13″
Solution 6.157
Angular impulse of mg is negligible during impact.
2
1
0:m sin
A
L
HmL
υθ ω
Δ= =
With L
υω
′′
= and
ω
from above,
4L
Solution 6.158
θ
Solution 6.159
Let
ω
o = true angular velocity of disk and armature
OA:
32.2 12
HI
ωωω
== =
22
15 4 15 9
2
π
ω
o
9″
Solution 6.160
motor
d
(mp+mr+mc)g
FdtT
po
22 2
−=− =
⎩
⎪
⎪
and
F
O
x
ωd
Solution 6.161
HH= (for system)
()
ws ws
Mt I
ωω
=+
Solution 6.162
Conservation of angular momentum about the vertical spin axis of the platform:
A
N
2
N
1
= 600 rev/min
Solution 6.163
(H1)disk =
lb-ft-sec
32.2 12 60
22 22
2
22
86 6 24600 86 6 25 ,
N
×+ ×+× =×+ ×+×
F
AB
m
A
,
k
A
,
d
A
A
M
+
B
y
ω
B
B
Solution 6.164
C
C
AB
ABC
d
dd
ωωω
== (4–5)
BC
H M dt H
AAA
t
t
A
z11
2
2
+=
⎧
⎨
⎪
∑
∫
0
H M dt H
FdtF dtmk
BBB
t
t
AB
B
BC
B
BB B
z11
2
22 2
2
+=
−+ =−
()
⎧
⎪
⎩
⎪
∑
∫
H M dt H
CCC
t
t
B
z11
2
23
2
+=
⎧
⎨
⎪
⎩
∑
∫
0
m
C
,
k
C
,
d
C
Solution 6.165
υ
′
2
22
22
22
1
2
BA
B
n
I
ω
ωυ
=
2
12cos
θ
+
3
υ‘
Solution 6.166
()
0:
AA
HmrhI
υω
Δ= −= ′
23 0.035 8.65
O
b
m
Solution 6.167
2
d
G
G’
A’
C
O
υ
A
‘
30°
30°
30°
O
1.2 m 1.2 m
+
Solution 6.168
222 2
υ
′
=
32.5 m s
A
sin30 23
dr=°=
−
12
222
O
22
0.700 1 1 0.7 0.7 0.7
5 9.81 cos30 5 0.7 5 5 9.81 sin30
υω
−°+=+−+°
′
and
υ
ω
== =
3.25 8.04 rad s
0.7 3
r
−=−
Solution 6.169
()
2
22
11
4 4 60 1.2 115.2 kg m
O
Im
== =⋅
:
OO
MI
α
=
Solution 6.170
1
Solution 6.171
r = 40 in.
a
Gn
a
Gt
D
A
A
x
A
y
G
CD
x
y
+
60°
32″
8″
θ
ω, α
16″
Solution 6.172
ω
=
2 0.04167 m
4.94 rad s
O
O
x
0.25 0.125
r
ω
υωω
== =
2
22 2
111 1
()
ω
=
=
2
39.01 rad s
6.25 rad s
r
υ
G
x
45°
υ
O
υ
O
υ
O
υ
O
y
r√—
2
Solution 6.173
2
Solution 6.174
(MBC is the resistive torque in shaft BC)
ω
A =
ω
C =
ω
BC
4500 lb
L
=
==
4500 lb
CD
F
=
0.733 ft sec
υ
Oy
FAB
AB
MMBC
ωA
ωBC
mA, kA
mB, kB
Solution 6.175
AA
Solution 6.176
For the entire spacecraft,
α
−
−
=×
6
12 2
6.67 10 rad s
xx
2
1
2
OO
tt
θθ ω α
=+ +
π
−
=
12 2
11
1206 s
t
Solution 6.177
P
()
:so,0
29 932
xxx
xG G G
FmaOma a
ππ
== =
−
2
2
18 4
π
=− +−
xP P
FF r
Oj kxiy j
()
ππ
π
−
2
2
2
2
33
932
PP
mr
x
y
+rG
F
P
m
y
x
4r
–––
3π
(x
P
, y
P
)
aa
rr
PGPG
O
=+× −
2
Rest
,
PG
Solution 6.178
O
BA
() ()
AO A B B B
Lb
mLbm Lbmb
b
υυυ
−
−= −+
22
0
OO
db Lb b Lb b
nn
== = =
−−
++
for
υ
B max
1,1
nL
Bn
BA
(before) (after)
υ
B
υ
O
n
60°
Solution 6.179
66
tt
Solution 6.180
Bar AB
max 16
fsA
y
a
B
B
x
B
y
B
mg
L
––
2