# Problem 4, Landau/Lifshitz, page 12

1. Apr 25, 2010

### fluidistic

1. The problem statement, all variables and given/known data
The problem can be found there, page 12, problem 4: http://books.google.com.ar/books?id...ntcover&dq=mechanics&cd=1#v=onepage&q&f=false.

2. Relevant equations
$$\vec v = \vec \omega \wedge \vec r$$.
$$L=L_1+L_2+L_3$$.

3. The attempt at a solution
I've made an attempt and then saw the solution given in the book and I don't get it. First, there is no mention of a gratitational field, hence why is there a "g" term in the solution?
My attempt: None of the 3 particles have potential energy, hence the Lagrangian is simply the sum of the kinetic energies of the particles.
For each one of the 2 particles $$m_1$$, I've found that $$T=\frac{m_1}{2} \Omega ^2 l^2 \sin ^2 (\phi)$$ since they describe a circular motion of radius $$l \sin \phi$$. So it only remains to find the kinetic energy of $$m_2$$.
Choosing the origin at point A, $$x=2l \cos \phi$$. Now to calculate $$\dot x$$, I think that only $$\phi$$ may vary thus $$\dot x = -2l \dot \phi \sin \phi$$. So I get $$T_2=2m_2 l^2 \sin ^2 (\phi )$$.
Which gives me $$L=l^2 \sin ^2 (\phi) (m_1 \Omega ^2 +2 \dot \phi ^2 m_2)$$.
I wonder if my answer is correct if I assume no external gravitational field. L&L didn't specify the field but in the answer there's a "g"...
Thanks for any kind of help.

2. Apr 26, 2010

### gabbagabbahey

Usually whenever the word "vertical" is used in a mechanics problem, it's safe to assume there is an earth-like gravitational field $g$ present.

Even without a gravitational field present, wouldn't there be an additional contribution to the kinetic energy of each $m_1$ when $\theta$ is changing?

And your kinetic energy for $m_2$ makes no sense to me...did you mean $$T_2=2m_2 l^2\dot{\theta}^2\sin^2\theta$$?

Edit: Maybe your version of the text uses different labels for the angles...in my version, $$\Omega=\dot{\phi}$$ and $\theta$ represents the angle the massless rods make with the axis of rotation.

3. Apr 26, 2010

### fluidistic

Ok thanks a lot. I will remember this.
Hmm. Ok a (or 2?) term is missing. I think I've calculated the KE the masses have with their circular motion. It would remain the term of the vertical (and horizontal?) KE. Unless I'm misunderstanding something.
Yes I meant this, I made a typo error.

I see. So when you said "theta" in the upper question, which notation did you use?

4. Apr 26, 2010

### gabbagabbahey

Not a factor of 2, but a term involving $\dot{\theta}$, corresponding to the motion of each $m_1$ when the rotating frame is stretched or squished.

The one where theta is the angle the upper supports make with the axis of rotaion/