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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..
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...
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.
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.
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.
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.
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.
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..
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.
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 .
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