Rotational Kinetic Energy

Equivalent forms

E_{\text{rot}} = \frac{L^2}{2I}

KE_{\text{total}} = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2

The rotational mirror of ½mv² — and the basis of every flywheel battery.

Symbol	Name	SI unit	Dimension	Range
$KE$	Rotational Kinetic Energyoutput Kinetic energy due to rotation	J	ML^2T^-2	0 – 100000
$I$	Moment of Inertia Rotational inertia about the spin axis	kg·m²	ML^2	0.01 – 10
$ω$	Angular Velocity Spin rate in radians per second	rad/s	T^-1	0 – 1500

Unit systems

Symbol	SI	CGS	Imperial
$KE$	J	erg	ft·lbf
$I$	kg·m²	g·cm²	slug·ft²
$ω$	rad/s	rad/s	rad/s

Where it holds

Valid for rigid rotation about a fixed axis (or principal axis through the center of mass). For rolling motion add translational

\tfrac{1}{2}mv^{2}

.

Dimensional analysis

KE \to [ML^{2}T^{-2}]

I\cdot \omega ^{2} \to [ML^{2}]\cdot [T^{-2}] = [ML^{2}T^{-2}]

Both sides are Joules.

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Discovery

Leonhard Euler · 1750

Euler developed rigid-body rotational dynamics in his Mechanica (1736-1765), introducing the moment of inertia and rotational energy concepts.

Try this

How much energy is stored in a flywheel spinning at 10000 RPM?

A 5 kg solid disk of radius 0.2 m spins at 10000 RPM. Find I = ½MR², then KE = ½Iω². Compare to a car battery (≈4 MJ).

Research status: stable

Real-world applications

Flywheel energy-storage batteries (grid-scale and racing)
Spinning hard-disk drives (now mostly replaced by SSDs)
Gymnastic somersaults and dive scoring
Reaction wheels for spacecraft attitude control

Common misconceptions

KE depends $on \omega ^{2}$ — doubling spin rate quadruples energy
A rolling object's total $KE = \tfrac{1}{2}mv^{2} + \tfrac{1}{2}I\omega ^{2}$ (translational + rotational)
Different geometries have different I formulas $(\tfrac{1}{2}MR^{2} for$ disk, ⅔ $MR^{2} for$ hollow sphere, etc.)

Experimental verification

Flywheel energy-storage systems (e.g., NASA's G2 flywheel) measure

\tfrac{1}{2}I\omega ^{2} to

better than 1% precision via spin-down power output.

Derivation

Sum kinetic energy over all mass elements:

KE = \Sigma \tfrac{1}{2}m

ᵢvᵢ

^{2}

with vᵢ

= r

ᵢ

\omega

.

Then

KE = \tfrac{1}{2}(\Sigma m

ᵢrᵢ

^{2})\omega ^{2} = \tfrac{1}{2}I\omega ^{2}

.

Equivalent forms:

KE = L^{2}/(2I)

using

L = I\omega

.

Limiting cases

\omega \to 0

⟶

KE \to 0

No rotation, no rotational energy.

\omega

doubles⟶

KE \to 4\times

Quadratic scaling — flywheels store energy best at high spin.

I \to 0

⟶

KE \to 0

Massless or pointlike-on-axis system stores no rotational energy.

What if…

What

if \omega

doubles?

Energy quadruples — that's why flywheel batteries push RPM as high as material strength allows.

What if the same mass is reshaped into a ring vs a disk?

Ring has $I = MR^{2}$ (twice the disk' $s \tfrac{1}{2}MR^{2})$ — more energy stored for the same $\omega$ , but more inertia to spin up.

What if the object both rolls and translates?

Total $KE = \tfrac{1}{2}mv^{2} + \tfrac{1}{2}I\omega ^{2}$ . For a solid sphere rolling without slipping, 5/7 of the energy is translational.

1

Flywheel battery

Given ·

M:: 5
R:: 0.2
rpm:: 10000

Find · KE

Steps

$I = \tfrac{1}{2}MR^{2} = \tfrac{1}{2} \times 5 \times 0.04 = 0.1 kg\cdot m^{2}$
$\omega = 10000 \times 2\pi /60 \approx 1047\,\mathrm{rad/s}$
$KE = \tfrac{1}{2} \times 0.1 \times 1047^{2} \approx 54800\,\mathrm{J} = 54.8\,\mathrm{kJ}$

Result ·

KE \approx 54.8\,\mathrm{kJ}

— about 1.4% of a car battery

2

Rolling ball down a ramp

Given ·

m:: 0.5
v:: 3
R:: 0.05

Find · KE_total

Steps

Translational: $\tfrac{1}{2} \times 0.5 \times 3^{2} = 2.25\,\mathrm{J}$
Solid sphere $I =$ ⅖ $mR^{2}$ , $\omega = v/R = 60\,\mathrm{rad/s}$
Rotational: $\tfrac{1}{2} \times (0.4 \times 0.5 \times 0.05^{2}) \times 60^{2} = 0.9\,\mathrm{J}$ → total 3.15 J

Result · KE_total

\approx 3.15\,\mathrm{J}

(translational 2.25 J + rotational 0.9 J)

Kinetic Energy→Angular Momentum→Torque→