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Introduction to Dynamics (N. Zabaras)
HW5 Solutions Kinematics Of Rigid Bodies
Prof. Nicholas Zabaras
Warwick Centre for Predictive Modelling
University of Warwick
Coventry CV4 7AL
United Kingdom
Email: nzabaras@gmail.com
URL: http://www.zabaras.com/
March 7, 2016
1
Introduction to Dynamics (N. Zabaras)
Compute wB at t = 5 s
considering that gear A
starts form rest and that
aA= ( 3t + 2 ) rad/s2
dd dt
dt
wa w a
(3 2)d t dtw
0 0
(3 2)
t
Ad t dt
w
w
2
51.5 2 47.5 /A t
t t rad sw
A A C Cr rw w
(47.5)(50) (50)Cw
47.5 /C rad sw
B B C Cr rw w
(75) 47.5(50)Bw
31.7 /B rad sw
Problem 1
2
Introduction to Dynamics (N. Zabaras)
rG = 80 mm
rC= rD = 40 mm
rE = rB = 50mm
rF= 70 mm
G In the reverse gear
transmission shown,
compute wB
G A C Cr rw w
80(40) 40 Cw
80 /C rad sw
80 /D C rad sw w
E E D Dr rw w
(50) 80(40)Ew
64 /E rad sw
64 /F E rad sw w
F F B Br rw w
64(70) 50 Bw
89.6 /B rad sw
Problem 2
3
Introduction to Dynamics (N. Zabaras)
2(0.4 ) /
? 3
? 3
A
C
t rad s
v when t s
start from rest
s in s
a
(0.4 )(0.3) (0.2) 0.6B Bt ta a
dd dt
dt
wa w a
3
2
0 0
0.6 0.3d t dt t
w
w w
23 : 0.3 3 2.70 /BAt t s rad sw
2.70(0.4) 1.08 /C Bv r m sw
dd dt
dt
w w
3
2 3
0 0
0.3 0.1td t dt
3 : 2.7At t s rad
2.7(0.4) 1.08s r m
Problem 3
4
A A B Br ra a
With the information
shown compute the
speed of C and the
distance it travels at
time t=3 secs.
Introduction to Dynamics (N. Zabaras)
Rest
aA= 2 rad/s2
uP=?
aP=?
B = 10 revolutions of B
Problem 4
5
The motor shown in the photo is used to turn a wheel and attached
blower contained within the housing. If the pulley A connected to the motor
begins to rotate from rest with a constant angular acceleration of aA= 2
rad/s2, determine the magnitudes of the velocity and acceleration of point P
on the wheel, after the pulley B has turned ten revolutions. Assume
the transmission belt does not slip on the pulley and wheel.
Introduction to Dynamics (N. Zabaras)
Rest
aA= 2 rad/s2
uP=?
aP=?
B = 10 revolutions of B
2 2( ) ( )
P B B
P P t P n
r
a a a
u w
2 0 2(2)[167.6 0]Aw 25.89 /A rad sw
A A B Br rw w
9.7 /B rad sw
( 9.7 )(0.4) 3.88 /P B Br m su w
2( 0.75)(0.4) 0.3 /t B ta r a m sa
A A B Br ra a
20.75 /B rad sa
2 2 2(9.7) (0.4) 37.636 /n B na r a m sw
2 2 2 2 2( ) ( ) (0.3) (37.636) 37.6 /P P t P na a a m s
A A B Br r [(10)(2 )](0.4)167.6
0.15A rad
Problem 4
6
2 2
0 02 ( )A A Aw w a
Introduction to Dynamics (N. Zabaras)
Problem 5
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) the number of
revolutions of the pulley in 2 s, (b)
the velocity and change in position of
the load B after 2 s, and (c) the
acceleration of the point D on the rim
of the inner pulley at t = 0.
SOLUTION:
• Due to the action of the cable, the
tangential velocity and acceleration
of D are equal to the velocity and
acceleration of C. Calculate the
initial angular velocity and
acceleration.
• Apply the relations for uniformly
accelerated rotation to determine
the velocity and angular position of
the pulley after 2 s.
• Evaluate the initial tangential and
normal acceleration components
of D.
7
Introduction to Dynamics (N. Zabaras)
Problem 5SOLUTION:
• The tangential velocity and acceleration of D are equal
to the velocity and acceleration of C.
srad4
3
12
sin.12
00
00
00
r
v
rv
vv
D
D
CD
w
w
2srad33
9
sin.9
r
a
ra
aa
tD
tD
CtD
a
a
• Apply the relations for uniformly accelerated rotation to
determine velocity and angular position of pulley after 2 s.
srad10s 2srad3srad4 20 taww
rad 14
s 2srad3s 2srad422
212
21
0
tt aw
revs ofnumber rad 2
rev 1rad 14
N rev23.2N
rad 14in. 5
srad10in. 5
w
ry
rv
B
B
in. 70
sin.50
B
B
y
v
8
Introduction to Dynamics (N. Zabaras)
Problem 5
• Evaluate the initial tangential and normal
acceleration components of D.
sin.9CtD aa
2220 sin48srad4in. 3 wDnD 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
9
Introduction to Dynamics (N. Zabaras)
Problem 6
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 rack R and point D 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 point P on the
gear may be written as
Evaluate the velocities of points B
and D.
APAAPAP rkvvvv
w
10
Introduction to Dynamics (N. Zabaras)
Problem 6
x
y
SOLUTION:
• The displacement of the gear center in one
revolution is equal to the outer circumference.
For xA > 0 (moves to right), w < 0 (rotates
clockwise).
1
22rx
r
xA
A
Differentiate to relate the translational and
angular velocities.
m0.150
sm2.1
1
1
r
v
rv
A
A
w
w
kk
srad8ww
11
Introduction to Dynamics (N. Zabaras)
Problem 6• For any point P on the gear, APAAPAP rkvvvv
w
Velocity of the upper rack is equal
to velocity of point B:
1.2 m s 8rad s 0.10 m
1.2 m s 0.8m s
R B A B Av v v k r
i k j
i i
w
ivR
sm2
Velocity of the point D:
1.2 m s 8rad s 0.150 m
D A D Av v k r
i k i
w
sm697.1
sm2.1sm2.1
D
D
v
jiv
12
Introduction to Dynamics (N. Zabaras)
Problem 7
The crank AB has a constant
clockwise angular velocity of 2000
rpm.
For the crank position indicated,
determine (a) the angular velocity
of the connecting rod BD, and (b)
the velocity of the piston P.
SOLUTION:
• Will determine the absolute velocity
of point D with
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 .BDv13
Introduction to Dynamics (N. Zabaras)
Problem 7
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 w
w
The velocity direction is as shown.
• The direction of the absolute velocity is
horizontal. The direction of the relative velocity
is perpendicular to BD. Compute the angle between
the horizontal and the connecting rod from the law of
sines.
Dv
BDv
95.13in.3
sin
in.8
40sin
14
Introduction to Dynamics (N. Zabaras)
Problem 7
• Determine the 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
w
w
sft6.43 DP vv
kBD
srad 0.62w
15
Introduction to Dynamics (N. Zabaras)
?
?
BC
wheel
w
w
v x rB AB Bw
v ( 30k) x (0.2cos 60 i 0.2sin 60 j)o o
B
v 5.2i 3.0 jB
/v v x rC B BC C Bw
i 5.20i 3.0 j ( k) x (0.2i)C BCu w
i 5.20i (0.2 -3.0) jC BCu w
5.2 /
15 /
C
BC
m s
rad s
u
w
5.252 /
0.1
cwheel rad s
r
uw
i
jk +
Problem 8
16
The bar AB of the linkage has a clockwise
angular velocity of 30 rad/s when = 60°.
Determine the angular velocities of member BC
and the wheel at this instant.
Introduction to Dynamics (N. Zabaras)
?Cu
Problem 9
17
The cylinder rolls without
slipping between the two moving
plates E and D. Determine the
angular velocity of the cylinder
and the velocity of its center C.
Since no slipping occurs, the
contact points A and B on the
cylinder have the same velocities as
the plates E and D. respectively.
Furthermore, the velocities vA and
vB are parallel, so that by the
proportionality ol' right triangles the
IC is located at a point on line A B.
Assuming this point to be a distance
x from B.
Introduction to Dynamics (N. Zabaras)
; 0.4B x xu w w
0.25 (0.25 )xw
0.154x m
0.42.6 /
0.154
B rad sx
uw
(0.25 )A m xu w
?Cu
Problem 9
18
/ 2.6(0.154 0.125) 0.075 /C C ICr m su w
The cylinder rolls without
slipping between the two moving
plates E and D. Determine the
angular velocity of the cylinder
and the velocity of its center C.
Introduction to Dynamics (N. Zabaras)
Problem 10
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 rack R
and point D of the gear.
SOLUTION:
• The 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 the angular velocity about C
based on the given velocity at A.
• Evaluate the velocities at B and D
based on their rotation about C.
19
Introduction to Dynamics (N. Zabaras)
Problem 10SOLUTION:
• The 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 the angular velocity about C based on
the given velocity at A.
srad8m 0.15
sm2.1
A
AAA
r
vrv ww
• Evaluate the velocities at B and D based on their
rotation about C.
srad8m 25.0 wBBR rvv
ivR
sm2
srad8m 2121.0
m 2121.02m 15.0
wDD
D
rv
r
sm2.12.1
sm697.1
jiv
v
D
D
20
Introduction to Dynamics (N. Zabaras)
Problem 11
The crank AB has a constant
clockwise angular velocity of 2000
rpm.
For the crank position indicated,
determine (a) the angular velocity
of the connecting rod BD, and (b)
the velocity of the piston P.
SOLUTION:
• Determine the velocity at B from the
given crank rotation data.
• The direction of the velocity vectors
at B and D are known. The
instantaneous center of rotation is at
the intersection of the perpendiculars
to the velocities through B and D.
• Determine the angular velocity about
the center of rotation based on the
velocity at B.
• Calculate the velocity at D based on
its rotation about the instantaneous
center of rotation.
21
Introduction to Dynamics (N. Zabaras)
Problem 11SOLUTION:
• From an earlier Sample Problem,
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
through B and D.
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 at B.
in. 10.14
sin.3.628
BC
v
BCv
BBD
BDB
w
w
• Calculate the velocity at D based on its rotation
about the instantaneous center of rotation.
srad0.62in. 44.8 BDD CDv w
sft6.43sin.523 DP vv
srad0.62BDw
22
Introduction to Dynamics (N. Zabaras)
Problem 12
Crank AG of the engine system has
a constant clockwise angular
velocity of 2000 rpm.
For the crank position shown,
determine the angular acceleration
of the connecting rod BD and the
acceleration of point D.
SOLUTION:
• The angular acceleration of the crod
BD and the acceleration of point D
will be determined from
nBDtBDBBDBD aaaaaa
• The acceleration of B is determined
from the given rotation speed of AB.
• The 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
and angular acceleration of the
connecting rod.
23
Introduction to Dynamics (N. Zabaras)
Problem 12
• The acceleration of B is determined from the given
rotation speed of AB.
SOLUTION:
• The angular acceleration of the connecting rod BD
and the acceleration of point D will be determined
from
nBDtBDBBDBD aaaaaa
22
1232
AB
sft962,10srad4.209ft
0
constantsrad209.4rpm2000
ABB
AB
ra w
a
w
jiaB
40sin40cossft962,10 2
24
Introduction to Dynamics (N. Zabaras)
Problem 12
• The directions of the accelerations
are determined from the geometry.
nBDtBDD aaa
and,,
From an earlier problem, wBD = 62.0 rad/s, = 13.95o.
22
1282 sft2563srad0.62ft BDnBD BDa w
jianBD
95.13sin95.13cossft2563 2
BDBDBDtBD BDa aaa 667.0ft128
The direction of (aD/B)t is known but the sense is not known,
jia BDtBD
05.76cos05.76sin667.0 a
iaa DD
25
Introduction to Dynamics (N. Zabaras)
Problem 12
nBDtBDBBDBD aaaaaa
• Component equations for acceleration of point D are
solved simultaneously.
x components:
95.13sin667.095.13cos256340cos962,10 BDDa a
95.13cos667.095.13sin256340sin962,100 BDa
y components:
ia
k
D
BD
2
2
sft9290
srad9940
a
26
Introduction to Dynamics (N. Zabaras)
Problem 13
In the position shown, crank AB has
a constant angular velocity w1 = 20
rad/s counterclockwise.
Determine the angular velocities
and angular accelerations of the
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
27
Introduction to Dynamics (N. Zabaras)
Problem 13
SOLUTION:
• The angular velocities are determined by
simultaneously solving the component equations for
BDBD vvv
ji
jikrv
DEDE
DEDDED
ww
ww
1717
1717
ji
jikrv BABB
160280
14820
w
ji
jikrv
BDBD
BDBDBDBD
ww
ww
123
312
BDDE ww 328017 x components:
BDDE ww 1216017 y components:
kk DEBD
srad29.11srad33.29 ww
28
Introduction to Dynamics (N. Zabaras)
Problem 13• The angular accelerations are determined by
simultaneously solving the component equations for
BDBD aaa
jiji
jijik
rra
DEDE
DE
DDEDDED
217021701717
171729.1117172
2
aa
a
wa
ji
jirra BABBABB
56003200
14820022
wa
jiji
jijik
rra
DBDB
DB
DBBDDBBDBD
2580320,10123
31233.293122
2
aa
a
wa
x components: 690,15317 BDDE aa
y components: 60101217 BDDE aa
kk DEBD
22 srad809srad645 aa
29
Introduction to Dynamics (N. Zabaras)
2
B/A /a a r rB A B Aa w
2cos 45 i sin 45 j 3cos 45 i 3sin 45 j ( k) x(10i) (0.283) (10i)o o o o
B Ba a a
2cos 45 3cos 45 (0.283) (10) [i]o o
Ba
sin 45 3sin 45 (10) [j]o o
Ba a
21.87 /Ba m s
20.344 /rod rad sa
Problem 14
30
The rod AB is confined to
move along the inclined
planes at A and B. If point A
has an acceleration of 3
m/s2 and a velocity of 2 m/s2,
both directed down the plane
at the instant the rod is
horizontal determine the
angular acceleration of the
rod at this instant.
0.283 /
?rod
rad s
at horizontal
w
a
Introduction to Dynamics (N. Zabaras)
2 2(4 / )(0.15 ) 0.6m/soa r rad s ma
2
B/O /a a r rB O B Oa w
2a 0.6i (4k) (0.15i) (6) (0.15i)B
2a { 6i 0.6 j} m/sB
2
A/O /a a r rA O A Oa w
2a 0.6i (4k) (0.15j) (6) (0.15j)A
2a { 1.2i 5.4 j} m/sA
aA=?
aB=?
Problem 15
31
Introduction to Dynamics (N. Zabaras)
2 2(4 / )(150mm) 600 /sGa r rad s mma
2
B/G /a a r rB G B Ga w
2600j ( 4k) (225j) (3) (225j)Ba
900 2625Ba i j
2 2 2(0.9) (2.625) 2.775m/sBa
1 2.625tan 71.1
0.9
o
?Ba
Problem 16
32
The spool "appears" to be rolling downward without
slipping at point A.
Introduction to Dynamics (N. Zabaras) 2rad/s95ABa
?CBa
?ABa
Problem 17
33
The collar C moves downward with an
acceleration of l m/s2.At the instant
shown, it has a speed of 2 m/s which
gives links CB and AB an angular
velocity wAB=wCB=10 rad/s.
Determine the angular accelerations of
CB and AB at this instant.
Introduction to Dynamics (N. Zabaras)
2
A B/A /a a r rB AB AB B Aa w
2a 0 ( k) ( 0.2 j) (10) ( 0.2 j)B ABa
a 0.2 i 20 jB ABa
2
B/C /a a r rB C CB CB B Ca w
20.2 i 20j 1j ( k) (0.2i 0.2 j) (10) (0.2i 0.2 j)AB CBa a
0.2 i 20j 1j 0.2 j 0.2 i 20i 20jAB CB CBa a a
0.2 0.2 20AB CBa a
20 1 0.2 20CBa 25rad/sCBa
2rad/s95ABa
?CBa
?ABa
Problem 17
34
Introduction to Dynamics (N. Zabaras)
ac =?
aBC= ?
Problem 18
35
The crankshaft AB turns with a
clockwise angular acceleration of
20 rad/s2.
Determine the acceleration of the
piston at the instant AB is in the
position shown. At this instant wAB =10
rad/s and wCB= 2.43 rad/s.
Introduction to Dynamics (N. Zabaras)
/ { 0.25sin 45 i 0.25cos 45 j} { 0.177i 0.177 j} fto o
B Ar
/ {0.75sin13.6 i 0.75cos13.6 j} {0.176i 0.729j} fto o
C Br
2
A B/A /a a r rB AB AB B Aa w
2a 0 ( 20k) ( 0.177i 0.177 j) (10) ( 0.177i 0.177 j)B
2a {21.2i 14.14j} ft/sB
2
C/B /a a r rC B BC BC C Ba w
2j 21.21i 14.14 j ( k) (0.176i 0.729j) (2.43) (0.176i 0.729j)C BCa a
0.17 18.45C BCa a 0 20.17 0.729 BCa
227.7 rad/sBCa 213.6 ft/sCa
ac =?
aBC= ?
Problem 18
36
Introduction to Dynamics (N. Zabaras)
Problem 19
In the Geneva mechanism, disk D
rotates with a constant counter-
clockwise angular velocity of 10
rad/s. At the instant when j =
150o, determine the 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 discussed in a
sample problem in the lecture notes.
• 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.
37
Introduction to Dynamics (N. Zabaras)
Problem 19SOLUTION:
• Absolute acceleration of the pin P may be expressed
as csPPP aaaa
• From the sample problem in the corresponding lecture:
jiv
k
sP
S
4.42sin4.42cossmm477
srad08.44.42 w
• Considering each term in the acceleration equation,
jia
Ra
P
DP
30sin30cossmm5000
smm5000srad10mm500
2
222w
jia
jira
jira
aaa
StP
StP
SnP
tPnPP
4.42cos4.42sinmm1.37
4.42cos4.42sin
4.42sin4.42cos2
a
a
w
note: aS may be positive or negative
38
Introduction to Dynamics (N. Zabaras)
Problem 19
• 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
velocity by 90o in the sense of wS.
ji
ji
jiva sPSc
4.42cos4.42sinsmm3890
4.42cos4.42sinsmm477srad08.42
4.42cos4.42sin2
2
w
• Equating components of the acceleration terms
perpendicular to the slot,
srad233
07.17cos500038901.37
S
S
a
a
kS
srad233a
39