332
7 Lagrangian and Hamiltonian Mechanics
Lagrange’s equation
d
dt
∂ L
∂ ˙
θ
−
∂ L
∂θ
= 0
yields
m(a
2 sin
2
θ + b
2 cos
2
θ) ¨
θ + m(a
2 sin θ cos θ − b
2 sin θ cos θ) ˙
θ
2
+ mgb cos θ + k(−a
2 sin θ cos θ + b
2 sin θ cos θ) = 0
or m(a
2 sin
2
θ + b
2 cos
2
θ) ¨
θ − (a
2
− b
2
)(k − m ˙
θ
2
) sin θ cos θ
+ mgb cos θ = 0
( 6 )
(ii) Equilibrium point is located where the force is zero, or ∂ V /∂θ = 0.
Differentiating (4)
∂ V
∂θ
= mgb cos θ + k(b
2
− a
2
) sin θ cos θ
(7)
Clearly the right-hand side of (7) is zero for θ = ±
π
2
Writing (7) as
[mgb + k(b
2
− a
2
) sin θ ] cos θ
(8)
Another equilibrium point is obtained when
sin θ =
mgb
k(a 2 − b 2 )
(9)
provided a > b.
(iii) An equilibrium point will be stable if
∂ 2 V
∂θ 2 > 0 and will be unstable if
∂ 2 V
∂θ 2 < 0. Differentiating (8) again we have
∂ 2 V
∂θ 2 = k(b
2
− a
2
)(cos
2
θ − sin
2
θ) − mgb sin θ
(10)
For θ =
π
2
, (10) reduces to
k(a
2
− b
2
) − mgb
(11)
Expression (11) will be positive if a 2 > b 2 +
mgb
k
, and θ =
π
2
will be
a stable point.