Solution 5.1
()()
()
ωω
α
ωω
ω
ωαθθ α
−−
== =
−
=+ ==
−
==
2
21
22
22 21
21
22
900 300 6000 rev min
660
2, 2
900 300 60 rev
2 6000
t
N
Solution 5.2
()
A
rkbihj
υω ω
=×= × −
Solution 5.3
ωα
==
2
4rads, 7rads,
kk
Solution 5.4
()
()
θ
ω
ω
θ
=− = = =
=− =−=
22
10
0
33
23
12 3 ; when 0, 4, 2
12 3 12 16 rad CW
ttts
tdt tt
Solution 5.5
22 2
Solution 5.6
Startup:
()( )
()
ωω α
πα
πθθ
θθ
=+
=+ −
−=
0
2
2
0
0
2
2
3450 2 60.2
60
1084 rad or 172.5 rev
t
O
()
ωω α
πα
θθ
=+
=+
−=
0
2
0
S
hutdown :
2
3450 32
60
60
5780 rad or 920 rev
t
O
First half of startup:
θ
2
2
11 6
()
()
θ
ωα
=+ = +− =
0
3450 11.29 4340 rad 690 rev
260222
tt
Solution 5.7
θθ ω
=
00
sin
t
Solution 5.9
θ
ω
αωω
θ
ω
=− =
2
0
w
d
kd
d
Solution 5.10
ω
α
==−
20 rad/s, 40 rad s
kk
Solution 5.11
()
()()
()
2
2
00
0.1
: 48.1 0.1493 lb
32.2
11
23
22
9 rad 516
FmaF
tt
θθ ω α
== =
=+ + =
=°
Solution 5.12
() ()
ω
== =
2
22
0.15 20 60 m s
An
ar
A
0.075
A
(a
A
)
t
(a
B
)
t
a
B
Solution 5.13
ω
α
==−
2
2 rad s, 5 rad s
kk
=− +
2
4.48 0.1465 m s
A
aij
Solution 5.14
C
ω
ωα
=−
0
rad
t
()
()
=
=
=
S
olving, 9.57 s
1
N
otes: 2 independent of .
PB
t
aa
trAB
Solution 5.15
()
()
()
αθθ
ω
ωαθ ωω θ θ
ωθθ
ω
=−
==−
=+ − =+=
== = =
2
2
50
20 2
22 2
0
1.8 rev s , in revolutions
: 1.8 0.06
521.8 0.03 254873revs
73 8.54 rev s or 8.54 60 513 rev min
k
ddd d
N
A
Solution 5.16
sin cos cos sin
rOBBA b ib jb ib j
θθθθ
=+=− − + −
21.6 28.7 in. sec
A
aij
Solution 5.17
Match of tangential speeds:
α
=
2
47.1 rad s
A
Match of tangential accelerations:
() ( ) ( )
() ()
() ( ) ( )
Β
Β
Β
ωω αθθ θ
ωω α
ωω αθθ θ
θ
=
−= − −= Δ
=+=+ =
−= − −= Δ
Δ=
2
22 22
00
0
22 22
00
75
200
17.67 rad s
a 2 :141.4 0 2 17.67
b 0 17.67 4 70.7 rad s
c 2 : 70.7 0 2 17.67
141.4 rad or 22.5 rev
A
AA BB B A
B
BB B B B
B
BB B
BB BBB
r
t
y
y
Solution 5.18
()
()
()
υω
α
=°
== =
== =
2
1.1 rad 63.0
0.2 2.1 0.42 m s
0.2 2 0.4 m s
t
r
ar
()()
=− °+ °+ − °− °
=− − 2
0.4 sin63.0 cos 63.0 0.882 cos 63.0 sin63.0
0.757 0.605 m s
ij ij
ij
a
n
O
A
θ
y
Solution 5.19
()
()()
()
θ
θ
ωωθ
α
ω
+= + =
== =
== =
22
0
2
2
22
,
0.2 2 2 0.8 m s
0.2 4.1 3.36 m s
O
t
t
n
d
ttdtd
dt
ar
a
r
a
()()
=− °+ °+ − °− °
=− 2
0.8 sin164.2 cos164.2 3.36 cos 164.2 sin164.2
3.02 1.683 m s
ij ij
ij
y
Solution 5.20
ω
ωωω
==
2
0
2
t
t
d
dt e
=− +0.0464 0.1403 m sij
()()
()()
θθ θθ
=− + +− −
=− °+ °+ − °− °
=− + 2
sin cos cos sin
0.296 sin 18.30 cos 18.30 0.1092 cos 18.30 sin 18.30
0.1965 0.246 m s
tn
a
aija ij
ij ij
ij
a
n
A
Oθ
Solution 5.21
()
θω
θ
ω
αθω θθωω
θ
ω
θθ
θ
ωω
θ
θ
== =
++
==
+
== °
222
2
2
22
2
2,2
1
12
ln
2
222
W
hen 2 s, 0.596 rad 34.2
O
O
t
OO OO
ddd
d
dt
d
dt t
t
Solution 5.22
ωα
Δ= =
6sec
2sec
Gear : area under “curve”
A
Adt
Solution 5.23
tan
s
θ
=
A
s
2
+ d
2
Solution 5.24
υ
υ
υ
υω
==
=+
0.4 m s, 0.2 m s
rad
0.5
0.400 CW
AB
PB
AB
BO
Solution 5.25
+=
222
AB
xyL
100 mm 200 mm
υ
A
υ
C
υ
B
ω
B
θ
Solution 5.26
θ
=+
cos
Sr d
A
θ
d
P
r
O
s
Solution 5.28
() ()
υ
υθθυ θ
υυ
θθ
=+= + + = +
=+= + =
00
2
22
22 2 2
00
1 sin cos 2 1 sin
cos sin toward 0
xy
a
xy rr
Solution 5.29
2
2
sin
sin
cos 1 sin
sin
1
bh
d
bh
d
θ
β
ββ
θ
+
=
=−
+
=−
2
1
22
cos cos
sin
cos 1
1sin sin
sin 1 2 cos
2
With 3.2 in., 4 in., 1.2 in., 30 , and 7 rad sec:
30.2 in. sec
xb d
bh
bd d
bh bhb
xb d ddd
bdh
θ
β
θ
θ
θθ
υθθ θθ
θθ
υ
−
=+
+
=+−
++
==− + − −
====°=−
=
h
A
x
d
b
B
θ
O
β
O
x
–y
θ
θ
a = r
υO2
A
O
υ
θ
90°
A
y
Solution 5.30
θθθ
θθ θθ
=+ =
=−
2
20 80 sin , 80 cos
80 cos 80 sin
yy
y
θβγβ γ
=− − =− −
F
rom Eq.1: cos
B
Solution 5.32
0.9 0.8 1.2 m s
OC
Solution 5.35
π
hx
Solution 5.36
()
υ
θ
υω
==
== =
30 4 120 mm s
40
A
lso 40 ,
B
B
B
ds d
A
ds
A
Solution 5.37
θ
β
θ
=
==°
sin sin
Law of sines: 2
24 2
With ft, 30 ,
12
b
()
θ
υ
=−
=
and 2 rad sec,
0.965 ft s right
A
Solution 5.38
From the solution to Prob. 5/37,
22
2
cos
sin 1 cos 1
11
21 sin 21 sin
424
sin 1
41 sin
4
ab b
b
θ
υθθ θθ
θθ
θθ
θ
==− + − +
−−
−
−
12
Solution 5.39
θ
s
=
2sin
y
b