CHAPTER 5

TRANSVERSE VIBRATIONS-III:

SIMPLE ROTOR SYSTEMS WITH GYROSCOPIC EFFECTS

In previous chapter, we considered rotor-bearing systems for a single mass rotor with different level

of complexity at supports. We analysed the rotor system for the translatory and rotary motions by

considering the respective inertias. However, we neglected an important effect on dynamic behaviours

of the rotor system so called the gyroscopic effect, which predominates especially for high speed

rotor. In the present chapter also we shall still be dealing with a single-mass rotor system; and again

with the assumption of rigid bearings. However, now we shall include the effect of gyroscopic effects

and will explore the motion of the rotor for the synchronous as well as the asynchronous whirl. For

the present case, we shall analyse the rotor system by two different approaches, firstly by the quasi-

static analysis (which gives a better physical insight into the effect of gyroscopic effects, however,

can be applied practically to simple systems only), and secondly by the dynamic analysis (which can

be easily extended to multi-DOF systems). An important aspect, which we will observe from the

present chapter, is that because of the gyroscopic effect the whirl natural frequency becomes

dependent on the rotor spin speed. Another interesting phenomenon that can be observed is that the

rotor can have the forward and backward whirling motion. Moreover, for the present case the

distinction among the rotor spin speed, the whirl natural frequency, and the critical speed will be

made clearer.

When a relatively large disc (or rotor) spins at a very high speed about its longitudinal (polar) axis,

then it has a large angular momentum and it is the main characteristics of the rotors to carry a

tremendous amount of rotary power. However, if it has precession (slow or fast) about its transverse

(diametral) axes, which comes due to flexibility of bearings or of shafts itself, then it develops change

in the angular momentum due to change in its direction. This leads to an inertia moment called the

gyroscopic moment. Basically the gyroscopic moment develops due to the Coriolis acceleration

component.

Figure 5.1(a) and (b) shows the motion of a disc in a simply supported shaft, when the disc is at the

mid-span and away from it, respectively. For the latter case, the precession of the disc about its

diametral axes takes place along with the spinning about the polar axis, which leads to the gyroscopic

moment on the disc and it depends upon the spin speed among other parameters. Rotor systems with a

point mass and with an appreciable mass moment of inertia are shown in Figures 5.1(c) and (d),

respectively. The critical speeds of both the rotor systems will not the same. This is due to the fact

207

that centrifugal forces of particles of the disc do not lie in one plane during motion and thus from a

moment tending to straighten the shaft and it will be discussed subsequently in more detail.

ω

(a) A simply supported shaft with a disc at the

mid-span

ω

(b) A simply supported shaft with a disc near the

bearing

(c) A cantilever shaft with a point disc at the

free end

(d) A cantilever shaft with a rigid disc at the free

end

Figure 5.1. The spinning and precession (whirling) motions of the disc

5.1 Angular Momentum

The angular velocity is a vector quantity. A change in the magnitude and direction of angular velocity

results in the angular acceleration. Let in Fig. 5.2, OA rotates about the z-axis in the x-y plane and OB

is the position it takes after an infinitesimal time interval. Let θ∆ be the infinitesimal angular

displacement of OA, it is the angular displacement vector along z-axis. Similarly, the angular velocity,

angular acceleration and angular momentum are also vector quantity. The gyroscopic moment can be

understood using the principle of angular momentum.

Figure 5.2. An angular displacement vector

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A particle of mass, m, is moving with a velocity, v. From Figure 5.3, the linear momentum, L, is

defined as

L = mv (5.1)

The direction and sense of the linear momentum are same as the linear velocity.

Figure 5.3. A particle in a linear motion

Now, referring to Figure 5.4(a) in which a point mass m is revolving at a radius r in a plane, the

angular momentum is defined as the moment of linear momentum

Angular momentum ( ) ( )2

pH mv r mr Iω ω= = = = (5.2)

where Ip is the polar mass of inertia of particle of mass, m, about it’s axis of rotation; and ω is the

angular velocity. The direction of the angular momentum will be same as angular velocity.

(a) A point mass in rotation

ω

(b) A flywheel in rotation

Figure 5.4 Angular momentum

Referring to Figure 5.4(b) in which a flywheel of the mass m and of the radius of gyration k is rotating

with an angular speed of, ω , the angular momentum is given as

2( ) ( ) pH mv k mk Iω ω= = = with 2

pI mk= (5.3)

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5.2 Gyroscopic Moments in Rotating Systems

In the present section, the concept of the gyroscopic moment will be introduced with the help of

simple rotating systems, e.g., discs and blades. These basic concepts you might have studied to some

extent in the subject of dynamics of machinery (Wrigley, et al., 1969; Bevan, 1984; Mabie and

Reinholts, 1987; Rao and Dukkipati, 1995).

5.2.1 Motion of a rotor mounted on two bearings

Let us assume that a rotor with a flexible massless shaft carrying a disc is constraint to move in a

vertical (single) plane. It is assumed that the constraint is not providing any friction forces during the

motion of the shaft that plane. A rotor is spinning with a constant angular velocity,ω ; the angular

momentum, H, will be given by pI ω . Let x-y-z be the rectangular coordinate system (see Figure 5.5),

where oz is the spin axis, ox is the precession axis, and oy is the gyroscopic moment axis.

Figure 5.5. A rotor mounted on two bearings with a single plane motion

Let the disc (or the spin axis) precession angle is ϕ∆ from z-axis as shown in Figure 5.5. The angular

momentum will change from H (i.e., OC) to H ′ (i.e., OD), which can be written as (see the triangle

∆OCD which is in y-z plane)

H H H′ = + ∆��� ��� �

(5.4)

210

where H∆ is the change in angular momentum (i.e., CD). It is due to change in the direction of H.

From ∆OCD, we have

( )CD OC ϕ= ∆ or ( )H H ϕ∆ = ∆ or pH I ω ϕ∆ = ∆

with

� � ���

Now the rate of change of the angular momentum can be written as

0

limp p

t

dHM I I

dt t

ϕω ων

∆ →

∆= = =

∆ (5.5)

where ν is the uniform angular velocity of precession, and M is the gyroscopic moment. The

gyroscopic moment will have same sense and direction as H∆ , i.e. CD����

. In the vector form, equation

can be written as

( )pM H Iν ν ω= × = ×�

(5.6)

From the right hand screw rule, we will get the direction of gyroscopic moment, i.e. along negative y-

axis (the clockwise direction when seen from above, see Figure 5.5). Whenever an axis of rotation or

spin axis changes its direction about another orthogonal axis then a gyroscopic moment acts about the

third orthogonal axis. This is active moment acting on the disc, which means disc will experience this

moment. In other word, if to a spinning disc a moment, M, is applied then precession take place in an

attempt to align the angular momentum vector, Ipω, with the applied moment vector.

If we consider the free body diagram of the disc, the reaction from the shaft on to the disc-hub will be

the active moment (i.e., in the negative y-axis direction). Hence, the shaft will experience a reactive

moment from the disc hub in the opposite direction as the active moment on to the disc (i.e., in the

positive y-axis direction). Let F be a force on the shaft from the bearing, then its direction due to

gyroscopic moment will be as shown in Figure 5.5. A reactive moment will be experienced by

bearings through the shaft in the opposite direction as the active gyroscopic moment on to the disc

(i.e., a couple due to –F forces).

5.2.2 Gyroscopic moments though Coriolis component of accelerations:

Figure 5.6 shows a disc that is spinning with a angular velocity, ω, and a precession angular velocity,

ν. Let z and y be the spin and precession axes, respectively. It is assumed here also that the disc has

211

precession about a single axis (i.e., y-axis). An infinitesimal mass, dm, at point P has coordinates

( ),r θ in polar coordinates or (x, y) in rectangular coordinates. From Figure 5.6(a), it can be seen that

the velocity of point P, i.e. rω , will be perpendicular to OP. The velocity component along x-axis will

be sinr yω θ ω= and whereas along the y-axis is cosr xω θ ω= . Particle P has the motion along the

x-axis (i.e., parallel to the x-axis) and simultaneously it is rotating about y-axis as shown in Figure

5.6(c). Hence, a Coriolis component of acceleration, i.e. 2( )yω ν , acts along positive z-axis direction

(Figures 5.6(a) and (b)).

Similarly for particle ,P′ the Coriolis acceleration component will be 2( )yω ν ; and it acts along

negative z-axis direction as shown in Figures 5.6(a and b). The force due to acceleration of the

particle P is given as

(2 )F dm yω ν= (5.7)

Figure 5.6 The gyroscopic moment on a rotating disc (a) x-y plane (b) y-z plane and (c) z-x plane

Let us first consider the moment about x-axis, and due to the particle P it is 2(2 )dm yω ν , hence the

total moment about x-axis will be

212

22 2xx xx pM y dm I Iων ων ων= = =∫ (5.8)

with

12

2xx pI y dm I= =∫ (for thin disc) (5.9)

where Ip (or Izz) is the polar moment of inertia. The component of the gyroscopic moment, Mxx, acts

along the positive x-axis direction (Fig. 5.6(a)). Now consider the moment about y-axis and due to the

particle P it is (2 )dm xyω ν , hence the total moment about y-axis will be

2 2yy xyM xydm Iων ων= =∫ (5.10)

with

0xyI xydm= =∫ (for symmetric disc) (5.11)

It should be noted that if the disc were not symmetric then 0xyI ≠ , so we would get and xx yyM M

both non-zero. Hence, accelerating forces arising out of these Coriolis acceleration components due to

motion of particles in x-direction, produce a net moment (or a couple) M, along positive x-axis

direction only. There is no Coriolis component of acceleration when we analyze the motion of particle

in the y-direction since it has no precession about x-axis, 0xω = (i.e., since the single plane

precession is assumed) as shown in Figure 5.6(b).

5.2.3 Gyroscopic moments in a rotating thin blade

In Figure 5.7, we have z-axis as the axis of spin and y-axis as the axis of precession. Let ξ and η are

the two orthogonal principal axes of the thin rod, with ξ making an angle of θ with the x-axis. Due to

the Coriolis component of acceleration the force at a point P, of the mass dm is given as

( ){ }sindF dm rν ω θ= 2 (5.12)

which acts along the spin axis (positive z-axis direction in Figure 5.7b). The moment due this force

about η-axis is given as (with the moment arm of r)

( )sindM dFr dm rη ων θ= = 22 (5.13)

The total moment of all particles above and below the η-axis is given as

( ) 2 22 sin 2 sin 2 sinM r dm r dm Iη ηων θ ων θ ων θ= = =∫ ∫ (5.14)

with

213

I r dmη = ∫2

(5.15)

Figure 5.7 A thin rod rotating about its centroid axis (a) x-y plane (b) y-z plane

From the parallel axis theorem, we have

pI I I Iξ η η= + ≈ (5.16)

where p zzI I= is the polar moment of inertia. Since the rod is thin, hence we have Iξ ≈ 0 . In view of

equation (5.16), equation (5.14) reduces to

2 sinpM Iη ων θ= (5.17)

which is along the negative η-axis direction as shown in Figure 5.7. Taking component of moment

Mη along the x- and y- axes, as

( )sin cosxx p pM I Iων θ ων θ= = −22 1 2 (5.18)

and

sin cos sinyy p pM I Iων θ θ ων θ= =2 2 (5.19)

Figure 5.8 A two-bladed propeller

214

There are two gyroscopic moments, respectively, about the x- and y-axes. This comes because of the

asymmetric body of revolution, i.e. I Iη ξ≠ . From equations (5.18) and (5.19) it can be seen that xxM

and yyM are varying with θ , i.e. xxM varies from 0 to 2 pI ων ; and yyM varies from pI ων− to pI ων .

The above analysis is applicable to the two-bladed propeller or the airscrew (Figure 5.8). The above

analysis can be extended to a multi-bladed propeller (e.g., Fig. 5.9).

5.2.4 Gyroscopic moments in a multi-bladed propeller: Let n be the number of blades ( 3n ≥ ) and

2 / nα π= is equally spaced angle between two blades. Let us consider one of the blades (designate as

1), which is inclined to an angle θ with x-axis as shown in Figure 5.10. Let the moment of inertia of

each blade about η-axis (i.e. perpendicular to the blade) be equal to Iη which in turn is equal to the

polar mass moment of inertia of blade 1 alone, i.e., 1pI ; since

1pI I I Iξ η η= + ≈ . Total polar moment

of inertia of the airscrew about the axis of rotation (z-axis) is: 1p pI nI= .

Figure 5.9 A three-bladed propeller Figure 5.10 One of the blade positions

Total moment about x-axis of blade 1 is given as

( )1 1 1

2 22 2 sin 1 cos2x p pM y dm I Iων ων θ ων θ= = = −∫ (5.20)

The location of other blades is given by the phase angle ( )1n α− (the phase angle of various blades

with respect to one of the reference blade would be ( )α1−n , where α is the phase difference between

two blades). Noting equation (5.20), on summing up the moments due to all n blades, we have

[ ] ( ) ( )1 1 1

1 cos 2 1 cos 2 1 cos 2 2 ...x p p pM I I Iων θ ων θ α ων θ α = − + − + + − + +

( ){ }1

1 cos 2 1pI nων θ α + − + − (5.21)

or

( ) ( ) ( ){ }{ }1

cos2 cos2 cos2 2 ... cos2 1x pM I n nων θ θ α θ α θ α = − + + + + + + + −

(5.22)

215

which can be simplified as

( ){ }

1

cos2 0.5 1 sin

sinx p

n nM I n

θ α αων

α

+ −= −

(5.23)

Since 2 / nα π= for 2n > , we have 2 /3, 2 / 4, α π π= � and sin 0α ≠ for all these values.

Moreover, since 2nα π= for all value of n, we have sin sin 2 0nα π= = . Hence for all values of

2n > , from equation (5.23) we can write

1x pM nI ων= or x pM I ων= (5.24)

For 2,n = we have sin sin 0α π= = and sin sin 2 0nα π= = , hence from equation (5.23), we get

( )1

1 cos 2x pM I ων θ= − (5.25)

The gyroscopic moment about y-axis for a blade as shown in Figure 5.10, which makes angle θ with

x-axis, is

1

sin 2y pM I ων θ= (5.26)

The total moment about y-axis for n blades with phase angles of ( 1)n α− for 1,2,n = � , is

( ) ( )( ){ }1

sin 2 sin 2 ... sin 2 1yy p

M I nων θ θ α θ α = + + + + + − (5.27)

The sine series in equation (5.27) will be zero for all values 2n > , hence

0yM = for 2n > (5.28)

So with equations (5.24) and (5.28), we can conclude that for the multi-bladed propeller with number

of blades 3 and above is equivalent to a plane disc with the polar mass moment of inertia 1p pI nI=

about the axis of rotation.

The objective of the present section was to have understanding of the gyroscopic moment in rotating

components especially the direction of application. Now we shall deal with effects of gyroscopic

moment, and the procedure of obtaining natural frequencies and critical speeds in simple single-mass

rotor systems with the synchronous and asynchronous whirls (Den Hartog, 1984).

216

5.3 Synchronous Motion

We are considering the case of perfectly balanced rotor, however, it is assumed to be whirling at its

critical speed in slightly deflected position. The angular whirl frequency, v, of the centre of the shaft is

assumed to be same as the angular velocity of rotation as of shaft ω (i.e., the spin speed). This implies

that a particular point of the disc which is outside will always be outside; the inside point will always

remains inside; it is called the synchronous motion. The shaft fibres in tension always remain in

tension while whirling, and similarly the compression fibres always remain in compression. Thus any

individual point of the disc moves in a circle in a plane perpendicular to the undistorted centre line of

the shaft. We will consider two representative cantilever (overhung) rotor cases, firstly a thin disc at

the free end and secondly a long stick at the free end.

5.3.1 A cantilever rotor with a thin disc

In the present section, a thin disc attached to a flexible cantilever shaft at its free end is considered. In

Fig. 5.11b, let points C and G are the shaft centre and the disc centre of gravity, respectively. Since no

unbalance in the rotor, hence points C and G are coincident. Let δ and ϕx are the linear and angular

displacements of the disc at the centre of shaft, C. The centrifugal force of a mass element dm at point

P with coordinate (r, θ) is ω2r1 dm and is directed away from point B (a point on the bearing axis as

shown in Figure 5.11b). It can be considered as two force vectors, the one when dm is assumed to be

rotated about shaft center C (along CP) and the second force when C itself is rotating about B (along

BC). Two similar triangles BCP and EFP can be constructed to represent these three forces (Fig.

5.11b). The component in the vertical direction is ω2δ dm and is directed vertical down. When these

component forces are added together will give a force and no moment (Fig. 5.12 (a and b)). Whereas,

the component in the radial direction is ω2r dm and is directed away from the disc centre C. When

these component forces are added together will give zero force and the moment will be non-zero (Fig.

5.12 (c and d)). These will be illustrated now. The force ω2δ dm for various masses add together will

give (Fig. 5.12 (a and b))

Fy = mω2δ

where m is the total mass of the disc. This force acts downward at C.

217

Figure 5.11 A cantilever rotor with centrifugal forces on the rigid disc

Figure 5.12 The centrifugal force on a particle of the disc

(a and b) due to pure spinning motion, and (c and d) due to pure whirling motion

218

The force ω2r dm are all radiate from the center of the disc C. From Figure 5.12 (c and d), the y-

component of the force, ω2r dm, is ω2

rsinθ dm = ω2y dm (since y = rsinθ) and the moment arm of this

elemental force is xyϕ . Thus the moment of the centrifugal force of a small particle dm is

dMyz = (ω2y dm) yϕx = ω2

y2ϕx dm

and it will act in the negative x-axis direction (Fig. 5.12c). The total moment Myz of centrifugal forces

is

2 2 2 2 2

yz x x x dM y dm y dm Iω ϕ ω ϕ ω ϕ= = =∫ ∫ (5.29)

where Id is the area moment of inertia of the disc about one of its diameter. The x-component forces

ω2rcosθ dm = ω2

x dm (since x = rcosθ) will balance themselves since these forces are on the plane of

the disc (Fig. 5.12(c and d)).

Thus in totality the end of the shaft is subjected to a force, mω2δ, and to a moment, Idω2ϕx, under the

influence of this it assumes a linear deflection δ and an angular deflection ϕx. This can happen only at

a certain speed ω, i.e. at the critical speed. Thus the calculation of critical speed is reduced to a quasi-

static problem (in which the dynamic forces are considered as time-independent), Now the objective

is to find at which value of ω a shaft will deflect δ and ϕx under the influence of Fy = mω2δ and Myz =

ω2Idϕx.

Figure 5.13 A cantilever beam with loadings at the free end

The linear angular displacement of the free end of the cantilevered (fixed-free end conditions) beam

as shown in Figure 5.13 will be (Timoshenko and Young, 1968)

( ) ( )2 3 2 223

3 2 3 2

d xy yz

m l I lM lF l

EI EI EI EI

ω δ ω ϕδ = − −= (5.30)

and

( ) ( )2 2 22

2 3

d xy yz

x

m l I lM lF l

EI EI EI EI

ω δ ω ϕϕ = − −= (5.31)

219

It should be noted here that the above relations can also be developed for other boundary conditions

such as simply supports, fixed-fixed, fixed-hinged etc. It requires calculation of influence coefficients

by using the deflection theory of strength of materials. Now the above relations can be used to find

critical speeds for the fixed-free boundary condition. Equations (5.30) and (5.31) can be rearranged as

3 22 21

3 20d x

l lm I

EI EIω ωδ ϕ

− −

+ = (5.32)

and 2

2 2 12

0d x

l lm I

EI EIω ωδ ϕ

− +

+ = (5.33)

This homogeneous set of equations can have a solution for δ and ϕx only when the determinant

vanishes

232 2

22 2

13 2

12

0

d

d

l lm I

EI EI

l lm I

EI EI

ω ω

ω ω

− −

− +

= (5.34)

which gives the frequency equation as

24 2

3 4

2 21212

03

d

d d

E IEI mlI

mI l mI lω ω

+ − − =

(5.35)

Defining the critical speed function, crω and the disc mass effect, µ , as

3

cr

ml

EIω ω= and

2

dI

ml

µ = (5.36)

Equation (5.35) can be written as

4 2 4 1212 0

cr crω ω

µ µ

+ − − =

(5.37)

With the solution

2

2 2 2 126 6crω

µ µ µ

= − ± − +

(5.38)

The positive sign will give a positive value for 2

crω or a real value for crω and the negative sign will

give a complex value of the critical speed, which has no physical significance. Plot of ���� versus µ is

given in Figure 5.14.

220

It should be noted that the critical speed of the rotor increases with the disc mass effect, µ . That

means the effective stiffness of the rotor system increases due to the thin disc rather than a point-mass

disc. This can be seen from Fig. 5.11 that the effect of centrifugal forces is to resist the tilting of the

disc thereby increasing the effective stiffness of the rotor system. Overall for the synchronous whirl

condition due to the gyroscopic effect the (forward) critical speed of the system increases. It will be

shown that for anti-synchronous whirl condition due to gyroscopic effect the (backward) critical speed

of the system decreases. Two limiting cases of Fig. 5.14 are discussed as follows:

Case 1: A disc having point-mass

For the disc effect µ = 0 (i.e., the concentrated mass of the disc) from equation (5.37), we have

2 24 12 0 3cr crω ω− = ⇒ = (5.39)

Noting equation (5.36), above equation gives

23

3

33 crcr

ml EI

mlEIω ω

= ⇒ = (5.40)

It gives the synchronous critical speed of disc having point-mass for the cantilever case.

Figure 5.14 Variation of the critical speed function with the disc effect

221

Case 2: A disc having infinite size

For µ → ∞ (i.e., a disc for which all the mass is concentrated at a relatively large radius, Id → ∞.) no

finite angular displacement ϕx is possible, since it would require an infinite torque, which the shaft

cannot furnish. The disc remains parallel to itself and the shaft is much stiffer than without the disc

effect (i.e., µ = 0). From equation (5.37) for µ → ∞, we get

( )2 2 2 2

3

1212 0 since 0 hence 12

cr cr cr cr cr

EI

mlω ω ω ω ω− = ≠ = ⇒ = (5.41)

It should be noted that for the present case one of the solution 2 0crω = is considered as not feasible.

However, when µ → - ∞ it is a feasible solution and the natural frequency of the system would be

zero. This particular issue will be considered in the next section by replacing the thin disc with a long

stick, which requires a separate analytical treatment; and it has been treated only for the pure

rotational motion.

5.3.2 A cantilever rotor with a long stick

For the present case, the disc at free end of a cantilever rotor has considerable amount of length as

shown in Figure 5.15. The couple of centrifugal forces for this case are such; it tries to push away the

rotor from the static equilibrium position angularly as shown in Figure 5.16. For the present case also

the synchronous whirl condition is assumed. For the thin disc case, the couple of centrifugal forces try

it to bring back to static equilibrium position angularly. In Figure 5.16 principal axes directions are (1)

and (2). The coordinate of a point mass dm is (y, z).

Figure 5.15 A cantilevered rotor with a long stick at the free end

222

Figure 5.16 Centrifugal forces in a cantilevered rotor with a long stick at the free end

For the present case, it is assumed that no unbalance is present in the rotor (i.e., the shaft geometrical

centre C and the centre of gravity G are coincident) and the centre of gravity G of the body is

positioned in the axis of rotation x (i.e., no linear displacement, δ = 0). Hence, there is no net

centrifugal force, mω2δ, and only a moment is present (where m is the mass of the long stick). The

force on a particle is ω2y dm and its moment arm about y-axis is z (Fig. 5.16), so that the moment is

dMyz = ω2yz dm (5.42)

For the thin disc we have z = yxϕ so above equation reduces to dMyz = ω2

y2

xϕ dm (see equation (5.29)

). Now for the whole body, we have

�� � ω ����

(5.43)

Let 1 and 2 be the principal axes along the longitudinal and transverse directions of the long stick

(Fig. 5.16), respectively. Let the mass moment of inertia about these principal axes be I1 and I2, which

are the polar and diametral mass moment of inertias, respectively. This set of axes is at an angle of xϕ

with respect to the y-z axes as shown in Figure 5.16. The product of inertia (it is identical to the shear

stress in the subject of the strength of materials) about y-z axes is defined as

( )1 2

1 2sin 2

2x x

I Iyzdm I Iϕ ϕ

−= ≈ −∫ (5.44)

Here the angular deformation, xϕ , is assumed to be small. For the thin disc I1 = Ip = 2Id and I2 = Id so

that equations (5.44) and (5.43) gives Myz = ω2 Id xϕ , which is same as equation (5.29). However, for

a disc of the diameter D and the thickness b (Fig. 5.16), we have for the present case

223

2 2 2

1 2and

8 16 12

mD mD mbI I += = (5.45)

On substituting equations (5.44) and (5.45) into equation (5.43), we get

�� � ���� � ���� � ������ � ���� ���� (5.46)

For moment of the centrifugal forces is to be zero, from above equation, we have

2 2 3 or 0.866

16 12 2

mD mbb D b D= ⇒ = =

and it becomes negative for b > 0.866D (i.e., for the long stick) and it is positive for b < 0.866D ( i.e.,

for the thin disc). Table 5.1 gives the summary of gyroscopic moment with its sign for different ratio

of b and D. Equation (5.46) can be written as

�� � ��ω�� with

2 2

16 12dI

mD mb =

− (5.47)

It can be found that the net resultant force will also be same for the long stick and the thin disc, when

we consider both the linear and angular motions simultaneously. Thus, the critical speed analysis of

the previous section will be valid for the present case also, when we consider both the linear and

angular motions simultaneously. Hence, Figure 5.14 of the previous case will still be applicable for

the range of thin disc. However, plot from the above equation will represent both the long stick and

thin disc cases as shown in Fig. 5.17, since the form of net moment equations (5.48) and (5.49) are

identical. From equation (5.38), we have

2

2 2 2 126 6crω

µ µ µ

= − ± − +

(5.50)

with

2

dI

ml

µ = and

2 2

16 12dI

mD mb = −

(5.51)

224

Figure 5.17 Variation of the critical speed function with the disc effect

The condition for which the square root term in equation (5.50) remain always positive is

2

2 126 0

µ µ

− + >

or ( )

23 1 3 0µ µ− + > (5.52)

From above equation it can be observed that both terms are always positive for positive value of µ,

since the first term is a square term and the second term is positive. For a negative value of µ, the first

term in the above inequality is always positive; and it is always greater than the second term. Hence

for all real value of µ, above inequality is true (for imaginary values of µ we may violate the

inequality however, it is not a feasible case in real systems) It means we will get always a real root

from equation (5.50) when we consider a positive sign in front of the square root. For 0µ → again

equation (5.40) would be valid. For µ → ±∞ from equation (5.50), critical speeds are 2 12crω → and

2 0crω → , respectively, for the positive and negative signs.

Table 5.1 Gyroscopic effects for different geometries of the disc/stick

Relation between b and D Sign of Id Gyroscopic moment sign

b = 0.87 D 0 0

b > 0.87D (Long stick) Negative Negative

b < 0.87 D (Thin disc) Positive Positive

225

For the long stick case, it is assumed that the shaft extend to the centre of the cylinder without

interference. If shaft is attached to the end of the cylinder, the elastic-influence coefficients are

modified. The phenomenon described in the present section is generally referred to as a gyroscopic

effect. Now with some numerical examples the calculation of critical speeds would be demonstrated.

Example 5.1 Obtain the transverse critical speed for the synchronous motion of a cantilever rotor as

shown in Figure 5.18. Take mass of the thin disc, m, as 1 kg with the radius, r, as 3 cm. The shaft is

assumed to be massless; and its length and diameter are 0.2 m and 0.01 m, respectively. Take shaft

Young’s modulus of the shaft material as E = 2.1×1011

N/m2.

Figure 5.18

Solutions: Case I: For the disc effect µ = 0 (i.e., the concentrated mass of the disc) from equation

(5.40), we have

11 10

3 3

3 2.1 10 4.909 10196.61 rad/s

1 0.2

3cr

EI

mlω

−× × × ×= =

×= Answer

with

4 10 4(0.01) 4.909 10 m

64I

π −= = × ; d = 0.01 m; m =1 kg; l = 0.2 m (a)

Case II: Considering the disc as rigid (i.e., 0µ ≠ ), from equation (5.38), we have

2 22 2 12 2 2 12

6 6 6 6 1.74310.0056 0.0056 0.0056

crωµ µ µ

= − + − + = − + − + =

(b)

with

1 2 2 2

4

2 2 22

0.030.0056

4 4 0.2

dmrI r

ml lml

µ = = = = =×

; (c)

Now, from equation (5.53), we have

226

3 3

11 10

1 0.20.00881

2.1 10 4.909 10cr cr cr cr

ml

EIω ω ω ω

−

×= = =

× × × (d)

On substituting value of ���� from equation (b) into equation (d), the critical speed is given by

1.7431

197.86 rad/s0.00881 0.00881

crcr

ωω = = = Answer

It should be noted that as compared to case I the critical speed is more, which is expected due to

increase in the effective stiffness while considering the diametral mass moment of inertia of the disc

For the present case 1 12 2 4

4 41 0.03 2.25 10dI mr

−= = × × = × kg-m2, which is very less that is why the

increase in the critical speed is marginal. Reader can check the change in the critical speed, for

example, with radius of the disc equals to 6 cm. The effect would be far more predominant in the

limiting case, when disc is very large that is µ → ∞ , the critical speed can be calculated from

equation (5.41), as

11 10

3 3

12 12 2.1 10 4.909 10393.22 rad/s

1 0.2cr

EI

mlω

−× × × ×= =

×= Answer

Example 5.2 Obtain the transverse critical speed for the synchronous motion of a rotor as shown in

Figure 5.19. The shaft is assumed to be fixed supported at one end. Take dimensions of the cylinder

(stick) as (i) D = 0.2 m, b = 0.0041 m (ii) D = 0.0547 m, b = 0.0547 m (iii) D = 0.0361 m, b = 0.1649

m and (iv) D = 0.0547 m, b = 0.1093 m. Parameters D and b are the diameter and the length of the

cylinder. The shaft is assumed to be massless and its length l and diameter d are 0.2 m and 0.01 m,

respectively. Take the Young’s modulus of the shaft material as 2.1×1011

N/m2

and the density of the

cylinder material as 7800 kg/m3.

Figure 5.19 A rotor with a long stick

227

Solution: For the long stick the critical speed is given as

3cr cr

EI

mlω ω=

(a)

with

2

2 2 126 6crω

µ µ µ

= − + − +

;

2

dI

ml

µ = ; and

2 2

16 12dI

mD mb =

− (b)

Above equation is valid for all value of µ except for 0µ = (i.e., the thin disc). For 0µ = we have the

relation given by equation (5.40) and it is given as

3

3cr

EI

mlω = so that

3

3 3

3

3 1.732cr

cr

EI

ml

EI EI

ml ml

ωω = = == (c)

with

4 4 90.01 7.854 10

64 64I d

π π −= = = × m4; 1649.34EI = N-m

2

1m = kg, 0.2l = m

Hence, for 0µ = (thin disc), we have

3 3

454.061649.34

1 0.2

EI

ml= =

× and

3 3786.45

3 3 1649.34

1 0.2cr

EI

mlω = =

×=

× (d)

Table 5.2 summarises the calculation of rotor parameters and critical speeds from the above

equations. It should be noted the choice of the D and b are such that the mass of the disc remains the

same for all cases. For the point mass parameters D and b are not defined. On comparing the cases of

point mass and thin disc, the increasing trend in the critical speed is observed. For short stick (with D

= b) the disc parameter, µ , becomes negative with a very low value. This leads to very high value of

critical speed. However, for the long stick again the trend is towards decrease in critical speed, where

the disc parameter, µ , remains negative with a relatively high value.

228

Table 5.2 Summary of rotor parameters and synchronous critical speeds

Type of

disc

D,

(m)

b,

(m)

m,

(kg)

Id,

(kg-m2) 2

dI

ml

µ = crω 3

EI

ml

(rad/s)

crω ,

(rad/s)

Thin disc 0.2000 0.0041 1 0.0025 0.0625 1.86 454.06 845.00

Point mass - - 1 0.0000 0.0000 1.73*

454.06 786.43

Short stick 0.0547 0.0547 1 -6.23×10-5

-0.0016 50.89 454.06 23 107.11

Long stick 0.0316 0.1649 1 -2.2×10-3 -0.0550 9.06 454.06 4113.78

* From equation (5.41)

5.4 Asynchronous Rotational Motion

In this section we will consider the asynchronous whirling motion of the spinning rotor. Consider a

rotor, which is suspended practically at its centre of gravity by three very flexible torsional springs as

shown in Figure 5.20. This will enable us to analyse the effect of rotational displacement of the rotor

on its whirling frequencies, without complicating with the general motion in which both the linear and

angular displacements take place simultaneously. Such general motion is quite complicated and will

be considered in subsequent sections.

The aim is to calculate natural frequencies of modes of motion for which the centre of gravity O

remain at rest and the shaft whirls about O in a cone of angle 2ϕx. Let the effective torsional stiffness

of the support is kt. The disc on the motor shaft rotates very fast, and as the springs on which the

motor is mounted are flexible, the whirling take place at a very slow rate than the shaft rotation. The

angular momentum, H, is given as

H = Ipω (5.54)

where Ip is the polar mass moment of inertia of the rotor (i.e., rotating parts of the motor and the disc).

In the case when the whirl is in the same direction as the rotation (i.e., the forward whirl), the time

rate of change of angular momentum will be directed from B to C (i.e., out of the plane of paper in

Figure 5.20(c)). This is equal to the moment experienced by the motor frame from the disc. The

reaction, i.e. the moment acting on the disc from the motor frame is pointing into the paper (in Figure

5.20(c)) and therefore tends to make ϕx smaller. This acts in an addition to the existing spring k.

Hence, it is seen that the whirl in the direction of rotation (forward whirl) makes the natural frequency

higher. In the same manner it can be reasoned that for the whirl opposite to the direction of rotation

(i.e. the backward whirl), the natural frequency is made lower by the gyroscopic effect (Fig. 5.20d).

To calculate the magnitude of the gyroscopic effect, we have

229

( ) BC BC AB

OB AB OB

p

x

p

d Idt

I

ων ϕ

ω= = = (5.55)

(a) A motor on torsional springs (b) A rotor system with motor suspensions

(c) A conical forward whirl of the rotor (d) A conical backward whirl of the rotor

Figure 5.20 A motor and a rotor system supported on torsional springs

Equation (5.55) can be rearranged as

( ) x pp

dI I

dtω νϕ ω= (5.56)

230

Equation (5.56) gives the gyroscopic moment. The elastic moment due to the springs k is equal to kϕx

and the total moment is equal to

(kt ± Ipων) ϕx (5.57)

where the positive sign for a whirl in the same sense as the rotation, and the negative sign for a whirl

in the opposite sense. In equation (5.57), the term in the parenthesis is the equivalent spring constant

and hence the natural frequency will be (since we have ν = ωnf)

� ! � "#$%&''()%* or � ! + %&'

%* � ! �"#%* � 0 (5.58)

where Id is the diametral mass of whole motor assembly (including frame and disc). The solution of

equation (5.58) can be given as

� ! � $ %&'%* $-�

%&'%*�

. "#%* (5.59)

From equation (5.59), it can be observed that the natural frequency of rotor system depends upon the

spin speed of rotor, ω. The ± sign before the square root, only the positive sign need to be retained

since the negative sign gives two values of ωnf, which are both negative, and equal and opposite to the

two positive roots obtained with positive sign before the square root. This is due to the fact that

natural frequencies remain same for the spining of rotor in either direction. Let us define non-

dimensional terms as the frequency ratio as �� ! � � !/� !(0(120#, where � !(0(120# � 345/�� be

the non-rotating shaft natural frequency, i.e. without the gyroscopic effect; and the spin ratio �� �60.5���/345��9. Equation (5.59) takes the following form

�� ! � $�� . √�� . 1 (5.60)

Figure 6.9 shows the variation of the non-dimensional natural frequency, nfω , with the non-

dimensional speed, ω , which govern by equation (5.60). It is seen that the natural frequency is split

into two frequencies on account of the gyroscopic effect (i) a slow one whereby the whirl is opposed

to the rotation (i.e. the backward whirl) and (ii) a fast one where the whirl and rotation directions are

the same (i.e., the forward whirl). It can be seen that we will have two critical speeds one

corresponding to the forward whirl and another backward whirl. These critical speeds can be obtained

231

by the condition that whenever the spin speed is equal to the natural frequency we will have critical

speeds. It will be illustrated in the following numerical example.

Figure 5.21 The natural frequency variation with the spin speed

Example 5.3 A long rigid symmetric rotor is supported at ends by two identical bearings. Let the

shaft has the diameter of 0.2 m, the length of 1 m, and the material mass density equal to 7800 kg/m3.

The bearing has dynamic parameters as follows: kxx = kyy = k = 1 kN/mm with other stiffness and

damping terms equal to zero. By considering the pure tilting motion and the gyroscopic effect, obtain

whirl natural frequencies of the system, if rotor is rotating at 10, 000 rpm. Obtain also the forward and

backward critical speeds of the rotor-bearing system. Compare the critical speeds of rotor without

considering the gyroscopic effect.

Solution: For the circular cylinder the polar moment of inertia, Ip, and the diameter moment of inertia,

Id, are given as

( )1 12 2 2

2 12, and 3p dI mr I m r l= = +

where m is the mass of the cylinder, r is the radius of the cylinder, and l is the length of the cylinder.

For the pure tilting motion of the rotor, we have the following rotor properties

1 12 2 2

2 2245.04 (0.1) 1.2252 kgmpI mr= × × ==

{ }1 12 2 2 2 2

12 12(3 245.04 3 (0.10) 1 21.0326 kgm)d m rI l × × + == + =

ϖ

υ

Forward

Backward

232

2

10,000 1047.198 rad/s60

πω

= =

The effective torsional stiffness due to bearings on the rotor can be obtained by considering Figure

5.22. A long rotor is supported at ends by two identical bearings; consider pure tilting of the rotor by

an angle xϕ , which gives 0.5 xlϕ compression of the left bearing and the same amount of extension of

the right bearing. This produces reaction forces at bearings with the magnitude of 0.5 xklϕ and the

direction as shown in Figure 5.22. The moment on to the rotor due these bearing forces would be

0.5x

kl ϕ2. Hence, the effective torsional stiffness would be 0.5kl

2 .

Figure 5.22 Free body diagram of the long rigid rotor supported on bearings

From equation (5.59), we have

3,4

2

2 2

p p eff

nf

d d d

I I k

I I I

ω ωω

= ± + +

(a)

with

1 2 6 2 5

20.5 1 10 1 5 10effk kl= = × × × = × N/m

where the positive sign for the forward whirl and the negative sign for the backward whirl. On

substituting values in equation (a), we get

[ ]3,4

251.2252 5 10 1.22521047.198 1047.198 30.50 157.17

2 21.0326 21.0326 2 21.0326nfω

× × ± × = ± + × ×

= +

which gives

3

187.67 rad/snfω = (forward whirl) and4

126.67 rad/snfω = (backward whirl). Answer.

233

It should be noted that these natural whirl frequencies change with the spin speed of the rotor. For

obtaining the forward and backward critical speeds, we have, respectively, following conditions, i.e.,

nfω ω= and nfω ω= − . On substituting these conditions in equation (a), one at a time, we get the

following expression for critical speeds

( )

,F B t

cr

d p

k

I Iω =

∓

(b)

where the negative sign is for the forward critical speed, F

crω , and the positive sign for the backward

critical speed, B

crω . It should be noted from equation (b) that for d pI I< , the term in side the square

bracket will be negative and there would not be any forward critical speed. For the present case, we

have d pI I> and the corresponding critical speeds are

( )

55 10158.88

21.0326 1.2252

F

crω×

= =−

rad/s

and

( )

55 10149.88

21.0326 1.2252

B

crω×

= =+

rad/s Answer.

Hence, the rotor is operating well above the critical speeds, since operating speed is 1047.2 rad/s. That

means if we consider perfectly balanced rotor that is rotating at 1047.2 rad/s and if we perturb the

rotor in forward whirl it will have whirl frequency equal to 187.67 rad/s. Moreover, if we perturb the

rotor in backward whirl it will have whirl frequency equal to 126.67 rad/s. While rotor is coasting up

from stationary position to operating speed then it will cross critical speeds, where large oscillations

are expected. An alternative solution of the present problem will be presented in subsequent section

based on dynamic analysis by considering the formulation of governing differential equations of the

rotor.

When no gyroscopic effect is considered, nfω1

and nfω4

remain the same; and from equation (a) for

pI = 0 , we get

,

55×10

154.184 rad/s21.0326

eff

nf

d

k

Iω = ± ==

3 4

Answer.

234

which is between the forward and backward whirl frequencies (187.67, 126.67) rad/s obtained by

considering the gyroscopic effect and these frequencies are now independent of the rotor spin speed.

If we consider the pure linear (translational) motion, then the critical speed will not experience any

gyroscopic moments, hence corresponding critical speeds can be obtained as

1 2

62 2 1 1090.34 rad/s

245.04cr cr

k

mω ω

× ×= = == Answer.

with

π π2 2

4 47800 (0.2) 1 245.04 kgm d lρ= = × × =

where d is the diameter of the cylinder. In this case the natural frequency does not change with the

spin speed of the rotor, and remains equal to the critical speeds (90.34 rad/s) as obtained above.

5.5 Asynchronous General Motion

In the previous analysis of the cantilever rotor in section 5.3, the rotor was whirling (the whirling is

defined as a circular motion of the deflected shaft centre line about its undeflected position with small

amplitude) and spinning at the same angular speed and in the same direction. Cases have been

observed where whirling and the spinning occur at different frequencies and sometimes in opposite

directions as described in previous section. The aim of the present section is to calculate natural

whirling frequencies, ν, of a shaft with a single disc on it at any speed of rotation, ω, in most general

manners as shown in Figure 5.23, where ω is the shaft spin speed about deflected centerline and ν is

the shaft whirling frequency about undeformed position OA. Since a general motion of the disc is

very difficult to visualize hence the following three cases have been considered that is relatively easier

to visualize, and the final motion will be superposition of some of these cases (i.e., cases I and II).

Figure 5.23 A cantilever rotor having a general motion of the spinning and the whirling

235

Case I: ν = ν0 with ω = 0 (Pure whirling motion)

The shaft does not spin about its deflected centerline, but deflected centerline OC whirls with ν0 about

bearing centre line OA.

Case II: ν = 0 with ω = ω0 (Pure spinning motion)

The shaft is in the deflected position OC, and it spins with ω0 about the deflected centerline OC.

Case III: ν = ω (Synchronous whirling and spinning motions)

The shaft fiber in tension remains in tension, and similarly shaft fiber in compression remains in

compression, i.e. the synchronous whirl. It has already been discussed in detail in section 5.3 and it

will not be discussed here again.

With this combined ν and ω motions (Cases I and II), our first aim would be to obtain its angular

momentum. For the case II when it does not whirl, but only spins, the angular momentum is equal to

Ipω (along AC as shown in Figure 5.23) where Ip is the polar moment of inertia of the disc.

(a) The pure whirling motion

(b) The angular momentum due to the pure whirling motion

Figure 5.24 A pure whirling motion of the rotor (case I)

236

For case I when no spinning ω = 0, but only a whirling, ν: 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 remarking 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 is

νy, where y is the linear displacement of the disc centre.

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

CC′ = (νy)dt (5.61)

The angle between two positions of line AC (i.e., AC and AC′ in Figure 5.24) is

CC

AC ACx

ydtdt

ννϕ

′= = with

ACx

yϕ = (5.62)

when ϕx is the angular displacement of the disc and it is considered to be small. From equation (5.63)

the angle of rotation of AC in time dt is equal to νϕx dt. Hence the angular speed of AC (and of the

disc) is equal to νϕx. 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). The angular

momentum vector of the disc due to whirl is Idνϕx and is shown in Figure 5.24(b). The direction 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 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 out side the plane

of the paper in that case 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 Idνϕx, which have been obtained from cases

I and II, respectively.

237

Fig. 5.25 Disc tilting during whirling disc is (a) into the plane of paper (b) outside the plane of paper

Figure 5.26 Angular momentums due to the whirling and the spinning of a rotor

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

( )2cos sin 2p x d x x d xI I Iω ϕ νϕ ϕ ω νϕ+ ≈ + (5.64)

and

( )sin cos 2p x d x x d xI I Iω ϕ νϕ ϕ ϕ ω ν− ≈ − (5.65)

where 2p dI I≈ for thin disc, and for a small angular displacement, ϕx, of the disc such that cos 1xϕ ≈

and sin x xϕ ϕ≈ .

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).

238

(a)

(b)

Figure 5.27 Angular momentum components (a) parallel (b) perpendicular to the undeflected position

of the shaft

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). At time t = 0 this vector

lies in the plane of paper, at time dt this vector moves behind the paper at an angle νdt (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, ( )2d xI ϕ ω ν− , multiplied by νdt. It is

given as

( )2d xI dtϕ ω ν ν− (5.66)

The rate of change of the angular momentum with time is then given as

( )2d xI ϕ ω ν ν− (5.67)

239

Hence, this 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.

Figure 5.28 The inertia force and moment acting from the disc on the shaft caused by the shaft

rotation, ω, and the shaft whirling, ν

Influence coefficients of the shaft can be defined as: α11 is the deflection y at the disc from 1 N force;

α12 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.

α21= α12 (by the Maxwell’s theorem of strength of materials); and α22 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

3 2

11 12 21 22, and3 2

l l l

EI EI EIα α α α= = = = (5.68)

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

11 12 12 22andx

y F M F Mα α ϕ α α= + = + (5.69)

It should be noted that the sign convention of M 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, F, and the moment, M, from Figure

5.28, we get

( ){ }2

11 12 2d xIy m y ϕ ν ω να ν α − −= + (5.70)

240

and

( ){ }2

12 22 2x d xIm yϕ ϕ ν ω να ν α − −= + (5.71)

which can be rearranged in matrix form as

2

11 12

2

12 22

1 (2 )0

1 (2 )

d

xd

ym I

m I

α ν α ν ω ν

ϕα ν α ν ω ν

− − =

− + − (5.72)

Equations (5.72) are homogeneous equations in y and ϕx, on putting determinant of the matrix equal

to zero, we get the frequency equation as

( ) ( ) ( ) ( )2 211 22 11 2212 12

4 3 2

22 11 222 2 2 1 0d d d d d dm I m I m I m I I m Iν α α α ν α α ω α ω ν α α ν α ω+− + − + + + − − = (5.73)

Equation (5.73) contain seven system parameters: ω, ν, m, Id, α11, α12, and α22, which makes a good

understanding of the solution very difficult. It is worthwhile to diminish the number of parameters as

much as possible by the dimensional analysis. Introducing four new variables

The dimensionless frequency: 11mν ν α= ;

The disc effect: 22

11

dI

m

αµ

α=

The elastic coupling:

2

12

11 22

αα

α α=

The dimensionless speed: 11mω ω α= (5.74)

With this new four non-dimensional variables, equation (5.73) becomes

4 3 21 2 12 0

( 1) 1 ( 1)

µ ων ων ν ν

µ α α µ α

+− + − − =

− − − (5.75)

Equation (5.75) is the fourth degree polynomial in ν , so for a given α , carrying a given disc µ , and

rotating at certain speed ω , there will be four natural frequencies of the whirl.

Case I: For a point mass of the disc, i.e., Id = 0 or µ = 0. Multiply equation (5.75) by µ and on

substituting µ = 0, we get