Solution B.21
()()
32
2
0
23 3
22
215m
23 so
15 2
15
22
a
z
a
a
bax a x ab ab
a
b
a
ρπ ρπ ρ π
πρπ
−
=−+= =
()
2
µµµµ
=−
−
ddx ad
ax a
z
y
b
π
2
dm = ρdv = ρx zdx
z
2
() () () ()
33 25 7 9
32 52 72 92
32
0
1331
3579
1331
3579
a
baaa
a
baax aax aax ax
a
ρπ µµµµ
ρπ
−
=−+−
−
=−−−+−−−
Solution B.22
2
15 2
2
Also 11
zz
zz
dI dm r
=
z
y
b
π
2
dm = ρdv = ρx zdx
z
2
() ()
332
2
0
4
75 7 5 35
a
ax
b
aaxaax a
µ
ρπ µ
µµ
=−
=− =−−−=
3
272
415 4
21 7
mb
π
a
Solution B.23
()
2 where mass unit areadm ya d
ρπ θ ρ
==
() () ()
()
0
2
33 3
33 cos 3 4
2
4cos 3
ara ara
ada
π
θ
θθ
=+ =+
==
()
44
22
243
59
and for 3, 2 6 2 0.865
27
3
0.865 ,0.890
1.0944
zz
zzz z
ra I a a
kIm ak a
πρ π πρ
== −=
== =
Solution B.24
2
2
so 4
yy
rr
dm y dy
h
h
dI dmz
πρ
=+
′′
=
12
use substitution
µ
32 0
2
32
24
use appendix 10 with 1, , and
4
yy yy
h
IdIzdm zds yy dy
hh
r
Cba xy
h
πρ
′′ ′
== = = +
′
== =
32
1.953
yy
m
Solution B.25
Use a circular “hoop” element.
222
2 2
00 0
22
hh h
rr r r
h y dy hy y dy y y dy
πρ
′′′′′′′′
=+−+++
y
y’
y, yc
zzc
z
r
dy
ds = dy‘1+ dz
dy‘
2
r2z2
r = 70 mm
h = 200 mm
ρ = 32 kg/m2
2
2212232
4
2
hh
h
ry
µµµ µ
+
=+
′
34 3 44
hy h h
hhh
′
=+= +−
(A)
4
hba xy
h
== =
′
32 32 32
22 2222
24
23 3
h
rr hrrrr
hy y h h
−
′′
=− − + = − + +
(B)
0
4with 1, , and
4
hba xy
h
== =
′
32
h
()
()
so 18.66 10 kg mhy dm −
′
−= ⋅
233
11
1.953
m
−−
Solution B.26
22 2
1
mr md md
+−
Solution B.27
()
22
22 2
2, 2
24
xx zz
yy
ImLImL
ImLLmL
==
=+=
Solution B.28
From Sample Problem B.3,
()
22
14
12
yy
Ima=+
()
2
10.80.1
+
r
2
r
1
z
L
––
2
L
––
2
Solution B.29
22
22 11
11
xx
Imrmr
=−
Solution B.30
(1): outer body; (2): central hole
22
()
1
2
22
2
2
22
2
22
412
412
15
43
12
L
rL
rL
rL m
rL rL
mrL
ρπ
ρπ
ρπ
=+
=+⋅
Solution B.31
From Sample Problem B.3, moment of inertia of complete cube about G is
22 32
11
3
()
So 6 1.898 lb-in.- sec
3 32.2 12
xx
I==
Solution B.32
222
22
111
:0
243
33
or
42
zz xx
I I mr mr mL
r
rL L
=+=+
==
Solution B.33
()()
()()
head
2
head
22 32
2
9000 0.020 0.1 1.131 kg
11
1.131 0.020 0.1 0.320
42
0.1169 kg m
xx
xx
m
I
π
−
==
=++
=⋅
Solution B.34
L
3
43 83
192 128
L
πρ
=+
Solution B.35
()
()
()
22 4 4
22 11 2 1
44
44
2
111 1
Rim : 222 2
1
17830 0.2 0.15 0.075
2
1.009 kg m
Imr mr rb rb
ρπ ρπ
π
=−= −
=−
=⋅
()
()
44
21
1
Hub : 2
Irrb
ρπ
=−
()
1.009 100% 97.8%
1.031
n
==
Solution B.36
()
2222
5.02 10 lb-ft-sec
141 61 4 6
xx
I
=
32
Note for -axis, part 4 can be
y
−
12
xy
6″
5″
4″
z″
z’
z
1
2
3
4
4″
b
y
b
1
Solution B.37
Part 1:
2
22
4
1
mm
mmb
=
()
2
22
1
2
11
2
xx
mm
m
Ibmb
=
==
232
11
xx
mmb
Ib bmb
=+ +=
Total:
22
111 3
xx
Imb mb
=++=
4″
z
Solution B.38
g 0.455 lb ft
ρ
=
G
Solution B.39
()
4 100
442.4 mm
33
r
r
ππ
== =
()
22
2
0.0369 7.38 0.060 0.0424 0.0424
0.1010 kg m
=+ + −
=⋅
Solution B.40
()
22
22
10.15 m 0.5 m 0.408
23 3
xx xx
bb
Ib Imb
=++→=
Solution B.41
22 22
11
77
Immxm
=+= +
22 2
44348
122
O
22 3
y
b
z
1
0.1 m
0.15 m
O
b
2
4
O
A
a
A
a
r
III
Solution B.42
II I=−
For II:
()
2
22 2
22 4
11 2
20.667
3
a
IIm ma ma ma
aa a
ρρ
=+ = + =
==
0.802 0.630
aa a
πρ ρ
==
4
0.0365
A
a
ρ
=
But 22 2
10.215
ma a a
ρ
π
ρ
=− =