# Simplifying Expression for Potential Energy of a Uniform Rod Attached by a Light String to a Fixed Point

I am self-studying through Mathematical Methods for Physics and Engineering and have come to a bit of a roadblock in chapter nine, on normal modes.

As pictured, the system is a uniform rod attached to a fixed point $P$ by way of a light string. The potential energy is given in the book as \begin{align} V &= Mlg\left[(1-\cos\theta_{1})+\frac{1}{2}(1-\cos\theta_{2})\right]\\ &\approx \frac{1}{4}Mlg\left(2\theta_{1}^{2}+\theta_{2}^{2}\right) \end{align}

My trouble is in transforming the first equation into the second; I have tried multiplying and dividing by the term in brackets but that didn't help. and I'm unsure of how to proceed.

Thank you!

Use the fact that $$1- \cos \theta = 2 \sin^2 \frac{\theta}{2} = 2 \left( \frac{\theta}{2} + O(\theta^2)\right)^2 = \frac{1}{2}\theta^2 + O(\theta^4) \textrm{ as } \theta \to 0$$ So for small $\theta_1, \theta_2$ $$(1-\cos \theta_1 ) + \frac{1}{2}(1-\cos \theta_2) \approx \frac{1}{2} \theta_1^2 + \frac{1}{4} \theta_2^2$$
• Taylor series expansion of $\sin \theta$. Commented Apr 12, 2017 at 1:12