g
, PROBLEMg9.1
Determinegbygdirectgintegrationgthegmomentgofginertiagofgthegshadedgarea
gwithgrespectgtogthegygaxis.
SOLUTION
Atg xg=g0,g g yg=gbg:g g bg=gkg(0g−g
)2 bg
kg =g yg=g 2 (gxg−ga) ;gdAg=g y
2g
2
then
b a a
dIgyg =g x2dAg=g x2gyg
dx agg 2g
Igyg =g 0gb a2 x xg−ga dx
2
g
ag g b
=g 0gg a x −g2ax +ga2g x
4 3 2
dx
g
2
bg g 1 a2gx3g
a
= 21 x15g −
a g g ax g 3
4
0
+
5
1g
Igy g =g a3bg
30
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, PROBLEMg9.2
Determineg byg directg integrationg theg momentgofg inertiagofg theg shadedg ar
eagwithgrespectgtogthegygaxis.
SOLUTION
yg=gkx1/3
Forg xg=ga:
bg 1/3
Thus: yg= x
a1/3
dIgyg =g x2dAg=g x2gyd
x
b bg
dIy =gx2 1/3 x1/3dxg= 1/3 x7/3dx
a a
b a b g 3 3g
=
a
Iy =g g dIyg g 1/3
0
x7/3dxg = 1/3 a10/3
a 10
Igy g =g a3bg
10
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, PROBLEMg9.3
Determineg byg directg integrationg theg momentgofg inertiagofg theg shadedg ar
eagwithgrespectgtogthegygaxis.
SOLUTION
hg
Bygobservation yg=g x
gb
gNow dIgyg =gx2 dAg=g x2[(hg−g y)dx]
xg
=ghx2 1g− dx
bg
b xg
Then
Igy g =g g dIyg g = hx 2 1g− gdx
0 bg
b
g1 3 x4g 1
=ghgg g x g or Igyg = b3hg◀
− g
3 4bg0 12
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