- 2072 Kartik · 4+8 marks
Parallel axis theorem; MOI with semicircular hole
Short answer
Parallel axis theorem: I=I_G+Ad². Trapezium (base 60 cm, top 30 cm, height 50 cm) minus semicircular hole r=15 cm; area=2046.57 cm², centroid ȳ=22.84 cm from base. I_xx about centroidal X-X = 2.722×10⁵ cm⁴.
State and prove the parallel axis theorem for moment of inertia. Determine the moment of inertia of the given composite area as shown in figure below about its centroidal X-X axis. [Figure: area bounded by x axis, left height 30 cm, right height 50 cm, base 30 cm + 30 cm, with a semicircular hole on the base, radius 15 cm centred at 30 cm from the left.]
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.
Moment of inertia about the centroidal X–X axis
Reading of the figure: origin at O (bottom-left). Left height 30 cm, right height 50 cm, base 30 + 30 = 60 cm, so the top edge slopes from (0, 30) to (60, 50). A semicircular hole of radius 15 cm has its diameter on the base, centre at x = 30 cm.
Main area (trapezium): A₁ = (30 + 50)/2 × 60 = 2400 cm², ȳ₁ = 20.42 cm. Own I_x of the trapezium about its centroidal axis = 359,583 cm⁴ (found by splitting it into a rectangle 60 × 30 and a triangle 60 × 20: rectangle 60·30³/12 + 1800 (15 − ȳ₁)², triangle 60·20³/36 + 600 (36.67 − ȳ₁)²).
Hole: A₂ = π × 15²/2 = 353.4 cm², ȳ₂ = 4 × 15/(3π) = 6.37 cm, I_x(own) = (π/8 − 8/9π) × 15⁴ = 5 556 cm⁴.
Centroid
| Part | A (cm²) | x (cm) | y (cm) | A·x | A·y |
|---|---|---|---|---|---|
| Trapezium (30, 50 × 60) | 2,400.00 | 32.50 | 20.42 | 78,000.0 | 49,000.0 |
| Semicircular hole r = 15 (hole, −) | -353.43 | 30.00 | 6.37 | -10,602.9 | -2,250.0 |
ΣA = 2,046.57 cm², ȳ = 22.84 cm above the base.
Moment of inertia about the centroidal X–X axis
I_xx = [I₁ + A₁(ȳ₁ − ȳ)²] − [I₂ + A₂(ȳ₂ − ȳ)²]
Trapezium: 359,583 + 2400 × (20.42 − 22.84)² = 373,713 cm4
Hole : 5,556 + 353.4 × (6.37 − 22.84)² = 101,508 cm4
I_xx = 272,205 cm⁴ ≈ 2.722 × 10⁵ cm⁴
✓ Answer checked by Claude Opus 5.5 (high effort)
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