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fluidistic
#1
Apr25-10, 05:14 PM
PF Gold
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P: 3,188
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
[tex]\vec v = \vec \omega \wedge \vec r[/tex].
[tex]L=L_1+L_2+L_3[/tex].

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 [tex]m_1[/tex], I've found that [tex]T=\frac{m_1}{2} \Omega ^2 l^2 \sin ^2 (\phi)[/tex] since they describe a circular motion of radius [tex]l \sin \phi[/tex]. So it only remains to find the kinetic energy of [tex]m_2[/tex].
Choosing the origin at point A, [tex]x=2l \cos \phi[/tex]. Now to calculate [tex]\dot x[/tex], I think that only [tex]\phi[/tex] may vary thus [tex]\dot x = -2l \dot \phi \sin \phi[/tex]. So I get [tex]T_2=2m_2 l^2 \sin ^2 (\phi )[/tex].
Which gives me [tex]L=l^2 \sin ^2 (\phi) (m_1 \Omega ^2 +2 \dot \phi ^2 m_2)[/tex].
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.
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