diapos de vibraciones

7/18/2019 diapos de vibraciones

1/38

VECTOR MECHANICS FOR ENGINEERS:

DYNAMICSDYNAMICS

Ninth EditionNinth Edition

Ferdinand P. BeerFerdinand P. Beer

E. Russell Johnston, Jr.E. Russell Johnston, Jr.

Lecture NotesLecture Notes

J. !alt "lerJ. !alt "ler

#e$as #ech %ni&ersit'#e$as #ech %ni&ersit'

CHAPTER

2010 The McGraw-Hill Companies, Inc. All rights reserved.

12Kinetics of Particles:Newtons Second Law


2/38

2010 The McGraw-Hill Companies, Inc. All rights reserve.

(ector Mechanics )or En*ineers D'na+ics(ector Mechanics )or En*ineers D'na+icsNinth

Contents

- -

Introduction

Newtons Second Law of Motio

n

Linear Momentum of a Particle

Sstems of !nits

E"uations of Motion#namic E"uili$rium

Sam%le Pro$lem &'(&

Sam%le Pro$lem &'()

Sam%le Pro$lem &'(*

Sam%le Pro$lem &'(+

Sam%le Pro$lem &'(,

An-ular Momentum of a Particle

E"uations of Motion in Radial .

Trans/erse Com%onents

Conser/ation of An-ular Momentum

Newtons Law of 0ra/itation

Sam%le Pro$lem &'(1Sam%le Pro$lem &'(2

Tra3ector of a Particle !nder a

Central 4orce

A%%lication to S%ace Mec5anics

Sam%le Pro$lem &'(6

Ke%lers Laws of Planetar Motion


3/38



Introduction

- /

Newtons first and third laws are sufficient for the study of bodies at

rest (statics) or bodies in motion with no acceleration.

When a body accelerates (changes in velocity magnitude or direction),

Newtons second law is required to relate the motion of the body to the

forces acting on it.

Newtons second law:

! "article will have an acceleration "ro"ortional to the magnitude of

the resultant force acting on it and in the direction of the resultant

force.

#he resultant of the forces acting on a "article is equal to the rate of

change of linear momentum of the "article.

#he sum of the moments about Oof the forces acting on a "article is

equal to the rate of change of angular momentum of the "article

about O.


4/38



Newtons Second Law of Motion

- 0

Newtons Second Law: $f the resultant forceacting on a

"article is not %ero, the "article will have an acceleration

proportional to the magnitude of resultantand in the

direction of the resultant.

&onsider a "article sub'ected to constant forces,

m

a

F

a

F

a

Fmass,constant

*

* =====

When a "article of mass mis acted u"on by a force

the acceleration of the "article must satisfy

,F

amF=

!cceleration must be evaluated with res"ect to aNewtonian frame of reference, i.e., one that is not

accelerating or rotating.

$f force acting on "article is %ero, "article will not

accelerate, i.e., it will remain stationary or continue on a

straight line at constant velocity.N


5/38



Linear Momentum of a Particle

- 1

+e"lacing the acceleration by the derivative of the

velocity yields

( )

"articletheofmomentumlinear=

==

=

L

dt

Ldvm

dt

d

dt

vdmF

Linear Momentum Conservation Principle:

$f the resultant force on a "article is %ero, the linear

momentum of the "article remains constant in bothmagnitude and direction.

N


6/38



Sstems of !nits

- 2

f the units for the four "rimary dimensions (force,

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

#he fourth must be com"atible with Newtons nd -aw.

International System of nits($ /nits): base units are

the units of length (m), mass (0g), and time (second).

#he unit of force is derived,

( ) s

m0g*

s

m*0g*N*

=

=

!S! Customary nits: base units are the units of force(lb), length (m), and time (second). #he unit of mass is

derived,

ft

slb*

sft*

lb*slug*

sft.

lb*lbm*

===

N


7/38 2010 The McGraw-Hill Companies, Inc. All rights reserve.


E"uations of Motion

- 3

Newtons second law "rovides

amF =

olution for "article motion is facilitated by resolving

vector equation into scalar com"onent equations, e.g.,

for rectangular com"onents,

"mFymF#mF

maFmaFmaF$a%aiam$F%FiF

"y#

""yy##

"y#"y#

====== ++=++

1or tangential and normal com"onents,

v

mFdt

dvmF

maFmaF

nt

nntt

==

==

N




#namic E"uili$rium

- 4

!lternate e2"ression of Newtons second law,

ectorinertial vamamF

3

=

With the inclusion of the inertial vector, the system

of forces acting on the "article is equivalent to

%ero. #he "article is in dynamic e&uili'rium.

4ethods develo"ed for "articles in static

equilibrium may be a""lied, e.g., co"lanar forces

may be re"resented with a closed vector "olygon.

$nertia vectors are often called inertial forcesas

they measure the resistance that "articles offer tochanges in motion, i.e., changes in s"eed or

direction.

$nertial forces may be conce"tually useful but are

not li0e the contact and gravitational forces found

in statics.

N




Sam%le Pro$lem &'(&

- 5

! 33lb bloc0 rests on a hori%ontal

"lane. 1ind the magnitude of the force

Prequired to give the bloc0 an accelera

tion or *3 ft5s

to the right. #he coefficient of 0inetic friction between the

bloc0 and "lane is$= 3.6.

-/#$N:

+esolve the equation of motion for the

bloc0 into two rectangular com"onent

equations.

/n0nowns consist of the a""lied force

Pand the normal reaction Nfrom the"lane. #he two equations may be

solved for these un0nowns.

)N




Sam%le Pro$lem &'(&

- 6

N

NF

g

(m

$

6.3

ftslb*.7

sft.

lb33

)

)

==

=

==

#

y

O

-/#$N:

+esolve the equation of motion for the bloc0into two rectangular com"onent equations.

:maF#=( )( )

lb*.7

sft*3ftslb*.76.33cos

=

= NP

:3= yF3lb333sin =PN

/n0nowns consist of the a""lied force Pand

the normal reactionN

from the "lane. #he twoequations may be solved for these un0nowns.

( ) lb*.7lb333sin6.33cos

lb333sin

=++=

PP

PN

lb*6*=P

( M h i ) E i D iN




Sam%le Pro$lem &'()

-

#he two bloc0s shown start from rest.

#he hori%ontal "lane and the "ulleyare frictionless, and the "ulley is

assumed to be of negligible mass.

8etermine the acceleration of each

bloc0 and the tension in the cord.

-/#$N:

Write the 0inematic relationshi"s for the

de"endent motions and accelerations of

the bloc0s.

Write the equations of motion for the

bloc0s and "ulley. &ombine the 0inematic relationshi"s

with the equations of motion to solve for

the accelerations and cord tension.

)( t M h i ) E i D iN




Sam%le Pro$lem &'()

- -

Write equations of motion for bloc0s and "ulley.

:))# amF =( ) )a* 0g*33*=

:++y amF =

( )( ) ( )( ) +

+

+++

a*

a*

am*gm

0g33N93

0g33sm;*.90g33

==

=

:3== CCy amF

3* =

**

-/#$N:

Write the 0inematic relationshi"s for the de"endentmotions and accelerations of the bloc0s.

)+)+ aa#y )*

)

* ==#

y

O

( M h i ) E i D i( t M h i ) E i D iN




Sam%le Pro$lem &'()

- /

( )

N*7;3

N;30g*33

sm3.

sm3.;

*

*

*

====

==

=

**

a*

aa

a

)

)+

)

&ombine 0inematic relationshi"s with equations of

motion to solve for accelerations and cord tension.

)+)+ aa#y *

* ==

( ) )a* 0g*33*=

( )

( ) ( ))+

a

a*

*

0g33N93

0g33N93

=

=

( ) ( ) 30g*330g*63N93

3 *

==

)) aa

**

#

y

O

( t M h i ) E i D i( t M h i ) E i D iN




Sam%le Pro$lem &'(*

- 0

#he *lb bloc0+starts from rest and

slides on the 3lb wedge), which is

su""orted by a hori%ontal surface.

Neglecting friction, determine ,a-the

acceleration of the wedge, and ,'-the

acceleration of the bloc0 relative to the

wedge.

-/#$N:

#he bloc0 is constrained to slide downthe wedge. #herefore, their motions are

de"endent.




Sam%le Pro$lem &'(*

- 1

-/#$N:

#he bloc0 is constrained to slide down thewedge. #herefore, their motions are de"endent.

)+)+ aaa +=

Write equations of motion for wedge and bloc0.

#

y

:))#

amF =

( ) ))

))

ag(N

amN

==

*

*

6.3

3sin

:3cos )+)+#+# aamamF ==

( )+=

=3sin3cos

3cos3sin

gaa

aag((

))+

)+)++

( ) :3sin == )+y+y amamF( ) = 3sin3cos

* )++ ag((N

( t M h i ) E i D i( t M h i ) E i D iN




Sam%le Pro$lem &'(*

- 2

( ) )) ag(N =*6.3

olve for the accelerations.

( )

( ) ( )

( )( )( ) ( ) +

=

+

=

==

3sinlb*lb3

3coslb*sft.

3sin

3cos

3sin3cos

3sin3cos

*

)

+)

+

)

)++))

)++

a

((

g(a

ag((ag(

ag((N

sft3=.6=)a

( ) ( ) +=+=

3sinsft.3cossft3=.6

3sin3cos

)+

))+

a

gaa

sft6.)3=)+a

( t M h i ) E i D i( t M h i ) E i D iNi




Sam%le Pro$lem &'(+

- 3

#he bob of a m "endulum describes

an arc of a circle in a vertical "lane. $f

the tension in the cord is .6 times the

weight of the bob for the "ositionshown, find the velocity and accel

eration of the bob in that "osition.

-/#$N:

+esolve the equation of motion for thebob into tangential and normal

com"onents.

olve the com"onent equations for the

normal and tangential accelerations.

olve for the velocity in terms of the

normal acceleration.




(ector Mechanics )or En*ineers D'na+ics(ector Mechanics )or En*ineers D'na+icsinth

Sam%le Pro$lem &'(+

- 4

-/#$N:

+esolve the equation of motion for the bob intotangential and normal com"onents.

olve the com"onent equations for the normal and

tangential accelerations.

:tt maF =

=

=

3sin

3sin

ga

mamg

t

t

sm9.=ta

:nn maF =( )=

=

3cos6.

3cos6.

ga

mamgmg

n

n

sm3.*7=na

olve for velocity in terms of normal acceleration.

( )( )

sm3.*7m=== nn avv

a

sm77.6=v




(ector Mechanics )or En*ineers D'na+ics(ector Mechanics )or En*ineers D'na+icsinth

Sam%le Pro$lem &'(,

- 5

8etermine the rated s"eed of a

highway curve of radius > 33 ft

ban0ed through an angle > *;o. #he

rated s"eed of a ban0ed highway curveis the s"eed at which a car should

travel if no lateral friction force is to

be e2erted at its wheels.

-/#$N:

#he car travels in a hori%ontal circular"ath with a normal com"onent of

acceleration directed toward the center

of the "ath.#he forces acting on the car

are its weight and a normal reaction

from the road surface.

+esolve the equation of motion for

the car into vertical and normal

com"onents.

olve for the vehicle s"eed.

( t M h i ) E i D i(ector Mechanics )or En*ineers D'na+icsNi



(ector Mechanics )or En*ineers D'na+ics(ector Mechanics )or En*ineers D'na+icsnth

Sam%le Pro$lem &'(,

- -6

-/#$N:

#he car travels in a hori%ontal circular

"ath with a normal com"onent of

acceleration directed toward the centerof the "ath.#he forces acting on the

car are its weight and a normal

reaction from the road surface.

+esolve the equation of motion for

the car into vertical and normal

com"onents.

:3= yF

cos

3cos

(.

(.

=

=

:nn maF =

sincos

sin

v

g

((

ag(. n

=

=

olve for the vehicle s"eed.

( )( ) ==

*;tanft33sft.

tan

)

) gv

hmi*.sft=.7 ==v

( t M h i ) E i D i(ector Mechanics )or En*ineers D'na+icsNi




An-ular Momentum of a Particle

- -

moment of momentumor the angular

momentumof the "article about O.

== /mr0O

8erivative of angular momentum with res"ect to time,

=

=

+=+=

O

O

M

Fr

amr/m//mr/mr0

$t follows from Newtons second law that the sum of

the moments about Oof the forces acting on the

"article is equal to the rate of change of the angular

momentum of the "article about O!

"y#

O

mvmvmv

"y#

$%i

0

=

is "er"endicular to "lane containingO0 /mr and

)

sin

mr

vrm

rm/0O

=

==

(ector Mechanics )or En*ineers D'na+ics(ector Mechanics )or En*ineers D'na+icsNin




E"s of Motion in Radial . Trans/erse Com%onents

- --

( )( )

rrmmaF

rrmmaF rr

)

+==

==

&onsider "article at rand , in "olar coordinates,

( )

( )( )

rrmF

rrrm

mrdt

dFr

mr0O

)

)

+=

+=

=

=

#his result may also be derived from conservation

of angular momentum,






- -/

When only force acting on "article is directed

toward or away from a fi2ed "oint O, the "article

is said to be moving under a central force.

ince the line of action of the central force

"asses through O, and3 == OO 0M

constant== O

0/mr

?osition vector and motion of "article are in a

"lane "er"endicular to .O0

4agnitude of angular momentum,

333 sin

constantsin

/mr

/rm0O

===

massunit

momentumangular

constant

)

)

===

==

hr

m

0

mr0

O

O

or






- -0

+adius vector OPswee"s infinitesimal area

drd) ))

*=

8efine ===

*

* r

dt

dr

dt

d)areal velocity

+ecall, for a body moving under a central force,

constant) == rh

When a "article moves under a central force, its

areal velocity is constant.





Newtons Law of 0ra/itation

- -1

@ravitational force e2erted by the sun on a "lanet or by

the earth on a satellite is an im"ortant e2am"le ofgravitational force.

Newtons law of universal gravitation two "articles of

massMand mattract each other with equal and o""osite

force directed along the line connecting the "articles,

9

*

slb

ft*3.

s0g

m*3=.77

ngravitatioofconstant

=

=

=

=

1

r

Mm1F

1or "article of mass mon the earths surface,

s

ft.

s

m;*.9 ==== gmg

.

M1m(



26/38



Sam%le Pro$lem &'(1

- -2

! bloc0+of mass mcan slide freely on

a frictionless arm O)which rotates in a

hori%ontal "lane at a constant rate .3

a) the com"onent vrof the velocity of+

along O), and

b) the magnitude of the hori%ontal force

e2erted on+by the arm O).

Anowing that+is released at a distance

r3from O2 e2"ress as a function of r

-/#$N:

Write the radial and transverseequations of motion for the bloc0.

$ntegrate the radial equation to find an

e2"ression for the radial velocity.

ubstitute 0nown information into thetransverse equation to find an

e2"ression for the force on the bloc0.



27/38



Sam%le Pro$lem &'(1

- -3

-/#$N:

Write the radial and transverse

equations of motion for the bloc0.

:

:

amF

amFrr

== ( )

rrmF

rrm

3

+= =

$ntegrate the radial equation to find an

e2"ression for the radial velocity.

=

==

====

r

r

v

rr

rr

rr

rrr

drrdvv

drrdrrdvv

dr

dvv

dt

dr

dr

dv

dt

dvvr

r

3

3

3

3

dr

dvv

dt

dr

dr

dv

dt

dvvr rr

rrr ====

3

3

rrvr =

ubstitute 0nown information into the

transverse equation to find an e2"ression

for the force on the bloc0.

( ) *3

3 rrmF =



28/38



Sam%le Pro$lem &'(2

- -4

! satellite is launched in a direction

"arallel to the surface of the earth

with a velocity of *;;3 mi5h from

an altitude of 3 mi. 8etermine the

velocity of the satellite as it reaches itma2imum altitude of 3 mi. #he

radius of the earth is 973 mi.

-/#$N:

ince the satellite is moving under a

central force, its angular momentum is

constant.


29/38



Sam%le Pro$lem &'(2

- -5

-/#$N:

ince the satellite is moving under a

central force, its angular momentum is

constant.


30/38


(ector Mechanics )or En*ineers D'na+ics(ector Mechanics )or En*ineers D'na+icsth

Tra3ector of a Particle !nder a Central 4orce

- /6

1or "article moving under central force directed towards force center,

( ) 3 ==+== FrrmFFrrm r

econd e2"ression is equivalent to from which,,constant)

==hr

==

rd

d

r

hr

r

h *and

!fter substituting into the radial equation of motion and sim"lifying,

ru

umh

Fu

d

ud *where

==+

$fFis a 0nown function of ror u2then "article tra'ectory may be

found by integrating for u 3 f,-, with constants of integration

determined from initial conditions.



31/38




- /

constant

*where

==+

====+

h

1Mu

d

ud

1Mmur

1MmF

ru

umh

Fu

d

ud

&onsider earth satellites sub'ected to only gravitational "ull

of the earth,

olution is equation of conic section,

( ) tyeccentricicos**

==+==

1M

hC

h

1M

ru

rigin, located at earths center, is a focus of the conic section.

#ra'ectory may be elli"se, "arabola, or hy"erbola de"ending

on value of eccentricity.

(ector Mechanics )or En*ineers D'na+ics(ector Mechanics )or En*ineers D'na+icsNint


32/38




- /-

( ) tyeccentricicos**

==+=

1MhC

h1M

r

#ra'ectory of earth satellite is defined by

hyper'ola, 4 * or C 4 1M5h. #he radius vector

becomes infinite for

=

==+

**** cos*cos3cos*

hC1M

para'ola, > * or C 3 1M5h. #he radius vector

becomes infinite for

==+ *;33cos*

ellipse, B * or C 6 1M5h. #he radius vector is finite

for and is constant, i.e., a circle, for B 3.



33/38




- //

$ntegration constant Cis determined by

conditions at beginning of free flight, >3, r 3 r72

( ) 333

3

3

**

3cos**

vr

1M

rh

1M

rC

1M

Ch

h

1M

r

==

+=

( )

3

3

33

or*

r

1Mvv

vr1Mh1MC

esc ==

=

atellite esca"es earth orbit for

#ra'ectory is elli"tic for v76 vescand becomes

circular for > 3 or C> 3,

3r

1Mvcirc =



34/38




- /0

+ecall that for a "article moving under a central

force, the areal velocityis constant, i.e.,

constant*

* === hr

dt

d)

Periodic timeor time required for a satellite to

com"lete an orbit is equal to area within the orbit

divided by areal velocity,

h

a'

h

a'

==

where ( )

*3

*3*

rr'

rra

=

+=



35/38



Sam%le Pro$lem &'(6

- /1

8etermine:

a) the ma2imum altitude reached by

the satellite, and

b) the "eriodic time of the satellite.

! satellite is launched in a direction

"arallel to the surface of the earth

with a velocity of 7,933 0m5h at an

altitude of 633 0m.

-/#$N:

#ra'ectory of the satellite is described by

cos*

C

h

1M

r+=

3.

8etermine the ma2imum altitude by

finding rat > *;3o.

With the altitudes at the "erigee and

a"ogee 0nown, the "eriodic time canbe evaluated.



36/38



Sam%le Pro$lem &'(6

- /2

-/#$N:

#ra'ectory of the satellite is described by

cos*

C

h

1M

r+=

3.

( )

( )( )

( )( )*

7

9

733

3

73

sm*39;

m*3=.7sm;*.9

sm*3.=3

sm*36.*3m*37.;=

sm*36.*3

s5h733

m50m*333

h

0m7933

m*37.;=

0m6337=3

=

==

=

==

=

=

=

+=

g.1M

vrh

v

r

( )*9

*

7

3

m*3.76

sm.=3

sm*39;

m*3;=.7

*

*

=

=

= h1M

rC



37/38


(ector Mechanics )or En*ineers D'na+ics(ector Mechanics )or En*ineers D'na+icsh

Sam%le Pro$lem &'(6

- /3

8etermine the ma2imum altitude by finding r*

at > *;3o.

( )0m77=33m*3=.77

m

**3.76

sm.=3

sm*39;*

7*

9

*

*

==

==

r

Ch

1M

r

( ) 0m73330m7=377=33altitudema2 ==

With the altitudes at the "erigee and a"ogee 0nown,

the "eriodic time can be evaluated.

( ) ( )

( )( )sm*3=3.

m*3*.m*37.;

h

m*3*.m*3=.77;=.7

m*37.;m*3=.77;=.7

9

77

77*3

77

*

*3*

==

===

=+=+=

a'

rr'

rra

min*h*9s*3.=3 ==



38/38

(ector Mechanics )or En*ineers D'na+ics(ector Mechanics )or En*ineers D'na+icsh

Ke%lers Laws of Planetar Motion

+esults obtained for tra'ectories of satellites around earth may also be

a""lied to tra'ectories of "lanets around the sun.

?ro"erties of "lanetary orbits around the sun were determined

astronomical observations by Cohann Ae"ler (*6=**73) before

Newton had develo"ed his fundamental theory.

*)

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