Physics · System Of Particles And Rotational Motion · NEET
Learn this short list (all about the axis shown). Ring about central axis: I = MR^2. Ring about a diameter: I = (1/2)MR^2. Disc about central axis: I = (1/2)MR^2. Disc about a diameter: I = (1/4)MR^2. Solid cylinder about its own axis: I = (1/2)MR^2 (same as a disc). Solid sphere about a diameter: I = (2/5)MR^2. Thin hollow sphere (shell) about a diameter: I = (2/3)MR^2. Thin rod about its centre, perpendicular to length: I = (1/12)ML^2. Thin rod about one end: I = (1/3)ML^2. Notice every formula is (a number) x M x (distance)^2.
In a ring every bit of mass sits at the rim, exactly distance R from the axis, so I = MR^2 with no fraction. In a disc the mass is spread from the centre (distance 0) out to R, so on average the mass is closer to the axis. That smaller average distance halves the value, giving (1/2)MR^2. Rule: mass farther from the axis means larger I.
A hollow sphere has all its mass pushed out to the surface (far from the centre), so it resists rotation more, giving the bigger fraction (2/3)MR^2. A solid sphere has mass filled inward toward the axis, so it is easier to spin, giving the smaller fraction (2/5)MR^2. Same idea: mass farther out means bigger I.
About the centre, half the rod lies on each side and the mass is on average close to the axis, giving I = (1/12)ML^2. About one end, the whole rod stretches away from the axis (up to full length L), so the mass sits much farther out, giving the larger I = (1/3)ML^2. You can also get the end value from the centre value using the parallel axis theorem: (1/12)ML^2 + M(L/2)^2 = (1/3)ML^2.
Compare the fractions: ring/hoop = 1, hollow sphere = 2/3 = 0.67, disc/solid cylinder = 1/2 = 0.5, solid sphere = 2/5 = 0.4. So the ring (hoop) has the maximum I, and the solid sphere has the minimum. This is why a solid sphere rolls down a slope fastest and a ring slowest.
The moment of inertia of a thin rod about an axis through its mid-point and perpendicular to its length is 2400 g.cm^2. The length of the 400 g rod is nearly:
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:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. NEET only tests you on applying them. NCERT itself states the derivations are beyond the syllabus, so just memorise the table and know which axis each formula belongs to.
Yes, about its own central axis a solid cylinder has I = (1/2)MR^2, exactly the same as a disc, because length along the axis does not change how far mass sits from that axis.
I = (1/2)MR^2 about a diameter. This follows from the perpendicular axis theorem: the two in-plane diameter values add up to the central axis value MR^2, so each diameter gives half, that is (1/2)MR^2.
Moment of inertia has units of mass times distance squared (kg.m^2). So every formula must be a pure number multiplied by M and by the square of a length (R or L). The pure number just depends on the shape and axis.
The ring or thin hoop, with I = MR^2, because all its mass sits at the maximum distance R from the axis. The solid sphere, at (2/5)MR^2, gives the smallest.
