Solution 8.87
For the bar,
2
22
1313
Imm m
=+ =
Let
θ
= 0 be the equilibrium position shown and choose V = 0 @
θ
= 0:
2
22
13 9
Vk k
θθ
==
max 0
12
3326
mm
θωθ
== +
Solution 8.88
222
111
2
00
Arr rr
222 2
31
dE Mr m r r r r mr r r
Small
θ
,
()
222
:0
32
Mr m r r rr
θθ θ
+=
++−
12
Solution 8.89
Take
θ
= 0 to be the position where V = 0.
2
2
11 mg 1 cos sin
dt = to obtain, for small angles, 2
Imr
+
So 2
n
Imr
ω
=+
Solution 8.90
L
So
ω
=+
63g
n
k
L/2
L/2
L/2
G
C
υ
G
θ
Solution 8.91
θ
≅− +
2
g2
mm
()
θθ
=− + +
+
2
2
max
max 1 2 max
2
12
g2
T
V
mm k
mm
Solution 8.92
2mg
e
c
Δ
=−
gmgVy
22 22 2
1
by by
()
π
2
s2
24 0.25
nn
Solution 8.93
3
23
lb
9 or 15,552 lb ft
in.
k
=
Solution 8.94
Thus 2
max max
mg
V
r
θ
π
=
2
22
11
GG
C
θ
r
T = O, V = V
max
V = O, T = T
max
Solution 8.95
2
24
30
RR
yA
yAR
′
==
′
whole hole
O
I
TI
=−
()
=− = − =
ωωωθθ
=+× =− + × −
15
GRR
Now, E = T + V
()
()
22
22 2 2 2
11 mg 1 cos
111
0.5135 1.001 cos mg 1 cos
2 2 15 30
R
R
EmRmRR
θθ θ θ
=+−+−
G
y
O
We now assume small
θ
:
θθ
+=
2
1
.448 0.0333 0
RgR
Solution 8.96
42
r
=
()
22
2
2
29 18
9162cos
6
GO G
CG C
mr
IImd I
ππ
πθ
π
=+ →= −
2
1and mg cos
2C
TI V r
θθ
==−
dTV
π
ττ
ω
=→=
24.44 g
n
r
O
Vg = O
Solution 8.97
τ
ωβ
β
==
+
12
2
24gsin
3
nT
k
mb b
θ
Solution 8.99
Linear momentum is conserved during impact
ω
π
τ
ω
==
==
297 17.23 rad s
20.365 s
n
n
Solution 8.100
22 2
15
AA
Imrmrmr
−
=+=
3
3000 x
O
(a)
r
A – A
Solution 8.101
dE
Solution 8.102
12 1 2
2
For the damped linear oscillator (case III, underdamped, of Art. 8/2b)
xe
4
2dm
m