# Why doesn't static friction decelerate a rolling body?

I know that static friction isn't the cause of deceleration of a rolling body. But if static friction is the only force in the horizontal direction, then shouldn't there be some acceleration produced in this direction? The body should decelerate or accelerate. So how come in this case, a force does not produce acceleration?

• A force can only perform work when there is also displacement in the same direction as the force. This is not the case when no slip occurs. – fibonatic Dec 1 '13 at 14:57
• Yes it does, and there is an effective coefficient of rolling friction, usually in the $\mu=0.005$ range. – ja72 Dec 2 '13 at 0:25
• @user34304 what is the difference between acceleration and decceleration? It is just a matter of sign. Are you asking why friction always opposes motion then? If so, then change the title of your question. – ja72 Dec 2 '13 at 1:31
• Ideally, static friction is zero for a rolling body. It only exists for an accelerating body. – ja72 Jun 26 '14 at 15:43

When a disk or other object is rotating on a horizontal surface with constant velocity, there is no static frictional force. Your logic is correct: if there were a horizontal force, the center of mass would be accelerating.

If the rolling object suddenly encounters a frictionless surface, it would continue to satisfy the rotating without slipping condition. Friction is not required to keep the object rotating. Once it gets rotating (which might indeed require friction), it'll continue to rotate unless a net external torque acts on the object.

• A rolling contact looses energy from the elastic deformation of the parts, and some other effects (rolling waves) which dissipate as heat. The effect is a small deceleration which can be described as "friction" as it depends on the contact force. To pronounce the effect, try to roll a bowling ball on sand, or on a trampoline and you will see what I am talking about. – ja72 Dec 2 '13 at 0:28
• I know about rolling friction. So If A rolling body we're moving at a constant velocity, static friction wouldn't alter its velocity, but it should right, because it is a force? – user34304 Dec 2 '13 at 1:34

Here's my explanation (hopefully it's right but I'm no expert):

Consider the simpler example of pushing a really heavy box; if we push on it there's a static friction force, but if we don't push on it there's no force. Similarly for a cylinder on a horizontal plane, if we push on it the static friction force causes it to rotate, but when we stop pushing there's no static friction force but it keeps moving because no other force is there to slow it down. The static friction only arises in response to our push.

Let's say you roll a ball (of mass $m$) down an inclined plane of angle of inclination $\theta$ and coefficient of static friction $\mu_{static}$. Then you know a force parallel to the inclined plane acts on the ball through its center of mass. Another force parallel to the surface acts in the opposite direction of motion as follows, The force $\vec F = mg\sin\theta$ , $\vec N$ is the nomal reaction and $\vec f$ is the frictional force.

$\vec F$ and $\vec f$ act in opposite directions as shown in the figure. These two forces produce a couple (or torque) which makes the ball rotate. However, irrespective of their point of application, they act in opposite directions on the center of mass and so may or may not produce an acceleration.

If $\vec f \ge\vec F$, then there will be no acceleration, otherwise there would be.

However as you are asking of the case when the only horizontal force is the frictional force, you should know that there won't be any friction at all if you are talking about pure rolling. This is because, the velocity of the lowermost point (or the point of contact) in case of pure rolling is $0$ and so there is no impending motion of the lowermost point. The velocity $\vec v$ and $\vec\omega R$ are equal and acting in opposite directions on the lowermost point. As a result, its velocity is zero resulting in zero friction. So in an ideal situation such as this, there will be no static friction, so clearly there will be no deceleration or acceleration.

You might ask what do I mean by an ideal situation. Well, in this case, we are considering pure rolling which happens only if there are no momentary deformations of the surface of contact. Now, this does happen in real life, so, there is a friction acting in real life and that is why the ball slows down.

• In the inclined plane case, you still have v = omega.r; this doesn't explain why there's no static friction force at the point of contact in the horizontal-plane case. In the inclined-plane case, I think the static friction is caused by the fact that the velocity of the CoM is not constant. – thedoctar Jun 26 '14 at 15:16
• @Ris97 I'm sorry but these two figures are poorly drawn (the normal force doesn't look at all normal, it's missing the right angle indication, the text is barely legible, etc.), and they fail to properly identify where the angle $\theta$ actually is, assuming the reader already knows. This may be more confusing than helpful to those who are not intimately familiar with inclined plane problems, please consider redrawing them. – Bryson S. Jun 26 '14 at 16:26