Solution 6.84
2
2
5
Imr=
ααα
=+ =−= =
25
:57
GOGO
a
aaa aar r r
αθαα
θ
=+ = + −
22
2
:mg sin cos
5
5gsin cos
O
M I ma d r mr mr ma r
ma Iα
mg
O
POO
G
a
O
=a
ma
Iα
mg mrα
G
G
OO
R
P
r
θ
Solution 6.85
()
2
1gsin cos
O
O
rar
k
dd
αθθ
θθ θθ
=+
=
()
2
1gsin cos
O
OO
drard
k
ω
θθ θ θ
θ
=+
2
1g 1 cos sin where
ωθθ
2
mrα
ma
o
a
o
x
O
O
x
y
G
O
Iα
α
= θ
··
ω
= θ
·
r
Solution 6.86
()()
ωυυ
ωπ
=+ = = =
== =
2232
; since
1.7 100 13.98 10 ft sec
AB AB AB AB AB
tn
aa a a AB O
ar
()
32
42.5 10 rad sec CW
=
()
3
32
13.98 10 tan
A
a
β
=
aA
A
(a
A/B
)
t
β = cos–1 1.7
4.3 = 66.7°
B
x
B
x
()
3
22 3
1.80
32.2
Rod; Eq. 6.3
1.20
BBB
MI ma
αρ
=+×
Solution 6.87
()
ω
πθ
== =°=
60 rev min 2 rad s constant , 45 , 7 kg
OA AB
m
Orientation:
190 80 cos 240 cos 200 cos 17.62
OC OA AB BC
rrrr
i
θφψ
φ
=++
=− = − −
=°
→
υθ ω ψ ω φ ω
→
=+ =
: cos 0.200 cos 0.240 cos 2.29 rad s CCW
ABCAB
BC
j
22
2
Bn AB AB
22
: sin cos 0.2 sin 0.2 cos 0.240 sin 0.240 cos
At An BC BC AB AB
ia a
θθα
ψ
ω
ψ
α
φ
ω
ψ
−− =− − + −
+
x
y
OA = 80
a
Bt
a
At
= 0
a
Bn
a
An
O
A
C
B
υ
B
ω
OA
, α
OA
υ
A
ψ
ψ
240
190
Dimensions in mm
70
200
ϕ
θ
θ
22
ψ
ψ
=−=
=−+=
:cos
:sinmg
xGxBC xGx
yGyBC Gy
FmaF Ama
FmaF Ayma
=− −
2
1.788 1.581 m s
G
aii
A
x
O
y
O
x
O
A
M
θ
Solution 6.88
ω
πθ
== ====°60 rev min 2 rad s , 7 kg, 1.2 kg, 1.8 kg, 45
OA AB OA BC
mm m
(Constant)
()
ωπ
== →=
2
22
0.080 2 3.16 m s
AN OA An
aOA a
Orientation:
rrrr=++
AB AB
υθ ω
ψ
ω
φ
ω
υθ ω ψ ω φ ω
−=− + =
→
=+ =
: sin 0.200 sin 0.240 sin 1.376 rad s CCW
: cos 0.200 cos 0.240 cos 2.29 rad s CCW
ABCAB
AB
i
j
2
22
AB AB
AB AB AB
+
x
a
Bt
a
Bn
a
O
C
υ
ω
OA
, α
OA
υ
A
190
Dimensions in mm
70
200
θ
ψ
2
1
CxyBCBCBC
C
()()
33, 3,
sin cos cos sin
0.671 0.465 m s
tn
GGG
aa i ja i j
ij
ψψ ψψ
=− + +− −
2
1, 1,
1, 1,
0
1
xx
yy
xGxxOAG
yGyOAyOAG
OO OA y x OA OA
()()
ωθθ πθθ
=−−= −−
1
2
2
1
2
cos sin 0.04 2 cos sin
GOA
aOG i i i j
ABCO
m
OA
g
OG
1
= G
1
A = 40 mm
A
y
O
y
θ
GO L/4
1
m
2m
: Inverted
2
1
Solution 6.89
υ
−
+=
112 2
22
17
TT
L
3g
14
L
=
=+ =
2
22
17
Note 12 4 48
O
L
ImLm mL
Solution 6.90
υ
−
+=
112 2
2
22
11
TT
L
Solution 6.91
υ
−′=Δ +Δ
12 g
22
140 mg51cos
TV
Solution 6.92
υυ
−
+=
+= =
112 2
2
22
11 0.3
S
o 0.955 95.2 10.62 , 3.01 m s
TU T
Solution 6.93
ω
== =3 kg, 4 rad s, 75 mmmr
rkImg r
TO
22
1
2
1
2
3 9 81 0 075
−+ = +
−
()
mg
..
(()
+
=
()( )
()
+
()
×
1
22
1
22 3 0 075 4 3 9 81 2 0 075
2
22
k
T
..
.
=⋅k
T
315.Nmrad
Solution 6.94
ω
=
2
2
32 g
Oy
Solution 6.95
g 7 lb, g 24 lb
BD
mm==
2
11020
gg
21212
OB D
OI m m
ω
=−−
Solving, 6.23rad sec CCW
ω
=
22
10 6.23 32.3 ft sec
== ↑
G
1
‘
G
2
‘
,
A’
G
1
G
2,
A
m
B
m
D
ω
V
g
= O
10″
6″
10″
O
Oy
Solution 6.96
UT
=Δ
B
Solution 6.97
The general length L of the spring is
22
220.6 20.6 cos90L
===
1.2 kg, 0.8 ,g 9.81m s
bar
mLm
ω
=
22
22
2.42 rad s CW
O
mg
A
Solution 6.98
′′
== = = =
m
g 60 lb, 300 ft–lb, 35.5 in., 3.17 , 35.5 in.
O
Erhk
=→=
=
mg 300 60
so 5ft.
Eh h
h
O
G
Solution 6.99
()
ω
ω
=−=
2
3g g
51, 0.861
5
bb
Solution 6.100
112 2
TT
υ
−
+=
()
()( )
πθ
θ
ωωω
−
−
=
== =
1
252
0.1
0.1
2
222
2
0
11
50 0.4 4
T
TI
b
mg
1
O
b
2bmg
2
5 b
2
Solution 6.101
2O
Substitute and solve,
ω
=4.40 rad s
:mgcos45
123.2 N
nGnC Gn
C
FmaN ma
N
=−°=
=
r
M
O’
G’
R
M
C
G
V
g
= O
υ = rω
ϕ = 45°
r
r
N
c
F
f
Solution 6.102
ω
θθ
′
======°=°
m
g 40 lb, g 12 lb, 4 ft, 6 in. 4.5 in., 60 , 45mLrk
222 222 222
11 1 1
221222
2
Lm L mL md m k r m k r
ω
Substitute and solve, 9.04 rad sec and 1.597 rad sec CW
AB
ωω ω
== =
m
ω
, k
r
L
A’
A
G
C
υ
A
0.25 m
Solution 6.103
Power
()
EnergydE
pdt t
Δ
==
2
22
2
22
11 2
π
()
7.457 10
Solution 6.104
For system,
Δ
+Δ = =
υ
=
2
0.735
A
()
()
υ
ω
Δ=× − =−
Δ= = ××
g
2
2
2
H
oop : 4 9.81 0.5 19.62
11
2 4 0.25
A
C
VJ
TI
AAA
M
8″
Solution 6.105
2
81
r
ππ
22323
222
8 1 3 8 32 g g 32
rmr rr
()
ππ
=+
−
128
mg 1 39 16
N
Solution 6.106
π
υθ
′== =
72 4.712 ft-lb
12 4
81
M
O
G
mg
8r
ω