I have this question which is confusing as it should be harder than it appears...(adsbygoogle = window.adsbygoogle || []).push({});

A uniform chain of length [tex]L[/tex] lays on a frictionless table top. Suppose one link just hangs over the table's edge so that the chain begins to fall.

Let [tex]x[/tex] be the amount of chain that has fallen. What is the chains vertical acceleration at this instant.

I started off trying to use conservation of energy

(with the zero potential line being a distance [tex]L[/tex] below the table top).

I went with

[tex]mgL = (L - x)mgL + xmg\left(L - \frac{1}{2}x\right) + \frac{1}{2}mv^2[/tex]

which led to

[tex]v^2 = gx^2 + 2gL(1 - L)[/tex]

Now firstly im not sure if this is correct. If it is then what is my next step?

Secondly, I started thinking is the problem maybe a pretty simple differential equation from Newtons second law?

Or can it be done both ways?

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# Homework Help: Chain on table

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