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0593_C08_fm  Page 250  Monday, May 6, 2002  2:45 PM





                       250                                                 Dynamics of Mechanical Systems


                                                                                               *
                       passing through an arbitrary point (say G) together with a couple with torque T . Then,
                              *
                        *
                       F  and T  are:
                                                         N      N
                                                             * ∑
                                                     F =    F =    m  R a  P i                  (8.6.3)
                                                      *
                                                        ∑ i         i
                                                         = i 1   = i 1
                       and
                                                      N           N
                                                 T =    r × F * = − ∑  m r × R a  P i           (8.6.4)
                                                   *
                                                     ∑ i    i        ii
                                                      = i 1       = i 1
                       where N is the number of particles of B. Recall that we already examined the summation
                       in Eqs. (8.6.3) and (8.6.4) in Section 7.12. Specifically, by using the definitions of mass
                                                                  *
                       center and inertia dyadic we found that F  and T  could be expressed as (see Eqs. (6.9.9),
                                                            *
                       (7.12.1), and (7.12.8)):
                                                          *
                                                         F =−M  RG                              (8.6.5)
                                                                 a
                       and
                                                   T =− I BG  ⋅ −αωω ×  I (  BG ⋅ )             (8.6.6)
                                                                       ωω
                                                            α
                                                     *
                       where M is the total mass of B.
                        Consider the form of the inertia torque: Suppose n , n , and n  are mutually perpen-
                                                                                 3
                                                                          2
                                                                       1
                       dicular unit vectors parallel to central principal inertia axes of B. Then, the inertia dyadic
                       I B/G  may be expressed as:
                                                 I BG  =  I  n n +  I  n n +  I  n n            (8.6.7)
                                                       11 1 1  22  2  2  33  3  3
                       Let the angular acceleration and angular velocity of B be expressed as:

                                                    αα = α n  + α n  2  + α n 3  = α n  i
                                                         1 1
                                                                          i
                                                                    3
                                                              2
                                                                                                (8.6.8)
                                               and  ωω = ω n  + ω n  2  + ω n 3  = ω n i
                                                                           i
                                                                     3
                                                               2
                                                         1 1
                                                                                     *
                       Then, in terms of the α , ω , I , and the n  (i = 1, 2, 3), the inertia torque T  may be expressed
                                                         i
                                             i
                                           i
                                               ii
                       as:
                                                    *
                                                  T = T  n + T  n + T  n = T n                  (8.6.9)
                                                       1 1   2  2  3  3  i  i
                       where the components T (i = 1, 2, 3) are:
                                             i
                                                               2 (
                                                   T =−α  I + ω ω  I − )                       (8.6.10)
                                                                       I
                                                    1    1 11    3  22  33
                                                               3 (
                                                   T =−α  I + ω ω  I − )                        (8.6.11)
                                                                       I
                                                    2    2 22    1  33  11
                                                               1 (
                                                   T =−α  I + ω ω  I − )                       (8.6.12)
                                                                       I
                                                    3    3 33    2  11  22
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