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 PF Gold P: 3,165 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=...page&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.