Friction force, what causes it? Unslipping coin on spinning turntable w/ constant vel

[InvisibleMan1]

Homework Statement

A coin is off-center on a turntable which is spinning at a constant velocity. The coin is not slipping. From what I have gathered, the radial acceleration of the coin is caused by the force of static friction on the coin. There is no tangential acceleration.

Homework Equations

The Attempt at a Solution

Exactly why is there a static friction force there? I've been trying to figure this out, but nothing makes sense. Needing to take in the rotation is goofing me up since I don't have much groundwork in rotating things (enough to come up with this problem, understand the problem, and word it, but not enough to solve it).

The coin has a tangent velocity. Shouldn't the friction force be tangent as well, but pointing in the other direction?

 

[Doc Al]

The coin has a tangent velocity. Shouldn't the friction force be tangent as well, but pointing in the other direction?

The coin has a tangential velocity, but not a tangential acceleration. It's acceleration that requires a force, not velocity.

 

[zhermes]

Exactly why is there a static friction force there?
Static friction is generally due to 'sticking' between the materials, and imperfections on each's surface colliding when a force attempts to move it.

Needing to take in the rotation is goofing me up

The rotation is simply providing a force for the coin to resist via friction. It's the same as if someone was pushing on it.

The coin has a tangent velocity. Shouldn't the friction force be tangent as well, but pointing in the other direction?

It doesn't have a tangential velocity relative to the surface however.

 

[InvisibleMan1]

Since the radial acceleration is from the friction force, doesn't that mean that the force which it is reacting to is pointing out? I don't know of a force which is doing that.

 

[zhermes]

The coin feels a centrifugal force directed radially outward.

 

[InvisibleMan1]

But isn't the centrifugal force fake? How can something which is fake create friction?

 

[Doc Al]

But isn't the centrifugal force fake? How can something which is fake create friction?

There's no need to invoke centrifugal force if you stick to an inertial frame. For the coin to move in a circle it must be centripetally accelerated, which requires a centripetal force. Static friction provides that force.

 

[Doc Al]

Since the radial acceleration is from the friction force, doesn't that mean that the force which it is reacting to is pointing out? I don't know of a force which is doing that.

The friction force on the coin acts radially inward. Of course, the reaction force of the coin on the turntable is outward, per Newton's 3rd law. (But who cares?)

 

[InvisibleMan1]

I care since it might explain where the friction force is coming from. I still don't understand how it is being generated.

 

[zhermes]

I care since it might explain where the friction force is coming from. I still don't understand how it is being generated.

Good answer. The friction force emerges at the location of the coin, so its appropriate to think in the rotating reference frame. The centrifugal force is not a 'fake' force at all, it's just a force that only exists in rotating reference frames. If you were the coin, you would feel a force directed radially outward (the centrifugal force); therefore, to keep rotating (i.e. to stay at rest in your reference frame), you need a responding force to appose the centrifugal force and keep you in place---this is provided by friction.

From your reference frame (you are the coin, facing the center of the wheel), it's exactly as if a force of constant magnitude was pulling you outward, and friction prevented you from moving.

In an inertial (outside) reference frame, when the disc rotates, the coin's inertia tends to keep going in a straight line---therefore there is a tendency towards a difference in velocities between the coin and the disc. This is apposed by the force of friction---which keeps the coin moving in a circle, and is therefore a centripetal force.

 

[InvisibleMan1]

The reference frame is only rotating because of the static friction though. To say that the static friction is needed because it is a rotating reference frame doesn't seem like a very solid proof, and actually just seems to go in circles (not intended as a pun).

I tried attacking the problem from the view of the turntable, but without much success. Here was my train of thought: The point under the coin has a force applied to it which causes the point to have a radial acceleration. Is this the force which causes the static friction on the coin? It doesn't make sense that it is, since both forces would be pointing in the same direction. The static friction force should be opposed to the force which it is preventing.

 

[Doc Al]

I care since it might explain where the friction force is coming from. I still don't understand how it is being generated.

Viewing things from a rotating reference frame, and thus invoking centrifugal forces, doesn't really gain you anything as far as understanding how friction is generated. (At least in my opinion.) To say that friction exists to oppose centrifugal force is no better than saying friction exists to provide the centripetal force. In fact it's a bit of a dodge, since you first have to explain where centrifugal force comes in. I recommend sticking to the usual inertial frame until you are ready to seriously consider rotating reference frames.

More to the point is to understand how friction is generated in general. Friction acts to oppose slipping between surfaces. Imagine that coin spinning with the turntable. If there were no friction, where would it tend to go? Per Newton's First law, the coin would take off in a tangential path. Since the turntable is rotating, relative to the turntable the initial tendency is for the coin to move outward. But friction acts to prevent such a motion, by exerting an inward force on the coin.

 

[Doc Al]

The reference frame is only rotating because of the static friction though. To say that the static friction is needed because it is a rotating reference frame doesn't seem like a very solid proof, and actually just seems to go in circles (not intended as a pun).

The turntable is rotating so it's perfectly OK to use a non-inertial frame of reference. But I wouldn't.

I tried attacking the problem from the view of the turntable, but without much success. Here was my train of thought: The point under the coin has a force applied to it which causes the point to have a radial acceleration. Is this the force which causes the static friction on the coin?

The static friction is the force applied to the coin.

It doesn't make sense that it is, since both forces would be pointing in the same direction. The static friction force should be opposed to the force which it is preventing.

Think of static friction as opposing the slipping of the coin, not as opposing some other force. The only force acting on the coin (at least horizontally) is the friction.

 

[InvisibleMan1]

Is there a force which causes the static friction? From your post, it sounds like only velocities are involved in the calculation.

Edit: This is in response to your earlier post, not your most recent one.

 

[InvisibleMan1]

The static friction is the force applied to the coin.

That's not the force I was referring to. Put in a possibly clearer way: "The point under the coin has a force applied to it which causes the point to have a radial acceleration. Static friction requires a force to counter in order to exist, so is this radial acceleration on the point the force which the static force on the coin is countering?"

Think of static friction as opposing the slipping of the coin, not as opposing some other force. The only force acting on the coin (at least horizontally) is the friction.

So a static friction force can exist without having a force to "counter"/"oppose"?

 

[Doc Al]

That's not the force I was referring to. Put in a possibly clearer way: "The point under the coin has a force applied to it which causes the point to have a radial acceleration.

And that force is the static friction.

Static friction requires a force to counter in order to exist

No, not always.

So a static friction force can exist without having a force to "counter"/"oppose"?

Sure, when acceleration is involved.

Here's another example (another typical textbook setup): Imagine a flatbed truck with a box resting on the floor in back. As the truck accelerates, static friction prevents the box from slipping along the bed of the truck. There's no other force acting horizontally on the box.

 

[InvisibleMan1]

I keep running into a problem with that truck example. It's going to take a lot more force to budge the truck than to budge the box. From what I have read, you would compare the force required to budge the truck with the force of static friction between the truck and the box. The force of static friction is directly proportional to the normal force on the box. That normal force could be large, but I don't see it being anywhere near enough to counter-act the large force being applied to the truck. Even if the truck isn't accelerating very fast, it's applied force will still be large. This doesn't seem to add up... What should be compared to the static friction is the acceleration of the truck, not the force causing the acceleration (the force is directly proportional to the mass of the truck, and is thus pretty big). I'm sure I missed something here... How do you correctly handle this?

 

[Doc Al]

I keep running into a problem with that truck example. It's going to take a lot more force to budge the truck than to budge the box.

Well sure, the truck has more mass.

From what I have read, you would compare the force required to budge the truck with the force of static friction between the truck and the box.

Why do you think this?

The force of static friction is directly proportional to the normal force on the box.

The maximum value of static friction is proportional to the normal force.

That normal force could be large, but I don't see it being anywhere near enough to counter-act the large force being applied to the truck.

The normal force is just the weight of the box.

Even if the truck isn't accelerating very fast, it's applied force will still be large.

What do you mean by "it's applied force"? The force needed to accelerate the truck may be large. Is that what you mean?

This doesn't seem to add up... What should be compared to the static friction is the acceleration of the truck, not the force causing the acceleration (the force is directly proportional to the mass of the truck, and is thus pretty big).

The box is small compared to the truck, therefore the force needed to accelerate the box is much less than that needed to accelerate the truck. Static friction is enough (as long as the acceleration isn't too great).

 

[InvisibleMan1]

Why do you think this?

That's what it seems to say here https://www.physicsforums.com/library.php?do=view_item&itemid=39 "it will always be equal and opposite to the total of the other forces on that body"

The normal force is just the weight of the box.

Yeah I know, that's why I said "could be large". It depends on what is in the box (and what the box is made of).

What do you mean by "it's applied force"? The force needed to accelerate the truck may be large. Is that what you mean?

Yes.

The box is small compared to the truck, therefore the force needed to accelerate the box is much less than that needed to accelerate the truck. Static friction is enough (as long as the acceleration isn't too great).

Exactly, but everything I have read about static friction talks about comparing it with a force, not an acceleration.

 

[Doc Al]

That's what it seems to say here https://www.physicsforums.com/library.php?do=view_item&itemid=39 "it will always be equal and opposite to the total of the other forces on that body"

That's what it says all right. It's wrong.

When the object is in equilibrium--no acceleration--the net force is zero. But not when it accelerates.

Exactly, but everything I have read about static friction talks about comparing it with a force, not an acceleration.

Well, keep reading!

 

[InvisibleMan1]

So how does the friction force work in the truck example?

 

[Doc Al]

So how does the friction force work in the truck example?

Static friction will act to prevent slipping--up to its maximum value, of course.

Let the acceleration of the truck be 'a'. In order for the box to have the same acceleration--and thus move with the truck without slipping--a force equal to 'ma' must be exerted on it. Friction provides that force. The maximum value of static friction is μN = μmg. So as long as that maximum force is enough (> ma), the box won't slip. Thus the maximum acceleration the truck can have without the box slipping would be given by: μmg = ma, thus a = μg.

Make sense?

 

[InvisibleMan1]

Alright, so in the truck example:
A force F1 is applied to the truck to give it a forward acceleration "a". Relative to the truck, the box wants to move backwards. To prevent this, the truckbed applies a force of static friction to the box, Fb. Fb is opposed to the attempted motion of the box, so it is directed forwards. Now that I think about it, the motion of the box relative to the truckbed is actually giving an acceleration of "a" to the box. This acceleration is directed backwards. This acceleration doesn't actually exist, but you can calculate an imaginary force which causes the imaginary acceleration. If the box has mass "m", then it has an imaginary force on it with magnitude ma and directed backwards. I'm not comfortable with claiming that an imaginary force can cause friction, so let's step back a bit. We looked at the box's motion relative to the truckbed in order to get the direction of the friction force. However, we still don't know what the magnitude of that force is. Because of this, we don't even know if we are supposed to apply static or kinetic friction. So let's try and find the magnitude of the force. We can attack this by calculating how much static friction would be required to keep the box at rest, and then compare it with the maximum static friction to find out if there is enough static friction to keep it at rest. If there isn't enough static friction, we will know that kinetic friction needs to be applied (which is easily calculated). So for the box to be at rest, it must have the same acceleration as the truckbed. Let's say that Ta is the acceleration of the truckbed and Ba is the acceleration of the box.
Ta = Ba
So what are the forces involved here? Using Newton's second law, we can easily calculate the net force on the box. Bf is the net force on the box and Bm is the mass of the box.
Bf = Bm*Ba
Bf = Bm*Ta

So the force required to keep the box at rest on the truckbed is the mass of the box times the acceleration of the truckbed. If Bf is lower than or equal to the maximum value of static friction between the box and the truckbed, Bf is the force which the static friction will oppose and be equal in magnitude to. If Bf is greater than the maximum static friction, the force of friction will be the kinetic friction between the truckbed and the box.

So to summarize: The force of friction on the box is in the same direction as the acceleration of the truckbed (this is surprising to me, but if you step through it you find that it is true). Its magnitude depends on whether the friction is kinetic or static. To check which it is, calculate Bf = Bm*Ta where Bf is the force required to keep the box at rest, Bm is the mass of the box, and Ta is the acceleration of the truckbed. If Bf <= Fs (where Fs is the maximum force of static friction between the truckbed and the box) then the force of friction is static and thus equal in magnitude to Bf. Otherwise, the force of friction is kinetic and equal in magnitude to Uk*N where Uk is the coefficient of kinetic friction and N is the force pushing the truckbed and the box together.

If I did this correctly, it appears as though it may be valid to use the truckbed as the frame of reference, and apply the imaginary force Bf to the box and generate the friction that way. I do not have a way to prove this for certain though.

Now, this becomes relevant to the original question when you consider this: What happens when the truckbed has an acceleration directed perpendicular to its velocity? You have a radial acceleration, and the truckbed starts rotating along a curve with radius r=(v^2)/a where v is the velocity of the truckbed and a is the radial acceleration. Is the force of friction still calculated the same way? I don't see why it wouldn't be. The force of friction is in the same direction as the acceleration, which means it is radial. Its magnitude is calculated in the same manner stated previously.

If you take a point under the coin in the first example in this thread, this is like your "truckbed". The coin is then like your box. The area is obviously different, but that shouldn't affect this conclusion much. The point under the coin is rotating along with the rest of the turntable at a constant velocity. This means that it has a radial acceleration. So the force of friction on the coin will be radial as well. Since the coin isn't slipping, we know that the friction is static, so its magnitude is m*a = m*((v^2)/r) where m is the mass of the coin, a is the radial acceleration of the point, and thus v is the velocity of the point and r is the distance between the point and the center of the turntable. The maximum force of static friction between the coin and the turntable must be greater than or equal to the force of friction on the coin.

This all seems to apply if you have a tangential acceleration in addition to a radial acceleration. So it all seems to check out.

Is this all correct? I will most likely use this thread as a reference if I later forget how this works, so I would like this to be as accurate as possible.

 

[Doc Al]

Good job!

 

[InvisibleMan1]

Thanks for the help :)