# Friction between two rotating cylinders

1. Apr 11, 2007

### Scooter057

I have two hollow cylinders of different sizes. The smaller cylinder is inside the larger cylinder but NOT concentrically. The OD of the smaller cylinder contacts the ID of the larger cylinder at the 6 o'clock position. Both cylinders rotate but in opposite directions. An actuator pushes the larger cylinder upwards applying a normal force to the smaller cylinder. I know the torque being applied to each cylinder, the constant rotational speeds of each cylinder and the force being applied to the actuator. I also know the weights of everything. How would i find the coefficient of friction of the two materials that are rubbing together?

2. Apr 11, 2007

### ShawnD

Maybe I'm under-thinking this but couldn't you work it like a linear problem and just change each term to be rotational? Force --> torque. speed (v) --> rotation speed (w).

(friction force) = (friction coefficient) * (normal force)

Speed is constant, so:
(friction force) = (applied torque)

Substitute that into the original equation:
(applied torque) = (friction coefficient) * (normal force)
(friction coefficient) = (applied torque) / (normal force)

You already know the torque on the small cylinder, and the normal force is just the net force acting upward on the small cylinder.

Please tell me how it goes

Last edited: Apr 11, 2007
3. Apr 11, 2007

### Gokul43201

Staff Emeritus
Is this a real life problem that you are trying to solve, or is it from a textbook/course?

4. Apr 11, 2007

### Scooter057

I work for a company that produced drill pipe. One of the engineers designed a machine that simulates the drill pipe wearing against the side of the hole. This particular machine uses two cylinders in contact rotating in opposite directions to measure the amount of wear to the surface of the drill pipe. I'm pretty new to the company and have been tasked with determining a formula for the coefficient of friction regardless of the two materials being worn. I have an idea about how to proceed I just want to get other opinions.

5. Apr 11, 2007

### Scooter057

The coefficient of friction is a unitless number. Dividing a torque by a force gives you a distance. (Nm/N = m)

6. Apr 11, 2007

### ShawnD

If that's the case, we need to know either the inside diameter of the big pipe or the outside diameter of the small pipe.

7. Apr 11, 2007

### Scooter057

If the radius of the big cylinder is r1 and the small cylinder r2, then I would think that the coefficient could be expressed as T1*r1/F where T1 is the torque on the big cylinder and F is the applied force. It would also be expressed as T2*r2/F. Does that sound right?

8. Apr 12, 2007

### lpfr

Yes. You are right. Your formulas are the good ones (I found the same).