sdac lec final revision

8/23/2019 Sdac Lec Final Revision

http://slidepdf.com/reader/full/sdac-lec-final-revision 1/197

IN THE NAME OFALLAH

THE MOST BENEFICENT

THE MOST MERCIFUL



SPACERAFTDYNAMICS AND CONTROL

http://www.amazon.com/gp/product/images/0894640690/ref=dp_image_0?ie=UTF8&n=283155&s=books

http://www.amazon.com/gp/product/images/9027712042/ref=dp_image_0?ie=UTF8&n=283155&s=books

http://www.amazon.com/gp/product/images/1881883159/ref=dp_image_z_0?ie=UTF8&n=283155&s=books

http://www.amazon.com/gp/reader/0486439259/ref=sib_dp_pt



[email protected] ; 0321-9595510

DR. QASIM ZEESHANBE, MECHANICAL ENGINEERING

NATIONAL UNIVERSITY OF SCIENCE AND TECHNOLOGY, NUST, PAKISTAN, 2000

MS, FLIGHT VEHICLE DESIGNBEIJING UNIVERSITY OF AERONAUTICS AND ASTRONAUTICS, BUAA, P.R.CHINA, 2006

PhD, FLIGHT VEHICLE DESIGNBEIJING UNIVERSITY OF AERONAUTICS AND ASTRONAUTICS, BUAA, P.R.CHINA, 2009



SPACECRAFT

DYNAMICS ANDCONTROL

LECTURE #

SPACECRAFT DYNAMICS



Cross Product

x y y x z x x z y z z y

z y x

z y x

babababababa

bbbaaa

ba

k ji

ba



Cross Product

z y x x z z

z x y x z y

z y y z x x

x y y x z x x z y z z y

bbabac

babbac

bababc

babababababa

0

0

0

bac

ba



February 24, 2011

Cross Product

The cross product of two vectors says something about

how perpendicular they are.

Magnitude:

is smaller angle between the vectors

Cross product of any parallel vectors = zero

Cross product is maximum for perpendicular vectors

Cross products of Cartesian unit vectors:

sin A

sin B

0ˆˆ ;0ˆˆ ;0ˆˆ

ˆˆˆ ;ˆˆˆ ;ˆˆˆ

k k j jii

ik j jk ik ji

y

x

z

i j

k

i

k j



February 24, 2011

Cross Product

Direction: C perpendicular to bothA and B (right-hand rule)

Place A and B tail to tail

Right hand, not left hand

Four fingers are pointed along the

first vector A

“sweep” from first vector A into

second vector B through the smaller

angle between them

Your outstretched thumb points the

direction of C

First practice

? A B B A

? A B B A



February 24, 2011

More about Cross Product The quantity ABsin is the area of the

parallelogram formed by A and B

The direction of C is perpendicular to the

plane formed by A and B

Cross product is not commutative

The distributive law

The derivative of cross productobeys the chain rule

Calculate cross product

- A B B A



Cross Product

z

y

x

x y

x z

y z

z

y

x

z y x x z z

z x y x z y

z y y z x x

b

b

b

aa

aa

aa

c

c

c

bbabac

babbac

bababc

0

0

0

0

0

0



Cross Product

0

0

0

ˆ

ˆ

0

0

0

x y

x z

y z

z

y

x

x y

x z

y z

z

y

x

aa

aa

aa

b

b

b

aa

aa

aa

c

c

c

a

baba



Derivative of a Rotating Vector

Let’s say that vector r is rotating around the origin,

maintaining a fixed distance

At any instant, it has an angular velocity of ω

rω

r

dt

d

rω

rω



Derivative of Rotating Matrix

If matrix A is a rigid 3x3 matrix rotating with

angular velocity ω

This implies that the a, b, and c axes must be rotating

around ω

The derivatives of each axis are ωxa, ωxb, and

ωxc, and so the derivative of the entire matrix is:

AωAω

Aˆ

dt

d



Product Rule

dt

dcabc

dt

dbabc

dt

da

dt

abcd

dt

dbab

dt

da

dt

abd

The product rule defines the derivative of products



Product Rule

It can be extended to vector and matrix products as

well

dt

d

dt

d

dt

d

dt

d

dt

d

dt

d

dt

d

dt

d

dt

d

BAB

ABA

bab

aba

bab

aba



DYNAMICS OF PARTICLES



Kinematics of a Particle

onaccelerati

ityveloc

position

2

2

dt

d

dt

d

dt d

xv

a

xv

x



Mass, Momentum, and Force

force

momentum

mass

ap

f

vp

m

dt

d

m

m



Moment of Momentum

The moment of momentum is a vector

Also known as angular momentum (the two termsmean basically the same thing, but are used in slightly

different situations) Angular momentum has parallel properties with linear

momentum

In particular, like the linear momentum, angular

momentum is conserved in a mechanical system

prL



Moment of Force (Torque)

The moment of force (or torque) about a point is the rate of change of the

moment of momentum about that point

dt

d Lτ



Moment of Force (Torque)

f rτ

f rvvτ

f rpvτ

prprLτ

prL

m

dt

d

dt

d

dt

d



Rotational Inertia

L=rxp is a general expression for the moment of momentum of a particle

In a case where we have a particle rotating around the origin while keeping

a fixed distance, we can re-express the moment of momentum in terms of it’s

angular velocityω



Rotational Inertia

rrI

ωIL

ω

rrL

ωrrrωrL

vrvrL

prL

ˆˆ

ˆˆ

m

m

mm

mm



Rotational Inertia

22

22

22

0

0

0

0

0

0

ˆˆ

y x z y z x

z y z x y x

z x y x z y

x y

x z

y z

x y

x z

y z

r r r r r r

r r r r r r

r r r r r r

m

r r

r r

r r

r r

r r

r r

m

m

I

I

rrI



Rotational Inertia

ωIL

I22

22

22

y x z y z x

z y z x y x

z x y x z y

r r mr mr r mr r mr r r mr mr

r mr r mr r r m



Rotational Inertia

The rotational inertia matrix I is a 3x3 matrix that isessentially the rotational equivalent of mass

It relates the angular momentum of a system to itsangular velocity by the equation

This is similar to how mass relates linear momentum tolinear velocity, but rotation adds additionalcomplexity

ωIL

vp m



Systems of Particles

momentumtalto

massof center of position

particlesallof massltota1

iiicm

i

ii

cm

n

i

itotal

m

m

m

mm

vpp

xx



Velocity of Center of Mass

cmtotal cm

total

cmcm

i

ii

i

ii

cm

i

iicmcm

m

m

m

m

m

dt

d m

m

m

dt

d

dt

d

vp

pv

vx

v

xxv



Force on a Particle

The change in momentum of the center of mass is

equal to the sum of all of the forces on the individual

particles

This means that the resulting change in the total

momentum is independent of the location of the

applied force

i

iicm

icm

dt

d

dt

d

dt

d f

ppp

pp




icmicm

iicm

pxxL

prL

The total moment of momentum around the center of mass is:



Torque in a System of Particles

iicm

iicm

iicmcm

iicm

dt d

dt

d

dt

d

f rτ

prτ

prLτ

prL




We can see that a system of particles behaves a lot like a

particle itself

It has a mass, position (center of mass), momentum, velocity,acceleration, and it responds to forces

We can also define it’s angular momentum and relate a

change in system angular momentum to a force applied to an

individual particle

iicm f rτ

icm f f



Internal Forces

If forces are generated within the particle system (sayfrom gravity, or springs connecting particles) they

must obey Newton’s Third Law (every action has anequal and opposite reaction)

This means that internal forces will balance out andhave no net effect on the total momentum of the

system As those opposite forces act along the same line of

action, the torques on the center of mass cancel out aswell



KINEMATICS OF RIGIDBODIES



Kinematics of a Rigid Body

For the center of mass of the rigid body:

2

2

dt

d

dt

d

dt

d

cmcmcm

cmcm

cm

xva

xv

x



Kinematics of a Rigid Body

For the orientation of the rigid body:

onacceleratiangular

locityangular ve

matrixnorientatio3x3

dt d ωω

ω

A



Offset Position

Let’s say we have a point on a rigid body

Ifr

is the world space offset of the point relative tothe center of mass of the rigid body, then the position

x of the point in world space is:

rxx cm



Offset Velocity

The velocity of the offset point is just the derivative of

its position

rωvv

rxxv

rxx

cm

cm

cm

dt

d

dt

d

dt

d



Offset Acceleration

The offset acceleration is the derivative of the offset

velocity

rωωrωaa

rωr

ωvva

rωvv

cm

cm

cm

dt

d

dt

d

dt

d

dt

d



Kinematics of an Offset Point

The kinematic equations for an fixed point on a rigid body are:

rωωrωaa

rωvv

rxx

cm

cm

cm



DYNAMICS OF RIGIDBODIES



Rigid Bodies

We treat a rigid body as a system of particles, where thedistance between any two particles is fixed

We will assume that internal forces are generated to hold therelative positions fixed. These internal forces are all balancedout with Newton’s third law, so that they all cancel out and

have no effect on the total momentum or angular momentum

The rigid body can actually have an infinite number ofparticles, spread out over a finite volume

Instead of mass being concentrated at discrete points, we willconsider the density as being variable over the volume



Rigid Body Mass

With a system of particles, we defined the total mass

as:

For a rigid body, we will define it as the integral of

the density ρ over some volumetric domain Ωd m

n

i

imm1



Rigid Body Center of Mass

The center of mass is:

d d

cmxx



Rigid Body Rotational Inertia

zz yz xz

yz yy xy

xz xy xx

y x z y z x

z y z x y x

z x y x z y

I I I

I I I

I I I

d r r d r r d r r

d r r d r r d r r

d r r d r r d r r

I

I

22

22

22



Diagonalization of Rotational Inertial

z

y

xT

zz yz xz

yz yy xy

xz xy xx

I

I

I

where

I I I

I I I

I I I

00

00

00

00 IAIAI

I



Derivative of Rotational Inertial

ωIIω

I

ωIIωωAIAIω

I

AωIAIω

I

AωIAAIAωI

AIAAI

AAIAI

ˆ

ˆˆ

ˆ

0

0

00

000

dt

d

dt

d

dt

d

dt d

dt

d

dt

d

dt

d

dt

d

T T T

T

T T

T

T T



Derivative of Angular Momentum

ωIωIωτ

ωIωωIωIωτ

ωIωωIIωτ

ω

Iω

ILτ

ωIL

ˆ

ˆ

dt

d

dt

d

dt

d



Newton-Euler Equations

ωIωIωτ

af m



Applied Forces & Torques

ωIωτIω

f a

f rτ

f f

1

1m

iicg

icg



Properties of Rigid Bodies

af

vp

a

v

x

m

m

m

ωIωIωf rτ

ωIL

ω

ω

A

I

C ti L f S t f



Conservation Laws for Systems of

Particles

Center of Mass

Consider a system made up of n particles. A typical particle, i, has mass mi,and, at the instant considered, occupies the position ri relative to a frame

xyz. We can then define the center of mass, G, as the point whose position

vector, rG, is such that,

Here, m is the total mass

It is important to note that the center of mass is a property of the system

and does not depend on the reference frame used. In particular, if wechange the location of the origin O, rG will change, but the absolute

position of the point G within the system will not.



Forces

In order to derive conservation laws for our system, we isolate it a little

more carefully, identify what mass particles it contains and what forces act

upon the individual particles.

We will consider two types of forces acting on the particles :

External forces arising outside the system. We will denote the resultant of

all the external forces N.

Internal forces due to pairwise particle interactions. This force could arisefrom gravitation attraction or from internal force due to the connections

between particles. It could also arise from collisions between individual

particles that, as we have seen, produce equal and opposite impulsive

forces that conserve momentum.



Conservation of Linear Momentum



These ideas also describe the conservation of linear momentum under

external and internal collisions. Since individual internal collisions between

particles in the system conserve momentum, the sum of their interactions alsoconserves momentum. If we consider an external particle imparting

momentum to the system, it could be treated as an external impulse.

Conversely we can consider the particle about to collide to be a part of thesystem, and include its momentum as part of total system momentum, which

is then conserved by Newton’s law.

Angular Momentum of a



Angular Momentum of a

Rigid Body

Angular momentum of a rotating rigid object

L has the same direction as

L is positive when object rotates in CCW

L is negative when object rotates in CW

Angular momentum SI unit: kgm2

/s

Calculate L of a 10 kg disc when = 320 rad/s, R = 9 cm = 0.09 m

L = I and I = MR2/2 for disc

L = 1/2MR2

= ½(10)(0.09)2

(320) = 12.96 kgm2

/s

L



A solid sphere and a hollow sphere have the same

mass and radius. They are rotating with the same

angular speed. Which one has the higher angular

momentum?

A) the solid sphere

B) the hollow sphere

C) both have the same angular momentumD) impossible to determine

Finding angular momentum



Linear Momentum and Force

Linear motion: apply force to a mass

The force causes the linear momentum to change

The net force acting on a body is the time rate of change

of its linear momentum



Angular Momentum and Torque

Rotational motion: apply torque to a rigid body

The torque causes the angular momentum to change

The net torque acting on a body is the time rate of change

of its angular momentum

and to be measured about the same origin

The origin should not be accelerating, should be an inertial

frame



Demonstration

Start from

Expand using derivative chain rule

dt

Ld net

dt

pd F F net

What about SYSTEMS of Rigid Bodies?



What about SYSTEMS of Rigid Bodies?

• i = net torque on particle “i”

• internal torque pairs are

included in sum

i L L sys

• individual angular momenta Li

• all about same origin

i

i

sys

dt

Ld

dt

Ld

i

BUT… internal torques in the sum cancel in Newton 3rd lawpairs. Only External Torques contribute to Lsys

Nonisolated System: If a system interacts with its environment in thesense that there is an external torque on the system, the net externaltorque acting on a system is equal to the time rate of change of itsangular momentum.

dt

Ld ii :body singleafor lawRotational

nd

2

Total angular momentumof a system of bodies:

net external torque on the systemnet,

i

ext i

sys

dt

Ld



a

a

Example: A Non-isolated System

A sphere mass m1 and a block

of mass m2 are connected by a

light cord that passes over a

pulley. The radius of the pulleyis R, and the mass of the thin

rim is M. The spokes of the

pulley have negligible mass.

The block slides on africtionless, horizontal surface.

Find an expression for the

linear acceleration of the two

objects.

gRmext 1

a



a

Masses are connected by a light cord

Find the linear acceleration a.

• Use angular momentum approach

• No friction between m2

and table

• Treat block, pulley and sphere as a non-

isolated system rotating about pulley axis. As

sphere falls, pulley rotates, block slides

• Constraints:

/ pulleyfor

sphere and block for sa' and sv'Equal

dv/dt αRa

dt d αωRv

• Ignore internal forces, consider external forces only

• Net external torque on system:

• Angular momentum of system:

(not constant) ω MRvRmvRm I ωvRmvRm L sys

2

2121

gRmτ MR)a Rm R(mα MRaRmaRmdt

dLnet

sys

121

2

21

wheelof center about1 gR mnet

21

1 mm M

g ma same result followed from earlier

method using 3 FBD’s & 2nd law

I



Isolated System

Isolated system: net external torque acting on a

system is ZERO

no external forces

net external force acting on a system is ZERO



Angular Momentum Conservation

where i denotes initial state, f is final state

L is conserved separately for x, y, z direction For an isolated system consisting of particles,

For an isolated system is deformable

f itot L L L

or constant



First Example

A puck of mass m = 0.5 kg is attachedto a taut cord passing through a smallhole in a frictionless, horizontal surface.The puck is initially orbiting with speed

vi = 2 m/s in a circle of radius ri = 0.2m. The cord is then slowly pulled frombelow, decreasing the radius of thecircle to r = 0.1 m.

What is the puck’s speed at the smaller

radius?

Find the tension in the cord at thesmaller radius.



Angular Momentum Conservation

m = 0.5 kg, vi = 2 m/s, ri = 0.2 m, rf

= 0.1 m, vf = ?

Isolated system?

Tension force on m exert zero torqueabout hole, why?



constant0 L τ axisabout z -

net

final

f f

initial

ii ω I ω I L Moment of inertia

changes

Isolated

System

H f t h ld th t d t i ?



How fast should the student spin?

L is constant… while moment of inertia changes

The student on a platform is rotating (no friction) with angular speed 1.2

rad/s.• His arms are outstretched and he holds a brick in each hand.

• The rotational inertia of the system consisting of the professor, the

bricks, and the platform about the central axis is 6.0 kg·m2.

By moving the bricks the student decreases the rotational inertia of the

system to 2.0 kg·m2.

(a) what is the resulting angular speed of the platform?

(b) what is the ratio of the system’s new kinetic energy to theoriginal kinetic energy?



f f i I I L

LL L

i axis fixed aabout...

initialfinal torqueexternal Zero

KE has increased!!

L is constant… while moment of inertia changes,

Ii = 6 kg-m2

i = 1.2 rad/s

If = 2 kg-m2

f = ? rad/s

Solution (a):rad/s 3.6 1.2

2

6 i

f

i f

I

I

Solution (b):3 )( )(

22

2

2

1

221

f

i

f

i

i

f

f

f

i

f

ii

f f

i

f

I

I

I

I

I

I

I

I

I

I

K

K

Controlling spin ( ) by changing I (moment of inertia)



In the air, net = 0L is constant

f f ii I I L

Change I by curling up or stretching out- spin rate must adjust

Moment of inertia changes

Example: A merry-go-round problem



p y g p

A 40-kg child running at 4.0

m/s jumps tangentially onto astationary circular merry-go-round platform whose radius is2.0 m and whose moment ofinertia is 20 kg-m2. There isno friction.

Find the angular velocity ofthe platform after the child

has jumped on.



The moment of inertia of thesystem = the moment of inertiaof the platform plus the moment ofinertia of the person.

Assume the person can betreated as a particle

As the person moves toward thecenter of the rotating platform themoment of inertia decreases.

The angular speed must increasesince the angular momentum isconstant.

The Merry-Go-Round

Solution: A merry go round problem



Solution: A merry-go-round problem

I = 20 kg.m2

VT = 4.0 m/s

mc = 40 kg

r = 2.0 m0 = 0

r vmr vm I I L T cT ciii 0

f c f f f ωr m I ω I L )( 2

r vmωr m I T c f c )( 2

rad/s78.124010244022r m I

r vmω

c

T c f



Radius of Gyration

It is common to report the moment of inertia of a rigid

body in terms of the radius of gyration, k.

This is defined as

can be interpreted as the root-mean-square of themass element distances from the axis of rotation. Since

the moment of inertia depends upon the choice of axis,

the radius of gyration also depends upon the choice of

axis.

Thus we write for the radius of gyration about

the center of mass, and for the radius of

gyration about the fixed point O.



Parallel Axis Theorem

Equations of Motion for a Rigid



Equations of Motion for a Rigid

Body • Consider a rigid body acted upon

by several external forces.

• Assume that the body is made of

a large number of particles.

• For the motion of the mass center

G of the body with respect to the

Newtonian frame Oxyz ,

am F

• For the motion of the body with

respect to the centroidal frame

Gx’y’z’ ,

GG H M

• System of external forces is

equipollent to the system

consisting of .and G H am

Angular Momentum of a Rigid



Angular Momentum of a Rigid

Body in Plane Motion

• Consider a rigid slab in

plane motion.

• Angular momentum of the slab may be

computed by

I

mr

mr r

mvr H

ii

n

i

iii

n

iiiiG

Δ

Δ

Δ

21

1

• After differentiation,

I I H G

• Results are also valid for plane motion of bodies

which are symmetrical with respect to the

reference plane.

• Results are not valid for asymmetrical bodies or

three-dimensional motion.

Plane Motion of a Rigid Body:



Plane Motion of a Rigid Body:

D’Alembert’s Principle

I M am F am F G y y x x

• Motion of a rigid body in plane motion iscompletely defined by the resultant and moment

resultant about G of the external forces.

• The external forces and the collective effective

forces of the slab particles are equipollent (reduceto the same resultant and moment resultant) and

equivalent (have the same effect on the body).

• The most general motion of a rigid body that is

symmetrical with respect to the reference plane

can be replaced by the sum of a translation and a

centroidal rotation.

• d’Alembert’s Principle: The external forces acting

on a rigid body are equivalent to the effective

forces of the various particles forming the body.

Axioms of the Mechanics of Rigid



Axioms of the Mechanics of Rigid

Bodies

• The forces act at different points on

a rigid body but but have the same magnitude,

direction, and line of action.

F F

and

• The forces produce the same moment about

any point and are therefore, equipollent

external forces.

• This proves the principle of transmissibility

whereas it was previously stated as an axiom.

Problems Involving the Motion of a



ob e s o g e o o o a

Rigid Body• The fundamental relation between the forces

acting on a rigid body in plane motion and the

acceleration of its mass center and the

angular acceleration of the body is illustrated

in a free-body-diagram equation.

• The techniques for solving problems of staticequilibrium may be applied to solve

problems of plane motion by utilizing

- d’Alembert’s principle, or

- principle of dynamic equilibrium

• These techniques may also be applied to

problems involving plane motion of

connected rigid bodies by drawing a free-

body-diagram equation for each body and

solving the corresponding equations of

motion simultaneously.



Sample Problem.1

At a forward speed of 30 ft/s, the truck

brakes were applied, causing the wheels

to stop rotating. It was observed that the

truck to skidded to a stop in 20 ft.

Determine the magnitude of the normalreaction and the friction force at each

wheel as the truck skidded to a stop.

SOLUTION:

• Calculate the acceleration during the

skidding stop by assuming uniform

acceleration.

• Apply the three corresponding scalar equations to solve for the unknown

normal wheel forces at the front and rear

and the coefficient of friction between the

wheels and road surface.

• Draw the free-body-diagram equation

expressing the equivalence of the

external and effective forces.



Sample Problem1

16 - 83

ft20s

ft300 xv

SOLUTION:

• Calculate the acceleration during the skiddingstop by assuming uniform acceleration.

ft202s

ft300

2

2

020

2

a

x xavv

s

ft5.22a

• Draw a free-body-diagram equation expressing

the equivalence of the external and effective

forces.• Apply the corresponding scalar equations.

0W N N B Aeff y y F F

699.02.32

5.22

g

a

a g W W

N N

am F F

k

k

B Ak

B Aeff x x F F



Sample Problem 1

16 - 84

W N W N B A 350.0

W N N Arear 350.021

21 W N rear 175.0

W N N V front 650.021

21 W N front 325.0

W N F rear k rear 175.0690.0

W F rear 122.0

W N F front k front 325.0690.0

W F front 227.0.0

• Apply the corresponding scalar equations.

W N g

aW a

g

W W N

am N W

B

B

B

650.0

45

12

45

12

1

ft4ft12ft5

eff A A M M



Sample Problem 2

16 - 85

The thin plate of mass 8 kg is held inplace as shown.

Neglecting the mass of the links,

determine immediately after the wire

has been cut (a) the acceleration of the

plate, and (b) the force in each link.

SOLUTION:

• Note that after the wire is cut, all

particles of the plate move along parallel

circular paths of radius 150 mm. The

plate is in curvilinear translation.




• Resolve into scalar component

equations parallel and perpendicular to

the path of the mass center.

• Solve the component equations and the

moment equation for the unknown

acceleration and link forces.



Sample Problem 2

16 - 86

SOLUTION:

• Note that after the wire is cut, all particles of the plate move along parallel circular paths

of radius 150 mm. The plate is in curvilinear

translation.• Draw the free-body-diagram equation

expressing the equivalence of the external

and effective forces.

• Resolve the diagram equation into

components parallel and perpendicular to

the path of the mass center.

eff t t F F

30cos

30cos

mg

amW

30cosm/s81.9 2a

2sm50.8a 60o



Sample Problem 2

16 - 87

2sm50.8a 60o

• Solve the component equations and themoment equation for the unknown acceleration

and link forces.

eff GG M M

0mm10030cosmm25030sin

mm10030cosmm25030sin

DF DF

AE AE F F

F F

AE DF

DF AE

F F

F F

1815.0

06.2114.38

eff nn F F

2sm81.9kg8619.0

030sin1815.0

030sin

AE

AE AE

DF AE

F

W F F

W F F

T F AE N9.47

N9.471815.0 DF F C F DF N70.8



Sample Problem 16.3

16 - 88

A pulley weighing 12 lb and having a radius

of gyration of 8 in. is connected to two

blocks as shown.

Assuming no axle friction, determine the

angular acceleration of the pulley and the

acceleration of each block.

SOLUTION:

• Determine the direction of rotation by

evaluating the net moment on the

pulley due to the two blocks.

• Relate the acceleration of the blocks tothe angular acceleration of the pulley.



external and effective forces on the

complete pulley plus blocks system.

• Solve the corresponding moment

equation for the pulley angular

acceleration.



Sample Problem 3

16 - 89

• Relate the acceleration of the blocks to the angular

acceleration of the pulley.

ft12

10

A A r a

ft12

6

B B r a

2

2

2

22

sftlb1656.0

ft12

8

sft32.2

lb12

k g W k m I note:

SOLUTION:

• Determine the direction of rotation by evaluating thenet moment on the pulley due to the two blocks.

lbin10in10lb5in6lb10G M

rotation is counterclockwise.



Sample Problem 3

16 - 90

• Draw the free-body-diagram equation expressingthe equivalence of the external and effective forces

on the complete pulley and blocks system.

2

126

2

1210

2

sft

sft

sftlb1656.0

B

A

a

a

I

eff GGM M

1210

1210

2.325

126

126

2.3210

1210

126

1210

126

1210

126

1656.0510

ftftftlb5ftlb10 A A B B amam I

• Solve the corresponding moment equation for the

pulley angular acceleration.

2srad374.2

2126 srad2.374ft

B B r a

2sft187.1 Ba

2

1210 srad2.374ft

A A r a

2sft978.1 Aa

Then,



Sample Problem 16.4

16 - 91

A cord is wrapped around a

homogeneous disk of mass 15 kg.The cord is pulled upwards with a

force T = 180 N.

Determine: (a) the acceleration of the

center of the disk, (b) the angular

acceleration of the disk, and (c ) the

acceleration of the cord.

SOLUTION:



external and effective forces on the disk.

• Solve the three corresponding scalar equilibrium equations for the horizontal,

vertical, and angular accelerations of the

disk.

• Determine the acceleration of the cord by

evaluating the tangential acceleration of

the point A on the disk.

S l P bl 4



Sample Problem 4SOLUTION:

• Draw the free-body-diagram equation expressingthe equivalence of the external and effective

forces on the disk.

eff y y F F

kg15

sm81.9kg15- N180 2

m

W T a

amW T

y

y

2sm19.2 ya

eff GG M M

m5.0kg15

N18022

2

21

mr

T

mr I Tr

2

srad0.48

eff x x F F

xam0 0 xa

• Solve the three scalar equilibrium equations.



Sample Problem 4

16 - 93

2sm19.2 ya

2srad0.48

0 xa

• Determine the acceleration of the cord by

evaluating the tangential acceleration of the point A

on the disk.

22 srad48m5.0sm19.2

t G At Acord aaaa

2sm2.26cord a



Sample Problem 5

16 - 94

A uniform sphere of mass m and radius

r is projected along a rough horizontal

surface with a linear velocity v 0

. The

coefficient of kinetic friction between the

sphere and the surface is k .

Determine: (a) the time t 1 at which the

sphere will start rolling without sliding,

and (b) the linear and angular velocities

of the sphere at time t 1.

SOLUTION:



external and effective forces on the

sphere.

• Solve the three corresponding scalar

equilibrium equations for the normal

reaction from the surface and the linear

and angular accelerations of the sphere.

• Apply the kinematic relations for uniformly accelerated motion to

determine the time at which the

tangential velocity of the sphere at the

surface is zero, i.e., when the sphere

stops sliding.



Sample Problem 5

16 - 95

SOLUTION:

• Draw the free-body-diagram equation expressingthe equivalence of external and effective forces

on the sphere.

• Solve the three scalar equilibrium equations.

eff y y F F

0W N mg W N

eff x x F F

mg

am F

k g a k

2

32 mr r mg

I Fr

k r

g k

2

5

eff GGM M

NOTE: As long as the sphere both rotates and

slides, its linear and angular motions are

uniformly accelerated.



Sample Problem 5

16 - 96

g a k

r

g k

2

5

• Apply the kinematic relations for uniformly accelerated

motion to determine the time at which the tangential

velocity of the sphere at the surface is zero, i.e., when

the sphere stops sliding.

t g vt avv k 00

t r g t k

2500

11025 t

r g r gt v k

k g vt k

01

72

g

v

r

g t

r

g

k

k k 011

7

2

2

5

2

5

r

v01

7

5

vr r v 011

50751 vv

At the instant t 1 when the sphere stops sliding,

11 r v



Constrained Plane Motion

16 - 97

• Most engineering applications involve rigid

bodies which are moving under givenconstraints, e.g., cranks, connecting rods, and

non-slipping wheels.

• Constrained plane motion: motions with

definite relations between the components of

acceleration of the mass center and theangular acceleration of the body.

• Solution of a problem involving constrained

plane motion begins with a kinematic analysis.

• e.g., given and , find P , N A, and N B.- kinematic analysis yields

- application of d’Alembert’s principle yields P ,

N A, and N B.

.and y x aa

Constrained Motion: Noncentroidal Rotation



Constrained Motion: Noncentroidal Rotation

16 - 98 • Noncentroidal rotation: motion of a body isconstrained to rotate about a fixed axis that does

not pass through its mass center.

• Kinematic relation between the motion of the mass

center G and the motion of the body about G,

2r ar a nt

• The kinematic relations are used to eliminate

from equations derived from

d’Alembert’s principle or from the method of

dynamic equilibrium.

nt aa and

Constrained Plane Motion: Rolling



Constrained Plane Motion: Rolling

Motion16 - 99

• For a balanced disk constrained to

roll without sliding,r ar x

• Rolling, no sliding:

N F s r a

Rolling, sliding impending:

N F s r a

Rotating and sliding:

N F k r a, independent

• For the geometric center of an

unbalanced disk,

r aO

The acceleration of the mass center,

nOGt OGO

OGOG

aaa

aaa

S l P bl 6



Sample Problem 6

16 - 100

The portion AOB of the mechanism is

actuated by gear D and at the instant

shown has a clockwise angular velocity

of 8 rad/s and a counterclockwiseangular acceleration of 40 rad/s2.

Determine: a) tangential force exerted

by gear D, and b) components of the

reaction at shaft O.

kg3

mm85

kg4

OB

E

E

m

k

m

SOLUTION:

• Draw the free-body-equation for AOB,



• Evaluate the external forces due to the

weights of gear E and arm OB and theeffective forces associated with the

angular velocity and acceleration.

• Solve the three scalar equations

derived from the free-body-equation for

the tangential force at A and the

horizontal and vertical components of

reaction at shaft O.

Sample Problem 6



Sample Problem 6

16 - 101

rad/s8

2srad40

kg3

mm85

kg4

OB

E

E

m

k

m

SOLUTION:

• Draw the free-body-equation for AOB.

• Evaluate the external forces due to the weights

of gear E and arm OB and the effective forces.

N4.29sm81.9kg3

N2.39sm81.9kg4

2

2

OB

E

W

W

m N156.1

srad40m085.0kg4 222 E E E k m I

N0.24

srad40m200.0kg3 2r mam OBt OBOB

N4.38

srad8m200.0kg322r mam OBnOBOB

m N600.1

srad40m.4000kg3 22

1212

121 Lm I OBOB

S l P bl 6



Sample Problem 6

16 - 102

N4.29

N2.39

OB

E

W

W

m N156.1 E I

N0.24t OBOB am

N4.38nOBOB am

m N600.1OB I

• Solve the three scalar equations derived from the

free-body-equation for the tangential force at Aand the horizontal and vertical components of

reaction at O.eff OO M M

m N600.1m200.0 N0.24m N156.1

m200.0m120.0 OBt OBOB E I am I F

N0.63 F

eff x x F F

N0.24t OBOB x am R

N0.24 x R

eff y y F F

N4.38 N4.29 N2.39 N0.63 y

OBOBOB E y

R

amW W F R

N0.24 y R



Sample Problem 8

16 - 103

A sphere of weight W is released with

no initial velocity and rolls without

slipping on the incline.

Determine: a) the minimum value of

the coefficient of friction, b) the velocity

of G after the sphere has rolled 10 ft

and c ) the velocity of G if the sphere

were to move 10 ft down a frictionless

incline.

SOLUTION:

• Draw the free-body-equation for the

sphere, expressing the equivalence of the


• With the linear and angular accelerations

related, solve the three scalar equationsderived from the free-body-equation for

the angular acceleration and the normal

and tangential reactions at C .

• Calculate the velocity after 10 ft of

uniformly accelerated motion.

• Assuming no friction, calculate the linear

acceleration down the incline and the

corresponding velocity after 10 ft.

• Calculate the friction coefficient required

for the indicated tangential reaction at C .

Sample Problem 8



Sample Problem 8

16 - 104

SOLUTION:

• Draw the free-body-equation for the sphere,expressing the equivalence of the external and

effective forces.

r a

• With the linear and angular accelerations related,

solve the three scalar equations derived from the

free-body-equation for the angular acceleration

and the normal and tangential reactions at C.

eff C C M M

2

2

52

52

sin

r g W r r

g W

mr r mr

I r amr W

r g 7sin5

7

30sinsft2.325

7

30sin5

2

g r a

2sft50.11a

S l P bl 8



Sample Problem 8

6 - 105

• Solve the three scalar equations derived from thefree-body-equation for the angular acceleration

and the normal and tangential reactions at C.

r g 7sin5

2sft50.11r a

eff x x F F

W W F

g

g

W

am F W

143.030sin7

27

sin5

sin

eff y y F F

W W N

W N

866.030cos

0cos

• Calculate the friction coefficient required for the

indicated tangential reaction at C .

W

W

N

F

N F

s

s

866.0

143.0165.0 s

Sample Problem 8



Sample Problem 8

16 - 106

r

g

7

sin5

2sft50.11r a

• Calculate the velocity after 10 ft of uniformly

accelerated motion.

ft10sft50.1120

2

2

020

2 x xavv

sft17.15v

eff GG M M 00 I

• Assuming no friction, calculate the linear acceleration and the corresponding velocity after

10 ft.

eff x x F F

22 sft1.1630sinsft2.32

sin

a

a

g

W amW

ft10sft1.1620

2

2

020

2 x xavv

sft94.17v

Sample Problem 9



Sample Problem 9

16 - 107

A cord is wrapped around the inner

hub of a wheel and pulled horizontally

with a force of 200 N. The wheel has

a mass of 50 kg and a radius of

gyration of 70 mm. Knowing s = 0.20

and k = 0.15, determine the

acceleration of G and the angular

acceleration of the wheel.

SOLUTION:

• Draw the free-body-equation for the wheel,

expressing the equivalence of the external

and effective forces.

• Assuming rolling without slipping andtherefore, related linear and angular

accelerations, solve the scalar equations

for the acceleration and the normal and

tangential reactions at the ground.

• Compare the required tangential reactionto the maximum possible friction force.

• If slipping occurs, calculate the kinetic

friction force and then solve the scalar

equations for the linear and angular

accelerations.

Sample Problem 9



Sample Problem 9

16 - 108

SOLUTION:

• Draw the free-body-equation for the wheel,.

Assume rolling withoutslipping,

m100.0

r a

2

22

mkg245.0

m70.0kg50k m I

• Assuming rolling without slipping, solve the scalar

equations for the acceleration and ground

reactions.

22

2

22

sm074.1srad74.10m100.0

srad74.10

mkg245.0m100.0kg50m N0.8

m100.0m040.0 N200

a

I am

eff C C M M

eff x x F F

N5.490sm074.1kg50

0

2mg N

W N

eff x x F F

N3.146

sm074.1kg50 N200 2

F

am F

Sample Problem 9



Sample Problem 9

16 - 109

N3.146 F N5.490 N Without slipping,

• Compare the required tangential reaction to the

maximum possible friction force. N1.98 N5.49020.0max N F s

F > F max , rolling without slipping is

impossible.

• Calculate the friction force with slipping and solve

the scalar equations for linear and angular accelerations. N6.73 N5.49015.0 N F F k k

eff GG M M

2

2

srad94.18

mkg245.0

m060.0.0 N200m100.0 N6.73

2

srad94.18

eff x x F F

akg50 N6.73 N2002sm53.2a



Sample Problem 10

16 - 110

The extremities of a 4-ft rod

weighing 50 lb can move freely and

with no friction along two straight

tracks. The rod is released with no

velocity from the position shown.

Determine: a) the angular

acceleration of the rod, and b) the

reactions at A and B.

SOLUTION:

• Based on the kinematics of the constrained

motion, express the accelerations of A, B,

and G in terms of the angular acceleration.

• Draw the free-body-equation for the rod,



• Solve the three corresponding scalar

equations for the angular acceleration and

the reactions at A and B.

S l P bl 10



Sample Problem 10

16 - 111

SOLUTION:

• Based on the kinematics of the constrained

motion, express the accelerations of A, B, and

G in terms of the angular acceleration.

Express the acceleration of B as

A B A B aaa

With the corresponding vector triangle

and the law of signs yields

,4 A Ba

90.446.5 B A aa

The acceleration of G is now obtained from

AG AG aaaa

2 where AGa

Resolving into x and y components,

732.160sin2

46.460cos246.5

y

x

a

a

Sample Problem 10



Sample Problem 10

16 - 112

• Draw the free-body-equation for the rod,

expressing the equivalence of the external andeffective forces.

69.2732.1

2.32

50

93.646.42.32

50

07.2sftlb07.2

ft4sft32.2

lb50

12

1

2

2

2

2

121

y

x

am

am

I

ml I

• Solve the three corresponding scalar equations

for the angular acceleration and the reactions at

A and B.

2srad30.2

07.2732.169.246.493.6732.150

eff E E M M

2srad30.2

eff x x F F

lb5.22

30.293.645sin

B

B

R

R

lb5.22 B R

45o

eff y y F F

30.269.25045cos5.22 A R

lb9.27 A R

Kinetic energy of a system of particles1



• Kinetic energy of a system of particles

• Introducing a center of mass:

• We can rewrite the coordinates in the center-of-mass coordinate system:

• Kinetic energy can be rewritten:

i

ii r mT 2)(2

1

i

i

i

ii

m

r m

R

Rr r ii

' Rr r ii

'

i

ii r mT 2)(2

1

i

iii Rr Rr m )'()'(2

1

i

i

i

ii

i

ii Rm Rr mr m 22 )(

2

1 )'()'(

2

1

M

r mi

ii

R M r mi

ii

Kinetic energy of a system of particles



• On the other hand

• In the center-of-mass coordinatesystem, the center of mass is at the

origin, therefore

i

i

i

ii

i

ii m Rr m Rr m 22 )(2

1 ')'(

2

1

M Rr mdt

d Rr m

i iii ii

22 )(2

1 ')'(

2

1

i

i

i

ii

i

ii Rm Rr mr mT 22 )(

2

1 )'()'(

2

1

R M r mi

ii

'' R M r mi

ii

M Rr mT i

ii

22 )(

2

1 )'(

2

1




• Kinetic energy of the system of particles consists of

a kinetic energy about the center of mass plus akinetic energy obtained if all the mass were

concentrated at the center of mass

• This statement can be applied to the case of a rigid

body: Kinetic energy of a rigid body consists of akinetic energy about the center of mass plus a kinetic

energy obtained if all the mass were concentrated at

the center of mass

• Recall Chasles’ theorem!

M Rr mT i

ii

22 )(

2

1 )'(

2

1




• Chasles: we can represent motion of a rigid body as

a combination of a rotation and translation

• If the potential and/or the generalized external

forces are known, the translational motion of center

of mass can be dealt with separately, as a motion of a

point object

• Let us consider the rotational part or motion

M Rr mT i

ii

22 )(

2

1 )'(

2

1

iii R r mT

2

)'(2

1

Rotational kinetic energy1 1

1



• Rate of change of a vector

• For a rigid body, in the rotating frame of reference,

all the distances between the points of the rigid body

are fixed:

• Rotational kinetic energy:

i

ii R r mT 2)'(2

1

i

iii r r m ''2

1

')'()'( ir i si r ωr r

0)'( r ir

')'( i si r ωr

i

iii r ωr ωm )'()'(2

1

iiii R r

ω

r ω

mT )'()'(2

1

i j ji jii

r ωr ωm3

1

)'()'(2

1

i j nm

nim jmn

l k

l ik jkl i r ωr ωm3

1

3

1,

3

1,

''2

1

Rotational kinetic energy

3



i j nm

nim jmn

l k

l ik jkl i R r ωr ωmT 3

1

3

1,

3

1,

''

2

1

i nml k j

nil imk jm n jkl i r r ωωm3

1,,,,

''2

1knlmnl km

j

jmn jkl

3

1

i nml k

nil imk knlmnl kmi r r ωωm3

1,,,

'')(

2

1

i l k

l l ik ik

l k

l ik i ωr r ωr ωm3

1,

3

1,

22 '')'()(2

1

3

1,

2 ]'')'[(21

l k

l ik ikl i

i

il k r r r mωω

]'')'[( 2

l ik ikl i

i

ikl r r r m I

3

1,21

l k

l kl k ω I ω

2~Iωω

Inertia tensor and moment of inertia~Iωω 2



• (3x3) matrix I is called the inertia tensor

• Inertia tensor is a symmetric matrix (only 6

independent elements):

• For a rigid body with a continuous distribution of

density, the definition of the inertia tensor is as

follows:

• Introducing a notation

• Scalar I is called the moment of inertia

2

Iωω

RT

lk kl I I

dV r r r I V

l k kl kl ])[( 2

nω ω 2

~Iωω

RT 2

~ωω Inn

2

2 I ω

Inn~ I

]'')'[( 2

l ik ikl i

i

ikl r r r m I

Inertia tensor and moment of inertia2I



• On the other hand:

• Therefore

• The moment of inertia depends upon the position

and direction of the axis of rotation

2

2 I ωT R

i

iii R r ωr ωmT )'()'(2

1

i

iii r nr nmω

)'()'(2

2

i

iii r nr nm I )'()'(

Parallel axis theorem



• For a constrained rigid body, the rotation may occur

not around the center of mass, but around someother point 0, fixed at a given moment of time

• Then, the moment of inertia about the axis of

rotation is:

Rr r ii

'

i

iii r nr nm I )()(0

i

iii Rr n Rr nm ))'(())'((

i

i

i

ii

i

ii

Rnm

Rnr nmr nm

2

2

)(

)()'(2)'(

CM I

M Rn Rnr mni

ii

2)()()'(2




• Parallel axis theorem: the moment of inertia about a

given axis is equal to the moment of inertia about a

parallel axis through the center of mass plus the

moment of inertia of the body, as if concentrated at

the center of mass, with respect to the original axis

2

0 )( Rn M I I CM




• Does the change of axes affect the ω vector?

• Let us consider two systems of coordinates defined

with respect to two different points of the rigid body:

x’1y’1z’1 and x’2y’2z’2

• Then

• Similarly

R R R

12

s s s R R R )()()( 12

Rω R R r s

11 )()(

s s s R R R )()()( 21

Rω R Rr s

22

)()(

Rω R R s s

112 )()(

RωRR s s

221 )()(

0)( 21 Rωω




• Any difference in ω vectors at two arbitrary pointsmust be parallel to the line joining two points

• It is not possible for all the points of the rigid body

• Then, the only possible case:

• The angular velocity vector is the same for all coordinate

systems fixed in the body

0)( 21 Rωω

21 ωω

Example: inertia tensor of a

homogeneous cube



homogeneous cube

• Let us consider a homogeneous cube of mass Mand side a

• Let us choose the origin at one of cube’s corners

• Then dV r r r I V

l k kl kl ])[( 2

a a a

dr dr dr r r r I

0 0

321

0

11

2

11 ])[(a a a

dr dr dr r r

0 0

321

0

2

3

2

2 ])()[(

a a

dr dr r r a0

32

0

2

3

2

2 ])()[(3

2 5a

3

2 2 Ma3322 I I

Example: inertia tensor of a

homogeneous cube



homogeneous cube

dV r r r I V l k kl kl ])[(

2

a a a

dr dr dr r r I 0 0

321

0

2112 ][

3

2

4

1

4

14

1

3

2

4

1

4

1

4

1

3

2

2 MaI

a a

dr dr r r a0

21

0

21 ][4

5a

4

2 Ma

322331132112 I I I I I I

Angular momentum of a rigid body



• Angular momentum of a system of particles is:

• Rate of change of a vector

• For a rigid body, in the rotating frame of reference,

all the distances between the points of the rigid bodyare fixed:

• Angular momentum of rigid body:

i

iii r r m L )(

ir i si r ωr r )()(

0)( r ir

i si r ωr )(

i

iii r ωr m L ))((

i l k nm

inmlmnik jkl i j r ωr m L3

1,

3

1,

i nml k

iminik lmn jkl mωr r 3

1,,,

Angular momentum of a rigid body3



• Rotational kinetic energy:

i nml k

iminik lmn jkl j mωr r L1,,,

i nmk

iminik km jnkn jm mωr r 3

1,,

)(

3

1

2 ])[(k

ik ij jk i

i

ik r r r mω

3

1k

k jk ω I

IωL

2

~Lω

2

~ωL

2

~Iωω

RT

Principal axes of inertia



• Inertia tensor is a symmetric matrix

• In a general case, such matrices can be

diagonalized – we are looking for a system of

coordinates fixed to a rigid body, in which the inertia

tensor has a form:

• To diagonalize the inertia tensor, we have to find the

solutions of a secular equation

3

2

1

00

00

00

I

I

I

I

0

333231

232221

131211

IIII

I I I I

I I I I




• Coordinate axes, in which the inertia tensor is

diagonal, are called the principal axes of a rigid body;the eigenvalues of the secular equations are the

components of the principal moment of inertia

• After diagonalization of the inertia tensor, the

equations of motion for rotation of a free rigid bodylook like

• After diagonalization of the inertia tensor, therotational kinetic energy a rigid body looks like

03

1,

k k j

k j

ijk ii I ωωω I

3

1

2

2

1

i

ii R ω I T




• To find the directions of the principal axes we have

to find the directions for the eigenvectorsω

• When the rotation occurs around one of the

principal axes In, there is only one non-zero

component ωn

• In this case, the angular momentum has only one

component

• In this case, the rotational kinetic energy has only

one term

knk k k ω I L

22

123

1

2 nn

i

iiin R

ω I ω I T

Stability of a free rotational motion



• Let us choose the body axes along the principal

axes of a free rotating rigid body

• Let us assume that the rotation axis is slightly off

the direction of one of the principal axes (α - small

parameter):

• Then, the equations of motion

332211ˆˆˆ iiiωω

03

1,

k k j

k j

ijk ii I ωωω I

0)(

0)(

0)(

122133

313122

233211

I I ωωω I

I I ωωω I

I I ωωω I

0)(

0)(0)(

122133

313122

233211

I I ω I

I I ω I I I ω I


0)(

IIωI 01

ω



0)(

0)(

0)(

122133

313122

233211

I I ω I

I I ω I

I I ω I

0

0

0

3

12213

2

31312

1

I

I I ω

I

I I

ω

ω

0))(( 3121

32

2

1)3(2)3(2

1

I I I I I I

ω

const ω

0)3(2)3(2 K ))(( 3121

32

2

1 I I I I I I

ω K


2ω



• The behavior of solutions of this equation depends

on the relative values of the principal moments of

inertia

• Always stable

• Exponentially unstable

0)3(2)3(2 K ))(( 3121

32

1 I I I I I I

ω K

0 K 2

K 3121 ; I I I I

3121 ; I I I I

0)3(2

2

)3(2

)cos( )3(2)3(2)3(2 t A

0 K 2 K

213 I I I

312 I I I

0)3(2

2

)3(2

t e A )3(2)3(2

t e A )3(2)3(2

Example: principal axes of a uniform

cube



cube

• Previously, we have found the inertia tensor for auniform cube with the origin at one of the corners,

and the coordinate axes along the edges:

• The secular equation:3

2

4

1

4

14

1

3

2

4

1 4

1

4

1

3

2

2 MaI

03

2

483

2

12

11 22422

22

I Ma Maa M

I Ma

I Ma

0

3

2

44

43

2

4

443

2

222

222

222

I Ma Ma Ma

Ma I

Ma Ma

Ma Ma I

Ma


cube



cube

• To find the directions of the principal axes we have

to find the directions for the eigenvectors

• Let us consider

03

2

483

2

12

11 22422

22

I

Ma Maa M

I

Ma

I

Ma

12

11 2

1

Ma I

6;

12

11 2

3

2

2

Ma I

Ma I

6

2

3

Ma I

33

23

13

3

ω

ω

ω

ω333 1ωIω I


cube



cube

13

2

33

2

23

2

13

2

6443

2

ω

Ma

ω

Ma

ω

Ma

ω

Ma

23

2

33

2

23

2

13

2

643

2

4ω

Maω

Maω

Maω

Ma

33

2

33

2

23

2

13

2

63

2

44ω

Maω

Maω

Maω

Ma

1

2

33

23

33

13

ω

ω

ω

ω

12

33

23

33

13

ω

ω

ω

ω

233

23

33

13

ω

ω

ω

ω

2313 ωω

3313 ωω332313 ωωω

Review of fundamental equations" t "

Given a system of particles



"system"

•

•

•

•

•

X

Y

Z

XYZ inertial

•

mi + 1

mncm

m2m1

m iFi

ri

rc

Given a system of particles

translating through space,

each particle i being actedupon by external force Fi, and

each particle located relative

to an inertial reference frame,

the governing equations are

Fc = m ac

M =

Mc = c

H

HWhat is an inertial

frame?

Review of fundamental equationswhere



m = total mass (sum over all mass particles)

ac = acceleration of center of mass (cm) of all mass particles

Fc = sum of external forces applied to system of particles as if

applied at cm

Hi = r i x mivi = angular momentum of particle i (also called moment

of momentum)

H, Hc = angular momentum summed over all particles, measured

about

inertial point, cm point, respectively

M, Mc = moment of all external forces applied to system of particles,measured about inertial point, cm point, respectively

Rigid bodies in general motion(translating and rotating)



( g g)

Z

X

Y

x

yz

The time rate of change of any vector Vcapable of being viewed

in either XYZ or xyz is

[ ]XYZ = [ ]xyz + x V

where is the angular velocity of a

secondary translating, rotating reference

frame (xyz).

dtdV

dtdV

Rigid bodies in general motion(translating and rotating)



(translating and rotating)

Z

X

Y

x

yzAnother common form of the

equation is:

= r + x V

and when applied to rate of change

of angular momentum becomes

M = = r + x H

which is referred to as Euler’s equation.

V V

HH

Rotating rigid bodyBy integrating the motion over the rigid body

z



By integrating the motion over the rigid body,

we can express the

angular momentum relative

to the xyz axes as

H = Hx i + Hy j + Hz k

= (Jxx x + Jxy y + Jxz z) i

+ (Jyx x + Jyy y + Jyz z) j

+ (Jzx x + Jzy y + Jzz z) k

P•

x

y

dm = d dV or r

or in matrix form

H = J

where J = inertia matrix

moments products

Rotating rigid bodyTaking the derivative of (6 20) and substituting into (6 16) also assuming



Taking the derivative of (6.20) and substituting into (6.16), also assuming

the body axes to be aligned with the principal axes, we get Euler’s moment

equations:

Mx = Jxx x + (Jzz - Jyy) y z

My = Jyy y + (Jxx - Jzz) x z

Mz = Jzz z + (Jyy - Jxx) x y

What are principal axes?

Acceleration relative to a non-

inertial reference frame




y

x

z

X

Y

Z

P

Rr






By taking two derivatives and applying (6.13) appropriately, the absoluteacceleration of point P can be shown to be

a = + x + x ( x + + 2 x

where

= acceleration of xyz origin

x = tangential acceleration

x ( x = centripetal acceleration

= acceleration of P relative to xyz

2 x = Coriolis acceleration

R r r r

R

r

r






For the special case of xyz fixed to rigid body and P a point in the body,

and reduces to

a = + x + x ( x

If P at cm, then

ac = + x c + x ( x c

0r r

R r

R

SPACECRAFT



DYNAMICS ANDCONTROL

LECTURE #

ATTITUDE DYNAMICS ANDKINEMATICS

Angular momentum and Inertia

Matrix



Matrix

Let us suppose that a rigid body is moving in an inertial frame. This motion

can be described by the translation motion of its center of mass (cm), together with

a rotational motion of the body about some axis through its center of mass.

which simply states that the rate of change of the vector A as observed in the fixed

coordinate system (I - "inertial" in our case) equals the rate of change of the vector

A as observed in the rotating coordinate system (B - "body" in our case) with angular

velocity w, plus the vector product w X A



Total momentum



Rotational Kinetic Energy of a Rigid

Body



Body



Moment 0f Inertia about a Selected

Axis in the Body Fram



Axis in the Body Fram



How to cater the product of inertia terms

Principal Axes of Inertia



p s

The problem at hand is to transform the general inertia matrix

into a diagonal one. Transformation of a nondiagonal real square symmetric matrix

into a diagonal one is a common procedure that is treated in linear algebra



Ellipsoid of Inertia and the Rotational

State of a Rotating Body



g y

With the body axes chosen to be principal axes,



Euler's Moment Equations



q

for the angular momentum vector h, we can write

This is the well-known Euler's moment equation. In this equation, the subscript "I"

indicates a derivative in the inertial frame, while the subscript "B" indicates a derivative

in the rotating body frame.

Assuming that XB, YB, ZB are the principal

axes of inertia and performing the

vector product, we obtain the three scalar

equations

These equations are nonlinear, so they do not have an analytical closed-form solution.

However, they can be solved under some relieving conditions

Stability of Rotation for Asymmetric

Bodies about Principal Axes



p

Previously we assumed an axisymmetric body now we want to find the condition

for stability about any principle axis with no external moments acting on the body

As no external moments are present so put Mx = My= Mz = O. To find stability for Z-axis take

where E is a small disturbance;



Characteristics of Rotational Motionof a Spinning Body



if the body is in rotational motion caused by

initial conditions but not under the influence of

external moments, then the norm of the

angular velocity remains constant, IwI = const

Moreover, since there are no applied moments

on the body, M = Ihi’I =0, meanshi = const; the momentum vector is constant in

inertial space.

Nutation of a Spinning Body



the angular momentum and the angular

velocity vectors will be resolved

into two components, one in the XB-YB

plane and another along the ZB body axis

(assume once more that the body is

symmetric, Ix = Iy): w = Wxy + Wz and h=

IxWxy + IzWz

Since W and b have components in the

same directions (Wxy and Wz), it is

evident that h, W, and Wz are coplanar. But

since the momentum vector h is constant, itsdirection is fixed in space. The vector Wxy

rotates in the XB-YB body plane, so the

angular velocity vector w must also rotate

about h;



also fixed in space. Whenever the body ZB

axis deviates from the momentum vector

h, the body is said to nutate. This nutationforces the spin axis to deviate from the

nominal desired direction. Keeping the

nutation angle small is one of the important

tasks of attitude control systems,

Attitude Kinematics Equations ofMotion for a Nonspinning Spacecraft



Frames of reference

Angular Velocity Vector of a Rotating Frame



Time Derivation of the Direction

Cosine Matrix



Time Derivation of the Quaternion

Vector



Derivation of the Velocity Vector W RI



Attitude Dynamics Equations ofMotion for a Non spinning Satellite



The attitude dynamics equations will be obtained from Euler's moment

equation present derivations we will allow for the existence of rotating

elements inside the satellite - known as momentum exchange devices - and

for other kinds of gyroscopic devices. The most common momentum

exchange devices are the reaction wheel. the momentum wheel. and the

control moment gyro.

M is the total external moment acting on the body, which is equal to the

inertial momentum change of the system. External inertial moments are

products of aerodynamic, solar, or gravity gradient forces, or of magnetic

torques or reaction torques produced by particles expelled from the body,such as hydrazine gas or ion particles

External Disturbance Torques

NOTE Th it d f th t i



Orbital Altitude (au)

T o r q u e ( a

u )

Solar

Press.

Drag

Gravity

Magnetic

LEO GEO

NOTE: The magnitudes of the torques is

dependent on the spacecraft design.

Internal Disturbing Torques



Examples

Uncertainty in S/C Center of Gravity (typically 1-3 cm)

Thruster Misalignment (typically 0.1 – 0.5 )

Thruster Mismatch (typically ~5%) Rotating Machinery

Liquid Sloshing (e.g. propellant)

Flexible structures

Crew Movement

Disturbing Torques



F r T

I H T

Gravity Gradient Torque



2sin2

33 y z g I I

RT

where:

verticalfromawaydeviationmaximum

inertiaof momentsmassS/C,

radiusorbit

parameter nalgravitatiosEarth'

gradientgravitymaximum

z y

g

I I

R

T

z

y

Magnetic Torque



where: B xmT m

metersradiusorbitmtesla107.96momentmagneticsEarth'

polestheabove pointsfor 2

equator theabove pointsfor

fieldmagneticsEarth'of strength

mAmpdipolemagneticresidualS/C

torqueedisturbancmagnetic

315

3

3

2

R M

R

M

R

M

B

m

T m

*Note value of m depends on S/C size and whether on-board compensation is used

- values can range from 0.1 to 20 Amp-m2

- m = 1 for typical small, uncompensated S/C

Aerodynamic Torque



where:

g paa cc F T

2

2

1 AvC F D

gravityof center C

pressurecatmospheriof center C

velocity

areasectional-cross

2.5-2arevaluesS/Ctypicaldragof tcoefficien

densitycatmospheritorqueedisturbanccaerodynami

g

pa

v

A

C

T

D

a

Solar Pressure Torque



where:

g ps srp cc F T

i Ac

F F s

s cos1

angleincidencesun

S/Cfor 0.6valuetypical1,0factor ereflectanc

surfacedilluminateof area

lightof speedm

Wdensityfluxsolar

gravityof center c

pressureradiationsolar of center

torqueedisturbanc presureradiationsolar

2

g

i

A

c

F

c

T

s

s

ps

srp

FireSat

Example



Equations 0/ Motion for Spacecraft Attitude



We shall break down the external torque T into two principal parts: Tc' the

control moments to be used for controlling the attitude motion of the

satellite; and Td , those moments due to different disturbing environmentalphenomena. The total torque vector is thus T = Tc+Td'



Equation 4.8.2 summarizes the full attitude dynamics that must be

implemented in the complete six-degrees-of-freedom (6-DOF) simulation

necessary for analyzing the attitude control system. Care must be taken inderiving the vector w, since it must be expressed in the correct coordinate

frame.

Gravity Gradient Moments



After doing linearinzation by small angle approximation and expansions

we have

Linearized Attitude DynamicsEquations 0/ Motion



For momentum-biased satellites a constant momentum bias hwyo isapplied along the YB axis to give inertial angular stability about the

YB axis of the sIc. For that case



h ' h h h f h h l h f



hwx' h wy, hwz are the momentum components of the wheels with axes of

rotation along the XB, VB, and ZB body axes of the satellite; hwx =

IwxWwx, hwy =IwyWwy + hwyo, and hwz = IwzWwz' where Iwx, Iwy,

Iwz are the moments of inertia of the individual wheels and Wwx,

Wwy' Wwz are the angular velocities of the wheels. The terms hwx, h

wy, hwz are the angular moments that the wheels exert on the s/c alongthe body axes. If Wwx is the angular acceleration of the XB axis

wheel, then hwx = IwxWwx is the negative of the angular moment that

the XB wheel exerts on the satellite about its XB axis. The same applies

for the YB and ZB axes wheel components. Attitude control of a s/ccan be achieved by controlling these angular accelerations, which are

internal torques exerted on the satellite. If, in addition, external

(inertial) torques such as magnetic or reaction torques are applied to

the satellite, they are incorporated in Tc the vector of control torques.

QUESTIONS



THANK YOU FOR YOUR INTEREST

sdac lec final revision

Documents