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Transcript of Unidad II Cinematica de Cuerpo Rígido Beer_dinamica_8e_presentaciones_c15
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VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Eighth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
J. Walt Oler
Texas Tech University
CHAPTER
2007 The McGraw-Hill Companies, Inc. All rights reserved.
15Cinemtica de
Cuerpo rigido
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2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 2
ContenidosIntroduccin
Translacin
Rotacin alrededor de un eje fijo: Velocidad
Rotacin alrededor de un eje fijo:
Aceleracin
Rotacin alrededor de un eje fijo: Placa
representativa
Ecuaciones que definen la rotacin de uncuerpo rgido alrededor de un eje fijo
Problema resuelto 5.1
Movimiento plano General
Velocidad absoluta y relativa en el
movimiento plano general
Problema resuelto 15.2
Problema resuelto 15.3
Centro de rotacin instantneo en el
movimiento plano
Problema resuelto 15.4
Problema resuelto 15.5
Aceleracin absoluta y relativa en el
movimiento plano
Analisis del movimiento plano en trminos de
un parmetro
Problema resuelto 15.6
Problema resuelto 15.7
Problema resuelto 15.8
Razn de cambio con respecto a un sistema dereferencia en rotacin.
Aceleracin de coriolis
Problema resuelto 15.9
Problema resuelto 15.10
Movimiento alrededor de un punto fijo
Movimiento general.
Problema resuelto 15.11
Movimiento tridimensional. Aceleracin de
Coriolis
Sistema de referencia en movimiento general.
Problema resuelto 15.15
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2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 4
Traslacin Considere un cuerpo rgido en traslacin:
- La direccin de toda lnea recta dentro de
un cuerpo es constante,
- Toda partcula que constituyen el cuerpo
se mueven a lo largo de lneas paralela.
Para dos partculas en el cuerpo,
ABAB rrr
Diferenciando con respecto al tiempo,
AB
AABAB
vv
rrrr
Las partculas tienen la misma velocidad.
AB
AABAB
aa
rrrr
Diferenciando con respecto al tiempo,
Las particulas tienen la misma aceleracin.EE
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2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 5
Rotacin Alrededor de un eje fijo. Velocidad
Considere rotacin de un cuerpo rgido
alrededor de un eje fijoAA
Vector Velocidad de la partculaP
tangente a la trayectoria de magnitud
dtrdv
dtdsv
sinsinlim
sin
0
rt
rdt
dsv
rBPs
t
locityangular vekk
rdt
rdv
El mismo resultado se obtiene de
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EE
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Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 7
Rotacin alrededor de un eje fijo. Placa representativa Considere el movimiento de una placa
representativa en un plano perpendicular al eje
de rotacin.
La Velocidad de un puntoPde la placa es,
rv
rkrv
La Acceleration de un puntoPde la placa,
rrk
rra
2
Resolviendo la aceleracin en componentes
tangencial y componentes normal,
22
rara
rarka
nn
tt
EE
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Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 8
Definiendo la ecuacin de rotacin de un cuerpo rgido
Alrededor de un eje fijo
El Movimiento de un cuerpo rgido rotandoalrededor de un eje fijo esta especificado por el tipo
de aceleracin angular.
dd
dtd
dtd
ddt
dt
d
2
2
or Recordando
Rotacin uniforme, = 0:
t 0
Rotacin uniformemente acelerado, = constant:
02
0
2
2
21
00
0
2
tt
t
V M h i f E i D iEE
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Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 9
Sample Problem 5.1
Cable Chas a constant acceleration of 9
in/s2and an initial velocity of 12 in/s,
both directed to the right.
Determine (a)the number of revolutions
of the pulley in 2 s, (b) the velocity and
change in position of the loadBafter 2 s,
and (c)the acceleration of the pointDon
the rim of the inner pulley at t= 0.
SOLUTION:
Due to the action of the cable, thetangential velocity and acceleration of
Dare equal to the velocity and
acceleration of C. Calculate the initial
angular velocity and acceleration.
Apply the relations for uniformlyaccelerated rotation to determine the
velocity and angular position of the
pulley after 2 s.
Evaluate the initial tangential and
normal acceleration components ofD.
V t M h i f E i D iEE
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Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 10
Sample Problem 5.1SOLUTION:
The tangential velocity and acceleration ofDare equal to the
velocity and acceleration of C.
srad4
3
12
sin.12
00
00
00
r
v
rv
vv
D
D
CD
2srad3
3
9
sin.9
r
a
ra
aa
tD
tD
CtD
Apply the relations for uniformly accelerated rotation to
determine velocity and angular position of pulley after 2 s.
srad10s2srad3srad4 20 t
rad14
s2srad3s2srad4 22
2
12
2
10
tt
revsofnumberrad2
rev1rad14
N rev23.2N
rad14in.5
srad10in.5
ry
rv
B
B
in.70
sin.50
B
B
y
v
V t M h i f E i D iEE
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Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 11
Sample Problem 5.1 Evaluate the initial tangential and normal acceleration
components ofD.
sin.9CtD aa
2220 sin48srad4in.3 DnD ra
22 sin.48sin.9 nDtD aa
Magnitude and direction of the total acceleration,
22
22
489
nDtDD aaa
2sin.8.48Da
9
48
tan
tD
nD
a
a
4.79
V t M h i f E i D iEE
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Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 12
Movimiento Plano General
Movimiento plano generalno es ni rotacin ni
traslacin.
El movimiento plano general puede considerares
como la suma de traslacin y rotacin.
El desplazamiento de partculasAandBaA2y
B2pueden ser divida en dos partes:
- traslacin deA2and
- rotacin de alrededor deA2
aB2
1B
1B
V t M h i f E i D iEE
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Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 13
Velocidad Absoluta y Relativa en el movimiento Plano
Cualquier movimiento plano puede ser reemplazado por una
traslacin mediante el movimiento de un punto de referencia
arbitrario A y una rotacin simultaneo alrededor de A.
ABAB vvv
rvrkv ABABAB
ABAB rkvv
V t M h i f E i D iEE
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Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 14
Velocidad Absoluta y Relativa en el movimiento Plano
Asumiendo que la velocidad vAdel extremoAes conocida, se propone determinar
la velocidad vBdel extremoBy la velocidad angular en termino de vA, l, and .
La direccin de vB
y vB/A
son conocidos. Completando el diagrama velocidad.
tan
tan
AB
A
B
vv
v
v
cos
cos
l
v
l
v
v
v
A
A
AB
A
V t M h i f E i D iE
E
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Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 15
Velocidad Absoluta y Relativa en el Movimiento plano
Seleccionando el puntoBcomo el punto de referencia y resolviendo para la
velocidad vAen el extremoAy la velocidad angular lleva a una velocidad
equivalente en el triangulo.
vA/Btiene la misma magnitud pero sentido opuesto al de vB/A. El sentido de la
velocidad relativa depende del punto de referencia escogido.
La velocidad angular de la varilla en su rotacin alrededor deBes la misma
que su rotacin alrededor deA. La velocidad angular no depende del punto de
referencia escogida.
V t M h i f E i D iEi
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Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 16
Sample Problem 15.2
The double gear rolls on the
stationary lower rack: the velocity of
its center is 1.2 m/s.
Determine (a)the angular velocity of
the gear, and (b)the velocities of the
upper rackRand pointD of the gear.
SOLUTION:
The displacement of the gear center in
one revolution is equal to the outer
circumference. Relate the translational
and angular displacements. Differentiate
to relate the translational and angular
velocities.
The velocity for any pointPon the gear
may be written as
Evaluate the velocities of pointsB andD.
APAAPAP rkvvvv
V t M h i f E i D iEi
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Vector Mechanics for Engineers: DynamicsEighth
Edition
15 - 17
Sample Problem 15.2
x
y
SOLUTION:
The displacement of the gear center in one revolution isequal to the outer circumference.
ForxA> 0 (moves to right), < 0 (rotates clockwise).
1
22rx
r
xA
A
Differentiate to relate the translational and angular
velocities.
m0.150sm2.1
1
1
rv
rv
A
A
kk srad8
Vector Mechanics for Engineers: DynamicsEi
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Vector Mechanics for Engineers: Dynamicsighth
dition
15 - 18
Sample Problem 15.2
For any pointPon the gear, APAAPAP rkvvvv
Velocity of the upper rack is equal to
velocity of pointB:
ii
jki
rkvvv ABABR
sm8.0sm2.1
m10.0srad8sm2.1
ivR
sm2
Velocity of the pointD:
iki
rkvv ADAD
m150.0srad8sm2.1
sm697.1
sm2.1sm2.1
D
D
v
jiv
Vector Mechanics for Engineers: DynamicsEi
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Vector Mechanics for Engineers: Dynamicsghth
dition
15 - 19
Sample Problem 15.3
The crankABhas a constant clockwise
angular velocity of 2000 rpm.
For the crank position indicated,
determine (a)the angular velocity of
the connecting rodBD, and (b)the
velocity of the pistonP.
SOLUTION:
Will determine the absolute velocity ofpointDwith
BDBD vvv
The velocity is obtained from the
given crank rotation data.Bv
The directions of the absolute velocity
and the relative velocity are
determined from the problem geometry.
Dv
BDv
The unknowns in the vector expression
are the velocity magnitudes
which may be determined from the
corresponding vector triangle.
BDD vv and
The angular velocity of the connecting
rod is calculated from .BDv
Vector Mechanics for Engineers: DynamicsEi
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Vector Mechanics for Engineers: Dynamicsghth
dition
15 - 20
Sample Problem 15.3SOLUTION:
Will determine the absolute velocity of pointDwith
BDBD vvv
The velocity is obtained from the crank rotation data.Bv
srad4.209in.3
srad4.209rev
rad2
s60
min
min
rev2000
ABB
AB
ABv
The velocity direction is as shown.
The direction of the absolute velocity is horizontal.
The direction of the relative velocity is
perpendicular toBD. Compute the angle between thehorizontal and the connecting rod from the law of sines.
Dv
BDv
95.13in.3
sin
in.8
40sin
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Vector Mechanics for Engineers: Dynamicsghth
dition
15 - 21
Sample Problem 15.3
Determine the velocity magnitudes
from the vector triangle.
BDD vv and
BDBD vvv
sin76.05
sin.3.628
50sin95.53sin
BDD vv
sin.9.495
sft6.43sin.4.523
BD
D
v
v
srad0.62
in.8
sin.9.495
l
v
lv
BDBD
BDBD
sft6.43 DP vv
kBD
srad0.62
Vector Mechanics for Engineers: DynamicsEig
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Vector Mechanics for Engineers: Dynamicsghth
dition
15 - 22
Centro Instantneo de Rotacin en Movimiento Plano
El movimiento plano de las partculas en una placa
puede ser reemplazada la traslacin de un puntoarbitrario A y su rotacin alrededor deA con una
velocidad angular que es independiente del puntoA.
La traslacin y rotacin de la velocidad deA son obtenida
y que permite que la placa rote con la misma velocidad
angular alrededor del punto C y es perpendicular a la
velocidad deA.
La velocidad de todas las dems partculas de una placa
son las mismas y estn originalmente definida por la
velocidad angular y la velocidad y traslacional deA.
Tal como las velocidades estn involucradas . La placa
parece rotar alrededor de un centro instantneo de
rotacin C.
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Vector Mechanics for Engineers: Dynamicsghth
dition
15 - 23
Centro Instantneo de Rotacin en Movimiento Plano
Si son conocidas la velocidad de dos partculasAyB,
el centro instantneo de rotacin depende de lainterseccin de las perpendiculares de los vectores
velocidades a travs de las partculas AyB.
Si los vectores velocidadAyBson perpendiculares a la
lneaAB, el centro instantneo de rotacin se encuentra
en la interseccin de la lneaABcon la lnea que se
junta con los extremos de los vectores velocidadAyB.
Si los vectores velocidad son paralelos, el centro
instantneo de rotacin es infinito y la velocidad
angular es cero.
Si las magnitudes de las velocidades son iguales, el
centro instantneo de rotacin es infinito y la velocidad
angular es cero.
Vector Mechanics for Engineers: DynamicsEig
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Vector Mechanics for Engineers: Dynamicsghth
dition
15 - 24
Centro Instantneo de Rotacin en Movimiento Plano
El centro instantneo de rotacin se determina por la
interseccin de la perpendicular de los vectores velocidadAyB
.
cosl
v
AC
v AA
tan
cossin
A
AB
v
l
vlBCv
La velocidad de todas las partculas de la varilla son las
mismas cuando rotan alrededor de C. La partcula con centro de rotacin en la placa la velocidad
es cero.
La partcula coincidente con el centro de rotacin cambian
con el tiempo y la aceleracin en el centro instantneo de
rotacin no es cero.
Las aceleracin de las partculas de la placa no pueden
determinarse como si la placa simplemente rotara en C.
El trazo en lugar del centro de rotacin del cuerpo es el
centrodo y en el espacio el centrodo espacial.
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Vector Mechanics for Engineers: Dynamicsghth
ition
15 - 25
Sample Problem 15.4
The double gear rolls on the
stationary lower rack: the velocity
of its center is 1.2 m/s.
Determine (a)the angular velocity
of the gear, and (b)the velocities of
the upper rackRand pointD of the
gear.
SOLUTION:
The point Cis in contact with the stationarylower rack and, instantaneously, has zero
velocity. It must be the location of the
instantaneous center of rotation.
Determine the angular velocity about C
based on the given velocity atA.
Evaluate the velocities atBandDbased on
their rotation about C.
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Vector Mechanics for Engineers: Dynamicsghth
ition
15 - 26
Sample Problem 15.4SOLUTION:
The point Cis in contact with the stationary lower rack
and, instantaneously, has zero velocity. It must be thelocation of the instantaneous center of rotation.
Determine the angular velocity about Cbased on the
given velocity atA.
srad8
m0.15
sm2.1
A
AAA
r
vrv
Evaluate the velocities atBandDbased on their rotation
about C.
srad8m25.0 BBR rvv
ivR
sm2
srad8m2121.0
m2121.02m15.0
DD
D
rv
r
sm2.12.1sm697.1
jiv
v
D
D
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2007 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamicsghth
ition
15 - 27
Sample Problem 15.5
The crankABhas a constant clockwise
angular velocity of 2000 rpm.
For the crank position indicated,
determine (a)the angular velocity of
the connecting rodBD, and (b)the
velocity of the pistonP.
SOLUTION:
Determine the velocity atBfrom thegiven crank rotation data.
The direction of the velocity vectors atB
andDare known. The instantaneous
center of rotation is at the intersection of
the perpendiculars to the velocitiesthroughB andD.
Determine the angular velocity about the
center of rotation based on the velocity
atB.
Calculate the velocity atDbased on its
rotation about the instantaneous center
of rotation.
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Vector Mechanics for Engineers: Dynamicsghth
tion
15 - 29
Aceleracin Absoluta y Relativa en Movimiento Plano
La aceleracin absoluta de una partcula en la placa,
ABAB aaa
La aceleracin relativa asociada con la rotacin alrededor deAincluye las componentes tangencial y normal,
ABa
ABnAB
ABtAB
ra
rka
2
2
ra
ra
nAB
tAB
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Vector Mechanics for Engineers: Dynamicshthtion
15 - 30
Aceleracin Absoluta y Relativa en Movimiento Plano
dadodetermine
,and AA va
.and
Ba
tABnABA
ABAB
aaa
aaa
El vector resultante del sentido y de las
magnitudes relativas denABA
aa andAa
Debe conocerse tambin la velocidad angular .
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Vector Mechanics for Engineers: Dynamicshthtion
15 - 31
Aceleracin Absoluta y Relativa en Movimiento Plano
xcomponentes: cossin0 2 llaA
ycomponentes: sincos2
llaB
Resolviendo para aB y.
Escribiendo en trminos de ecuacin de dos
componentes,ABAB aaa
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Vector Mechanics for Engineers: Dynamicshthtion
15 - 32
Anlisis de Movimiento Plano en Trminos de un Parametro
En algunos casos, este es una ventaja para determinar
la velocidad y aceleracin absoluta directamente deun mecanismo.
sinlxA coslyB
cos
cos
l
l
xv AA
sin
sin
l
l
yv BB
cossincossin
2
2
llll
xa AA
sincossincos
2
2
llll
ya BB
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Vector Mechanics for Engineers: Dynamicshthtion
15 - 33
Sample Problem 15.6
The center of the double gear has a
velocity and acceleration to the right of
1.2 m/s and 3 m/s2, respectively. The
lower rack is stationary.
Determine (a) the angular acceleration
of the gear, and (b)the acceleration of
pointsB, C, andD.
SOLUTION:
The expression of the gear position as a
function of is differentiated twice to
define the relationship between the
translational and angular accelerations.
The acceleration of each point on thegear is obtained by adding the
acceleration of the gear center and the
relative accelerations with respect to the
center. The latter includes normal and
tangential acceleration components.
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Vector Mechanics for Engineers: Dynamicshthtion
15 - 34
Sample Problem 15.6SOLUTION:
The expression of the gear position as a function of
is differentiated twice to define the relationshipbetween the translational and angular accelerations.
11
1
rrv
rx
A
A
srad8m0.150sm2.1
1 r
vA
11 rraA
m150.0
sm3 2
1 r
aA
kk 2srad20
Vector Mechanics for Engineers: DynamicsEigh
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Vector Mechanics for Engineers: Dynamicshthtion
15 - 35
Sample Problem 15.6
jiijjki
rrka
aaaaaa
ABABA
nABtABAABAB
222
222
2
sm40.6sm2sm3m100.0srad8m100.0srad20sm3
222 sm12.8sm40.6m5 BB ajisa
The acceleration of each point
is obtained by adding the
acceleration of the gear centerand the relative accelerations
with respect to the center.
The latter includes normal and
tangential acceleration
components.
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Vector Mechanics for Engineers: Dynamicshthion
15 - 36
Sample Problem 15.6
jii
jjki
rrkaaaa ACACAACAC
222
222
2
sm60.9sm3sm3
m150.0srad8m150.0srad20sm3
jac 2sm60.9
iji
iiki
rrkaaaa ADADAADAD
222
222
2
sm60.9sm3sm3
m150.0srad8m150.0srad20sm3
222 sm95.12sm3m6.12 DD
ajisa
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Vector Mechanics for Engineers: Dynamicshthion
15 - 37
Sample Problem 15.7
CrankAGof the engine system has a
constant clockwise angular velocity of
2000 rpm.
For the crank position shown,
determine the angular acceleration of
the connecting rodBDand the
acceleration of pointD.
SOLUTION:
The angular acceleration of theconnecting rodBDand the acceleration
of pointDwill be determined from
nBDtBDBBDBD aaaaaa
The acceleration ofBis determined fromthe given rotation speed ofAB.
The directions of the accelerations
are
determined from the geometry.nBDtBDD
aaa
and,,
Component equations for acceleration
of pointDare solved simultaneously for
acceleration ofDand angular
acceleration of the connecting rod.
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Vector Mechanics for Engineers: Dynamicshthion
15 - 38
Sample Problem 15.7
The acceleration ofBis determined from the given rotation
speed ofAB.
SOLUTION:
The angular acceleration of the connecting rodBDand
the acceleration of pointDwill be determined from
nBDtBDBBDBD aaaaaa
221232
AB
sft962,10srad4.209ft
0
constantsrad209.4rpm2000
ABB
AB
ra
jiaB
40sin40cossft962,10 2
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Vector Mechanics for Engineers: Dynamicshthion
15 - 39
Sample Problem 15.7
The directions of the accelerations are
determined from the geometry.nBDtBDD
aaa
and,,
From Sample Problem 15.3, BD= 62.0 rad/s, = 13.95o.
221282 sft2563srad0.62ft BDnBD
BDa
jia nBD
95.13sin95.13cossft2563 2
BDBDBDtBD BDa 667.0ft128
The direction of (aD/B)tis known but the sense is not known,
jia BDt
BD
05.76cos05.76sin667.0
iaa DD
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Vector Mechanics for Engineers: Dynamicsthon
15 - 40
Sample Problem 15.7
nBDtBDBBDBD aaaaaa
Component equations for acceleration of pointDare solved
simultaneously.
xcomponents:
95.13sin667.095.13cos256340cos962,10 BDDa
95.13cos667.095.13sin256340sin962,100 BD
y components:
iak
D
BD
2
2
sft9290
srad9940
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Vector Mechanics for Engineers: Dynamicsthon
15 - 41
Sample Problem 15.8
In the position shown, crankABhas a
constant angular velocity 1= 20 rad/s
counterclockwise.
Determine the angular velocities and
angular accelerations of the connecting
rodBDand crankDE.
SOLUTION:
The angular velocities are determined by
simultaneously solving the component
equations for
BDBD vvv
The angular accelerations are determined
by simultaneously solving the component
equations for
BDBD aaa
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Vector Mechanics for Engineers: Dynamicsthon
15 - 42
Sample Problem 15.8SOLUTION:
The angular velocities are determined by simultaneously
solving the component equations for
BDBD vvv
ji
jikrv
DEDE
DEDDED
1717
1717
ji
jikrv BABB
160280
14820
ji
jikrv
BDBD
BDBDBDBD
123
312
BDDE 328017 xcomponents:
BDDE 1216017 ycomponents:
kk DEBD
srad29.11srad33.29
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Vector Mechanics for Engineers: Dynamicsthon
15 - 43
Sample Problem 15.8 The angular accelerations are determined by
simultaneously solving the component equations for
BDBD aaa
jiji
jijik
rra
DEDE
DE
DDEDDED
217021701717
171729.111717 2
2
ji
jirra BABBABB
56003200
148200 22
jiji
jijik
rra
DBDB
DB
DBBDDBBDBD
2580320,10123
31233.29312 2
2
xcomponents: 690,15317 BDDE
y components: 60101217 BDDE
kkDEBD
22 srad809srad645
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Vector Mechanics for Engineers: Dynamicsthon
15 - 44
Rapidez de cambio con respecto a un Sistema de Rotacion
Sistema fijo OXYZ.
Sistema Oxyz rotando
alrededor del eje fijoOA
con una velocidad angular
La funcin del vector
vara en direccin y
magnitud.
tQ
kQjQiQQ zyxOxyz
Con respecto al sistema fijo OXYZ ,
kQjQiQkQjQiQQ zyxzyxOXYZ
Rapidez de
cambio con respecto al sistema de rotacin. Oxyzzyx QkQjQiQ
Si donde esta fijo dentro de Oxyz
entonces esto es equivalente a la velocidad de unpunto en un cuerpo rgido Oxyzy
OXYZQ
QkQjQiQ zyx
Q
Con respecto al sistema de rotacin Oxyz
kQjQiQQ zyx
Con respecto al sistema fijo OXYZ ,
QQQ OxyzOXYZ
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Vector Mechanics for Engineers: Dynamicsthon
15 - 46
Coriolis Acceleration
FPP
OxyP
vv
rrv
Absolute acceleration for the particlePis
OxyOXYP r
dt
drra
OxyOxyP rrrra
2
OxyOxyOxy
OxyOXY
rrrdt
d
rrr
but,
OxyP
P
ra
rra
F
Utilizing the conceptual pointPon the slab,
Absolute acceleration for the particlePbecomes
22
2
F
F
F
POxyc
cPP
OxyPPP
vra
aaa
raaa
Coriolis acceleration
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Vector Mechanics for Engineers: Dynamicshon
15 - 47
Coriolis Acceleration Consider a collarPwhich is made to slide at constant
relative velocity ualong rod OB. The rod is rotating at
a constant angular velocity . The pointAon the rodcorresponds to the instantaneous position ofP.
cPAP aaaa
F
Absolute acceleration of the collar is
0 OxyP ra
F
uavacPc
22 F
The absolute acceleration consists of the radial and
tangential vectors shown
2rarra AA
where
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ecto ec a cs o g ee s y a cshon
15 - 48
Coriolis Acceleration
uvvtt
uvvt
A
A
,at
,at
Change in velocity over tis represented by the
sum of three vectors
TTTTRRv
2rarra AA
recall,
is due to change in direction of the velocity of
pointA on the rod,
AAtt arrtvt
TT
2
00 limlim
TT
result from combined effects of
relative motion ofPand rotation of the rod
TTRR and
uuu
t
r
tu
t
TT
t
RR
tt
2
limlim00
uavacPc
22 F
recall,
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g yhon
15 - 49
Sample Problem 15.9
Disk D of the Geneva mechanism rotates
with constant counterclockwise angular
velocity D= 10 rad/s.
At the instant when = 150o, determine
(a) the angular velocity of disk S, and (b)
the velocity of pinPrelative to disk S.
SOLUTION:
The absolute velocity of the pointPmay be written as
sPPP vvv
Magnitude and direction of velocity
of pinP are calculated from the
radius and angular velocity of diskD.Pv
Direction of velocity of pointP on
Scoinciding withPis perpendicular to
radius OP.
Pv
Direction of velocity of Pwithrespect to Sis parallel to the slot.
sPv
Solve the vector triangle for the
angular velocity of Sand relative
velocity ofP.
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g yhon
15 - 50
Sample Problem 15.9
SOLUTION:
The absolute velocity of the pointPmay be written as
sPPP vvv
Magnitude and direction of absolute velocity of pinP are
calculated from radius and angular velocity of diskD.
smm500srad10mm50 DP Rv
Direction of velocity ofPwith respect to Sis parallel to slot.
From the law of cosines,
mm1.37551.030cos2 2222 rRRllRr
From the law of cosines,
4.42742.0
30sinsin
30sin
R
sin
r
6.17304.4290
The interior angle of the vector triangle is
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g yhon
15 - 51
Sample Problem 15.9
Direction of velocity of pointP on Scoinciding withPis
perpendicular to radius OP. From the velocity triangle,
mm1.37
smm2.151
smm2.1516.17sinsmm500sin
ss
PP
r
vv
ks
srad08.4
6.17cossm500cosPsP vv
jiv sP
4.42sin4.42cossm477
smm500Pv
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g yhn
15 - 52
Sample Problem 15.10
In the Geneva mechanism, diskD
rotates with a constant counter-clockwise angular velocity of 10
rad/s. At the instant when j= 150o,
determine angular acceleration of
disk S.
SOLUTION:
The absolute acceleration of the pinPmaybe expressed as
csPPP aaaa
The instantaneous angular velocity of Disk
Sis determined as in Sample Problem 15.9.
The only unknown involved in the
acceleration equation is the instantaneous
angular acceleration of Disk S.
Resolve each acceleration term into the
component parallel to the slot. Solve for
the angular acceleration of Disk S.
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g yhn
15 - 53
Sample Problem 15.10SOLUTION:
Absolute acceleration of the pinPmay be expressed as
csPPP aaaa
From Sample Problem 15.9.
jivk
sP
S
4.42sin4.42cossmm477
srad08.44.42
Considering each term in the acceleration equation,
jiaRa
P
DP
30sin30cossmm5000
smm5000srad10mm500
2
222
jia
jira
jira
aaa
StP
StP
SnP
tPnPP
4.42cos4.42sinmm1.37
4.42cos4.42sin
4.42sin4.42cos2
note: Smay be positive or negative
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15 - 54
The relative acceleration must be parallel to
the slot.sPa
Sample Problem 15.10
sPv
The direction of the Coriolis acceleration is obtained
by rotating the direction of the relative velocityby 90o in the sense of S.
jiji
jiva sPSc
4.42cos4.42sinsmm3890
4.42cos4.42sinsmm477srad08.42
4.42cos4.42sin2
2
Equating components of the acceleration terms
perpendicular to the slot,
srad233
07.17cos500038901.37
S
S
kS
srad233
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15 - 55
Motion About a Fixed Point The most general displacement of a rigid body with a
fixed point Ois equivalent to a rotation of the body
about an axis through O. With the instantaneous axis of rotation and angular
velocity the velocity of a particlePof the body is,
rdt
rdv
and the acceleration of the particlePis
.dt
drra
Angular velocities have magnitude and direction and
obey parallelogram law of addition. They are vectors.
As the vector moves within the body and in space,
it generates a body cone and space cone which are
tangent along the instantaneous axis of rotation.
The angular acceleration represents the velocity of
the tip of .
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15 - 56
General Motion For particlesAandBof a rigid body,
ABAB vvv
ParticleAis fixed within the body and motion of
the body relative toAXYZis the motion of a
body with a fixed point
ABAB rvv
Similarly, the acceleration of the particlePis
ABABAABAB
rra
aaa
Most general motion of a rigid body is equivalent to:
- a translation in which all particles have the same
velocity and acceleration of a reference particleA,and
- of a motion in which particleAis assumed fixed.
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15 - 57
Sample Problem 15.11
The crane rotates with a constantangular velocity 1= 0.30 rad/s and the
boom is being raised with a constant
angular velocity 2= 0.50 rad/s. The
length of the boom is l= 12 m.
Determine: angular velocity of the boom,
angular acceleration of the boom,
velocity of the boom tip, and
acceleration of the boom tip.
Angular acceleration of the boom,
21
22221
Oxyz
Velocity of boom tip,
rv
Acceleration of boom tip,
vrrra
SOLUTION:
With
Angular velocity of the boom,
21
ji
jirkj
639.10
30sin30cos1250.030.0 21
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15 - 58
Sample Problem 15.11
jir
kj
639.10
50.030.0 21
SOLUTION:
Angular velocity of the boom,
21 kj
srad50.0srad30.0
Angular acceleration of the boom,
kj
Oxyz
srad50.0srad30.021
22221
i 2srad15.0
Velocity of boom tip,
0639.10
5.03.00
kji
rv
kjiv
sm12.3sm20.5sm54.3
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g yn
15 - 59
Sample Problem 15.11
jir
kj
639.10
50.030.0 21
Acceleration of boom tip,
kjiik
kjikji
a
vrrra
90.050.160.294.090.0
12.320.53
50.030.00
0639.10
0015.0
kjia 222 sm80.1sm50.1sm54.3
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15 - 60
Three-Dimensional Motion. Coriolis Acceleration
With respect to the fixed frame OXYZand rotating
frame Oxyz,QQQ OxyzOXYZ
Consider motion of particlePrelative to a rotatingframe Oxyzor Ffor short. The absolute velocity can
be expressed as
FPP
OxyzP
vv
rrv
The absolute acceleration can be expressed as
onacceleratiCoriolis22
2
F
F
POxyzc
cPp
OxyzOxyzP
vra
aaa
rrrra
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Vector Mechanics for Engineers: DynamicsE
ighth
E
dition
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Sample Problem 15.15
For the disk mounted on the arm, the
indicated angular rotation rates are
constant.
Determine:
the velocity of the pointP,
the acceleration ofP, and
angular velocity and angular
acceleration of the disk.
SOLUTION:
Define a fixed reference frame OXYZat Oand a moving reference frameAxyzor F
attached to the arm atA.
WithPof the moving reference frame
coinciding withP, the velocity of the point
Pis found from
FPPP vvv
The acceleration ofPis found from
cPPP aaaa
F
The angular velocity and angular
acceleration of the disk are
F
FD
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Sample Problem 15.15SOLUTION:
Define a fixed reference frame OXYZat Oand a
moving reference frameAxyzor Fattached to thearm atA.
j
jRiLr
1
k
jRr
D
AP
2
F
WithPof the moving reference frame coinciding
withP, the velocity of the pointPis found from
iRjRkrv
kLjRiLjrv
vvv
APDP
P
PPP
22
11
FF
F
kLiRvP
12
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Sample Problem 15.15 The acceleration ofPis found from
cPPP aaaa
F
iLkLjraP 2
111
jRiRk
ra APDDP
2222
FFF
kRiRj
va Pc
2121 22
2
F
kRjRiLaP
2122
21 2
Angular velocity and acceleration of the disk,
FD
kj
21
kjj
F
i