A
y
B
β
Solution 5.113
From the solution to Prob. 5.82:
β
ω
=° =44.4 , 4.37 rad s CW
AB
2
aaa a r r
αω
=+ =+ × − (1)
Terms in Eq. 1:
2
BB
aai
arr
αω
=
=×−
()
CW
(left)
A
O
0.8 m
PB
Solution 5.114
υ
==
2
2m s, 5m s
OO
a
P is ICR, so
ω
==
0.48
O
OP
2
13.89 3.33 m sec
ij
=− +
Solution 5.115
2
1.25
R
19.36 ft sec
B
υ
()()
αω
=+
+=+×−
2
BABA
BA BA
BBA
tn
a
aa
aaa r r
B
R/2
a
β
β
x
Solution 5.116
From
212
Orr
From 2
4
,24rad/secCCW
212
O
O
a
ar r
αα
====
2
24 270 ft sec
DO DO
DODOO
ij
aaa a r r
αω
=− −
=+ =+× −
x
y
O
A
D
P
10″
2″
4 ft/sec
2
3 ft/sec
B
A
60°
x
y
Solution 5.117
ω
ω
=
=
15.18 rad s CW
11.79 rad s CCW
OB
2
BA
BABAAAB AB B A
aaa a r r
αω
=+ =+ × − (1)
()
CW
CCW
A
y
Solution 5.118
From the solution to Prob. 5.90:
Note that
α
of the sector is zero.
Numbers:
22
1.5 1.5
0.18 0.3
BD AB
With the usual solution process:
() ()
α
α
==−
22
7.44 rad s , 14.38 rad s
CCW CW
BD AB
Solution 5.119
Carry out vector algebra and equate coefficients:
()
CW
y
Solution 5.120
Using C as the instant center for AB gives
6
0.1
BAO
()()()()
()()
+=++ +
22
AABBABAB
nt nt nt
aaaaa a
Substitute and equate i and jcoefficients and get
() ()()
()
=−=− =−+
22
150
225 320 480 m s , 120 4 3 m s
150
D
Btn
DD
t
aiiaij
Solution 5.121
BO
0.429 rad sec
AC
ω
2
2
(1)
AB AB
ABABB AB AB
aaa a r r
αω
=+ =+ × −
Solution 5.122
ω
=
10 rad s CCW
O
()
both CCW
C
B
15°
Solution 5.123
12.84
=°
2
18
157.9 36.9 cos 12.84 8.41 sin 12.84
195.8 in. sec
A
AB
a
=+ °+ °
=
A
υ
υ
A
18″
4″
4″
βω = 60 rev/min
2
2
D
0.2
Solution 5.124
()
()
()
()
ω
αα
=− = −
=×= −−
22
2
0.0816 0.3 0.05 m s
0.3 0.05 m s
BA
AB AB
n
BA
AB AB AB
t
ar ij
ar ji
Substituted into *, equate like coefficients, and obtain
α
α
=− =−
22
0.0519 rad s , 1.186 rad s
OA AB
y
υ
A
β
Solution 5.125
Equation 5.12: relAB r
υυω υ
=+×+
=
+× +
=
20.1250.1
Ok i i
tan 0.1
β
=
Solution 5.126
Attach Bxy to the sector with origin and axes as shown in the textbook figure.
rel
COR rel
AB
r
υυω υ
=+×+
x
+
Solution 5.127
For the coordinates, the no-slip constraints are 0r
υω
=− and 0
ar
α
=− . So
ω
α
=− =− =−
2
0
516.67 rad s
0.30
a
r
Use the frame Oxy as disk-fixed.
()
0rel
5.12 :
A
r
υυω υ
=+×+
υ
=− =
=− −
0
2
0
2
rel
22
3ms 0.24 m
70.24
irj
=− 2
2 0.667 m s
A
a
ij
Solution 5.128
Attach axes Bxy to the rider at B as shown in the textbook figure.
rel
25
AB r
υυω υ
=+++
x
100 m
= 20 m/s
Solution 5.129
vv rv
ω
=+×+
ω
=
==−
so 0.2 rad s
a
nd 30 m
AB
k
rr j
Solution 5.130
Refer to figure and solution for Prob. 5.129 where
ω
=−
=
rel 46 m s
0.2 rad s const.
vi
k
()
()()
ω
ω
×=
×= ×− =−
=− − − + =
2
rel
2
rel
0
2 2 0.2 46 18.4 m s
So 4 4 1.2 18.4 9.2 m s
r
vkij
ajjjjj
Solution 5.131
rel
cor 2
22
a
kuj ui
ωυ
ωω
=×
=×=−
yx
u
d
L
Solution 5.132
()
1.44 0.658
ijR
=− +
Attach axes x–y to locomotive at B.
rel
512: AB r
υυω υ
=+×+
υ
rel 18.33 ft secj
5.14: 2
rel rel
2
AB
oo
aa rr a
ωωωυ
=+×− +×+
2
2
2
RR
ka
υυ
υυ
+− × +
B
υ
B
Rcos 20° +
2
(R – Rsin 20°)
20°
x
y
z
Solution 5.133
()
()
=−=−
540 1000 150 m s
B
v
ii
ρ
400
AB
() ()
ω
ω
××= ×− =
2
337.5 14.06 m s
8
rj i k
rel
Solution 5.134
Let the axes Bxy be attached to the sector.
rel
rel
512:
B
A
AO
OC
r
rOkBAi i
υυω υ
ωωυ
=+×+
×=+× +
Solution 5.135
rel
rel 0.577 m s (from above), and
Solve to obtain
()
α
=
=−
2
2
rel
5.70 rad s CCW
6.03 m sa
y
x
30°
A
OA = 0.5 m
C
z
υ
Solution 5.136
Solution 5.137
rel 2.71 0.259 m sij
π
rel 0.864 0.0642 m s