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Falling Chain

Date: 12/31/97 at 15:16:19
From: ilya
Subject: Falling chain

Here is my problem. A chain (1 meter length) is lying on a table. Part 
of it (25 cm long) hangs in the air. The chain starts to fall down
without friction. How mach time will the chain fall?

Thanks a lot!

Date: 01/01/98 at 12:33:21
From: Doctor Anthony
Subject: Re: Falling chain

I will use letters to represent lengths so that you can see how the 
equations are derived.  

Suppose the total length of the chain is L and that at any time t 
after the beginning of motion there is a length x hanging down from 
the table. Let m = mass per unit length, and if v is the velocity of 
the chain at time t and T the tension exerted by the hanging portion 
on the portion still on the table, we get

  m(L-x)v.dv/dx = T   .......(1)   and

     mx.v.dv/dx = mgx - T  .....(2)

adding (1) and (2) we get

      mLv.dv/dx = mgx

          Lv.dv = gx.dx   and integrating

           Lv^2 = gx^2 + C  ...........(3)

      sqrt(L).v = sqrt(gx^2+C)

     sqrt(L/g) dx
     ------------  = dt

In this problem  L=100 and at the start v=0 when x = 25, so we can 
find C from equation (3)

           0 = 625g + C   and so  C = -625g  and C/g = -625

So now we have 

     sqrt(L/g) dx
     ------------  = dt

Integrating  sqrt(100/g)cosh^(-1)[x/25] = t + const.

At t=0  x=25 and cosh^(-1)(1) = 0  so the constant is 0.

We get  t = sqrt(100/g) cosh^(-1)(x/25) 

The time to when the chain leaves the table is found by putting 
x = 100 in this equation.

        t = 10/sqrt(g) cosh^(-1)(4)

          =  6.5914 seconds

-Doctor Anthony,  The Math Forum
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Associated Topics:
High School Physics/Chemistry

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