A uniform solid cylinder with radius R and length L has moment of inertia , about the axis of the cylinder. A concentric solid cylinder of radius and length is carved out of the original cylinder. If is the moment of inertia of the carved out portion of the cylinder then = ___________.
(Both and are about the axis of the cylinder)
(Both and are about the axis of the cylinder)
Show answer and solution
Answer: 32
Solution
The moment of inertia of a solid cylinder about its own axis is (1/2)MR².
Mass of original cylinder: M₁ ∝ R²L (since M = ρπR²L)
Mass of carved portion: M₂ = ρπ(R/2)²(L/2) = ρπR²L/8 = M₁/8
I₁ = (1/2)M₁R²
I₂ = (1/2)M₂(R/2)² = (1/2)(M₁/8)(R²/4) = M₁R²/64
Therefore I₁/I₂ = [(1/2)M₁R²] / [M₁R²/64] = 64/2 = 32
Mass of original cylinder: M₁ ∝ R²L (since M = ρπR²L)
Mass of carved portion: M₂ = ρπ(R/2)²(L/2) = ρπR²L/8 = M₁/8
I₁ = (1/2)M₁R²
I₂ = (1/2)M₂(R/2)² = (1/2)(M₁/8)(R²/4) = M₁R²/64
Therefore I₁/I₂ = [(1/2)M₁R²] / [M₁R²/64] = 64/2 = 32
[JEE Main 2023 · 24 Jan · Shift 2]