Kinetic Friction & Rotation: Questions & Answers

[Grunfeld]

I have a question regarding rotation and kinetic friction that I can't find anywhere else on the web.

Say you have a ball of radius r and mass m rotating on a flat surface without any forces acting on it but the force of friction. How does the deceleration of the ball relate to the frictional coefficient of the surface?

Is there a simple equation that describes the force of kinetic friction acting on the ball?

 

[Grunfeld]

Nobody have any ideas? I just need to know what knowledge is required to solve this problem.

 

[Hootenanny]

Is there a simple equation that describes the force of kinetic friction acting on the ball?

Yes,

$$F=\mu R$$

 

[Grunfeld]

What is R? And how to calculate the deceleration of the ball?

 

[Hootenanny]

R is the normal reaction force. One can calculate the deceleration of the ball by applying Newton's Second law (and it's rotational analogy).

 

[Grunfeld]

So what is the difference between the ball sliding and the ball rotating? How is the deceleration diffenent?

 

[Hootenanny]

So what is the difference between the ball sliding and the ball rotating? How is the deceleration diffenent?

If the motion is rotational with no sliding, then there will be no acceleration, assuming that both the ball and surface a rigid. If the ball is sliding there will be both linear and angular acceleration unit the no-slip condition is met (i.e. the velocity of the point of the ball in contact with the surface is zero relative to the surface), at which point it will cease sliding and the ball will roll with a constant angular and linear velocity indefinitely.

 

[Grunfeld]

say the frictional coefficient, the radius, the mass of the ball, and an initial force was given. How can the distance that the ball travel be calculated?

 

[Hootenanny]

say the frictional coefficient, the radius, the mass of the ball, and an initial force was given. How can the distance that the ball travel be calculated?

Quite easily using Newton's second law (and it's rotational analogue).

 

[Grunfeld]

I don't know what the rotational analogue is. What is it?

 

[Hootenanny]

For a rigid body,

$$\sum_i\underline{OP}_i\times\underline{F}_i = \frac{d}{dt}\underline{L}_0$$

In words: The sum of the external torques about the point O is equal to the time derivative of the angular momentum about that same point.

 

[Grunfeld]

I can find the angular and linear deceleration of the ball, but what do they represent? Which one do I use to find the time it takes for the ball to stop?

Also, the force of friction for both the angular and linear deceleration are the same?

 

[Hootenanny]

I can find the angular and linear deceleration of the ball, but what do they represent? Which one do I use to find the time it takes for the ball to stop?

The angular acceleration represents the change in angular velocity and the linear acceleration represents the change in linear velocity, as one would expect. However, at some point the ball will stop accelerating and have a constant angular and linear velocity. If the ball and surface are rigid, then it will never stop. See my post #7 above.

Also, the force of friction for both the angular and linear deceleration are the same?

Yes.

If your solving a specific question, why don't you post it so we can help you out?

 

[Grunfeld]

However, at some point the ball will stop accelerating and have a constant angular and linear velocity. If the ball and surface are rigid, then it will never stop. See my post #7 above.

I don't know what you mean by this.

I am not trying to solve a specific problem, I am just trying to figure out the relation ship between the friction coefficient and the time it takes for the ball to stop rolling.

 

[Hootenanny]

I am not trying to solve a specific problem, I am just trying to figure out the relation ship between the friction coefficient and the time it takes for the ball to stop rolling.

Simply put, if the surface is horizontal and both the ball and surface do not deform, then the ball will never stop.

 

[Grunfeld]

You mean the ball will roll on forever even though a friction force is acted on it?

 

[Hootenanny]

You mean the ball will roll on forever even though a friction force is acted on it?

Yes, once the ball has finished sliding the frictional force is a static frictional force and does no work on the ball since the point of contact on the ball doesn't move relative to the surface.

This is only the case for a rigid ball and surface.

 

[Grunfeld]

What happens in real life? Eg. When a basketball or pool ball is rolled on a smooth surface.
It obviously stops rolling at some point. How is that time calculated?

 

[Shooting Star]

Say you have a ball of radius r and mass m rotating on a flat surface without any forces acting on it but the force of friction.

Sorry to intrude, Hoot, but right in the first post the OP says there's no force acting on the ball. I presume he meant an earthly scenario where the weight of the ball was acting as the reaction R. If there is actually no force pressing the ball to the surface, there will be no friction.

You mean the ball will roll on forever even though a friction force is acted on it?

In reality, rolling friction will slow the ball down, as Hootenanny will explain.

 

[Hootenanny]

What happens in real life? Eg. When a basketball or pool ball is rolled on a smooth surface.
It obviously stops rolling at some point. How is that time calculated?

Again using Newton's second law (in rotational form), however to take into account the deformation of the ball and surface one must include the action of an additional force, the so-called rolling resistance, which is usually of the form,

$$F = K_rR$$

and acts in the same direction as the frictional force. More information can be found http://webphysics.davidson.edu/faculty/dmb/PY430/Friction/rolling.html" .

 

[Hootenanny]

Sorry to intrude, Hoot, but right in the first post the OP says there's no force acting on the ball. I presume he meant an earthly scenario where the weight of the ball was acting as the reaction R. If there is actually no force pressing the ball to the surface, there will be no friction.

Excellent point Shooting Star

 

[Grunfeld]

What i meant was no other horizontal force.

So the time it takes for the ball to stop is just the initial rotational velocity divided by the rotational deceleration (which is obtained by rotational friction / moment of inertia)?

 

[Hootenanny]

So the time it takes for the ball to stop is just the initial rotational velocity divided by the rotational deceleration (which is obtained by rotational friction / moment of inertia)?

Provided that the ball is rolling and there is no slipping, yes.

 

[Grunfeld]

If the ball does not slip, why does it move forwards? Also, why does the linear motion stop when the rotational motion stops?

 

[Shooting Star]

Yes, once the ball has finished sliding the frictional force is a static frictional force and does no work on the ball since the point of contact on the ball doesn't move relative to the surface.

This is only the case for a rigid ball and surface.

Once the ball has finished sliding and the no-slip condition has been reached, that is v=rw, there is no frictional force acting anywhere. You might as well change the surface to a frictionless one, like making it ice from one point onward in its path, and still the ball will roll on and v will be equal to rw.

The static frictional force will act if the ball is accelerating and rolling without slipping, like rolling down a plane before it starts to slide. The sliding starts when the maximum static frictional force cannot make the ball stop sliding.

This is valid for rigid ball and surface, as Hootenanny has said.

 

[Grunfeld]

Ok, i get the general idea of it, but I still am unsure of the specifics.

Say on a flat surface, a ball with mass 1 kg and radius 5 cm initially starts rolling with a velocity of 1 m/s. The ball is not slipping. The the coefficient of rolling resistance is 0.01. How far will the ball travel?

Also, what's w in the equation v = rw

 

[Hootenanny]

Once the ball has finished sliding and the no-slip condition has been reached, that is v=rw, there is no frictional force acting anywhere. You might as well change the surface to a frictionless one, like making it ice from one point onward in its path, and still the ball will roll on and v will be equal to rw.

Not quite, because if the surface was frictionless, the ball wouldn't be rolling at all. Static friction is required for rolling whether the ball is accelerating or traveling at a constant angular velocity.

 

[Shooting Star]

Not quite, because if the surface was frictionless it wouldn't be rolling. Static friction is required for rolling whether the ball is accelerating or traveling at a constant angular velocity.

Friction is necessary to achieve the state of rolling without slipping, but once that state has been achieved, the rigid ball doesn't give a hoot (NPI ) whether the surface underneath is rigid and with static friction, or rigid without any friction. The speed v equals [itex]r\omega[/tex] in both cases.<br /> <br /> In fact, if there is static friction acting at <b>any</b> point on the body, even if only at the point of contact, and if its direction is forward to oppose the point of contact of the ball from moving backward, it would accelerate the body linearly, but the torque would tend to roll it backward. And vice versa -- you get the drift. These are avoided when there is an external force acting on the body, where the static friction plays a crucial role.<br /> <br /> Please think about it for a while.[/itex]

 

[Grunfeld]

I am more interested to the method to solve the problem I proposed, how to do so?

 

[Hootenanny]

Friction is necessary to achieve the state of rolling without slipping, but once that state has been achieved, the rigid ball doesn't give a hoot (NPI ) whether the surface underneath is rigid and with static friction, or rigid without any friction.

This will teach me to read posts more carefully before replying. I failed to notice that you said,

Once the ball has finished sliding and the no-slip condition has been reached...

and I thought you were referring to the whole scenario. Oops