Double pendulum

1. Mar 21, 2007

zell99

1. The problem statement, all variables and given/known data
A double pendulum consists of light, inextensible strings, AB and BC each of length l. It is fixed at one end A and carries two particles, each of mass m, which hang under gravity. The pendulum is constrained to move in a vertical plane. The angle between the vertical and AB is $\theta$, which the angle between BC and the vertical is $\phi$. Show for smll angles about the equilibirium position:
$d^2\theta/dt^2 +g/l(2\theta -\phi)=0$
$d^2\phi/dt^2 +2g/l(\phi - \theta)=0$

2. Relevant equations
Newton's second law.
I shouldn't need to use Langrangian mechanics.

3. The attempt at a solution
I've managed to derive the first equation, first by assuming the tension in BC is mg (small angle approximation) then resolving forces about the top mass, and using the small angle approximation for sin. But I just can't get the second result out. I've tried to do the same this (resolving forces) but I must be making a mistake somewhere. Any help would be appreciated. Thanks

Last edited: Mar 22, 2007
2. Mar 21, 2007

e(ho0n3

Strange. I get the following for the force in the tangential direction of the mass at the end of BC:

$$ma = m \frac{dv}{dt} = m l \, \frac{d^2 \phi}{dt^2} = mg \sin \phi \approx mg \phi$$

Rearranging gives me:

$$\frac{d^2 \phi}{dt^2} - \frac{g}{l} \, \phi = 0$$

This does not look like what you want to show. Hmm...

Last edited: Mar 21, 2007
3. Mar 22, 2007

zell99

Thnaks for the reply. I think what you're missing, and what I can't get the right value for, is that now the tension is no longer perpendicular to the gravitational force, due to the combination of strings, for the bottom mass. This will lead to an extra term but I can't get it to come out correctly.
If anyone knows I'd really appreciate it.
Thanks

4. Mar 22, 2007

e(ho0n3

Who said the tension is perpendicular to the graviational force? The tension is always parallel to the string.

5. Mar 22, 2007

zell99

Sorry my fault, should have been more accurate with words, (I don't think what I meant was right either). I'm still confused so if someone is able to derive or give me a hint towards the second equation I'd be really grateful.
Thanks

6. Mar 23, 2007

zell99

7. Mar 23, 2007

J77

It's hard to say where you've made the mistake.

There's obviously more work involved using Newton's equations than Lagrange's -- therefore, you may have to post all your working for help...

(At soem point, you should have 4 equations for $$\ddot\theta, \dot\theta^2, \ddot\phi, \dot\phi^2$$ from which the tensions in terms of $$\ddot\theta, \ddot\phi, \theta, \phi$$ can be found...)

8. Mar 23, 2007

zell99

Thanks for the reply. I gave up in the end and used the Langrange.