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Classical Mechanics, Second Edition, 2nd Edition
book

Classical Mechanics, Second Edition, 2nd Edition

by Tai L. Chow
May 2013
Intermediate to advanced content levelIntermediate to advanced
639 pages
23h 12m
English
CRC Press
Content preview from Classical Mechanics, Second Edition, 2nd Edition
100 Classical Mechanics
© 2010 Taylor & Francis Group, LLC
Solution:
Using spherical coordinates as indicated in Figure 4.9, we see that the kinetic energy of the par-
ticle is
Tmll=+
()
1
2
22 22 2
θφ θsin
.
If we take the support point as the reference level for the potential energy, then we have
V = mglcos θ
and the Lagrangian of the particle is
Lmll mgl=+
()
1
2
22 22 2
θφ
θθ
si
nc
os .
There are two degrees of freedom; θ and φ are the two independent generalized coordinates.
From Lagrange’s equation for coordinate θ, we obtain
ml ml mgl
222
0

θφθθ θ
−−
=sincos sin
. (4.22)
Similarly, we have the equation of motion for coordinate φ:
ml ml
22 2
20

φθ θφ
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Publisher Resources

ISBN: 9781466569980