5.6 Static properties of matter 151
The position of the centre of mass is given by (see equation (5.40))
(5.46)
and the moment of inertia with respect to an arbitrary pointx
0on the line is
(5.47)
We note that, for example, the total mass can be interpreted as the ‘area under the
curve’ of the graph of the mass density functionρ(x).
EXAMPLE 5.14Find the total mass, centre of mass, and moment of inertia of the
linear distribution of mass of length aand densityρ(x) 1 = 1 b(a 1 − 1 x), 0 1 ≤ 1 x 1 ≤ 1 a.
The total mass is equal to the area a
2b 22 of the triangle
shaded in Figure 5.19. Thus
The position of the centre of mass is
The moment of inertia with respect to an arbitrary pointx
0on the line is
The value ofx
0for whichIhas its minimum value is given by
so thatx
01 = 1 a 23. The moment of inertia therefore has its smallest value when computed
with respect to the centre of mass; this is a general result.
0 Exercise 51
dI
dx
ba
ax
0200
12
== ()− + 412
Ixxxdxbaxxxdx
ba
aa=−=−−=ZZ
002002212
ρ()() ()() (()aaxx
2002−+ 46
X
M
xxdx
a
axxdx
a
aa==−=
12
3
020ZZρ() ( )
Mxdxbaxdx
ab
aa==−=ZZ
0022
ρ() ( )
Ixxxdx
l=−Z
002ρ()( )
MX x x dx X
xxdx
xdx
lll=, =Z
Z
0 Z
00ρ
ρ
ρ
()
()
()
or
x
ρ(x)
...................................................................................................................................................................................................................................................................................................................................................................................................
...
..
...
...
..
........
......
........
......
......ab
0 a
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Figure 5.19