PROBLEM 9.162
Thin aluminum wire of uniform diameter is used to form the figure
shown. Denoting by
m
′
the mass per unit length of wire, determine the
products of inertia
,
xy
I
,
yz
I
and
zx
I
of the wire figure.
SOLUTION
First compute the mass of each component. Have
m
mL
m
L
L
′
==
Then
15
1
1
22
mmm
R
m
R
ππ
′
′
=
( )
24
2
1
RR
′
−
322
mmR
m
R
′
′
Now observe that because of symmetry the centroidal products of inertia,
,,
x
yy
z
I
I
′
′′
′
and
,
zx
I
′ ′
of components 2 and 4 are zero and
( )
( )
( )
( )
1
13
3
00
xy
yz
II
=
=
( )
( )
5
5
0
=
=
Also
12
2 34
45
0
xx
y
y
y
z
z
=
=
=
Using the parallelaxis theorem [Equations (9.47)], it follows that
x
zz
x
I
for components 2 and 4.
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 07/11/2011 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.
 Spring '08
 Jenkins
 Statics

Click to edit the document details