Experts pls ..can u check this...? this takes some imagination...it violates law of consv of energy....?

Question

now imagine theres a point A at height of 40 metres . now from that point a thin non elastic lace (thread) starts to materialize(magically orignate) and run down and hits the ground due to gravity. the lace keeps on appearing and its like endless. now when it appears at point A its velocity is zero but it gets jerked up to a velocity . now what would be the vellocity of the lace when hits the ground after runnin down 40 metres. AIR RESISTANCE IS TAKEN AS ZERO. here gravity is acting on the vertical length of the lace which is forty metres. according to me the lace can have a maximum velocity of only 20 m/s. which is less than 28 m/s when u drop something from a height of 40 metres..

now for 1000 metres of lace (thread ) to run down originating from that point A, how long will it take..>? hope ppl get my question

heres anotherway of pttin the question.......

ok forget that that point A. take some 1000 metres of chain or lace to a table which is forty metres high . and now throw down one end of lace or chain. remember the lace is neatly arranged and not etangled . now one starts running down ad hits the ground and begins to pull down the rest of the chain which is at a height of 40 metres on top the table. it ill take some time for the whole length(1000m) to run down. relate both the descriptions to get a clear picture . could you answer my question now.?

sorry the length of the chain is 1000 metres.......(say) and the chain IS THIN AND AIR RESISTANCE AND FRICTION ARE ZERO....

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;no . the chin is not stretched along the surface of the table . it is kept in a heap but not entanged.

Answer 1

I see where you are going . . . you just won't get there.

The first link should reach a max velocity of ~20m/s (19.8, I think). From there, the chain will continue to accelerate because the force of gravity on the dangling chain exceeds the brake force cause by the acceleration of the link at the edge bringing it up to speed right before it falls. (The brake force is directly related to change in velocity for the link, which in turn is directly related to the velocity of the falling thread)

Pulling each link off of the elevation will act as a brake on the thread falling and it will be equal to the force of gravity on the falling thread when the string reaches a speed of 28 m/sec which will be its terminal speed. You can figure this out by balancing the impulse force F=m1 delta v/delta t and the force of gravity = m *g.

I'll leave it to you to solve the problem from this angle because I don't want my head to hurt. This is because the easy way is to calculate the max speed from the energy equations:

The chain will max out at 28.01 m/s (using g of 9.81). At this point, the KE of the chain link hitting the bottom will be equal to its potential energy drop from the point a to point b. That is except for the last link which will accelerate from its starting velocity of 28 m/sec to finally hit the ground at 40 m/sec.

(I am assuming that the thread will fold out -- like one of those emergency firehose you see in buildings -- rather than accelerating all of the chain behind it. If you have the thread on a wheel or laid out in a line, the initial acceleration will be slower).

:-)

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I've thought more about your problem. If the thread/link at the top edge is forced to accelerate horizontal by the initial impulse from the falling thread, the velocity will be limited to 20 m/sec down. You will also get a horizontal component so that the thread/chain will look like a waterfall going over the edge.

But if the thread/chain is initially pulled down then you can get the 28 m/sec. Hope this makes sense.

Answer 2

Take M = total mass of the chain; so that its potential energy PE = Mgh; where h = 40 meters. I assume the length of the chain L = 40 meters; so when the first link touches the floor (h = 0) the last link is still at h = 40 meters.

Total energy TE = PE(h = 40) = Mgh before the chain unfurls from height h = 40 meters.

Total energy TE(h/2) = PE(h/2) + KE(h/2) = Mgh/2 + 1/2 Mv^2; where h/2 is the height of the center of mass half way up the height. h/2 is the point when the first link is about to strike the floor and the last link is still at h = 40 meters.

From the conservation of energy, the two total energies are equal. Thus, TE(h) = Mgh = Mgh/2 + 1/2 Mv^2 = TE(h/2).

Therefore Mgh (1 - 1/2) = Mgh/2 = Mv^2/2 and gh = v^2 or v = sqrt(gh) and that is the velocity of the chain as the first link is about to strike the floor. That would put the velocity v ~ 20 mps as I eyeball the last equation.

And as you can see from my total energy approach, the reason v ~ 20 mps is because some of the total energy is still potential energy. There will be some PE as long as there is at least one link above h = 0.