M and R be the mass and radius of a disc. A small disc of radius R / 3 is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis A B passing through the centre O and perpendicular to the plane of disc is 4/x MR². The value of x is ____________.

Solution

Physics sir · 3 steps

Treat the hole as negative mass: full disc minus the small disc's moment of inertia about O.

1. Mass removed. Mass is spread uniformly over the area, so it scales as radius squared. The small disc has radius R/3, so its mass is m = M × (1/3)² = M/9. From the figure it touches the rim, so its centre O' is at d = R - R/3 = 2R/3 from O.
2. Small disc about O. About its own centre it has 1/2·M/9·R²/9 = MR²/162. The parallel axis theorem adds md² = M/9·4R²/9 = 8MR²/162, so the total is 9MR²/162 = MR²/18.

   I_(remaining) = MR²/2 - MR²/18 = 8MR²/18 = 4/9MR²

3. So. Comparing with 4/xMR² gives x = 9.

The common slip: 12. Taking the removed mass as M/3 (scaling with radius instead of area) gives 1/3MR² = 4/12MR², so x = 12.

The idea to keep. For a disc with a hole, subtract the removed piece's moment of inertia about the same axis. Shift it with the parallel axis theorem first, and find its mass from the area ratio.

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