# Small angle approximation.

1. Jan 19, 2007

### quasar987

In the context of a Lagrangian mechanics problem (a rigid pendulum of lenght l attached to a mass sliding w/o friction on the x axis), I found the following equations of motion and now I must solve them in the small oscillation limit. (I know the equations are correct)

$$(m_1+m_2)\ddot{x}+m_2l\ddot{\theta}\cos(\theta)-m_2l\dot{\theta}^2\sin(\theta)=0$$

$$l\ddot{\theta}+\ddot{x}\cos(\theta)+g\sin(\theta)=0$$

I know that small thetas mean $\cos\theta \approx 1$ and $\sin\theta\approx \theta$ but why can we say that $\dot{\theta}^2\approx 0$? The angle can be small and nevertheless vary furiously fast. What indicates that if theta is small, the so is its derivative?

2. Jan 19, 2007

### marcusl

Well for one thing the potential energy at the top of the swing must be small if theta is small, so the KE (0.5*m*v^2) at the bottom must be small as well.

3. Jan 19, 2007

### quasar987

But there is also the kinetic energy of the sliding block in the expression of the energy:

$$E=-m_2gl\cos\theta+\frac{1}{2}m_1\dot{x}^2+\frac{1}{2}m_2(\dot{x}^2+2l\dot{x}\dot{\theta}\cos\theta+(l\dot{\theta})^2)$$

Can one extract $\theta\approx 0 \Rightarrow \dot{\theta}^2\approx 0[/tex] from that? 4. Jan 19, 2007 ### marcusl I think so. theta is a function of time, and as with all oscillators the energy sloshes between PE and KE so PE_max=KE_max. You have $$PE_m=-m_2gl\cos\theta_m$$ is small, so $$KE_m=\frac{1}{2}m_1\dot{x}^2+\frac{1}{2 }m_2(\dot{x}^2+2l\dot{x}\dot{\theta}\cos\theta+(l \dot{\theta})^2)$$ is also small. The KE of m1 makes KE of the pendulum that much smaller. You already know that cos(theta) is one, so with the energy in the first terms the last term implies that theta_dot^2 must also be small. EDIT: corrected last sentence EDIT2: can't get equation to render right.. Last edited: Jan 19, 2007 5. Jan 19, 2007 ### D H Staff Emeritus Are [itex]x$ and $\theta$ truly independent variables, as implied by your original post? Rather than assume $\ddot\theta\approx0$ (which is not correct, otherwise you would not get harmonic motion), eliminate one of either $x$ or $\theta$ by use of the small angle approximation.

6. Jan 20, 2007

### SGT

Since $$\theta \approx sin \theta$$, $$\dot{\theta} \approx cos \theta \approx 1$$. I don't see how $$\dot{\theta}^2 \approx 0$$

7. Jan 20, 2007

### arildno

Let us use the small angle approximation, and formulate a 2 order diff. eq for the angle. We get:
$$\ddot{\theta}+b\theta\dot{\theta}^{2}+\omega^{2}\theta=0, \omega^{2}=\frac{(m_{1}+m_{2})g}{m_{1}l}, b=\frac{m_{2}}{m_{1}}$$
Suppose that we DO chop away the term including the first derivative.
The typical solution will therefore be of the form: $\theta(t)=A\cos(\omega{t}), A<<1$
where the inequality is the small angle approximation.
Note now that the order of magnitude of the retained terms in the diff.eq is $A\omega^{2}$ whereas the order of magnitude of the chopped away term is $A^{3}\omega^{2}<<A\omega^{2}$
that is, the chopping procedure yields a CONSISTENT dominant balance picture.

The other two possible two-term dominant balance procedures won't give a consistent picture.