In the following problem, a disk of radius R is attached to a slender shaft of length L, which is pinned to another vertical shaft a point A. The disk and slender shaft swing freely about pin A. The disk rotates about the longitudinal axis of the slender shaft at a constant speed ω. The vertical shaft has a precession rate of  and the swinging shaft and disk have a nutation rate  from the vertical shaft. The following will develop the equations of motion for the motion of the system using Lagrange’s equations.

 

 

 

           

 

Transformations between the fixed frame A and reference frame B of the shaft and disk are:

 

  or  

 

Taking the process in reverse yields the transformation from frame B to A

 

 

The kinetic energy for the disk is the sum of the translational energy and the rotational energy. For the disk, the kinetic energy is

 

 

First, we calculate the total angular velocity of the disk taking advantage of the coordinate transformations above such that all base vectors are resolved into the frame of the disk

                                                                (1.1)

 

The linear velocity of the disk is simply the velocity at point B.

       Where,  

 

Since the velocity at point A is zero, we have

 

                                                                                        (1.2)

 

We can now put together the total kinetic energy of the disk

                                       (1.3)

 

Using the principle moments of inertia of the disk

Substituting these moments of inertia into equation (1.3) simplifies the kinetic energy for the disk to

                                                                              

             (1.4)

 

 

Now, we examine the kinetic energy of the shaft. Since the shaft is not rotating about its own longitudinal axis and only rotating about the Z-axis of the vertical shaft and about the pin at point A, the total angular velocity of the shaft is simply       

                                                                     (1.5)

                                                                              

The linear velocity of the mass center of the shaft is

 

                                                                                                               (1.6)

 

Now having the total angular velocity of the shaft and its total linear velocity, we can now obtain the total kinetic energy of the shaft

 

 

Assuming a slender rod, the moments of inertia of the rod for rotation in frame B are

 

Collecting all the terms and making the substitutions, the kinetic energy of the shaft simplifies to

                                                                                                           (1.7)

 

_{Summing the energy of the shaft and disk, the total kinetic energy of the system can now be written as}

                                 (1.8)

 

 

In order to simplify the equations we make the following substitutions

 and  

 

Equation (1.8) can now be written as

                                                                     (1.9)

 

Now, we need to derive the total potential energy of the system. Using point A as the datum, the potential energy is simply the sum of individual potentials of the shaft and disk

                                                                 (1.10)

 

We now have everything needed to form the Lagrangian Function, L. Using the Lagrangian we can systematically write the final equations of motion for each generalized coordinate.

 

                                     (1.11)

 

 

First, let’s start with  

                                (1.12)

 

Since there are no terms that are functions of ψ, .

For the θ coordinate we have the following

                                                                                                                           (1.13)

 

                                  (1.14)

 

Since there are no non-conservative forces acting on the system, the virtual work done in the δ and δθ directions are zero

 

We now have everything we need to assemble the equations of motion. The corresponding Lagrangian equations are:

 

 

for the ψ and θ coordinates we have

 

                                                         (1.15)

                                               (1.16)