3 
b 
4 
a — a 
point O. When O is the 
~centre of gravity of the figure 
MOMENTS AND CENTROIDS 57 
EF about the axes OP and OQ respectively. It is required to find the 
relations between P and Q, and A, B, and @. 
Consider a small element of the figure EF at L, the area of this 
element being a. Draw LM perpendicular to OA, LNK and MH per- 
pendicular to OP, and MK parallel to OP. Let OM=z, and LM=y. 
LN=LK- KN =LK-MH=y cos 0-2 sin 0, and 
LN?=,? cos? 0+ 2° sin? 0—2ay sin 0 cos 0 
=y* cos? 0+ x? sin? 0—axy sin 20. 
The moment of inertia of the element at L about the axis OP is 
equal to ay? cos? 0+ ax? sin? @-—azxy sin 26, and the moment of inertia 
of the whole figure EF about OP is 
P= az’ cos? 6 + Laz? sin? 6 —- Lary sin 26, therefore 
P=A cos? 6+B sin? 0—C sin 20, where C = Lazy. 
Changing @ into 90° + 6, Q=A sin? 6+ B cos? 0+C sin 20. 
Hence P— Q=(A-B) cos 20-20 sin 20. 
If the moment of inertia of the figure is a maximum about the axis 
OP, then P will be a maximum and Q a minimum, also P—Q will be a 
maximum. 
Differentiating, ieee — 2(A —B) sin 20-4C cos 20, and when 
P—Q isa maximum, — 2(A — B) sin 20—4C cos 20=0, and 
2C = —(A-B) tan 20. 
Hence when P — Q is a maximum 
P-—Q=(A-B) cos 20+(A—-B) tan 20 sin 20, therefore 
A-B=(P-Q) cos2¢. But A+B=P+Q, therefore 
har( ites) + Q(-=as "ap cos? 04+ Q sin? 0, and 
2 
p=P(i- 2) +Q (98%) -P sin? 6+ Q cos? 0. 
The axes OP and OQ, about which the moments of inertia are a 
maximum and a minimum respectively, are called the principal axes of 
inertia of the figure for the 
the axes OP and OQ are then 
called the principal axes of 
inertia of the figure. 
If a plane figure is sym- 
metrical about an axis in its 
plane, it is obvious that that 
axis is one of the principal 
axes, and if the figure is 
symmetrical about two _per- 
pendicular axes in its plane, 
these will be the principal 
axes. 
73. Inertia Curves and 
Momental Ellipse.—Let OP 
and OQ (Fig. 64) be the 
principal axes of inertia of the plane figure shown by dotted lines. 
7 
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