View Single Post
Apr25-10, 05:14 PM
PF Gold
fluidistic's Avatar
P: 3,203
1. The problem statement, all variables and given/known data
The problem can be found there, page 12, problem 4:

2. Relevant equations
[tex]\vec v = \vec \omega \wedge \vec r[/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.
Phys.Org News Partner Science news on
What lit up the universe?
Sheepdogs use just two simple rules to round up large herds of sheep
Animals first flex their muscles