Page 237 - Biomedical Engineering and Design Handbook Volume 1, Fundamentals
P. 237

214  BIOMECHANICS OF THE HUMAN BODY

                       Substituting Eqs. (8.79) and (8.80) into the Lagrangian, Eq. (8.3), results in
                                                    1  2   2  1  2
                                               −
                                           L =  T V =  mr θ  +  I θ  +  mgr cos θ 1
                                                       G 1 1
                                                             G 1 1
                                                    2       2        G 1                  (8.81)
                       Substituting Eq. (8.81) into Eq. (8.2) with k =θ results in the equation of motion for the upper arm
                                                         1
                       (assuming that the torso is fixed), as shown in Eq. (8.82). Q would include all nonconservative or
                                                                  θ
                       externally applied torques.

                                                 2
                                               mr θ +  I θ +  mgr sin θ =  Q θ            (8.82)
                                                      G 1 1
                                                                  1
                                                G 1 1
                                                             G 1
                       where I = mr 2 G 1  + I , which can be found by using the parallel axis theorem.
                            B
                                     G 1
                         In a similar manner to the derivation of the three-segment system of Fig. 8.5, a coordinate trans-
                       formation matrix is defined in order to convert between coordinate systems.  A transformation
                       between the b frame and the inertial frame for a single rotation, θ , about the b axis in matrix form
                                                                                3
                                                                      1
                       is shown in Eq. (8.83).
                                                b 1  sinθ 1  −cosθ 1  0  i
                                                b 2  = cosθ 1  sinθ 1  0  j               (8.83)
                                                b 3    0     0    1  k
                       This results in the following equations:
                                                  b = sinθ  i − cosθ  j
                                                   1     1     1
                                                  b = cosθ 1 i + sinθ 1 j
                                                   2
                                                  b =  k                                  (8.84)
                                                   3
                                                   i = sinθ 11  12
                                                         b + coosθ b
                                                   j =− cosθ b + sinθ b
                                                          11
                                                                 1 2
                       In transferring a velocity or acceleration from one table to the next, a conversion between frames, as
                       shown in Eq. (8.68), or a conversion from the b frame to the inertial frame as in Eqs. (8.83) and
                       (8.84), and then from the inertial into the c frame, may be used.
           8.5.2  Single Elastic Body with Two Degrees of Freedom
                       Table 8.4 is the kinematics table for the single elastic body pendulum shown in Fig. 8.3. The absolute
                       velocity of the center of mass  G is required to complete the Lagrangian approach, and the
                                                1
                       Newtonian approach utilizes the absolute acceleration of point G , Eq. (8.86), in F = ma G 1  for
                                                                        1
                       problems of constant mass.
                         The absolute velocity and acceleration of G are expressed in terms of the body-fixed coordinate
                                                        1
                       system, b , b , b , which is fixed to the pendulum and rotates about the b axis as before. Although
                                                                            3
                              1
                                2
                                   3
                       it is equivalent to expressing within the inertial frame of reference, i, j, k, the body-fixed coordinate
                       system, b , b , b , uses fewer terms. The velocity and acceleration for G are respectively as follows:
                             1  2  3                                      1
                                                   v  =  r   b  +  r θ    b               (8.85)
                                                    G 1  G 1 1  G 1 1 2
                                                                    G 1 1)

                                                               θ
                                                 r −
                       and                  a G 1  = (    G 1  r θ 2    b )  1 +  r (  G 1 1  +  r 2  θ    b 2  (8.86)
                                                     G 1 1
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