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FeDeX_LaTeX

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## Homework Statement

http://desmond.imageshack.us/Himg339/scaled.php?server=339&filename=53171696.jpg&res=medium

## Homework Equations

GPE = mgh, EPE = λx^2 / 2l

## The Attempt at a Solution

I'm taking the horizontal line through A as the zero-level for potential energy. Clearly, the GPE of the rod is mgacosθ + constant. To find the elastic potential energy, I noted that:

[tex]EPE = \frac{0.5mgx^2}{4a}[/tex]

I said that the extension x is given by the length of BC minus the natural length. To get BC, I used the cosine rule;

BC

^{2}= 16a

^{2}+ 4a

^{2}- 16a

^{2}cosθ

So BC = 2a√(5 - 4cosθ).

Then,

[tex]EPE = \frac{mg(2a \sqrt{5 - 4 \cos \theta} - 2a)^2}{8a}[/tex]

Which, to me, simplifies to:

[tex]EPE = mga(3 - 2 \cos \theta - \sqrt{5 - 4 \cos \theta})[/tex]

The total potential energy V can thus be expressed as:

[tex]V = -mga(\cos \theta - 3 + \sqrt{5 - 4 \cos \theta}) + \mathbb{constant}[/tex].

This is almost identical to what they've got, except without the 3. Where have I gone wrong?

Thanks.

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