Environmental Engineering Reference
In-Depth Information
resort to the approximation that the matter in the body is continuous and specify
a function,
ρ(
r
)
, to tell us how that matter is distributed in space. We now need
to write down the version of Eq. (4.25) appropriate for a continuous body. To that
end, we imagining breaking the body up into an infinity of tiny volume elements.
The mass of the element at
r
=
(x,y,z)
is
d
m
=
ρ(
r
)
d
V
=
ρ(x,y,z)
d
x
d
y
d
z,
(4.26)
=
where d
V
d
x
d
y
d
z
is the volume of the element. The sum in Eq. (4.25) now
becomes a triple integral:
(x
2
y
2
)ρ(x, y, z)
d
x
d
y
d
z
I
=
+
(4.27)
and we have used Pythagoras' Theorem to write the square of the distance from
the rotation axis (the
z
-axis) as
d
2
y
2
. Performing the integral is generally
not so easy and we underline the importance of looking for symmetries to make
the calculation easier.
x
2
=
+
Example 4.3.1
Calculate the moment of inertia of a uniform, thin, straight beam
of length l and mass M, for rotations about an axis that passes through one end of
the beam and runs perpendicular to it. Calculate also the moment of inertia when
the axis runs through the centre of the beam and is perpendicular to it.
Solution 4.3.1
This is a one-dimensional problem with a uniform mass distribution
and we can write the element of mass
M
l
d
m
=
d
x
with x as shown in Figure 4.4(a). The moment of inertia is
l
x
2
M
l
I
=
d
x.
0
Doing the integral gives
l
3
3
M
l
1
3
Ml
2
=
=
I
(4.28)
R
l
d
x
r
x
d
r
(a)
(b)
Figure 4.4
Calculating the moments of inertia of: (a) a beam and (b) a solid disc.
