Comparing the rate at which two chains fall in different scenarios

[ImaginalDisc]

I'm not terribly good at math, so I need help with a comparision of two different kinematic systems, please. It's been puzzling me.

There are two chains at the very edges of two different precipices. Each chain is 1000 meters long. Chain A is just over a 1000 meter drop. Chain B is just over a 10 meter drop. The entire system is completely frictionless, and the chains are assumed not to stack, but to just lie on the ground like limp noodles.

Here is where the kinematics kicks in:

The first link of each chain is nudged over their respective precipices.

Does each chain slip over the edge at the same rate, or does the chain slipping over the shorter drop slip over the edge at a lower rate?

 

[Dogdoctor]

I assume that the chain with the larger drop would increase in speed as it dropped and pull it over the edge faster unless there are other forces involved.

 

[jj]

Consider. At maximum extension, all 1000 meters of the chain will be accilerating the last link on the long drop.

And only 10 meters of chain for the short drop.

Seems obvious they won't be the same speed, eh?

 

[ImaginalDisc]

Consider. At maximum extension, all 1000 meters of the chain will be accilerating the last link on the long drop.

And only 10 meters of chain for the short drop.

Seems obvious they won't be the same speed, eh?

I thought so too, but I considered that when chain B's first link hits the groud, the link 10 meters behind it is falling off of the ledge with exactly the same rate of speed, correct? (It's a chain, the links 10 meters back has to be falling at the same speed as the first link after the first link falls 10 meters.) When that link 10 meters back hits the ground, the link 10 meters behind it will also be falling at B's rate.

Perhaps I don't understand the math well enough, but I thought that the last link of each chain would fall over the edge at the same rate of speed.

 

[Dogdoctor]

The way I see it the ten meter drop chain would increase in speed till it reached 10 meters and then would fall at a constant rate. The other chain would increase in speed till it hit the ground (without friction) making it's speed much faster.

 

[ceptimus]

You can run an empirical test with a fine necklace chain hanging over the edge of a smooth table. Time it dropping to the floor, then time it again with a tray catching the chain a few inches beneath the table.

 

[ceptimus]

Consider the case when 500 metres of chain have fallen. In the first case you have the weight of 500 metres accelerating the full 1000 metres of chain. In the second case you have only the weight of 10 metres of chain accelerating 510 metres of remaining chain.

The horizontal velocity imparted to the chain still on the level doesn't transfer into vertical velocity unless there is some kind of pipe guiding the chain around the edge.

I suspect the analysis of the situation would be quite complex: once the chain gets going at a reasonable speed, a curve would form in the top part of the chain - its horizontal inertia would move it away from the cliff . A wave would then propogate down the falling chain, tending to move lower portions of the chain further out from the cliff face.

 

[ImaginalDisc]

The horizontal velocity imparted to the chain still on the level doesn't transfer into vertical velocity unless there is some kind of pipe guiding the chain around the edge.

I'm sorry, I tried to make it clear that it is a frictionless environment. If it makes it easier, we could imagine the chains wrapped around rollers.

 

[ceptimus]

That doesn't affect my argument. If a chain is going at (say) 20 mph horizontally towards the edge of a cliff, then its inertia will carry it away from the cliff as it falls. The faster the chain goes, the further it will tend to fly from the cliff face. Lower portions of the chain that went over the edge with less horizontal velocity will be closer to the cliff than the top part 'wants' to be. This will result in a downward wave propogation (I think).

A guide pipe at the cliff edge would prevent this by turning the chain through a right angle. This could be a frictionless pipe. A roller doesn't help as that's just the same as a frictionless cliff edge.

 

[roger]

Consider. At maximum extension, all 1000 meters of the chain will be accilerating the last link on the long drop.

And only 10 meters of chain for the short drop.

Seems obvious they won't be the same speed, eh?

Are you arguing that heavier things fall faster than light things?

 

[Dogdoctor]

I don't think the chain moving horizontally will affect how fast it moves vertically other than the sooner the pieces are over the edge the sooner the pieces will start to fall.

 

[SteveGrenard]

Maybe these researchers can help you:

Found in the BMJ: BMJ 2003;327:1459-1461 (20 December), doi:10.1136/bmj.327.7429.1459



http://bmj.bmjjournals.com/cgi/content/full/327/7429/1459








Parachute use to prevent death and major trauma related to gravitational challenge: systematic review of randomised controlled trials

Gordon C S Smith, professor1, Jill P Pell, consultant2
1 Department of Obstetrics and Gynaecology, Cambridge University, Cambridge CB2 2QQ, 2 Department of Public Health, Greater Glasgow NHS Board, Glasgow G3 8YU
Correspondence to: G C S Smith gcss2@cam.ac.uk

Abstract here:

http://bmj.bmjjournals.com/cgi/content/full/327/7429/1459

 

[jj]

Are you arguing that heavier things fall faster than light things?

No, Roger, I'm not. I'm considering the ratio of the falling vs. non-falling part of the chain.

The whole chain isn't falling.

 

[ImaginalDisc]

Steve, I'm sorry, I don't see how an article about parachutes relates.

No, Roger, I'm not. I'm considering the ratio of the falling vs. non-falling part of the chain.

The whole chain isn't falling.

JJ, but the chain is continuously falling off of the precipice. I'm not clear why it matters what happens to chain that has already slipped off.

 

[Jimbo07]

I thought so too, but I considered that when chain B's first link hits the groud, the link 10 meters behind it is falling off of the ledge with exactly the same rate of speed, correct? (It's a chain, the links 10 meters back has to be falling at the same speed as the first link after the first link falls 10 meters.) When that link 10 meters back hits the ground, the link 10 meters behind it will also be falling at B's rate.

Perhaps I don't understand the math well enough, but I thought that the last link of each chain would fall over the edge at the same rate of speed.

After the 1st 10 metres, both chains will be moving at exactly the same speed.

Since the entire system is frictionless, we can assume acceleration is uniform. All internal forces are conservative, and normal forces on the ledge balance the force of gravity from the links still on the ledge. The only net force is that due to gravity. Regardless of how many links go over the edge, the acceleration will be a=g=9.81 m/s. Again, we take this acceleration to be uniform along the length, because there are no 'net' forces of friction holding it back. In a frictionless scenario, it is exactly like the entire chain is falling.

To pile up on the floor (in a 0 height pile), every single link must accelerate through the difference in height. Every point on the chain is accelerating at g. Therefore, if the chains are laid out, the last link in the 10 m chain accelerates through 1010 m, whereas the last link in the 1000m chain accelerates through 2000 m! The velocity of the last link of the 1000m high chain will be falling much faster than the last link of the 10 m chain, when each hits.

When will each hit and at what speed? We need d=1/2at^2 and v=at.

Chain 1: d = 1010 m, t=14.35s, v=140.8 m/s
Chain 2: d = 2000 m, t=20.19s, v=198.0 m/s

Now, in any scenario where friction is a factor, no uniform 1000 m chain of, say, 10cm links would budge, if only the first link were dropped over the edge.

 

[ImaginalDisc]

After the 1st 10 metres, both chains will be moving at exactly the same speed.

Since the entire system is frictionless, we can assume acceleration is uniform. All internal forces are conservative, and normal forces on the ledge balance the force of gravity from the links still on the ledge. The only net force is that due to gravity. Regardless of how many links go over the edge, the acceleration will be a=g=9.81 m/s. Again, we take this acceleration to be uniform along the length, because there are no 'net' forces of friction holding it back. In a frictionless scenario, it is exactly like the entire chain is falling.

To pile up on the floor (in a 0 height pile), every single link must accelerate through the difference in height. Every point on the chain is accelerating at g. Therefore, if the chains are laid out, the last link in the 10 m chain accelerates through 1010 m, whereas the last link in the 1000m chain accelerates through 2000 m! The velocity of the last link of the 1000m high chain will be falling much faster than the last link of the 10 m chain, when each hits.

When will each hit and at what speed? We need d=1/2at^2 and v=at.

Chain 1: d = 1010 m, t=14.35s, v=140.8 m/s
Chain 2: d = 2000 m, t=20.19s, v=198.0 m/s

Now, in any scenario where friction is a factor, no uniform 1000 m chain of, say, 10cm links would budge, if only the first link were dropped over the edge.

Wow Jimbo, I really appreciate the work you did, and I can certain see that the last links would each hit the ground at a different rate of speed, but I asked at what rate of speed the last link would slip off the edge of the precipice at.

 

[69dodge]

I thought so too, but I considered that when chain B's first link hits the groud, the link 10 meters behind it is falling off of the ledge with exactly the same rate of speed, correct? (It's a chain, the links 10 meters back has to be falling at the same speed as the first link after the first link falls 10 meters.)

Yes, this is correct.

When that link 10 meters back hits the ground, the link 10 meters behind it will also be falling at B's rate.

Not sure about this, though. What do you mean by "B's rate"? The chain speeds up as time passes, because gravity continuously pulls on it. So it doesn't have a rate; it has different rates at different times.

 

[jj]

Steve, I'm sorry, I don't see how an article about parachutes relates.



JJ, but the chain is continuously falling off of the precipice. I'm not clear why it matters what happens to chain that has already slipped off.

How much chain is being accelerated downwar by gravity in each case? (as opposed to just laying on the ground, sliding sideways) Once there is more than 10 meters of chain out, is the answer not different?

 

[Jimbo07]

but I asked at what rate of speed the last link would slip off the edge of the precipice at.

You're right, I guess I answered the whole thing, but the same analysis holds.

As each link on the 10 m high chain goes over it is moving at the same velocity as the one before. However, it continues to accelerate for another 10 m. Again, the acceleration is uniform, so the last link will move as if accelerated constantly for 1000m, as will the first.

In this case, their velocities, as you put it should be the same. Ultimately, I guess I was arguing your argument, at the point to the precipice, and in the end that is the answer to the OP.

At the precipice

both, d=1000m, t=14.27s, v=140.0

Notice there is not much more difference for the 10 m drop, and only a few seconds, but much more speed for the last link of the 1000m drop.

ETA: jj, the amount of chain over the edge should only make a difference where friction is concerned. A large amount of chain would need to be over for the force of gravity to overcome the force of friction holding it back. In a no-friction case, the downward force of the moon's gravity would get it going. Of course, your constant of acceleration would be different.

 

[69dodge]

After the 1st 10 metres, both chains will be moving at exactly the same speed.

Yes.

Since the entire system is frictionless, we can assume acceleration is uniform.

No.

All internal forces are conservative, and normal forces on the ledge balance the force of gravity from the links still on the ledge. The only net force is that due to gravity.

Yes.

Regardless of how many links go over the edge, the acceleration will be a=g=9.81 m/s. Again, we take this acceleration to be uniform along the length, because there are no 'net' forces of friction holding it back. In a frictionless scenario, it is exactly like the entire chain is falling.

No.

You just said, correctly, that some of the downward gravitational force on the chain is cancelled out by the upward force of the ledge on the chain. So why do you still think that the acceleration of the chain will be g? It will be less, because there's less force acting on it than on a freely falling chain.

Google for "Atwood machine".