Tenth Editioneng.sut.ac.th/me/2014/document/EngDynamics/12_Lecture_ppt.pdf · enth Vector Mechanics for Engineers: Dynamics dition 2 ) •, > • •, •, S .

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Tenth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Phillip J. Cornwell

Lecture Notes:

Brian P. SelfCalifornia Polytechnic State University

CHAPTER

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

12Kinetics of Particles:

Newton’s Second

Law


Vector Mechanics for Engineers: Dynamics

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Contents

12 - 2

Introduction

Newton’s Second Law of

Motion

Linear Momentum of a Particle

Systems of Units

Equations of Motion

Dynamic Equilibrium

Sample Problem 12.1

Sample Problem 12.2

Sample Problem 12.3

Sample Problem 12.4

Sample Problem 12.5

Angular Momentum of a Particle

Equations of Motion in Radial &

Transverse Components

Conservation of Angular Momentum

Newton’s Law of Gravitation

Sample Problem 12.6

Sample Problem 12.7

Trajectory of a Particle Under a

Central Force

Application to Space Mechanics

Sample Problem 12.8

Kepler’s Laws of Planetary Motion



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Kinetics of Particles

12 - 3

We must analyze all of the forces

acting on the wheelchair in order

to design a good ramp

High swing velocities can

result in large forces on a

swing chain or rope, causing

it to break.



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Introduction

12 - 4

• Newton’s Second Law of Motion

m F a

• If the resultant force acting on a particle is not

zero, the particle will have an acceleration

proportional to the magnitude of resultant

and in the direction of the resultant.

• Must be expressed with respect to a Newtonian (or inertial)

frame of reference, i.e., one that is not accelerating or rotating.

• This form of the equation is for a constant mass system



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Linear Momentum of a Particle

12 - 5

• Replacing the acceleration by the derivative of the velocity

yields

particle theof momentumlinear

L

dt

Ldvm

dt

d

dt

vdmF

• Linear Momentum Conservation Principle:

If the resultant force on a particle is zero, the linear momentum

of the particle remains constant in both magnitude and direction.



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Systems of Units

12 - 6

• Of the units for the four primary dimensions (force,

mass, length, and time), three may be chosen arbitrarily.

The fourth must be compatible with Newton’s 2nd Law.

• International System of Units (SI Units): base units are

the units of length (m), mass (kg), and time (second).

The unit of force is derived,

22 s

mkg1

s

m1kg1N1

• U.S. Customary Units: base units are the units of force

(lb), length (m), and time (second). The unit of mass is

derived,

ft

slb1

sft1

lb1slug1

sft32.2

lb1lbm1

2

22



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Equations of Motion

12 - 7

• Newton’s second law amF

• Can use scalar component equations, e.g., for

rectangular components,

zmFymFxmF

maFmaFmaF

kajaiamkFjFiF

zyx

zzyyxx

zyxzyx



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Dynamic Equilibrium

12 - 8

• Alternate expression of Newton’s second law,

ectorinertial vam

amF

0

• With the inclusion of the inertial vector, the system

of forces acting on the particle is equivalent to

zero. The particle is in dynamic equilibrium.

• Methods developed for particles in static

equilibrium may be applied, e.g., coplanar forces

may be represented with a closed vector polygon.

• Inertia vectors are often called inertial forces as

they measure the resistance that particles offer to

changes in motion, i.e., changes in speed or

direction.

• Inertial forces may be conceptually useful but are

not like the contact and gravitational forces found

in statics.



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Free Body Diagrams and Kinetic Diagrams

12 - 9

The free body diagram is the same as you have done in statics; we

will add the kinetic diagram in our dynamic analysis.

2. Draw your axis system (e.g., Cartesian, polar, path)

3. Add in applied forces (e.g., weight, 225 N pulling force)

4. Replace supports with forces (e.g., normal force)

1. Isolate the body of interest (free body)

5. Draw appropriate dimensions (usually angles for particles)

x y225 N

FfNmg

25o



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12 - 10

Put the inertial terms for the body of interest on the kinetic diagram.

2. Draw in the mass times acceleration of the particle; if unknown,

do this in the positive direction according to your chosen axes

1. Isolate the body of interest (free body)

x y225 N

FfNmg

25o

may

max

m F a



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12 - 11

Draw the FBD and KD for block A (note that the

massless, frictionless pulleys are attached to block A

and should be included in the system).



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12 - 12

1. Isolate body

2. Axes

3. Applied forces

4. Replace supports with forces

5. Dimensions (already drawn)

x

y

mg

Ff-1N1

TT

T

T

T

Ff-B

NB

may = 0

max

6. Kinetic diagram

=



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12 - 13

Draw the FBD and KD for the collar B. Assume

there is friction acting between the rod and collar,

motion is in the vertical plane, and q is increasing



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12 - 14

1. Isolate body

2. Axes

3. Applied forces

4. Replace supports with forces

5. Dimensions

6. Kinetic diagram

mg

Ff

N

mar

maqeq er

=q

q



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Sample Problem 12.1

12 - 15

An 80-kg block rests on a horizontal

plane. Find the magnitude of the force

P required to give the block an

acceleration of 2.5 m/s2 to the right. The

coefficient of kinetic friction between

the block and plane is mk 0.25.

SOLUTION:

• Resolve the equation of motion for the

block into two rectangular component

equations.

• Unknowns consist of the applied force

P and the normal reaction N from the

plane. The two equations may be

solved for these unknowns.



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Sample Problem 12.1

12 - 16

80 kg

0.25

k

m

F N

N

m

=

=

=

x

y

O

SOLUTION:

• Resolve the equation of motion for the block

into two rectangular component equations.

:maFx

Pcos30°- 0.25N = 80 kg( ) 0.25 m/s2( )= 200N

:0 yF

N -Psin30°- 785N = 0

• Unknowns consist of the applied force P and

the normal reaction N from the plane. The two

equations may be solved for these unknowns.

N = Psin30°+ 785N

Pcos30°- 0.25 Psin30°+ 785N( ) = 200N

535P N=



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Sample Problem 12.2

12 - 17

The two blocks shown start from rest.

The horizontal plane and the pulley are

frictionless, and the pulley is assumed

to be of negligible mass. Determine

the acceleration of each block and the

tension in the cord.

SOLUTION:

• Write the kinematic relationships for the

dependent motions and accelerations of

the blocks.

• Write the equations of motion for the

blocks and pulley.

• Combine the kinematic relationships

with the equations of motion to solve for

the accelerations and cord tension.



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Sample Problem 12.2

12 - 18

• Write equations of motion for blocks and pulley.

:AAx amF

AaT kg1001

:BBy amF

B

B

BBB

aT

aT

amTgm

kg300-N2940

kg300sm81.9kg300

2

22

2

:0 CCy amF

02 12 TT

SOLUTION:

• Write the kinematic relationships for the dependent

motions and accelerations of the blocks.

ABAB aaxy21

21

x

y

O



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Sample Problem 12.2

12 - 19

N16802

N840kg100

sm20.4

sm40.8

12

1

221

2

TT

aT

aa

a

A

AB

A

• Combine kinematic relationships with equations of

motion to solve for accelerations and cord tension.

ABAB aaxy21

21

AaT kg1001

A

B

a

aT

21

2

kg300-N2940

kg300-N2940

0kg1002kg150N2940

02 12

AA aa

TT

x

y

O



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Sample Problem 12.3

12 - 20

The 6-kg block B starts from rest and

slides on the 15-kg wedge A, which is

supported by a horizontal surface.

Neglecting friction, determine (a) the

acceleration of the wedge, and (b) the

acceleration of the block relative to the

wedge.

SOLUTION:

• The block is constrained to slide down

the wedge. Therefore, their motions are

dependent. Express the acceleration of

block as the acceleration of wedge plus

the acceleration of the block relative to

the wedge.


wedge and block.

• Solve for the accelerations.



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Sample Problem 12.3

12 - 21

SOLUTION:

• The block is constrained to slide down the

wedge. Therefore, their motions are dependent.

ABAB aaa

• Write equations of motion for wedge and block.

x

y

:AAx amF

N1 sin30° =mAaA

0.5N1 =mAaA

:30cos ABABxBx aamamF

-WB sin30° =mB aA cos30°- aB A( )aB A = aA cos30°+ gsin30°

:30sin AByBy amamF

N1 -mBgcos30° = - mB( )aA sin30°



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Sample Problem 12.4

12 - 22

10.5 a AN m a=

• Solve for the accelerations.

N1 -mBgcos30° = -mBaA sin30°

2mAaA -mBgcos30° = -mBaA sin30°

aA =mBgcos30°

2mA +mB sin30°

aA =6 kg( ) 9.81m/s2( )cos30°

2 15kg( ) + 6 kg( )sin30°

21.545m sAa =

aB A = aA cos30°+ gsin30°

aB A = 1.545m s2( )cos30°+ 9.81m s2( )sin30°

26.24m sB Aa =



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Group Problem Solving

12 - 23

The two blocks shown are originally at

rest. Neglecting the masses of the pulleys

and the effect of friction in the pulleys and

between block A and the horizontal

surface, determine (a) the acceleration of

each block, (b) the tension in the cable.

SOLUTION:



the blocks.


blocks and pulley.

• Combine the kinematic relationships

with the equations of motion to solve for

the accelerations and cord tension.

• Draw the FBD and KD for each block



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12 - 24

xA

yB

const nts3 aA Bx y L

3 0A Bv v

3 0A Ba a

3A Ba a

SOLUTION:



the blocks.

This is the same problem worked last

chapter- write the constraint equation

Differentiate this twice to get the

acceleration relationship.

(1)



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12 - 25

: x A AF m a

A BT m a

3 A BT m a

y B BF m a

3(3 )B A B B Bm g m a m a

229.81 m/s

0.83136 m/s30 kg

1 91 925 kg

BA

B

ga

m

m

22.49 2.49 m/sA a

23 30 kg 0.83136 m/sT

74.8 NT

• Draw the FBD and KD for each block

mAg

T

NA

maAx

mBg

2T T

maBy

• Write the equation of motion for each block

==

From Eq (1)(2) (3)

3B B BW T m a

(2) (3)

BA

• Solve the three equations, 3 unknowns

+y

+x



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Concept Quiz

12 - 26

The three systems are released from rest. Rank the

accelerations, from highest to lowest.

a) (1) > (2) > (3)

b) (1) = (2) > (3)

c) (2) > (1) > (3)

d) (1) = (2) = (3)

e) (1) = (2) < (3)

(1) (2) (3)



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Kinetics: Normal and Tangential Coordinates

12 - 27

Aircraft and roller coasters can both experience large

normal forces during turns.



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Equations of Motion

12 - 28

• For tangential and normal components,

t t

t

F ma

dvF m

dt

• Newton’s second law amF

2

n n

n

F ma

vF m



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Sample Problem 12.4

12 - 29

The bob of a 2-m pendulum describes

an arc of a circle in a vertical plane. If

the tension in the cord is 2.5 times the

weight of the bob for the position

shown, find the velocity and accel-

eration of the bob in that position.

SOLUTION:

• Resolve the equation of motion for the

bob into tangential and normal

components.

• Solve the component equations for the

normal and tangential accelerations.

• Solve for the velocity in terms of the

normal acceleration.



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Sample Problem 12.4

12 - 30

SOLUTION:

• Resolve the equation of motion for the bob into

tangential and normal components.

• Solve the component equations for the normal and

tangential accelerations.

:tt maF

30sin

30sin

ga

mamg

t

t

2sm9.4ta

:nn maF

30cos5.2

30cos5.2

ga

mamgmg

n

n

2sm03.16na

• Solve for velocity in terms of normal acceleration.

22

sm03.16m2 nn avv

a

sm66.5v



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Sample Problem 12.5

12 - 31

Determine the rated speed of a

highway curve of radius = 20 m

banked through an angle q = 18o. The

rated speed of a banked highway curve

is the speed at which a car should

travel if no lateral friction force is to

be exerted at its wheels.

SOLUTION:

• The car travels in a horizontal circular

path with a normal component of

acceleration directed toward the center

of the path.The forces acting on the car

are its weight and a normal reaction

from the road surface.

• Resolve the equation of motion for

the car into vertical and normal

components.

• Solve for the vehicle speed.



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Sample Problem 12.5

12 - 32

SOLUTION:

• The car travels in a horizontal circular

path with a normal component of

acceleration directed toward the center

of the path.The forces acting on the

car are its weight and a normal

reaction from the road surface.

• Resolve the equation of motion for

the car into vertical and normal

components.

:0 yF

q

q

cos

0cos

WR

WR

:nn maF

q

q

q

2

sincos

sin

v

g

WW

ag

WR n

• Solve for the vehicle speed.

v2 = gr tanq

= 9.81m s2( ) 120m( ) tan18°

19.56m s 70.4km hv = =



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12 - 33

The 3-kg collar B rests on the

frictionless arm AA. The collar is

held in place by the rope attached to

drum D and rotates about O in a

horizontal plane. The linear velocity

of the collar B is increasing according

to v= 0.2 t2 where v is in m/s and t is

in sec. Find the tension in the rope

and the force of the bar on the collar

after 5 seconds if r = 0.4 m.

v SOLUTION:


collar.

• Combine the equations of motion with

kinematic relationships and solve.

• Draw the FBD and KD for the collar.

• Determine kinematics of the collar.



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12 - 34

SOLUTION: • Given: v= 0.2 t2, r = 0.4 m

• Find: T and N at t = 5 sec

Draw the FBD and KD of the collar

manT N

et

en

mat

Write the equations of motion

=

n nF ma t tF ma

2vN m

dvT m

dt



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12 - 35

manT N

et

en

mat

=2 2

2562.5 (m/s )

0.4n

va

20.4 0.4(5) 2 m/st

dva t

dt

Kinematics : find vt, an, at

2 20.2 0.2(5 )=5 m/stv t

Substitute into equations of motion

(3.0)(62.5)N (3.0)(2)T

187.5 NN 6.0 NT

q

n nF ma t tF ma



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12 - 36

manT N

et

en

mat

=How would the problem

change if motion was in the

vertical plane?

mg

qYou would add an mg term

and would also need to

calculate q

When is the tangential force greater than the normal force?

Only at the very beginning, when starting to accelerate.

In most applications, an >> at



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Concept Question

12 - 37

D

B

CA

A car is driving from A to D on the curved path shown.

The driver is doing the following at each point:

A – going at a constant speed B – stepping on the accelerator

C – stepping on the brake D – stepping on the accelerator

Draw the approximate direction of the car’s acceleration

at each point.



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Kinetics: Radial and Transverse Coordinates

12 - 38

Hydraulic actuators and

extending robotic arms are

often analyzed using radial

and transverse coordinates.



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Eqs of Motion in Radial & Transverse Components

12 - 39

qq

q

qq

rrmmaF

rrmmaF rr

2

2

• Consider particle at r and q, in polar coordinates,



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Sample Problem 12.6

12 - 40

A block B of mass m can slide freely on

a frictionless arm OA which rotates in a

horizontal plane at a constant rate .0q

a) the component vr of the velocity of B

along OA, and

b) the magnitude of the horizontal force

exerted on B by the arm OA.

Knowing that B is released at a distance

r0 from O, express as a function of r

SOLUTION:

• Write the radial and transverse

equations of motion for the block.

• Integrate the radial equation to find an

expression for the radial velocity.

• Substitute known information into the

transverse equation to find an

expression for the force on the block.



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Sample Problem 12.6

12 - 41

SOLUTION:

• Write the radial and transverse

equations of motion for the block.

:

:

qq amF

amF rr

qq

q

rrmF

rrm

2

0 2

• Integrate the radial equation to find an

expression for the radial velocity.

r

r

v

rr

rr

rr

rrr

drrdvv

drrdrrdvv

dr

dvv

dt

dr

dr

dv

dt

dvvr

r

0

20

0

20

2

q

qq

dr

dvv

dt

dr

dr

dv

dt

dvvr r

rrr

r

20

220

2 rrvr q

• Substitute known information into the

transverse equation to find an expression

for the force on the block.

2120

2202 rrmF q



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12 - 42


collar.

• Combine the equations of motion with

kinematic relationships and solve.

• Draw the FBD and KD for the collar.

• Determine kinematics of the collar.

SOLUTION:

The 3-kg collar B slides on the frictionless arm AA. The arm is attached to

drum D and rotates about O in a horizontal plane at the rate where

and t are expressed in rad/s and seconds, respectively. As the arm-drum

assembly rotates, a mechanism within the drum releases the cord so that the

collar moves outward from O with a constant speed of 0.5 m/s. Knowing that

at t = 0, r = 0, determine the time at which the tension in the cord is equal to

the magnitude of the horizontal force exerted on B by arm AA.

q = 0.75t q



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12 - 43

mar

T N

eq

er

maq

=

r rF ma BF m aq q

Draw the FBD and KD of the collar

Write the equations of motion

SOLUTION: • Given:

• Find: time when T = N

-T =m(r - rq 2) N =m(rq +2rq)

q = 0.75t

r = 5 m/s

(0) 0r



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12 - 44

3 2: (3 kg)( 0.28125 ) m/sr rF ma T t

2: (3 kg)(1.125 ) m/sBF m a N tq q

0 00.5

r t

dr dt (0.5 ) mr t

r = 0

q = (0.75t) rad/s

q = 0.75 rad/s2

ar= r - rq 2 = 0-[(0.5t) m][(0.75t) rad/s]2 = -(0.28125t3) m/s2

aq

= rq + 2rq = [(0.5t) m][0.75 rad/s2]+ 2(0.5 m/s)[(0.75t) rad/s]

= (1.125t) m/s2

Kinematics : find expressions for r and q

Substitute values into ar , aq

Substitute into equation of motion

3(0.84375 ) (3.375 )t t2 4.000t

2.00 st

Set T = N

r = 5 m/s



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Concept Quiz

12 - 45

Top View

e1

e2

w

v

The girl starts walking towards the outside of the spinning

platform, as shown in the figure. She is walking at a constant

rate with respect to the platform, and the platform rotates at a

constant rate. In which direction(s) will the forces act on her?

a) +e1 b) - e1 c) +e2 d) - e2

e) The forces are zero in the e1 and e2 directions



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12 - 46

Satellite orbits are analyzed using conservation

of angular momentum.



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Eqs of Motion in Radial & Transverse Components

12 - 47

qq

q

qq

rrmmaF

rrmmaF rr

2

2

• Consider particle at r and q, in polar coordinates,

qq

qq

q

q

q

q

rrmF

rrrm

mrdt

dFr

mrHO

2

22

2

2

• This result may also be derived from conservation

of angular momentum,



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12 - 48

• moment of momentum or the angular

momentum of the particle about O.

VmrHO

• Derivative of angular momentum with respect to time,

O

O

M

Fr

amrVmVVmrVmrH

• It follows from Newton’s second law that the sum of

the moments about O of the forces acting on the

particle is equal to the rate of change of the angular

momentum of the particle about O.

zyx

O

mvmvmv

zyx

kji

H

• is perpendicular to plane containingOH

Vmr

and

q

q

2

sin

mr

vrm

rmVHO



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12 - 49

• When only force acting on particle is directed

toward or away from a fixed point O, the particle

is said to be moving under a central force.

• Since the line of action of the central force passes

through O, and 0 OO HM

constant OHVmr

• Position vector and motion of particle are in a

plane perpendicular to .OH

• Magnitude of angular momentum,

000 sin

constantsin

Vmr

VrmHO

massunit

momentumangular

constant

2

2

hrm

H

mrH

O

O

q

q

or



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12 - 50

• Radius vector OP sweeps infinitesimal area

qdrdA 221

• Define qq 2

212

21 r

dt

dr

dt

dAareal velocity

• Recall, for a body moving under a central force,

constant2 qrh

• When a particle moves under a central force, its

areal velocity is constant.



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Newton’s Law of Gravitation

12 - 51

• Gravitational force exerted by the sun on a planet or by

the earth on a satellite is an important example of

gravitational force.

• Newton’s law of universal gravitation - two particles of

mass M and m attract each other with equal and opposite

force directed along the line connecting the particles,

4

49

2

312

2

slb

ft104.34

skg

m1073.66

ngravitatio ofconstant

G

r

MmGF

• For particle of mass m on the earth’s surface,

222 s

ft2.32

s

m81.9 gmg

R

MGmW



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Sample Problem 12.7

12 - 52

A satellite is launched in a direction

parallel to the surface of the earth

with a velocity of 30,000 km/h from

an altitude of 400 km. Determine the

velocity of the satellite as it reaches it

maximum altitude of 4000 km. The

radius of the earth is 6370 km.

SOLUTION:

• Since the satellite is moving under a

central force, its angular momentum is

constant. Equate the angular momentum

at A and B and solve for the velocity at B.



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Sample Problem 12.7

12 - 53

SOLUTION:

• Since the satellite is moving under a

central force, its angular momentum is

constant. Equate the angular momentum

at A and B and solve for the velocity at B.

( )( )( )

sin constant

6370 400 km30,000km h

6370 4000 km

O

A A B B

AB A

B

rmv H

r mv r mv

rv v

r

j = =

=

=

+=

+

19,590km hBv =



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Trajectory of a Particle Under a Central Force

12 - 54

• For particle moving under central force directed towards force center,

022 qqqq FrrmFFrrm r

• Second expression is equivalent to from which,,constant 2 hr q

rd

d

r

hr

r

h 1and

2

2

2

2

2 qq

• After substituting into the radial equation of motion and simplifying,

ru

umh

Fu

d

ud 1where

222

2

q

• If F is a known function of r or u, then particle trajectory may be

found by integrating for u = f(q), with constants of integration

determined from initial conditions.



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12 - 55

constant

1where

22

2

2

2222

2

h

GMu

d

ud

GMmur

GMmF

ru

umh

Fu

d

ud

q

q

• Consider earth satellites subjected to only gravitational pull

of the earth,

• Solution is equation of conic section,

tyeccentricicos11 2

2

GM

hC

h

GM

ru q

• Origin, located at earth’s center, is a focus of the conic section.

• Trajectory may be ellipse, parabola, or hyperbola depending

on value of eccentricity.



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12 - 56

tyeccentricicos11 2

2

GM

hC

h

GM

rq

• Trajectory of earth satellite is defined by

• hyperbola, > 1 or C > GM/h2. The radius vector

becomes infinite for

2

1111 cos

1cos0cos1

hC

GM

qq

• parabola, = 1 or C = GM/h2. The radius vector

becomes infinite for

1800cos1 22 qq

• ellipse, < 1 or C < GM/h2. The radius vector is finite

for q and is constant, i.e., a circle, for < 0.



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12 - 57

• Integration constant C is determined by conditions

at beginning of free flight, q =0, r = r0 ,

2000

20

2

20

11

0cos11

vr

GM

rh

GM

rC

GM

Ch

h

GM

r

00

200

2

2

or 1

r

GMvv

vrGMhGMC

esc

• Satellite escapes earth orbit for

• Trajectory is elliptic for v0 < vesc and becomes

circular for = 0 or C = 0,

0r

GMvcirc



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12 - 58

• Recall that for a particle moving under a central

force, the areal velocity is constant, i.e.,

constant212

21 hr

dt

dAq

• Periodic time or time required for a satellite to

complete an orbit is equal to area within the orbit

divided by areal velocity,

h

ab

h

ab

2

2

where

10

1021

rrb

rra



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Sample Problem 12.8

12 - 59

Determine:

a) the maximum altitude reached by

the satellite, and

b) the periodic time of the satellite.

A satellite is launched in a direction

parallel to the surface of the earth

with a velocity of 36,900 km/h at an

altitude of 500 km.

SOLUTION:

• Trajectory of the satellite is described by

qcos1

2C

h

GM

r

Evaluate C using the initial conditions

at q = 0.

• Determine the maximum altitude by

finding r at q = 180o.

• With the altitudes at the perigee and

apogee known, the periodic time can

be evaluated.



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Sample Problem 12.8

12 - 60

SOLUTION:

• Trajectory of the satellite is described by

qcos1

2C

h

GM

r

Evaluate C using the initial conditions

at q = 0.

2312

2622

29

3600

3

0

6

0

sm10398

m1037.6sm81.9

sm104.70

sm1025.10m106.87

sm1025.10

s/h3600

m/km1000

h

km36900

m106.87

km5006370

gRGM

vrh

v

r

1-9

22

2312

6

20

m103.65

sm4.70

sm10398

m1087.6

1

1

h

GM

rC



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Sample Problem 12.8

12 - 61

• Determine the maximum altitude by finding r1

at q = 180o.

km 66700m107.66

m

1103.65

sm4.70

sm103981

61

9

22

2312

21

r

Ch

GM

r

km 60300km6370-66700 altitudemax

• With the altitudes at the perigee and apogee known,

the periodic time can be evaluated.

sm1070.4

m1021.4m1036.82

h

2

m1021.4m107.6687.6

m1036.8m107.6687.6

29

66

6610

66

21

1021

ab

rrb

rra

min31h 19s103.70 3



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Kepler’s Laws of Planetary Motion

12 - 62

• Results obtained for trajectories of satellites around earth may also be

applied to trajectories of planets around the sun.

• Properties of planetary orbits around the sun were determined

astronomical observations by Johann Kepler (1571-1630) before

Newton had developed his fundamental theory.

1) Each planet describes an ellipse, with the sun located at one of its

foci.

2) The radius vector drawn from the sun to a planet sweeps equal

areas in equal times.

3) The squares of the periodic times of the planets are proportional to

the cubes of the semimajor axes of their orbits.

Documents

Tenth Editioneng.sut.ac.th/me/2014/document/EngDynamics/12_Lecture_ppt.pdf · enth Vector Mechanics for Engineers: Dynamics dition 2 ) •, > • •, •, S .