Deriving the small-x approximation for an equation of motion

[Abhishek11235]

Homework Statement

The problem is taken from Morin's book on classical mechanics. I found out Lagrangian of motion. Now to solve, we need small angle and small x approximation. The small angle approximation is easy to treat. But how to solve small x approximation i.e how do I apply it?

Homework Equations

Given $x/l \lt\lt 1$ we need to solve:
$$Ml^2\ddot\theta +ml(l\ddot\theta + \ddot x)+mx^2\ddot\theta+2mx\dot\theta\dot x=-(M+m)glsin\theta - mgxcos\theta$$

The Attempt at a Solution

[/B]
I tried to divide whole expression and since ##x\lt\lt

l##,I ignored these terms. But the answer was wrong. How do I solve it?

 

[Dr. Courtney]

Defining your variables, perhaps including a picture, are necessary.

I suspect one can simply relate x, l, and theta in the case of sufficiently small x.

 

[Ray Vickson]

Homework Statement

The problem is taken from Morin's book on classical mechanics. I found out Lagrangian of motion. Now to solve, we need small angle and small x approximation. The small angle approximation is easy to treat. But how to solve small x approximation i.e how do I apply it?

Homework Equations

Given $x/l \lt\lt 1$ we need to solve:
$$Ml^2\ddot\theta +ml(l\ddot\theta + \ddot x)+mx^2\ddot\theta+2mx\dot\theta\dot x=-(M+m)glsin\theta - mgxcos\theta$$

The Attempt at a Solution

I tried to divide whole expression and since ##x\lt\lt View attachment 237939 l##,I ignored these terms. But the answer was wrong. How do I solve it?[/B]

If $x$ and $\theta$ (and their derivatives) are small, you can neglect products of them, so you have $x^2 \ddot \theta \approx 0$ and $x \dot \theta \dot x \approx 0$. Don't forget that $\sin \theta \approx \theta$ and $x \cos \theta \approx x.$

 

[Abhishek11235]

If $x<<l$ implies $/dot x<<$?

If $x$ and $\theta$ (and their derivatives) are small, you can neglect products of them, so you have $x^2 \ddot \theta \approx 0$ and $x \dot \theta \dot x \approx 0$. Don't forget that $\sin \theta \approx \theta$ and $x \cos \theta \approx x.$

 

[Ray Vickson]

If $x<<l$ implies $/dot x<<$?

No, obviously not. The function $x(t) = 0.05 \sin(1000 t)$ satisfies $|x| \leq 0.05,$ but $x'(t) = 200 \cos(1000 t)$ is not always small, and that is even more true for $x''(t) = -200,000 \sin(1000 t).$ So, no: what I said was "If $x$ and $\theta$ (and their derivatives) are small", then … Somehow you need to get some decent bounds on the derivatives as well, so the problem is not yet solved.