dynamics 12 lecture

7/28/2019 Dynamics 12 Lecture

1/62

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICS

Tenth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Phillip J. CornwellLecture Notes:

Brian P. SelfCalifornia Polytechnic State University

CHAPTER

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

12Kinetics of Particles:

Newtons Second Law


2/62


Vector Mechanics for Engineers: DynamicsTenth

Edition

Contents

12 - 2

Introduction

Newtons Second Law of

Motion

Linear Momentum of a Particle

Systems of Units

Equations of MotionDynamic Equilibrium

Sample Problem 12.1

Sample Problem 12.3

Sample Problem 12.4Sample Problem 12.5

Sample Problem 12.6

Angular Momentum of a Particle

Equations of Motion in Radial &

Transverse Components

Conservation of Angular Momentum

Newtons Law of Gravitation

Sample Problem 12.7Sample Problem 12.8

Trajectory of a Particle Under a

Central Force

Application to Space Mechanics

Sample Problem 12.9

Keplers Laws of Planetary Motion


3/62



Edition

Kinetics of Particles

2 - 3

We must analyze all of the forcesacting 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.

T


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Edition

Introduction

12 - 4

Newtons 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 aNewtonian (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

TE


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Edition

Linear Momentum of a Particle

12 - 5

Replacing the acceleration by the derivative of the velocity

yields

particletheofmomentumlinear

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.

TE


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Edition

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 Newtons 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

TE


7/62 2013 The McGraw-Hill Companies, Inc. All rights reserved.


Edition

Equations of Motion

12 - 7

Newtons second lawamF

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

rectangular components,

zmFymFxmF

maFmaFmaF

kajaiamkFjFiF

zyx

zzyyxx

zyxzyx

TE




Edition

Dynamic Equilibrium

12 - 8

Alternate expression of Newtons second law,

ectorinertial vamamF

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 tochanges 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.

fTE




Edition

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 lb 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

FfN mg

25o

V t M h i f E i D iTE




Edition


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

FfN mg

25o

may

max

m F a





Edition


2 - 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).





Edition


2 - 12

1. Isolate body

2. Axes3. Applied forces

4. Replace supports with forces

5. Dimensions (already drawn)

x

y

mg

Ff-1N1

T

T

T

T

T

Ff-B

NB

may = 0

max

6. Kinetic diagram

=

V t M h i f E i D iTe

E




Edition


2 - 13

Draw the FBD and KD for the collar B. Assumethere is friction acting between the rod and collar,

motion is in the vertical plane, and qis increasing


E




Edition


2 - 14

1. Isolate body

2. Axes3. Applied forces

4. Replace supports with forces

5. Dimensions

6. Kinetic diagram

mg

Ff

N

mar

maqeq er

=q

q


E




Edition

Sample Problem 12.1

12 - 15

A 200-lb block rests on a horizontal

plane. Find the magnitude of the forceP required to give the block an

acceleration of 10 ft/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 theplane. The two equations may be

solved for these unknowns.


Ed




Edition

Sample Problem 12.1

12 - 16

N

NF

g

Wm

k

25.0

ftslb21.6

sft2.32

lb200

2

2

m

x

y

O

SOLUTION:

Resolve the equation of motion for the blockinto two rectangular component equations.

:maFx

lb1.62

sft10ftslb21.625.030cos 22

NP

:0 yF

0lb20030sin PN

Unknowns consist of the applied force P and

the normal reaction N from the plane. The two

equations may be solved for these unknowns.

lb1.62lb20030sin25.030cos

lb20030sin

PP

PN

lb151P


Ed



Vector Mechanics for Engineers: Dynamicsenth

Edition

Sample Problem 12.3

12 - 17

The two blocks shown start from rest.

The horizontal plane and the pulley arefrictionless, 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.

Vector Mechanics for Engineers D namicsTe

Ed




dition

Sample Problem 12.3

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

0212

TT

SOLUTION:

Write the kinematic relationships for the dependentmotions and accelerations of the blocks.

ABAB aaxy 21

21

x

y

O

Vector Mechanics for Engineers: DynamicsTeEd




dition

Sample Problem 12.3

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 aaxy 21

21

AaT kg1001

AB

a

aT

21

2

kg300-N2940

kg300-N2940

0kg1002kg150N2940

02 12

AA aa

TT

x

y

O





dition

Sample Problem 12.4

12 - 20

The 12-lb blockB starts from rest and

slides on the 30-lb wedgeA, 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 downthe 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.





dition

Sample Problem 12.4

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

AA

AA

agWN

amN

1

1

5.0

30sin

:30cos ABABxBx aamamF

30sin30cos

30cos30singaa

aagWW

AAB

ABABB

:30sin AByBy amamF

30sin30cos1 ABB

agWWN




Vector Mechanics for Engineers: Dynamicsnthdition

Sample Problem 12.4

12 - 22

AAagWN

15.0

Solve for the accelerations.

30sinlb12lb302

30coslb12sft2.32

30sin2

30cos

30sin30cos2

30sin30cos

2

1

A

BA

BA

ABBAA

ABB

a

WW

gWa

agWWagW

agWWN

2sft07.5Aa

30sinsft2.3230cossft07.5

30sin30cos

22AB

AAB

a

gaa

2sft5.20ABa





Group Problem Solving

2 - 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.

Write the equations of motion for theblocks 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

Vector Mechanics for Engineers: DynamicsTen

Ed





2 - 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)


Ed


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Vector Mechanics for Engineers: Dynamicsnthition


2 - 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/s

30 kg1 91 9

25 kg

BA

B

ga

m

m

22.49 2.49 m/s

A 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)

B A

Solve the three equations, 3 unknowns

+y

+x


Ed


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Concept Quiz

2 - 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)


Edi


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Kinetics: Normal and Tangential Coordinates

2 - 27

Aircraft and roller coasters can both experience largenormal forces during turns.


Edi


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Equations of Motion

12 - 28

For tangential and normal components,

t t

t

F ma

dvF mdt

Newtons second law amF

2

n n

n

F ma

vF m


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Vector Mechanics for Engineers: Dynamicsnthtion

Sample Problem 12.5

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 positionshown, find the velocity and accel-

eration of the bob in that position.

SOLUTION:

Resolve the equation of motion for thebob 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 12.5

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

2

sm03.16na

Solve for velocity in terms of normal acceleration.

22

sm03.16m2 nn avv

a

sm66.5

v


Edit


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Sample Problem 12.6

12 - 31

Determine the rated speed of a

highway curve of radius= 400 ft

banked through an angle q= 18o. The

rated speed of a banked highway curveis 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 circularpath 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.


Edit


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Vector Mechanics for Engineers: Dynamicsthtion

Sample Problem 12.6

12 - 32

SOLUTION:

The car travels in a horizontal circular

path with a normal component of

acceleration directed toward the centerof 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

agWR n

Solve for the vehicle speed.

18tanft400sft2.32

tan

2

2 qgv

hmi1.44sft7.64 v

Vector Mechanics for Engineers: DynamicsTent

Edit


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2 - 33

The 3-kg collarB rests on the

frictionless armAA. The collar is

held in place by the rope attached to

drumD and rotates about O in a

horizontal plane. The linear velocityof 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 ifr= 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.


Edit


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2 - 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 2v

N m

dv

T m

dt


Edit


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2 - 35

manT N

et

en

mat

=2 2

25 62.5 (m/s )0.4

n

va

20.4 0.4(5) 2 m/stdv

a tdt

Kinematics : find vt, an, at2 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


Edit


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Vector Mechanics for Engineers: Dynamicsthion


2 - 36

manT N

et

en

mat

=How would the problemchange if motion was in the

vertical plane?

mg

q

You would add an mgterm

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


Edit


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Concept Question

2 - 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:

Agoing at a constant speed Bstepping on the acceleratorCstepping on the brake Dstepping on the accelerator

Draw the approximate direction of the cars acceleration

at each point.


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Kinetics: Radial and Transverse Coordinates

2 - 38

Hydraulic actuators and

extending robotic arms are

often analyzed using radial

and transverse coordinates.


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Vector Mechanics for Engineers: Dynamicshion

Eqs of Motion in Radial & Transverse Components

12 - 39

qqq

qq

rrmmaF

rrmmaF rr

2

2

Consider particle at rand q, in polar coordinates,


Editi


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Vector Mechanics for Engineers: Dynamicshon

Sample Problem 12.7

12 - 40

A blockB of mass m can slide freely ona frictionless arm OA which rotates in a

horizontal plane at a constant rate .0q

a) the component vrof the velocity ofB

along OA, and

b) the magnitude of the horizontal force

exerted onB by the arm OA.

Knowing thatB is released at a distance

r0

from O, express as a function ofr

SOLUTION:

Write the radial and transverseequations 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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Vector Mechanics for Engineers: DynamicshonSample Problem 12.7

12 - 41

SOLUTION:

Write the radial and transverse

equations of motion for the block.

:

:

qq amF

amFrr

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

202

q

qq

dr

dvv

dt

dr

dr

dv

dt

dvvr rr

rrr

20

220

2rrvr q

Substitute known information into the

transverse equation to find an expression

for the force on the block.

21202202 rrmF q


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Vector Mechanics for Engineers: DynamicshonGroup Problem Solving

2 - 42


collar.

Combine the equations of motion withkinematic relationships and solve.

Draw the FBD and KD for the collar.

Determine kinematics of the collar.

SOLUTION:

The 3-kg collarB slides on the frictionless armAA. The arm is attached to

drumD and rotates about O in a horizontal plane at the rate whereand tare 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 onB by armAA.

0.75tq q


Editio


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2 - 43

mar

T N

eqer

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

2( )T m r r q ( 2 )N m r rq q

0.75tq

5 m/sr

(0) 0r


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44/62



2 - 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 00.5r tdr dt

(0.5 ) mr t

0r

2

(0.75 ) rad/s

0.75 rad/s

tq

q

2 2 3 20 [(0.5 ) m][(0.75 ) rad/s] (0.28125 ) m/sra r r t t t q

2

2

2 [(0.5 ) m][0.75 rad/s ] 2(0.5 m/s)[(0.75 ) rad/s]

(1.125 ) m/s

a r r t t

t

q q q

Kinematics : find expressions for r and q

Substitute values into ar , aq

Substitute into equation of motion3(0.84375 ) (3.375 )t t

2 4.000t

2.00 st

SetT = N

5 m/sr


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Vector Mechanics for Engineers: DynamicshonConcept Quiz

2 - 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 aconstant 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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Vector Mechanics for Engineers: DynamicshonAngular Momentum of a Particle

2 - 46

Satellite orbits are analyzed using conservation

of angular momentum.


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Vector Mechanics for Engineers: DynamicshonEqs of Motion in Radial & Transverse Components

12 - 47

qqq

qq

rrmmaF

rrmmaF rr

2

2

Consider particle at rand 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 conservationof angular momentum,


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ecto ec a cs o g ee s y a cshonAngular Momentum of a Particle

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 Newtons 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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g yonConservation of Angular Momentum

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, and0 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


Editio


50/62


g yonConservation of Angular Momentum

12 - 50

Radius vectorOPsweeps infinitesimal area

qdrdA 221

Define qq

2212

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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g yonNewtons 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 ofgravitational force.

Newtons law of universal gravitation - two particles of

massMand 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

ngravitatioofconstant

G

r

MmGF

For particle of mass mon the earths surface,

222 s

ft2.32

s

m81.9 gmg

R

MGmW


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g ynSample Problem 12.8

12 - 52

A satellite is launched in a direction

parallel to the surface of the earth

with a velocity of 18820 mi/h from

an altitude of 240 mi. Determine the

velocity of the satellite as it reaches it

maximum altitude of 2340 mi. The

radius of the earth is 3960 mi.

SOLUTION:

Since the satellite is moving under a

central force, its angular momentum is

constant. Equate the angular momentum

atA andB and solve for the velocity atB.


Edition


53/62



12 - 53

SOLUTION:

Since the satellite is moving under a

central force, its angular momentum is

constant. Equate the angular momentum

atA andB and solve for the velocity atB.

mi23403960mi2403960

hmi18820

constantsin

B

AAB

BBAA

O

r

rvv

vmrvmrHvrm

hmi12550Bv


Edition


54/62


g ynTrajectory 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,,constant2

hr q

rd

d

r

hr

r

h 1and

2

2

2

2

2

q

q

After substituting into the radial equation of motion and simplifying,

ru

umh

Fu

d

ud 1where

222

2

q

IfFis a known function ofroru, then particle trajectory may be

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

determined from initial conditions.


Edition


55/62


g ynApplication to Space Mechanics

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 earths center, is a focus of the conic section.

Trajectory may be ellipse, parabola, or hyperbola depending

on value of eccentricity.


Edition


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12 - 56

tyeccentricicos112

2 GMhC

hGM

rq

Trajectory of earth satellite is defined by

hyperbola, > 1 orC > GM/h2. The radius vector

becomes infinite for

21111 cos1cos0cos1

hCGM

qq

parabola, = 1 orC = GM/h2. The radius vector

becomes infinite for

1800cos122

qq

ellipse, < 1 orC < GM/h2. The radius vector is finite

forqand is constant, i.e., a circle, for< 0.


Edition


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12 - 57

Integration constant Cis determined by conditions

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

20002

0

2

20

11

0cos11

vr

GM

rh

GM

rC

GM

Ch

h

GM

r

0

0

200

2

2

or1

r

GMvv

vrGMhGMC

esc

Satellite escapes earth orbit for

Trajectory is elliptic forv0 < vesc and becomes

circular for= 0 orC= 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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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 rat q= 180o.

With the altitudes at the perigee and

apogee known, the periodic time canbe evaluated.


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12 - 60

SOLUTION:

Trajectory of the satellite is described by

qcos12

Ch

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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n

Sample Problem 12.9

12 - 61

Determine the maximum altitude by finding r1

at q= 180o.

km66700m107.66

m

1103.65

sm4.70

sm103981

61

9

22

2312

21

r

Ch

GM

r

km60300km6370-66700altitudemax

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

66

10

66

21

1021

ab

rrb

rra

min31h19s103.70 3

Vector Mechanics for Engineers: DynamicsT

enth

E

dition


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Keplers Laws of Planetary Motion

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 itsfoci.

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 tothe cubes of the semimajor axes of their orbits.

dynamics 12 lecture

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