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1109
21–1.
SOLUTION
However,,where ris the distance from the origin Oto dm. Since
is constant, it does not depend on the orientation of the x,y,zaxis. Consequently,
is also indepenent of the orientation of the x,y,zaxis. Q.E.D.Ixx +I
yy
+Izz
ƒrƒ
x2+y2+z2=r2
=2L
m
(x2+y2+z2)dm
Ixx +Iyy +Izz =L
m
(y2+z2)dm +L
m
(x2+z2)dm +L
m
(x2+y2)dm
Show that the sum of the moments of inertia of a body,
, is independent of the orientation of the
x,y,zaxes and thus depends only on the location of its
Ixx +Iyy +Izz
origin.
21–2.
Determine the moment of inertia of the cone with respect to
avertical axis passing through the cone’s center of mass.
What is the moment of inertia about a parallel axis that
passes through the diameter of the base of the cone? The
cone has a mass m.
y¿
y
SOLUTION
However,
Hence,
Using the parallel axis theorem:
Ans.
Ans.=m
2
0
(2h2+3a2)
=3m
80 (h2+4a2)+mah
4b2
Iy'=Iy+md2
Iy=3m
80 (h2+4a2)
3m
20 (4h2+a2)=Iy+ma3h
4b2
Iy=Iy+md2
Iy=3m
20 (4h2+a2)
m=L
m
dm =rpa2
h2Lh
0
x2dx =rpa2h
3
Iy=LdIy=rpa2
4h4(4h2+a2)Lh
0
x4dx =rpa2h
20 (4h2+a2)
=rpa2
4h4(4h2+a2)x4dx
=1
4
B
rpa2
h2x2dx
R
aa
hxb2
+
¢
rpa2
h2x2
≤
x2dx
dIy=1
4dmy2+dmx2
x
y
a
yy¿
–
1111
21–3.
SOLUTION
Ans.=69.33(p)(12) =2614 slug #ft2
Iy=LdIy=rp L4
0
(1
4x2+x3)dx =69.33 pr
=rp(1
4x2+x3)dx
=1
4[rpxdx](x)+(rpxdx)x2
dIy=1
2dmy2+dmx2
Determine moment of inertia of the solid formed by
revolving the shaded area around the xaxis. The density of
the material is r=12 slug/ft3.
Iy
2ft
y
x
y2=x
Iy=2614 slug
#
ft
2
*21–4.
SOLUTION
Ans.
Ans.Iy=LdIy=L2
0
10y2dy =26.7 slug #ft2
dIy=1
2dmz2=2rpy2dy =10y2dy
Ix=LdIx=L2
0
A
5y2+10y3
B
dy =53.3 slug #ft2
=
A
5y2+10y3
B
dy
=1
4[2rpydy](2y)+[2rpydy]y2
dIx=1
4dmz2+dm
A
y2
B
20 =4rp
r=5
pslug/ft3
m=20 =L
m
dm =L2
0
2rpydy
Determine the moments of inertia and of the paraboloid
of revolution. The mass of the paraboloid is 20 slug.
Iy
Ix
2ft
z
y
z22y
1113
21–5.
Determine by direct integration the product of inertia for
the homogeneous prism. The density of the material is .
Express the result in terms of the total mass mof the prism.
r
I
yz
SOLUTION
Using the parallel axis theorem:
Ans.Iyz =rh2
2La
0
(ay -y2)dy =ra3h2
12 =1
6ara2h
2b(ah)=m
6ah
=rh2
2(ay -y2)dy
=rh2
2xydy
=0+(rhxdy)(y)ah
2b
dIyz =(dIy¿z¿)G+dmyGzG
m=L
m
dm =rhLa
0
(a-y)dy =ra2h
2
a
y
z
a
h
Ans:
Iyz =
m
6
ah
1114
21–6.
SOLUTION
Using the parallel axis theorem:
Ans.=ra4h
24 =1
12
a
ra2h
2ba2=m
12a2
Ixy =rh
2La
0
(y3-2ay2+a2y)dy
=rh2
2(y3-2ay2+a2y)dy
=rh2
2x2ydy
=0+(rhxdy)ax
2b(y)
dIxy =(dIx¿y¿)G+dmxGyG
m=L
m
dm =rhLa
0
(a-y)dy =ra2h
2
Determine by direct integration the product of inertia for
the homogeneous prism. The density of the material is .
Express the result in terms of the total mass mof the prism.
r
Ixy
a
y
z
a
h
12
1115
21–7.
Determine the product of inertia of the object formed
by revolving the shaded area about the line
Express the result in terms of the density of the material, r.
x=5 ft.
Ixy
SOLUTION
Thus,
The solid is symmetric about y, thus
Ans.Ixy =636r
=0+5(1.055)(38.4rp)
Ixy =Ixy¿+xym
Ixy¿=0
y=40.5rp
38.4rp =1.055 ft
=40.5rp
=rp L3
0
(5 -x)(3x)dx
L3
0
y
~dm =r2pL3
0
y
2(5 -x)ydx
L3
0
dm =r2pL3
0
(5 -x)ydx=r2pL3
0
(5 -x)23xdx =38.4rp
3ft2ft
y
x
y
2
3x
I
xy
=636
r
1116
*21–8.
SOLUTION
Mass of body;
Ans.
Also,
=2rpL3
0
(5 -x)(3x)1/2 dx
=2rpL3
0
(5 -x)ydx
m=L3
0
dm
=466.29 rp
=2rp L3
0
(5 -x)3(3x)1/2 dx
=L3
0
(5 -x)2r(2p)(5 -x)ydx
Iy¿=L3
0
r2dm
Iy=4.48(103)r
=1426.29 rp
Iy=466.29 rp +(38.4 rp)(5)2
=38.4 rp
=L3
0
rp(5 -y2
3)2dy -12 rp
m=L3
0
rp(5 -x)2dy -m¿
Iy¿=490.29 rp -1
2(12 rp)(2)2=466.29 rp
m¿=rp(2)2(3) =12 rp
=490.29 rp
=1
2rpL3
0a5-y2
3b4
dy
L3
0
1
2dm r2=1
2L3
0
rp(5 -x)4dy
Iy¿=L3
0
1
2dm r2-1
2(m¿)(2)2
Determine the moment of inertia of the object formed by
revolving the shaded area about the line Express
the result in terms of the density of the material, r.
x=5ft.
Iy
3ft2ft
y
x
y
2
3x
21–9.
Determine the moment of inertia of the cone about the
axis.The weight of the cone is 15 lb, the height is
and the radius is r=0.5 ft.h=1.5 ft,
z¿
SOLUTION
Using Eq. 21–5.
Ans.Iz¿z¿=0.2062( 15
32.2)=0.0961 slug #ft2
=0.2062m
=0+[cos(108.43°)]2(1.3875m)+[cos(18.43°)]2(0.075m)
Iz¿z¿=u2
z¿xIxx +u2
z¿yIyy +u2
z¿zIzz
Ixy =Iyz =Izx =0
Iz=3
10 m(0.5)2=0.075 m
Ixx =Iyy =1.3875 m
Ixx =Iyy =[3
80m{4(0.5)2+(1.5)2}] +m[1.5 -(1.5
4)]2
u=tan-1(0.5
1.5)=18.43°
z¿
z¿
z
r
h
1118
21–10.
Determine the radii of gyration and for the solid
formed by revolving the shaded area about the yaxis.The
density of the material is r.
ky
kx
SOLUTION
However,
Hence, Ans.
For :
Hence, Ans.
kx=Ix
m=78.99r
24.35r=1.80 ft
Ix=I¿x+I¿¿x=28.53r+50.46r=78.99r
=50.46r
I¿¿x=1
4
C
rp(4)2(0.25)
D
(4)2+
C
rp(4)2(0.25)
D
(0.125)2
I¿x=LdI¿x=rp L4
0.25
A
1
4y4+1
B
dy =28.53r
=rp
A
1
4y4+1
B
dy
=1
4crpdy
y
2daI
y2b+arpdy
y
2by2
dI¿
x=1
4dmx2+dmy2
0.25 ft 6y…4ftkx
ky=AIy
m=A134.03r
24.35r=2.35 ft
m=L
m
dm =rp L4
0.25
dy
y2+r
C
p(4)2(0.25)
D
=24.35r
=134.03r
Iy=LdIy=1
2rp L4
0.25
dy
y4+1
2
C
r(p)(4)2(0.25)
D
(4)2
2dmx2=1
2
C
y2
DA
1
y2
B
2rpdy
y4
y2
4ft
0.25 ft
0.25 ft
y
xy 1
Ans:
k
y
=2.35 ft
kx=1.80 ft
21–11.
Determine the moment of inertia of the cylinder with
respect to the a–a axis of the cylinder.The cylinder has a
mass m.
a
a
*21–12.
SOLUTION
Vertical plates:
Using Eq. 21–5,
Thus,
Ans.Ixx =0.0155 +2(0.00686) =0.0292 slug #ft2
=0.00686
Ixx =(0.707)2(0.001372) +(0.707)2(0.01235)
=0.01235
Iy¿y¿=(6(1
4)(122)
32.2 )( 1
12)[(1
4)2+(122)2]+(6(1
4)(122)
32.2 )(1
8)2
Ix¿x¿=1
3(6(1
4)(122)
32.2 )(1
4)2=0.001372
lxx¿=0.707,
lxy¿=0.707,
lxz¿=0
Determine the moment of inertia of the composite plate
assembly.The plates have a specific weight of 6lb>ft2.
Ix
0.5 ft
0.5 ft
0.5 ft
0.5 ft
z
y
0.25 ft
1121
21–13.
SOLUTION
0.5 ft
0.5 ft
0.5 ft
0.5 ft
z
y
0.25 ft
Determine the product of inertia of the composite
plate assembly. The plates have a weight of 6 lb>ft
2.
Iyz
Ans:
I
yz
=0
21–14.
Determine the products of inertia ,,and ,of the
thin plate.The material has a density per unit area of
.50 kg>m2
Ixz
Iyz
Ixy
SOLUTION
200 mm
400 mm
z
21–15.
SOLUTION
Ans.=0.0595 kg #m2
=0.415(-0.3162)2+0+0.02(0.9487)2-0-0-0
Iz¿=Ixu2
x+Iyu2
y+Izu2
z-2Ixy uxuy-2Iyz uyuz-2Izx uzux
ux=cos (90° +18.43°) =-0.3162
uz=cos (18.43°) =0.9487,
uy=cos 90°=0,
Iz=1
2(4)(0.1)2=0.02 kg #m2
=0.415 kg #m2
Iy=Ix=
B
1
4(4)(0.1)2+4(0.3)2
R
+1
3(1.5)(0.3)2
Determine the moment of inertia of both the 1.5-kg rod
and 4-kg disk about the axis.z¿
300 mm z
z'
100 mm
1124
*21–16.
The bent rod has a mass of
3 kg>m
. Determine the moment
of inertia of the rod about the O–a axis.
SOLUTION
The bent rod is subdivided into three segments and the location of center of mass for
each segment is indicated in Fig. a. The mass of each segments is
m1=3(1) =3 kg
,
m2=3(0.5) =1.5 kg
and
m3=3(0.3) =0.9 kg
.
Ixx =
c1
12
(3)
(
1
2
)
+3
(
0.52
)
d
+
3
0+1.5
(
1
2
)
4
+
c1
12
(0.9)
(
0.32
)
+0.9
(
0.152+1
2
)
d
=
3.427 kg
#
m
2
Iyy =0+
c
1
12
(1.5)
(
0.52
)
+1.5
(
0.252
)
d
+
c
1
12
(0.9)
(
0.32
)
+0.9
(
0.152+0.52
)
d
=0.377 kg #m2
Izz =
c1
12
(3)
(
1
2
)
+3
(
0.52
)
d
+
c1
12
(1.5)
(
0.52
)
+1.5
(
1
2+0.252
)
d
+
3
0+0.9
(
1
2+0.52
)
4
=
3.75 kg
#
m
2
Ixy =[0 +0] +[0 +1.5(0.25)(-1)] +[0 +0.9(0.5)(-1)] =-0.825 kg
#
m
2
Iyz =[0 +0] +[0 +0] +[0 +0.9(-1)(0.15)] =-0.135 kg
#
m
2
I
zx =
[0
+
0]
+
[0
+
0]
+
[0
+
0.9(0.15)(0.5)]
=
0.0675 kg
#
m
2
The unit vector that denes the direction of the
Oa
axis is
UOa
=
0.5i-1j+0.3k
2
0.52
+
(
-
1)2
+
0.32
=
0.5
2
1.34
i-
1
2
1.34
j+
0.3
2
1.34
k
Thus, ux=
0.5
21.34
uy=-
1
21.34
uz=
0.3
21.34
xy
0.5 m
0.3 m
1 m
a
O
z
*21–16. Continued
21–17.
SOLUTION
Ans.
Ans.
Ans.
Ans.
=0.0427 slug #ft2
+1
12 a1.5
32.2 b(1)2+a1.5
32.2 b(0.3333)2
Iz¿=2c1
12 a1.5
32.2 b(1)2+a1.5
32.2 b(0.52+0.16672)d
=0.0155 slug #ft2
Iy¿=2c1
12 a1.5
32.2 b(1)2+a1.5
32.2 b(0.667 -0.5)2d+a1.5
32.2 b(1 -0.667)2
=0.0272 slug #ft2
Ix¿=2ca1.5
32.2 b(0.5)2d+1
12 a1.5
32.2 b(1)2
x=©xW
©w=(-1)(1.5)(1) +2
C
(-0.5)(1.5)(1)
D
3
C
1.5(1)
D
=-0.667 ft
The bent rod has a weight of Locate the center of
gravity G() and determine the principal moments of
inertia and of the rod with respect to the
axes.z¿
y¿,x¿,Iz¿
Iy¿,Ix¿,
y
x,
1.5 lb
>
ft.
y
z
y¿
z¿
1ft
1ft
G
A
_
x
_
y
1127
21–18.
Determine the moment of inertia of the rod-and-disk
assembly about the xaxis.The disks each have a weight of
12 lb.The two rods each have a weight of 4 lb, and their
ends extend to the rims of the disks.
SOLUTION
For a disk:
Thus.
Ans.Ix=2(0.04141) +2(0.1863) =0.455 slug #ft2
Ix=a1
2ba 12
32.2 b(1)2=0.1863 slug #ft2
Ix=0+0+(0.08282)(-0.7071)2=0.04141 slug #ft2
Ix¿y¿=Iy¿z¿=Ix¿z¿=0
Iy¿=0
Ix¿=Iz¿=a1
12 ba 4
32.2 b
C
(2)2+(2)2
D
=0.08282 slug #ft2
ux¿=cos 90° =0, uy¿=cos 45° =0.7071,
uz¿=cos (90° +45°) =-0.7071
2ft
1ft
x
1ft
Ans:
I
x=
0.455 slug
#
ft
2
21–19.
SOLUTION
Ans.Iaa =1.13 slug #ft2
Iaa =0+(0.707)2(1.6975) +(0.707)2(0.559)
Ixx =Iyy =1.6975 slug #ft2
Ixx =Iyy =1
12(20
32.2)[3(1)2+(2)2]+2[0.259( 10
32.2)(1)2+10
32.2(11
8)2]
=0.5590 slug #ft2
Izz =1
2(20
32.2)(1)2+2[2
5(10
32.2)(1)2]
uay =0.707
uax =0
uaz =0.707
Determine the moment of inertia of the composite body
about the aa axis.The cylinder weighs 20 lb, and each
hemisphere weighs 10 lb.
2ft
2ft
a
a
