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10–94.
SOLUTION
The mass moment of inertia of this element about the yaxis is
.
Mass: The mass of the solid can be determined by integrating dm.Thus,
The mass of the solid is .Thus,
Mass Moment of Inertia: Integrating ,
Substituting into Iy,
Ans.Iy=
32p
7a375
pb=1.71(103)kg#m2
r=
375
p
kg>m3
Iy=LdIy=L4m
0
rp
512 y6dy =
rp
512
¢
y7
7
≤
`
4m
0
=
32p
7r
dIy
1500 =4pr
r=
375
p
kg>m3
m=1500 kg
m=Ldm =L4m
0
rp
16 y3dy =
rp
16
¢
y4
4
≤
`
4m
0
=4pr
dIy=
1
2dmr2=
1
2
A
rpr2dy
B
r2=
rp
2r4dy =
rp
2a1
4y3>2b4
dy =
rp
512 y6dy
4y3>2b2
16 y3dyr =z=
4y3>2
Determ
i
ne t
h
e mass moment of
i
nert
i
a of t
h
e so
lid
formed by revolving the shaded area around the axis.The
total mass of the solid is .1500 kg
y
Iy
y
x
z
4m
2m
z
2
y
3
1
––
16
O
10–95.
The slender rods have a mass of 4 kg
>
m. Determine the
moment of inertia of the assembly about an axis
perpendicular to the page and passing through point A.
A
200 mm
*10–96.
The pendulum consists of a 8-kg circular disk A, a 2-kg
circular disk B, and a 4-kg slender rod. Determine the radius
of gyration of the pendulum about an axis perpendicular to
the page and passing through point O.
SOLUTION
O
0.5 m
1 m
0.4 m 0.2 m
A
B
10–97.
Determine the moment of inertia of the frustum of the
cone which has a conical depression. The material has a
density of 200 kg>m3.
I
z
SOLUTION
-
3
10 [1
3p(0.4)2(0.6)(200)](0.4)2
-
3
10 [1
3p(0.2)2(0.8)(200)](0.2)2
Iz=
3
10[1
3p(0.4)2(1.6)(200)](0.4)2
z
0.8 m
0.6 m
0.2 m
0.4m
10–98.
SOLUTION
Ans.=4.45 kg #m2
=1
12 (3)(2)2+3(1.781 -1)2+1
12 (5)(0.52+12)+5(2.25 -1.781)2
IG=©IG¿+md2
©ym
The pendulum consists of the 3-kg slender rod and the 5-kg
thin plate.Determine the location of the center of mass G
of the pendulum; then find the mass moment of inertia of
the pendulum about an axis perpendicular to the page and
passing through G.
y
G
2m
1m
0.5 m
y
O
10–99.
Determine the mass moment of inertia of the thin plate
about an axis perpendicular to the page and passing
through point O.The material has a mass per unit area of
.20 kg>m2
SOLUTION
Composite Parts: The plate can be subdivided into the segments shown in Fig. a.
Here, the four similar holes of which the perpendicular distances measured from
their centers of mass to point Care the same and can be grouped as segment (2).
150 mm
O
50 mm
50 mm
150 mm
*10–100.
SOLUTION
Ans.
k
O=
A
IO
m
=
A
4.917
0.4969
=3.15ft
m
=a4
32.2b+a12
32.2b=0.4969 slug
=4.917slug#ft2
=1
12 a4
32.2b(5)2+a4
32.2b(0.5)2+1
12a12
32.2b(12+12)+a12
32.2b(3.5)2
I
O=©IG+md2
T
he pendulum consists of a plate having a weight of 12 lb
a
nd a slender rod having a weight of 4 lb.Determine the
r
adius of gyration of the pendulum about an axis
p
erpendicular to the page and passing through point O.
O
2ft
3ft
1ft
1ft
10–101.
SOLUTION
distance of from point Ocan be grouped as segment (2).
Mass Moment of Inertia: First, we will compute the mass moment of inertia of the
wheel about an axis perpendicular to the page and passing through point O.
The mass moment of inertia of the wheel about an axis perpendicular to the page and
passing through point Acan be found using the parallel-axis theorem ,
where and .Thus,
Ans.IA=84.94 +8.5404(42)=221.58 slug #ft2=222 slug #ft2
d=4ftm=100
32.2 +8
¢
20
32.2
≤
+15
32.2 =8.5404 slug
IA=IO+md2
=84.94 slug #ft2
IO=
¢
100
32.2
≤
(42)+8
B
1
12
¢
20
32.2
≤
(32)+
¢
20
32.2
≤
(2.52)
R
+
¢
15
32.2
≤
(12)
a1+3
2bft =2.5 ft
If the large ring,small ring and each of the spokes weigh
100 lb,15 lb,and 20 lb,respectively,determine the mass
moment of inertia of the wheel about an axis perpendicular
to the page and passing through point A.
O
1ft
4ft
10–102.
Determine the mass moment of inertia of the assembly
about the zaxis.The density of the material is 7.85 Mg m3.
SOLUTION
Mass: The mass of each segment is calculated as
Mass Moment of Inertia: Since the zaxis is parallel to the axis of the cone and the
hemisphere and passes through their center of mass,the mass moment of inertia can be
computed from and .Thus,
Ans. =29.4 kg #m2
=3
10(158.9625p)(0.32)+2
5(141.3p)(0.32)-3
10(5.8875p)(0.12)
Iz=©(Iz)i
3
10 m3r2
3
(Iz)1=3
10 m1r2
1, (Iz)2=2
5 m2r2
2,
m3=rV3=ra1
3pr2hb=7.85(103)c1
3p(0.12)(0.225) d=5.8875p kg
m2=rV2=ra2
3pr3b=7.85(103)c2
3p(0.33)d=141.3p kg
m1=rV1=ra1
3pr2hb=7.85(103)c1
3p(0.32)(0.675) d=158.9625p kg
0.45 +z=0.1
0.3
>z
y
x
300 mm
100 mm
1116
10–103.
Each of the three slender rods has a mass m. Determine the
moment of inertia of the assembly about an axis that is
perpendicular to the page and passes through the center
point O.
O
a
a
2
*10–104.
SOLUTION
distances measured from the centroid of each segment to the yaxis are indicated in
Fig. a.The mass moment of inertia of each segment about the yaxis can be
determined using the parallel-axis theorem.
Ans.=0.144 kg#m2
=2c1
12 (1.6)(0.42)+1.6(0.22)d-2c1
4(0.1p)(0.12)+0.1p(0.22)d
Iy=©
A
Iy
B
G+md2
The thin plate has a mass per unit area of .
Determine its mass moment of inertia about the yaxis.
10 kg
>
m
2
200 mm
200 mm
200 mm
z
100 mm
100 mm
10–105.
The thin plate has a mass per unit area of .
Determine its mass moment of inertia about the zaxis.
10 kg>m
2
SOLUTION
moment of inertia of each segment about the zaxis can be determined using the
parallel-axis theorem.
200 mm
200 mm
200 mm
z
100 mm
100 mm
=1
12 (1.6)(0.42)+c1
12 (1.6)(0.42+0.42)+1.6(0.22)d-1
4(0.1p)(0.12)-c1
2(0.1p)(0.12)+0.1p(0.22)d
Iz=©
A
Iz
B
G+md2
1119
10–106.
SOLUTION
Ans.
=
118 slug
#
ft2
+1
2ca 90
32.2 bp(2)2(0.25) d(2)2-1
2ca 90
32.2 bp(1)2(0.25) d(1)2
I
G=1
2ca 90
32.2 bp(2.5)2(1) d(2.5)2-1
2ca 90
32.2 bp(2)2(1) d(2)2
Determine the moment of inertia of the assembly about an
axis
that is perpendicular to the page and passes through
the
center of mass G.The material has a specific weight of
.
g
=90 lb>ft3
O
1ft
2ft
G
0.25 ft
Ans:
I
G=
118
slug
#
ft
2
10–107.
Determine the moment of inertia of the assembly about an
axis that is perpendicular to the page and passes through
point O.The material has a specific weight of .g=90 lb>ft3
SOLUTION
O
1ft
2ft
0.5 ft
G
0.25 ft
*10–108.
SOLUTION
Ans.
Ans.IG=5.61 kg #m2
-1
2[p(0.1)2(12)(0.1)2-[p(0.1)2(12)](1.8 -0.8878)2
+1
2[p(0.3)2(12)(0.3)2+[p(0.3)2(12)](1.8 -0.8878)2
+1
12 (1.5)(3)(1.5)2+1.5(3)(0.75 -0.8878)2
IG=1
12 (0.8)(3)(0.8)2+0.8(3)(0.8878)2
=0.8878 m =0.888 m
y=1.5(3)(0.75) +p(0.3)2(12)(1.8) -p(0.1)2(12)(1.8)
1.5(3) +p(0.3)2(12) -p(0.1)2(12) +0.8(3)
T
h
e pen
d
u
l
um cons
i
sts of two s
l
en
d
er ro
d
s AB an
d
OC
which have a mass of The thin plate has a mass of
Determine the location of the center of mass G
of the pendulum, then calculate the moment of inertia of
the pendulum about an axis perpendicular to the page and
passing through G.
y
12 kg>m2.
3kg>m.
AB
0.1 m
0.3 m
C
G
O
1.5
m
0.4 m 0.4 m
_
y
10–109.
Determine the moment of inertia of the frustrum of the
cone which has a conical depression. The material has a
density of 200 kg>m3.
I
z
SOLUTION
The mass moment of inertia of each cone about zaxis can be
determined using
Ans.=34.2 kg#m2
-3
10 cp
310.22210.6212002d10.222
-3
10 cp
310.22210.333212002d10.222
I
z=©1I
z2i=3
10 cp
310.82211.333212002d10.822
I
z=3
10 mr2.
z=0.333 m.
z
z
200 mm
600 mm
400 mm
