# Torque due to friction on a series of disks

Recently I was looking at two situations involving friction and torque.

The first situation seemed pretty straightforward at first. A disk of mass $$m$$ and radius $$r$$, with a coefficient of static friction $$\mu _s$$ with the ground, is given a force $$F$$ originating at its center. I have drawn a free body diagram below.

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For this situation, the applied force is exactly equal to the frictional force, that is to the coefficient of static friction is high enough, and that the applied force is low enough, such that $$\mu_sN = F$$, we can pretty trivially show that $$\Sigma_x F=0$$ and $$\Sigma_y F=0$$. However , when solving for the net torque, we arrive at $$\Sigma \tau = \mu _srF$$, which means a non-zero net torque is applied to the disk.

This paradoxically, at least to me, means that the ball is spinning in place without actually moving. How could this be?

The second situation is very similar to the first, except that a second disk, with the same mass and radius, has been placed directly next to the first disk. The coefficient of static friction between the two disks is $$\mu_{sb}$$. I have drawn another free body diagram below.

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The situation is very similar to the first, however I hypothesize that there is a torque ($$f$$) due to friction between the first and second disks. I think that the direction of the force points downward, as when the the first disk tries to rotate due to the friction with the table, the second disk resists this change, thus causing a force opposite the motion of the spin.

However, I am at a complete loss as to how to calculate this force new force.

Any help at all would be appreciated, thank you!

• In your first scenario why are you saying the friction force is equal to the applied force? Also, static friction is at most equal to $\mu_sN$, but in general it is less than or equal to this. – BioPhysicist Jan 30 '19 at 4:11
• In your second scenario I don't see why there should be friction between the two balls, since they are not trying to slide vertically past each other. – BioPhysicist Jan 30 '19 at 4:14
• I should have clarified, sorry about that! I was stating that to mean the applied force is not sufficient enough to move the disk to the left, therefore $\mu_s N$ is equal to the applied force. – Inothernews1 Jan 30 '19 at 4:15
• There isn't a way to apply the force and have the ball not move – BioPhysicist Jan 30 '19 at 4:17
• See this related question and my answer here: physics.stackexchange.com/q/452309/179151 – BioPhysicist Jan 30 '19 at 4:21

For this situation, the applied force is exactly equal to the frictional force

Unfortunately this is impossible for the reason you give in your answer. If you were somehow able to apply a force equal to the static friction force$$^*$$, then you have no net force with a net torque about the center of the disk. Therefore you have sliding between the disk and the ground, which means there is no static friction. We have arrived at a contradiction. Therefore in this scenario the static friction force will never be able to be equal to our applied force. The static friction force will actually be less than the applied force (until the applied force becomes too large and slipping occurs).

If you want to apply a force equal to $$\mu_sN$$ then that is fine. But then for the disk, to have rolling without slipping, the friction force will be equal to $$\frac13\mu_sN$$ (Derivation here showing that for an object with moment of inertia $$I=\gamma mr^2$$, in order to have rolling without slipping due to applying a force $$F$$ at the center of the object, the static friction force must be equal to $$\frac{\gamma}{1+\gamma}F$$ ).

It turns out that for the disk as long as $$F<3\mu_sN$$ then you will have rolling without slipping in the direction of the applied force.

$$^*$$ Quick clarification, the static friction force is not always equal to $$\mu_sN$$. In fact, this is such a specific case that really you should write $$f_s<\mu_sN$$. Better yet, you should really never even think about $$\mu_sN$$ when static friction is involved unless the work you are doing requires you to think about when static friction is right up against this boundary. Essentially you should very rarely ever think that $$f_s=\mu_sN$$, which it seems like most of the confusion lies here.

• For all of the rotational motion problems I have dealt with, I guess I've just always assumed rolling without slipping. So this situation depicts one where there isn't rolling without slipping, and thus the link between $a$ and $\alpha$ is broken. – Inothernews1 Jan 30 '19 at 5:17
• And as for the second situation, as there is no normal force between the two, even if the first disk starts spinning in place, am I correct in assuming that there is no frictional force between the two? – Inothernews1 Jan 30 '19 at 5:19
• @Inothernews1 You are still mistaken about the first scenario. In this situation there is no slipping. If you apply a force equal to $\mu_sN$ then the friction force ends up being $\frac13\mu_sN$ and there is still a net force. By assuming static friction is at play you have to have rolling without slipping by definition of static friction – BioPhysicist Jan 30 '19 at 5:22
• In the problem, I really just wanted to express that the applied force was equal to the frictional force. So, me stating that the frictional force involved $\mu_s$ was in error as $F$ exceeded the limit for rolling without slipping? – Inothernews1 Jan 30 '19 at 5:43
• @Inothernews1 No, the error is not because you are exceeding the maximum static friction force. Please look at my very first paragraph. It is impossible for the static friction force to be equal to your applied force. In this scenario the static friction force will always be less than the applied force (until the applied force is too large and slipping does occur) – BioPhysicist Jan 30 '19 at 11:03

Assuming that the coefficient of static friction is high enough such that $$μ_sN=F...$$, we can pretty trivially show that $$Σ_xF=0$$ and $$Σ_yF=0$$.

Actually, I'm not sure you can. $$\mu _sN$$ gives you the maximum possible force of static friction, not necessarily the actual force. To find the actual force you'd need to relate the acceleration of the disk with the angular acceleration. Only a particular frictional force will provide the necessary linear and angular acceleration.

Your second force will be difficult to calculate. You don't know the normal force, and the second disk must either slip against the first or slip against the table.

• This is right. The specific friction force has to have a magnitude of $\frac{\gamma}{1+\gamma}F$ where $\gamma$ is the factor in the moment of inertia of the form $I=\gamma mr^2$ – BioPhysicist Jan 30 '19 at 4:27
• If $F$ is the applied force, you can certainly postulate that it also happens to be $\mu_s N$, the maximum static friction. Perhaps the OP will tell us if that was their intention. – cms Jan 30 '19 at 4:47
• @Inothernews1, imagine a block sitting still on a flat table. What is the force of static friction? What is μsN? Are they equal? – BowlOfRed Jan 30 '19 at 4:51
• Exactly. But μsN does not change. So the two are not equal. – BowlOfRed Jan 30 '19 at 5:02
• @AaronStevens fair enough :) I guess that happened while I was busy mashing all those extra exclamation marks lol – cms Jan 30 '19 at 5:09

Lets write the equations for a free body diagram and see if we can determined the force $$F_r$$

Left disk

$$-m\,a_1+F-F_c-\mu\,N_1=0\tag 1$$ $$-m\,g-F_r-N_1=0\tag 2$$ $$-m\,r^2\,\alpha_1+\mu\,N_1\,r+F_r\,r=0\tag3$$

right disk

$$-m\,a_2+F_c-\mu\,N_2=0\tag 4$$ $$m\,g+F_r-N_2=0\tag 5$$ $$-m\,r^2\,\alpha_2+\mu\,N_2\,r-F_r\,r=0\tag 6$$

Constraint Equation

$$x1-x2=0\quad \Rightarrow\quad$$ $$a_1=a_2\tag 7$$

and

$$F_r-\mu_r\,Fc=0\tag 8$$

Where:

$$F_c\quad N_1\quad N_2$$ are constrain forces

$$\mu$$ and $$\mu_r$$ are the friction coefficients

$$a_1$$ and $$a_2$$ are the translation accelerations

$$\alpha_1$$ and $$\alpha_2$$ are the rotational accelerations

We have 8 equations for the 8 unknowns: $$a_1,\alpha_1,a_2,\alpha_2\,N_1,N_2,F_c,F_r$$

Result

$$\boxed{F_r=\frac{1}{2}\,\frac{\mu_r\,F}{1-\mu_r\,\mu}}$$