# Why are angular velocities of double pendulum small in small angle approximation? [duplicate]

In the lagrangian for double pendulum for small angles, the term $\dot{\theta}_1\dot{\theta}_2 \left [ 1-\frac{(\theta_1-\theta_2)^2}{2} \right ]$ is replaced with $\dot{\theta}_1\dot{\theta}_2$, because $\dot{\theta}_1\dot{\theta}_2 \frac{(\theta_1-\theta_2)^2}{2}$ is neglected. The product $\dot{\theta}_1\dot{\theta}_2$ has the second order of smallness, but why? This comment explains for simple pendulum, but says that it is more complicated for double pendulum. What is explanation for double pendulum?

Edit: Correct me if I am wrong, but I think I found the answer. If double pendulum starts oscillating from rest, the potential energy at that moment is $E_{pm}=m_1gh_{1i}+m_2gh_{2i}$, where $h_{1i}$ and $h_{2i}$ are lengths from reference line to centres of masses of two pendulums. In small angles aproximation these heights are $h_{1i}=l_1+l_2-l_{cm1}+\frac{\theta_{1i}^2}{2}l_{cm1}$ and $h_{2i}=l_2-l_{cm2}+l_1\frac{\theta_{1i}^2}{2}+l_{cm2}\frac{\theta_{2i}^2}{2}$. Kinetic energy of first pendulum is $E_{k1}=E_{pm}-E_{k2}-E_{p1}-E_{p2}=\frac{m_1v_1^2}{2}+\frac{I_1\dot{\theta_1}^2}{2}=\frac{m_1l_{cm1}^2+I_1}{2}\dot{\theta_1}^2$. Kinetic energy of second is $E_{k2}=\frac{m_2v_2^2}{2}+\frac{I_2\dot{\theta_2}^2}{2}=\frac{m_2l_{cm2}^2+I_2}{2}\dot{\theta_2}^2$. $E_{k1}=\frac{g}{2}((\theta_{1i}^2-\theta_{1}^2)(m_1l_{cm1}+m_2l_1)+m_2l_{cm2}(\theta_{2i}^2-\theta_{2}^2))-E_{k2}$ $\dot{\theta_1}=\sqrt{\frac{g((\theta_{1i}^2-\theta_{1}^2)(m_1l_{cm1}+m_2l_1)+m_2l_{cm2}(\theta_{2i}^2-\theta_{2}^2))-2E_{k2}}{m_1l_{cm1}^2+I_1}}=\sqrt{\frac{2E_{k1}}{m_1l_{cm1}^2+I_1}}$
$\dot{\theta_2}=\sqrt{\frac{g((\theta_{1i}^2-\theta_{1}^2)(m_1l_{cm1}+m_2l_1)+m_2l_{cm2}(\theta_{2i}^2-\theta_{2}^2))-2E_{k1}}{m_2l_{cm2}^2+I_2}}=\sqrt{\frac{2E_{k2}}{m_2l_{cm2}^2+I_2}}$
Masses and lengths have influence, but $\dot{\theta_1}$ and $\dot{\theta_1}$ should be small because terms $(\theta_{1i}^2-\theta_{1}^2)$ and $(\theta_{2i}^2-\theta_{2}^2)$ are small. Also when $\dot{\theta_1}$ is maximal, $E_{k1}$ is maximal, so $\dot{\theta_2}$ will be smaller. Similar is with $\dot{\theta_1}$. $\dot{\theta_1}$ and $\dot{\theta_1}$ will not be maximal in same time, so their product will be small.

## marked as duplicate by Kyle Kanos, ZeroTheHero, Jon Custer, user259412, heatherAug 6 '17 at 18:30

• Is it ignored because $\dot\theta_1\dot\theta_2\ll1$ or $\theta_1\approx\theta_2$? – Kyle Kanos Aug 6 '17 at 12:10
• Possible duplicate of Small oscillations of the double pendulum – Kyle Kanos Aug 6 '17 at 12:12
• Because $\dot\theta_1\dot\theta_2\ll1$. – LEM Aug 6 '17 at 12:13

I think that $\dot{\theta}$'s are not neccesarily small, but actually what is small is the term $\frac{(\theta_1-\theta_2)^2}{2}$.
So I guess that the brackets are $\simeq [1+\ \sim 0 ]\approx 1$, and consequently you only have the double $\omega$ product.
The simplest way to keep track of "smallness" is to introduce a dummy parameter $\epsilon$ and replace $\theta\to \epsilon\theta$ everywhere. It then follows that $$\dot\theta_i\to \epsilon\dot\theta_i\, .$$ Using this substitution in the Lagrangian, and expanding to quadratic order in $\epsilon$ will produce the correct linearlized equations of motion. (The assumption is that the equilibrium position is at $\theta_i=0$.)
With this, for instance, the term $\dot\theta_1\dot\theta_2\left(1-\frac{1}{2}(\theta_1-\theta_2)^2\right)$ in the Lagrangian becomes \begin{align} \dot\theta_1\dot\theta_2\left(1-\frac{1}{2}(\theta_1-\theta_2)^2\right) &\to \epsilon^2\dot\theta_1\dot\theta_2\left(1-\frac{1}{2}\epsilon^2(\theta_1-\theta_2)^2\right)\, ,\\ &= \epsilon^2\dot\theta_1\dot\theta_2 \end{align}