GREPhysics.NET: GR9277 #66

GR9277 #66

Problem

GREPhysics.NET Official Solution

\prob{66}
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

100 J
200 J
1,000 J
2,000 J
10,000 J

Mechanics $\Rightarrow$ }Work

Work is defined by $W=\vec{F} \cdot d\vec{l}$ .

The force here is just due to gravity, thus $F=\rho y g$ , where $\rho=2kg/m$ is the density of the chain. The chain is wound upwards, so work is $W=\int_0^{10} \rho g y dy = \frac{2}{2}(gx^2)^{10}_0=10\times 100 =1000J$ , as in choice (C). (The approximation $g\approx 10m/s^2$ is made.)

Alternate Solutions

TheXDestroyer 2007-10-02 22:36:46	Something is not right in the official (supposed to) solution, 1st $dF$ must equal $dF= \rho (dy) gy$ to get the unit of force ( $dy$ is the unit length, second thing is the the integral must be from 0 to 5 not to ten after multiplying the hole integral by 2, because you are raising only 1 half of the chain (the down half) while upper half is not touched, and multiply by 2 because every mass unit $(\rho dm)$ is being raised twice the distance from the center !!!!!!!!!!! this gives the answer 500 J !!!! this is the only correct explanation !! I've shared the problem with physics forums you can see it on the site where you can find my solution as PDF: http://www.physicsforums.com/showthread.php?p=1452586 Reply to this comment
hamood 2007-04-04 13:01:26	If we assume conservation of energy: Initial PE = mgh = 20kg * g * 5m = 1000 J (average h =5m) Final PE = mgh = 20g10= 2000 J So work done = 2k - 1k =1000 J Reply to this comment

Comments

mdornfe1
2008-11-06 17:51:19

Think about pulling the chain up all at once. The work required to do this is W=mgL, where m is the mass of the chain and L is the length. So W=2*10*10*10=2000J. Here we approximated g as 10. This is an over estimate. So we choose the next lowest answer C.

carle257
2010-04-10 01:01:32

This is wrong because you only move the bottom of the chain 10m. The halfway point only moves 5m and so on. You must integrate or look at the energy of the center of mass to get the correct answer.

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tensordyne
2008-11-04 19:46:20

In order to put this problem to bed please consider the following solution. The ammount of force the chain exerts on the axel is gh, where h is the length of chain at some point in time that is hanging and g is gravity ~ $10 m/s^2$ .

The work done can be evaluated using an integral of the distance the work is done and the force exerted.
If x is the length of chain wound around the axel then $h(x) = 10 - x$ . The work is then

$W = \int F \cdot dx = 10 \int^{10}_{0} (10 - x) dx = 1000 J$ .

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balla
2008-07-14 16:04:01

Similar to previous explanations, I concentrated on the center of mass. It is a uniform chain, so the COM is initially 1/2 way along the chain (5m from pivot). When rolled up, the COM is the pivot; the change in position is 5m, the total mass is 2kg --> increase in energy = 2kg*10m/s^2 *5m = 1000J

balla
2008-07-14 16:04:50

Correction...2kg/m, so total of 20kg.

Reply to this comment

grae313
2007-10-14 15:06:48

The method I used is essentially equivalent to Yosun's, but I approached it a bit differently. To approximate the work, I imagined the hanging chain as being composed of 10 links, each a meter long with a mass of 2 kg. Then the total work done is comparable to the work needed to lift the bottom link up 10 meters, plus the work needed to lift the second to last link up 9 meters, plus the work needed to lift the third to last link up 8 meters plus..... until you finally lift the first link up 1 meter.

It works out to W = F*d = m*g*(10+9+8+7+...) where m is the mass of each link, or 2 kg, and g is approximately 10, so

W = 2*10*(10+9+8+7+6+5+4+3+2+1) = 20*(55) = 1100, so choice (c) is it.

I like hamood's solution better! :)

Reply to this comment

TheXDestroyer
2007-10-02 22:36:46

Something is not right in the official (supposed to) solution, 1st $dF$ must equal $dF= \rho (dy) gy$ to get the unit of force ( $dy$ is the unit length, second thing is the the integral must be from 0 to 5 not to ten after multiplying the hole integral by 2, because you are raising only 1 half of the chain (the down half) while upper half is not touched, and multiply by 2 because every mass unit $(\rho dm)$ is being raised twice the distance from the center !!!!!!!!!!! this gives the answer 500 J !!!! this is the only correct explanation !!

I've shared the problem with physics forums you can see it on the site where you can find my solution as PDF:

http://www.physicsforums.com/showthread.php?p=1452586

marten
2007-10-24 12:14:51

I'm not sure what you mean by the down half. I think that you're confused about how the problem is setup. Both halves are "touched". It might to clarify that while the chain winds around the axle, the entire chain is lifted.

Describing physical setups in words can be confusing, perhaps a diagram would have been more helpful here.

Marten

Albert
2009-10-21 07:49:53

I saw your solution, and its really strange. The problem is very straight forward, you may not be reading this site anymore, I am writing for everyone.

The center of mass lies at the center of the chain, which is at 5 meters from either side, right? So we are talking about taking that part up on height. And the total mass is 20Kg.

deltaPE = work done.

W= m*g*h

W= 20*9.8*5 = 1000J (approximately)
And there is no need for calculus.

There you go big guy!

Reply to this comment

hamood
2007-04-04 13:01:26

If we assume conservation of energy:
Initial PE = mgh = 20kg * g * 5m = 1000 J (average h =5m)
Final PE = mgh = 20*g*10= 2000 J
So work done = 2k - 1k =1000 J

kroner
2009-10-06 22:02:05

or even quicker, by setting h=0 to be at the top:
Initial PE = mgh = 20kg * g * (-5m) = -1000 J
Final PE = 0

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