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Tables give you the moment of inertia of a rod, ring or disc about one or two standard axes. Exam questions almost never use those axes. The two axis theorems are how you get from the table to the axis in the question. This post explains when each theorem applies, works through the standard cases, lists the errors examiners see most and gives a practice routine.
Are the axis theorems still in the syllabus?
Yes for the entrance exams. The NTA Syllabus for JEE (Main) 2026 lists "parallel and perpendicular axes theorems and their applications" in Unit 5, Rotational Motion. The JEE (Advanced) 2026 Information Brochure lists "Rigid body, moment of inertia, parallel and perpendicular axes theorems, moment of inertia of uniform bodies with simple geometrical shapes".
The current NCERT Class 11 chapter on systems of particles and rotational motion (Reprint 2026-27) still defines moment of inertia and radius of gyration and gives Table 6.1 of standard values, but it no longer has a section on the two theorems. So you will need to learn them from your class notes or another book. Check your exam year's syllabus, because NTA publishes a new one each cycle.
What does the parallel axis theorem say?
For any rigid body of mass M:
I = I_cm + Md²
where I_cm is the moment of inertia about an axis through the centre of mass, and d is the perpendicular distance from that axis to a parallel axis. Since Md² is never negative, the moment of inertia is smallest about the axis through the centre of mass.
Worked example 1: rod about one end. A uniform rod has M = 0.6 kg and L = 1.0 m. From NCERT's Table 6.1, about a perpendicular axis through its midpoint, I_cm = ML²/12.
I_cm = 0.6 × 1.0² ÷ 12 = 0.05 kg m².
The end is d = L/2 = 0.5 m from the centre.
I_end = 0.05 + 0.6 × 0.5² = 0.05 + 0.15 = 0.20 kg m².
That matches the general result ML²/12 + M(L/2)² = ML²/3, and 0.6 ÷ 3 = 0.20.
Worked example 2: the trap. Find I for the same rod about an axis a quarter of the way along, given only I_end = ML²/3. A common mistake is to add M(L/4)² to I_end. That is wrong, because the theorem only shifts from the centre of mass. Go back to the centre first:
The axis is L/4 from the end, so it is L/2 − L/4 = L/4 from the centre.
I = ML²/12 + M(L/4)² = ML²/12 + ML²/16 = (4ML² + 3ML²)/48 = 7ML²/48.
For M = 0.6 kg, L = 1.0 m: 7 × 0.6 ÷ 48 = 0.0875 kg m².
What does the perpendicular axis theorem say?
For a flat object (a lamina) lying in the x-y plane:
I_z = I_x + I_y
where the x and y axes lie in the plane and the z axis is perpendicular to it, all three meeting at one point. It does not work for a solid sphere, cylinder or cube.
Worked example 3: disc about a diameter. NCERT's table gives MR²/2 for a disc about the axis through its centre, perpendicular to its face. By symmetry, every diameter has the same moment of inertia, I_d. Two perpendicular diameters give I_z = I_d + I_d, so
MR²/2 = 2I_d, and I_d = MR²/4.
That matches the table's MR²/4. For a ring, the same steps take MR² to MR²/2 about a diameter.
Worked example 4: combining both. A disc about a tangent in its own plane: start from the diameter, MR²/4, and shift by R: MR²/4 + MR² = 5MR²/4. About a tangent perpendicular to its plane: MR²/2 + MR² = 3MR²/2.
Worked example 5: square plate about a diagonal. A square plate of side a has I = Ma²/6 about the perpendicular axis through its centre (a standard result, derived from Ma²/12 about each of the two in-plane axes parallel to the sides). The two diagonals are perpendicular, lie in the plane, and by symmetry have equal moments of inertia, so each is (Ma²/6) ÷ 2 = Ma²/12. Same as about a line through the centre parallel to a side.
What mistakes cost marks with the axis theorems?
Shifting from a non-centre axis. Covered in worked example 2.
Using the perpendicular axis theorem on 3D bodies. It is only for laminas.
Axes that do not meet. The x, y and z axes must pass through the same point.
Wrong d. d is the perpendicular distance between the two parallel axes, not the distance to a point on the body.
Mixing up which table entry you started from. Write "about centre, perpendicular to face" next to each value you use.
Research on student difficulties backs up the need for care. In a study of rotational and rolling motion concepts (Rimoldini and Singh, 2005), students from introductory to upper-level courses showed similar difficulties, and greater mathematical skill did not seem to help them understand more deeply. Getting the axis right is a conceptual step, not an algebra step. Our post on rotational motion mistakes covers torque and rolling errors that often follow.
A practice routine
Derive, using the theorems only, I for a ring about a tangent in its plane (answer 3MR²/2) and perpendicular to it (answer 2MR²).
Find I for a rod about an axis L/3 from one end. Check: ML²/12 + M(L/6)² = ML²/9.
Do five past-paper rotational questions, and before solving each, write which axis and which theorem you will use.
Make a one-page sheet of every derived result, each with the axis drawn.
If you cannot see why a step works, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can compare your choice of axis with a full solution.
Before your next session, derive the disc's diameter result from MR²/2 without looking at this page.
