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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.
PROBLEM B.50 (Continued)
0
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PROBLEM B.51
Complete the derivation of Eqs. (9.47), which express the parallel-axis theorem for mass products of
inertia.
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.
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
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
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.
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
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.
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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