Problem 4, Landau/Lifshitz, page 12

[fluidistic]

Homework Statement

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.

Homework Equations

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

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.

 

[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.

 

[fluidistic]

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.

Ok thanks a lot. I will remember this.

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?

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.

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

Yes I meant this, I made a typo error.

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.

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

 

[gabbagabbahey]

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.

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.

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

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