Physics · System Of Particles And Rotational Motion · NEET
You SUBTRACT it. MOI adds up for parts of a body, so the full body's MOI is the remaining part plus the removed part: I_full = I_remaining + I_removed. Rearranging gives I_remaining = I_full − I_removed. So you always subtract the hole's contribution.
Use the area (or volume) ratio, because the body is uniform. Mass is proportional to area. If a disc of radius R has mass M, a hole of radius R/2 has area (R/2)² / R² = 1/4 of the full area, so its mass is m = M/4. For a 90° sector removed from a ring, that is 90/360 = 1/4 of the ring, so mass removed = M/4.
The removed piece's own centre is usually NOT on the axis you are asked about. The simple formula (like ½mr²) gives MOI about the piece's OWN centre. To get its MOI about the far-away common axis O you must shift it: I_hole about O = I_hole about its centre + m·d², where d is the distance between the hole's centre and O.
You can only subtract MOIs if both I_full and I_removed are measured about the exact same axis. If I_full is about the centre O, then I_removed must also be about O (not about the hole's own centre). Forgetting this is the most common mistake.
In a ring every bit of mass is at the same distance R from the central axis, so I = (mass)·R² for any part. Removing a 90° arc just removes M/4 of the mass, all still at radius R. So I_remaining = (3M/4)·R² = (3/4)MR². No shifting is needed because nothing moves closer or farther from the axis.
From a disc of mass M and radius R, a circular hole of radius R/2 whose rim passes through the centre is cut out. The moment of inertia of the remaining disc about an axis perpendicular to the plane and passing through the centre O is:
From a circular ring of mass M and radius R, an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part about an axis through the centre and perpendicular to the plane is K·MR². Then K is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
I_remaining = I_full − I_removed, with both moments of inertia taken about the SAME axis. Find the removed mass from the area/volume ratio, then use the parallel axis theorem if the removed piece is off the axis.
Yes. Because the body is uniform, mass is proportional to area (2D) or volume (3D). You must first compute the removed mass, e.g. a hole of radius R/2 in a disc of radius R removes M/4 of the mass.
Only when the removed piece's own centre lies exactly on the axis you are using, or when every element of the shape is at the same distance from the axis (like a ring). Otherwise you must add m·d².
It is a very high-yield, repeat pattern — NEET has asked disc-with-hole and ring-sector problems directly. It combines three ideas (area ratio for mass, standard MOI formulas, and the parallel axis theorem) in one question, so it tests whether you truly understand rotational inertia.
Forgetting the m·d² term for the off-centre hole. That gives 15MR²/32 instead of the correct 13MR²/32 for the classic disc problem, and 15/32 is always placed as a tempting option.
