A
O
16″
6″
4″
υ
A
γ
Solution 5.159
υ
B = 10 in./sec, constant
ABAB
υυυ
=+
46
AC
+
Solution 5.160
Law of cosines gives
()
222
2cos120
Rh R r Rr
+=+− °
Rel. velocity from nonrotating system x–y at O is
υ
A –
υ
B. But
υ
A =
υ
B +
ω
× r +
υ
rel
()
()
rel
so
AB
r
ωυυυ
×= − −
B
A
h
r
x
30°
υB
υA
Solution 5.161
ω
α
==
=
0
0
222
0.2 m, 10 rad/s
0
r
AC
r
y
ωO
1
2
0
–20
–10
40 80 120 160 200 240 280 320 360
θ, deg
Solution 5.162
()
θθ
θ
θ
=−
2
100 1 cos rad s
θ
θ, deg
–10 0 40 80 120 160 200 240 280 320 360
Solution 5.163
Bo Ao BA
γγγ
=+
Also, BCoBC
γγ γ
=+
B
O
x
γ
β
(mm)
70
80
240
200
θ
Solution 5.164
See the solution to Prob. 5/163 for the determination of the positions and angular velocities
of bars AB and BC. In addition:
αθ
==°
2
max
112.2 rad s at 182.1
BC
0
–150
–50
0
50
100
40 80 120 160
θ, deg
α
BC
200 240 280 320 360
Solution 5.165
ωθ
==°
max
11.83 rad s at 216
BC
0
–15
–5
0
5
10
40 80 120 160
θ, deg
ω
BC
200 240 280 320 360
Solution 5.166
r2 = h2 + R2 – 2hR cos
β
()()
()()
2
sin 90 cos 90
cos 90 sin 90
A
A
RIJ
aR I J
υβ β β
ββ β
=−−°+ −°
=−−°+−°
Substitute the above transformation equations into the expressions for
υ
A and aA and
simplify to obtain (with c = cos, s = sin)
()() ()()
{
}
()() ()()
{}
2
c s 90 s c 90 s s 90 c c 90
c c 90 s s 90 s c 90 c s 90
A
A
Rij
aR i j
υ β θβ θβ θβ θβ
βθβ θβ θβ θβ
=−−°+−°+−°+ −°
=−−°−−°+−°+ −°
Equations 5.12 and 5.14, Bxy attached to BD: rel rel
rel rel
Note : ,
,,
rri i
aai k k
υυ
ωω αα
==
===
rel x
A
r
rel x
A
rel
y
A
r
ji
i
J
I
I
θ
θ
+
A
y
O
r
h = 0.25 m (90-θ)
(β–90°)
R = 0.5 m
X
υ
A
Y
θ
β
=–cos θjsin θ
Program above equations and plot:
0.5
–1.5
–2 0 30 60 90 120 150
180 210 240 270 300 330 360
0.3
0.1
–0.3
–0.4 0 30 60 90 120 150
β, deg
180 210 240 270 300 330 360
υrel
B
y
Solution 5.167
cos cos
xr
β
θ
=+
AA
θθ
−
Solution 5.168
From the results of Prob. 5.167, we may write
()
()
22
22
22
sin cos
sin
1sin2
12
sin cos 2 sin 2
2sin
AA
d
arr
dt r
r
θθ
υωθω
θθ
θθ θ θ
== +
−
−−
which reduces to
2
AA
θθ
+−
–8
–6
–4
–2
0
2
4
6
8
10
12
14
80
72.3°