Parallel and perpendicular axes theorems

What do the two theorems say?

In plain words: once you know a body’s moment of inertia about one axis, these two theorems give it about many other axes without any new integration. The perpendicular axes theorem connects three axes through one point in the plane of a flat body; the parallel axes theorem shifts an axis sideways from the centre of mass.

Both theorems are listed in the JEE Main 2027 and NEET (UG) 2026 syllabuses (Unit 5; in the JEE Main wording, “parallel and perpendicular axes theorems and their applications”), but the 2026–27 NCERT textbook has no section on them. Its Table 6.1 gives moments of inertia about specific axes only; this page shows how to get the rest. The moment of inertia itself is I = Σmr^², with r the perpendicular distance from the axis, and the kinetic energy of a body rotating with angular speed ω about a fixed axis is ½Iω^².

Perpendicular axes theorem

Take a flat body in the xy-plane. A particle of mass m at (x, y) is at distance √(x^² + y^²) from the z-axis, at distance |y| from the x-axis and at distance |x| from the y-axis. So

The first sum is I_x and the second is I_y:

The proof needs every particle to have z = 0, so the theorem holds only for a flat body, with the x- and y-axes in its plane and all three axes through one point. It does not apply to a sphere or a cylinder. The point need not be the centre: at a point on the rim of a ring, 2MR^² (perpendicular to the plane) = 3MR^²/2 (the tangent in the plane) + MR^²/2 (the diameter through that point).

Ring and disc about a diameter. A thin ring of radius R has I_z = MR^² about its central perpendicular axis. Any two perpendicular diameters are alike by symmetry, so I_x = I_y = MR^²/2. In the same way a disc, with I_z = MR^²/2, has MR^²/4 about a diameter. Both agree with NCERT’s Table 6.1.

Rectangular plate. For a uniform plate with sides a (along x) and b (along y), the axes through the centre parallel to the sides give I_y = Ma^²/12 and I_x = Mb^²/12 (cut the plate into thin strips parallel to x: each is a rod of length a turning about the y-axis through its middle, giving I_y = Ma^²/12; strips parallel to y give I_x = Mb^²/12), so about the perpendicular axis through the centre

Parallel axes theorem

Let both axes be parallel to z, the first through the centre of mass C. For each particle, let r′ be its position from C with the z-component left out, so its distance from the first axis is r′. Let d be the perpendicular vector from the first axis to the second. The particle is then |r′ − d| from the second axis, and |r′ − d|^² = r′^² − 2d · r′ + d^². Then

The first sum is I_cm. The last term is Σm × d^² = Md^², since d is the same for every particle. The middle sum is zero, because Σmr′ = 0 when positions are measured from the centre of mass. So

This holds for any body, flat or not, but one of the two axes must pass through the centre of mass. It also shows that, of all parallel axes, the one through the centre of mass gives the smallest moment of inertia.

Using the theorems for rods, rings and discs

Starting from NCERT’s Table 6.1, the two theorems give:

For example, a rod about one end: the end is L/2 from the centre, so I = ML^²/12 + M(L/2)^² = ML^²/3. A ring about a tangent in its plane: the tangent is parallel to a diameter and R from it, so I = MR^²/2 + MR^² = 3MR^²/2.

Solved examples

Common mistakes to avoid

Quick revision summary

• Perpendicular axes theorem: for a flat body, I_z = I_x + I_y, with x and y in the plane and all three axes through one point.
• Parallel axes theorem: for any body, I = I_cm + Md^², where one axis passes through the centre of mass and d is the distance between the parallel axes.
• The centre-of-mass axis gives the smallest moment of inertia of a family of parallel axes.
• Standard results: rod about an end ML^²/3; ring about a diameter MR^²/2; disc about a diameter MR^²/4; disc about a tangent in its plane 5MR^²/4; sphere about a tangent 7MR^²/5.

Questions students ask

What is the difference between the parallel and perpendicular axis theorems, and where does each apply?

The parallel axis theorem, I = I_cm + Md^², shifts an axis sideways: it applies to any body, flat or solid, but one of the two parallel axes must pass through the centre of mass. The perpendicular axis theorem, I_z = I_x + I_y, adds the moments of inertia about two axes instead: it applies only to a flat body (a lamina), with the x- and y-axes in its plane and all three axes through one point. Many standard results need both, one after the other.

Are the parallel and perpendicular axis theorems in the NCERT Class 11 book?

No, the 2026–27 NCERT Class 11 Physics Part I has no section on the parallel and perpendicular axes theorems; its Table 6.1 gives moments of inertia about particular axes only. Both theorems are still in Unit 5, Rotational Motion, of the JEE Main 2027 and NEET (UG) 2026 syllabuses, which list the parallel and perpendicular axes theorems and their applications, so they are needed for both exams.

What is the moment of inertia of a rectangular plate about an axis through a corner, perpendicular to the plate?

The moment of inertia of a uniform rectangular plate of mass M and sides a and b, about an axis through a corner perpendicular to the plate, is M(a^² + b^²)/3. About the parallel axis through the centre it is M(a^² + b^²)/12 (from the perpendicular axis theorem), and the corner is half a diagonal, √(a^² + b^²)/2, from the centre, so the parallel axis theorem adds M(a^² + b^²)/4. For a square plate of side a this gives 2Ma^²/3.

What is the moment of inertia of a thin hollow sphere about a tangent?

The moment of inertia of a thin hollow sphere of mass M and radius R about a tangent is 5MR^²/3. About a diameter it is 2MR^²/3, and a tangent is parallel to a diameter at distance R, so the parallel axis theorem gives 2MR^²/3 + MR^² = 5MR^²/3. For example, a hollow ball of mass 0.3 kg and radius 0.1 m has I = 5 × 0.3 × 0.1^²/3 = 0.005 kg m^² about a tangent.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. A thin ring of mass 0.5 kg and radius 0.2 m. Find its moment of inertia about (a) a diameter and (b) a tangent perpendicular to its plane.

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   MR^² = 0.5 × 0.04 = 0.02 kg m^². (a) About a diameter: MR^²/2 = 0.01 kg m^². (b) About a tangent perpendicular to the plane: MR^² + MR^² = 2MR^² = 0.04 kg m^².

2. Q2. A solid sphere of mass 5 kg and radius 0.1 m. Find its moment of inertia about a tangent.

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   I = 2MR^²/5 + MR^² = 7MR^²/5 = 7 × 5 × 0.01/5 = 0.07 kg m^².

3. Q3. The moment of inertia of a body about an axis through its centre of mass is 2 kg m^², and its mass is 3 kg. Find its moment of inertia about a parallel axis 0.5 m from the centre of mass, and about another parallel axis 1 m from the centre of mass.

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   I = 2 + 3 × 0.5^² = 2.75 kg m^², and I = 2 + 3 × 1^² = 5 kg m^². (If both axes are on the same side of C they are 0.5 m apart, but 2.75 + 3 × 0.5^² = 3.5 kg m^² is wrong: always shift from the centre-of-mass axis.)