Solution 5.40
For vertical motion only of B, its horizontal coordinate remains constant so
()
()
2
1cos
2tan
tan 1
1cos cos 2
dLx Lx x
b
Lx Lx
c
bb
θ
θθδ
++
==
++
c
b
B
bD
h
Lx
Solution 5.41
()
()
So
So and
so
absolute angle
t
Rr r
r
rar
θβ
υβ ωβ
υω α
β
−=
==
==
=
()
()
so
So
where
So so
( absolute angle)
t
Rr r
rRr
rar
θ
β
θ
β
θ
θβ θ
ωβθ
υω α
βθ
=+=
+=+
=+
==
+=
Solution 5.42
Belt velocity is the same for both pulleys
122
11 2 2 12 12
22 1
2
22
so
rr
rr rrrr
rr
ωω
ωω
αω ω
=
−−
== =
θ
(R–r)θ
β
θ
R
r
O
O’
θ
Rθθ
(R+r)θ
β
R
r
O
O’
Solution 5.43
22
2
22
2
3sin, cos
22
3
so cos , sin
2222
42 24
But 2 sin cos
22
1
so 2 ,
4cos 2
c
B
BB
cB c
xb yb
b
xb y
dbb
dt
θθ
θθ
θθ
θθ
υθ
υυ
υυ υ
θ
==
==−
=− =−
=+ = 2
8sec
22
θ
+
b
b
b
B
O
C
x
y
y
θ
υ
B
x
Solution 5.44
()
()
γγγ γ βββ β
βγ
=− + + +
=
22
0 300 sin cos 200 sin cos
A
lso: per problem statement
S
o we have seven
N
equations in seven unknowns:
OA
A
O
200 mm
β
Solution 5.45
()()
F
βπθβ θβ
F
θ
==
−− +
−
=−
2
sin sin sin
2 0.9397
a
nd for 2 rad s, Eq. b g
bb b
ives
Solution 5.46
2
2
8tan , 8 sec
8sec sincos
4
But for const. , 2
A
AA A
xx
xaax
θυ θ θ
θθ θθ
υ
===
=+
== =
() ()
22
2
sin 3 5 for 6 in.
46 34
4 8 5 4 , 0.1408 rad sec
255
So 0.1408 rad sec CCW
x
θ
θθ
α
==
=+ =−
=
b = 0.2 mO1O2
P
θβ
=
2
b
√2
0.2 m
√
ℓ
B
Ax
υA
Solution 5.47
υ
υω
=+×
=− +
1.386 1.2 m s
BG
BG r
ij
Solution 5.48
AG
AGAGG
r
υυυ υω
=+ =+×
7.5″7.5″
y
C
y
υo = υ
υ
υ
A
B
Solution 5.51
υ
=
107,257 km hr
O
j
()
1672 107,257 km hr
107,257 1672 105,585 km hr
1672 107,257 km hr
107,257 1672 108,929 km hr
AoAo
BoBo
CoCo
DoDo
ij
jj j
ij
jj
υυυ
υυυ
υυυ
υυυ
=+ =− +
=+ = − =
=+ = +
=+ = + =
Solution 5.52
υ
υ
υ
=
=
0.8 8
0.4 m s
0.533 m s
B
C
y
x
Solution 5.53
AO
AO r
υυω
=+×
Solution 5.54
υυυ υυ
=+ =+
Solution 5.55
AO
AOAOO
r
υυυ υω
=+ =+×
So
()
υυ
=+°=
22
1
2 2 1 sin30 , 6.93 m s
oO
And
υ
ω
=− =− =−
6.93 21.3 rad s
0.325r
or
ω
=21.3 rad s CW
υ
B
= 3 m/s
υ
C/B
30°
30°
y
x
Solution 5.56
832ftsec
B
Solution 5.57
ββ
== ==
90 3 120 4
sin , cos
C
45
υ
(υ
B
)
t
υ
A
x
y
A
y
Solution 5.58
BA
BABAA
r
υυυ υω
=+ =+×
tan15 0.268 CCW
AA
AB LL
υυ
ωω
== =
Solution 5.59
222
υ
Solution 5.60
()
2,since 0; CW
C
rr
υ
υωυω
===
ω
υω
−=−
3.23 rad s
3
: 1.174
2
B
j
Solution 5.62
22
1.366 4.37 4.57 ft sec
A
υ
=+=
C
υC = 14.48°
υA
200
mm
ωo
υB/C
α
Solution 5.64
υ
υυ υ
=+ =
, 300 mm s
CACA
A
39.7
=°
x
y
Solution 5.66
()
0.4 sin30 cos 30
AB
ABABB
A
r
kij
υυυ υω
ω
=+ =+×
+×− °+ °
: 0.483 0.346
iO
ω
=−
y
A
θ
Solution 5.67
BABA
(Alternatively and more simply)
υ
==
21.031 m/s
B
Solution 5.68
BABAA BA
r
υυυ υω
=+ =+×
15.36 rad sec CCW
ω
x
O
R
B
β
υ
B
(⊥ OB)
υ = 2.5 ft/sec
R = 16 in.
(
(
R/2
–1
1.25 R
Solution 5.69
y
A
C
O
B
200 mm
240 mm
80 mm
60°
15°
ω
o
ω
o
= 10 rad/s CCW
A
Solution 5.70
Let E be a point on D coincident with pin C.
υυυ
=+
()
68
AC
AC AC
kij
ω
=×+
0.5 6 3 in. sec
BO
BOB
rkji
υω
=×= ×=−
x
6″6″
C (ICR)
Solution 5.72
υ
υ
υ
π
ω
ωπ
=− =−
=− + =
=− + =
2rad
or rad s
2s
0.1 0.2 m, 0.05 m
0.3 0.05 m, 0.6 ,
CB
CB
OA CB
BA OD
k
rijrj
rijrj
E D OD OA
Solution 5.73
y
D
0.6
Solution 5.74
B
βγ
C
β
Solution 5.75
Solution 5.76
31.5 ft
B
BC
υ
===
AAC
30°
30°
160 mm
120 mm
A
υ
A
= 2.8 m/s
ω = 12 s
rad