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0593_C11_fm  Page 382  Monday, May 6, 2002  2:59 PM





                       382                                                 Dynamics of Mechanical Systems



                                                                      n  1  θ            n  z



                                                                            n               n
                                                            θ                2 θ              2
                                                             1
                                                                     n  1r
                                                                                     n
                       FIGURE 11.10.4                                  θ              1
                                                                     2         n
                       Unit vector geometry for the double-                      2r
                       rod pendulum.

                                     n = cos θ 2 (  − θ  n )  + sin  2 (θ  − θ  n )  = cosθ  n + sinθ  n
                                       2r         1  1r          1  1θ     2  1    2  2
                                                                                              (11.10.14)
                                     n 2θ  = sin  2 (θ  − θ 1  n )  1r  + cos  2 (θ  − θ 1  n )  1θ  =−sinθ 2 n +cosθ  2 n  2
                                                                              1

                        Let the inertia forces on B  and B  be represented by forces  F *  *  passing through
                                                1     2                        1  and  F 2
                                                                                   *
                                                                                      *
                                                                                  F
                       G  and G  together with couples having torques  T *  and  T * . Then  ,  ,  T * , and  T *  are:
                        1      2                                    1      2       1  F 2  1    2
                                                                           ˙˙
                                             F =−  ma G r  =  m(l  2)θ 2 ˙  n  −  m(l  2) n
                                                                           θ
                                               *
                                              1                 1  r 1      1 1θ
                       or
                                                                                ˙˙
                                                      ˙˙
                                    *
                                                                 m
                                       m
                                   F = (l  2)( θ 2 ˙  1 cosθ 1 + θ 1 sinθ 1) n + (l  2)( θ 1 2 ˙  sinθ 1 − θ 1 cosθ 1) n 2  (11.10.15)
                                                               1
                                    1
                                                          ˙˙
                                                                                   ˙˙
                                    *
                                                                m
                                                                             m
                                   F =−m  a G 2  = mlθ 2 ˙  n − mlθ  n + (l  2)θ 2 ˙  n − (l  2)θ  n
                                    2             1  1r    1 1θ        2  2r        2  2θ
                       or
                                                                           2 ˙
                                     F = ml θ 1 [  sinθ  + θ 2 cosθ 1 ( θ 2 )  2 ( θ 2 )  2] n
                                                                        +
                                                                 2 sinθ
                                                            +
                                                              ˙˙
                                                                             2 cosθ
                                            ˙˙
                                      *
                                      2           1  1                                1       (11.10.16)
                                               1 [      2 ˙                  2 ˙     2]
                                                                          +
                                                                               2 sinθ
                                              ˙˙
                                                                ˙˙
                                                                   2 cosθ
                                                              −
                                         + ml  −θ  cosθ 1  + θ 1 sinθ 1 ( θ 2 )  2 ( θ 2 )  n  2
                                                                   ˙˙
                                                       *
                                                      T =−(ml  2  12) θ  n                    (11.10.17)
                                                       1            1  3
                       and
                                                                   ˙˙
                                                       *
                                                      T =−(ml  2  12) θ  n                    (11.10.18)
                                                       2            2  3
                        From Eq. (11.10.12) the partial velocities of G  and G  and the partial angular velocities
                                                                1      2
                       of B  and B  are (see also Eqs. (11.4.23)):
                          1      2
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