Quasi07 Jun 27 at 14:02
  • $\begingroup$ @Quasi07 My answer explains why there is no static friction. There is no force attempting to slide the surfaces past each other. It is equivalent to asking "Why does a book sitting in an upward accelerating elevator not have static friction on it? The book has the external force of the elevator acting upwards on it." $\endgroup$ – Aaron Stevens Jun 27 at 14:14

There are two different points here:

  1. The friction at the base of a rolling ball is static if we are in the case of a pure rolling. In fact the base point is istantaneously stationary, the image should be eloquent on this.

enter image description here 2. Static friction makes no work, because there is no displacement, so making no work it can't alter the speed of a system.

Sorry for the oversized image!


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