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Appendix: All Solutions to the Problems
The Lagrange function is
L = T − V =
3
4
m(R − r )
2 ˙
θ
2
+ mg(R − r ) cos θ
The corresponding functional is
J =
t 1
t 0
3
4
m(R − r )
2 ˙
θ
2
+ mg(R − r ) cos θ
dt
According to the Hamilton’s principle, there is
δ J =
t 1
t 0
Ldt = 0
The Euler equation of the functional is
3
2
m(R − r )
2 ¨
θ + mg(R − r ) sin θ = 0
or
3
2
(R − r ) ¨
θ + g sin θ = 0
Of course, this equation can also be directly obtained by the results of Example
9.6.6.
When the circular cylinder is in the small vibration, there is sin θ = θ , from the
above equation, we get
¨
θ +
2g
3(R − r )
θ = 0
This is the differential equation of small vibration for the circular cylinder. Its
vibration period is
T = 2π
3(R − r )
2g
9.13 As shown in problem 9.13 figure, find the differential equation of motion of a
spherical pendulum of length L.
Solution Choosing the spherical coordinates, regarding the spherical pendulum as
a particle. The kinetic energy, potential energy and Lagrange function of the particle
are respectively