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8/14/2019 Ch15-Kinematics of Rigid Bodies
1/64
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ninth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
J. Walt Oler
Texas Tech University
CHAPTER
2010 The McGraw-Hill Companies, Inc. All rights reserved.
15Kinematics of
Rigid Bodies
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2010 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: DynamicsNinth
Edition
Contents
15 - 2
Introduction
Translation
Rotation About a Fixed Axis: Velocity
Rotation About a Fixed Axis:
Acceleration
Rotation About a Fixed Axis:
Representative Slab
Equations Defining the Rotation of aRigid Body About a Fixed Axis
Sample Problem 5.1
General Plane Motion
Absolute and Relative Velocity in Plane
Motion
Sample Problem 15.2
Sample Problem 15.3
Instantaneous Center of Rotation in
Plane Motion
Sample Problem 15.4
Sample Problem 15.5
Absolute and Relative Acceleration in
Plane Motion
Analysis of Plane Motion in Terms of a
Parameter
Sample Problem 15.6
Sample Problem 15.7
Sample Problem 15.8
Rate of Change With Respect to aRotating Frame
Coriolis Acceleration
Sample Problem 15.9
Sample Problem 15.10
Motion About a Fixed Point
General Motion
Sample Problem 15.11
Three Dimensional Motion. Coriolis
Acceleration
Frame of Reference in General Motion
Sample Problem 15.15
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NE
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2010 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: DynamicsNinth
Edition
Rotation About a Fixed Axis. Velocity
15 - 5
Consider rotation of rigid body about a
fixed axisAA
Velocity vector of the particlePis
tangent to the path with magnitude
dtrdv
dtdsv
sinsinlim
sin
0
rt
rdt
dsv
rBPs
t
locityangular vekk
rdt
rdv
The same result is obtained from
NE
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Vector Mechanics for Engineers: DynamicsNinth
Edition
Rotation About a Fixed Axis. Acceleration
15 - 6
Differentiating to determine the acceleration,
vr
dt
d
dt
rdr
dt
d
rdt
d
dt
vda
kkk
celerationangular acdt
d
componentonacceleratiradial
componenonacceleratiltangentia
r
r
rra
Acceleration ofPis combination of two
vectors,
NE
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Vector Mechanics for Engineers: DynamicsNinth
Edition
Rotation About a Fixed Axis. Representative Slab
15 - 7
Consider the motion of a representative slab in
a plane perpendicular to the axis of rotation.
Velocity of any pointPof the slab,
rv
rkrv
Acceleration of any pointPof the slab,
rrk
rra
2
Resolving the acceleration into tangential and
normal components,
22
rara
rarka
nn
tt
NE
8/14/2019 Ch15-Kinematics of Rigid Bodies
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Vector Mechanics for Engineers: DynamicsNinth
Edition
Equations Defining the Rotation of a Rigid Body About a Fixed Axis
15 - 8
Motion of a rigid body rotating around a fixed axis isoften specified by the type of angular acceleration.
dd
dtd
dtd
ddt
dt
d
2
2
or Recall
Uniform Rotation, = 0:
t 0
Uniformly Accelerated Rotation, = constant:
02
0
2
2
21
00
0
2
tt
t
fNE
8/14/2019 Ch15-Kinematics of Rigid Bodies
9/64 2010 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: DynamicsNinth
Edition
Sample Problem 5.1
15 - 9
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 iNE
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Vector Mechanics for Engineers: DynamicsNinth
Edition
Sample Problem 5.1
15 - 10
SOLUTION:
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 iNE
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Vector Mechanics for Engineers: DynamicsNinth
Edition
Sample Problem 5.1
15 - 11
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 iNE
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Vector Mechanics for Engineers: DynamicsNinth
Edition
General Plane Motion
15 - 12
General plane motionis neither a translation nor
a rotation.
General plane motion can be considered as the
sumof a translation and rotation.
Displacement of particlesAandBtoA2andB2
can be divided into two parts:
- translation toA2and
- rotation of aboutA2
toB2
1B
1B
V t M h i f E i D iNE
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Vector Mechanics for Engineers: DynamicsNinth
Edition
Absolute and Relative Velocity in Plane Motion
15 - 13
Any plane motion can be replaced by a translation of an
arbitrary reference pointAand a simultaneous rotation
aboutA.
ABAB vvv
rvrkv ABABAB
ABAB rkvv
V t M h i f E i D iNE
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Absolute and Relative Velocity in Plane Motion
15 - 14
Assuming that the velocity vAof endAis known, wish to determine the
velocity vBof endBand the angular velocity in terms of vA, l, and .
The direction of vB
and vB/A
are known. Complete the velocity diagram.
tan
tan
AB
A
B
vv
v
v
cos
cos
l
v
l
v
v
v
A
A
AB
A
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Vector Mechanics for Engineers: DynamicsNinth
Edition
Sample Problem 15.2
15 - 17
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
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Vector Mechanics for Engineers: Dynamicsinth
dition
Sample Problem 15.2
15 - 18
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
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Vector Mechanics for Engineers: Dynamicsinth
dition
Sample Problem 15.3
15 - 19
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
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Vector Mechanics for Engineers: Dynamicsnth
dition
Sample Problem 15.3
15 - 20
SOLUTION:
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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Sample Problem 15.3
15 - 21
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
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Instantaneous Center of Rotation in Plane Motion
15 - 23
If the velocity at two pointsAandBare known, the
instantaneous center of rotation lies at the intersectionof the perpendiculars to the velocity vectors throughA
andB.
If the velocity vectors atAandBare perpendicular to
the lineAB, the instantaneous center of rotation lies at
the intersection of the lineABwith the line joining the
extremities of the velocity vectors atAandB.
If the velocity vectors are parallel, the instantaneous
center of rotation is at infinity and the angular velocity
is zero.
If the velocity magnitudes are equal, the instantaneous
center of rotation is at infinity and the angular velocity
is zero.
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Instantaneous Center of Rotation in Plane Motion
15 - 24
The instantaneous center of rotation lies at the intersection of
the perpendiculars to the velocity vectors throughAandB.
cosl
v
AC
v AA
tan
cossin
A
AB
v
l
vlBCv
The velocities of all particles on the rod are as if they were
rotated about C.
The particle at the center of rotation has zero velocity.
The particle coinciding with the center of rotation changes
with time and the acceleration of the particle at the
instantaneous center of rotation is not zero.
The acceleration of the particles in the slab cannot be
determined as if the slab were simply rotating about C.
The trace of the locus of the center of rotation on the body
is the body centrode and in space is the space centrode.
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Sample Problem 15.4
15 - 25
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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Sample Problem 15.4
15 - 26
SOLUTION:
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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Sample Problem 15.5
15 - 27
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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ition
Sample Problem 15.5
15 - 28
SOLUTION:
From Sample Problem 15.3,
95.13
sin.3.628sin.3.4819.403
BB vjiv
The instantaneous center of rotation is at the intersection
of the perpendiculars to the velocities throughB andD.
05.7690
95.5340
D
B
sin50
in.8
95.53sin05.76sin
CDBC
in.44.8in.14.10 CDBC
Determine the angular velocity about the center of
rotation based on the velocity atB.
in.10.14
sin.3.628
BC
v
BCv
BBD
BDB
Calculate the velocity atDbased on its rotation about
the instantaneous center of rotation.
srad0.62in.44.8 BDD CDv
sft6.43sin.523 DP vv
srad0.62BD
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Absolute and Relative Acceleration in Plane Motion
15 - 29
Absolute acceleration of a particle of the slab,
ABAB aaa
Relative acceleration associated with rotation aboutAincludestangential and normal components,
ABa
ABnAB
ABtAB
ra
rka
2
2
ra
ra
nAB
tAB
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Vector Mechanics for Engineers: Dynamicsthtion
Absolute and Relative Acceleration in Plane Motion
15 - 30
Givendetermine
,and AA va
.and
Ba
tABnABA
ABAB
aaa
aaa
Vector result depends on sense of and the
relative magnitudes ofnABA
aa andAa
Must also know angular velocity .
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Absolute and Relative Acceleration in Plane Motion
15 - 31
xcomponents: cossin0 2 llaA
ycomponents: sincos2
llaB
Solve for aBand .
Write in terms of the two component equations,ABAB aaa
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Analysis of Plane Motion in Terms of a Parameter
15 - 32
In some cases, it is advantageous to determine the
absolute velocity and acceleration of a mechanismdirectly.
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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Sample Problem 15.6
15 - 33
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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Sample Problem 15.6
15 - 34
SOLUTION:
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
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Sample Problem 15.6
15 - 35
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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Sample Problem 15.6
15 - 36
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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Sample Problem 15.7
15 - 37
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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Sample Problem 15.7
15 - 39
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: DynamicshonSample Problem 15.7
15 - 40
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: DynamicshonSample Problem 15.8
15 - 41
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: DynamicshonSample Problem 15.8
15 - 42
SOLUTION:
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: DynamicshonSample Problem 15.8
15 - 43
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: DynamicshonRate of Change With Respect to a Rotating Frame
15 - 44
Frame OXYZis fixed.
Frame Oxyz rotates about
fixed axis OAwith angular
velocity
Vector function varies
in direction and magnitude.
tQ
kQjQiQQ zyxOxyz
With respect to the fixed OXYZ frame,
kQjQiQkQjQiQQ zyxzyxOXYZ
rate of change
with respect to rotating frame.
Oxyzzyx QkQjQiQ
If were fixed within Oxyz then is
equivalent to velocity of a point in a rigid body
attached to Oxyzand
OXYZQ
QkQjQiQ zyx
Q
With respect to the rotating Oxyz frame,
kQjQiQQ zyx
With respect to the fixed OXYZ frame,
QQQ OxyzOXYZ
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Vector Mechanics for Engineers: DynamicshonCoriolis Acceleration
15 - 45
Frame OXYis fixed and frame Oxyrotates with angular
velocity .
Position vector for the particlePis the same in both
frames but the rate of change depends on the choice of
frame.
Pr
The absolute velocity of the particlePis
OxyOXYP rrrv
Imagine a rigid slab attached to the rotating frame Oxyor Ffor short. LetPbe a point on the slab which
corresponds instantaneously to position of particleP.
OxyP rv F velocity ofPalong its path on the slab
'Pv
absolute velocity of pointPon the slab
Absolute velocity for the particle P may be written as
F
PPP
vvv
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Vector Mechanics for Engineers: DynamicshonCoriolis Acceleration
15 - 46
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: DynamicsonCoriolis Acceleration
15 - 47
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 csonCoriolis Acceleration
15 - 48
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 yonSample Problem 15.9
15 - 49
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 yonSample Problem 15.9
15 - 50
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 yonSample Problem 15.9
15 - 51
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 ynSample Problem 15.10
15 - 52
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 ynSample Problem 15.10
15 - 53
SOLUTION:
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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g ynSample Problem 15.10
15 - 54
The relative acceleration must be parallel to
the slot.sPa
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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g ynMotion About a Fixed Point
15 - 55
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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g ynGeneral Motion
15 - 56
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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g ynSample Problem 15.11
15 - 57
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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g ynSample Problem 15.11
15 - 58
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 ynSample Problem 15.11
15 - 59
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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g ynThree-Dimensional Motion. Coriolis Acceleration
15 - 60
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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n
Frame of Reference in General Motion
15 - 61
Consider:
- fixed frame OXYZ,
- translating frameAXYZ, and
- translating and rotating frameAxyzor F.
With respect to OXYZandAXYZ,
APAP
APAP
APAP
aaa
vvv
rrr
The velocity and acceleration ofPrelative to
AXYZcan be found in terms of the velocityand acceleration ofPrelative toAxyz.
FPP
AxyzAPAPAP
vv
rrvv
cPP
AxyzAPAxyzAP
APAPAP
aaa
rr
rraa
F
2
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Sample Problem 15.15
15 - 62
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 15.15
15 - 63
SOLUTION:
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