Does small angle mean small angular velocity?

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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?

Screen Shot 2016-08-04 at 5.14.16 pm.png
 

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  • #2
wrobel
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For this to be fourth order, it seems like we must have θ=˙θ\theta=\dot{\theta}. Why is this true?
why quantities of different dimension must be equal? They must not. David Morin wrote a good book but definitely not for high school
 
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why quantities of different dimension must be equal? They must not. David Morin wrote a good book but definitely not for high school

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

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.
 
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wrobel
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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
 
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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.
 

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