For case I when no spinning ω = 0, but only a whirling,v: The disc wobbles in the space (about its diameter) and it is difficult to visualize its (wobbling) angular speed. The visualization can be made easier by noting that at the point C the shaft is always perpendicular to the disc, so that we can study the motion of a shaft segment near C instead of the disc. The line CA is tangent to the shaft at the point C. The piece ds of the shafting at the disc moves with the line AC, describing a cone with the point A as an apex as shown in Figure 5.24(a). The velocity of the point C for a whirling in the count clockwise direction, as seen from the right, is perpendicular and into the paper and its value isvy, where y is the linear displacement of the disc centre andv is the whirl frequency.
The line AC lies in the paper at time t = 0 (see Figure 5.24(a)), but at time dt later, point C is behind the paper by
 |
(5.59) |
The angle between two positions of line AC (i.e., AC and AC' in Figure 5.24) is
 |
(5.60) |
when φx is the angular displacement of the disc and it is considered to be small. From equation the angle of rotation of AC in time dt is equal to vφx dt. Hence the angular speed of AC (and of the disc) is equal to vφx. The disc rotates about a diameter in the plane of the paper and perpendicular to AC at C, so that the appropriate moment of inertia is Id (= ½ Ip for the thin disc). Hence, the angular momentum vector of the disc due to whirl is Idvφx and is shown in Figure 5.24(b). The direction as shown in Fig. 5.24(b) of the angular momentum can be obtained by considering the tilting of the disc when point C is moving to C/. During this motion disc will try to tilt such that its left hand side face would be visible to the observer. It will be more clearer when the disc centre will occupy position along the line OB and the disc is inside the plane of the paper (Fig. 5.25a). At this instant it tilts about its diameter, hence the observer will see motion of the disc tilting in the clockwise direction when looking the disc from the bottom along the diagonal. The disc centre will occupy same position on line OB when it is outside the plane of the paper and in this case the observer will be able to see the right hand side face of the disc (Fig. 5.25b). The total angular momentum is the vector sum of Ipω and Idvφx (see Fig. 5.26), which have been obtained from cases I and II, respectively.
Now our aim is to calculate the rate of change of the angular momentum vector, which has been obtained for cases I and II, individually. For this purpose we resolve the vector into components parallel (the direction from O to A as positive) and perpendicular (the direction from B to C as positive) to line OA as shown in Figure 5.26; and are given, respectively, as
 |
(5.61) |
and
 |
(5.62) |
where Ip ≈ 2Id for the thin disc, and for a small angular displacement, φx, of the disc such that cosφx ≈ 1 and sinφx ≈ φx.
Components parallel to line OA rotates around line OA in a circle with a radius y and keeps the magnitude, and the direction of the angular momentum constant during the process so that its rate of change is zero as shown in Figure 5.27(a).
Angular momentum components perpendicular to line OA is a vector along the direction of line BC, and it rotates in a circle with the center as point B as shown in Figure 5.27(b). Let at time t = 0 this vector lies in the plane of paper, and at time dt this vector moves inside the paper at an angle vdt (see Figure 5.27b). The increment in the vector is CC', which is directed perpendicular to the paper and into it, with the magnitude equal to the length of the vector itself, Id φx(2ω - v), multiplied by an angle of vdt. It is given as
 |
(5.63) |
The rate of change of the angular momentum with time (i.e., the moment) is then given as
 |
(5.64) |
For the synchronous case v = ω, we have from above equation moment as Id ω2φx, which is same as equation (5.29). Earlier, it was obtained by the concept of the centrifugal force. Hence, the above expression is an active moment the disc would experience or in other words this is the moment exerted on the disc by the shaft (i.e., by action). The reaction moment exerted by the disc on the shaft is the equal and opposite, i.e. a vector directed out of the paper and perpendicular to it at C. Beside this moment there is a centrifugal force mω2y acting on the disc from case II as shown in Figure 5.28.
Influence coefficients of the shaft can be defined as: 
is the deflection y at the disc from 1 N force; 
is the angle φx at the disc from 1 N force or is the deflection y at the disc from1 N-m moment, i.e. 
(by the Maxwell’s theorem of strength of materials); and 
is the angle φx at disc from 1N m moment. For the cantilever beam (Timoshenko and Young, 1968) with a concentrated load F and the moment M as shown in Figure 5.28, we have
 |
(5.65) |
It should be remembered that other boundary conditions can also be used; however, we have to obtain the relevant influence coefficients. The linear and angular deflections can be expressed as
 |
(5.66) |
It should be noted that the sign convention of Myz used for obtaining the influence coefficient is clockwise, however, the reactive moment from the disc to the shaft is counter clockwise, and hence the negative sign in the moment term. On substituting the force, Fy, and the moment, Myz, from Figure 5.28, we get