Physics · System Of Particles And Rotational Motion · NEET
Moment of inertia is I = Σ m·r², so mass placed far from the axis counts much more (r is squared). In a ring, ALL the mass sits at the rim, at distance R from the centre axis, so I = MR². In a disc the same mass is spread from the centre out to R, so on average it is closer to the axis, giving a smaller I = (1/2)MR². Same mass, same radius, but different mass distribution, so the ring wins.
Yes, about the central (geometric) axis, a disc and a solid cylinder of the same mass M and radius R both have I = (1/2)MR². A solid cylinder is just a thick disc, and the length of the cylinder does not enter the formula for the axis along its own length. NEET often swaps 'disc' for 'cylinder' in a question to test if you know they are equal about this axis.
The hollow (thin shell) sphere has more. Solid sphere: I = (2/5)MR² = 0.40 MR². Hollow shell: I = (2/3)MR² ≈ 0.67 MR². In the hollow shell all the mass is pushed out to the surface at distance R, while in the solid sphere much of the mass is near the centre. Farther mass means larger I, so hollow > solid.
Just compare the numerical coefficients in front of MR². Rank from largest: ring MR² (1.0) > hollow sphere (2/3 ≈ 0.67) > disc = solid cylinder (1/2 = 0.5) > solid sphere (2/5 = 0.4). Anything that pushes mass outward raises the coefficient. This ranking directly decides which body has more rotational KE, more angular momentum, or needs more work to stop, when ω is the same.
Yes, always state the axis. The values MR², (1/2)MR², (2/5)MR² are for the standard symmetry axis (ring/disc: axis perpendicular through centre; sphere: any diameter). About a diameter a ring is (1/2)MR² and a disc is (1/4)MR². If the axis shifts away from the centre of mass you must use the parallel axis theorem, so the same body can have many different I values.
Three bodies A (solid sphere), B (disc) and C (ring), each of mass M and radius R, spin about their symmetry axes with the same angular speed ω. The work needed to stop them satisfies:
The ratio of the radius of gyration of a thin uniform disc about an axis through its centre and normal to its plane to that about its diameter is:
The ratio of the radius of gyration of a solid sphere (about its own axis) to that of a thin hollow sphere of the same mass and radius (about its axis) is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For the same M and R about the symmetry axis, largest to smallest: ring (MR²) > hollow sphere (2/3 MR²) > disc = solid cylinder (1/2 MR²) > solid sphere (2/5 MR²). The ring is largest because its whole mass is at the rim.
In a solid sphere a lot of mass lies close to the central axis. Since I = Σm·r² weighs distant mass much more heavily (r is squared), keeping mass near the axis keeps I low, giving the smallest coefficient (2/5)MR².
No. About a diameter, disc I = (1/4)MR² and ring I = (1/2)MR². The ring is still double the disc, because the ring's mass stays at the rim while the disc's mass spreads inward toward the axis.
About its own central axis, no. A solid cylinder has I = (1/2)MR² whatever its length, exactly like a disc. Length only matters for axes that are perpendicular to the cylinder's long axis.
NEET frequently disguises one question as another: work to stop, rotational KE (1/2)Iω², angular momentum Iω, or torque τ = Iα all depend on I. If you know the ranking of I for standard bodies, you can answer these instantly without deriving anything.
