Solution 6.130
222
=+ =
233
22
madx m madx
()
=++=+ +
123
gggg
0mg mg
dx
d
VdVVV dx
=−
33g
82
P
am
dx
P
1
υ
Solution 6.131
Radius to each weight is r = 0.25 + 1.5 sin
θ
in.
θ
+
22
12 0.25 1.5 sin
16 32.2 12
But
δ
r = 1.5 cos
θ
δθ
in.
so 30.625sin15
cos 0.9461, 18.90
3
θθ
===°
()
ωδθ
−
+° °
=−
32
0.25 1.5 sin18.90 1.5 cos18.90
0.3378 10 ft-lb
()
()
()
()
()
21.5 21.5
512 1 cos 1 cos
12 12
e
Vkxx
δδ θδ θ
== − −
Solution 6.132
Sector:
2 32.2 12 32.2
dT I d d d
αθ αθ αθ
== =
32.2 80
Solution 6.133
Upper arm: dU = dT
1.2
9.81
mg = 200(9.81) N
ℓ = 6 m
a
A
B
D
0.1 m
dθ
–
ds
–
D
= 0.3dθ
Solution 6.134
g
dU dT dV
′=+
angular displacement of
dOA
θ
()( )
22
0.3 0.3 0.3
14 0.3 0.03 kg m
12
I
==⋅
0.3 0.3 0.3
α
=2
27.3 rad s
Solution 6.135
2
2
11
=⋅
2
0.373 kg m s CW
Ob
Solution 6.136
2
t
ω
=
32
0
3
1.811 rad s
Solution 6.137
+
H
–
b
O0.3 m
Solution 6.138
π
2
22
Solution 6.139
()
[
()
υ
=Δ
−× = −−
360 35 9.81 5 35 27
Fdt G
ω
=Δ
=
2
56.0 rad s CW
GG
Mdt H
160 N
15(9.81) N
20(9.81) N
8 rad/sec
mm
+
+
200 N
ω
υ
N
F
G
1680
Solution 6.140
1
8
22
66
80 8 6
32.2 12 12
23.4 rad sec
ω
ω
=+
=
mυ1
mgΔt
L/4
L/4
Solution 6.142
G is system mass center.
2
2
32 2 4
11
O
II md mR m
mR
=− = −
=
16 2 4 4
37
O
OO
HI md mR m R
ωυ υ
=+ = +
y
F
T
Oy
T
Solution 6.144
υ
=
0.778 m s
Oi O
OFrtTrtmk
+−=
⎩
⎪
GyFdtGy
yz
t
t
1
1
2
+=
∑
∫
0
OyΔtL/3
Solution 6.145
2
()
2
MmL
+
+
l
O
(1)
r
+Z
I
z
Ω
Solution 6.146
Approximate the diver’s body as a uniform slender bar in the first case and as a sphere in
()( )
=
12
2
2
2
2
25
120.7
20.3
252
N
Solution 6.147
32
2
rev
12
412 2
OO
HH
=
Solution 6.148
12
2
112
22
91.7%
n
=
Solution 6.149
2
1.2 120
t
=+
=
=+
0.02750 m s
0.750 0.01719 N
r
Tt
B
x
ℓ
A
Solution 6.150
2.5 lb ft
W
=
4
xy
==
()
=+=
2
23
15
22
G
ww w
I
Solution 6.151
Let represent the 120-mm distance
d
I1 N1 = I2 N2 (working in rev/min)
2
Solution 6.152
21
Subsequent rotation:
1
mυ1mgΔt
AyΔt
IAω2
GG
AA
+=
2
mg mg
G
G
AA
ω
2
23
υ
–
Solution 6.153
12 2 3
B
Immxmmxmx
=+−=−+
Solution 6.154
From Prob. B.30,
2
2
15
B
yy
mb
IBr
=+
(4) Rod A just outside B.
Note 2
3
AB
mm=
2
22 2 2
7
yy A A A
rb b r b
Im m m
=++ =+
234
y
b
4
b
4
BA
b
4
b
4
1
Solution 6.155
()
00mgcos
t
yyy
o
t
Fdt m N
υυ θ
=−==
Solution of Eqs. 1–3:
()
µ
θθ
υµ
υµθ
=
−
=
−
g7 cos 2sin
5
72tan
o
k
ok
k
t
k
7
k
x
+
y
Solution 6.156
t
ko
Slipping ceases when
υ
= r
ω
(3)
ω
2
o
r
()
k