
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
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
22 22 2
θφ
si
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
θφθθ θ
=sincos sin
. (4.22)
Similarly, we have the equation of motion for coordinate φ:
ml ml
22 2
φθ θφ