Physics · System Of Particles And Rotational Motion · NEET
No. The theorem only works for flat (planar, two-dimensional) bodies whose whole mass lies in one plane, like a thin ring, thin disc, or thin square plate. A sphere, cube, or solid cylinder has thickness, so its mass is spread in three dimensions and the theorem fails. For those, use the parallel axis theorem or standard formulas from the table. In NEET, always check first: is the body flat? Only then apply Iz = Ix + Iy.
The proof uses the fact that for any small mass in the x-y plane, its distance from the z-axis follows Pythagoras: r-squared = x-squared + y-squared. This is only true when every mass point has z = 0, meaning the whole body sits in one plane. If the body has thickness (z is not zero), the equation r-squared = x-squared + y-squared no longer holds, so the theorem breaks. That is why it is a planar-body-only rule.
They answer different questions. Perpendicular axis theorem relates three axes at right angles for a flat body: Iz = Ix + Iy (all axes meet at the same point, one is perpendicular to the plane). Parallel axis theorem shifts one axis to a parallel new position: I = I_cm + M d-squared (used to move an axis away from the centre of mass by distance d). Use perpendicular for flat bodies; use parallel to change the axis location.
For a disc, the moment of inertia about the central perpendicular axis (z-axis) is Iz = (1/2) M R-squared. By symmetry, the two in-plane diameters give equal values, so Ix = Iy = I_diameter. Apply the theorem: Iz = Ix + Iy = 2 I_diameter. So I_diameter = Iz / 2 = (1/4) M R-squared. This is the classic use of the theorem and appears often in NEET numericals.
No, they must all pass through the same point O, but O can be any point on the flat body, not only the centre. The two in-plane axes (x and y) and the perpendicular axis (z) must intersect at that one common point O. Choosing the centre is common because symmetry makes Ix = Iy, but the theorem itself works at any point as long as all three axes meet there.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For a flat body in the x-y plane, the moment of inertia about the perpendicular (z) axis through a point O equals the sum about two in-plane axes through O: Iz = Ix + Iy.
Only planar (flat, thin) bodies whose mass lies in one plane, such as a ring, disc, or square plate. Not for spheres, cubes, or solid cylinders.
For a ring, Iz = M R-squared about the central perpendicular axis. Since Ix = Iy by symmetry, Iz = 2 I_diameter, so I_diameter = (1/2) M R-squared.
The two in-plane axes (x and y) must be perpendicular to each other and lie in the body's plane, and the third (z) axis is perpendicular to the plane. All three meet at one point O.
It quickly gives the moment of inertia about a diameter of a disc or ring from the easy central value, saving derivation time in fast MCQs.
