978-0073398242 Appendix B Solution Manual Part 10

subject Type Homework Help
subject Pages 9
subject Words 1430
subject Authors Brian Self, David Mazurek, E. Johnston, Ferdinand Beer, Phillip Cornwell

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page-pf1
PROBLEM B.50
Brass wire with a weight per unit length w is used to form the figure
shown. Determine the mass products of inertia I
xy
, I
yz
, and I
zx
of the wire
figure.
page-pf2
PROBLEM B.50 (Continued)
0
page-pf3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM B.51
Complete the derivation of Eqs. (9.47), which express the parallel-axis theorem for mass products of
inertia.
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PROBLEM B.52
For the homogeneous tetrahedron of mass m shown, (a) determine by
direct integration the mass product of inertia I
zx
, (b) deduce I
yz
and I
xy
from
the result obtained in part a.
page-pf5
PROBLEM B.52 (Continued)
(b) Because of the symmetry of the body,
x
y
I
and
y
z
I
can be deduced by considering the circular
permutation of
(, ,)
xyz
and
(,,)
abc
.
First divide the tetrahedron into a series of thin horizontal slices of thickness
dy
as shown.
ay

bb

The mass
dm
of the slab is
2
11
y

Now
,areazx zx
dI tdI
where tdy
22
1
Then
44
1
() 1 1
24
yy
dIzx dy a c
bb












page-pf6
PROBLEM B.52 (Continued)
Alternative solution for part a:
The equation of the included face of the tetrahedron is
1
xyz
abc

x
z

dm dV dydxdz

(1 / ) (1 / / )
00 0 ()
zx
I zxdm zx dydxdz


(1 / )
0
6
ca zc xz
ab z dz
ccc




345
c
or 1
20
zx
I
mac
page-pf7
PROBLEM B.53
The homogeneous circular cone shown has a mass m.
Determine the mass moment of inertia of the cone with respect
to the line joining the origin O and Point A.
page-pf8
PROBLEM B.54
The homogeneous circular cylinder shown has a mass m. Determine the mass
moment of inertia of the cylinder with respect to the line joining the origin O and
Point A that is located on the perimeter of the top surface of the cylinder.
222
OA x x y y z z
II I I


22
222
22 22
11
(3 4 )
212
ha
ma m a h
ha ha
 

 
 

 
or
22
222
1103
12
OA ha
Ima
ha
Note: For Point A located at an arbitrary point on the perimeter of the top surface,
OA
is given by
22
1(cos sin )
OA
aha
ha


ij k
which results in the same expression for
.
OA
I
page-pf9
PROBLEM B.55
Shown is the machine element of Problem 9.141. Determine its
mass moment of inertia with respect to the line joining the
origin O and Point A.
page-pfa
PROBLEM B.55 (Continued)
From the solution to Problem 9.141, we have
32
13.98800 10 kg m
x
I

13
OA λij
Substituting into Eq. (9.46)
222
222
OA x x y y z z xy x y yz y z zx z x
II I I I I I


22
32
23
22
2(0.39460) 10 kg m
 



OA
0 0
0

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