I Does small angle mean small angular velocity?

1. Aug 4, 2016

Happiness

Why is the term $(\sin\theta\,\dot{\theta})^2$ fourth order in the small $\theta$, as claimed by the sentence below (5.108)?

By small-angle approximation, $(\sin\theta\,\dot{\theta})^2\approx\theta^2\,\dot{\theta}^2$.

For this to be fourth order, it seems like we must have $\theta=\dot{\theta}$. Why is this true? What are the conditions for this to be true?

2. Aug 4, 2016

wrobel

why quantities of different dimension must be equal? They must not. David Morin wrote a good book but definitely not for high school

3. Aug 4, 2016

Happiness

So is he right or wrong?

4. Aug 4, 2016

wrobel

My advice is that you should first think about your own understanding whether it is right or wrong.

5. Aug 4, 2016

Happiness

My understanding is that $(\sin\theta\,\dot{\theta})^2$ is second order in $\theta$. But that would mean that it cannot be ignored, since (5.109) contains second-order terms.

6. Aug 4, 2016

wrobel

The degree of the term $\theta^2\dot \theta^2$ is equal to 2+2=4. It is a degree of polynomial $P(\theta,\dot \theta)= \theta^2\dot \theta^2$. Both quantities $\theta,\dot \theta$ are assumed to be small. The approximation of the Lagrangian up to the second order terms corresponds to linearization of the Lagrange equations in the vicinity of equilibrium

Last edited: Aug 4, 2016
7. Aug 4, 2016

Staff: Mentor

Here it is important that θ is small, but the timescale is not. $\dot \theta$ is "something with θ divided by time", which is small of θ is small.