Solution 8.47
2
10.15
1.5
X
===
Solution 8.48
,
iB iB
xx xxxx=+ =−
2
2
sin , cos
So 2 sin cos
BB
d
mmm m
xb txb t
kc
xxxbtbt
ωωω
δω ω ω ω ω
==
++= +
Solution 8.51
2
2
2
π
() ()
22
But or 1500
11
nn
kx
ωω ωω
==
−−
ω
=
2
Solution 8.52
For seismic response,
2
Solution 8.53
1mg
42
kk
ckm
Solution 8.54
For steady-state motion,
()
2
n
x
ωω
=
n
Solution 8.55
0
2
24
cx
cx
ωω
πω
=+
=
Solution 8.56
υ
=
km
4.23 m/s or 15.23 h
c
Solution 8.57
Dynamic forces:
b
b
Solution 8.60
2
Solution 8.61
:
oo
MI
θ
=
cos
To
oTo
Mk I
MtkI
θθ
ωθθ
−=
−=
or cos
oT o
Ik M t
θθ ω
+=
Assume
θ
P = (H) cos
ω
t:
()
2cos
o
IHt
ωω
−
()
()
cos
T
KH t
ω
+
()
cos
o
Mt
ω
=
(H) 222
oo
ToT
MM
kIkk
ωω
==
−−
(H)
()
2
2
80.1271 rad
20 6
160 25 32.2 12
==
−
+
M = Mo cos ωt
A
kTθ
Solution 8.62
+
2
1
:12
G
JG
MI m
L
θθ θ
=− =
ππ
τ
τ
ω
=→=
22
= 1.003 s
6.26
n
mass m
Solution 8.64
:
OO O yy
MI II
α
==
22
2
23
n
rh
+
Solution 8.65
L/2
3
N
O
O
C
A
m
g
θ
Solution 8.66
Let L be the 0.8-m rod length
131.2
So
()
2
4 1, 0.558 m
250 0.8
π
Solution 8.67
200 mm=
1
4
mg 3
ρ
=
Small
θ
so sin
θ
θ
3
5
:mgsin 4
OAA
MI r
θθ
ρθ
−
=− =
ρθ ρ θ
θ
θ
+=
+=
3
53
0
43
43g 0
5
()
()
ω
ω
== →=
4 3 9.81
43g 8.24 rad s
550.200
nn
+
L/2
L
2
k1.2 kg
sin θ
O
–
O
y
mg
Solution 8.68
() ()
2
: sin cos cos cos
1
OO
MI kb b ca a
θθθ θθθ
=− −
2
33
ac k
cr 2
3
Solution 8.69
Dynamic forces, small
θ
:
2
22
2
o
n
kb
kgr
m
π
τπ
ω
== +
Solution 8.70
Dynamic forces:
2
2
OO
0.43
12
0.16 2
360
0.43 60
3820 N m
n
k
k
π
ω
==
=
θ
a
O
y
O
θ
kbθ
Solution 8.71
O
y
mg
4
mg
Solution 8.73
,,
B
BB
x
xbb
xxxx
ab a a
xx
===
()
2
22
102 120
22
2
2
0
1
2
nn
mkma mmka
cb a
k
km m
a
δ
++
=
+
O
2
a
O
1
A
Solution 8.74
Arc AC = Arc BC: r(
β
+
θ
) = R
θ
R
Solution 8.75
Solution 8.77
oo
and with
cos 1 ,
o
o
θ
θ
=− +
222 22
111.5 0.320 0.0256
J
θω θω
=×× =
22
n
ππ
Solution 8.78
The forces associated with static equilibrium are not shown.
2
222
11 9
22 2
(for small
θ
)
22 22
95
ETV mb kb
θθ
=+= +
θ
o
m = 1.5 kg
k = 120 N/m
k (2b sin θ)
Solution 8.79
V = 0 @ θ = 0;
()
0max 0
mg 1 cos
VV
θ
θ
== −
22
()
max 0
2
2
2
11
C
TI
θ
ω
==
ω
23g
n
n
Solution 8.80
Choose V = 0 @ x = 0
1
2
2
2
22
11
2
k
mx
r
=+
2
22
2
11
1
22
k
ETV m x kx
r
=+= + +
()
22
0:
1:2
n
n
n
kkm
k
mkr kr km
ω
ωω
==
=
+==
Eq. pos.
θ
ℓ
C
Solution 8.81
22
n
ππ
Solution 8.83
3
2
p
mr
rm
ρ
+
== =
ρ
3
17
3
17
n
Solution 8.84
Let y be the downward displacement from the equilibrium position where V = Ve + Vg is
taken to be zero.
π
2
nn
V
g
= O
O
k
T
Solution 8.85
()( )
22
11
mg 1 cos
22
G
ETV m I Rr
υω θ
=+= + + − −
ω
g
n
Solution 8.86
Let
θ
= 0 be the angular position of static equilibrium and choose V = 0 there.
222 22
11 1
2
(R–r) cos θ