Stick-Slip of Box on a Moving Belt

Question:

I am a student from Slovenia. I am doing some research on mechanical system which can be described by Rayleigh's equation. I'd like to describe the stick-slip motion of a mass sitting on a constant velocity conveyor belt and connected via linear spring to a fixed point. The friction force which acts on the mass is dependent on the velocity of the mass attached to the spring. The problem is that I know how the equation looks like but I don't know how to explain the friction force which acts on the mass. The equation of motion is: x"-ax'+b(x')³+x=0. One idea is to expand the resultant friction force in a Taylor series about velocity of the belt. But how can I explain that the square term is absent in the equation of motion? Is it because the friction force always opposes the mass displacement? How can I prove that? I have tested your program DynLab and it is really great. I hope I'll be able to order it in the nearest future. (ran out of money) Thank you very much in advance.

Answer:

The picture I get of the physical situation from your description and equation is that you have a horizontal belt with a block of wood or something on it, attached to a fixed point with a Hooke's law spring. In the absence of a frictional interaction with the belt we could expect simple harmonic behavior from the block.

The motion of the belt under the block and the friction between the two complicates the issue. Once there is relative velocity between the block and the belt your equation comes into play to describe the motion of the block relative to the spring attachment point. Here I have a difficulty. Since x is taken to be the displacement from the attachment point, then x' must be the velocity relative to that point, not the velocity relative to the belt. In that case the belt velocity does not appear explicitly in your equation.

Let's look at the startup situation. Initially the block and the belt are motionless and the spring is relaxed. In this situation the coefficient of friction is the static coefficient. As the belt begins to move, the block moves with it until the tension in the spring is sufficient to break the static friction. At this point the frictional force is the normal force times the kinetic coefficient of friction which is smaller than the static coefficient. The spring force then exceeds the force of kinetic friction, drawing the block back toward the attachment point of the spring. I anticipate some sort of oscillatory motion

If we re-write your equation a bit so that: x''=-x-x'(bx'²-a) we have a simple harmonic oscillator with an energy loss when bx'² greater than a and an energy gain when bx'² less than a. For small x', out near the ends of the oscillatory motion, the belt couples energy into the system. Near the center of the oscillatory motion the block looses energy to the belt. That is what your equation suggests.

This leaves me with the fundamental problem that the energy loss-gain seems not to depend on the velocity of the belt.

Please let me know if I have misinterpreted the situation or the equation. I would be happy to discuss it further if these thoughts prompt any questions from you.

Question:

You are totally right in the interpretation of the model. You are also right that the frictional force is not dependent on the velocity of the belt. I think that it depends on the relative velocity of the mass. I read in an explanation of that equation that the friction force could be expanded as F(dx/dt-v), where v is the velocity of the belt. If I do that I get: F(x') = F(-v) + (x'+v)dF(-v)/dx' + (x'+v)² d²F(-v)/dx'2² + (x+v)³(...). In that explanation is written that by noting that friction force always opposes the mass displacement x(t), one can find that x"-ax'+b*(x')³+x=0 approximately, where x now equals to x+F(v).

The first term is eliminated from the equation of motion by this substitution. The second and the fourth terms are somehow present with a little imagination. What about the third term in such explanation. Maybe this explanation is not the most appropriate. I'd just like to know from where all the terms come. Otherwise we can say - the equilibrium point is unstable and we have a repellor. Therefore we need negative dumping for small absolute values of the velocity of the mass and positive dumping for higher velocities. How can we get that? By applying combined damping -ax'+b(x')³. The result is limit cycle.

Answer:

It appears that the explanation you read invokes a physical argument to clean up the mathematics. Any terms in even powers of x' are going to result in a uni-directional friction force bias in the +x direction, contrary to what we experience physically. Using that notion to drop all the even power terms in the expansion strikes me as a bit questionable but perhaps it works.

The expression which we are left with does generate a limit cycle, driving large oscillations down and small oscillations up for a and b greater than 0. I ran it through DynaLab-Pro. The vector field view is seen below. The green line is the velocity nullcline. The dark blue line is the acceleration nullcline, the red track is the system trajectory in phase space starting at (1.5,0). The light blue track is the trajectory starting at (0.11,0). We see an unstable spiral fixed point at the origin enclosed by the limit cycle on which the trajectories settle out. Hope this helps.