# A chain sliding down an incline

1. Oct 16, 2006

### Cronomius

My first post :D
I have an incline of angle (θ) to the horizontal, and a chain of length (l) of a uniform mass (m), I place the chain so that one end is at the bottom of the incline, the entire system is frictionless.
I need to find what the velocity is as the last end leaves the incline, and how long this took.

Since the system is frictionless then am i right in assuming the the length (let call it x) that has already left the incline no longer has any significant influence on the system with regards to finding the velocity?
If so then can i simple sett it up as

0,5mv^2=mglsin(θ)
and get
v=√2glsin(θ)

and then find the time to be
t=[√(2glsin(θ))]/[gsin(θ)]

but that just seems to simple...

2. Oct 16, 2006

### Chi Meson

Yes it is too simple. The links that are not yet at the bottom will continue to accelerate the system (the chain is the system), but the inertia of the links that are at the bottom will continue to oppose the chang in motion.

Essentially, it is a Newton's second law question. The total mass is constant, but the net force changes with time. You are doing calculus I assume?

3. Oct 16, 2006

### Cronomius

I'm confused :P
Yes i am taking calculus I

4. Oct 16, 2006

### Chi Meson

OK, velocity (starting from zero) is acceleration times time, right? But the rate of acceleration changes with time. Can you think of a expression for the net force on the chain as a function of the length that is NOT yet at the bottom?

And, sorry, I can't get back tot he computer for a few hours from now...anyone else take it from here?

5. Oct 16, 2006

### Cronomius

So using F=ma
i can use a=gsin(θ) and then the mass that is still on the incline as m(b-x)/b?
so
F=g(m(b-x)/b)sin(θ)?

6. Oct 16, 2006

### OlderDan

Since Chi Meson is away, I'll jump in here and go back to the beginning. First, if energy were conserved in this problem you would need to fix your potential energy term. You have calculated the potential energy as if all the mass were at the top of the length of chain. You would need to use the location of the center of mass of the chain to calculate the PE, so the height would be only half of what you used.

If you can't use conservation of energy, there has to be a mechanism for removing mechanical energy from the system. I think you can see that what you have here is a whole series of inelastic collisions. The chain on the horizontal surface (I assume there is a horizontal surface) is being bumped by a faster piece of chain arriving at the bottom. If the collisions were elastic, the chain would separate. Since it is all stuck together, the collisions must not be elastic. This event should warm up the chain. At the end it would be worthwhile to compare the result you get to one based on a correct conservation of energy calculation.

If b is the length of the chain, and x is the distance the chain has moved, that looks good.

By the way, I think we are assuming the chain stays in the original vertical plane that contained it. In reality, this would not happen.

Last edited: Oct 16, 2006
7. Oct 16, 2006

### Cronomius

Yeah b was suppose to be l there :P sorry
Energy is conserved in the system.

When i used F=g(m(l-x)/l)sin(θ)
I get the acceleration to be: a=g((l-x)/l)sin(θ)
Then using v=v0+a(t-t0) where v0=0 and t0=0 i get:
v=gt((l-x)/l)sin(θ)

But i can solve it using conservation of energy as well? so i let PE =mgsin(θ)l/2? but do i need to treat one of them as having a variable(changing) mass?

8. Oct 16, 2006

### OlderDan

Your velocity equation is only valid for constant acceleration. You do not have constant acceleration.

9. Oct 16, 2006

### Cronomius

Ok, so then i need to take the integral of the acceleration wquation i got? or do i need to redo that as well?

10. Oct 16, 2006

### OlderDan

If you have force as a function of time, you could integrate over time to find the impulse and use that to find the change in momentum. It's not clear to me that you can easily find F(t), but I have not tried too hard. I think you do need to focus on the horizontal component of the net force. At the bottom of the ramp every link of the chain is experiencing an additional normal force that is just enough to stop its vertical motion (another inelasticity), so only the horizontal component is ultimately adding to the momentum.

Last edited: Oct 16, 2006
11. Oct 16, 2006

### Cronomius

And then i'm confused again :P
At the end of the ramp the normal force changes from mgcosθ to just mg, but seing as this is only in a vertical componet as you said it has no impact on the velocity, this i get.

12. Oct 16, 2006

### Cronomius

Earlier you made it seem like it was possible to solve this problem using conservation of energy as well, but i needed to use the center of mass of the chain.

could i then use mglsin(θ)/2 = mgl((l-x)/l)sin(θ)/2 + .5 (mx/b)v^2 ?

13. Oct 16, 2006

### OlderDan

I'm going to have to think more about this, but I'm starting to think this is just a conservation of energy problem, if you think of the chain as a continuous uniform distribution of mass. If you make the assumption that all forces other than gravity acting on the chain are perpendicualr to its motion, including the one at the turning point at the bottom of the plane, and that all points in the chain have the same velocity at all times, then mechanical energy is conserved. With the corrected PE expression, you have it.

It is not uncommon to make this assumption for ropes and chains. You might want to look at some of the problems here.

http://electron9.phys.utk.edu/phys513/Modules/module3/problems3.htm

Last edited: Oct 16, 2006
14. Oct 16, 2006

### Cronomius

Ok, seing as this isn't suppose to be all that complicated making those assumptions seems like a fair bet :D
And i take it that my equation using conservation of energy was wrong? even when using those assumptions?

15. Oct 16, 2006

### Cronomius

and will look at those problems now :D
thanks :D

16. Oct 16, 2006

### OlderDan

Your revised equation is OK. Your original PE was twice what it should have been.

17. Oct 16, 2006

### Chi Meson

Ha, jeez, yeah. I thought of this while I was teaching. I've got calculus on the brain, and I missed the much easier conservation of energy solution. If the bottom surface is frictionless, it's quite the way to go.

18. Oct 16, 2006

### OlderDan

Yeah, and I let you bend my thinking a bit at the beginning.:yuck: It's a worthwile calculus problem to calculate the work done by the force by integrating F(x) over x. Of course it turns out to be the PE you had to start with.

And of course for a real chain everything I said earlier is true. There is no way a real chain could behave like this unless it was constrained by something like a frictionless tube with a gradual bend at the turning point.

19. Oct 16, 2006

### Cronomius

Thanks for all the help :D
I like this forum :D Having a place to go to get the help I need is a great asset :D

20. Oct 17, 2006

### Cronomius

Ok so using my equation above, and then letting x=l since the chain has traveld down the entire slope the equation simplifies down to
.5mglsinθ = .5mv^2
which gives me a velocity of
v=√glsinθ

But then i also had to find how long this took, but since the system has a variable acceleration i can't use v=at, leaving me a little confused as to how to find the time :P