Find out Moment of Inertia of circular area:
Find out Moment of Inertia of circular area of radius a = 10 cm about its centroidal axis OX as illustrated in Figure
Solution
Let a thin strip of width B_{y} at distance y from axis OX where,
y = a sin θ and B_{y } = 2 (a cos θ)
∴ dy = a cos θ d θ
dA = B_{y} dy = (2 a cos θ) a cos θ d θ
All elements ought to be considered for values of θ ranging from - 90^{o }to + 90^{o} + 90^{o}
Now that
cos 2 θ = cos^{2} θ - sin^{ 2} θ = 2 cos^{2} θ - 1
∴ 2 cos^{2} θ = 1 + cos 2 θ
and,
cos 4 θ = 1 - 2 (sin 2 θ)^{2}
∴ 2 (sin 2 θ)^{2} = 1 - cos 4 θ
∴ I_{x} = a^{4}/2 ∫ 2 [ sin 2 θ]^{2} d θ = a^{4}/2 ∫ (1 - cos 4 θ) d θ
= π a ^{4} / 4
= π D ^{4}/64 , where D = diameter of circle = 2 a
∴ I _{x} = π× 10^{ 4} /4 = 7854 cm4 .
This is to be noted down that circular cross-section is axis-symmetric, that means it is symmetric w.r.t. both x and y axes or any other radial direction and also has similar nature of shape regarding all centroidal axes.
∴ I_{y } = I_{x} = πa^{4} /4= π a^{2} (a^{2}/4) = A (a^{2}/4)
where, A = Area of the circle of radius a.
Letting perpendicular axis theorem, we have following :
I _{z } = I _{x} + I _{y} = A (a^{2}/2)