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Physics · Systems of Particles and Rotational Motion

Theorems of Parallel & Perpendicular Axes

Standard tables list the moment of inertia of a body about only one or two privileged axes. Real problems — and NEET — keep asking about other axes: a disc about its diameter, a rod about its end, a ring about a tangent. Two short theorems bridge that gap. The perpendicular-axis theorem relates the moments of inertia of a plane lamina about three mutually perpendicular axes; the parallel-axis theorem shifts a known centre-of-mass result to any parallel axis. This deep-dive states both, justifies them, contrasts their scope, and applies them to derive the non-tabulated results NEET expects on demand.

Why two theorems are needed

The moment of inertia of a body is not a single number — it depends on the position and orientation of the axis of rotation. NCERT states this plainly: the moment of inertia "is not a fixed quantity but depends on distribution of mass about the axis of rotation, and the orientation and position of the axis of rotation with respect to the body as a whole." A thin rod has $I = ML^2/12$ about a perpendicular axis through its midpoint, but a different value about an axis through its end.

Computing $I=\sum m_i r_i^2$ from scratch for every new axis would be tedious. The two theorems of moment of inertia avoid that. Each connects the wanted moment of inertia to one we already know — and in particular, NIOS notes, "one of which is passing through the CM of the body." Many entries in the standard tables are themselves computed this way.

The perpendicular-axis theorem

Choose three mutually perpendicular axes such that the $x$ and $y$ axes lie in the plane of the body and the $z$-axis is perpendicular to that plane, all three meeting at one point. The perpendicular-axis theorem states that the sum of the moments of inertia about the $x$ and $y$ axes equals the moment of inertia about the $z$-axis:

$$I_z = I_x + I_y$$

plane lamina x y z I​z = I​x + I​y O dm r
Perpendicular-axis theorem. For a flat lamina, every mass element lies in the x–y plane, so its distance from the z-axis satisfies $r^2 = x^2 + y^2$. Summing over the body gives $I_z = I_x + I_y$.

Justification. For a plane lamina, every mass element $dm$ lies in the $x$–$y$ plane, so its perpendicular distance from the $z$-axis obeys the Pythagorean relation $r^2 = x^2 + y^2$, where $x$ and $y$ are its distances from the $y$- and $x$-axes respectively. Multiplying by $dm$ and integrating over the body,

$$\int r^2\,dm = \int x^2\,dm + \int y^2\,dm \;\Rightarrow\; I_z = I_x + I_y.$$

The whole argument hinges on $r^2 = x^2 + y^2$, which is only true when the third coordinate is zero — that is, when the body is flat. This is exactly why the theorem is restricted to plane laminae.

The parallel-axis theorem

Suppose a body rotates about an axis through a point $P$ that is not its centre of mass. The parallel-axis theorem lets you find that moment of inertia from the moment of inertia about a parallel axis through the centre of mass. As stated by NIOS, "the moment of inertia about an axis parallel to the axis passing through its centre of mass is equal to the moment of inertia about its centre of mass plus the product of mass and square of the perpendicular distance between the parallel axes":

$$I = I_{cm} + Md^2$$

Here $I_{cm}$ is the moment of inertia about the centre-of-mass axis, $M$ is the total mass of the body, and $d$ is the perpendicular distance between the two parallel axes. Unlike the perpendicular-axis theorem, this result carries no restriction on the shape or dimensionality of the body.

rigid body, mass M CM axis C axis P P d I = I​cm + M d²
Parallel-axis theorem. The axis through P is parallel to the axis through the centre of mass C, separated by perpendicular distance $d$. The moment of inertia about P exceeds the centre-of-mass value by $Md^2$.

Justification. Take the centre-of-mass axis along $z$ through $C$, and a parallel axis through $P$ at perpendicular distance $d$. For a mass element with in-plane position $(x,y)$ measured from $C$, its squared distance from the $P$-axis is $(x-d)^2 + y^2$ when $P$ is offset along $x$. Expanding and integrating,

$$I = \int\!\big[(x-d)^2 + y^2\big]\,dm = \int (x^2+y^2)\,dm - 2d\!\int x\,dm + d^2\!\int dm.$$

The first integral is $I_{cm}$; the last is $Md^2$. The middle integral $\int x\,dm$ is the first moment of mass about the centre of mass, which is zero by the definition of the centre of mass. Hence $I = I_{cm} + Md^2$. That vanishing cross term is the entire reason the reference axis must pass through the centre of mass.

One consequence is worth stating outright: because $Md^2 \ge 0$, the moment of inertia about a centre-of-mass axis is the smallest of all parallel axes in that direction. Moving the axis away from the centre of mass can only raise $I$.

The two theorems compared

The two theorems answer different questions and carry different restrictions. The card below contrasts them on statement, condition, and formula.

Perpendicular-axis
What it relates
Three mutually perpendicular axes of a flat body — two in its plane, one normal to it.
Statement
The moment of inertia about the normal ($z$) axis equals the sum about the two in-plane axes.
Formula
$I_z = I_x + I_y$
Applicability
Plane laminae only (ring, disc, thin sheet). Fails for 3-D bodies.
Axes
All three intersect at one point; need not pass through the centre of mass.
Parallel-axis
What it relates
Two parallel axes — one through the centre of mass, one offset from it.
Statement
The moment of inertia about the offset axis exceeds the centre-of-mass value by $Md^2$.
Formula
$I = I_{cm} + Md^2$
Applicability
Any body — laminae and 3-D bodies alike. No shape restriction.
Axes
Must be parallel; one of them must pass through the centre of mass.

The two are frequently used together: the perpendicular-axis theorem first produces a moment of inertia about an in-plane axis through the centre, then the parallel-axis theorem shifts that axis to the rim or some other offset position. The ring-about-a-tangent derivation below shows the combination in action.

Deriving the disc about a diameter

The standard table gives a thin circular disc of mass $M$ and radius $R$ a moment of inertia $I_z = MR^2/2$ about the central axis perpendicular to its plane. Its moment of inertia about a diameter is not tabulated separately, but the perpendicular-axis theorem produces it at once.

Worked Example 1 · Perpendicular-axis

Find the moment of inertia of a uniform thin disc of mass $M$ and radius $R$ about any of its diameters.

Set up the axes. A disc is a plane lamina. Let $z$ be the central axis perpendicular to the disc, and let $x$ and $y$ be two perpendicular diameters lying in the plane of the disc, all meeting at the centre.

Use symmetry. By the symmetry of the disc, the moment of inertia is the same about every diameter, so $I_x = I_y = I_d$.

Apply the perpendicular-axis theorem. $I_z = I_x + I_y = 2I_d$. With $I_z = MR^2/2$, $$\frac{MR^2}{2} = 2I_d \;\Rightarrow\; \boxed{I_d = \frac{MR^2}{4}}.$$ This matches the table entry for a circular disc about a diameter, $MR^2/4$.

The same logic applied to a thin ring of moment of inertia $MR^2$ about its central axis gives $MR^2 = 2I_d$, so the ring about a diameter is $I_d = MR^2/2$ — exactly the result NIOS derives for a hoop.

i
Foundation

These derivations build on the table of standard moments of inertia — review them in moment of inertia before drilling theorem problems.

Deriving the rod about its end

A thin uniform rod has $I_{cm} = ML^2/12$ about a perpendicular axis through its midpoint, which coincides with its centre of mass. The parallel-axis theorem shifts this to a perpendicular axis through one end.

Worked Example 2 · Parallel-axis

A thin uniform rod of mass $M$ and length $L$ has $I = ML^2/12$ about a perpendicular axis through its centre. Find its moment of inertia about a perpendicular axis through one end.

Identify the shift. The centre-of-mass axis is at the midpoint; the new axis is at the end. The perpendicular distance between the two parallel axes is $d = L/2$.

Apply the parallel-axis theorem. $$I_{\text{end}} = I_{cm} + Md^2 = \frac{ML^2}{12} + M\!\left(\frac{L}{2}\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{4}.$$

Combine. $\dfrac{ML^2}{12} + \dfrac{3ML^2}{12} = \dfrac{4ML^2}{12} = \boxed{\dfrac{ML^2}{3}}.$ The end axis carries more inertia because more mass lies far from it.

Deriving the ring about a tangent

The tangent case showcases both theorems working in sequence, and it has two distinct answers depending on the tangent's orientation. NIOS works the example for a hoop (a thin ring) of mass $M$ and radius $R$.

Worked Example 3 · Both theorems

A thin ring of mass $M$ and radius $R$ has $I = MR^2$ about its central axis (perpendicular to its plane through the centre). Find its moment of inertia about (a) a tangent in the plane of the ring, and (b) a tangent perpendicular to the plane of the ring.

(a) Tangent in the plane. First find the moment of inertia about a diameter using the perpendicular-axis theorem: $MR^2 = 2I_d$, so $I_d = MR^2/2$. A tangent in the plane is parallel to this diameter and offset by $d = R$. Apply the parallel-axis theorem: $$I_{\text{tan}} = I_d + MR^2 = \frac{MR^2}{2} + MR^2 = \boxed{\frac{3MR^2}{2}}.$$

(b) Tangent perpendicular to the plane. Now the reference is the central axis ($I = MR^2$), and the tangent is parallel to it, offset by $d = R$. NIOS gives this directly: $$I_{\text{tan}} = MR^2 + MR^2 = \boxed{2MR^2}.$$

Read the difference. Both tangents touch the rim, but the in-plane tangent shifts a diameter ($MR^2/2$) while the perpendicular tangent shifts the central axis ($MR^2$). Identifying which centre-of-mass axis is parallel to the tangent is the whole game.

Quick recap

Both theorems in one breath

  • Perpendicular-axis: $I_z = I_x + I_y$ — plane laminae only ($x,y$ in the plane, $z$ normal), because $r^2 = x^2 + y^2$ holds only for flat bodies.
  • Parallel-axis: $I = I_{cm} + Md^2$ — any shape, reference axis through the centre of mass, $d$ the perpendicular distance between the parallel axes.
  • The centre-of-mass axis gives the minimum moment of inertia among all parallel axes; $Md^2$ is always an increase.
  • Disc about a diameter $= MR^2/4$; ring about a diameter $= MR^2/2$ (both from perpendicular-axis theorem).
  • Rod about its end $= ML^2/3$ (parallel-axis shift of $L/2$ from the midpoint).
  • Ring about an in-plane tangent $= 3MR^2/2$; about a perpendicular tangent $= 2MR^2$ (both theorems combined).

NEET PYQ Snapshot — Parallel & Perpendicular Axes

Three PYQs that lean on these theorems. Each reduces to picking the right reference axis and applying one formula.

NEET 2022

The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is:

  1. $\sqrt{2}:1$
  2. $4:1$
  3. $1:\sqrt{2}$
  4. $2:1$
Answer: (1) √2 : 1

Perpendicular-axis driven. Normal axis: $I_1 = MR^2/2$, so $k_1 = R/\sqrt{2}$. Diameter (from $I_z = 2I_d$): $I_2 = MR^2/4$, so $k_2 = R/2$. Ratio $k_1:k_2 = \dfrac{R/\sqrt 2}{R/2} = \sqrt 2 : 1$.

NEET 2016

From a disc of radius $R$ and mass $M$, a circular hole of diameter $R$, whose rim passes through the centre, is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis passing through the centre?

  1. $13MR^2/32$
  2. $11MR^2/32$
  3. $9MR^2/32$
  4. $15MR^2/32$
Answer: (1) 13MR²/32

Parallel-axis driven. The removed hole has radius $R/2$ and mass $M/4$. Its own moment of inertia about its centre is $\tfrac12(M/4)(R/2)^2 = MR^2/32$. Its centre sits at $d=R/2$ from the disc centre, so by the parallel-axis theorem $I_{\text{hole}} = MR^2/32 + (M/4)(R/2)^2 = MR^2/32 + MR^2/16 = 3MR^2/32$. Remaining $= MR^2/2 - 3MR^2/32 = 13MR^2/32$.

NEET 2023

The ratio of the radius of gyration of a solid sphere of mass $M$ and radius $R$ about its own axis to the radius of gyration of a thin hollow sphere of the same mass and radius about its axis is:

  1. $5:2$
  2. $\sqrt{3}:\sqrt{5}$
  3. $5:3$
  4. $2:5$
Answer: √3 : √5 (option as printed: 2)

Radius-of-gyration contrast. Solid sphere $I = \tfrac25 MR^2 \Rightarrow k_s = \sqrt{2/5}\,R$; hollow sphere $I = \tfrac23 MR^2 \Rightarrow k_h = \sqrt{2/3}\,R$. Ratio $k_s:k_h = \sqrt{2/5}:\sqrt{2/3} = \sqrt 3:\sqrt 5$. Note: neither theorem applies to a sphere — both are 3-D bodies, so the perpendicular-axis theorem is off-limits here, which is precisely the trap the question tests.

FAQs — Parallel & Perpendicular Axes

The theorem questions NEET aspirants most often get wrong.

When can the perpendicular-axis theorem be applied?
Only for a plane lamina — a flat, two-dimensional body of negligible thickness such as a ring, disc, or thin rectangular sheet. The theorem requires that the x and y axes lie in the plane of the body and the z axis be perpendicular to that plane through their intersection. It does not apply to three-dimensional bodies like a solid sphere or a thick cylinder, because for those a particle has a non-zero coordinate along all three axes.
What is the parallel-axis theorem and what does d mean?
The parallel-axis theorem states I = I_cm + Md², where I_cm is the moment of inertia about an axis through the centre of mass, M is the total mass, and d is the perpendicular distance between the given axis and the parallel axis through the centre of mass. Crucially, d is measured between the two parallel axes, and the reference axis must pass through the centre of mass — not through any other convenient point.
Why must one axis pass through the centre of mass in the parallel-axis theorem?
The simple form I = I_cm + Md² holds only when the reference axis passes through the centre of mass. The derivation produces a cross term that vanishes precisely because the position vectors are measured from the centre of mass, where the mass-weighted sum of coordinates is zero. If you try to transfer between two arbitrary parallel axes — neither through the centre of mass — the formula fails; you must first route through the centre-of-mass axis.
How do you find the moment of inertia of a disc about its diameter?
Use the perpendicular-axis theorem. A disc is a plane lamina, so Iz = Ix + Iy where z is the central axis perpendicular to the disc and x, y are two perpendicular diameters. By symmetry Ix = Iy = Id. Since Iz = MR²/2, we get MR²/2 = 2Id, so Id = MR²/4, the moment of inertia of a disc about any diameter.
How do you get the moment of inertia of a ring about a tangent?
For a tangent in the plane of the ring (a diameter shifted to the rim), combine both theorems. First, the perpendicular-axis theorem gives the moment about a diameter as MR²/2. Then the parallel-axis theorem shifts that diameter by a distance R to the tangent: Itan = MR²/2 + MR² = 3MR²/2. For a tangent parallel to the central axis, instead use Itan = MR² + MR² = 2MR².
Does the moment of inertia increase or decrease when the axis moves away from the centre of mass?
It always increases. Because I = I_cm + Md² and the term Md² is never negative, any parallel axis not through the centre of mass gives a larger moment of inertia than the centre-of-mass axis. The moment of inertia about a centre-of-mass axis is therefore the minimum among all parallel axes in that direction.
Can the parallel-axis theorem be applied to any shape?
Yes. Unlike the perpendicular-axis theorem, the parallel-axis theorem has no restriction on shape or dimensionality. It works for laminae and three-dimensional bodies alike, provided the reference axis passes through the centre of mass and the second axis is parallel to it. Many entries in standard moment-of-inertia tables are computed using this theorem.
Related Topics
Systems of Particles & Rotational Motion — Parent chapter Full chapter map: centre of mass, torque, moment of inertia, rolling. Moment of Inertia Definition, standard table, radius of gyration — the inputs these theorems shift. Centre of Mass The reference point the parallel-axis theorem requires. Dynamics of Rotational Motion τ = Iα; where moment of inertia about the rotation axis enters. Kinematics of Rotational Motion Angular displacement, velocity, acceleration; the rotational kinematic equations. Torque & Angular Momentum τ = r × F; L = Iω; conservation of angular momentum. Rolling Motion Combined translation and rotation; energy split using moment of inertia. Equilibrium of a Rigid Body Net force and net torque conditions; principle of moments.