Suppose there is a uniform circular disc of mass M kg and radius r m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by x/256Mr². The value of x is ____ .

Solution

Physics sir · 4 steps

Treat each hole as a negative disc: find its MI about A with the parallel axis theorem, then subtract.

1. Mass of each hole. The disc is uniform, so mass goes with area. Each hole has radius r/4, so its area is 1/16 of the disc's area and its mass is M/16.
2. One hole about A. About its own centre, a hole has 1/2·M/16(r/4)². Its centre is 3r/4 from A, so add M/16(3r/4)² by the parallel axis theorem.

   Iₕₒₗₑ = M/16(r²/32 + 9r²/16) = M/16·19r²/32 = 19/512Mr²

3. Angle does not matter. Moment of inertia about A depends only on each hole's distance from A, not on the direction. The 135° angle plays no part, and the two holes together remove 38/512Mr² = 19/256Mr².
4. So. I = 1/2Mr² - 19/256Mr² = (128 - 19)/256Mr² = 109/256Mr², so x = 109.

The common slip: 127. Leaving out the parallel axis term M/16(3r/4)² removes only 1/256Mr² in total and gives 127.

The idea to keep. For a disc with holes cut out, subtract each hole's MI about the same axis: own-centre MI plus md². Only the distance from the axis matters, never the angle between the holes.

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