a tricky problem involving physics...

vibgyor · 2007-11-28 03:05:16

http://in.answers.yahoo.com/question/index;_ylt=AhV.MITwGtI3jO5OYJXKF1GQHQx.;_ylv=3?qid=20071125231521AAgrxmJ

pls can you check the above link . and i need ur answers for the problem. thanx .

NullRoot · 2007-11-28 03:14:06

Sorry, Vibgyor, but I don't get what you're trying to describe or how you're arriving at your velocities? Can you try and explain it again and include some workings?

vibgyor · 2007-11-28 05:12:27

sorry about that. i ahd go away for a while. im copy pasting a part it.

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 that starts running down and 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. (the chain is not stretched . but its kept like rolled up fire hose). its going to take more than 30 secs for the entire length of the chain to hit the ground.

nowthe chain cant keep on accelerating rgt ..? what will be the terminal velocity..?

air resistance and friction taken as zero. take g as 10m/s^2 . hope you get what im tryin to describe . thanx.

Last edited by vibgyor (2007-11-28 08:40:52)

NullRoot · 2007-11-28 11:19:07

The short answer is No.

It's easier if you imagine the chain rather than the rope. I'm assuming you basically want to ignore friction of any sort (like the chain against the table) not just air resistance?
Each link will have an initial velocity of 0 as it goes over the edge of the table. It doesn't matter that ones have come before, because the entire length of chain that is in the process of falling is accelerating at the same rate. The acceleration cannot increase because gravity is constant.
So the link that's about to hit the floor is going faster than the one that's just gone over the edge table, but their velocity is changing at the same rate: ~9.8m/s². Once it hits the floor, it will stop accelerating and it's velocity will return to 0. As will each link in the chain after that.

So the rate at which another link is fed into the freefall will increase until the first link hits the floor; then it will equalise because the average velocity of the length of chain that is in freefall will remain constant.

Hope that makes sense?

vibgyor · 2007-11-28 19:24:52

thanx nullroot.

let me pose this question in another way. lets reduce this between two points A and point B

point A is onthe surface of the ground and point B is directly above it . distance AB lets take as 10 metres. now , imagine at point B, theres an unending supply of chain or lace which is neatly rolled up like a firehose(not entangled )
let us name every metre of chain or lace as L1 ,L2,L3,L4.....and so on till 1000 th metre
is named as L1000. now pull down L1 so that it touches point A on the floor. from now on gravity takes over and the chain or lace starts to flow down like waterfalls from pointB.

now how long will it take for first 200 metres of chain or lace to hit the ground (pointA)..? take acc due to gravity as 10 m/s^2. i assume the chain has to flow down at a constant velocity after a few seconds..

NullRoot · 2007-11-28 21:34:30

Let's assume you don't pull down the first 10 meters (effectively connecting A to B with a length of chain).
So we start off the process and link 1 falls. Just as it touches the floor we freeze time so we can see what's going on and work out some equations.

At point A there is a link touching the ground that has fallen there from a height of 10 meters, beginning from rest. At point B there is now another link (the first link of the next 10 meters of chain) which is at rest, waiting to fall.

For the end of the chain (that has fallen to A) to start from rest and reach the floor 10 meters away, the time taken is determined by:
displacement = initial velocity * time + 0.5 * acceleration * time²
So:
10m = 0m/s * t + 0.5 * 10m/s² * t²
10 = 0 + 5 * t²
2 = t²
t = 1.414s (to 3 decimal places)

EDIT: Amended a mathematical mistake; subtracted 5 instead of dividing

So you now know that it takes 1.414 seconds for 10 meters of chain to fall from A to B. It will also take 1.414 seconds for the link at point B to fall to point A.

After the first 1.414 seconds, though, only the first link has hit the floor; there isn't any chain on the ground. For the first 10 meters of chain to be on the floor, the link at point B must hit the floor, right? Now because it also starts from rest, the time it takes to hit the floor is also 1.414 seconds.

For all 200 meters to be on the floor at point A, it will take 1.414s (for the first link to fall) and then 1.414s for each length of 10 meters of chain. So I make it 1.414 + 1.414*20, which is about 24 seconds.

Last edited by NullRoot (2007-11-29 04:02:12)

vibgyor · 2007-11-28 22:20:40

t = 2.2 s no check. 10 = 5t^2; t= 1.414.

here links at B are suddenly jerked up from rest to a certain velocity at point B and should continue to travel down with that constant velocity . i wanna know what that velocity is. besides i think u cant use that simple s= ut + .5 at^2 eqn.

NullRoot · 2007-11-28 22:39:55

Yes, sorry, it's √2 seconds. Don't know why I subtracted the 5 instead of dividing. I'll go back and edit that to avoid future confusion.

They begin accelerating at 10m/s² towards the ground from their rest state and the acceleration is constant. They will not have a constant velocity because they are accelerating due to gravity. Once it has been pulled into the effects of gravity, the links below it cannot put an effect on it because it is being pulled down just as fast as they are: 10m/s²

I think s=ut+0.5at² is perfectly acceptable for this scenario.

vibgyor · 2007-11-28 23:40:55

man i really appreciate the effort you have put into your replies. but. no you cant use that equation here. i woud greatly appreciate the opinion of janefairfax in this problem coz she has already solved some mechanics problems.

NullRoot · 2007-11-28 23:45:49

Any particular reason you believe the equation is irrelevant?

vibgyor · 2007-11-29 02:51:51

becoz after a few sconds, velocity of links at point A is same as velocity at point B. an links get jerked up instantly at point b when they start to descend. u can chec that with a long lace . or check the link i posted initially. THE PONT OF POSTING THIS WHOLE THING IS THAT THIS PROBLEM SEEMS TO VIOLATES LAW OF CONSERVATION OF ENERGY.

NullRoot · 2007-11-29 04:22:58

Let me try a visual then to see if maybe I'm intepreting something wrong? Anyone's free to point out where I might be incorrect because now I'm worried that I've lost touch with Applied Maths.

B
56789
4
3
2
1
---------A

So the numbers here represent the individual links in the chain and the order in which they will hit the floor. A is the floor and B is the part where the links crest the edge.

A) The acceleration exerted on links 1 to 5 is 10m/s² in the direction of A. It's gravitational acceleration, so it's a constant.
B) Links 1 to 5 have a positive vertical velocity in the direction of A, albeit they each have a different velocity.
C) Links 6 to 9 have a positive horizontal velocity in the direction of link 5, because they are being pulled along by the fall of the links before.
D.1) The exact moment link 6 is pulled over the edge of the table, it has 0 vertical velocity. The table will prevent it from gaining a vertical velocity until that point.
D.2) The exact moment link 6 is pulled over the edge of the table, it will experience the same gravitational acceleration (10m/s²) in the direction of A as all other links in the chain. (In order for it to have a greater acceleration, there must be something with acceleration >10m/s² acting on it, right?)

E) The velocity of an object is calculated as: initial velocity plus acceleration times the amount of time the object has experienced that acceleration, ie (V=U+AT).

So if the link that crests the table has had a total vertical acceleration in the direction of A of 0m/s² until that point, then how can it's velocity be that of the link at the bottom which has been accelerating at 10m/s² for an amount of time greater than 0s?

vibgyor · 2007-11-29 05:39:48

nullroot , u should try it with a short piece of chain(may be three metres) from a small height .4 metres or so... when link hits point A the connecting links upto point B will have the same vel as that of the link hitting point A.

luca-deltodesco · 2007-11-29 05:53:04

im failing to see where conservation of energy is broken?

vibgyor · 2007-11-29 06:16:23

hi luca deltodesco

when u take the entire lenght of the chain to a height of 40 metres and drop it as whole . it will hit the ground at 28.28 m/s(g= 10 units) . here p.e = k.e. but in the described(see posts) method will the individual links hitting the point A reach 28.28 m/s...? hope you read my 2nd and third posts and get the clear and right picture.

luca-deltodesco · 2007-11-29 06:36:11

if you have a chain parallel to ground at 40m, it will indeed hit ground at 28 (g = 9.81).

assuming the chain is 10 links long, with each link being 0.1m, and with mass 0.1kg. then the total gpe of above at 40m horizontal will be 40*10*0.1*9.81 = 392J (3sf), and kinetic energy 0.5*10*0.1*28² = 392J (3sf) energy conserved.

if you then have the chain perpendicular to ground, with centre of bottom link (treating links as particles so it will fall 40m) at 40m from ground.

then the total gpe at t = 0, is

= 10*0.1*9.81*40 + 45*0.1*9.81*0.1 = 437J(3sf)

the velocity of link i (starting at i = 0) when it hits the ground will be √(2*9.81*(40+0.1i)), so the kinetic energy when it hits ground is:

0.5*0.1*2*9.81*(40+0.1i) = 0.1*9.81*(40+0.1i), this has to be summed from i = 0 to i = 9... oh wait, without going any further you can see this is the same sum as was done for gpe. so they are equal, energy is conserved.

at anyone point in time, while all of them are moving, they will have the same velocity since they all start at 0 and accelerate at same rate. however the ones lower down reach the ground first and so the ones higher up reach the ground with a larger velocity -> the difference in Ke of the ones above to the ones below is the same as the difference in initial gpe of the ones above to the ones below

i did this with a chain of 10 links with links 0.1m long, but the exact same result is reached if you have a string of near infinite number of particles over a finite space

Last edited by luca-deltodesco (2007-11-29 06:45:49)

NullRoot · 2007-11-29 06:51:33

Right. It was the second half of B where I was confused.

The chain in my diagram from link 1 to 5 will be going some velocity. The process of pulling link 6 over the edge of the table means that the downward velocity of everything attached to it momentarily becomes 0. It then accelerates back up to some speed until the point at which it needs to change the direction of motion on link 7 and so on.

This might be the part where you're getting confused. Link 6 won't "shoot up" to the velocity of 2 to 5, 2 to 5 will slow down to the velocity of 6. At that point, the whole lot speeds up again.
The thing is, it's only a moment in time in which the velocity is 0 so you won't notice it, especially on something like a string.

So what you have is a part of the downward acceleration dedicated to changing a horizontal velocity to a vertical one.

Last edited by NullRoot (2007-11-29 06:53:15)

vibgyor · 2007-11-29 21:15:40

so what do you think the velocity of the first link will when hitting ground...? when the lenght of chain is taken as 40 metres the first link should hit the ground at 20 m/s...

vibgyor · 2007-12-01 22:06:30

in this case when height is taken as 40 metres and chain length 40 m , the first link should hit the floor with velocity 20m/s.

but when the chain is much longer every link has to hit the floor with a velocity of 28.28 m/s. the force of gravity acting on 40 metres of chain that is vertically running between the floor and the plane as to be balanced at some point and at this point there will be no further acceleration.rgt.? now gravity acting on (g =10m/^2) 40 metres of chain balances out when 28.2 metres chain hit the floor per sec. (since v = 28.2). ,that is when the law holds good..

luca-deltodesco · 2007-12-02 00:45:16

the only time each link of the chain will hit the ground at the same speed, is when the chain is lying flat in the air and every link hits at the same time, if the chain is in rest vertical above ground, the links higher up will hit ground at a higher speed, the ones closer to ground, hit ground at a lower speed

Math Is Fun Forum

#1 2007-11-28 03:05:16

a tricky problem involving physics...

#2 2007-11-28 03:14:06

Re: a tricky problem involving physics...

#3 2007-11-28 05:12:27

Re: a tricky problem involving physics...

#4 2007-11-28 11:19:07

Re: a tricky problem involving physics...

#5 2007-11-28 19:24:52

Re: a tricky problem involving physics...

#6 2007-11-28 21:34:30

Re: a tricky problem involving physics...

#7 2007-11-28 22:20:40

Re: a tricky problem involving physics...

#8 2007-11-28 22:39:55

Re: a tricky problem involving physics...

#9 2007-11-28 23:40:55

Re: a tricky problem involving physics...

#10 2007-11-28 23:45:49

Re: a tricky problem involving physics...

#11 2007-11-29 02:51:51

Re: a tricky problem involving physics...

#12 2007-11-29 04:22:58

Re: a tricky problem involving physics...

#13 2007-11-29 05:39:48

Re: a tricky problem involving physics...

#14 2007-11-29 05:53:04

Re: a tricky problem involving physics...

#15 2007-11-29 06:16:23

Re: a tricky problem involving physics...

#16 2007-11-29 06:36:11

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#17 2007-11-29 06:51:33

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#18 2007-11-29 21:15:40

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#19 2007-12-01 22:06:30

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