COSMOS: Complete Online Solutions Manual Organization System
Chapter 9, Solution 197.
m = ρV = ρπ a 2 L
For the cylinder
dm = ρπ a 2dx
For the element shown
=
dI z = dI z + x 2dm
and
=
Then
m
dx
L
1 2
a dm + x 2dm
4
L
1 2
1
m m 1
I z = ∫ dI z = ∫
a + x 2 dx = a 2 x + x3
3 0
4
L L 4
L
0
=
m1 2
1 3
a L+ L
L4
3
or I z =
1
m 3a 2 + 4L2
12
(
)
COSMOS Chapter 9 Solution 197
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