Parallel Axis Theorem

Equivalent forms

I_P = I_{cm} + M d^2

A two-term decomposition — central spin plus orbital displacement — that always exactly equals the shifted-axis inertia.

Symbol	Name	SI unit	Dimension	Range
$I$	Moment of inertia about shifted axisoutput Rotational inertia about parallel axis	kg·m²	ML^2	0 – 100
$I_cm$	Moment of inertia about COM Rotational inertia about COM axis	kg·m²	ML^2	0 – 50
$m$	Mass Mass of the body	kg	M	0.1 – 20
$d$	Axis offset Distance from COM axis to parallel axis	m	L	0 – 5

Unit systems

Symbol	SI	CGS	Imperial
$I$	kg·m²	g·cm²	slug·ft²
$I_cm$	kg·m²	g·cm²	slug·ft²
$m$	kg	g	slug
$d$	m	cm	ft

Where it holds

Rigid bodies, axes parallel; if axes are not parallel, use the perpendicular-axis theorem or general inertia tensor.

Dimensional analysis

I \to [ML^{2}]

I_cm + m d^{2} \to [ML^{2}] + [M][L^{2}] = [ML^{2}]

Both sides

kg\cdot m^{2}

.

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Discovery

Christiaan Huygens / Jakob Steiner · 1673

Huygens used the result in his pendulum-clock analysis (Horologium Oscillatorium, 1673); Steiner later generalized and named it.

Try this

Why is a door so much harder to spin around its hinge than around its center?

Find the moment of inertia of a rod about a parallel axis at distance d from its center of mass, given I_cm.

Research status: stable

Real-world applications

Designing flywheels (offset hubs)
Pendulum-clock balance wheels
Rigid-body dynamics in game engines
Aerospace inertia tensor calculations

Common misconceptions

Theorem applies to any axis — only PARALLEL axes
Works for non-rigid bodies — only rigid
I always grows linearly with d — it grows quadratically

Experimental verification

Physical-pendulum period

T = 2\pi \surd (I/(mgd))

verifies the

I = I_cm + md^{2}

dependence as the pivot is moved.

Derivation

I = \Sigma m_i (r_i + d)^{2}

where r_i is measured from COM.

Expand:

I = \Sigma m_i r_i^{2} + 2d\cdot (\Sigma m_i r_i) + d^{2}\cdot \Sigma m_i

.

By definition of COM the middle term vanishes; the first is I_cm; the last is

m d^{2}

.

Limiting cases

d = 0

⟶

I = I_cm

Axis at the COM — minimum possible moment of inertia.

d ≫

\sqrt{I_cm/m}

⟶

I \approx m d^{2}

Far from COM, body behaves as a point mass at distance d.

Point mass⟶

I_cm = 0

;

I = m d^{2}

Reduces to single-particle formula.

What if…

What if the body isn't rigid?

Theorem fails — masses move relative to each other; use a time-dependent formulation.

What if I rotate about a perpendicular axis instead?

Use the perpendicular-axis theorem for planar bodies: $I_z = I_x + I_y$ .

What if d is huge?

$I \to m d^{2}$ ; the body acts like a point mass orbiting the axis.

1

Rod about end (length L=1, m=1)

Given ·

I cm:: 0.0833
m:: 1
d:: 0.5

Find · I

Steps

I_cm of uniform rod about center $= mL^{2}/12 = 1/12 \approx 0.0833$
$d = L/2 = 0.5\,\mathrm{m}$
$I = 0.0833 + 0.25 = 1/3 kg\cdot m^{2}$ (classic rod-about-end result)

Result ·

I = 0.0833 + 1\cdot 0.25 = 0.333 kg\cdot m^{2}

2

Disc shifted by d

Given ·

I cm:: 0.125
m:: 1
d:: 0.4

Find · I

Steps

$I_cm = \tfrac{1}{2}mR^{2}$ with $R = 0.5 \to 0.125$
$md^{2} = 1\cdot 0.16$
$I = 0.285 kg\cdot m^{2}$

Result ·

I = 0.125 + 0.16 = 0.285 kg\cdot m^{2}

Moment of Inertia (Point Mass)→Rotational Kinetic Energy→Torque→Simple Pendulum Period→