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791
17–1.
SOLUTION
Thus,
Ans.Iy=1
3ml
2
m=rAl
=1
3rAl
3
=Ll
0
x2(rAdx)
Iy=L
M
x2dm
Determine the moment of inertia for the slender rod. The
rod’s density and cross-sectional area Aare constant.
Express the result in terms of the rod’s total mass m.
r
Iy
y
z
A
l
3
17–2.
The solid cylinder has an outer radius R, height h, and is
made from a material having a density that varies from its
center as where kand aare constants.
Determine the mass of the cylinder and its moment of
inertia about the zaxis.
r=k+ar2,
SOLUTION
Ans.
Ans.Iz=phR
4
2[k+2aR2
3]
Iz=2ph[kR
4
4+aR6
6]
Iz=2phLR
0
(kr
3+ar
5)dr
Iz=LR
0
r2(k+ar2)(2prdr)h
dI =r2dm =r2(r)(2prdr)h
m=phR
2(k+aR2
2)
m=2ph(kR2
2+aR4
4)
m=LR
0
(k+ar2)(2prdr)h
R
h
z
17–3.
Determ
i
ne t
h
e moment of
i
nert
i
a of t
h
e t
hi
n r
i
ng a
b
out the
zaxis.The ring has a mass m.
SOLUTION
Thus,
Ans.I
z
=mR
2
m=L2p
0
rARdu=2prAR
Iz=L2p
0
rA(Rdu)R2=2prAR
3
x
y
R
*17–4.
SOLUTION
Ans.kx=AIx
m=A50
3(200) =57.7 mm
=rpa50
2b(200)2
=rp(50)c1
2x2d200
0
m=Ldm =L200
0
pr(50x)dx
=rpa502
6b(200)3
=rpa502
2bc1
3x3d200
0
Ix=L1
2y2dm =1
2L200
0
50 x{pr(50x)} dx
dm =rpy2dx =rp(50x)dx
The paraboloid is formed by revolving the shaded area
around the xaxis. Determine the radius of gyration .The
density of the material is .r=5Mg>m3
kx
y
x
y250x
100 mm
17–5.
SOLUTION
Ans.kx=AIx
m=A86.17r
60.32r
=1.20 in.
m=L8
0
prx2/3 dx =60.32r
Ix=L8
0
1
2prx4/3 dx =86.17r
dI
x=1
2(dm)y2=1
2pry4dx
dm =rdV =rp y2dx
Determine the radius of gyration of the body. The
specific weight of the material is g=380 lb>ft3.
kx
y
x
2in.
y
3
=x
8in.
796
17–6.
The sphere 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 sphere.The
material has a constant density r.
Ix
Thus,
Ans.Ix=2
5mr
2
=4
3rpr3
m=Lr
-r
rp(r2-x2)dx
=8
15 pr r5
Ix=Lr
-r
1
2rp(r2-x2)2dx
dIx=1
2rp(r2-x2)2dx
dm =rdV =r(py2dx)=rp(r2-x2)dx
x
y
x2+y2=r2
5
797
17–7.
The frustum is formed by rotating the shaded area around
the xaxis. Determine the moment of inertia and express
the result in terms of the total mass mof the frustum. The
frustum has a constant density .r
Ix
SOLUTION
Ans.Ix=93
70
mb2
m=L
m
dm =rp La
0
A
b2
a2x2+2b2
ax+b2
B
dx =7
3rpab2
=31
10rpab4
Ix=LdIx=1
2rp La
0
A
b4
a4x4+4b4
a3x3+6b4
a2x2+4b4
ax+b4
B
dx
dIx=1
2rp
A
b4
a4x4+4b4
a3x3+6b4
a2x2+4b4
ax+b4
B
dx
dIx=1
2dmy2=1
2rpy4dx
dm =rdV =rpy2dx =rp
A
b2
a2x2+2b2
ax+b2
B
dx
y
x
2b
b
–
axb
y
a
z
b
70
798
*17–8.
The hemisphere is formed by rotating the shaded area
around the yaxis. Determine the moment of inertia and
express the result in terms of the total mass mof the
hemisphere.The material has a constant density .r
Iy
Thus,
Ans.Iy=2
5mr
2
=rp
2cr4y-2
3r2y3+y5
5dr
0
=4rp
15 r5
Iy=L
m
1
2(dm)x2=r
2Lr
0
px4dy =rp
2Lr
0
(r2-y2)2dy
=rpcr2y-1
3y3dr
0
=2
3rp r3
V
0
0
x2y2r2
y
x
5
799
17–9.
SOLUTION
Thus,
Ans.Iy=m
6(a2+h2)
m=rV=1
2abhr
=1
12 abhr(a2+h2)
=abr[a2
3h3(h4-3
2h4+h4-1
4h4)+1
h(1
3h4-1
4h4)]
Iy=abrLk
0
[a3
3(h-z
h)3+z2(1 -z
h)]dz
=[b(a)(1 -z
h)dz](r)[a2
3(1 -z
h)2+z2]
=dm(x2
3+z2)
=1
12 dm(x2)+dm(x2
4)+dmz2
dIy=dIy+(dm)[(x
2)2+z2]
Determine the moment of inertia of the homogeneous
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–yplane and having a
thickness dz.
z
–h
––
a(x–a)z=
h
Ans:
Iy=
m
6
(
a2+h2
)
800
17–10.
The pendulum consists of a 4-kg circular plate and a
2-kg slender rod. Determine the radius of gyration of the
pendulum about an axis perpendicular to the page and
passing through point O.
Ans:
kO=2.17 m
O=
G+
=
c1
12
(2)
(
22
)
+2
(
1
2
)
d
+
c1
2
(4)
(
0.52
)
+4
(
2.52
)
d
=
28.17 kg
#
m
2
Thus, the radius of gyration is
kO=
AI
O
m
=
A28.17
4+2
=2.167 m =2.17 m Ans.
O
2 m
17–11.
The assembly is made of the slender rods that have a mass
per unit length of 3 kg
>
m. Determine the mass moment of
inertia of the assembly about an axis perpendicular to the
page and passing through point O.
O
0.8 m
0.4 m
0.4 m
SOLUTION
*17–12.
SOLUTION
Ans.=5.64 slug #ft2
=c1
2p(0.5)2(3)(0.5)2+3
10 a1
3bp(0.5)2(4)(0.5)2-3
10 a1
2bp(0.25)2(2)(0.25)2da490
32.2 b
Ix=1
2m1(0.5)2+3
10 m2(0.5)2-3
10 m3(0.25)2
D
etermine the moment of inertia of the solid steel assembly
about the xaxis. Steel has a specific weight of
.gst =490 lb>ft3
2ft3 ft
0.5 ft
0.25 ft
x
803
17–13.
SOLUTION
Ans.=7.67 kg #m2
=c2c1
12 (4)(1)2d+10(0.5)2d+18(0.5)2
IA=Io+md3
The wheel consists of a thin ring having a mass of 10 kg and
four spokes made from slender rods, each having a mass of
2kg. Determine the wheel’s moment of inertia about an
axis perpendicular to the page and passing through point A.
500 mm
Ans:
I
A=7.67
kg #m2
17–14.
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.I
A=
84.94
+
8.5404(42)
=
221.58 slug
#
ft2
=
222 slug
#
ft2
d=4ftm=100
32.2 +8a20
32.2 b+15
32.2 =8.5404 slugIA=IO+md2
=84.94 slug #ft2
IO=a100
32.2 b(42)+8c1
12 a20
32.2 b(32)+a20
32.2 b(2.52)d+a15
32.2 b(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
17–15.
SOLUTION
Ans.IO=1.5987 +4.7233(1.4 sin 45°)2=6.23 kg #m2
m=50(1.4)(1.4)(0.05) -50(p)(0.15)2(0.05) =4.7233 kg
IO=IG+md2
=1.5987 kg #m2
IG=1
12
C
50(1.4)(1.4)(0.05)
DC
(1.4)2+(1.4)2
D
-1
2
C
50(p)(0.15)2(0.05)
D
(0.15)2
Determine the moment of inertia about an axis perpendicular
to the page and passing through the pin at O. The thin plate
has a hole in its center. Its thickness is 50 mm, and the
material has a density r =50 kg>m3.
150 mm
O
Ans:
*17–16.
SOLUTION
inertia of the plate about an axis perpendicular to the page and passing through point
Ofor each segment can be determined using the parallel-axis theorem.
Ans.=0.113 kg #m2
=c1
2(0.8p)(0.22)+0.8p(0.22)d-c1
12 (0.8)(0.22+0.22)+0.8(0.22)d
IO=©IG+md2
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
200 mm
200 mm
O
807
17–17.
Determine the location
y
of the center of mass G of the
assembly and then calculate the moment of inertia about
an axis perpendicular to the page and passing through G.
The block has a mass of 3 kg and the semicylinder has a
mass of 5 kg.
Ans:
I
G=0.230 kg #m2
SOLUTION
G
400 mm
300 mm
200 mm
O
y
–
17–18.
Determine the moment of inertia of the assembly about an
axis perpendicular to the page and passing through point O.
The block has a mass of 3 kg, and the semicylinder has a
mass of 5 kg.
G
400 mm
300 mm
200 mm
O
y
–
SOLUTION
1
809
17–19.
Determine the moment of inertia of the wheel about an
axis which is perpendicular to the page and passes through
the center of mass G. The material has a specific weight
g
=
90 lb
>
ft
3
.
Ans:
I
G=118 slug #ft2
SOLUTION
1
1
G
O
0.5 ft
0.25 ft
0.25 ft
1 ft
2 ft
*17–20.
Determine the moment of inertia of the wheel about an axis
which is perpendicular to the page and passes through point O.
The material has a specific weight g
=
90 lb
>
ft
3
.
G
O
0.5 ft
1 ft
0.25 ft
0.25 ft
1 ft
2 ft
SOLUTION
90
