Solution B.43
1
0.0128 kg
ρ
=
12
3
80 40
OO OA
G
G
2
G
3
Solution B.44
()
2
32.2 12 in.
1
4
mr a
π
ρ
π
ρ
==
()
223
1
23
11 1
122 2
11
233
xx
Imr ar a
Ima a
πρ πρ
ρ
== =
==
()
()()
43 2
5.57 1.438 10 8 0.410 lb-in.-sec
−
==
r
x
2
8″
8″8″
r1
r2
Solution B.45
(2): Entire cone; (1): “Missing” top
hhh
+
111 1 1
21
10 10 3
ρπ
==
−
Imr r r
rr
21 21
33
21
21
1
3
rr rr
rr
−−
−
21
21
21
55
33
21
3
10
ρπ
−
Solution B.146
()
2
33
11
12
Total 2 52 3 23.62
Ir r
ρ
π
ρ
−
=+=
()
()
22 22
22
111
Axis 2 2; Shell; 2 6
262
Imr rr
ρπ
−
−=+= +
()
322
22
So 23.62 0.857
rr r
ρρ
π
=+
Solution B.47
()() ()()
2
2
2
xy
yz
Im m m
=−+−=−
y
Solution B.48
2
0
xy
I
=
Solution B.49
Let
ρ
= mass/unit length
222
2
2
11
sin 2 sin 2
224
xy
sds
Isdsm
ρθ
ρ
θθ
−
=−
=− =−
z
y
dm
r
Solution B.50
Let
ρ
= mass per unit length of rod
2
ρ
dm r d
π
ρ
θ
=
Solution B.52
2
bb b
I
ρπ
=−
Solution B.53
2
44
033
yz
xz
r
Imr mr
ππ
ππ
=+ − =−
y
zz
x
z
0
z
0
GG
r r
h
h
r=
4
r
–––
3π
Solution B.54
2
00
222
22
hb
h
h
h
b
m
y
x
Thickness
t
dm = ρdV
= ρtdxdy
dy
dx
x = ky
2
= y
2
b
–
h2
y
z
y°
z
o
x
o
x
6″
8″
Solution B.55
12 3 3 3
xy
′′ ′′
34
xy
′′ ′′
Solution B.56
12
50 1 0.1294 lb-in -sec
32.2 12
m−
==
()()
()( )
2
2
2
0 0.1294 3 2 0.776 lb-in. sec
0 0.1294 4 2 1.035 lb-in. sec
OO
OO
OO
xy x y x y
yz y z y z
xz x z x z
II mdd
II mdd
=+ =+ −−=− −
=+ =+ −=− −
y
y’, y″
y
x
2
r
θ
Solution B.57
1
2
Part 1: 4
1
mm
mb
=
3
0, 0
xy xz
II
==
88
xz
Solution B.58
12
;
xy xy
II= Let
ρ
= mass per unit length
y
Solution B.59
Eq. B.10: 222
222
AA xx yy zz xy xz yz
IIlImInImInImn=+ +− − −
2
2
0.0853 2.30 2.39 kg m
0.1920 0.576 0.768 kg m
yy
I
I
=+= ⋅
=+= ⋅
AA
Solution B.60
y
β
dβ
r
r
M
1
2
Solution B.61
22
2
zz
I
O
Numerical solution of cubic:
1
1
0.408
n
=
Solution B.62
y
1.5b
2b
2b
2b
222
123
Solving for yields 2.22 , 24.0 , 26.2
I I mb I mb I mb
===
()
z
°+−
Solution B.63
() ()
()
2
422222
22 2
22 2 2 2
4
24 4
11
11
22
12 4 4 3
10
3
1,0,
22 4
zz
xy xz yz
b
bb
Ibbbb bb
b
bb
Ib bI Ib
ρρ ρ
ρ
ρρ ρ
=++++
=
====
Substitute in Eq. B.11 letting I = IO
ρ
b4 and get
OO O
Solve by computer program or algebraic formula and get
4
3
3.78
3.61
Ib
Ib
ρ
ρ
=
=
Solution B.64
12.56 0.4
yy
222
22
111
400
z
y
x
2
1.502 0.0768 0.0341
−−
22
3
0.875 3.35 3.35 0
and 0.406 kg m
III
I
=⋅=⋅
1111
2
2222
For 1.431 kg m : 0.0433 0.439 0.897
Imn
= ⋅ =− =− =