Potential Energy of System (Uniform Rod and String)

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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:

EPE = \frac{0.5mgx^2}{4a}

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;

BC² = 16a² + 4a² - 16a²cosθ

So BC = 2a√(5 - 4cosθ).

Then,

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

Which, to me, simplifies to:

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

The total potential energy V can thus be expressed as:

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

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

Thanks.

 

[FeDeX_LaTeX]

Sigh... just realized in the mark scheme they've counted the +3 as 'a constant' so negligible. Ugh. Sorry to whoever was reading this thread.