Problems 7/75 Page 575, 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).  The angular speed of the shaft is a constant.

Find:  The forces exerted by the bearings on 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, constant.
g = Local acceleration of gravity.

1. Mechanical system: Shaft assembly as it rotates.

2. Free Body Diagram:  Note that the free body diagram is drawn for a typical shaft assembly orientation (compare the shaft assembly orientation in the free body to that shown in the problem definition).  At the instant shown the x axis attached to the shaft makes the angle q with the vertical (the axes are shown in light blue, and in the position shown the angle q is roughly 45 degrees).  Note that this angle changes with time as the shaft assembly rotates.  All of the forces are shown in red.  As the two spheres are of equal mass and are placed equidistant from either side of the shaft center, the mass center of the system is the center of the shaft.  The weight force is equal to the total mass multiplied by the local acceleration of gravity and acts through this mass center.  However as the weight force is always vertically downward (as shown), it has both x and y components depending upon the value of the angle q.  Both bearings resist movement in the x and y directions.  It will be assumed that the bearing at O carries any thrust (z axis) load.  Thus the bearing at O exhibits forces in all three coordinate directions (shown in red and labeled by location, O, and direction, x, y, or z).  The bearing at A exhibits forces only in the x and y directions (shown in red labeled by location, A, and direction, x or y).

3. Equations:
a. Kinetics:
S F = ( Ox + Ax - (M + 2m)g cos(q) ) i  + ( Oy + Ay + ( M + 2m) g sin(q) ) j + Oz k
S M_O = L/2 k X (M + 2m) g { -cos(q) i + sin(q) j }  + L k X { Ax i + Ay j }
        = -L/2 (M + 2m) g { cos(q) j + sin(q) i } + L { Ax j - Ay i }
b. Kinematics:
The mass center of the shaft assembly is at the shaft center.   This point is on the axis of rotation and never moves.  Thus the linear momentum of the mass center is always zero.  As the linear momentum is always zero, then its time derivative must be zero.
dG/dt|_ground = 0
To evaluate the time derivative of the angular momentum, we recall the expression for the angular momentum developed in problem 7-56.
H_O =   m R L/3 w  j + m(4/5 r² + 2 R²) w  k 
Note that the angular momentum vector has been expressed in the xyz axis system attached to the shaft assembly.  This was the motivation for summing the forces and moments using unit vectors from that coordinate system.  We require the time derivative of the angular momentum relative to the ground coordinate system.  However the vector has been expressed relative to rotating unit vectors.  Once more the second fundamental formula comes to our rescue.
d(H_O)/dt|_ground = d(H_O)/dt|_shaft + w_shaft/ground X H_O
Using the expression for the angular momentum and recalling from problem 7-56 the angular velocity of the shaft:
w_shaft/ground = w k
d(H_O)/dt|_ground = d( m R L/3 w  j + m(4/5 r² + 2 R²) w  k  )/dt|_shaft
                         + w k X { m R L/3 w  j + m(4/5 r² + 2 R²) w  k }
The time derivative in the above expression can be seen to be zero.  The unit vectors used are fixed in the shaft system so they do not change relative to that system.  Clearly the masses of the spheres and the shaft do not change.  The shaft length doesn't change.  The radii of the spheres do not change.  The distances of the spheres from the shaft axis do not change.  We are given that the shaft angular speed is a constant and does not change.  Thus the components of our vector in the shaft system are constants and the derivative in the shaft system must be zero.  The remaining cross product term is straightforward as all of the vectors are expressed in the xyz system.
d(H_O)/dt|_ground = -m R L/3 w² i 

4. Solve:
Ox + Ax - (M + 2m)g cos(q)  = 0
Oy + Ay + ( M + 2m) g sin(q) = 0
Oz = 0
-L Ay - L/2 (M + 2m) g sin(q) = 0
L Ax -  L/2 (M + 2m) g cos(q) = -m R L/3 w²

The above provides five equations in the five unknown bearing forces.  The fifth equation can be solved for Ax.  The fourth equation can be solved for Ay.  The third equation can be solved for Oz.  Having determined Ay, the second equation can be solved for Oy.  Having determined Ox, the first equation can be solved for Ox.
Ax = -1/3 m R w² + 1/2 (M + 2m) g cos(q)
Ay = -1/2 (M + 2m) g sin(q)
Oz = 0
Oy = -1/2 (M + 2m) g sin(q)
Ox = 1/3 m R w² + 1/2 (M + 2m) g cos(q)
These are the forces exerted by the bearings on the shaft for any shaft orientation.  The forces on the bearings are just the opposite of these.
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