Solution B.1
sin
2
22
12
sin
12 12
L
xx
Iydm y dy
L
β
ρ
β
ρ
==
For planar bodies,
12
zz xx yy
m = ρL
d
x
=
x
y
β
––
2
dx
ρ
cos
β
Solution B.2
()
()
22
23
0
2
1
0.408
6
h
xx xx
xx
x
b
h
b
IdI thyydy
mh
b
Ih
kh
m
ρ
== −
=
==
By analogy, 2
1,
yy y
b
Imbk==
22
1
z
m
hb
y
y
x
O
h
b
(thickness t)
Solution B.3
;
r
yx
=
y‘
Solution B.4
11
1
52
2
32
2
23
2
77
3
0.655
h
xx
x
hh
h
m
tbh mh
tbh
I
kh
m
ρ
ρ
−
==
==
k =
(x,y)
y
dy xx
b
–
2
b
–
2y = kx
2
⇒
h = k
2
b
–
2
b
–
2h
,
=
:
4h
b2
2
y
Solution B.5
by y
0
12
20
0.224
yy yy
yy
IdI tbhydy
I
ky b
m
ρ
==
==
y
b
–
b
–
Solution B.6
x = ky2
y’
y
= h
2
by2
Solution B.7
1
mabt
ρπ
=
y= 1
y2
b2
x2
a2+
Solution B.8
2
32
3
4
6
6
1
1
h
h
xx xx
r
mdm xdx rh
r
IdI xdx
ρπ ρπ
ρπ
== =
==
x
y
h
r
y
y = kx
3/2
r = kh
3/2
:
k = and
r
h3/2
x
y = 3/2
r
h3/2
h3
Solution B.9
3
0
2
42
65
4
4
1
7
yy y y
mdm xdx rh
h
dI dI dmx
hh h
rr
ρπ ρπ
′′
== =
=+
3
, by symmetry.
zz yy
II
=
y
r
y
y’
x
y =
3/2
r
h
3/2
a
a
2a
z
Solution B.10
1
22
rz
aa
+=
()
2
45
22
225
πρ
πρ
=−
=−
zz a
a
Irardr
ar r
2
39 39
T
hus, and
20 2 5
===
zz
zz
Ia
Ima k
m
Solution B.11
zz zz
IdI
=
2
2
236
zz
Irh
h
ρπ ρπ
=⋅=
42
11
So 1
zz
m
Irh mr
ρπ
==
z
Solution B.12
2
2
2
1
4
h
zy
rr
zdz z z
hh
ρπ
=
=+
0
yy yy
IdI zzdz
ρπ
== +
2
22
2
23
rh r m
ρπ
Solution B.13
1
n
a
2
1
yy
dI dmr=
+
+
+
21
2
22
22
0
0
2
21
21
a
n
nn
aa
n
m
4
222 4 4
11
aa
n
b
ρ
π
ρ
π
y = kzn
z
a
y
dy
Solution B.14
4
1
ydx
ρπ
=
5
2
53
960
R
ρπ
=
The mass of the body is
222
R
3
5
960 200
24
xx
R
ρπ
Solution B.15
For elemental cylindrical shell shown in section,
()
() ( ) ( ) ( )
32 2322
433
4 1234
RRxRxxaxdx
I
ρπ
ρπ
=−+−−
=−+−
()
23
322
12
a
a
aR
Raxdx
mV
π
ρ
−
=−=
=
x
dx
a
R
y
Solution B.16
2
2
1
xx
L
dI dmy
=
()
2
1.129 mb
2b
x
dx
y
y’
y
L
–
Solution B.17
Consider the right half: from previous solution,
15
mLb
ρπ
=
22
1
yy
Solution B.18
1
x
ta
=−
yy
2
2
0
0
2
1 , density
1,
34 5 30
1mh
10
yy
yy
hh
x
ab dx
I ab x dx ab abh
hhh
I
ρρ
ρρ ρ
=− =
=− =−+=
=
t
h
z
z’
x
dx
Solution B.19
Let
ρ
= mass per unit surface area. For elemental ring of cross section a d
θ
(t) and
circumference
()
2cosRa
πθ
+:
()
0
2
22 2
sin2
33 cos 3 3
Ra Ra R
π
θθ
θπ
==+=
22
00
222
3
2
a
m
π
t
Solution B.20
0
42
3
42
3
xx zz