Chapter Outline
15 - 2
1. Introduction
2. Translation
3. Rotation About a Fixed Axis
• Velocity
• Acceleration
• Representative Slab
• Equations Defining Rotation of
a Rigid Body About a Fixed
Axis
4. General Plane Motion
• Absolute and Relative Velocity
• Instantaneous Center of
Rotation
• Absolute and Relative
Acceleration
• Analysis of Plane Motion in
Terms of a Parameter
5. Rate of Change With
Respect to a Rotating Frame
6. Coriolis Acceleration
7. Motion About a Fixed Point
8. General Motion
9. 3-D Motion. Coriolis
Acceleration
10.Frame of Reference in
General Motion
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Introduction
15 - 3
• Kinematics of rigid bodies (RB):
relations between time and the positions,
velocities, and accelerations of the
particles forming a rigid body.
• Classification of RB motions:
General motion
Motion about a fixed point
General plane motion
Rotation about a fixed axis
• curvilinear translation
• rectilinear translation
Translation:
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Translation
15 - 4
RB in translation:
direction of any straight line inside
body is constant,
all particles forming RB move in
parallel lines.
For any two particles in the body,
ABAB rrr
Differentiating with respect to time,
AB
AABAB
vv
rrrr
All particles have same velocity.
AB
AABAB
aa
rrrr
Differentiating with respect to time again,
All particles have same acceleration.
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Rotation About a Fixed Axis
Velocity
15 - 5
Consider rotation of RB about a fixed
axis AA’
Velocity vector of the
particle P is 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
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Rotation About a Fixed Axis
Acceleration
15 - 6
Differentiating to determine the acceleration,
vrdt
d
dt
rdr
dt
d
rdt
d
dt
vda
kkk
celerationangular acdt
d
componenton accelerati radial
componenton accelerati l tangentia
r
r
rra
Acceleration of P is combination of two
vectors,
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Rotation About a Fixed Axis
Representative Slab
15 - 7
Velocity of any point P of the slab,
rv
rkrv
Acceleration of any point P of the slab,
rrk
rra
2
Resolving the acceleration into
tangential and normal components,
22
rara
rarka
nn
tt
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Equations Defining Rotation of a RB About a
Fixed Axis
15 - 8
d
d
dt
d
dt
d
ddt
dt
d
2
2
or• Recall
• Uniform Rotation, = 0:
t 0
• Uniformly Accelerated Rotation, = constant:
020
2
221
00
0
2
tt
t
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 5.1
15 - 9
Cable C has a constant acceleration of 9 in/s2
and an initial velocity of 12 in/s, both directed to
the right. Determine
a. number of revolutions of pulley in 2 s,
b. velocity and change in position of load B
after 2 s
c. acceleration of point D on the rim of the
inner pulley at t = 0.
SOLUTION:
Due to the action of the cable,
tangential velocity &
acceleration of D are equal to
velocity & acceleration of C.
Calculate initial angular velocity
& acceleration.
Apply relations for uniformly
accelerated rotation to determine
velocity and angular position of
the pulley after 2 s.
Evaluate initial tangential and
normal acceleration components
of D. Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 5.1 15 - 10
SOLUTION:
Tangential velocity and acceleration of D are equal to
velocity and acceleration of C.
srad4
3
12
sin.12
00
00
00
r
v
rv
vv
D
D
CD
2srad33
9
sin.9
r
a
ra
aa
tD
tD
CtD
Apply relations for uniformly accelerated rotation to
determine velocity & angular position of pulley after 2 s.
srad10s 2srad3srad4 20 t
rad 14
s 2srad3s 2srad422
212
21
0
tt
revs ofnumber rad 2
rev 1rad 14
N rev23.2N
rad 14in. 5
srad10in. 5
ry
rv
B
Bin. 70
sin.50
B
B
y
v
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 5.1
15 - 11
Evaluate initial tangential and normal acceleration
components of D.
sin.9CtD aa
2220 sin48srad4in. 3 DnD ra
22 sin.48sin.9 nDtD aa
Magnitude and direction of total acceleration,
22
22
489
nDtDD aaa2sin.8.48Da
9
48
tan
tD
nD
a
a 4.79
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Home Work Assignment # 15.1
1, 8, 14, 21, 28, 35
Due Sunday 8/4/2012
Saturday, April 07, 2012
Dr. Mohammad Abuhaiba, PE
2 - 12
General Plane Motion
15 - 13
General plane motion is neither a translation nor a rotation.
General plane motion can be considered as
the sum of a translation and rotation.
Displacement of particles A and B to A2
and B2 can be divided into two parts:
1. translation to A2 and
2. rotation of about A2 to B2 1B
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Absolute and Relative Velocity in
Plane Motion
15 - 14
Any plane motion can be replaced by a translation of an
arbitrary reference point A and a simultaneous rotation about A.
ABAB vvv
rvrkv ABABAB
ABAB rkvv
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Absolute and Relative Velocity in
Plane Motion
15 - 15
Assuming that the velocity vA of end A is known, determine the velocity
vB of end B and 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
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Absolute and Relative Velocity in
Plane Motion
15 - 16
Selecting point B as the reference point and solving for velocity vA of end
A and angular velocity leads to an equivalent velocity triangle.
vA/B has same magnitude but opposite sense of vB/A. Sense of relative
velocity is dependent on choice of reference point.
Angular velocity of rod in its rotation about B is the same as its rotation
about A. Angular velocity is not dependent on choice of reference point.
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.2
15 - 17
The double gear rolls on the
stationary lower rack: the velocity
of its center is 1.2 m/s. Determine
a. angular velocity of gear
b. velocities of upper rack R and
point D of gear.
SOLUTION:
The displacement of gear center in
one revolution is equal to the outer
circumference. Relate translational
and angular displacements.
Differentiate to relate translational
and angular velocities.
Velocity for any point P on gear
may be written as
Evaluate the velocities of points B
and D.
APAAPAP rkvvvv
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.2
15 - 18
x
y
SOLUTION:
The displacement of gear center in one revolution is equal to outer
circumference.
For xA > 0 (moves to right), < 0 (rotates clockwise).
1
22rx
r
xA
A
Differentiate to relate translational and angular velocities.
m0.150
sm2.1
1
1
r
v
rv
A
A
kk
srad8
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.2
15 - 19
For any point P on the gear, APAAPAP rkvvvv
Velocity of upper rack is equal to
velocity of point B:
ii
jki
rkvvv ABABR
sm8.0sm2.1
m 10.0srad8sm2.1
ivR
sm2
Velocity of the point D:
iki
rkvv ADAD
m 150.0srad8sm2.1
sm697.1
sm2.1sm2.1
D
D
v
jiv
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.3
15 - 20
The crank AB has a constant cw
angular velocity of 2000 rpm. For
the crank position indicated,
determine
a. angular velocity of connecting
rod BD
b. velocity of piston P
SOLUTION:
Absolute velocity of point D:
BDBD vvv
Velocity is obtained from the
given crank rotation data. Bv
• Directions of the absolute velocity
and relative velocity are
determined from geometry.
Dv
BDv
• Unknowns in the vector expression
are velocity magnitudes
which may be determined from the
corresponding vector triangle.
BDD vv and
• Angular velocity of connecting rod
is calculated from .BDv
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.3
15 - 21
SOLUTION:
Will determine the absolute velocity of point D with
BDBD vvv
The velocity is obtained from the crank rotation
data. Bv
srad 4.209in.3
srad 4.209rev
rad2
s60
min
min
rev2000
ABB
AB
ABv
The velocity direction is as shown.
Direction of absolute velocity is horizontal.
Direction of relative velocity is perpendicular
to BD. Compute angle between horizontal and
connecting rod from the law of sines.
Dv
BDv
95.13in.3
sin
in.8
40sin
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.3
15 - 22
• Determine velocity magnitudes from the vector triangle. BDD vv and
BDBD vvv
sin76.05
sin.3.628
50sin95.53sin
BDDvv
sin.9.495
sft6.43sin.4.523
BD
D
v
v
srad 0.62
in. 8
sin.9.495
l
v
lv
BDBD
BDBD
sft6.43 DP vv
kBD
srad 0.62
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Home Work Assignment 15.2
38, 46, 53, 59, 66, 72
Due Tuesday 10/4/2012
Saturday, April 07, 2012
Dr. Mohammad Abuhaiba, PE
2 - 23
Instantaneous Center of Rotation in
Plane Motion
15 - 24
• Plane motion of all particles in a slab can always be
replaced by the translation of an arbitrary point A
and a rotation about A with an angular velocity that
is independent of choice of A.
• Same translational & rotational velocities at A are
obtained by allowing slab to rotate with same
angular velocity about point C on a perpendicular to
velocity at A.
• Velocity of all other particles in slab are the same as
originally defined since the angular velocity and
translational velocity at A are equivalent.
• As far as velocities are concerned, the slab seems to
rotate about the instantaneous center of rotation C.
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Instantaneous Center of Rotation in
Plane Motion
15 - 25
• If the velocity at two points A and B are known,
the instantaneous center of rotation lies at
intersection of perpendiculars to velocity vectors
through A and B .
• If velocity vectors at A and B are perpendicular to
line AB, instantaneous center of rotation lies at
intersection of line AB with line joining
extremities of velocity vectors at A and B.
• If velocity magnitudes are equal, instantaneous
center of rotation is at infinity and angular
velocity is zero.
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Instantaneous Center of Rotation in
Plane Motion
15 - 26
instantaneous center of rotation lies at intersection of
perpendiculars to velocity vectors through A and B .
cosl
v
AC
v AA
tan
cossin
A
AB
v
l
vlBCv
velocities of all particles on rod are as if they were
rotated about C.
particle at center of rotation has zero velocity. particle coinciding with center of rotation changes
with time and acceleration of particle at instantaneous
center of rotation is not zero.
acceleration of particles in the slab cannot be
determined as if the slab were simply rotating about
C. trace of locus of center of rotation on the body is the
body centrode and in space is the space centrode.
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.4
15 - 27
The double gear rolls on the
stationary lower rack: the
velocity of its center is 1.2 m/s.
Determine:
a. angular velocity of the gear
b. velocities of the upper rack R
and point D of the gear.
SOLUTION:
Point C is in contact with the
stationary lower rack and,
instantaneously, has zero velocity. It
must be the location of the
instantaneous center of rotation.
Determine angular velocity about C
based on the given velocity at A.
Evaluate velocities at B and D based
on their rotation about C.
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.4
15 - 28
SOLUTION:
point C is in contact with stationary lower rack
and, instantaneously, has zero velocity. It must be
the location of instantaneous center of rotation.
Determine angular velocity about C based on the
given velocity at A.
srad8m 0.15
sm2.1
A
AAA
r
vrv
Evaluate velocities at B and D based on their
rotation about C.
srad8m 25.0 BBR rvv ivR
sm2
srad8m 2121.0
m 2121.02m 15.0
DD
D
rv
r
sm2.12.1
sm697.1
jiv
v
D
D
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.5
15 - 29
The crank AB has a constant cw
angular velocity of 2000 rpm. For
the crank position indicated,
determine:
a. angular velocity of connecting
rod BD
b. velocity of piston P
SOLUTION:
Determine velocity at B from the
given crank rotation data.
direction of velocity vectors at B
and D are known. The
instantaneous center of rotation is
at intersection of perpendiculars
to velocities through B and D.
Determine angular velocity about
center of rotation based on
velocity at B.
Calculate velocity at D based on
its rotation about instantaneous
center of rotation.
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.5
15 - 30
From Sample Problem 15.3,
95.13
sin.3.628sin.3.4819.403
BB vjiv
instantaneous center of rotation is at intersection
of perpendiculars to velocities through B and D.
05.7690
95.5340
D
B
sin50
in. 8
95.53sin05.76sin
CDBC
in. 44.8in. 14.10 CDBC
Determine angular velocity about center of
rotation based on velocity at B.
in. 10.14
sin.3.628
BC
v
BCv
BBD
BDB
Calculate the velocity at D based on its rotation
about the instantaneous center of rotation.
srad0.62in. 44.8 BDD CDv
sft6.43sin.523 DP vv
srad0.62BD
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Home Work Assignment 15.3
74, 80, 88, 95
Due Tuesday 17/4/2012
Saturday, April 07, 2012
Dr. Mohammad Abuhaiba, PE
2 - 31
Absolute and Relative Acceleration
in Plane Motion
15 - 32
• Absolute acceleration of a particle of the slab,
ABAB aaa
• Relative acceleration associated with rotation about A includes
tangential and normal components,
ABa
ABnAB
ABtAB
ra
rka
2
2
ra
ra
nAB
tAB
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
15 - 33
Given
determine
, and AA va
. and
Ba
tABnABA
ABAB
aaa
aaa
Vector result depends on sense of and the
relative magnitudes of nABA aa and Aa
Must also know angular velocity .
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Absolute and Relative Acceleration in
Plane Motion
Absolute and Relative Acceleration
in Plane Motion
15 - 34
x components: cossin0 2 llaA
y components: sincos2 llaB
Solve for aB and .
Write in terms of the two component equations, ABAB aaa
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Analysis of Plane Motion in Terms of a
Parameter
15 - 35
In some cases, it is advantageous to determine absolute
velocity and acceleration of a mechanism directly.
sinlxA coslyB
cos
cos
l
l
xv AA
sin
sin
l
l
yv BB
cossin
cossin
2
2
ll
ll
xa AA
sincos
sincos
2
2
ll
ll
ya BB
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.6
15 - 36
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. angular acceleration of the gear
b. acceleration of points B, C, D
SOLUTION:
• The expression of gear position as
a function of is differentiated
twice to define the relationship
between translational and angular
accelerations.
• The acceleration of each point on
gear is obtained by adding
acceleration of the gear center and
relative accelerations with respect
to the center. The latter includes
normal and tangential acceleration
components.
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
15 - 37
SOLUTION:
The expression of the gear position as a function of is differentiated twice to
define the relationship between the translational and angular accelerations.
11
1
rrv
rx
A
A
srad 8m 0.150
sm2.1
1
r
vA
11 rraA
m 150.0
sm3 2
1
r
aA
kk 2srad20
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.6
15 - 38
jii
jjki
rrka
aaaaaa
ABABA
nABtABAABAB
222
222
2
sm40.6sm2sm3
m100.0srad8m100.0srad20sm3
222 sm12.8sm40.6m5 BB ajisa
Acceleration of each point is obtained by adding acceleration of gear center and
relative accelerations with respect to the center. The latter includes normal and
tangential acceleration components.
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.6
15 - 39
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
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.6
Sample Problem 15.7
15 - 40
Crank AG of the engine system has
a constant cw angular velocity of
2000 rpm. For the crank position
shown, determine angular
acceleration of the connecting rod
BD and the acceleration of point D.
SOLUTION:
Angular acceleration of rod BD
and acceleration of point D will be
determined from
nBDtBDBBDBD aaaaaa
Acceleration of B is determined
from given rotation speed of AB.
Directions of the accelerations
are
determined from the geometry.
nBDtBDD aaa
and,,
Component equations for
acceleration of point D are solved
simultaneously for acceleration of
D & angular acceleration of
connecting rod.
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
15 - 41
Acceleration of B is determined from the given
rotation speed of AB.
SOLUTION:
Angular acceleration of connecting rod BD and
acceleration of point D will be determined from
nBDtBDBBDBD aaaaaa
22
1232
AB
sft962,10srad4.209ft
0
constantsrad209.4rpm2000
ABB
AB
ra
jiaB
40sin40cossft962,10 2
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.7
15 - 42
Directions of accelerations are determined from geometry. nBDtBDD aaa
and,,
From Sample Problem 15.3, BD = 62.0 rad/s, = 13.95o.
22
1282 sft2563srad0.62ft BDnBD BDa
jianBD
95.13sin95.13cossft2563 2
BDBDBDtBD BDa 667.0ft128
The direction of (aD/B)t is known but the sense is not known,
jia BDtBD
05.76cos05.76sin667.0
iaa DD
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.7
15 - 43
nBDtBDBBDBD aaaaaa
Component equations for acceleration of point D are solved simultaneously.
x components: 95.13sin667.095.13cos256340cos962,10 BDDa
95.13cos667.095.13sin256340sin962,100 BDy components:
ia
k
D
BD
2
2
sft9290
srad9940
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.7
Sample Problem 15.8
15 - 44
In the position shown, crank AB has
a constant angular velocity 1 = 20
rad/s ccw. Determine angular
velocities and angular accelerations
of connecting rod BD and crank DE.
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
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
15 - 45
SOLUTION:
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 x components:
BDDE 1216017 y components:
kk DEBD
srad29.11srad33.29
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.8
15 - 46
Angular accelerations are determined by simultaneously solving the component
equations for BDBD aaa
jiji
jijik
rra
DEDE
DE
DDEDDED
217021701717
171729.1117172
2
ji
jirra BABBABB
56003200
14820022
jiji
jijik
rra
DBDB
DB
DBBDDBBDBD
2580320,10123
31233.293122
2
x components: 690,15317 BDDE
y components: 60101217 BDDE
kk DEBD
22 srad809srad645
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.8
Home Work Assignment 15.3
105, 112, 119, 126, 133, 141
Due Sunday 22/4/2012
Saturday, April 07, 2012
Dr. Mohammad Abuhaiba, PE
2 - 47
Frame OXYZ is fixed.
Rate of Change With Respect to a Rotating
Frame
15 - 48
Frame Oxyz rotates
about fixed axis OA
with 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 Oxyz and
OXYZQ
QkQjQiQ zyx
Q
With respect to the rotating Oxyz frame,
kQjQiQQ zyx
With respect to the fixed OXYZ frame,
QQQ OxyzOXYZ
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Coriolis Acceleration
15 - 49
Frame OXY is fixed and frame Oxy rotates with
angular velocity .
Position vector for the particle P is the same
in both frames but the rate of change depends on
the choice of frame.
Pr
The absolute velocity of the particle P is
OxyOXYP rrrv
Imagine a rigid slab attached to the rotating frame Oxy or F for short. Let P’ be a point on
the slab which corresponds instantaneously to
position of particle P.
OxyP rv F velocity of P along its path on
the slab 'Pv
absolute velocity of point P’ on the slab
Absolute velocity for particle P may be written as
FPPP vvv
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
15 - 50
FPP
OxyP
vv
rrv
Absolute acceleration for particle P is
OxyOXYP rdt
drra
OxyOxyP rrrra
2
OxyOxyOxy
OxyOXY
rrrdt
d
rrr
but,
OxyP
P
ra
rra
F
• Utilizing the conceptual point P’ on the slab,
• Absolute acceleration for the particle P becomes
22
2
F
F
F
POxyc
cPP
OxyPPP
vra
aaa
raaa
Coriolis acceleration
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Coriolis Acceleration
15 - 51
Consider a collar P which is made to slide at
constant relative velocity u along rod OB. The
rod is rotating at a constant angular velocity w.
The point A on the rod corresponds to the
instantaneous position of P.
cPAP aaaa
F
Absolute acceleration of the collar is
0 OxyP ra F
uava cPc 22 F
The absolute acceleration consists of the radial
and tangential vectors shown
2rarra AA
where
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Coriolis Acceleration
15 - 52
uvvtt
uvvt
A
A
,at
,at
Change in velocity over t is represented by
sum of three vectors
TTTTRRv
2rarra AA
recall,
is due to change in direction of
velocity of point A on the rod,
AAtt
arrt
vt
TT
2
00limlim
TT
result from combined
effects of relative motion of P and rotation of
the rod
TTRR and
uuu
t
r
tu
t
TT
t
RR
tt
2
limlim00
uava cPc 22 F
recall,
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Coriolis Acceleration
Sample Problem 15.9
15 - 53
Disk D of the Geneva mechanism
rotates with constant ccw angular
velocity D = 10 rad/s. At the instant
when = 150o, determine:
a. angular velocity of disk S
b. velocity of pin P relative to disk S
SOLUTION:
The absolute velocity of point P
may be written as
sPPP vvv
Magnitude and direction of
velocity of pin P are
calculated from radius and
angular velocity of disk D.
Pv
Direction of velocity of point
P’ on S coinciding with P is
perpendicular to radius OP.
Pv
Direction of velocity of P
with respect to S is parallel to
the slot.
sPv
Solve the vector triangle for
angular velocity of S and
relative velocity of P. Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
15 - 54
SOLUTION:
Absolute velocity of point P may be written as
sPPP vvv
Magnitude and direction of absolute velocity of pin P
are calculated from radius and angular velocity of
disk D. smm500srad 10mm 50 DP Rv
Direction of velocity of P with respect to S is parallel
to slot. From the law of cosines,
mm 1.37551.030cos2 2222 rRRllRr
From the law of sines,
4.42742.0
30sinsin
30sin
R
sin
r
6.17304.4290
The interior angle of the vector triangle is
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.9
15 - 55
Direction of velocity of point P’ on S coinciding with P
is perpendicular to radius OP. From the velocity
triangle,
mm 1.37
smm2.151
smm2.1516.17sinsmm500sin
ss
PP
r
vv
ks
srad08.4
6.17cossm500cosPsP vv
jiv sP
4.42sin4.42cossm477
smm 500PvDr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.9
Sample Problem 15.10
15 - 56
In the Geneva mechanism, disk D
rotates with a constant ccw
angular velocity of 10 rad/s. At
the instant when j = 150o,
determine angular acceleration of
disk S.
SOLUTION:
The absolute acceleration of the pin
P may be expressed as
csPPP aaaa
The instantaneous angular velocity
of Disk S is 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.
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
15 - 57
SOLUTION:
Absolute acceleration of the pin P may be
expressed as csPPP aaaa
• From Sample Problem 15.9.
jiv
k
sP
S
4.42sin4.42cossmm477
srad08.44.42
• Considering each term in the acceleration eq,
jia
Ra
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: S may be positive or negative
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.10
15 - 58
• The relative acceleration must be parallel to
the slot. sPa
sPv
Direction of Coriolis acceleration is
obtained by rotating direction of the
relative velocity by 90o in the sense of
S.
ji
ji
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
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.10
Home Work Assignment 15.4
150, 158, 166, 172, 179,
Due Tuesday 24/4/2012
Saturday, April 07, 2012
Dr. Mohammad Abuhaiba, PE
2 - 59
Motion About a Fixed Point
15 - 60
The most general displacement of a RB with a fixed
point O is equivalent to a rotation of body about an
axis through O. With the instantaneous axis of rotation and angular
velocity the velocity of a particle P of the body is ,
rdt
rdv
acceleration of particle P is
.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.
Angular acceleration represents the velocity of the
tip of
.
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
General Motion
15 - 61
For particles A and B of a RB,
ABAB vvv
Particle A is fixed within the body and
motion of the body relative to AX’Y’Z’ is the
motion of a body with a fixed point
ABAB rvv
Similarly, acceleration of particle P is
ABABA
ABAB
rra
aaa
Most general motion of a RB is equivalent to:
a translation in which all particles have the same
velocity and acceleration of a reference particle A, and
of a motion in which particle A is assumed fixed.
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.11
15 - 62
The crane rotates with a constant
angular 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:
a. angular velocity of boom,
b. angular acceleration of boom,
c. velocity of boom tip
d. acceleration of 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
jir
kj
639.10
30sin30cos12
50.030.0 21
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
15 - 63
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
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.11
15 - 64
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
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.11
Home Work Assignment 15.5
184, 190, 197, 204, 210
Due Sunday 29/4/2012
Saturday, April 07, 2012
Dr. Mohammad Abuhaiba, PE
2 - 65
3-D Motion. Coriolis Acceleration
15 - 66
With respect to the fixed frame OXYZ and
rotating frame Oxyz,
QQQ OxyzOXYZ
Consider motion of particle P relative to a
rotating frame Oxyz or F for short. Absolute
velocity can be expressed as
FPP
OxyzP
vv
rrv
Absolute acceleration can be expressed as
onaccelerati Coriolis 22
2
F
F
POxyzc
cPp
OxyzOxyzP
vra
aaa
rrrra
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Frame of Reference in General
Motion
15 - 67
Consider:
- fixed frame OXYZ,
- translating frame AX’Y’Z’, and
- translating and rotating frame Axyz or F.
• With respect to OXYZ and AX’Y’Z’,
APAP
APAP
APAP
aaa
vvv
rrr
• The velocity and acceleration of P
relative to AX’Y’Z’ can be found in
terms of the velocity and acceleration
of P relative to Axyz.
FPP
AxyzAPAPAP
vv
rrvv
cPP
AxyzAPAxyzAP
APAPAP
aaa
rr
rraa
F
2
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.15
15 - 68
For the disk mounted on the arm,
the indicated angular rotation
rates are constant. Determine:
a. velocity of point P
b. acceleration of P
c. angular velocity and angular
acceleration of the disk
SOLUTION:
Define a fixed reference frame
OXYZ at O and a moving reference frame Axyz or F attached to the arm
at A.
With P’ of the moving reference
frame coinciding with P, the
velocity of the point P is found
from FPPP vvv
The acceleration of P is found from
cPPP aaaa
F
The angular velocity and angular
acceleration of the disk are
F
FD
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
15 - 69
SOLUTION:
Define a fixed reference frame OXYZ at O and a moving reference frame Axyz or F
attached to the arm at A.
j
jRiLr
1
k
jRr
D
AP
2
F
With P’ of the moving reference frame
coinciding with P, the velocity of the point P
is found from
iRjRkrv
kLjRiLjrv
vvv
APDP
P
PPP
22
11
FF
F
kLiRvP
12
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.15
15 - 70
• The acceleration of P is found from
cPPP aaaa
F
iLkLjraP
2111
jRiRk
ra APDDP
2222
FFF
kRiRj
va Pc
2121 22
2
F
kRjRiLaP
21
22
21 2
• Angular velocity and acceleration of the disk,
FD
kj
21
kjj
211
F
i
21
Dr. Mohammad Abuhaiba, PE
Saturday, April 07, 2012
Sample Problem 15.15