Better approximations with two sticks

How approximation ought to be done always confuses me. Considering this problem in a textbook:

Two massless sticks of length $2r$, each with a mass $m$ fixed at its middle, are hinged at an end. One stands on top of the other. The bottom end of the lower stick is hinged at the ground. They are held such that the lower stick is vertical, and the upper one is tilted at a small angle $ε$ with respect to the vertical. They are then released. At this instant, what are the angular accelerations of the two sticks? $ε$ is very small.

Short vision: It seemed to me, in the solution, the author (David Morin) has done approximation to the Lagrange with first order approximation to the Kinetic energy term, and then second order approximation to the potential energy term.

I have two questions, and listed them at the bottom of this post.

Long vision: the kinetic energy of the Lagrangian explicitly is:

$$\frac{1}{2}mr^2[(2\dot\theta_1cos\theta_1-\dot\theta_2cos\theta_2)^2+(-2\dot\theta_1sin\theta_1-\dot\theta_2sin\theta_2)^2]+\frac{1}{2}mr^2\dot\theta_1^2$$

To my understanding,

With 1st order approximation $^{\dagger}$, we obtain:

$$\frac{1}{2}mr^2(2\dot \theta_1 − \dot \theta_2)^2+\frac{1}{2}mr^2\dot\theta_1^2$$

The potential energy is:

$$mgr(3cos\theta_1 + cos\theta_2)$$

with 2nd order approximation:

$$mgr(4− \frac{3}{2}θ_1^2 − \frac{1}{2}θ_2^2)$$

Hence obtain EOMs: $$5\ddot\theta_1 − 2\ddot\theta_2 = \frac{3g}{r}\theta_1$$ $$-2\ddot\theta_1 + 1\ddot\theta_2 = \frac{g}{r}\theta_1$$

With the initial conditions $\theta_1=0$ and $\theta_2=\varepsilon$

$$\ddot θ_1=\frac{2g\varepsilon}{r}$$ $$\ddot θ_2=\frac{5g\varepsilon}{r}$$

I used a different approach of approximation -- I used the small angle approximation right at the beginning, and assuming at the very instant the upper stick just behaved like a mass $m$ on top of a long stick with length $3r$ -- then obtained an answer in a same mathematical form, but with a difference by one order of magnitude, ($\ddot \theta_1 = g\varepsilon/3r$).

Now, I am so profoundly confused in two sense: one technically, another methodically.

1. Technically, can we make different orders-approximations in one single equation?
2. Methodically, how do I know which one is a better approximation? (In other words, how to make better approximation?)

$\dagger$ He explained:

We can simplify this, using the small-angle approximations. The terms involving sin θ are fourth order in the small θ’s, so we can neglect them. Also, we can approximate cos θ by 1, because this entails dropping only terms of at least fourth order. So the top mass’s kinetic energy becomes $\frac{1}{2}mr^2(2\dot\theta_1 − \dot\theta_2)^2$

It rarely makes sense to make different order approximations in a single equation, because either the crude approximation is sufficient, then use it everywhere, or it is not, then don't use it anywhere. There are, however, subtleties where it may seem like the orders are different, but in getting to the result they are not.

In this specific example, they are all 3rd order approximations, i.e. the equations need a ${}+\mathcal{O}(\theta^4)$, because of the squares in the kinetic energy expression. When he says 1st order, that is technically not correct, since it is first order in a term that is then squared, making it second order (furthermore the 3rd order terms disappear, making it effectively 3rd order, as mentioned above).

I really can't follow your approach. There is clearly a hinge that will influence the motion, even at the first instant, so you can't just approximate it as being stiff. This would be a 0th order approximation in $\ddot{\theta}_2$, since you basically say $\ddot{\theta}_2 = 0$. However, we know nothing about $\ddot{\theta}_2$, which makes it a bad variable to approximate. But we know $\theta_1(0)$ and $\theta_2(0)$, which makes them good variables for approximations.

A good rule of thumb is to make your approximations as late as possible (i.e. when you run out of math skill or patience). Start out with exact expressions and when you are sure of all the places $\theta$ will finally appear in, make your approximations to whatever order you wish. Be careful to only approximate the variable that you actually know to be small, and do not imply anything about other variables or time derivatives.

• thanks! just to clarify: the words "1st order" and "2nd order" are my words, probably my misunderstanding. The author's words are in the quotation at the bottom . // I think I made a bad approximation though, I seemed assumed a bit too many things. The errors can be added up to quite high. – Shing May 24 '17 at 13:55

Having read Lawrence Krauss's Fear of Physics, I found some words of his will shed some light on my second question about how to make better approximations. Although he is mostly referring to modeling, since modeling and approximating are closely related to each other, I will share Krauss' words here:

……Before doing anything else [in working of science], abstract out all the irrelevant!

There are two operatives [about modeling] here: abstract and irrelevant. Getting rid of irrelevant details is the first step in building any model of the world, and we do it subconsciously from the moment we are born. Overcoming the natural desire to not throw out unnecessary information is probably the hardest and the most important part of learning physics. In addition, what may be irrelevant in a given situation is not universal but depends in most case on what interests you. This leads us to the second operative word - abstraction. Of all the abstract thinking required in physics, the most challenging probably lies in choosing how to approach a problem. The mere description of movement along a straight line required enough abstraction that it largely eluded some pretty impressive intellects until Galileo……

After a few pages, he wrote on...

How do you know in advance what is essential from what you can safely throw out? Often you don’t. The only way to find out is to go ahead as best you can and see if the result make sense. In words of Richard Feynman, “Damn the torpedoes, full speed ahead!”