AERO 310 Lecture Notes
Suman Chakravorty*
Contents
1 Angular Momentum of a Rigid Body
2
2 Properties of the Inertia Matrix
4
3 Principal Moments of Inertia
5
4 Parallel Axis Theorem
8
5 Eulerian Mechanics: Symmetric Top Example
10
6 Conservation of Energy/ Angular Momentum
14
7 Application of Conservation of Energy: The Symmetric Top Revisited
15
8 Torque-Free Motion: Symmetric Body
19
9 Precessing Top on a Smooth Surface
22
10 Rolling Disk
25
11 Linearization of Nonlinear ODEs
28
12 Linear Dynamical Systems
34
* Department
of Aerospace Engineering, Texas A&M University, College Station, TX
1
Rigid Body Dynamics
Previously, the discussion of dynamics was focused on particles, which were infinitesimally
small bodies located at a single point. Rigid bodies are collections of particles which do
not deform when subjected to loads and moments. What this means is that the distance
between two points on the rigid body will remain constant while loads and moments are
being applied to the body.
1
Angular Momentum of a Rigid Body
Figure 1:
Recall Euler’s equation, which states that the time rate of change of the angular momentum
about a point, O, is equal to the moment about that same point.
˙
HO = MO
(1)
2
In the following, we shall find a formula for H O , the angular momentum, in the case of a
rigid body.
Z
X
˙
˙
HO =
R p · dm · R p ≈ R p · R p dm
(2)
dm
B
From Figure 1.1, R p = R C + ρ , and equation 2 becomes
Z
HO =
˙
(R C + ρ ) · (R C + ρ̇ ) dm
B
Z
=
˙
(R C · R C ) dm +
Z
(R C · ρ̇ ) dm +
B
B
B
B
˙
= R c · mR c +
˙
(ρ · R C ) dm +
Z
(ρ · ρ̇ ) dm
B
0
0Z
Z
Z
> ˙
>
dm) + R C · ( ρ
dm) + (ρ · ρ̇ ) dm
dm + R C · ( ρ̇
Z
˙
= (R C · R C )
Z
Z
B
B
B
(ρ · ρ̇ ) dm
B
The final expression for the angular momentum about any arbitrary point, "O", can be simplified more, and is given in terms of the angular momentum about the center of mass.
HO =
˙
R · mR
| c {z }c
AM of CM about O
+
Hc
|{z}
AM of B about C
Let us now simplify
R the angular momentum of B about the point, "C". Starting with the
expression H c = (ρ · ρ̇ ) dm.
B
B
Note that ρ̇ =
TT
N d
dt (ρ ) =
d >
(
ρ
)
dt
| {z }
0
R
+ ( N ωB × ρ ) = ω × ρ . Thus, H c = ρ · (ω × ρ ) dm
B
B
Since B is a rigid body
Let ρ = (ρ 1 bˆ1 , ρ 2 bˆ2 , ρ 3 bˆ3 ) and ω = (ω1 , ω2 , ω3 ).
With those definititions of ρ and ω we can see that ρ · (ω × ρ ) =

ρ 22 + ρ 23
− ρ1ρ2
−ρ 1 ρ 3

ω1




 −ρ 2 ρ 1 ρ 2 + ρ 2 −ρ 2 ρ 3   ω2 
1
3



ω3
−ρ 3 ρ 1 − ρ 3 ρ 2 ρ 21 + ρ 22
3
Combining everything, the final result is H c =
Z
Z
Z

2
2

ρ
+
ρ
−
ρ
ρ
−
ρ
ρ
1 2
1 3 
2
3

 Z
 ω1
Z
Z



 −ρ 2 ρ 1


ρ 21 + ρ 23
−ρ 2 ρ 3 

  ω2 
 Z

Z
Z
 ω3

2
2
−ρ 3 ρ 1
−ρ 3 ρ 2
ρ1 + ρ2
= [ I ]c ω
(3)
[ I ] c ≡ the moment of inertia matrix about the center of mass "c". The inertia matrix [ I ] c is
independent of the motion of the rigid body. However, it is dependent on the body fixed axes
0
B
0
B, i.e., [ I ]B
c is different from [ I ] c when B 6= B.
2
Properties of the Inertia Matrix
1.
Let B’ and B be two body fixed frames with inertia matrices [I’] and [I].
Then: [ I 0 ] = [B0 B][ I ][BB0 ]...(∗)
Recall that [B’B] is the direct cosine matrix from B → B0 .
2.
There exists a unique reference frame B∗ such that

I 1∗ 0

∗
[ I ]∗ = 
 0 I2
0 0
3.
0


0

I 3∗
Finding B* (Principal Axes):
Since [I] is a symmetric matrix (why?), we can always write that [ I ] = V [ I ]∗ V T ...(∗∗),
where V = [v1 , v2 , v3 ], V T = V −1 , and [ I ]v i = I ∗i v i .
Note: ( I ∗i , v j ) are the eigenvalue/ eigenvectors of [I] for i = 1,2,3.
Compare (*) with (**), we get:
[ I ]∗ = |{z}
V T [ I ] |{z}
V
[B*B]
[BB*]
and thus, V T = [B ∗ B] = [v1 , v2 , v3 ]T
( I ∗j v i ) are called principle moments and the principle directions of inertia, i = 1, 2, 3.
4
3
Principal Moments of Inertia
Figure 2:
Consider a slender uniform rod of length "L" and mass "M". Find the moments of inertia of
the rod about the axes shown. Also find the principal axes.
R
Recall from the inertia matrix that I 11 = (ρ 22 + ρ 23 ) dm. In this case dm = ρ dl , ρ 2 = lsin(α),
ρ 3 = 0, ρ 1 = l cos(α). (Please note that the coordinates are ρ 1 , ρ 2 , ρ 3 while density is ρ )
Z
Thus: I 11 =
L
2
−L
2
( l 2 sin2 (α) + 02 )ρ dl =
2
Z
Similarly; I 12 = −
¯L
¯2
M l3
2
L [ 3 sin (α)] ¯ −L
L
2
−L
2
(ρ 1 ρ 2 ) dm = −
Z
L
2
−L
2
=
ML2
2
12 sin (α).
( l 2 sin(α) cos(α))ρ dl =
Further, note that I 23 = I 13 = I 31 = I 32 = 0 (Since
slender and uniform rod is:

sin2 (α) − sin(α) cos(α)
ML2 
 − sin(α) cos(α) cos2 (α)
[ I ]c =
12 
0 0
5
ML2
12 sin(α) cos(α).
ρ 3 = 0). Thus the inertia matrix for this
0


0

1
Next, Let us find the principal axes, such that the off diagonal terms of the moment of
inertia matrix above are zero. For the sake of solving the resulting eigenvalue problem,
assume that ML2 = 12 and α = 45◦ . Plugging in those values, the matrix becomes:

1

 2

[ I ]c =  − 1

 2
0

1
−
2
1
2
0
0


0


1
Now, we find the eigenvalues and eigenvectors of the inertia matrix above. To do this, we
use the equation [ I ]v = λv, where v is the eigenvector and λ is the eigenvalue.

1

 2
 1
−

 2
0
1
−
2
1
2
0

 
 
0 v
v1
1
 
 
 

v = λ  v2 

0
 2 
 v3
v3
1
Eigenvalues are found using the generic formula det( A − λ I ) = 0, where the I term is an
indentity matrix, and in our case A is the inertia matrix.
Thus eigenvalues are found with: |λ I − [ I ]| = 0
¯
¯
¯
¯
¯
¯ λ− 1 1
0
¯
¯
2 2
¯
¯
¡
1
1 ¢
¯
¯
1
1
=
0
→
(
λ − 1) (λ − )2 − ( )2 ) = 0 → λ = 1, λ = 0, λ = 1
¯
¯
λ−
0¯
¯
2
2
|
{z
}
¯
¯
2
2
¯
¯
principal moments
¯
0 0
λ−1 ¯
There is an eigenvalue at 0, and a repeated eigenvalue at 1.
Now that we have the λ values we can solve for the eigenvectors, v, with the previously
shown equation [ I ]v = λv.
Case 1, λ = 0:

1

 2
 1
−

 2
0
The
1
−
2
1
2
0



0 v
v − v = 0
 1 
1
2
 
v
=0→
0
 2 
 v3 = 0
 v3
1



1
p 
 2


1 
→ v1 = v2 and v3 = 0 → v1 = 
p 
 2


0
p
2 serves to normalize the eigenvector, i.e., make its length equal to one.
6
Case 2, λ = 12 : There are two eignevectors for this eignevalue

1

 2
 1
−

 2
0
1
−
2
1
2
0

 
 

0 v
v1
 1 
 v1 − v2 = v


1
 
2


v = 1  v2  → 2
0
 2 
 v3 = v3
 v3
v3
1

v = −v
1
2
→
 v3 = v3




1



 p 



2








1
v2 =  −



p




2


→
0


 



0


 




v3 = 


0



1
The three eigenvectors, v1 , v2 , and v3 are representative of the principal axes for the slender
rod pictured in figure 2. Using the principal axes will result in a principal moment of inertia
matrix where all off-diagonal terms of the matrix are zero.
Figure 3:
7
...(∗)
4
Parallel Axis Theorem
The parallel axis theorem allows us to transfer the moments of inertia to a parallel set of
axes.
Figure 4:
Z
r 22 + r 23

 Z

[ I ]O = 
 −r 2 r 1
 Z

−r 3 r 1
Z
Z
−r 1 r 2
Z
r 21 + r 23
Z
−r 3 r 2

−r 1 r 3 


−r 2 r 3 


Z

2
2
r1 + r2
Z
Let us examine the first term of the I O matrix.
R
O
I 11
= ( r 22 + r 23 ) dm
8
R
= ( (∆2 + ρ 2 )2 + (∆3 + ρ 3 )2 ) dm
R
c
= M (∆22 + ∆23 ) + (ρ 22 + ρ 23 ) dm = I 11
+ M (∆22 + ∆33 )
O
C
Similarly, I 12
= I 12
− M ∆1 ∆2 . Thus;


∆22 + ∆23 − ∆1 ∆2 −∆1 ∆3




[ I ]O = [ I ]C + M  −∆2 ∆1 ∆21 + ∆23 −∆2 ∆3 


−∆3 ∆1 − ∆3 ∆2 ∆21 + ∆22
We can see that the parallel axis theorim transfers the moment of inertia matrix from point
C on the rigid body to the arbitrary point O. Let’s use the following as an example.
Figure 5:
Let us say that:


It 0
0



[ I ]C = 
 0 I t 0
0 0 Is
(Why is the above true?)
Then, using the parallel axis theorem:


I t + ML2 0
0


[ I ]O = 
0 I t + ML2 0


0 0
Is
Note that in this example ∆ = (0, , 0 , L)
9
5
Eulerian Mechanics: Symmetric Top Example
Figure 6:
In Figure 6, the coordinate system is a 3-1-3 Euler angle system (ψ, θ , φ)
Assume that:

It 0

[ I ]C = 
 0 It
0 0
0


0

Is
We fix two sets of coordinates: ( q̂, n̂, b̂ 3 ) and ( b̂ 1 , b̂ 2 , b̂ 3 ). Note that the ( b̂ 1 , b̂ 2 , b̂ 3 ) coordinate system is a "body-fixed frame" while ( q̂, n̂, b̂ 3 ) is not, meaning it is not spinning (does
not have the "φ" part in Figure 6).
10
The free body diagram of the spinning top:
Figure 7:
Note that if we do Euler’s equation about point O, the reaction forces, R 1 , R 2 , and R 3 do not
matter.
H O = [ I ]O ω

 
I t + ML2 0
0
I 0t 0

 
0

[ I ]O = 
0 I t + ML2 0

 =  0 It
0 0
0 0
Is
HO:
.
.
.
.
.
.
0


0
 (F ound using parall el axis theorem)
Is
.
ω = ψ n̂ 3 + θ n̂ + φ b̂ 3 = −ψ sin θ q̂ + θ n̂ + (φ + ψ cos θ ) b̂ 3
From figure 1.6b: q̂ = − sin φ b̂ 1 − cos φ b̂ 2 ; n̂ = cos φ b̂ 1 − sin φ b̂ 2 . . . . (1)
.
.
.
.
.
.
ω = ψ sin θ sin φ b̂ 1 + ψ sin θ cos φ b̂ 2 + θ cos φ b̂ 1 − θ sin φ b̂ 2 + (ψ cos θ + φ) b̂ 3
.
.
.
.
→ ω = (ψ sin θ sin φ + θ cos φ) b̂ 1 + (ψ sin θ cos φ − θ sin φ) b̂ 2 + Ω b̂ 3
.
.
Where Ω = ψ cos θ + φ and is defined as the "total spin"
11
Therefore:

 .
. 
0 ψS θ S φ + θ C φ b̂ 1
 .
. 
0
b̂
ψ
S
θ
S
φ
−
θ
S φ

 2

Is
Ω b̂ 3
I 0t
0


H O = [ I ]O ω =  0 I 0t
0 0
.
.
.
.
→= I 0t (ψS θ S φ + θ C φ) b̂ 1 + I 0t (ψS θ S φ − θ S φ) + I s Ω
.
.
= I 0t ψS θ (S φ b̂ 1 + C φ b̂ 2 ) + I 0t θ (C φ b̂ 1 − S φ b̂ 2 ) + I s Ω
{z
}
{z
}
|
|
− q̂
n̂
.
.
⇒ H O = − I 0t ψ sin θ q̂ + I 0t θ n̂ + I s Ω b̂ 3 . . . . (2)
Note the above result is in the non-spinning frame
.
Finding the time rate of change of the angular momentum: H O :
.
TT B d
dt ( H O ) + (ω × H O )
HO =
IMPORTANT: The ω in the above equation corresponds to the co-ordinate system in which
we have represented H O
Thus,
.
..
..
.
..
H O = (− I 0t ψ sin θ − I 0t ψθ cos θ ) q̂ + ( I 0t θ ) n̂ + I s Ω b̂ 3
.
.
.
.
.
+ (−ψ sin θ q̂ + θ n̂ + ψ cos θ b̂ 3 ) × (− I 0t ψ sin θ q̂ + I 0t θ n̂ + I s Ω b̂ 3 )
{z
}
|
.
Why? Note there is no φ
.
..
..
.
..
.
.
.
⇒ H O = (− I 0t ψ sin θ − 2 I 0t ψθ cos θ + I s θ Ω) q̂ + ( I 0t θ + I s ψΩ sin θ − I 0t ψ sin θ cos θ ) n̂ + I s Ω b̂ 3 .
.(3)
M O : From figure 1.6, it is clear that:
M O = r × mg(− n̂ 3 ) = mgL sin θ n̂ . . . . (4)
.
H O = M O : Using equations (3) and (4), we obtain
..
..
0
..
.
I 0t ψ sin θ + 2 I 0t ψθ cos θ + I s θ Ω = 0 .
.
I t θ + I s ψΩ sin θ −
.
I sΩ = 0 .
.
.
I 0t ψ2 sin θ cos θ
.
.
.
.
. ( q̂)
= mgL sin θ .
. ( n̂) ....(5)
.
. ( b̂ 3 )
.
.
Equation (5) is the equation of motion of a symmetric top
12
.
It is immediately clear from the equations of motion that Ω = 0, thus the total spin,Ω, is a
constant. Note, however, that this is true only in the axisymmetric case. WHY? (Hint: see
.
the derivation of H O )
Steady State Precession:
.
..
.
Suppose now that θ .= 0 . Then, we see that from equation
. 5( q̂) that ψ = 0 → ψ = constant.
.
But Ω = ψ cos θ .+ φ = constant. Since θ = constant, φ = constant as well. Thus, if θ =
.
constant, ψ and φ are also constant. Then equation 5( n̂) becomes:
.
.
sin
θ
sin
θ cos θ = mgl
sin
θ − I 0t ψ2
I s ψΩ
⇒
.
.2
.
I s ψΩ − I 0t ψ2 cos θ = mgl
.
..
I s ψ cos θ − I 0t ψ2 cos θ + I s ψφ = mgl
....(6)
.
Equation (6) is. the steady-state precession equation. Note that this is a quadratic in ψ.
.
Given θ , and φ, one can find ψ. However it is not clear from just (6) which of the two
.
solutions is the right ψ! Why??
13
6
Conservation of Energy/ Angular Momentum
The kinetic energy of a rigid body is given by the expression:
T =
1
1 T
ω [ I ] c ω + mv c · v c
|2 {z }
|2 {z }
Rotational
Translational
Where the first part of the equation corresponds to the rotational kinetic energy of the body
and the translational part due to the movement (translation) of the body.
In addition, if the body is in pure rotation (remember the precessing top), the expression
can be simplified to:
T = 12 ωT [ I ]O ω
Where the [ I ]O is the moment of inertia about O in terms of the parallel set of axes B’.
In essence, the translational kinetic energy is subsumed into the translated moment of
inertia [ I ]O .
If B was a principal set of axes (as will always be the case in this class), then:
Trot =
1
2
2 I 1 ω1
+
1
2
2 I 2 ω2
+
1
3
2 I 3 ω1
And similarly using parallel axis theorem:
T =
1 0 2
2 I 1 ω1
+
1 0 2
2 I 2 ω2
+
1 0 3
2 I 3 ω1 ,
Where ( I 10 , I 20 , I 30 ) are the parallel set of axes.
14
7
Application of Conservation of Energy: The Symmetric Top Revisited
Figure 8:
A heavy symmetric top is precessing as shown above. In the ensuing. motion, the top
. reaches
.
.
a maximum nutation angle, θmax . Given that θ (0) = θ o ; ψ(0) = 0; θ (0)
=
0;
and
φ(0) = φ o ,
.
.
find the value of θmax and the associated precession and spin values φmax , ψmax at the maximum deflection.
.
Note
that we have 3 unknowns in this problem
. that we need to solve for: θmax , ψmax and
.
φmax . Why? Note that at θmax , by definition, θmax = 0 since the top is changing its direction
of motion.
Our goal in this problem is to find 3 "algebraic" equations that we can solve for the 3 unknowns. The following is what we know:
1. We know that the total spin is conserved (Ω = constant)
2. Note that the reaction forces are not doing any work and gravity is conservative⇒
Energy is conserved
3. Note that the moment due to gravity at O is in the n̂ direction ⇒
The angular momentum in the n̂ 3 direction is conserved
15
Applying what we know:
1. Total Spin Conserved:
.
.
From previous sections, Ω = ψ cos θ + φ = constant. The "constant of motion" is determined
by the initial conditions.
.
.
.
.
θo
*0
.
φo
.
.
.
.
*
= φ and hence ψ
cos
Then, ψ cos θ + φ = ψ(0)
θ (0)
+
φ(0)
o
max cos θmax + φmax = φ o . . . . (1)
>
2. Conservation of Energy:
We have T + V = constant. First we will find the "general expression" for T and V, and
only then, substitute the boundary conditions!!
It can be seen from Figure 1.8 that V = mgL cos θ
Recall that the translated moments of inertia of the top about O was taken to be ( I 0t , I 0t , I s ).
.
.
.
.
ω = (−ψ sin θ ) q̂ + θ n̂ + (ψ cos θ + φ) b̂ 3
|
{z
}
Ω
and hence ⇒ T =
1 0 .
2
2 I t (ψ sin θ )
+
.
1 0
2
2 I t (θ )
+
1
2
2 I s (Ω)
Recall: T + V = constant. Again to get the "constant of motion" we substitute the initial
conditions.
.
⇒ T o + Vo = mgL cos θ (0) +
= mgL cos θ o +
. 2
*0
1 0
ψ
2 I t
.
(0) sin2 θ (0) +
. *0
0 2
1
θ
2 I t
(0) +
>
1
2
Ω
2 I s
φ2o
1
2
2 I s φo
At the maximum deflection point:
T max + Vmax = mgL cos θmax +
⇒ = mgL cos θmax +
1 0 .2
2
2 I t ψmax sin θmax
1 0 .2
2
2 I t ψmax sin θmax
Since T o + Vo = T max + Vmax :
.
1 2
⇒ mgL cos θ o + 2 I s φ o = mgL cos θmax +
⇒ mgL cos θ o = mgL cos θmax +
+
* 0 (why?)
1 0 2
θmax + 12 I s Ω2
2 I t
+
1
2
2 I sΩ
1 0 .2
2
2 I t ψmax sin θmax
2
1
+
sΩ
2 I
1 0 .2
I t ψmax sin2 θmax . . . . (2)
2
16
3. Conservation of Angular Momentum About the n̂ 3 Axis:
Figure 9:
The angular momentum of the top about O is given by
.
..
..
..
.
H O = (− I 0t ψ sin θ − I 0t ψθ cos θ ) q̂ + ( I 0t θ ) n̂ + I s Ω b̂ 3
We have H O · n̂ 3 = HO,3 = constant. Again, we find the "general expression", and only then
substitute the boundary conditions.
.
HO,3 = I 0t ψ sin θ cos 90 − θ + I s Ω cos θ (why? see figure 1.9)
.
⇒ =; I 0t ψ sin2 θ + I s Ω cos θ
Now we substitute the initial conditions to get the constant.
.
.
φo
*
HO,3 (0) = I 0t
ψ(0) sin2 θ (0) + I s
Ω
(0) cos θ (0) = I s φ o cos θ o
. *0
17
At the maximum deflection point, we get
.
HO,3 max = I 0t ψmax sin2 θmax + I S Ω cos θmax
Equating the maximum and initial angular momentum equations we get
.
.
⇒ I 0t ψmax sin2 θmax + I s Ω cos θmax = I s ψ o cos θ o . . . . (3)
.
.
Equations (1)-(3) can be used to solve for θmax , ψmax , and φmax now.
Lets now rearrange equations (2) and (3)
(2) ⇒ mgL(cos θ o − cos θmax ) =
1 0 .2
2
2 I t ψmax sin θmax
.
(3) ⇒ I s Ω(cos θ o − cos θmax ) = I 0t ψmax sin2 θmax
Thus:
1 0 .2
2
It ψmax
sin
θmax
2
=
mgL
.
⇒ ψmax =
2mgL
I sΩ
(Why is Ω used here?)
.
2
I0t ψmax
sin
θmax
I sΩ
. . . . (4)
We can substitute (4) into (3) to obtain:
I 0t
2mgL
I sΩ
.
sin2 θmax + I s Ω cos θmax = I s ψ o cos θ o
2mgL
.
⇒ I 0t
(1 − cos2 θmax ) + I s Ω cos θmax = I s ψ o cos θmax
I Ω
| {zs }
C
Divide by C:
2
1 − cos θmax +
( I s Ω )2
2mgL I 0t
cos θmax =
( I s Ω )2
2mgL I 0t
cos θ o
( I s Ω )2
Let
=D
2mgL I 0t
cos2 θmax − D cos θmax + (D cos θ o − 1) = 0
| {z }
|
{z
}
t
E
⇒ t2 − Dt + E = 0 . . . . (5)
This is now a quadratic equation that we can use to solve for θmax . There will be two values
of θmax , why?
.
.
ψmax and θmax can be obtained from (4) and (5) thus φmax can be obtained with:
.
.
.
⇒ φmax = ψ o − ψmax cos θmax . . . . (6)
18
8
Torque-Free Motion: Symmetric Body
Next we consider the case of torque-free motion.
Figure 10:
The geometry of the problem can be shown to be as in figure 1.10 above. See the justification
in the free body diagram below.
Figure 11: No forces (best FBD ever!!)
.
Note that H C = M C but, M = 0 here, and thus
.
H C = 0 ⇒ H C = contsant
19
Note that the direction as well as the magnitude of H C is fixed! Hence, we take the direction
the symmetry
of H C as the "inertially fixed" n̂ 3 direction. Next, we define the angle between
.
axis b̂ 3 and n̂ 3 as the "nutation angle θ ". It’s clear that the spin φ has to be in the b̂ 3
.
direction. Further, ψ is the "precession rate" and about the n̂ 3 axis. Thus, the geometry
is given by the 3-1-3 Euler angles and identical to the top, except here the inertially fixed
direction n̂ 3 is the same as the direction of H c .
Now, let us derive the equations of motion for torque-free motion.
Let | H C | = H . Then, it’s clear that H C is always in the ( q̂, b̂ 3 ) plane.
H c = H cos θ b̂ 3 − H sin θ q̂. . . . (1)
Thus:
.
.
.
.
But, H C = − I t ψ sin θ q̂ + I t θ n̂ + I s (ψ cos θ + φ) b̂ 3 . . . . (2)
.
.
.
.
This is because ω = −ψ sin θ q̂ + θ n̂ + (ψ cos θ + φ) b̂ 3 , just as in the spinning top case, and
given that the moments of inertia are ( I t , I t , I s ) about C.
.
Using (1) and (2) it can be seen that θ = 0, i.e., there is no nutation, the nutation angle is a
fixed value!
Also:
.
q̂ : − H
sin
θ = − I t ψ
sin
θ
.
.
b̂ 3 : H cos θ = I s (ψ cos θ + φ)
.
From the q̂ equation: ⇒ ψ =
H
It
. . . . (3)
Then, substitute (3) into the b̂ 3 equation
H cos θ = I s
H (1 −
Is
It
H
It
.
cos θ + I s φ
.
.
) cos θ = I s φ ⇒ φ = (
It − Is
It Is
) . . . . (4)
Thus, it’s rather straightforward
to derive the equations of motion, in particular, the spin
. .
and precession rates, φ, ψ, in terms of H and the nutation angle θ .
20
Two cases:
1. Prolate Body e.g. cylinder, cone, etc.
Figure 12:
.
It > Is ⇒ φ > 0
Prograde precession
2. Oblate body e.g. disk, coin, etc.
Figure 13:
.
It < Is ⇒ φ < 0
Retrograde precession
21
9
Precessing Top on a Smooth Surface
Figure 14:
For a spinning top on a smooth floor, find the equations of motion of the top. Assume there
is no "horizontal velocity".
Again, we can use the 3-1-3 angles ψ−θ −φ to describe the orientation of the top. DOFs: ψ, θ , φ
Figure 15: FBD
The free body diagram is as shown above. Why are there no horizontal reactions?
.
Euler’s Equation: Note that O is accelerating and thus, H O 6= M O . In this case, we want
.
to use Euler’s equation about C, i.e., H C = M C . Note that Euler’s equation "always holds"
about the center of mass, even when it’s accelerating.
22
.
HC = MC :
.
.
.
.
ω = −ψ sin θ q̂ + θ n̂ + (ψ cos θ + φ) b̂ 3
.
.
.
.
⇒ H C = − I t ψ sin θ q̂ + I t θ n̂ + I s (ψ cos θ + φ) b̂ 3
{z
}
|
Ω
.
..
.
..
..
.
.
.
⇒ H C = (− I t ψ sin θ − 2 I t ψθ cos θ + I s θ Ω) q̂ + ( I t θ + I s ψΩ sin θ − I t ψ sin θ cos θ ) n̂ + I s Ω b̂ 3
.
Note that H C is the same as before!
M C = NL sin θ n̂ (since mg does not create a moment about C)
Thus the equations of motion are:
..
..
..
.
I t ψ sin θ + 2 I t ψθ cos θ + I s θ Ω = 0 .
.
.
.
.2
I t θ + I s ψΩ sin θ − I t ψ sin θ cos θ = NL sin θ .
.
I sΩ = 0 .
.
.
.
.
.
. ( q̂)
. ( n̂) ....(1)
.
. ( b̂ 3 )
The equations of motion are exactly the same as the precessing top except instead of the
translated moments of inertia I 0t = I t + mL2 in the previous case, it is simply I t .
However, note that we are not done! The reaction force N is unknown! Hence, here we will
need to use Newton’s second law to solve further.
Figure 16:
*Note that for Newton’s law, we are using ( n̂ 1 , n̂ 2 , n̂ 3 ) which is different from the ( q̂, n̂, b̂ 3 )
frame for Euler’s equation. This is perfectly fine since we want to use convenient reference
frame for Newton/Euler.
zC = L cos θ ⇒ r C = L cos θ n̂ 3
.
.
..
.
r C = −Lθ sin θ n̂ 3 ⇒ aC = (−Lθ sin θ − Lθ 2 cos θ ) n̂ 3
F = (N − mg) n̂ 3
..
.
Thus: N − mg = −mLθ sin θ − mLθ 2 cos θ
..
.
⇒ N = mg − mLθ sin θ − mLθ 2 cos θ . . . . (2)
23
Substituting (2) back into (1) ⇒
..
..
..
.
I t ψ sin θ + 2 I t ψθ cos θ + I s θ Ω = 0 .
.
.
..
.2
.
. ( q̂)
.
I t θ + I s ψΩ sin θ − I t ψ sin θ cos θ = (mg − mLθ sin θ − mLθ 2 cos θ )L sin θ .
.
I sΩ = 0 .
.
.
.
.
.
.
. ( n̂) ....(3)
. ( b̂ 3 )
Thus (3) is the equations of motion of a top on a smooth surface.
Steady Precession: Suppose now θ = constant
.
.
Then, just as before, ψ = constant from (3, q̂) and φ = constant from (3, b̂ 3 )
Finally, the (3, n̂) equation becomes:
0
0
0
..
.
. 7
θ 2 cos θ )L
sin
θ
I tθ + I s ψΩ
sin
θ − I t ψ2
sin θ cos θ = (mg − mLθ sin θ − mL
..
.
.
.
I s ψΩ − I t ψ2 cos θ = mgL
.
.
or, substituting for Ω = ψ cos θ + φ ....(4)
.
..
( I t − I s )ψ2 cos θ − I s ψφ + mgL = 0
Note that equation (4) is identical to the steady state precession equation for the top, except
the I 0t is replaced by I t , and the top precesses about it’s center of mass.
24
10
Rolling Disk
.
.
A. disk rolls without slip as shown. Given ψ = constant, find the value of ψ and associated
φ.
Figure 17:
The reference frame here is as shown. Note that here it is the "3-2-1 ψ − θ − φ" angles
Figure 18:
Note that the three reaction forces at point P. Also note that point P is accelerating.
.
H = MC
.
.
.
.
.
ω = ψ sin θ q̂ + θ n̂ + (ψ cos θ + φ) b̂ 3
.
.
.
H = I t ψ sin θ q̂ + I t θ n̂ + I s (ψ cos θ + φ) b̂ 3
{z
}
|
. .
.
Ω
Note here that ψ, φ, and θ are all constant.
25
.
.
.
.
H = (ψ sin θ q̂ + θ n̂ + ψ cos θ b̂ 3 ) × H C
{z
}
|
Therefore;
Why?
0
.
.
.2
.
0
.
7
.
.
⇒ H = I sθ Ω q̂ + ( I t ψ sin θ cos θ − I s ψΩ sin θ ) n̂ + I s
Ω b̂ 3
.
( I t ψ2 sin θ cos θ − I s ψΩ sin θ )
H =
n̂
or, substituting for Ω
.
..
.
( I t − I s )ψ2 sin θ cos θ − I s ψφ sin θ
H =
M C = (Nr cos θ − Fr sin θ ) n̂
n̂
(From the FBD)
Therefore:
..
.
( I t − I s )ψ2 cos θ − I s ψφ sin θ = Nr cos θ − Fr sin θ . . . . (1)
Note again that N and F are not known, and hence we need to use Newton’s second law.
Figure 19:
.
aC = −(R − r cos θ )ψ2 ŝ ⇒ why?
Note that the ( ŝ, t̂, n̂ 3 ) frame, a polar coordinate centered at C’, is the best to study the
center of masses motion. Since C is undergoing circular motion about C’
F = −F ŝ + FR t̂ + (N − mg) n̂ 3
.
− F = −m(R − r cos θ )ψ2
ŝ
FR = 0
t̂
N − mg = 0
n̂ 3
26
.
⇒ F = m(R − r cos θ )ψ2 ;
N = mg . . . . (2)
Now, we have to solve for the unknowns in (1), and substituting (2) into (1) we get
.
..
.
⇒ ( I t − I s )ψ2 sin θ cos θ − I s ψφ sin θ = mgr cos θ − mr(R − r cos θ )ψ2 sin θ . . . . (3)
.
.
.
Are we done? No! Since we have two unknowns, ψ and φ and only one equation (Eq. 3)
.
Now, we use the no slip condition.to relate ψ and φ. This, you have already done in Problem
21-67 of HW 6. Substituting for φ from no-slip in Eq. (3) gives us the final answer.
General Rolling Without Slip
For the general equations of motion of a disk rolling without slip, see the end of chapter 5,
section 9, in your textbook.
27
11
Linearization of Nonlinear ODEs
Thus far in this class, we have derived the equations of motion for various different systems.
Typically, these equations of motion are nonlinear, coupled, second order ordinary differential equations (ODE). In general, we can only find numerical solutions to these equations,
given the initial conditions. Now, we will introduce the powerful concept of linearization
that allows us to convert these equations into "linear ODEs" under certain assumptions.
The advantage of linear ODEs is that we can obtain analytical solutions for them. The example systems we will consider are the following.
1. Simple Pendulum:
Figure 20:
Equation of Motion:
..
θ+
g
L
sin θ = 0
2. Gyroscope:
Figure 21:
28
Equations of Motion:
.
..
..
..
0
I 0t ψ sin θ + 2 I 0t ψθ cos θ + I s θ Ω = 0
.
.
I t θ + I s ψΩ sin θ − I 0t ψ2 sin θ cos θ = mgL sin θ
.
I sΩ = 0
State Space Form
It is always possible to convert any n th order ODE into a first order ODE. In particular, let
us show how to do this for the example systems above.
Simple Pendulum
.
Let θ = ω, then the pendulum equations of motion can be written as:
.
θ= ω
.
ω=
g
L
sin θ
⇒ this is a first order ODE, albeit in 2 variables(θ , ω)
 
θ
State =  
ω
Gyroscope
.
.
Let ψ = ω p , θ = ωn
Then, the equations of motion may be written as:
.
.
ψ = ωp
θ = ωn
.
ω p = −2ω p ωn cot θ −
.
ωn = −
I sΩ
I 0t

I sΩ
I 0t
ωn csc θ
ω p sin θ + ω2p sin θ cos θ −
ψ
mgL
I 0t
sin θ



 θ


State = 

 ωp 


ωn
This is a first order ODE, only now in the variables (θ , ψ, ωn , ω p ). i.e., by defining the state
.
.
variables ωn = θ and ω p = ψ, we converted the equations above into first order ODEs or
"state space form"
29
In general, the state space form is written as:
. 
 . 



λ1


f 1 (λ1 , λ2 , . . λn )


f 1 (λ)


 f 2 (λ1 , λ2 , . . λn )  

 λ2 

 
.


.




 .  ≡ λ = F ( x) = 
. ≡ 
 ....(1)



 
.






 .
.

. 
f n (λ)
f n (λ1 , λ2 , . . λn )
λn
Where λ ∈ Rn
.
Next let us suppose that λ eq is an equilibrium of the system, i.e., λ = F(λ eq ) = 0: the system
does not move when at an equilibrium.
Now we shall consider "small deviations" δλ from the equilibrium λ eq , i.e. :
Let λ = λ eq + δλ, plugging back into (1)
0
.
7
⇒λ
+
δλ
= F(λ eq + δλ)
eq
.
¯δλ
¯
∂F ¯
⇒ λ eq
+ ¯
+ H.O.T.
∂λ ¯
0
.
λ eq
δλ = Aδλ where:

¯
∂F ¯¯
A=
¯
∂λ ¯
λ eq
∂f1

 ∂λ1

 ∂f2

=
 ∂λ1

:


 ∂f n
.
∂λ1
¯
¯
¯
Note that A = ∂F ¯
∂λ ¯
∂f1
∂λ2
∂f2
∂λ2
..
...
∂f1


∂λn 

∂f2 ¯¯
¯
...
∂λn 
¯¯ . . . . (2)
 λ eq


∂f n 
...
∂λn
is an n×n matrix, and that eq. (2) is a linear ODE! Thus, the equations
λ eq
of motion of small deviations about an equilibrium are "linear"!
30
Example 1: Simple pendulum
Figure 22:
λ eq for the simple pendulum is (θ , ω) = (0, 0)
Consider now small deviations


δθ 
0
δλ = 
from
δω
0
From eq (2),
∂ω
∂ω


∂θ


A=
 −g
∂
∂ω
¯
¯
¯
¯
¯
 (0, 0)

l sin θ
∂θ

0


A=
 −g
l cos θ
 . 

 δθ 
0

 

 = −g
 . 
l
δω
−g
∂ l sin θ
∂ω


1 ¯
¯
0
¯
=  −g
¯
¯
l
0 (0, 0)
1

0


δθ


0
δω
1

31
Example 2: Gyroscope
Figure 23:
.
.
The equilibrium is steady state precession where ψ = ω p , and θ are constant while θ =
ωn = 0. Note that ψ does not appear in any of the equations, and thus we can eliminate the
.
ψ = ω p equation!
Thus the equations of motion are:
.
θ = ωn
≡ f 1 (θ , ω p , ωn )
.
ω p = −2ω p ωn cot θ −
.
ωn = −
¯
¯
¯
Thus A = ∂F ¯
∂λ ¯
I sΩ
I 0t
I sΩ
I 0t
ωn csc θ
ω p sin θ + ω2p sin θ cos θ −
mgL
I 0t
λ eq
∂ f1
∂ f1
∂ f1

 ∂θ

 ∂ f2
A= 
 ∂θ

 ∂f
 3
∂θ
∂ω p
∂ f2

∂ωn ¯

∂ f 2 ¯¯
¯
∂ωn 
¯(θ∗ , ω∗ , ω∗ = 0)
p
n
∂ f3 


≡ f 2 (θ , ω p , ωn )
∂ω p
∂ f3
∂ω p

∂ωn
32
sin θ ≡ f 3 (θ , ω p , ωn )


0







A=







0
1






I
Ω
I
Ω
s
s

2
2ω p ωn csc θ + 0 ωn csc θ cot θ
−2ωn cot θ
−2ω p cot θ − 0 csc θ | ∗ ∗ ∗
 (θ , ω p , ωn = 0)
It
It





I sΩ
mgL
I sΩ

2
2
2
− 0 ω p cos θ + ω p (cos θ − sin θ ) −
cos
θ
−
sin
θ
+
2
ω
sin
θ
cos
θ
0
p
0
0
It
It
It


0







A=







0
0
−
I sΩ
I 0t
ω∗p cos θ ∗ + ω∗p2 cos 2θ ∗ −
0
mgL
I 0t
cos θ ∗ −
I sΩ
33
I 0t
sin θ ∗ + 2ω∗p sin 2θ ∗
1
−2ω∗p cot θ ∗ −
0
I sΩ
I 0t
csc θ ∗















12
Linear Dynamical Systems
We want to find the solution to the linear first-order vector differential equations:
.
x = A x;
given x(0)
Where x ∈ Rn×1 is an n-dimensional vector, x(0) ∈ Rn×1 is an n-dimensional vector, and A ∈
Rn×n is an n × n matrix.
Intuition:
The scalar case of the above is:
.
x = a x, x(0) given, where x, x(0) and a are all scalars.
The answer to this equation is given by:
dx
= ax
dt
x(t)
Z
x(0)
dx
=
x
⇒
Z
0
t
dx
= adt
x
¶
x( t)
(adt) ⇒ log
= at
x(0)
µ
⇒ x(t) = eat x(0)
We may want to mimic the above for the vector case by writing the solution as:
x(t) = eAt x(0),
where A is an n × n matrix, x(t) and x(0) are n-dimensional vectors.
What is eAt ?
Since, A is a matrix, it is not immediately clear what eAt is!
However, we do know that:
x2
x3
e =1+x+
+
+ . . ., when x is a scalar
2!
3!
x
Thus, we may mimic the above scalar definition in the vector case as:
eAt = I + At +
A 2 t2
A3 t3
+
+ ...
2!
3!
where, note now that I, At, A2 t2 etc. are all n × n matrices, thereby making eAt an n × n
matrix as well.
The above is all good, but how do we evaluate it?
34
Suppose we had a diagonal matrix


0 
 λ1




λ2



 then:
Λ=

.
.


.




0
λn

e
Λt
0
 λ1 t


λ2 t

= I+

..

.


0
λn t

t2
λ1 2!
 (1 + λ1 t +



=




0

 e



=





 
 
 
 
+
 
 
 
 

t2
λ1 2!
0
2
t
λ2 2!
..
.
2
t
λn 2!
0




 + ...





+ . . .)
0
2
t
(1 + λ2 t + λ2 2!
+ . . .)
..
.
2
t
(1 + λn t + λn 2!
+ . . .)










λ1 t
0
e
λ2 t
..
0

.
eλn t









Thus, the matrix exponential of a diagonal matrix is simply the exponentiation of
its diagonal elements!
35
Eigenvalue Problem to the Rescue
Now that we know the matrix exponential of a diagonal matrix, it becomes somewhat obvious that the eigenvalue problem might have a central role in finding the general
matrix exponential eAt .
This is confirmed as follows:
Let
A = V ΛV −1 , where

 λ1


λ2

£
¤
V = v1 v2 . . v n , Λ = 




0

0 



,

..

.


λn
where (λ i , v i ) are the i th eigenvalue-eigenvectorpairs for matrix A, i.e.,
Av i = λ i v i
Note that each v i is an n-dimensional vector, each λ i is a (complex) scalar, and V, the stacked
eigenvector matrix is an n × n matrix. A further note is that if λ is a complex eigenvalue
with eigenvector v, Av = λv. Taking the complex conjugate on both sides :
A ∗ v∗ = λ∗ v∗ ⇒ Av∗ = λ∗ v∗
Thus, if (λ, v) is an eigenpair, so is (λ∗ , v∗ ).
This leads us to the following fact: Complex eigenvalues and eigenvectors occur in
complex conjugate pairs.
However, continuing with our derivation:
e
At
= I + At + A
2t
2
2!
+ ...
= V V −1 + (V ΛV −1 ) t + (V ΛV −1 )(V ΛV −1 )
= V IV −1 + V (Λ t)V −1 + V (
⇒ e At = V ( I + Λ t +
t2
+ ...
2!
Λ2 t2 −1
)V + . . .
2!
Λ2 t 2
+ . . .)V −1
2!
⇒ e At = V eΛ t V −1 . . . . (1)
36
Let us expand Eq. (1):

Let V −1




=U =





u1 


u2 
 , where U is an n-dimensional row vector
.. 
. 


un

 e



£
¤
At
Λ t −1
Then: x( t) = e x(0) = V e V x(0) = v1 v2 . . . vn 





 e


£
¤
⇒ x ( t ) = v1 v2 . . . v n 




λ1 t
0
eλ2 t
..
.
eλn t
0
λ1 t











u1 


u2 
 x(0)
.. 
. 


un
0
eλ2 t
..
0


  u 1 x(0)


  u 2 x(0)


..

.


u n x(0)




 . . . . (2a)




.
eλn t
In Eq. (2a), note that u i x(0) are complex scalars since the u i terms are a row vector and x(0)
is a column vector, thereby making u i x(0) a scalar.
Continuing our work, we find:
x( t) =
n
X
eλ i t v i ( u i x(0) )
i =1
Where recall that

Av i = λ i v i ,
i = 1 . . n,




U =





u1 


u2  £
¤
 = v1 v2 . . . vn −1
.. 
. 


un
The above is called a modal decomposition!
1. Note that the response can be thought of as the superposition of "n" separate responses,
i.e.,
P
x( t) = ni=1 x i ( t)
Where x i ( t) = eλ i t v i ( u i x(0) ) may be thought as the response due to the i th eigenvalueeigenvector, or i th mode. Note that the eigenresponses are independent of each other.
37
2. The i th modal response:
x i ( t) = eλ i t v i ( u i x(0) )
looks like the scalar solution e at x(0) except that the initial condition x(0) is projected via u i ,
and the mode v i is invariant to time. The temporal variation of the modal response is solely
determined by the eigenvalue "λ i ".
Thus, the response can be thought as the linear combination of the different eigenvectors
v i , i.e.,
x( t) =
n
X
α i ( t) v i
i =1
where α i ( t) = eλ i t ( n i x(0) ), i.e., the "modal initial condition" is u i x(0) = Z i , and the "temporal" variation is given by eλ i t .
Further Simplifications:
The eigenvalues λ i can either be real or occur in complex conjugate pairs (λ, λ∗ ) and (v, v∗ ).
Let us now look at the response due to each type of eigenvalue.
Real Eigenvalue Case:
In this case, x i ( t) = eλ i t ( u i x(0) )v i , and λ i , u i , v i are all real. Dropping the subscript i for
the notational simplicity:
x( t) = eλ t ( ux(0) )v ;
If λ is real . . . . (∗)
Complex Eigenpairs Case:
Recall again that if (λ, v) is an eigenpair, so is (λ∗ , v∗ )! Now it’s not really clear why the response due to the nodes (λ, v) and (λ∗ , v∗ ) should not be complex, and how is that physically
realistic? (In the following, we will use complex phasors, please refer to the notes towards
the end of this section for more details)


j ψ1
 v1 e




j ψ2 

¯ ¯
 v2 e
 , ux(0) ≡ Z = ¯ Z ¯ e jφ
Suppose that λ = σ + j ω, v = 


.
..






v n e j ψn
The eigenvectors and the modal initial conditions ux(0) are in general complex, and hence,
written in the phasor form above.
38
Thus, the response due to (λ, v) and (λ∗ , v∗ ) is:
¯ ¯
∗ ¯ ¯
∗
x( t) = eλ t ( ux(0) ) v + eλ t ( u∗ x(0) ) v∗ = eλ t ¯ Z ¯ e jφ v + eλ t ¯ Z ¯ e− jφ v∗
| {z }
| {z }
Z∗
Z

j ψ1
 v1 e
¯ ¯ (σ + jω)t jφ 
..

= ¯Z¯ e
e 
.


v n e j ψn

j(ω t + φ + ψ1 )
 v1 e

¯ ¯
..

= ¯ Z ¯ eσ t 
.


vn e j(ω t + φ + ψn )



− j ψ1

 v1 e
 ¯ ¯

..
 ¯ ¯ (σ − jω)t − jφ 
e
+ Z e

.




v n e− j ψn


− j(ω t + φ + ψ1 )

 v1 e
 ¯ ¯

..
 ¯ ¯ σt 
+ Z e 
.




vn e− j(ω t + φ + ψn )

 v1 cos (ω t + φ + ψ1 )


¯ ¯ σ t  v2 cos (ω t + φ + ψ2 )
¯
¯
=2Z e 

..

.


vn cos (ω t + φ + ψn )























Therefore, the response due to a complex eigenpair is still real and given by:

 v1 cos (ω t + φ + ψ1 )

¯ ¯ σt 
 v2 cos (ω t + φ + ψ2 )
x ( t ) = 2¯ Z ¯ e 

..

.


vn cos (ω t + φ + ψn )





 . . . . (∗∗)





j ψ1
 v1 e

¯ ¯ jφ
..

¯
¯
where: Z = ux(0) = Z e , λ = σ + j ω, and v = 
.


v n e j ψn







Therefore, note that the response due to a complex eigenpair is oscillatory at the frequency
"ω" = increasing part of the eigenvalues, and amplified temporally by the real part σ, exponentially on eσ t .
σ < 0 ⇒ Stable response since eσ t approaches 0 as t approaches in f t y
σ > 0 ⇒ Unstable response since eσ t approaches ∞ as t approaches in f t y
σ = 0 ⇒ Oscillatory sustained response since eσ t = 1
39
Qualitatively, the responses due to a real eigenvalue and a complex eigenpair are very different
Figure 24:
40
Collating everything in one place:
the response of the linear dynamical system:
.
x = Ax, x(0) given, is:
x( t) =
n
X
eλ i t ( u i x(0) )v i
|
{z
}
i =1
x i (t)





£
¤−1
where, Av i = λ i v i and v1 v2 ...vn
=
|
{z
} 


V


u1 


u2 
 . . . . ( M.1)
.. 
. 


un
Further, in the above (dropping the subscript ’i’ after this for notational convenience)
If λ is real:
x( t) = eλ t ( ux(0) )v
If λ, λ∗ is a complex pair:


1
 v cos(ω t + φ ψ1 )
¯ ¯ σt 
..

x ( t ) = 2¯ Z ¯ e 
.


v n cos(ω t + φ ψn )







1 j ψ1
 v e

¯ ¯ jφ
..

where ux(0) = Z = ¯ Z ¯ e , λ = σ + j ω, and v = 
.


v n e j ψn
41




 . . . . ( M.2)


First, let us look at complex numbers as phasors, since this will be highly convenient for our
development. Consider a complex number z = x + j y :
¯ ¯
¯ ¯
¯The
¯ phasors notation is Z = Z cos φ +
¯ Z ¯ j sin φ
¯ ¯
= ¯ Z ¯ (cos φ + j sin φ)
¯ ¯
= ¯ Z ¯ e jφ
¯ ¯
⇒ Z = ¯ Z ¯ e jφ
The complex conjugate of Z is Z ∗
¯ ¯
⇒ Z ∗ = ¯ Z ¯ e− jφ
Figure 25:
¯ ¯
How do we find ¯ Z ¯ and φ?
¯ ¯ q
¯ Z ¯ = x2 + y2

tan−1 ( y ) if x ≥ 0 . . . . (a)
x
φ=
π + tan−1 ( y ) if x < 0 . . . . ( b)
x
y
Note that (b) is required since tan−1 ( x ) only
varies between −2π and π2 , whereas φ can
vary between −π and π.
Figure 26:
42
EXAMPLES:
Example 1. Consider the spring mass system:
Figure 27:
..
k
The EOM of the system is x + m
x = 0. Let
 
  
.
 x   0 1  x 
 
 . =
−1 0
v
v
|
{z
}
k
m
= 1. Then, writing in state space form:
A
Eigenvalue problem for A:
¯
¯
Eigenvalues: ¯λ I − A ¯ = 0
¯
¯
¯
¯
¯ λ −1 ¯
¯
¯
¯
¯ = 0 ⇒ λ2 + 1 = 0 ⇒ λ = ± j
¯
¯
¯ 1 λ ¯
Eigenvectors: Av = λv

For eigenvalue λ = + j :



 0 −1   v1 
 v1 


 = j

−1 0
v2
v2
v2 = jv1



1  1 
⇒ v= p  
2 j
43
The
p1
2
¯ ¯2
factor is required to normalize the complex eigenvector since |1|2 + ¯ j ¯ = 2.
Similarly, the eigenvector corresponding to λ = − j is


1  1 
v∗ = p 

2 −j



¤−1
 u1  £
U =
 = v1 v2
u2
1
 p
 2


=

 j

p
2


− j −1

 p
 2 p2 


1


=
 −j

 p p1 − p1 pj


 −j
2 2
1  2 2

p
p
2
2


−j
1
 p
 

 2 p2 



  u 
U =
=



u∗
 1

j 

p
p
2
2

"
⇒ u=
1
p
2
−j
p
2
#
"
, u∗ =

−1
1
p 
2 




−j 

p
2

−1
p 
2 
 1


 −j
1 

p
2
−j
 p
 2




 −j

p
2
1
p
2
j
p
2
#

 1 
Suppose that x(0) =   i.e., there is a unit displacement at initial time.
0
Then, let us find all the quantities required in Eq. (M), noting that λ is complex, so we need
to use Eq. (M.2) only as there are no real eigenvalues.
44
Initial condition in phasor form:
"
Z = ux(0) =
1
p
2
−j
p
2
#


1
1 j(0)
 1 
 = p = p e
2
2
0
¯ ¯
1
⇒ ¯Z¯ = p , φ = 0
2
Eigenvector in phasor form:



v=


1
p
2
−j
p
2


 
 
=
 
 
1
p e j(0)
2
−π
1 j(
)
p e 2
2

1

v1 = p ;

2
⇒

1

v2 = p ;
2
ψ1 = 0
ψ2 =
−π
2
Eigenvalue:
λ = j = 0 + j (1) ⇒ σ = 0, ω = 1
Now, we have all the ingredients needed to solve this problem giving us:


¯ ¯
 v1 cos (ω t + φ + ψ1 ) 
x ( t ) = 2¯ Z ¯ e σ t 

v2 cos(ω t + φ + ψ2 )


1 (0)t 
x( t) = 2p e 


2
1
p cos(1 t + 0 +
2
1
p cos(1 t + 0 −
2


0) 


π 

)
2

cos( t)

⇒ x( t) = 



π 
cos( t − )
2
Thus, this is a sustained oscillatory response as is to be expected since we have a
complex eigenpair and σ = 0.
45
Example 2. The lateral equations of motion of a business jet are given by:

 .  

∂v
∂v



−0.1567 0.0632
−0.998 0.096  
 v  

v 

 .  


  −2.5194 −1.1767 0.1823


0 
∂
p

 ∂p  


=





 .  

0 
 ∂ r   1.7442 −0.0112 −0.0927

  ∂r 
 




 . 
0
1
0.0634
0
∂φ
∂φ
Where ∂v is the lateral velocity, v is the equilibrium velocity, ∂ p and ∂ r are the pitch and
roll rates, and ∂φ is the roll angle (all perturbed about steady level flight). Find response to
a one degree perturbation in φ, i.e.,







x(0) = 




0 


0 


0 


1 
2π
Eigenvalue Problem:


−0.998 0.096
 −0.1567 0.0632


0
 −2.5194 −1.1767 0.1823
A =

 1.7442 −0.0112 −0.0927
0


0
1
0.0634
0









This is usually difficult/impossible to solve by hand beyond a 2 × 2 matrices.
We use MATLAB and with:
£
¤
V , Λ = eig(A) to get the following:


0
0
0
+ 1.3895}i
 −
| 0.1158{z

λ1



0
−
− 1.3895}i
0
0

| 0.1158{z

∗
λ

1
Λ=


0
0
0
−
2034}
| 1.{z


λ2



0
0
0
|0.0089
{z }
λ3
46

























V =








v1∗
v1
v2
v3

}|
{ z
}|
{ z }| { z }| { 
z
−
0
.
2834
−
0
.
3085
i
−
0
.
2834
+
0
.
3085
i −0.0078 0.0056 

{z
}
|

v11


0.6076
−0.7692 0.0028 
0

{z }
| .6076

v12


+ 0.3670}i −0.3812 − 0.3670 i 0.0045 0.0950 
|−0.3812{z


v13



−
0
.
0181
−
0
.
4184
i
−
0
.
0181
+
0
.
4184
i
0
.
6389
0
.
9955
|
{z
}
v14


 0.0101 + 0.0082 i


 0.0101 − 0.0082 i
−1
U x(0) = V x(0) = 


0.0164


0.1429


  Z1 
 

  ∗ 
  Z1 
=

 

  Z2 
 

 

Z3
The various quantities of interest are identified for use in Eq. (M) above.
Qualitatively, there is a complex eigenpair (dutch roll mode, (λ1 , λ∗1 ) ), a real stable mode
(roll mode, λ2 ), and a real unstable mode (spiral mode, λ3 ). Thus, the response is:


 |v11 | cos(ω1 t + φ1 + ψ11 ) 




¯ ¯ σ t  |v12 | cos(ω1 t + φ1 + ψ12 ) 
1 
¯
¯
+
eλ2 t Z2 v2
+
eλ3 t Z3 v3
x( t) = 2 Z 1 e 

|
{z
}
|
{z }
 |v13 | cos(ω1 t + φ1 + ψ13 ) 

 Response due to roll mode Response due to spiral mode


|v14 | cos(ω1 t + φ1 + ψ14 )
{z
}
|
Response due to dutch roll mode
The various quantities of interest may now be read off as follows:
Dutch roll mode (λ1 , λ∗1 ):
λ1 = −0.1158 + 1.3895 i ⇒ σ1 = −0.1158 ω1 = 1.3895
¯ ¯
Z1 = 0.0101 + 0.0082 i ⇒ ¯ Z1 ¯ = 0.0130 φ1 = 0.6819 rads
¯ ¯
¯ ¯
Similarly, you may find ¯v11 ¯ , ψ1 , ¯v12 ¯ , ψ12 ... etc.
For instance, v11 = −0.2834 − 0.3085 i
¯ ¯
⇒ ¯v11 ¯ = 0.4189, ψ11 = −2.3189 rads
47
The real modes are less tedious:
Roll mode (λ2 ) = −1.2034, Z2 = 0.0164
Spiral mode (λ3 ) = 0.0089, Z3 = 0.9955
The final answer to this problem is:

 0.4189 cos(1.3895 t + 0.6819 − 2.3189)

..

.

x( t) = 2(0.0130) e−0.1158t 

..

.


..
.

 −0.0078


 −0.7692
−1.2034t
+e
(0.0164) 

 0.0095


0.6389




 0.0056 






0
.
0028



0.0089t
+e

(0.9955) 




 0.0950 






0.9955
48





 ...



