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## Homework Statement

The statement of the question is:

A chain of uniform linear mass density ##\rho##, length ##b## and mass ##M## hands as shown in the figure below. At time t=0, the ends A and B are adjacent, but end B is released. Find the tension in the chain at point A after end B has fallen a distance x by using energy conservation.

Refer to the figure:

http://i.stack.imgur.com/RGmud.jpg

I can solve this example (from Marion and Thornton Classical Dynamics 5e) easily from assuming a free fall and summing force. I however do not follow the example's solution using conservation of energy.

The example states:

We treat the chain as one-dimensional motion ignoring the small horizontal motion. Let the potential energy ##U## be measured relative to the fixed end of the chain so that the initial potential energy ##U(t=0)= U_0 = -\rho g b^2 /4##. A careful geometric construction shows that the potential energy after the chain has dropped a distance ##x## is

##U=-\frac{1}{4} \rho g (b^2 +2bx - x^2)##

I can reproduce the initial potential energy via integration, that is to say, with ##U(x=0)=0##

##U=2mgh=2g\rho \int_{\frac{b}{2}}^{0} x dx =- \rho g b^2 /4##

but am still unable to reproduce the falling chain potential. Can anyone help me derive the expression for the potential energy as the chain falls,

##U=-\frac{1}{4} \rho g (b^2 +2bx - x^2)##

Thank you.

## Homework Equations

##U=mgh##

## The Attempt at a Solution

Let ##U(x=0) = 0##. The initial potential is derived above. When the chain falls by an amount x on the B side, the same length ##x## is added to the A side.

On the ##A## side:

##U=mgh = g \rho \int_{\frac{b}{2} + x}^{0} x dx = -\frac{g\rho (x+b/2)^2}{2}##

Similarly, on the ##B## side:

##U=mgh = g \rho \int_{\frac{b}{2} - x}^{x} x dx =\frac{g \rho [ x^2 - (b/2 -x)^2 ]}{2}##

We next sum these two expressions yielding:

##U = \frac{-g\rho (x+b/2)^2 + g \rho [ x^2 - (b/2 -x)^2 ]}{2}##

which, after expanding we see the ##bx## terms cancel, and therefore does not reproduce the solution

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