Calculate work when winding chain (GRE Problem 66 - Exam 9277)

[TheDestroyer]

This problem is in a GRE exam which I think might have big mistake in forming the answers:

A thin uniform steel chain is 10 meters long with a mass density of 2 kg per meter. One end of the chain is attached to a horizontal axle having a radius that is small compared to the length of the chain. If the chain initially hangs vertically, the work required to slowly wind it up on the axle is closest to


1. 100 J
2. 200 J
3. 1,000 J
4. 2,000 J
5. 10,000 J

The solution I made is attached in the post or you can download it from this web site:

http://www.samer.ws/ChainProblem.pdf

IF THE LINK DIDN'T WORK COPY IT AND PASTE IT IN YOUR BROWSER AND TRY AGAIN


and the solution people are sharing is here:

http://grephysics.net/ans/9277/66

Please check the attachment, the answer should be 500 joules, GRE said it must be 1000J and people did every trick to reach it, Any suggestions?

 

[Loren Booda]

The shared solution from GRE seems right without tricks. (Your attachment is still pending approval.) I will be interested to see where you disagree.

 

[TheDestroyer]

Well, If the attachment is still pending let me tell you with what i disagree in GRE Physics Forums solution,

The first solution integrates from 0 to 10 while this is wrong because only half of the chain should be raised to wind the chain

the second solution offered by Hamood (with all due respect) he said that the chain mass we are going to raise is 20 kg while we are going to raise only half of the chain which is 10 kg!

Am I wrong with what I'm saying? please recheck this in the link i provided with my first post

thank you for reading

 

[TheDestroyer]

I've added a link for the problem solution I made because I think the attachment is making problems, Please comply as fast as you can I have GRE exam in 2 days

IF THE LINK DIDN'T WORK COPY IT AND PASTE IT IN YOUR BROWSER AND TRY AGAIN

Thanks

 

[Doc Al]

Please check the attachment, the answer should be 500 joules, GRE said it must be 1000J and people did every trick to reach it, Any suggestions?

I agree with the GRE. The mass of the chain is 20 kg. When you wind it, its center of mass is raised 5 m. So the change in gravitational PE = mgh = (20)(10)(5) = 1000 J.

(I'll take a look at your attachment.)

 

[Doc Al]

your solution

OK, I looked at your solution. It's a bit hard to follow, but you seem to think that lifting a half-length chain up twice is equivalent to lifting the full chain up once. Not so!

If you require an integral, why not just this:

$$W = \int_0^L \lambda g z dz = \lambda g L^2/2 = 1000 {J}$$

(I see that that's just what the GRE folks say; they are correct.)

 

[TheDestroyer]

OK, Let's talk in another way, raising half of the chain means equivalently we are going to CUT the chain to 2 halves and raise the lower half to 5 meters right? think about it, it's of course equivalent, so:

W= m g h =Lambda * L * g * h = 2 * 5 * 10 * 5 = 500 J

How is it? Am I right?, the integral you made has the same mistake the website made, you shouldn't integrate for all the chain while you raise only half of it and while the resulting function isn't linear ;)

and about the center of mass, we are not going to raise the center of mass, we are going to raise only the center of mass of the lower half which gives also the result I wrote above !

I don't know what to do ! VERY CONFUSING !

 

[Doc Al]

Why are you talking in terms of raising half the chain?? The problem is to wind up the entire chain from its hanging position so that all of it is at the same height.

OK, Let's talk in another way, raising half of the chain means equivalently we are going to CUT the chain to 2 halves and raise the lower half to 5 meters right?

Not sure what you mean by raising half of the chain. Do you mean winding up the chain half-way?

think about it, it's of course equivalent, so:

W= m g h =Lambda * L * g * h = 2 * 5 * 10 * 5 = 500 J

Sure, if you cut the chain in half and raised the lower half up 5m, that would require W = m g h = 10*10*5 = 500 J. But that's not equivalent to winding up the full chain. You haven't even wound up either half!

You have to now wind those two chain halfs. Winding up each 5m chain will take 250 J, so once again the total for winding the chain equals 1000 J.

How is it? Am I right?, the integral you made has the same mistake the website made, you shouldn't integrate for all the chain while you raise only half of it and while the resulting function isn't linear ;)

Again, I don't know where you got this "raise only half" business. You may want to reread the problem.

and about the center of mass, we are not going to raise the center of mass, we are going to raise only the center of mass of the lower half which gives also the result I wrote above !

You have the wrong picture of what the problem is asking for. Imagine the axle being on a table at a height of 10 m above the floor. Initially, you have a 10 m chain hanging down to the floor: it's center of mass is at h = 5 m. Once you wind it up onto the table, the entire chain (and thus its center of mass) is at height 10 m. The center of mass rises 5 m.

 

[TheDestroyer]

Thanks Doc Al, I thought the problem what's to put the 2 ends together only, not wind until the whole chain is rounded around the axle,

Thanks :) I Appreciate your efforts :)