Page 74 - Singiresu S. Rao-Mechanical Vibrations in SI Units, Global Edition-Pearson (2017)
P. 74

1.8  mass or inertia elements   71
                                                        Pinion, mass moment of inertia J 0



                                                       u
                                                                   R

                                     Rack, mass m                         x

                                    FiGure 1.37  Translational and rotational masses in a
                                    rack-and-pinion arrangement.



                                    Case 2: Translational and Rotational Masses Coupled Together.  Let a mass m having
                                                       #
                                    a translational velocity x be coupled to another mass (of mass moment of inertia J ) having
                                                     #
                                                                                                     0
                                    a rotational velocity u, as in the rack-and-pinion arrangement shown in Fig. 1.37.
                                    These two masses can be combined to obtain either (1) a single equivalent translational
                                    mass m  or (2) a single equivalent rotational mass J , as shown below.
                                          eq
                                                                              eq
                                    1.  Equivalent translational mass. The kinetic energy of the two masses is given by
                                                                    1  #   1   #
                                                                       2
                                                               T =    mx +   J  u 2                    (1.22)
                                                                    2      2  0
                                    and the kinetic energy of the equivalent mass can be expressed as
                                                                       1    #
                                                                  T  =   m x 2                         (1.23)
                                                                   eq
                                                                       2  eq eq
                                         #     #    #  #
                                    Since x eq  = x and u = x>R, the equivalence of T and T  gives
                                                                                eq
                                                                                 #
                                                                        #
                                                                #
                                                            1  m x =  1  mx +  1  x  2
                                                                 2
                                                                        2
                                                                              0
                                                            2  eq    2      2  J ¢ ≤
                                                                                 R
                                    That is,
                                                                            J
                                                                  m eq  = m +  0                       (1.24)
                                                                            R 2
                                                                   #    #    #    #
                                    2.  Equivalent rotational mass. Here u eq  = u and x = uR, and the equivalence of T and
                                       T  leads to
                                        eq
                                                                #
                                                                         #
                                                            1  J  u =  1  m1uR2 +  1  # 2
                                                                           2
                                                                 2
                                                                                 0
                                                            2  eq    2         2  J  u
                                    or
                                                                 J eq  = J + mR 2                      (1.25)
                                                                       0
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