# B Friction Between Two Discs

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1. Jul 26, 2017

### person123

Suppose there are two discs held in contact such that they rotate around the same axis of rotation. A torque is applied to one of the discs, and due to friction the other disc accelerates. Using calculus I found the torque applied due to friction for a disc to be $\frac {2μF_N r} {3}$, meaning the acceleration is $\frac {4μF_N } {3mr}$.

What confuses me is that the expression doesn't include the torque applied to the first disc. Intuitively, I would think that the greater the angular acceleration of the first disc is, the greater the acceleration of the second disc would be. Is my intuition wrong, or did I make an error when solving for acceleration? Thanks in advance.

2. Jul 26, 2017

### Dr.D

I presume you are saying that you determined the torque transferred between the two disks due to friction (I have not worked through that part at all; I'll take your word for it at the moment).

If we presume that the two disks are moving independently, then they are not rotating together and there is slip between them; that seems to be consistent with your friction torque expression.

We have to understand that the speed of the driving disk is greater than that of the driven disk; otherwise, the sense of the friction torque is reversed. The acceleration of the driving disk is not a factor, but its speed certainly is, as is the speed of the driven disk. The torque applied externally to the driving disk must keep it moving faster than the driven disk, whether by accelerating it or simply by keeping it is rotational equilibrium.

In your final expression for the (angular) acceleration of the driven disk, you have evidently made some assumption about the mass moment of inertia for this disk. Be aware that not all such situations involve simple, uniform disks, but often disks of variable thickness. For a disk with variable thickness, the simple expression for the MMOI will likely not be valid.

3. Jul 26, 2017

### person123

So the equation is valid as long as there is slip between the two discs (the driving disc is rotating faster than the driven disc)?

Yes, I forgot to mention that I'm assuming the discs are of uniform density and thickness.

4. Jul 27, 2017

### CWatters

Consider a car with enough power to spin it's tyres. The maximum acceleration is limited by the friction between tyre and road not the engine power. You can see this in action at any drag strip or motor race, particularly at the start.

In general static friction is higher than kinetic/dynamic friction which is why traction control is so effective in improving max acceleration.

5. Jul 27, 2017

### person123

It doesn't seem unreasonable I guess; maybe testing it for myself would give it more intuitive sense.

First I stated that $F_f=μPA$ where $P$ is pressure and $A$ is area. For one strip of the disc $F_f=μP2πrdr$ meaning $τ_f=μP2πr^2dr$. That means the torque from friction is $∫μP2πr^2dr=\frac {μP2πr^3}{3}=\frac {2μF_Nr}{3}$.

Last edited: Jul 27, 2017
6. Jul 29, 2017

### Arjan82

Ok, wait... with friction you have two cases: One in which there is no relative movement (no slip) and in that case $F_f=μPA$, with $μ$ the static friction coefficient, will tell you what the maximum force is that you can apply before slip occurs. In that case the acceleration of both disks are equal. And this formula will tell you nothing about how much force is actually applied.

But if there is slip, then $F_f=μPA$, with $μ$ this time the dynamic friction coefficient, will give you the actual force that is impaired onto the second disk. According to this somewhat limited theory this force does not change, in other words the amount of force you can impart on the second disk is capped by this friction coefficient. So you're intuition is, well, wrong in the sense that this way modelling of friction does not allow for a greater force than $μPA$. However, reality might be different...

7. Jul 29, 2017

### Dr.D

This seems rather confused, first and foremost because of all the talk about forces; the concern here is with torques. Secondly, the idea is to calculate the torque due to the force on a differential element of area and then integrate to find the total; none of this seems to fit your discussion.

8. Jul 30, 2017

### Arjan82

A torque is nothing but a force times a distance. Since in this discussion the disks do not change, the distance does not change either. In other words, what applies to a force also applies to the torque. In other words still, my discussion is about the force of the differential element as defined by person123: $Ff=μP2πrdr$. The integration of this force to a torque does not change.

9. Jul 30, 2017

### Dr.D

My understanding is that we are talking about two flat disks, face to face. In that case, the "distance" (radius) does vary from zero to the maximum value at the outer edge. To say that the "distance" does not change simply fails to reflect the facts of the situation.

To say, "what applies to a force also applies to the torque" is simply to confuse the two. Most well informed folks, when speaking about a torque will refer to it as a torque (or a moment), not as a force.

The statement, "Ff=μP2πrdrFf=μP2πrdr" is nonsense. It looks like perhaps a double copy, but even then it is incorrect. It equates a finite form on the left (Ff) to a differential form on the right (μP2πrdr). That will get you in trouble more times than not.

It is not reasonable to expect the reader to read your mind. We are each expected to mean what we say, not what we expected someone else to imagine.

10. Jul 30, 2017

### Arjan82

Hi Dr.D

You are right, I am a bit imprecise in my words. I hoped it would be clear from the context what I exactly meant. But let me do a last attempt in getting some of the confusion out of the way.

$F_f=μP2πrdr$ is indeed nonsense, I copied it without really reading it. It should be $dF_f=μP2πrdr$, sorry for the missing $d$. So the frictional force equals the frictional coefficient $μ$, times pressure $P$ times area $2πrdr$ applied to an infinitesimal small annulus of the disk. To get the torque you'll have to multiply this annulus with the arm, $r$, and integrate it from the center to the maximum radius. The total integral looks like this $T_f = ∫μP2πr^2dr$ and the solution is this $∫μP2πr^2dr={μP2πr^3}/3={2μF_Nr}/{3}$ where $F_N = PA = Pπr^2$. All of this was however already pointed out in the post of person123.

Where I said 'the distance does not change' I meant that the integral doesn't change since $F_f$ is not a function of $r$ and can be taken out of the integral.

So, only in this case, what is true for the force is also true for the torque, in that it is capped to a maximum... So there is a maximum force that you can transfer between two objects if you apply this model for friction, and there is also a maximum torque that you can transfer by applying this model for friction.

I'm sorry that I'm not meeting your standards of being well informed. However I'm just trying my best to answer questions in the hopes that it is useful for anyone. If it's not, I'm happy to elaborate.

You are right! I'm just trying! No harm was intended!

11. Aug 4, 2017

### person123

Sorry for late response—I was on vacation.

True. By writing that it was for a strip of the disc, I meant it was the derivative, but my notation was sloppy.

Last edited: Aug 4, 2017