A uniform solid cylinder with radius R and length L has moment of inertia I₁, about the axis of the cylinder. A concentric solid cylinder of radius R'=R/2 and length L'=L/2 is carved out of the original cylinder. If I₂ is the moment of inertia of the carved out portion of the cylinder then I₁/I₂= __________.
(Both I₁ and I₂ are about the axis of the cylinder)

Solution

Physics sir · 3 steps

The carved piece is smaller in mass and in radius, so find its mass first and then square its radius.

1. Original cylinder. A solid cylinder about its own axis has I₁ = 1/2MR². The length does not appear in this formula, but it does decide the mass.
2. Mass of carved piece. Density is the same, so mass scales with volume π r² l. The new volume is π (R/2)² L/2 = 1/8π R² L, so the carved mass is m = M/8.

   I₂ = 1/2(M/8)(R/2)² = MR²/64

3. So. I₁/I₂ = (MR²/2)/(MR²/64) = 32.

The common slip: 16. Halving only the radius when finding the mass makes it M/4 instead of M/8, which gives 16. The halved length also cuts the mass in half.

The idea to keep. When a piece is cut from a uniform body, find its mass from the volume ratio first. Then use its own dimensions in the formula. Length can hide inside the mass even when it is missing from I.

Answer from the NTA final answer key. Solution by PrepWiser.