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I was going through Example 9.2 in Thornton and Marion's Classical Dynamics, and I am stuck on the Potential Energy part of the Question. How do they get the term at the top of the page on the right? I understand how to reach U0, but getting from there to U is really confusing me. P = linear mass density "M/l", length "l" and mass "M", with "x" being distance traveled by the falling end of the chain. enter image description here

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  • $\begingroup$ What do you think the expression for $U$ should be? What exactly is confusing you? $\endgroup$ – sammy gerbil Dec 6 '20 at 20:58
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In sketch (a), if we assume, b, is the total length of the chain, then the gravitational potential energy is mgh, where m = ρb, and h = - b/4 (measured down with the zero potential chosen at the top). This is consistent with his formula if you set, x = 0. For sketch (b), The total length (down and up) is, (b+x). The left side P.E. = -ρ[(b+x)/2]g(b+x)/4. The right side P.E. = -(ρ){[(b+x)/2]-x}g[({[(b+x)/2]-x}/2)+x]. You must add these two.

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