Page 168 - The Mechatronics Handbook
P. 168

0066-frame-C09  Page 42  Friday, January 18, 2002  11:01 AM









                       When this form is expanded for either case into x, y, z components, then

                              h =  h x i + h y j +  h z k =  ∫  ( xi +  yj + zk) ×  ( [  ω x i +  ω y j + ω z k) ×  ( xi + yj +  zk)]dm
                                                            ˆ
                                                                        ˆ
                                                                   ˆ
                                    ˆ
                                                        ˆ
                                                    ˆ
                                        ˆ
                                            ˆ
                                                                            ˆ
                                                                                         ˆ
                                                                                      ˆ
                                                                                  ˆ
                                                 m
                       which can be expanded to
                                                        2
                                                           2
                                  h x i + h y j + h z k =  ω x ∫  ( y +  z )dm ω y ∫  xy dm ω z ∫  xz dm i ˆ
                                             ˆ
                                        ˆ
                                    ˆ
                                                                –
                                                                            –
                                                     m               m          m
                                                                     2
                                              =  – ω x ∫  xy dm +  ω y ∫  ( x +  z )dm ω z ∫  yz dm j ˆ
                                                                        2
                                                                             –
                                                      m           m               m
                                                                                2
                                                                                   2
                                              =  – ω x ∫  xy dm ω y ∫  zy dm ω z ∫  ( x +  y )dm k ˆ
                                                             –
                                                                        –
                                                      m          m           m
                       The expression for moments and products of inertia can be identified here, and then this expression
                       leads to the three angular momentum components, written in matrix form
                                                          –     I
                                                      I xx  I xy – xz  ω x
                                                h =  –  I yx  I yy  I – yz    ω y  = Iω          (9.25)
                                                     –  I zx  –  I zy  I zz  ω z
                       Note that the case where principal axes are defined leads to the much simplified expression
                                                  h =  I xx ω x i +  I yy ω y j +  I zz ω z k ˆ
                                                           ˆ
                                                                  ˆ
                       This shows that when the body rotates so that its axis of rotation is parallel to a principal axis, the angular
                                      h,
                       momentum vector,   is parallel to the angular velocity vector. In general, this is not true (this is related
                       to the discussion at the end of the section “Inertia Properties”).
                         The angular momentum about an arbitrary point, Case 3, is the resultant of the angular momentum
                       about the mass center (a free vector) and the moment of the translational momentum through the mass
                       center,
                                                p =  mV x i +  mV y j +  mV z k =  mV
                                                                      ˆ
                                                        ˆ
                                                               ˆ
                       or
                                                        h =  h G + r ×  p

                            r
                       where   is the position vector from the arbitary point of interest to the mass center, G. This form can
                       also be expanded into its component forms, as in Eq. (9.25).
                       Kinetic Energy of a Rigid Body
                       Several forms of the kinetic energy of a rigid body are presented in this section. From the standpoint of
                       a bond graph formulation, where kinetic energy storage is represented by an  I element, Eq. (9.25)
                       demonstrates that the rigid body has at least three ports for rotational energy storage. Adding the three
                       translational degrees of freedom, a rigid body can have up to six independent energy storage “ports.”


                       ©2002 CRC Press LLC
   163   164   165   166   167   168   169   170   171   172   173