Parallel axis theorem; centroid of area between parabolas

Parallel axis theorem

Statement: the moment of inertia of an area about any axis equals its moment of inertia about a parallel axis through its centroid plus the product of the area and the square of the distance between the two axes.

I = I_G + A d²

Proof: let G be the centroid of the area A, and let X′ be the centroidal axis. Take a parallel axis X at distance d from X′. For a small element dA at distance y′ from X′, its distance from X is y = y′ + d.

I_X = ∫ y² dA = ∫ (y′ + d)² dA
    = ∫ y′² dA + 2d ∫ y′ dA + d² ∫ dA

• ∫ y′² dA = I_G, the moment of inertia about the centroidal axis.
• ∫ y′ dA = A·ȳ′ = 0, because the distance of the centroid from the centroidal axis is zero.
• ∫ dA = A.

So I_X = I_G + A d². Similarly, I_Y = I_Gy + A d_x². Since A d² is never negative, the moment of inertia about the centroidal axis is the least of the moments of inertia about all parallel axes.

Centroid of the area between x = ky² and y = kx²

Reading of the figure: the two parabolas meet at O and at (500, 500) mm, so for both curves k = 1/500 mm⁻¹ (500 = k × 500²). The area lies between them, above y = kx² and to the right of x = ky² (i.e. below y = √(x/k)). Let a = 500 mm.

Upper curve y_u = √(x/k) = √(a x), lower curve y_l = kx² = x²/a.

Area A = ∫₀ᵃ (√(a x) − x²/a) dx = √a (2/3) a^{3/2} − a³/(3a) = (2/3)a² − (1/3)a² = a²/3 = 500²/3 = 83 333 mm²

x̄ = (1/A) ∫ x (y_u − y_l) dx ∫₀ᵃ x √(a x) dx = √a (2/5) a^{5/2} = (2/5) a³ ∫₀ᵃ x · x²/a dx = a³/4 x̄ = (3/a²) · a³ (2/5 − 1/4) = (3/a²) · (3/20) a³ = 9a/20 = 9 × 500/20 = 225 mm

ȳ: the figure is symmetrical about the line y = x (the two curves are mirror images), so ȳ = x̄ = 225 mm.

(Check by integration: ȳ = (1/2A) ∫ (y_u² − y_l²) dx = (3/(2a²)) (a³/2 − a³/5) = (3/(2a²)) (3a³/10) = 9a/20 ✓.)

Answer: centroid at (225 mm, 225 mm) from O.