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10–77. Continued
I
min =
20.4 in
1091
10–78.
SOLUTION
Ans.
u
=6.08°
-2Ixy
The area of the cross section of an airplane wing has the
f
ollowing properties about the xand yaxes passing through
t
he centroid C:
D
etermine the orientation of the principal axes and the
prin
cipal moments of inertia.
Ixy =138 in4.Iy=1730in4,Ix=450 in4,
y
x
C
Ans:
1092
10–79.
SOLUTION
tan 2u=
-2Ixy
Ix-Iy
=
-2(138)
450-1730
y
x
C
Solve Prob. 10–78 using Mohr’s circle.
Ans:
1093
*10–80.
Determine the moments and product of inertia for the
shaded area with respect to the u and v axes.
SOLUTION
Moment And Product of Inertia About x and y Axes. Since the x axis is an axis
of symmetry,
I
xy
=0
. Also, the x axis passes through the centroids of the two
segments, Fig. a
Ans:
y
x
v
u
10 mm
10 mm
10 mm
20 mm
120 mm
120 mm
60
1094
10–81.
Solve Prob. 10–80 using Mohr’s circle.
SOLUTION
Moment And Product of Inertia About x and y Axes. Since the x axis is an axis
of symmetry,
I
xy
=0
. Also, the x axis passes through the centroids of the two
segments, Fig. a
Ix=
1
12
(20)(1203)+
1
12
(120)(203)=2.96(106) mm4
y
x
v
u
10 mm
10 mm
10 mm
20 mm
120 mm
120 mm
60
10–82.
Determine the directions of the principal axes with origin
located at point
O, and the principal moments of inertia for
the area about these axes
.
1096
10–83.
Solve Prob. 10–82 using Mohr’s circle.
SOLUTION
2 in.
y
2in.
2in.
1 in.
Ans:
1097
*10–84.
Determine
the moment of inertia of the thin ring about the
z
axis.The ring has a mass m.
x
y
R
Ans:
I
z=
m R
2
10–85.
Determine the moment of inertia of the ellipsoid with respect
to the xaxis and express the result in terms of the mass mof
the ellipsoid.Thematerial has aconstant density r.
SOLUTION
dm=Lpy2dx
y
x
a
b
1
b
2
y
2
a
2
x
2
10–86.
SOLUTION
Differential Disk Element: The mass of the differential disk element is
.The mass moment of inertia of this element
i
ne t
h
e ra
di
us of gyrat
i
on of t
h
e para
b
o
l
o
id
.T
h
e
density of the material is r=5Mg>m3.
kx
y
x
100mm
y
2
50 x
200mm
Ans:
10–87.
SOLUTION
The paraboloid is formed by revolving the shaded area
around the xaxis. Determine the moment of inertia about
the xaxis and express the result in terms of the total mass m
of the paraboloid. The material has a constant density .r
y
x
a
a2
–
hxy2=
1101
*10–88.
Determine the moment of inertia of the homogenous
triangular prism with respect to the yaxis. Express the
result in terms of the mass mof the prism. Hint: For
integration, use thin plate elements parallel to the x-y plane
having a thickness of dz.
SOLUTION
Differential Thin Plate Element: Here,The mass of the
ssam ehT si tnemele etalp niht laitnereffid
moment of inertia of this element about yaxis is
dm =rdV =rbxdz =raba1-z
hbdz.
x=aa1-z
hb.x
y
z
–h
a(x–a)z=
h
a
b
10–89.
Determ
i
ne t
h
e moment of
i
nert
i
a of t
h
e sem
i
-e
lli
pso
id
w
i
t
h
respect to the xaxis and express the result in terms of the
mass mof the semiellipsoid. The material has a constant
density r.
SOLUTION
Differential Disk Element: Here, .The mass of the differential disk element is
Mass Moment of Inertia: Performing the integration,we have
Ix=LdIx=La
0
rp b4
2ax4
a4-
2x2
a2+1bdx
y2=b2a1-x2
a2b
y
x
b
a
1
b
2
y
2
a
2
x
2
10–90.
Determine the radius of gyration kx of the solid formed by
revolving the shaded area about x axis. The density of the
material is
r.
SOLUTION
Differential Disk Element. The mass of the differential disk element shown
shaded in Fig. a is
dm =
r
dv =
rp
y2dx
. Here y=
1
n
x 1
n. Thus, dm =rp
a
h
1
n
x1
n
b2
Total Mass. Perform the integration,
m=
Lm
dm =
La
0
rph
2
a
2
n
(x 2
ndx)
y
x
h
a
yn x
a
hn
10–91.
The concrete shape is formed by rotating the shaded area
about the yaxis.Determine the moment of inertia The
specific weight of concrete is g=150 lb>ft3.
Iy.
SOLUTION
dI
y=
1
2(dm)(10)2-
1
2(dm)x2
y
x
8in.
6in. 4in.
2
9x
2
y
Ans:
1105
*10–92.
Determine the moment of inertia of the sphere and
express the result in terms of the total mass mof the sphere.
The sphere has a constant density r.
Ix
SOLUTION
dI
x=
y2dm
2
x
y
r
x
2
y
2
r
2
10–93.
The right circular cone is formed by revolving the shaded
area around the xaxis. Determine the moment of inertia
and express the result in terms of the total mass mof the
cone.The cone has a constant density .r
Ix
SOLUTION
m=Lh
0
r(p)
¢
r2
h2
≤
x2dx =rp
¢
r2
h2
≤
a1
3bh3=
1
3rp r2h
dm =rdV =r(py2dx)
y
x
r
r
–
hxy
h
