Axis theorems

A spinning disc resists being spun about its centre one way and about a point on its rim a very different way — same object, same mass, wildly different moment of inertia, just because the axis moved. Calculating that new value from scratch, by integrating over every particle again, is slow and error-prone. The parallel-axis theorem and the perpendicular-axis theorem are shortcuts: two clean rules that let you reuse a moment of inertia you already know about one standard axis to find it about a new axis, with no integration at all. One rule works for any rigid body and any parallel shift; the other works only for flat, two-dimensional laminas relating in-plane and perpendicular axes. Together they turn a handful of memorised standard results (ring, disc, rod, sphere, cylinder) into dozens of usable answers.

The core insight

The parallel-axis theorem shifts a known centre-of-mass moment of inertia to any parallel axis via $I=I_{cm}+Md^2$ , while the perpendicular-axis theorem, valid only for plane laminas, relates two in-plane axes to the perpendicular one via $I_z=I_x+I_y$ .

The core idea

Why you need an axis theorem at all. Moment of inertia $I=\sum m_ir_i^2$ depends on how mass is distributed relative to a specific axis. Move the axis and every $r_i$ changes, so $I$ changes too — even though the body and its mass $M$ are unchanged. Rather than re-integrating for every new axis a problem throws at you, the two axis theorems give an algebraic shortcut whenever the new axis is either parallel to a known centre-of-mass axis, or perpendicular to a lamina whose in-plane axes you know.

Parallel-axis theorem (any rigid body). If $I_{cm}$ is the moment of inertia about an axis through the centre of mass, the moment of inertia $I$ about any axis parallel to it, at perpendicular distance $d$ , is

I = I_{cm} + Md^2

where: $M$ is the total mass of the body and $d$ is the perpendicular distance between the two parallel axes. Because $Md^2\ge 0$ always, shifting away from the centre of mass can only increase $I$ — the centre-of-mass axis gives the smallest moment of inertia among all parallel axes. This holds for any shape, provided one of the two axes passes through the centre of mass; you cannot jump directly between two arbitrary off-centre parallel axes.

Perpendicular-axis theorem (plane laminas only). For a flat, effectively two-dimensional body, if $x$ and $y$ are mutually perpendicular axes lying in the plane and meeting at point $O$ , and $z$ is the axis through $O$ perpendicular to the plane, then

I_z = I_x + I_y

where: $I_x, I_y$ are moments of inertia about the two in-plane axes and $I_z$ is the moment of inertia about the perpendicular axis, all three concurrent at $O$ . The in-plane axes need not be symmetry axes — only mutually perpendicular and meeting $z$ at the same point — though using symmetry axes usually simplifies the algebra a lot. The ISC syllabus asks you to state and apply both theorems; no derivation of either is examinable. The figure below shows the geometry that the parallel-axis theorem describes.

Figure

A disc with its centre of mass axis and a parallel tangent axis at perpendicular distance d, both perpendicular to the page

Derivation AWhy the parallel-axis theorem works

This is a plausibility check with two point masses, not required for the ISC exam since the syllabus asks only for the statement and application of both theorems, but it makes the $Md^2$ term trustworthy rather than a formula to memorise blindly. Consider a body idealised as two equal point masses $m$ placed symmetrically about a centre-of-mass axis, each at perpendicular distance $a$ from it, so $I_{cm}=2ma^2$ ; the goal is to check that shifting to a parallel axis a distance $d$ away really does add exactly $Md^2$ , with $M=2m$ , and that no cross term survives once the first axis passes through the centre of mass.

Step 1. Place the centre-of-mass axis at the origin along one direction, with the two masses at perpendicular distances $+a$ and $-a$ from it. By definition $I_{cm}=m(a)^2+m(-a)^2=2ma^2$ .

Step 2. Shift the reference axis by $d$ , so the two masses now sit at distances $(a+d)$ and $(-a+d)$ from the new axis. The new moment of inertia is $I=m(a+d)^2+m(a-d)^2$ .

Step 3. Expand both squares: $I=m(a^2+2ad+d^2)+m(a^2-2ad+d^2)=2ma^2+2md^2$ . The two cross terms $2mad$ and $-2mad$ cancel exactly because the masses are placed symmetrically about the centre of mass — this cancellation is precisely what makes $I_{cm}$ , and not any other axis, the one that appears cleanly in the formula.

Step 4. Recognise the result: $I=2ma^2+2md^2=I_{cm}+Md^2$ since $M=2m$ . The same cancellation happens for any mass distribution once the first axis passes through the centre of mass, which is why the theorem requires that condition and why it fails if you try to shift between two arbitrary non-central axes directly.

Worked example

A uniform disc of mass $M$ and radius $R$ has $I_{cm}=\tfrac12MR^2$ about its central axis, perpendicular to its plane. Find its moment of inertia about a tangent axis in its own plane (i.e. a line touching the rim, lying flat in the plane of the disc, not perpendicular to it).

Step 1. This needs both theorems together. First find the moment of inertia about an in-plane diameter, using the perpendicular-axis theorem, since the disc is a plane lamina. By symmetry the two perpendicular in-plane diameters give equal values, $I_x=I_y=I_{dia}$ , so $I_z=I_x+I_y=2I_{dia}$ , giving $I_{dia}=\dfrac{I_z}{2}=\dfrac{MR^2/2}{2}=\dfrac14MR^2$ .

Step 2. The diameter passes through the centre of mass, so treat $I_{dia}=\tfrac14MR^2$ as the " $I_{cm}$ " for a parallel-axis shift within the plane. The tangent line in the plane is parallel to a diameter, at perpendicular distance $d=R$ from it (centre to rim).

Step 3. Apply the parallel-axis theorem: $I_{tangent}=I_{dia}+MR^2=\tfrac14MR^2+MR^2=\tfrac54MR^2$ .

Final answer: the moment of inertia of a uniform disc about a tangent lying in its own plane is $I=\tfrac54MR^2$ , larger than about a central diameter ( $\tfrac14MR^2$ ) but smaller than about the tangent axis perpendicular to the disc ( $\tfrac32MR^2$ , from the standard rim-tangent case), since the in-plane tangent axis passes closer to the bulk of the mass than the perpendicular one does relative to their respective reference axes.

Common traps

Using the perpendicular-axis theorem on a solid three-dimensional body (a solid sphere, a solid cylinder) — it applies only to plane laminas, not to genuinely 3-D shapes.
Applying the parallel-axis theorem with $I_{cm}$ replaced by the moment of inertia about some other, non-centroidal axis — the theorem needs one of the two axes to specifically pass through the centre of mass; you cannot hop directly between two arbitrary off-centre parallel axes.
Misreading $d$ in $Md^2$ — it is the perpendicular distance between the two parallel axes, not a radius or any other convenient-looking length in the figure.
Assuming the perpendicular-axis theorem's two in-plane axes must be symmetry axes — they only need to be mutually perpendicular and meet the $z$ -axis at the same point, though symmetry axes (as used above) usually make the arithmetic easiest.

Ask the tutor: want to work through one more combined-theorem problem, such as finding a ring's moment of inertia about a tangent lying in its own plane?

Why does the perpendicular-axis theorem only work for flat laminas and not for solid 3-D bodies?

The derivation of that theorem (beyond this course, but useful for intuition) relies on every particle of the lamina having zero coordinate along the perpendicular direction, so that $r^2=x^2+y^2$ splits cleanly into the two in-plane contributions that add up to $I_z$ . A genuinely three-dimensional body has mass spread out of the plane too, so that clean split fails and $I_z\ne I_x+I_y$ in general. Restricting the theorem to laminas is exactly what keeps that identity true.

Can I use the parallel-axis theorem to go directly from one rim axis to another rim axis on the opposite side of a disc, skipping the centre?

No — you must route through the centre-of-mass axis. Compute the distance from each rim axis to the central axis, apply the theorem once to get to the centre (or, going the other way, subtract $Md^2$ to get back to $I_{cm}$ ), and only then apply it again to reach the second axis. The formula's $Md^2$ term is only guaranteed correct when $I_{cm}$ genuinely refers to a centre-of-mass axis, so skipping that step gives a wrong answer even if the arithmetic looks similar.

Do I need to memorise the proofs of these two theorems for the ISC exam?

No. The syllabus explicitly states that derivation of neither theorem is required — you are expected to know their statements, know when each applies, and use them to compute unknown moments of inertia from standard ones you already know (ring, disc, rod, sphere, cylinder). The plausibility argument in this article, using two point masses, is there to build confidence in the $Md^2$ term, not because you will be asked to reproduce it.

Axis theorems

The core idea

Worked example

Common traps

Before this

Where this leads