Solution 5.77
Solution 5.78
A
CO
8 rad/s
υ
A
B
υ
B
= 0.5 m/s
15°
60°
0.8
BC
1
O
υ
A
υ
B
10″
rad
sec
Solution 5.80
()
1
1
10 0.8 8 in. sec
20 cos30 17.32 in.
cos 20
6.30 10 16.30 in.
Br
AO
AC AO
CB
υω
== =
=°=
=°
=+=
()
ω
υ
υω
===
== =
1
1
80.491 rad s
16.30
18.43 0.491 9.04 in. sec
AB B
AAB
BC
AC
C
ω
Solution 5.81
=
0.416 m s right
Solution 5.82
υ
A
A
P
0.5 m
45°
Solution 5.83
()( )
=× °+ °=
22
1 cos 45 0.5 sin 45 0.791 m
C
P
Solution 5.84
3 cos30 3 sin30 tan15 2.20 in.
BC
=°−°°=
′
Solution 5.85
Point P is the instantaneous center
()
υ
υ
ωω
υ
ω
==
== + =
22
3
,212
210
18 15.30 ft sec
O
C
OP
CP
Solution 5.86
()
22
103 8 0.45 46.5 ft sec
B
BC
υω
==+ =
AB
D
100′
(not to scale)
C
8‘
υ
B
A
10″
υ
υ
C
υ
B
Solution 5.87
υ
1.414 11.79 rad s CCW
0.12
BAB
O
B
Solution 5.89
υB
υA
0.36 m
A
B
A
C
4″
20.9
υ
S
rad
sec
=
Solution 5.90
()
υ
==
0.4
A
Solution 5.91
99
Solution 5.92
ω
== =
0.1732
A
D
AC
C
D
B
P
1.5 m/s
υ
A
A
Solution 5.93
1
15.00 tan 17.14
64.63
CO
AC
=°
+
Solution 5.94
C
D
B
υ
B
υ
υ
D
υβ
β
αγ
90 mm
120 mm
60 mm
535.6
DA
CA
Solution 5.96
17.5 34.9
16.6 1.10 rad/sec CW
15
E
AE
AE
υ
==
D
E
A
C2
C1
O
B
υB
60°
Solution 5.97
()
=
2
3.2 m s
AO t
a
AB
a
B
a
O
= 8 m/s
2
(a
A/O
)
n
(a
B/O
)
n
(a
B/O
)
t
2
Solution 5.98
0.667 ft sec
A
a=
2
10 10.67
A
a=+
y
a
C
Solution 5.100
In the coordinates shown, the no-slip kinematic constraints are ,.
oo
ra r
υω α
=− =−
υ
3
0.4
o
r
υ
υυ υω
=+ =+×
=− +
2
16.25 2.5 m s
Ao
AoAoo
r
ij
Solution 5.101
CAB
A
Solution 5.102
From the solution to Prob. 5/66:
A
D
O
B
a
υ
x
y
r
A
B
L
15°
υ
A
, a
A
x
y
Solution 5.104
CW
2
CW
2
r
a
r
υ
ω
α
=
=
2
2
22
2
BA BA
BA
arr
rr
iaj
r
αω
υ
=× −
=− −
Solution 5.105
From the solution to Prob. 5.58:
tan15 CCW
A
L
υ
ω
°
=
2
BA
BABAA BA
aaa a r r
αω
=+ =+× −
B
(a
A/B
)
n
= r 2ω
45°
Solution 5.106
AB
Solution 5.107
ω
=+ ×
AB
AB AB
vv r
AB A
y
Solution 5.108
From Prob. 5.63, 10.93
ω
= rad/sec CCW
()()
()
2
2
10.93 0.25 cos30 sin30
CB CB
CBCBB
C
aaa a r r
ij
αω
=+ =+× −
−°−°
Carry out cross product and equate coefficients to obtain
α
=− =−
22
126.0 rad s , 43.1 ft sec
C
a
Then 2
AB AB
ABABB
aaa a r r
αω
=+ =+× −
()
()
()
()
2
2
22 2
1.2 126.0 0.25 cos30 sin30
10.93 0.25 cos30 sin30
41.6 13.55 ft sec
41.6 13.55 43.8 ft sec
A
jk ij
ij
ij
a
=+− × − °+ °
−−°+°
=+
=+=
2
15°
A
x
3″
AB
200 mm 150 mm
C
O
ω
2
υ
C
= 0.4 m/s
Solution 5.109
12
0.3 0.3 0.3 3
x
0.2 m
Solution 5.111
22
2
a
C
r
a
rj
Rr
ω
ω
=−
+
Solution 5.112
AB
2
BA BA
BABAAAB AB
aaa a r r
αω
=+ =+ × −
Carry out the cross product and equate like coefficients to obtain
()
α
=−
=
2
2
10.47 m s
9.98 rad s
CCW
B
AB
a
(a)
C