# How to calculate the frictional force required to “stick” two rotating objects?

Say you had two fixed, non-identical objects rotating around one axis at different speeds relative to one another, and they can be coupled together by friction (like how a clutch in a car works). How would I calculate the frictional force required for them to stick together instantaneously?

For example:

Object A and B are two solid cylinders rotating about a single axis (imagine how they would rotate if they were rolling) Object A is rotating at 10000 RPM, and object B is rotating at 1000 RPM; Object B has a larger radius, is made of a denser material, and has more mass and a larger moment of inertia. The two object come into contact with each other (picture the clutch of a car).

If the frictional force is sufficient, they should "stick" and the angular momentum should be conserved (I1ω1+I2ω2=Inetω); but if it isn't, they should slip.

What would be the frictional force required to "stick" the two objects together instantaneously?

• Related post by OP – Aaron Stevens Oct 4 at 18:42
• In a clutch, the inner object's centrifugal force produces the normal force. What is the source of the normal force here? – Acccumulation Oct 4 at 18:47
• Why are you asking the same question you asked 16 hours ago with an accepted answer? – Bob D Oct 4 at 19:05
• Darn I guess I am missing something then, I thought that the normal force is just the "clamping" force applied by the spring or whatever. Also, this question is sort of a continuation of my other question, they are related but I am asking how to calculate two different things. I thought it would make more sense for it to have it's own post rather than ask in the comments. – Bureto Oct 4 at 20:47

Forces cause accelerations. The rotational analogue of this is that torques cause rotational accelerations. "Instantaneous" doesn't exist. You need to figure out the maximum amount of time you want it to take.

Say you determine that before and after coupling, one object will have some change in angular velocity, $$\Delta \omega$$. You will have already calculated or measured the moment of inertia, $$I$$. Further, you decide that for the purposes of your device, "instantly" means in less than 0.1 seconds.

$$\Delta \omega = \alpha t$$ $$\Delta \omega = \frac {\tau}{I} t$$ $$\frac{\Delta \omega I}{t} = \tau$$

So you'll plug in for the change in rotational velocity, moment of inertia, and time required and you'll get the necessary torque. The faster you need it to couple, the higher torques/forces you'll have to generate.

Given the geometry of how you're applying it, you can turn the torque into a force. If it's driven by friction, you can use that and the coefficient of friction to determine the required normal force to produce that much friction.

Say the problem is that of two disks that are pressed together with some axial force $$F$$ and their surfaces have a coefficient of friction $$\mu$$.

How much torque can they transfer from one to the other? I think this is what you are asking.

The answer depends on the pressure distribution between the disks. Once the radial shape of the pressure distribution is decided, then the following force equation is used to find the pressure $$F = \int \limits_{r_1}^{r_2} P(r) \,2 \pi r\, {\rm d} r \tag{1}$$ as well as the torque capacity $$T = \int \limits_{r_1}^{r_2} r \mu P(r) \,2 \pi r\, {\rm d} r \tag{2}$$

1. With constant distribution $$P = P_0$$ across the radius $$r = r_1 \ldots r_2$$, and applied load $$F$$, the pressure is $$P_0 = \frac{F}{\pi ( r_2^2 -r_1^2 ) } \tag{3}$$ which makes the torque $$T = \frac{2 \mu F (r_1^2 + r_1 r_2 + r_2^2)}{3 ( r_1+r_2) } \tag{4}$$

2. Another popular pressure distribution is $$P = P_1 /r$$ which corresponds to equal wear along the contact surfaces. The pressure coefficient is found from the total force $$P_1 = \frac{F}{2 \pi (r_2-r_1)} \tag{5}$$ making the torque capacity $$T = \frac{\mu F (r_1+r_2)}{2} \tag{6}$$

The first one is higher than the second one, except for the limiting case where $$r_1 \rightarrow r_2$$ the disk is very thin, like a hoop.