Page 232 - Biomedical Engineering and Design Handbook Volume 1, Fundamentals

BIODYNAMICS: A LAGRANGIAN APPROACH 209

c 1 cos(θ 2 − θ 1 ) sin(θ 2 − θ 1 ) 0 b 1
c = −sin(θ − θ ) cos(θ − θ ) 0 b (8.68)
θ
2 2 1 2 1 2
c 3 0 0 1 b 3
From Eq. (8.68), the velocity at point G becomes
2

θ
v = l θ sin( θ − θ ) c + ⎡ l θ cos( θ − θ ) + l 4 ⎤ c
G 2 31 2 1 1 ⎢ ⎣ 31 2 1 2 2 2 ⎥ ⎦ 2
⎡ l 4 ⎤
− Ω( sinl θ + )l + Ωl + Ωsinθ c (8.69)
⎢ ⎣ 3 1 2 2 2 2 ⎥ ⎥ ⎦ 3
Substituting Eqs. (8.61), (8.62), and (8.69) into Eq. (8.63), the equation for the kinetic energy of
segment CD, and solving gives
1 2 2 1 1 2 2
T = m l θ + m l l θ θ cos( θ 2 − θ ) + m l θ
1
234 1 2
2 24 2
2
2 3 1
2 2 6
1 2 2 2 1 2 2
+ ml Ω sin θ + ml l Ω sin θ sinθ + mll Ω sinθ 1
n
2 3 4
2
1
23
1
2 2 3
2 2
1 2 2 1 2 1 2 2 2
+ ml Ω + m l l Ω sinθ + ml Ω sin θ (8.70)
2 12 2 224 2 2 6 2 4 2
and the potential energy of segment CD is determined to be
l 4
V =− m gl cosθ − m g cosθ (8.71)
2 2 3 1 2 2
2
The total kinetic and potential energy of the system can be determined by summation of
T = T + T (8.72)
1 2
and V = V + V (8.73)
1 2
respectively, where the Lagrangian for the system is given by
L = (T + T ) − (V + V ) = T − V (8.74)
1 2 1 2
It is left to the reader to apply Lagrange’s equation (8.11), using the two generalized coordinates of
the two-segment system, q =θ and q =θ , to work through the mathematics involved in solving
1 1 2 2
Eq. (8.11) to yield the equations of motion. For the first generalized coordinate, q =θ , the equation
1 1
of motion is solved to be
1 2 1 1
2
)l θ
( m + m 2 3 1 + m l l cos(θ 2 − θ 1 )θ 2 − ( m + m l ) Ω 2 sinθ 1 cosθ 1
m
2 3 4
1
1
2 3
3 2 3
1 2 1
+
− ( m + m l l Ω cosθ + ml l θθ sin θ cos θ 2
)
1
2
234 1 2
1
2 3
1
2 2
1 2 1
Ω
− ml l θθ( + Ω )cosθ sinθ + ( m + m )gl sinθ = 0 (8.75)
2 234 1 2 1 2 2 1 2 3 1
For the second generalized coordinate, q =θ , the equation of motion is solved to be
2 2
1 2 1 1
ml θ + ml l cos( θ − θ θ + ml l θθ cos θ sin θ 2
)
1
2 34
2
24 2
1
2
2 34 12
3 2 2
1 2 1 2 2
− ml l ( θθ + Ω )sin θ coosθ − ml Ω sinθ 2 cosθ 2
2 4
1 2
1
234
2
2 3
1 2 1
− mll Ω cosθ + mgl sinθ= 0 (8.76)
224
2
4
2
2
2 2 2