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! Ilya
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 sqrt(x^2+C/g) 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 sqrt(x^2-25^2) 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 Check out our web site! http://mathforum.org/dr.math/
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