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Appendix: All Solutions to the Problems
or
r ¨
θ − r ω
2 sin θ cos θ − g sin θ = 0
Of course, the problem can directly obtain the differential equation of motion for
the particle by the Lagrange equation, it does not have to establish the functional.
9.15 Given a spring-mass system composed of the linear spring of mass m and
stiffness k, as shown in problem 9.15 figure. Find the differential equation of the
system.
Solution The kinetic energy and potential energy of the system are respectively
T =
1
2
m ˙
x
2
, V =
1
2
kx
2
The Lagrange function is
L = T − V =
1
2
m ˙
x
2
−
1
2
kx
2
Substituting the above into the Lagrange equation, we get
∂ L
∂ x
−
d
dt
∂ L
∂ ˙
x
= −kx − m ¨
x = 0
namely the differential equation of motion of the system is
m ¨
x + kx = 0
9.16 A system consists of the simple pendulum of length l, mass m and the block
of mass 2m, the simple pendulum swings at point O of the block, as shown in
problem 9.16 figure. The block can frictionlessly slide on the horizontal plane. Find
the equation of motion of the system.
Solution The kinetic energy of the system is the sum of the kinetic energy of the
block and the kinetic energy of the single pendulum, namely
T = m ˙
x
2
+
1
2
m( ˙
x + l ˙
θ)
2
=
3
2
m ˙
x
2
+ ml ˙
x ˙
θ +
1
2
ml
2 ˙
θ
2
The potential energy of the system is gravitational potential energy, it can be
expressed as
V = mgl(1 − cos θ)
The Lagrange function is