9.7 The Hamilton’s Canonical Equations
583
From Eq. (6), the generalized velocity expressed by the generalized momentum
can be obtained
˙
θ =
5
7m(a + b) 2 p
(7)
The Hamiltonian function expressed by the Hamiltonian variables θ and p is
H = p ˙
θ − L =
1
2
×
5
7m(a + b) 2 p
2
− mg(a + b)(1 − cos θ)
(8)
Substituting Eq. (8) into one of the canonical equations ˙
p = −
∂ H
∂θ
, we get
˙
p = mg(a + b) sin θ
(9)
Equations (7) and (9) are the two canonical equations.
Substituting Eq. (6) into Eq. (9) gets the second order differential equation of the
generalized coordinate θ
¨
θ =
5g
7(a + b)
sin θ
(10)
The tangential acceleration of the center of the ball is
a τ = (a + b) ¨
θ =
5
7
g sin θ
(11)
Example 9.7.5 The center of mass of the compound pendulum is at point C, point O
is the hanging axis, let OC = a, the moment of inertia of the compound pendulum
to the O axis is J O , as shown in Fig. 9.16. Find the differential equation of motion
of the compound pendulum with the canonical equation.
Fig. 9.16 Compound
pendulum diagram
O
a
C
mg
ϕ