150 Chapter 5Integration
whereR 1 = 1 r
11 + 1 r
2is the distance between the masses. The moment of inertia is then
where the quantity μ 1 = 1 m
1m
22 (m
11 + 1 m
2)is the reduced massof the system of two
masses (see also Example 1.18). The moment of inertia about the centre of mass of a
system of two masses is therefore the same as the moment of inertia of a single mass μ
at distance Rfrom the centre of mass.
0 Exercise 50
The continuous case
Consider a continuous mass distribution such as a straight rod of matter of length l
(Figure 5.18).
Let the mass in the segment xtox 1 + 1 ∆xbe∆m, so that∆m 2 ∆xis the mass per unit
length in the segment. If the mass is distributed evenly over the length, then∆m 2 ∆x
is independent of the choice of segment. It is then the density(or, more correctly, the
linear mass density) ρof the body, a constant throughout the length. In this case, the
total mass isM 1 = 1 ρl. If the mass of the rod is notevenly distributed over its length,
then ∆m 2 ∆xis the average densityin the segmentxtox 1 + 1 ∆x, and its value depends
on the position of the segment and on its length. As this length is reduced to zero, the
ratio approaches the limit
(5.43)
The value of the functionρ(x)at each point is the density at that point, and the
differential quantitydm 1 = 1 ρ(x)dxis the mass of a segmentdxatx. The total mass of
the body is then
(5.44)
Therefore, when a discrete mass distribution is replaced by a continuous distribution,
the sums of the discrete case are replaced by integrals. Thus, the average density is
(5.45)
ρρ
==
M
l
xdx
dx
llZ
Z
00()
Mdm xdx
Ml==ZZ
00ρ()
ρ( ) limx
m
x
dm
dx
x
=
=
→∆
∆
∆
0
Imr mr
mm
mm
=+ = RR
=
112222
121222μ
................................................................................................................................................. .....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................0 x x+∆ x l
∆ m
x
...
.......
.......
.......
........
.......
.......
.......
......
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Figure 5.18