Methods of Applied Dynamics - CiteSeerX

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NASA -_-__ Reference _-_ Publication 1262

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M ay 1 991

Methods of Applied Dynamics E

M. H. Rheinfurth and

H. B. Wilson

c _./31

"

_:

iJ'--

t

NASA Reference Publication 1262

1991

Methods of Applied Dynamics

M. H. Rheinfurth George

C. Marshall

Marshall

Space

Space Flight

H. B. Wilson

National Aeronautics and Space Administration Office of Management Scientific and Technical Information Division

University

of Alabama

Tuscaloosa,

Alabama

Flight

Center,

Center Alabama

Contents

1

Kinematics

5

1.1

Vectors

1.2

Angular

1.3

Vector

1.4

General

Dynamics

3

4

5

.................................. Velocity

5

. ............................

Derivative

in a Rotating

Motion

in a Moving

10

Frame Frame

.................

...................

11 12

of a Particle

21

Laws

21

2.1

Newton's

2.2

D'Alembert's

2.3

Work;

2.4

Applications

.............................. Principle

Kinetic

and

(1747)

Potential

...................... Energy

of D'Alembert's

Principle

24

. .................

25

.................

30

Dynamics of a System of Particles 3.1 Translation and Rotation ........................ 3.2 Linear and Angular Momentum .....................

39

3.3

Kinetic

Energy

47

3.4

Variable

Mass

3.5

Impact

and

Equations

Vector-Dyadic/Matrix

4.3

Orientation

4.4

Moment

4.5

Free

4.6

Forced

4.7

Rheonomic

Lagrangian 5.1

Constraint

Body

69

Notation

Kinematics of Inertia

Body

of a Rigid

Systems

69 .....................

.........................

Properties:

of a Rigid

Motion

55

.............................

4.2

......

Body

70 75

......................

103

. .....................

112

. ....................

126

.....................

139 149

Dynamics Equations

43 49

.............................

of a Rigid

Motion

........................

...............................

Dynamics

Dynamics 4.1 Euler's

Work

39

...........................

150

6

5.2 5.3

Principle of Virtual Generalized Forces

5.4

Classical

5.5

Lagrange

Modal

Lagrange Equations

Synthesis

Work (Bernoulli ............................ Equations With

Boltzmann-Hamel

6.2

Component

6.3

Applications

Lagrange

References

Forces

......................

............................

to Aerospace

167 177

Equations

Modes

156 156 159

...............

Technique

6.1

The

...............

...................... Reaction

Systems

177 183

...................

186

205

Appendix A

1717)

Multiplier

Method

205

213

Methods

of Applied

Dynamics

Description: This of the ning

monograph principles

with

the

discussion method

is designed of dynamics

basic

of the

concepts dynamics

of dynamic

formulations

of the

plied

dynamic

to the

technique.

to give the practicing with

emphasis

of kinematics of a system

analysis Lagrange modeling

A list of references

special

and

dynamics

in full

equations

including

of aerospace at the

a clear

on their

of particles.

is treated

is given

engineer the The

detail.

applications. course

Both

structures

using

the

end

monograph.

to the

(Lagrangian)

classical are

Begin-

proceeds

analytical

constraints of the

understanding

and

modern

discussed

and

ap-

modal

synthesis

Chapter

1

Kinematics

Kinematics motion.

1.1

relates

to the

geometry

disregarding

the

forces

causing

the

Vectors

A vector

has

Physical

types

1. free

direction

and

magnitude

vector:

force

3.

vector:

position,

NOTE: properties Addition" formed

force,

etc.)

velocity

vector:

bound

(velocity,

of vectors:

2. sliding

gram

of motion

on rigid force

All mathematical of magnitude The

sum

by the

two

body on elastic

operations

and

with

body

vectors

involve

only

their

free

vector

direction.

of two vectors vector

is represented

by the

diagonal

of a parallelo-

sides.

A

5

PRL=CEDING

PAGE BLANK

NOT

FILMED

NOTE:

Vectors

Vector

are

denoted

addition

by bolding

is commutative

the

and

symbol.

associative.

Unit Vector: A unit vector has unit magnitude (length). Vectors are often conveniently expressed in terms of unit A = Ale1 where

A1, A_, A3 are

Often

unit

vectors

known

are

used

+ A2e2 + .4ae3

as (scalar) which

vectors:

components

are

of the

(mutually)

vector

ortho_onal:

A.

i,j,k

Z

A = A.

i + Auj

+ Az k

Here the components .4,, Au, Az are the orthogonal projections of A onto

NOTE:

The

unit

vector

in the

direction ea

the

coordinate

of the

= A/A

vector

Scalar

("Dot")

to use suggestive

symbols

Product

A.

x, y,z.

A is identified

as:

A = IAt

P.S. #1: The German word for unit is "EINHEIT." use of the letter e for the unit vector. P.S. #2: It is important mass; M = large mass.

axes

B = A B

6

cos

This

is the origin

for physical

of the common

quantities:

m = small

where 0

is the (smaller) vectors el, e2, ea:

unit

A-

angle

between

A and

B = (A1 el + ,42 e2 + As ea).

=A1

B.

Expressed

in terms

of ortho_onal

(B1 el + B_ e2 + Ba ea)

Bl+.42B_+AaBa

NOTE: el

• el

:

e2

• e2

---- e3

el

.e2

-:

el

"e3

---- e2"e3

Vector

("Cross")

• e3

----

1

(UNIT

LENGTH)

(ORTHOGONALITY)

---- 0

Product A x B=ABsin0N

where

0 is the

such

that

smaller

angle

A, B, N form

Expressed

in terms

AxB

0 o Therefore: Similar

111 + I_2 > 133

relationships

B) Theorem

of Parallel

Let O be the origin is defined

are

and

C.M.

Q.E.D,

obtained Axes

(Steiner

of the reference be the

by cyclic

mass

of the

indices.

Theorem)

frame center

permutations

about of the

103

which rigid

the moment

body.

of inertia

dyadic

In the figure: d = distance of C.M. from 0 ri = location of masselement ml. Using vector-dyadic notation the inertia dyadic about 0 can be written as:

Zo= [R E-(R, R,)]m, Zo = Z [(d + r,)_E -(d

+ r,) (d + r,_)]rn,

70 = E [(d_ + 2d.r; + r_)E -(d d + dr; + r;d + r, r;] m; For the mass center Er; m; = 0 and _d

m; = Md

where M

= E rn,.

Therefore:

2-0 = Z(r_ c - r; r_.) m, + M(d2f

- d d)

or

2-0 = 2-C.M. + M ( d2 E - d d

Converting

to matrix

form.

we introduce:

d = die1 It can be easily L = -t) _

verified

that

the

+ d2e2 + daea

dyadic

£ = a_2- -

aa

corresponds

where

a3

0

0

--a2

Therefore

the

matrix

form

of the

--_/1

-a3

a2

inertia

dyadic

al

above

Io = Ic.M.-

104

0

M _2

1

2-0 is:

to the

matrix

In component form:

and

111

=

l_l + M(d_

+ d_)

I_

=

l_2 + M(d_

+ d_)

Ia3

=

l_3 + M(d_

+ d_)

also:

It can

be

results

in an increase

seen

of inertia

may

C) Theorem Here in two

that

=

I_-

Ila

=

I_a -- hldldz

l_z

=

l_a -

a translation of the

increase

we establish

of the

moments

or decrease

of Rotated

different

I1_

Mdld_ Md2dz

axes

away

of inertia. depending

on the

and

the

relation

reference

energy,

being

the other

between

frames.

the

a scalar,

the

To do

this

unprimed

has

system.

rotational

moment

kinetic

to be the

transformation

energy

mass

center

hand,

always

the

products

situation.

same

of inertia

we calculated

for both

wTIw

is:

_wl T(AZ FA)w

105

=

_wT(I)

that

coordinate

is therefore:

T=

matrices

the

origin is we call one

to' = Aw The

the other

particular

It is apparent

T=-_ 1 t_tTi_t_ t = The

the

Axes

energy for the two different systems whose common chosen at the mass center. For convenience sake system

from

On

_

system

the

expressed

rotational the

rotational

systems.

kinetic primed kinetic

The

two bracketed

terms

must

be identical,

i.e.,

AT I'A = I By proper the primed

pre-and post-multiplication moment of inertia matrix:

with

the

I' = AIA This is the (unprimed)

desired transformation to a new (primed)

D) Principal The matrix by

analysis

a reference axes

of inertia that

frame

an)" real

moments

is a real

symmetric

all

matrix

eigenvalues

system

The

diagonal

such

and

the

terms

their

the

principal

matrix

from

an old

associated

to form the

of the

are

three

It is a theorem

that

to diagonal we can

zero.

This

mutually

of this diagonal

a principal

problem

axes

are

the

eigenvectors of the

a matrix components

X 1 _

the

for

form

always

find

reference

orthogonal inertia

of

frame

coordinate

matrix

are called

of inertia.

of this

and

Designating

going

be reduced

means

of inertia

axes

(I - AE)z

together

when

matrix.

can

This

products

axes

axes.

The problem of finding eigenvalue problem

The

matrix

symmetric

transformation.

in which

as principal

as principal

principal

A, we can solve

T

law of the inertia system by a rotation.

matrix

of an orthogonal

is known

matrix

Axes

moment

means

rotation

column

reference

X21 Xll X31

vectors

principal

are

eigenvectors

X2 1

moments The the

(unit)

eigenvectors eigenvectors

vectors can z,,z,,

the

and

[x12] [ 13] Z22

X3

---_

modal

106

_23

Z33

matrix

is

along

be grouped

as

X32

called

to solving

of inertia

geometrically

frame.

columns

of the

is equivalent

= 0

represent

_ whose

frame

z a.

@ Xll X31

Let

also

The

A1 = 11, A2 = I_,

diagonalization

performed

*T12 _32

X13 _33

1

Aa = la.

of the

inertia

matrix

I using

the

modal

#

can then

be

as follows:

I#

=

[lzl

=

[I_z_

I

Ix2 I

I

lz_]

I2z_

I

I3z3] 0 0

The I_1

matrix

0

/2 0 0 13

second step in the above equation is a consequence = A1 zt = It _1. Premultiplying by _r yields:

of the

eigenvalue

solution

or finally:

Io=_r Comparing this transformation matrix under rotation we see matrix

A simply

law with the law governing that the moda] matrix _

rotation

as principal

the change of the inertia is related to the rotation

as:

A=_

The

l#

matrix

is the

transposed

axes

transformation.

For the

Gemini

spacecraft

origin

at the

T

modal

matrix.

This

rotation

is also

known

axis

system

Example:

with

mass

center

the

inertia

matrix

was: 107

referred

to the

control

I=

We want

to make

l) Solve

the

31.2

4560.7 -43.4

a principal

eigenvector

axis

4545.0

-270.7

31.2 -270.7

-43.4

slg ft 2

1567.4

transformation:

problem (I -

AE)_

= 0

Eisenvalues:

Modal

11

=

4530.11944

I_

=

4600.531811

la

=

1542.448749

Matrix:

0.74693 -0.66305

0.66477 0.74319

0.04964

-0.07584

0.01339 0.08965 0.99588

]

J

NOTE: The

eigenvectors

2) Check

the

sign

are

normalized

to unit

of the

determinant

of 4i.

length.

I_li 1= 0.9999997403 If[

_ti I turns

3) Find

out

to be negative

minimum

rotation

we change

angle

we have

to change

the

sign on one

6:

trY= To do this

the

sign

1+2cos6 on an)' _

108

rows.

row or column.

tr

_x

=

+0.74693+0.74319+0.99588

tr

_2

=

-0.74693-

0.74319+

0.99588

_

6 = 138.34 °

tr

_a

=

-0.74693+

0.74319-

0.99588

---,

6 = 178.88 °

tr

'1_4

=

+0.74693-

0.74319-

0.99588

_

6 = 174.19 °

transpose

of the

4) The

rotation

having

the

matrix

minimum

A is obtained rotation

The

orientation

be expressed

of the in terms

by taking

angle

A = _

=

of the

the

6=42.01

°

modal

matrix

6 :

0.66477 0.74693 0.01339

principal

_

0.74319 -0.66305 0.08965

axis system classical

Euler

relative angles

-0.07584 0.04964 0.99588 to the

control

or in terms

axis system

of the

modern

can Euler

angles. Example: Determine to the

the direction

corresponding

of the control

minimum axis.

principal

moment

of inertia

axis

e3 relative

We obtain:

0 = cos -1 Az3 = cos -1 0.99588

0 = 5.2 °

Al____aa _ A2a - tan-1

¢ = tan-_ This

angle

lies in the

second

quadrant.

¢ = (180 E) Ellipsoid

of Inertia

The

properties

inertial

surface

which

of the

rotation

kinetic

energy

of a rigid

is in essence axis

body

33.206)

body

a plot

direction.

of a rigid

0.04964 -0.07584

of the Going

° ---* ¢ = 146.8 °

can be conveniently moment back

we have 109

of inertia

to the

matrix

depicted of the form

by an ellipsoidal body

as a function

for the

rotational

1 T _w lw

T= where The

the single

by letting axes.

Now

Inserting

scalar

IA is the

scalar the

expression

angular

of inertia

for the

velocity

let us define

this

moment

4.81

about

kinetic

p having

above,

pZlp=

(4.81)

the

energy

w coincide

a vector

in Equation

1 2 = _IA_ instantaneous

on the

momentarily the

same

right

with direction

rotation

side

can

axis.

be verified

one of the

coordinate

as w such

that:

we obtain

Ix1

1 wherep=

y

Z

Consider now p to be a positive vector then we can write the scalar equation:

drawn

from

the

origin

0 to a point

Illx _ + I_2y _ + I33z _ + 2112x_ + 2113xz + 2I_3yz This

ellipsoidal

The

coordinate

form

is, of course,

surface

centered

transformation exactly

The moment of inertia magnitude of the vector

where p is the length surface of the inertia

A quantity defined by

closely

the about p.

about

0 is called

the

which

brings

inertia

principal an','

of the straight ellipsoid.

related

to the

axis

the

rotation

axis

line drawn

moment

110

can

from

of inertia

= 1

ellipsoid

of inertia.

ellipsoid

transformation

in its

previously be found

the

origin

is the

(x, y, z)

standard

discussed.

directly

from

0 to a point

radius

the

on the

of gyration

ka

Ia = Mk2a In terms

The

of the

inertia

having

Roughly

mass

They

are

ellipsoid

inertia

is fixed can

have

p can

with

the

quite

called

dynamically

the

of the

rigid

body.

ellipsoid.

standard

+ where

I1, I_ and

I3 are

the

principal

and

rotates

shapes

and

the

inertia

For instance, The

body

as:

equivalent

of revolution,

shape

be written

different

is a figure

speaking

corresponding (oblate)

of gyration

ellipsoid

equal

of inertia. inertia

radius

ellipsoid

(oblate)

of the

111

rigid

is similar inertia

+ ]3 I = 1 moments

it.

Two

have

of inertia.

rigid

rigid

identical

or equimomental.

corresponding

a prolate form

with still

bodies

ellipsoids

Although body

need

to the

shape

body

ellipsoid

has is

the

not

be.

of the

a prolate

4.5

Free

The

free

of the

Motion

motion

is characterized

reference

dynamically (Poinsot

frame

method

There

1834) the

system. There of the reference

fixed

and

the

problem,

ways

This

method

unsymmetrical energy

and

(Euler

case

frame

are

obtained

from

mass this

1758)

center.

involved

12_

+ (I, -

cases

- the

are

geometric

to be a principal in this

4.12

particular

and

read

axes choice

in scalar

(4.82)

a:a = 0

I1)_vl.v2

origin

way.

Equation

+ (Ia - I_)._2_a = 0

The Both

problem

is chosen

I1_

la)_

moments.

= 0

(1834) represents

rigid the

or at the

no loss of generality

of motion

Method

of external

of treating

reference

la&a + (I: A) Poinsot

absence

analytic

the

Body

in space

are two

is, of course, frame.

equations

by the

is either

identical.

To simplify

The form:

of a Rigid

a geometric

body.

angular

With

solution

no external

momentum

of the

torque

acting

torque-free on the

motion body,

the

of an kinetic

are conserved.

Therefore:

2T = 11% 2 + Iu,,'_ + Ia._'_ This

equation

represents

kinetic energy ellipsoid. scale factor x/_.

geometrically It differs

an

from

ellipsoid

the

standard

(4.83)

known inertia

as Poinsot ellipsoid

Ellipsoid only

by

or the

Also: H 2

This

ellipsoid

is the

angular

2 2 + I_2 11%

momentum

2

2 3 2 + Ia_

ellipsoid.

112

2TD

(4.84)

The

quantity

D is defined

as: H 2

D -

2T

It has the dimension of a moment as a matter of convenience. The

Poinsot

only

the

This

trace

method

path

Poinsot

traced

it is given

the and

by the

ensuing

not

by the

is called

ellipsoid

For the

does

yield

and

the angular

instantaneous and

angular

simultaneous discussion,

CONSTANT

of inertia

polhode the

-

(4.85)

is used

in the subsequent

velocity

w as a function

rotation

is the

curve

momentum solution

(spin)

obtained

without

on

by the

ellipsoid.

of Equation

we assume

axis

discussion

of time

the

rigid

intersection

and

speaking,

Equation

loss of generality

4.84.

that

11 < 12 < la It can

then

be shown

that

for this

case,

the

(4.86)

constant

D must

lie in the

range

Ii _< D _< la As a consequence,

a polhode

than

the

corresponding

trate

the

geometrical

It is observed moment the

vicinity

ellipsoid

case become

stable,

stable.

The

polhodes about

The

is always

more

following

two

elongated

figures

illus-

method.

closed

motion.

moment

a principal

axis.

the

furnish

the

curves On

about

the

of inertia

Poinsot

of revolution

spin

whereas

neutrally

form

symmetry,

ellipsoids to the

Poinsot

a stable

ellipsoid

ellipsoid.

the smallest

other

hand,

and

the

I_ have

hyperbolic

and

angular

largest

polhodes

in

character

motion.

of axial

perpendicular

polhodes

intermediate

an unstable

In the

motion

the

(4.87)

momentum

Poinsot

of the

reflecting

of the

indicating

more

aspects

that

of inertia

angular

polhode

body. of the

Mathematically

4.83

but

The

rotational

(spheroids).

in the

The

the

polhodes

rotation

about

the

motion

about

a transverse

an argument axis

ellipsoid

for the presence

axis

momentum

become

of symmetry

circles becomes

principal

stability

behavior

of the

of internal

energy

dissipation.

axes rotational

Since H 2

I, < (D = _-_) from while

Equation the

4.87 it is observed

angular

momentum

that

< la

in this case the

H remains

constant. 113

(4.88) kinetic

energy

As a consequence,

T must

decrease

the

quantity

is

_---Y

t

(-'_1

L

6"_3 f

Figure

4.1"

Intersection

of Poinsot

and

Momentum

6.9 2

Figure

4.2:

Pollodes

on Inertia

114

Ellipsoid

Ellipsoid

D is increasing minimum the

moment

principal Thus,

librium

dissipation

tation. The

Poinsot

or wobble write

can

method

Poinsot

(Equation

4.84)

by internal

in terms

of the

angular

I--7+

and

H_

Equation

combine

on the

both

4.89 equations

momentum

trace

D and

the

or feasible

the

of the

A: Spin

of the

rigid

body.

To

-

momentum

D) Ix

+H_(1-

vector

about

ll-axis

D) 12

: H 3

mutational this

momentum

end,

we

ellipsoid

Thus,

Sphere

(4.90)

by definition

of the

angular

H on the

D) 13

+H_(1momentum

a-

D/I_

la (D - I_)

1-

D/I3

I_ (13i la

X

The

(4.89)

(D < 12)

H_

OMA

en-

in orien-

2T

For OMI N : H2 = O(w_ = O)

For

antennas.

limits

that

path

actually

for the

changes

the

figure. Case

large

Momentum

observing

to obtain

source

components.

+ I-7=

was

was spin-stabilized

The

angular

equi-

H 2 = 2TD momentum

we

vector

sphere: H_(1

The

by

of

about

of unstable

behavior

without

to calculate

H_ + H i + H_ = H 2

can

motion

momentum

H_

axis

to a rotation

which

of inertia.

fluid

4.83)

H_

Multiplying

This

torque-free

(Equation

principal

is one

I satellite

satellite

be used

unsymmetrical

ellipsoid

of inertia

moment

long a spinning

of the

go over

dissipation.

the Explorer

also

vicinity (D _ Ia).

moment

of minimum

can

of an

of inertia

energy

be provided

motion

the

moment

notably

is how

in the

will gradually

of minimum

axis

question

I1),

of internal

satellites

longitudinal

important

a motion

(D _

axis

presence

for some the

that

of maximum

principal

in the

about

means

of inertia

axis

the

observed ergy

which

:

0(0,.)3

:

O)

115

D)

D - I1

= 0

sphere

(4.91) is illustrated

in the

)

Figure

4.3:

I12

Trace

__

of H on Momentum

1-

D/I1

12

D-

Sphere

ll

= -(T(__tj)

_go_,_._ = _ T_(I-77-5_) Spin

about

13-axis

(D > I_)

For 0_11r," " H_ -- O(w2 --- 0)

H____

I-

D/I3

1_ I_-

D,

H_- -T- D/_, - L (-5-_-_) 11 I_t9 OMeN = '_]_ _,i-_-

For O_lax " H_ = 0(_

D l_ j

= O) 13 -

tg o_,_x = I_ 116

D,

(-b--__l_) I3 -17,

B) Analytical The body

Method

analytical was

first

components There

solution obtained

become are,

of the by

elliptic

of course,

Equation

Euler

in 1758.

functions

also

4.82

three

for the

torque-free

He showed

that

the

unsymmetrical angular

velocity

of time.

very

simple

particular

solutions

of Equation

4.82

namely:

These are

solutions only

The

derivation

The

elliptic

elliptical

w2 = wa = 0

w2 = CONSTANT

w_ = _a = 0

wa = CONSTANT

wI = _

represent

the

The

wL = CONSTANT

axes

steady

about

which

of the

of the

means

case

first

u is a function

foy

about

body

appearing

u = that

the

Euler

functions integral

rotations

will spin

will not in the

kind

elliptic

function

the

of inertia.

steadily.

be given

solution

axes

are

only

the

defined

solution. as follows:

dy

V/(1 _ y2)( 1 _ k2y 2)

of y and

is then

principal

is:

k:

u = F(y, The

the

= 0

inverse

k)

function

y=F-l(u,k)=Sn(u,k) where The

k is called solution

the

for the

modulus angular

(0 < k < 1) velocity

components

can

then

(D- 13) w_= H I I1D(I_la) l)n (At,k) 117

be written

as:

These

"_3 = H where

the

Cn and

7)n

functions

I

(I1- D)

s,_ (At,k)

i

(I_- --13) D) 13D(Iz

cn (_., k)

are

related

Cn_z Dn_z The

constants

A and

k are

given

A= H

above

condition

solution w2(0)

For small functions:

corresponds

Sn

function

by:

= 1 -Sn2z

= 1-

k_Sn_z

by:

(11 - I2)(D - la) (I_I213D)

I3)(I_ k = I(I2(11- - I2)(DThe

to the

- 12) D)

(4.92)

0 < k < 1

to a rotation

about

the

(4.93) II-axis

with

the

initial

= 0.

values

of k(12

_-, /3)

the

elliptic

Sn(AI,O)

Cn(M,

functions

= sin

At

O) = cos

At

approach

the

trigonometric

vn(,v, 0) = 1 The

determination

course,

not

orientation

of the

angular

velocity

the

solution

of the

complete of the

body

relative

done,

in principle,

by substituting

which

yields

differential

these pursued

equations any

three

in this further.

general However,

to an

components problem,

inertial

observer

the solutions equations form

for the

presents

it is important 118

as functions

because

at any

into

Euler's

three

Euler

a formidable to notice,

of time

we still

have

time.

kinematical angles.

The

problem

and

that

although

does,

to find This

of the

can

be

equations, solution

of

will not

be

the

angular

velocities are periodic functions of time, the motion of the rigid body as viewed from an inertial observeris no longer periodic. C)

Perturbation

Method

In many from

practical

a nominal

behavior a nominal

direction.

of a system condition.

To illustrate about the

applications

the/i-axis. Poinsot

the

If the first

such

that

the

second

and

third

an

motion

of these

not

insight equations

unsymmetrical

rigid

small

(i.e.

4.82

greatly

the

dynamic

of motion

body

/2 _/3)

to the

deviate

into

the

is confined

of Equation

must

valuable

is rather

or

motion

by linearizing

asymmetry

equation

body

situations

we consider

I1_1=0 The

rigid

be obtained

method

construction

Therefore

In

can

the

the

about

which

spins

we observe

from

vicinity

of the

spin

axis.

is approximately

_I=_0=CONSTANT

equations

can

then

be written

as:

_2 + A1_3 = 0

(4.94)

&3 -

(4.95)

= 0

A_

where

Al -

Because are

of the

smaller

than

We assume

In order obtained

triangle the

13

inequality

rule of the

spin

equations

will be solved

_o

moment I_ is the

approximate

we assume

_2(0) The

that

the

section,

11 - I_

of inertia

both

[ At [ and

] A2 [

w0.

discussing

to compare

preceding

A2 -

_o

nominal

in the ensuing

to be able

in the

11 -

=0fort

by the

the

solution same

initial

moment

with

the

exact

condition:

=0

Laplace 119

intermediate

transformation

method.

of inertia.

solution

Denoting

the

Laplace

transforms

of the

angular

velocities

by capital

letters:

c {_(t)} = _(_) we obtain:

_(_)

+ _1_3(_) = 0

(4.96)

s_3(._)- ,,.,3(0) - _,(._) = o where

_3(0)

is the

Solving

initial

for the

angular

Laplace

velocity

transforms

along

of the

A_w3(0) _2(S)

The corresponding time:

=

inverse

the/a-axis.

angular

f)a(s)

transforms

velocities

(4.98) s _' +

yield

the

we obtain

-

2 -t- AIA2

--s

(4.97)

angular

A1A2

velocities

as a function

of

A1_3(0)

w2(t)

--

sin _/A1A2t

(4.99)

_3(t) = _3(o)cos v_ A_t For a spin close Equation

4.93

to the/1-axis it is seen

angular

velocities

verified

that

the

It is also z_ -

clockwise This

The symmetry modern

same

of the

plane. sense

is also

that

if A_ > 0(I_

interest

or spin axis, Euler

angle

true

path

of course,

in the

as seen

for the

amplitude

ratio

the

angular

velocity

projection which

of the

is traversing sense

also

in of the

be easily

of w3 and

in the polhode the

w2.

equatorial onto zl-axis

the in a

if A_ < 0(11 < 13).

construction. motion

by an inertial

which

frequencies It can

in a counterclockwise

Poinsot

system

the angular

approximation

solution.

path

the

this

exact

of the

is actually

> 13) and

means

the

an elliptical

with

lies,

the

This

We obtain

that

with holds

to plot

body.

D _ 11. Inserting

A = _v/_l_,

agreement

in agreement

real

quantity

in agreement

of interest

z3 plane

equatorial

are

the

(4.100)

observer.

is particularly

120

of the body, suited

especially

To this end

the

axis of

we introduce

for perturbation

studies.

the

Repeating

the

We now use the 0 and

_.

corresponding

kinematical

equations

wl

=

_-0sin0

co_

=

0cosq_+0cos0sin4_

w3

=

_cos0cosg_-0sin4_

approximation

wl _ w0 = ¢_ and

we have:

restrict

the

motion

to small

angles

Thus: co_ = 0 cos wot + _.sin wot cos = -t_ sin Wot + _ cos coot

CASE

A: Spin

about

Minimum

Moment

of Inertia

=

cos(0)

(Rod:

,/1 < 0; ll < I_ < 13)

Set:

= The

equations

of motion

are

co_

=

cs = s(O)

then: c2sinAt

=t}cosw0t+_sinco0t

cos = cs cos At = -t_ sin coot + _ cos coot We now

introduce

a complex

cone

(4.101)

an_;le:

(4. 02)

a=_+iO Adding

the

two

equations

of motion

ws + ico2 = cscosM Solving

for the

complex

cone

angle

in quadrature

+ ic, sin,lt

this

equation

into a more

= (_ + iO)ei'_°t = &e _'_°t

(4.103)

yields:

c_ = (c3 cosM To bring

we obtain:

+ ic2sinM)e

convenient

cs = B1 + B2 and

121

form

-i'°°t for integration

c2 = B1 -

B_

(4.104) we introduce: (4.105)

Thus:

&

Integrating

with

Converting

back

we finally

arrive

=

[(B1 + B2)cosM

=

[(Bl(COSAt

=

Bte-i(,oo-_)

B2) sin),tle

+ isin M)+

-i'°*

B2(cosM-

isinM)le

-i''o'

(4.106)

t + B_e-;("o+'_) t

respect

to time

c_=i

.P_ ("00-A)

to the

+ i(Bl-

original

furnishes

the

-;(,,0-_)t

B2 + (`00+A)e

e constants

complex

cone

-i(,_0+_)t

by observing

Bt -

ca+c2 2

and

a=

i[Ale

-'(_°-_)'

B2-

angle:

from

4.105

(4.107) that:

ca-c:_ 2

at:

(4.1o8)

+ .42e -'('`°+_)t]

where

.41-

ca+c2 2(.00 - A)

It is seen that the complex cone angle different amplitudes and frequencies. In order prove

the

to make following

a more two

definite

#

A_-

ca-c2 2(,00 + A)

is represented

statement

by two rotating

aboul

the

motion

vectors

we have

having

first

to

inequalities:

,_= _ Proof

and

< "00a,nd

"_ > 1

1:

A_A_ = (11-

I3)(I1 I_I3

(11 -- I3)(I1

-

I2) wo < "00

-- I2) < I213

122

(4.109)

121- IlI_ - ILI3 + Ida < I213

The

final inequality

tia.

As a consequence

Proof This

holds both

11(11-

I_-13)

because

of the

vectors

< 0 Q. E. D. triangle

in Equation

inequality 4.108

moments

clockwise

of inerrotations.

#2: inequality

Because

ensures

It < /3 and

us that

2-2 is indeed

Adding

the

term

(I1 - 13)13

_2

(11 - I2)I_

11 < 12 the

I_I3 to both

the

_1

(h

inequality

-

I1)I3

sides

can

> (I_ -

of the

intermediate

According

to the

triangle

inequality

rule

be rewritten

body-fixed nominal

can

be made

zl-axis

(spin

spin

of inertia.

as:

11)12

inequality

I_ + 13 -

gives:

11)12 Ii > 0 and

therefore,

Q.E.D.

4>4 motion

moment

>1

(12 + h - 11)13 > (12 + 13 -

The

for the

represent

visible axis)

by projecting

onto

the

inertial

direction. Y

123

the

tip of the

Y - Z plane

unit which

vector

along

is normal

the

to the

The

coning

motion

is a superposition

and a small amplitude about the spin axis. confined

within

_ The

of a large

frequency spin itself

an annular

ring.

The

minimum

,4_ + ,42 = aMlx

The maximum

value

is obtained

is in agreement

in general,

not

Case

B:

Spin

about This

because

the

exact

Maximum

Moment

final

angle

is the

initial

rotation rotation motion is one:

la_3(O)

--

I1 wo

- 11) I1) o,_tl:,_" = II2(13 13(I_ -solution

on page

is obtained

by changing

complex

cone

angle

(Disk:

the sign steps

115 and

116.

The

motion

is,

A1 > 0;11 > I2 > la)

in the

remain

9'2 equation

the

of Equation

4.101

same.

is:

o: where

cone

of Inertia

(72 < 0 for A1 > 0. All other

The

frequency

periodic.

the

case

with

small

for (A2 < 0!)

,41 - ,42 = aMAX This

amplitude

rotation both having a clockwise is, of course, also clockwise. The

+

(4.1101

now

ca + c2

ca -- c2

.4__ 2(_0+ _) and .% - 2(_0- A) The coning motion consists now of a superposition rotation and a small amplitude small frequency clockwise

sense

of rotation.

This

is graphically

124

of large amplitude rotation both having illustrated

in the

_ frequency again the same

following

figure

Y f

S The

motion

angles

are

is again

confined

identical

to the

to the

previous

annular

region.

Maximum

and

minimum

cone

ones.

NOTE: For a symmetric of

satellite

I_ = la the

coning

motion

becomes

steady

with

a frequency

ll

wp = w0 + A = I_ w0 This

precessional

regardless ertia.

of whether

One

can

a symmetric attention case

motion

to the

about

the

maximum

or minimum

the

preceding

steps

12 = I3 which

leads

to c2 = es.

this

of A which

It is negative

is about

("forward")

from

sign

disk.

spin

clockwise

result

obtain

body

of a flat

the

is always

Precession

changes for the

when

going

former

case

by

the

from

of inthat

to pay

a long slender positive

axis

moment

observing

It is important and

spin

careful

rod

for the

for

to the

latter.

Sometimes the geometric axes or control axis system deviate slightly from the principal axis system. Let us assume that the control axis has a small angle 13 with the

spin axis.

dynamics angle

This

equations.

of Equation

clockwise with for this case

misalignment It can 4.108

the

spin

is strictly

be simply

or Equation frequency

_0,

a = i[Aie-'('o°-_')t+

a geometric

taken

care

4.110

a vector

is an example A:_e -i(''°+)')t

125

effect

and

of, by adding of magnitude . Equation + Be -'_°t]

does

to the

not

enter

complex

/7 which 4.108

the cone

rotates

would

read

(4.111)

4.6

Forced

Unlike which

of a Rigid

the case of the torque-free is acted upon by external

special

cases.

of motion fixed

Motion

It was

can

point

treatment

be used

to describe

mathematical systems.

The

depend

on the

will

the

velocities by

essentially case.

rotational gravity

We will not perturbation

body

from

or not

the

with

this

methods

a

rather

which

condition.

of spinning

on whether

field

discuss

a nominal

a variety

equations

can The

("gyroscopic") external

torques

body.

equations

one

we consider

ment

a constant

of motion

of time.

principal

axes torque

of motion

are

relative

kinematical

in the

an axially

be directly

motion

the

used

can

The

integrating

as the

The

equations

the

torques)

misalignment.

The

for

depending

of the

Euler

same

will present of a rigid

that

in a uniform

top").

be applicable

as functions

As an example produces

again

will differ

subsequently

the

body

("heavy

deviations

orientation

In this case obtained

instead

methods

(1788)

for a symmetric

small

I (Body-fixed

angular

by Lagrange

of symmetry

but models

rigid body the equations of motion of a rigid body torques can only be solved analytically for very

shown

be integrated

on its axis

lengthy

Method

first

Body

equations.

preceding

paragraph

symmetric

system

spinning

is aligned

about

the

integrated to inertial

such

The for the

missile

that

to find space

the

the

is then

method

is

torque-free

with

thrust

a thrust misa[ign-

w-axis.

then

I1 &l = 0

12 d:_ + (I1 - h)

_t _3 =/;:

13 _a+(l_-From

the

first

of these

about

the

spin

axis

equations

is constant.

11)_1

we see

that

We will denote

11 -

12

126

_

I: = h

: the

0 angular

it by _0.

velocity Introducing

component the

quantity:

_

we can

write

the

other

two equations

as:

L2

d_3This

set

of equations

and yields

the

can

solution

be solved

for the

$ w_ = 0

again

initial

by the

condition

w2(0)

of Laplace

+ sin I AII2

lAIr

L2A w3(t) where

w3(0)

is the

For the

initial

symmetric

a spin

about

because

the

Therefore,

= wz(0)cos

the change

I ,kit-

angular

case,

there

minimum

and

a spin

for both

approximate

introduced

integration

to distinguish

about

-

kinematical

in Equation

the

cone

the

a spin

motion A) which rotational

about

consists

the

equations

4.102

angle

of the

minimum regular

is superimposed vectors

revolve

maximum

between

moment

performed

by the

of inertia

sine function.

on page

_]

e-;('_°+_)t

is obtained

moment

the

complex

cone

+ _e

-i''°t

(4.112)

iL2 e_,_ot AI2 w0

(4.113)

as

' +

of inertia

low frequency the

114 and

we get:

by a small about

mathematically

cos M +

o_ = i[w3(O)rz/(Al2)]e_i(,,o+X) (w0 + A) For

I3-axis.

[w3(0)-_-_ L2]sinM

& = [wa(O) Upon

the

t + A2I-----_

cases:

=

angle

[_l

in sign of )_ can be automatically

we obtain

the

about

is no need

w2(t)---

Using

AL2

i2_--S cos

velocity

transformation

= 0:

L2

Awa(O) sin I lt

2(t) -

method

precession

amplitude spin axis

127

(A