Dynamics of Mechanical Systems 2009 Part 11 pptx

If Ω2 > g/r, a small disturbance about the equilibrium position may be expressed as:14.3.11Then, to the first order in θ*, sinθ and cosθ and may be approximated as: From the definition o

482 Dynamics of Mechanical Systems

Equation (14.2.15) shows that the small disturbance gets larger and larger Thus, θ = π

is an unstable equilibrium position In the next several sections, we will use the foregoing

technique to explore the stability of several other mechanical systems

14.3 A Particle Moving in a Vertical Rotating Tube

Consider the system consisting of a particle free to move in the smooth interior of a vertical

rotating tube as depicted in Figure 14.3.1 (we considered the kinematics and dynamics of

this system in Section 8.5) If the angular speed Ω of the tube is prescribed, the system

has one degree of freedom represented by the angle θ From Eq (8.5.15) we see that the

equation of motion is then:

(14.3.1)

where r is the tube radius

If the particle P has reached an equilibrium position, will be zero The equilibrium

angle will then satisfy the equation:

(14.3.2)

It is readily seen that the solutions to this equation are:

(14.3.3)

Thus, there are three equilibrium positions In the following paragraphs, we consider the

stability of each of these

A vertical rotating tube with a smooth

interior surface and containing a particle P.

Ω

θ

By referring to the solutions of Eqs (14.2.5) and (14.2.12), we see that if [(g/r) – Ω2] is

positive, the solution of Eq (14.3.6) may be expressed in terms of trigonometric functions

and thus will be bounded and stable Alternatively, if [g/r – Ω2] is negative, the solution

of Eq (14.3.6) will be expressed in terms of exponential or hyperbolic functions and thus

will be unbounded and unstable Hence, the equilibrium position θ = 0 is stable if:

(14.3.9)

By substituting from Eqs (14.3.8) and (14.3.9) into (14.3.1) we obtain:

(14.3.10)

In this case, the coefficient of θ* is negative for all values of Ω Therefore, the solution of

Eq (14.3.10) will involve exponential or hyperbolic functions; thus, the equilibrium

posi-tion is unstable

Case 3: θθθθ = θθθθ 3 = cos –1(g/rΩ2 )

Finally, consider the equilibrium position θ = cos–1(g/rΩ2) Observe that this equilibrium

position will not exist unless g/rΩ2 is smaller than 1 That is, Ω2 must be greater than g/r

for equilibrium at θ = cos–1(g/rΩ2)

If Ω2 > g/r, a small disturbance about the equilibrium position may be expressed as:

(14.3.11)Then, to the first order in θ*, sinθ and cosθ and may be approximated as:

From the definition of θ3 in Eq (14.3.3), we see that:

and

(14.3.14)Hence, the coefficient of θ* in Eq (14.3.13) becomes:

θ θ= 3+θ*

sinθ=sin(θ3+θ*)=sinθ3+θ*cosθ3

cosθ=cos(θ3+θ*)=cosθ3−θ*sinθ3

3 2

(specifically, Ω2 < g/r), there are only two equilibrium positions: θ = θ1 = 0 and θ = θ2 = π,with θ = θ1 = 0 being stable (see Eq (14.3.7)) and θ = θ2 = π being unstable (see Eq (14.3.10)).For fast tube rotation (specifically, Ω2 > g/r), there are three equilibrium positions:

θ = θ1 = 0, θ = θ2 = π, and θ = θ3 = cos–1(g/rΩ2), with θ = θ1 = 0 and θ = θ2 = π being unstable(see Eqs (14.3.7) and (14.3.10)) and θ = θ3 = cos–1(g/rΩ2) being stable (see Eq (14.3.16)).That is, the third equilibrium position does not exist unless the tube rotation is such that

Ω2 > g/r, but if it does exist, it is stable.

14.4 A Freely Rotating Body

Consider next an arbitrarily shaped body B that is thrown into the air, rotating about one

of its central principal axes of inertia Our objective is to explore the stability of that motion;that is, will the body continue to rotate about the principal inertia axis or will it be unstable,wobbling away from the axis?

To answer this question, consider a free-body diagram of B as in Figure 14.4.1, where

G is the mass center of B; m is the mass of B; k is a vertical unit vector; F* and T* are theinertia force and couple torque, respectively, of a force system equivalent to the inertia

forces on B; –mgk is equivalent to the gravitational forces on B, with g being the gravity

constant; and R is an inertial reference frame in which B moves In the free-body diagram,

we have neglected air resistance; thus, the gravitational (or weight) force –mgk is the only

applied (or active) force on B.

From Eqs (7.12.1) and (7.12.8), we recall that the inertia force F* and couple torque T*

may be expressed as:

(14.4.1)

where a is the acceleration of G in R; ωω and αααα are the angular velocity and angular

acceleration, respectively, of B in R; and I is the central inertia dyadic of B (see Sections

7.4 to 7.9)

From the free-body diagram we then have:

(14.4.2)and

By inspection of Eqs (14.4.1) and (14.4.2), we have:

(14.4.4)

Thus, G moves as a projectile particle having a parabolic path (see Section 8.7) Note further that points of B not lying on the central principal inertia axis of rotation will not have a parabolic path That is, as B rotates it rotates about the central inertia axis.

From an inspection of Eqs (14.4.1) and (14.4.3) we have:

where I11, I22, and I33 are the central principal moments of inertia

Equations (14.4.8), (14.4.9), and (14.4.10) form a set of three coupled nonlinear ordinarydifferential equations for the three ωi (i = 1, 2, 3) To use these equations to determine the stability of rotation of the body, let B be thrown into the air such that B is initially rotating

about the central principal inertia axis parallel to n1 That is, let B be thrown into the air

such that its initial angular velocity components ωi are:

(14.4.11)(14.4.12)

(14.4.13)

a= −gk

I⋅ + × ⋅αα ωω (I ωω)=0

ωωαα

ω2=0

By inspection, we see that the ωi (i = 1, 2, 3) of Eqs (14.4.11), (14.4.12), and (14.4.13) are

solutions of Eqs (14.4.8), (14.4.9), and (14.4.10) To test for the stability of this solution, letsmall disturbances to the motion occur such that:

(14.4.14)(14.4.15)(14.4.16)

where as before the (*) quantities are small Then, by substituting these expressions intoEqs (14.4.8), (14.4.9), and (14.4.10), we have:

(14.4.22)Equation (14.4.20) has a solution:

(14.4.25)

Inspection of Eq (14.4.25) shows that the disturbance will remain small and the

motion of B will be stable if the coefficient of is positive, and that this will occur if I11

is either a maximum or minimum movement of inertia If I11 is an intermediate valued

moment of inertia (that is, if I33 < I11 < I22 or I22 < I11 < I33 ), the motion will be unstable.

14.5 The Rolling/Pivoting Circular Disk

Consider again the rolling circular disk (or “rolling coin”) as discussed earlier in Sections

4.12 and 8.13 and as shown in Figure 14.5.1 As before, D is the disk, with radius r, mass

m, mass center G, contact point C, and orientation angles θ, φ, and ψ Recall that the

condition of rolling requires that the contact point C of D has zero velocity relative to the rolling surface Recall further that by setting the moments of forces on D about C equal

to zero, we obtained the governing equations:

I I

ω2*ω2*

Case 1: Straight-Line Rolling

Recall from Eq (8.13.20) that if D is rolling in a straight line with constant speed the angles

θ, φ, and ψ are (see Figure 14.5.1):

(14.5.12)

where is the small constant As before, the motion is stable if the coefficient

of θ* is positive That is, the motion is stable if:

(14.5.13)

θ=0, φ φ= 0, a constant, ψ ψ˙ = ˙ ,0 a constant

˙ψ

Thus, if the angular speed of D exceeds , D will remain erect and continue to roll

in a straight line If the angular speed of D is less than , the motion is unstable D

will wobble and eventually fall

Case 2: Rolling in a Circle

Next, suppose D is rolling in a circle with uniform speed such that θ, , and are (seeFigure 14.5.1):

To test the stability of this motion, let the disk encounter a small disturbance such that

θ, , and have the forms:

(14.5.16)where, as before, the quantities with the (*) are small Then, sinθ and cosθ are:

(14.5.17)and

(14.5.18)

By substituting from Eqs (14.5.16), (14.5.17), and (14.5.18) into Eqs (14.5.1), (14.5.2), and(14.5.3) and by neglecting quadratic (second) and higher order terms in the (*) terms weobtain:

sinθ=sin(θ0+θ*)=sinθ0+θ*cosθ0

cosθ=cos(θ0+θ*)=cosθ0−θ*sinθ0

(14.5.23)where and are constants Solving Eq (14.5.23) for we have:

(14.5.24)Then, by substituting into Eq (14.5.22), we have:

(14.5.25)

Finally, by solving for and by substituting for and into Eq (14.5.19) (withoutthe first three terms) we have:

(14.5.26)where λ and κ* are defined as:

(14.5.27)

and

(14.5.28)

We recall from our previous analyses that stability will occur if the coefficient λ of θ* in

Eq (14.5.26) is positive; as a corollary, instability will occur if λ is negative (λ = 0 represents

a neutral condition, bordering between stability and instability.) Recall also from Eq.(14.5.15) and by inspection of Figure 14.5.1, that if 0 and 0 are positive, then θ0 must

be negative Hence, from Eq (14.5.27) we see that λ is positive; thus, stability occurs, if:

( )ψ˙ +φ˙ >( g r)cosθ −( )φ ψ˙ ˙ sinθ

Therefore, by making 0 and 0 sufficiently large, we can attain stability Finally, notethat in the limiting case when 0 = 0 and θ0 = 0, we have rolling in a straight line andthen Eq (14.5.29) becomes identical with Eq (14.5.27).

(14.5.32)(14.5.33)(14.5.34)Equations (14.5.33) and (14.5.34) may be integrated, leading to:

(14.5.35)and

(14.5.36)where, as before, and are small constants By substituting from Eq (14.5.35) into(14.5.32), eliminating , we obtain:

As before, the motion is stable if the coefficient of θ* is positive That is, the pivoting motion

is stable if:

(14.5.38)

14.6 Pivoting Disk with a Concentrated Mass on the Rim

For a generalization of the foregoing analysis consider a pivoting circular disk D with a concentrated mass at a point Q on the rim of the disk as depicted in Figure 14.6.1 Let the mass at Q be m, and let the mass of D be M As before, let the radius of D be r, and let the orientation of D be defined by the angles θ, φ, and ψ as in Figure 14.6.1.

The presence of the mass at Q makes it necessary to know the kinematics of Q The kinematics of Q, however, are directly dependent upon the kinematics of D Therefore, in developing the kinematics of Q it is helpful to review and summarize the kinematics of

D as previously developed in Section 4.12.

In developing the kinematics and dynamics of D and Q it is helpful to introduce various

unit vector sets as shown in Figure 14.6.1 As before, let N1, N2, and N3 be mutually

perpendicular unit vectors fixed in an inertial frame R; similarly, let d1, d2, and d3 be unit

vectors fixed in D; let n1, n2, and n3 be, as before, unit vectors parallel to principal inertia

axes of D; and, finally, let 1 and 2 be mutually perpendicular unit vectors parallel and

perpendicular to the surface on which D pivots as indicated in Figure 14.6.1.

To obtain relations between these unit vector sets it is helpful to use a configurationgraph (see Section 4.3) as in Figure 14.6.2 where and are reference frames containing

the ni and the (i = 1, 2, 3) respectively As before, it is convenient to express the

ˆ

ˆ

kinematical quantities in terms of n1, n2, and n3 To this end, the configuration graph may

be used to obtain the relations:

(14.6.1)

(14.6.2)

and

(14.6.3)

Using the procedures of Chapter 4 and the configuration graph of Figure 14.6.2, we

readily find the angular velocity of D in R to be (see Eq (4.7.6)):

(14.6.4)

Then, by using the third expression of Eq (14.6.2) to express n3 in terms of n2 and n3,

RωD becomes:

(14.6.5)

Observe in Eq (14.6.4) that in computing the angular acceleration of D in R we will

need to compute the derivatives of the unit vectors n1, n2, and n3 Observe further from

the configuration graph of Figure 14.6.1 that the ni (i = 1, 2, 3) are fixed in reference frame

Then, also from the configuration graph, we see that:

cos sin cos sin sin

Using Eq (14.6.7) we can now differentiate Eq (14.6.5) to obtain the angular acceleration

of D in R as:

(14.6.8)

Recall that because D is pivoting (a special case of rolling; see Section 4.11) on a surface

in the X–Y plane, the velocity of the center G of D in R is (see Eq (4.11.5)):

(14.6.9)

Similarly, because Q is fixed in D, the velocity of Q in R is:

(14.6.10)Then, by using Eqs (14.6.1) and (14.6.6), we obtain:

(14.6.11)

Finally, by differentiating in Eqs (14.6.9) and (14.6.11) and by using Eq (14.6.7), we

obtain the accelerations of G and Q in R to be:

˙ sin cos ˙ cos

˙ ˙ sin sin ˙ ˙ cos cossin

˙ ˙ cos cos ˙ cos sin cos

˙˙ cos ˙˙ sin cos

˙˙ cos cos ˙ ˙ sin

˙ ˙ sin ˙ cos sin ˙ ˙ cos sin

˙ cos ˙ cos ˙˙ sin ˙˙ sin sin

Next, consider the forces acting on D and Q The applied (or active) forces may be

represented by a contact force C acting through the contact point together with weight

forces w and W acting through Q and G The inertia (or passive) forces may be represented

by forces and and a couple with torque Figure 14.6.3 contains a free-bodydiagram depicting these forces and the torque Analytically, they may be expressed as:

(14.6.14)(14.6.15)

(14.6.16)

(14.6.17)and

(14.6.18)

where as before, I D is the central inertia dyadic of D (Regarding notation, the superscript

[*] in Eqs (14.6.15), (14.6.16), and (14.5.17) designates inertia forces and not “small” tities.)

quan-The unit vectors n1, n2, and n3 are parallel to principal inertia directions for D (see Figure

14.6.1) In terms of the ni (i = 1, 2, 3), the inertia torque may be expressed as (see Eqs.(8.6.10), (8.6.11), and (8.6.12)):

where the αi and the ωi (i = 1, 2, 3) are the n i components of RααααD and RωD and where I11,

I22, and I33 are:

(14.6.23)Consider next the free-body diagram itself in Figure 14.6.2 By setting moments of the

forces, about C, equal to zero we have:

(14.6.24)Then, by substituting from Eqs (14.6.13) through (14.6.18), we obtain the scalar equations:

(14.6.25)

(14.6.26)

and

(14.6.27)where the and the are the ni (i = 1, 2, 3) components of RaG and RaQ Then, bysubstituting from Eqs (14.6.12), (14.6.13), (14.6.25), (14.6.26), and (14.6.27), we have:

(14.6.28)

(14.6.29)

22 2

sin ˙˙ ˙ sin cos ˙ ˙ cos

cos cos ˙ ˙ cos ˙ sin cos ˙˙

˙˙ sin cos ˙ ˙ cos cos

˙˙ ˙˙ sin ˙ ˙ cos ˙˙ ˙ ˙ cos

˙˙ sin ˙ ˙ cos cos ˙ ˙ sin

˙ ˙ sin ˙ cos ˙ ˙ cos sin

˙ cos ˙˙ sin ˙˙ sin sinsin cos

(14.6.30)

Equations (14.6.28), (14.6.29), and (14.6.30) are the governing equations of motion of the

rim-weighted disk Observe that if we remove the mass on the rim (that is, let m = 0) and

if we let I11, I22, and I33 have the values as in Eq (14.6.23), then Eqs (14.6.28), (14.6.29), and(14.6.30) reduce to:

(14.6.31)

(14.6.32)and

(14.6.33)

After simplification, these equations are seen to be identical to Eqs (14.5.1), (14.5.2), and(14.5.3), the governing equations for the uniform rolling/pivoting circular disk

Next, observe that two steady-state solutions of Eqs (14.5.28), (14.6.29), and (14.6.30)

occur when the disk D is pivoting with the mass at Q in either the upper-most or most position Specifically, for Q in the upper- or top-most position, we have:

lower-(14.6.34)

where 0 is the steady-state spin speed Similarly, for Q in the lower- or bottom-most

position, we have:

(14.6.35)

In the following paragraphs we examine the stability of these two positions

14.6.1 Rim Mass in the Uppermost Position

Let there be a small disturbance to the equilibrium position with the rim mass in its most position, as defined by Eq (14.6.34) Specifically, let θ, , and ψ have the forms:

0

sin ˙ ˙ cos ˙ sin cos ˙˙ cos

˙˙ sin cos ˙ ˙ cos cos ˙ ˙ sin

sin ˙˙ ˙ sin cos ˙ ˙ cos ˙˙

˙ ˙ cos ˙ ˙ sin ˙ cos

where, as before, the quantities with the (*) are small Then, Eqs (14.6.28), (14.6.29), and(14.6.30) become (after simplification):

22

2 0 2

22 33

2 0 2

22 2 0 2

0 2

0 2

φφ

a b

a b

a b

1 1

2 2

3 3

00

00

00

,,

φφ

where by comparison of Eqs (14.6.42) and (14.6.44) the c i (i = 1,…, 8) are seen to be:

(14.6.45)

Note that each c i is positive

To solve Eqs (14.6.40) and (14.6.41), let θ* and ψ* have the forms:

(14.6.46)Then, Eqs (14.6.40) and (14.6.41) become:

(14.6.47)and

(14.6.48)or

(14.6.49)and

(14.6.50)

Equations (14.6.49) and (14.6.50) are simultaneous, homogeneous, linear, algebraic tions similar to those we obtained in the eigenvalue inertia problem (see Section 7.7, Eqs.(7.7.10) and (7.7.11)) There is a nonzero solution to these equations only if the determinant

equa-of the coefficients is zero That is,

(14.6.51)or

(14.6.52)or

θ*=Θ*eλt and ψ*=Ψ*eλt

1 2

where A, B, and C are (see Eqs (14.6.43)):

(14.6.54)or

(14.6.55)and

(14.6.56)Finally, by solving Eq (14.6.53) for λ2, we have:

(14.6.57)

Observe from Eq (14.6.46) that the solutions θ* and ψ* for the motion following thedisturbance will be bounded (that is, the motion is stable) if λ is either negative orimaginary Expressed another way, the motion is stable if λ does not have any positivereal part Thus, from Eq (14.6.57) we see that λ2 will be negative, and there will bestability if:

(14.6.58)

The first of these conditions is satisfied if A and B have the same signs From Eq (14.6.54)

we see that A is positive because c1 and c5 are positive Therefore, the first inequality in

Eq (14.6.58) is satisfied if B is positive From Eq (14.6.55), we see that B will be positive if:

(14.6.59)and

(14.6.60)

Because each of the c i (i = 1,…, 8) is positive (see Eq (14.6.45)), the inequality of Eq (14.6.60)

will be satisfied if the spin rate 0 is sufficiently large and the inequality of Eq (14.6.59)

is satisfied That is, the first inequality of Eq (14.6.58) will be satisfied if 0 is sufficientlylarge and if:

1 242

From Eq (14.6.45) we see after computation and simplification that:

(14.6.62)

Therefore, the inequality of Eq (14.6.61) is inherently satisfied

Next, the second inequality of Eq (14.6.58) will be satisfied if AC is positive By tion of Eqs (14.6.54) and (14.6.56) we see immediately that both A and C are positive, thus

inspec-the second inequality of Eq (14.6.58) is satisfied Therefore, inspec-the motion of inspec-the pivotingdisk with the rim mass in the upper-most position is stable if 0 is sufficiently large tosatisfy Eq (14.6.60) From Eq (14.6.45) we see that:

(14.6.63)Hence, from Eqs (14.6.60), (14.6.62), and (14.6.63) we have the stability criterion:

(14.6.64)

Finally, observe that if m = 0, the stability criterion reduces to:

(14.6.65)which is identical to Eq (14.5.38)

14.6.2 Rim Mass in the Lowermost Position

Next, let there be a small disturbance to the equilibrium position with the rim mass in itslower-most position as defined by Eq (14.6.35) That is, let θ, , and ψ have the forms:

(14.6.66)where, as before, the quantities with the (*) are small The governing equations, Eqs.(14.6.28), (14.6.29), and (14.6.30), then become (after simplification):

(14.6.67)

(14.6.68)and

Note that in the development of these equations we have neglected all products of smallquantities (those with a [*]) Also, we have approximated trigonometric functions as:

(14.6.72)

where the a i and b i (i = 1, 2, 3) are:

and

(14.6.73) Following the procedure of the foregoing case we can solve Eqs (14.6.67) and (14.6.68)

by letting θ* and ψ* have the forms:

(14.6.74)Then, Eqs (14.6.67) and (14.6.68) become:

(14.6.75)and

(14.6.76)or

(14.6.77)and

As with Eqs (14.6.49) and (14.6.50), Eqs (14.6.77) and (14.6.78) are simultaneous, geneous, linear, algebraic equations Thus, there is a nonzero solution to these equationsonly if the determinant of the coefficients is zero That is,

homo-(14.6.79)or

(14.6.80)or

(14.6.81)where , , and are:

(14.6.82)

(14.6.83)

(14.6.84)Solving for λ2 we have:

(14.6.85)

As before, the solutions θ* and ψ* of Eq (14.6.74) are stable if λ does not have any positivereal part From Eq (14.6.85), this means that stability occurs if (see Eq (14.6.58)):

(14.6.86)From Eqs (14.6.73), (14.6.82), (14.6.83), and (14.6.84), we see that , , and are:

(14.6.87)

(14.6.88)and

(14.6.89)

3 1 2

By inspection of Eqs (14.6.87) and (14.6.88), we see that the first condition of Eq (14.6.86)

is satisfied if 0 is sufficiently large The second condition of Eq (14.6.86), however, willnot be satisfied if is positive and is negative independent of the magnitude or sign

of 0 Thus, for all pivoting speeds, the disk with the rim mass in the lower position isunstable

Finally, observe that with m = 0, Eq (14.6.89) shows that = 0, thus stability can be

attained by having > 0 This in turn will occur if:

(14.6.90)which is identical to Eqs (14.6.65) and (14.5.38)

14.7 Discussion: Routh–Hurwitz Criteria

Recall that in the previous section we solved the governing equations for the disturbancemotion by seeking to simultaneously solve the equations Consider again Eqs (14.6.50)and (14.6.41):

(14.7.1)

(14.7.2)

Instead of seeking simultaneous solutions to these equations we could have eliminatedone of the dependent variables, leaving a single equation of higher order for the othervariable Specifically, suppose we solve Eq (14.7.1) for , giving us:

(14.7.3)Then, by differentiating, we have:

(14.7.4)Also, by differentiating Eq (14.7.2), we have:

(14.7.5)Finally, by substituting from Eqs (14.7.3) and (14.7.4), we obtain (after simplification):

(14.7.6)

To solve Eq (14.7.6), let θ* have the form:

(14.7.7)

˙φˆ

Then, we obtain:

(14.7.8)

Equation (14.7.8) is identical to Eq (14.6.52), which was obtained by the simultaneoussolution of Eqs (14.6.40) and (14.6.41)

It happens, in infinitesimal stability procedures in the analysis of small disturbances of

a system from equilibrium, that we can generally convert the governing simultaneouslinear differential equations into a single differential equation of higher order, as Eq.(14.7.6) Then, in the solution of the higher order equation, we generally obtain a polyno-mial equation of the form:

(14.7.9)

From our previous analyses we see that a solution to the governing equations in theform of θ* in Eq (14.7.7) is stable if (and only if) λ does not have a positive real part.Necessary and sufficient conditions such that Eq (14.7.9) will produce no λi (i = 1,…, n)

with positive real parts may be determined using the Routh–Hurwitz criteria (see ence 14.1) Specifically, let determinants ∆i (i = 1,…, n) be defined as:

Refer-(14.7.10)

Then, the criteria, such that the solutions λi (i = 1,…, n) have no positive real part, are that

each of the ∆i (i = 1,…, n) must be positive and that each of the a k (k = 0,…, n) must have

the same sign

The above statements asserting that the a i and the ∆i need to be positive for the λi to

have no real parts are commonly called the Routh–Hurwitz stability criteria It is seen that