Moment of Inertia (Point Mass)

Equivalent forms

I = \sum_i m_i r_i^2

I = \int r^2 \, dm

I_{\text{disk}} = \tfrac{1}{2} M R^2

The quadratic dependence on radius is what makes flywheels, satellites, and gymnasts so geometry-sensitive.

Symbol	Name	SI unit	Dimension	Range
$I$	Moment of inertiaoutput Rotational inertia of a point mass about a perpendicular axis	kg·m²	ML^2	0 – 50
$m$	Point mass Mass of the particle	kg	M	0.1 – 20
$r$	Radius Perpendicular distance from the rotation axis	m	L	0 – 3

Unit systems

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

Where it holds

Exact for an idealized point mass. For extended bodies use the integral

I = \int r^{2} dm

or tabulated formulas (disk, rod, sphere).

Dimensional analysis

I \to [ML^{2}]

m\cdot r^{2} \to [M]\cdot [L^{2}] = [ML^{2}]

Both sides are

[ML^{2}] = kg\cdot m^{2}

.

\checkmark

Discovery

Leonhard Euler · 1765

Euler introduced the moment of inertia in 'Theoria Motus Corporum Solidorum seu Rigidorum', generalizing Newton's laws to rotating rigid bodies.

Try this

Why is it harder to spin a barbell with weights at the ends than near the center?

A 2 kg point mass orbits a fixed axis at radius 0.5 m. Compute its moment of inertia I = mr² and explain how doubling the radius affects rotational difficulty.

Research status: stable

Real-world applications

Flywheels for energy storage
Tuning the swing weight of bats, rackets, and golf clubs
Vehicle wheel design (low rotational inertia for acceleration)
Satellite design and stabilization

Common misconceptions

Moment of inertia depends on the chosen axis — there is no single 'I' for an object
It is not the same as mass: I has units of $kg\cdot m^{2}$ , not kg
Even identical masses can have very different I depending on their geometry

Experimental verification

Verified by torsion-pendulum experiments where the period of oscillation depends on I, and by spinning rigid bodies on calibrated bearings.

Derivation

Starting from

K = \tfrac{1}{2}mv^{2} and v = r\omega for

circular motion, we get

K = \tfrac{1}{2}m(r\omega )^{2} = \tfrac{1}{2}(mr^{2})\omega ^{2}

.

Identifying the coefficient

of \tfrac{1}{2}\omega ^{2}

yields

I = mr^{2}

.

Limiting cases

r \to 0

⟶

I \to 0

A mass at the rotation axis offers no rotational inertia.

m \to 0

⟶

I \to 0

Massless points contribute nothing to inertia.

r doubles⟶ I quadruplesQuadratic scaling: moving mass twice as far multiplies I by 4.

What if…

What if mass triples?

I triples linearly. A 6 kg point at 0.5 m has $I = 1.5 kg\cdot m^{2}$ .

What if I rotate about a different axis?

I changes because r changes. Use the parallel-axis theorem: $I = I_cm + Md^{2} to$ shift axes.

What if the body is extended?

Replace the sum with an integral: $I = \int r^{2} dm$ — geometry now matters, not just total mass.

1

Point mass on a string

Given ·

m:: 2
r:: 0.5

Find · I

Steps

Square the radius: $0.5^{2} = 0.25$
Multiply by mass: $2 \times 0.25 = 0.5$
$I = 0.5 kg\cdot m^{2}$

Result ·

I = mr^{2} = 2 \times 0.25 = 0.5 kg\cdot m^{2}

2

Doubling the radius

Given ·

m:: 2
r:: 1

Find · I

Steps

Square the new radius: $1^{2} = 1$
Multiply by mass: $2 \times 1 = 2$
Compare: $2.0 / 0.5 = 4\times$ increase

Result ·

I = 2 \times 1^{2} = 2.0 kg\cdot m^{2}

— four times the previous value

Angular Momentum→Rotational Kinetic Energy→Torque→