Page 229 - Biomedical Engineering and Design Handbook Volume 1, Fundamentals

206 BIOMECHANICS OF THE HUMAN BODY

Substitution of Eqs. (8.38), (8.43), and (8.44) into Eq. (8.42) yields
b + Ω sinθ
b + θ
v =− l Ω b + − Ω( cosθ 1 1 1 2 1 3 l 3 b 1 (8.45)
b ×)
2
3
2
G 1
which is solved to be
l 3 l ⎛ sin θ ⎞
b − Ω
v = θ 12 3 1 + l 2 b 3 (8.46)
G 1 ⎝ ⎠
2 2
Therefore, Eq. (8.41) can be written by using Eqs. (8.38), (8.39), (8.40), and (8.46):
2
1 1 2 2 2 1 2 2⎞ 1 ⎧ l l ⎛ ⎪ 3 ⎞ 2 ⎡ l ⎛ sin θ ⎞ ⎤ ⎪ ⎫
⎛

T = m l Ω sin θ 1 + m l θ + m ⎨ θ 1 +− Ω 3 1 l + 2 ⎥ ⎬ ⎬ (8.47)
1 3
1 3 1
1
⎝
1
2 12 12 ⎠ 2 ⎩ ⎪ ⎝ 2 ⎠ ⎢ ⎣ ⎝ 2 ⎠ ⎦ ⎪ ⎭
and simplified to the final form of the kinetic energy for the system:
1 2 1 2 2 1 2 l sinθ ⎞ 2
⎛
2
2
2
Ω
T = m l (Ω sin θ 1 + θ 1) + m l θ + m Ω 3 1 + l 2 (8.48)
1
1 3
1
13 1
24 8 2 ⎝ 2 ⎠
The potential energy of the system is simply
l
V =− m g 3 cosθ 1 (8.49)
1
1
2
The Lagrangian for segment BC, L , is subsequently determined to be
1
1 2 1 2 2
2
2
2
L = m l (Ω sin θ 1 + θ 1) + m l θ
1 3
1
13 1
24 8
⎛
1 2 l sinθ ⎞ 2 l
Ω
+ m Ω 3 1 l + 2 + m g 3 cosθ 1 (8.50)
1
1
2 ⎝ 2 ⎠ 2
Applying Lagrange’s equation (8.11) and using the only generalized coordinate for this system, θ ,
1
q =θ (8.51)
1 1
each term of Eq. (8.11) can easily be identified by differentiation of Eq. (8.50). The derivative of L
1
with respect to θ is solved accordingly:
1
∂L 1 1
2
2
2
2
1 = ml Ω sin θ cos θ + ml Ω sin θ coosθ
∂θ 1 12 13 1 1 4 1 3 1 1
1 2 1
+ ml l Ω cosθ − m gl 3 sinθ 1 (8.52)
1
1
12 3
2 2
which reduces to
1
1
∂L 1 = ml Ω sin θ cos θ + ml l Ω cos θ − mgl sinθ
1 1
2
2
2
∂θ 3 13 1 1 2 1 2 3 1 2 1 3 1 (8.53)
1
The derivative of L with respect to θ is
1 1
1
∂L 1 = 1 ml θ + ml θ = ml θ
1
2
2
2
∂θ 1 12 13 1 4 1 3 1 3 1 3 1 (8.54)