Problems 7/56 Page 565, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: The system shown.  The system consists of a slender shaft of mass M and length L (blue).  Welded to the shaft are two massless arms of length R(brown).  These arms are attached to the shaft at points a distance L/3 from the ends of the shaft.  The two arms run in opposite directions.  Welded to the end of arms are spheres of mass m and radius r (green).  The combination of the slender shaft, the two arms, and the two spheres will be referred to as the shaft assembly.   The coordinate system shown (light blue) is attached to the shaft assembly.  The z axis runs along the shaft direction.  The x and y axes are parallel to the shaft axis.  The y axis is in the direction of the arms.  The shaft is supported by bearings (purple) at its two ends so it can rotate about the z axis.  The end points of the shaft (bearing locations) are labeled O and A.  The rate of rotation of the shaft assembly about the z axis is w (yellow).

Find:  The angular momentum of the shaft assembly about the fixed pivot point O.  The kinetic energy of the shaft assembly.
 
Parameter Values:
M = Shaft mass
L = Shaft Length
m = Sphere Mass
r = Sphere Radius
R = Arm Length
w = Shaft assembly angular speed about z axis.

In order to evaluate the angular momentum about point O, we need the angular mass matrix (inertia tensor) about point O.  As this system consists of three objects with mass (shaft and the two spheres), the angular mass matrix of the shaft assembly about point O must be equal to the sum of the angular mass matrices about point O of the shaft and the two spheres.  In order to evaluate these individual mass matrices about point O we must first evaluate the mass matrices about the center of each piece and then transfer them to the desired point.  As the shaft is a slender rod, we know that its mass matrix about its center is diagonal.  The diagonal elements about the axes perpendicular to the shaft (x and y) are given by 1/12 the mass of the shaft multiplied by its length squared.  The diagonal element about the axis parallel to the shaft (z axis) is zero.  Thus:
                        1/12 M L²     0         0
I_G(Shaft) =             0      1/12 M L²    0
                             0              0          0
The center of the shaft is located at a distance of L/2 along the z axis from O.  Thus the shaft center location is (0, 0, L/2).  Thus:
                                      M (L/2)²         0           0
I_O(Shaft) = I_G(Shaft)  +          0          M (L/2)²    0
                                           0                0           0
                     1/3 M L²       0             0
I_O(Shaft) =           0        1/3 M L²      0
                           0             0             0

Having evaluated the angular mass matrix of the shaft about point O, we must next evaluate the angular mass matrix of one of the spheres.  The angular mass matrix of a sphere about its center is diagonal.  All three diagonal elements are equal with value 2/5 the mass of the sphere multiplied by its radius.  Thus for either of the two spheres:
                      2/5 m r²       0            0
I_G(Sphere) =        0        2/5 m r²       0
                          0              0       2/5 m r² 
The location of the sphere closest to O is given by (0,R,L/3) in the x, y, and z directions.  The location of the other sphere relative to O is given by (0,-R,2L/3).  Thus:
                                            m(R² + (L/3)²)     -0            -0
I_O(Sphere1) = I_G(Sphere1) +           -0             m(L/3)²     -m R L/3
                                                    -0           -m R L/3        m R²

                        m(2/5 r² + R² +L²/9)       0                     0
I_O(Sphere1) =                0             m(2/5 r² +L²/9)      -m R L/3
                                    0                -m R L/3              m(2/5 r² + R²)

                                              m(R² + (2L/3)²)     -0            -0
I_O(Sphere2) = I_G(Sphere2) +               -0             m(2L/3)²     -m -R 2L/3
                                                        -0           -m -R 2L/3        m R²

                        m(2/5 r² + R² +4L²/9)       0                     0
I_O(Sphere2) =                0             m(2/5 r² +4L²/9)      m R 2L/3
                                    0                m R 2L/3              m(2/5 r² + R²)

To determine the angular mass matrix (inertia tensor) about the point O we must sum the contributions from the three objects.  Thus:
        1/3 M L² + m(4/5 r² + 2 R² +5L²/9)          0                                     0
I_O =           0                                    1/3 M L² +  m(4/5 r² +5L²/9)      m R L/3
                  0                                                 m R L/3                         m(4/5 r² + 2 R²)

The angular momentum is given by the mass matrix mutliplied by the angular velocity vector.  The angular velocity vector of the shaft assemply relataive to the ground is given by:
w_shaft/ground = w k
Thus the angular momentum of the shaft assembly relative to the ground about point O is given by:
H_O = I_O  w_shaft/ground
         1/3 M L² + m(4/5 r² + 2 R² +5L²/9)          0                                     0                              0
H_O =           0                                    1/3 M L² +  m(4/5 r² +5L²/9)      m R L/3                        0
                  0                                                 m R L/3                         m(4/5 r² + 2 R²)              w
Performing the required matrix multiplication:
              0
H_O =   m R L/3 w
           m(4/5 r² + 2 R²) w
Note that the components of the angular momentum vector given above are given relative to the same axes used in evaluating the angular mass matrix and the angular velocity.  Thus the angular momentum vector is given in terms of the unit vectors attached to the x, y, z system which rotates with the shaft assembly.  Thus:
H_O =   m R L/3 w  j + m(4/5 r² + 2 R²) w  k 
 
The kinetic energy of the system is given by one half of the dot product of the angular velocity vector with the angular momentum vector.
Kinetic Energy = 1/2 w_shaft/ground   . H_O
                 = 1/2 w k .  { m R L/3 w  j + m(4/5 r² + 2 R²) w  k  }
 Performing the required dot product:
Kinetic Energy = 1/2 m(4/5 r² + 2 R²) w²