Solution 6.34
90 rad
π
θ
=°= m = 5 kg, 15 N m rad
k=⋅ , r = 150 mm
2
2
:mg2
y
yyy G
OO T
r
MI k mr
αθ α
π
=−=
Solving, 62.9 N
x
O
=
=
()
=
2
2
12.59 m s right
x
G
a
Solution 6.35
2
IImd
=+
32.2
x
+
r
Solution 6.36
()
()
2
22
15
2
O
Imbb mb=+=
Oy
At
Solution 6.37
Solution 6.39
(a) IO = 0, aA = aB = a
(b)
()( )
2
22
11
5 0.25 0.1562 kg m
22
O
Imr== = ⋅
Solution: a = 1.570 m/s2, 49.4 N
45.5 N
B
T
=
α
== 2
1.570 6.28 rad s CCW
0.25
O
R
Mf
15″
Solution 6.40
α
=
32g
7
b
Solution 6.41
()
2
20 15 12 25 lb-ft
f
M
==
Solution 6.42
223
aa
α
==
1
Minimum to rotate :m
211
LLL
33 363 333333
Mass:
11
y
yB C
Solving,
()
α
=
=−
2
91.1 N
0.701 rad s CW
B
T
2
x
y
+
A
NA
OyTB
B
O
L/3
L/6 L/6
1/3 mg
2/3 mg
L/3
TBTB
C
1.5 N∙m
r
Solution 6.43
ω
422
4
23
Mrtrr
τ
=+++
Solution 6.44
2
Solution 6.45
Solution 6.46
2222
Kinematics:
()
()
9gRight
176
135 gDown
176
18 g CW
11
x
y
G
G
a
L
α
=
−
=
−
=
()
9mg Right
176
x
O
=
+
O
y
OG
L/4
mg
m
y
(torques only)
Solution 6.47
2
0.0101 lb-ft-sec
=
Spin down (Mmot = 0 !)
3450 60
Solution 6.48
2
12 12
22
g
1
12
x
x
α
=
+
()
22
1gg 2
xxx
+−
G
O
xmg m, ℓ
Solution 6.49
We all assume each spoke is 320-mm long and stops at the shaft, and at the sphere.
T = 20 N
()( ) ( ) ( ) ()( )
22 2 2 2
21
4 3 0.04 3 0.4 5 0.060 2 0.04 0.025
O
I
=+++++
()( )
4 1.5 0.320 1.5 0.040 2.24 kg m
+++=⋅
OO
T
O
100 mm
40 mm
400 mm
mD, kO
mS
Solution 6.50
2
22 2
11
; 312 2
O
b
mbcI mb mbm b
ρ
==+++
53g
23 10
OO
b
For each plate, 23
11
12 12
Imb bC
ρ
==
αρ
==
39
g
220
mb bc
:2mg sin
zzz
FmaO md
α
β
=−=−
Oz
b
mg
z
Solution 6.51
In vertical position Rt = 0 if impact occurs at center of percussion so
n
22
A
F
N
F
N
M = 12 N•m
Solution 6.52
0.24 m
0.15 m
A
A
A
r
k
=
=
0.1m
10 kg
B
k
m
=
=
()
α
=
25.5 rad s CCW
B
30°
N
30°
5.30 lb
W
B
B
10″
Solution 6.53
Power unit C:
α
3.84 rad sec
α
=
Steady-state speed:
AA BB
rr
ωω
=
::+=
BBO
B
tt .
..sec==
384 34 9
30°300(9.81) N
T
t
n
12 m
16 m 8 m
α
Solution 6.54
So 53
22
s
Fk k
δ
== −
48
()
α
α
α
=°−−=
=− −
2
53 7
:mg sin30
422248
6g 12 k 53
77m
Oo
MI k m
Solution 6.55
()
2 1300
,
2 0.600
4333 N
I
MrT==
=
α
=
;
OO
MI
() ()( )
; 4333 sin30 300 9.81 cos30 300 12 0.071 637 N
ttt t
FmaO O
=+ °− °= =
O30°
ℓ
4
ℓ
4
Solution 6.56
2
1
Solution 6.58
2
64.4
Solution 6.61
[Weight and static normal (distributed ⊥ paper)] m = 1.2 kg
()
α
=−
2
48.8 rad s CW
GG
x
+
+
rG
m/2
P = 6 N
100 mm
500 mm
m/2
G
y