Physics · Rotation

Rotational kinetic energy of a rigid body

Rigid rotation about a fixed axis has kinetic energy $E_{\mathrm{rot}}=\frac12I\omega^2$ . Inertia belongs to that exact axis; squaring makes energy independent of rotation direction.

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Formula

Calculate

E_{ro t} = \frac{1}{2} I ω^{2}

Rotational energy in J, inertia in kg·m², angular velocity in rad/s. Signed rotation rate in revolutions per minute gives $\omega=\frac{2\pi n_{\mathrm{rpm}}}{60\,\mathrm{s\,min^{-1}}}$ .

Variables & units

Variables & units
Symbol	Meaning	Unit
$E_{\mathrm{rot}}$	Kinetic energy of the specified rotation in the chosen reference frame	J
$I$	Mass moment of inertia about the exact rotation axis; use ICM in the center-of-mass decomposition	kg·m²
$ω$	Angular velocity about this axis; its sign specifies rotation direction	rad/s

LaTeX

E_{rot}=\frac{1}{2}I\omega^2

QUANTA STUDYquanta-study.de

Rolling-disk energy balance: an upper bar shows 13.2435 joules available gravitational energy. The equally long lower bar splits two to one into translation and rotation about the centre of mass. — Uniform disk from the original rolling example: M = 3.00 kg, R = 0.200 m, h = 0.450 m. Without slipping or losses, 8.829 J translation plus 4.4145 J rotation equals Mgh = 13.2435 J. Bars use the same energy scale.

Rotational kinetic energy calculator

Find the rotational energy, moment of inertia or angular speed of a rigid body turning about a fixed axis.

Calculation model

Find

Moment of inertia about the specified axis I

Moment of inertia about exactly this axis, e.g. 0.5 kg·m² for a small flywheel.

Angular velocity ω with direction

Angular velocity with its direction, e.g. 20 rad/s; the sign does not change the energy.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for E, substitute, evaluate

1. Rearrange for E
$E=\frac12 I\omega^2$
2. Substitute values (in SI units)
$\frac{0.5\cdot20^2}{2}=100$
3. Result
E = 100 J

Result: Energy E

E = 100 J

With I = 0.5 kg·m² and ω = 20 rad/s, E = 100 J.

Quick tips

Twice the rotation rate, four times the energy: E grows with ω². Halve ω and only a quarter of the energy is left.
By hand, convert rotation rates to rad/s first: 1 rpm = 2π/60 rad/s ≈ 0.105 rad/s. Then kg·m² times (rad/s)² gives joules.

Assumptions and limits

Rigid body in an inertial frame, rotation about a specified axis; I and ω refer to that same axis. This mode does not add centre-of-mass translation in general motion. Energy determines only |ω|, not rotation direction.

Derivation & background

Each particle has $v_i=r_{\perp i}|\omega|$ . Summing kinetic energies gives $E_{\mathrm{rot}}=\frac12\left(\sum_i m_ir_{\perp i}^2\right)\omega^2$ . The bracket is mass moment of inertia, continuously $I=\int r_\perp^2\,dm$ . Centre motion adds translation through $E_{\mathrm{kin}}=\frac12Mv_{\mathrm{CM}}^2+\frac12I_{\mathrm{CM}}\omega^2$ . The stationary-ground rolling contact-axis form with $I_C=I_{\mathrm{CM}}+MR^2$ already includes both parts. Fixed-axis energy requires no principal axis; generally $E_{\mathrm{rot}}=\frac12\vec\omega\cdot\mathsf I\vec\omega$ . Deformation and internal relative motion need an expanded balance.

Exam blueprint

Validity range

Classical rigid rotation about an axis fixed in the chosen inertial frame, using I about that exact axis. A principal axis is not required for $\frac12I_{zz}\omega_z^2$ . For translation plus rotation use center-of-mass inertia $I_{\mathrm{CM}}$ ; general spatial rotation needs the inertia tensor. Constant I is additionally required for ΔErot = ½I(ωf²−ωi²).

Derivation steps

Add particle energies of shared rigid rotation and factor out their common angular-speed term.

1Choose an inertial frame and fixed z axis; all particles share rigid angular velocity $\omega_z$ .
2At perpendicular axial distance $r_{\perp i}$ , particle $m_i$ has speed $v_i=r_{\perp i}|\omega_z|$ .
3Substitute into $E_i=\frac12m_iv_i^2$ , giving $E_i=\frac12m_ir_{\perp i}^2\omega_z^2$ .
4Add every contribution and factor out the shared $\frac12\omega_z^2$ .
5Define $I_{zz}=\sum_i m_ir_{\perp i}^2$ , or $I_{zz}=\int r_\perp^2\,dm$ for continuous matter, about the same axis.
6Thus $E_{\mathrm{rot}}=\frac12I_{zz}\omega_z^2$ . For a moving centre, use $I_{\mathrm{CM}}$ and add $\frac12Mv_{\mathrm{CM}}^2$ ; the stationary-ground rolling contact-axis form already includes both parts.

Rearrangements

Angular-speed magnitude from energy

|\omega|=\sqrt{\frac{2E_{\mathrm{rot}}}I}

With $I>0$ and $E\ge0$ , the square root gives angular-speed magnitude only. Energy does not specify rotation direction.

Task variant

A disk has $I=0.25\,\mathrm{kg\,m^2}$ and $\omega=8\,\mathrm{rad\,s^{-1}}$ . Find Erot.

Erot=½·0.25·8²=8.0 J.

Common mistakes

Using inertia about another axis, omitting center-of-mass translation, or adding translation again to contact-axis energy.

The center-of-mass decomposition is $E_{\mathrm{kin}}=\frac12Mv_{\mathrm{CM}}^2+\frac12I_{\mathrm{CM}}\omega^2$ . For pure rolling on a stationary surface, $\frac12I_{\mathrm{contact}}\omega^2$ already gives total energy, with $I_{\mathrm{contact}}=I_{\mathrm{CM}}+MR^2$ .

Exam context

Rotating disks, flywheels and energy conservation in rolling motion.

Formula cluster

Rotation and energy

Moment of inertia is the rotational counterpart of mass.

I = Σ m_ir_i²

Moment of inertia of a particle system

L = Iω

Angular momentum of a rigid body

Ekin = ½mv²

Kinetic Energy

M = F·r·sin(θ)

Torque

Worked example

With $I=0.40\,\mathrm{kg\,m^2}$ and $\omega=10\,\mathrm{rad\,s^{-1}}$ , $E_{\mathrm{rot}}=\frac12I\omega^2=20\,\mathrm J$ . The same energy results at −10 rad/s; doubling angular speed at fixed inertia gives 80 J. Dimension $\mathrm{kg\,m^2\,s^{-2}}$ is the joule.

Applications

Balance flywheel and rotor energy, check braking work and power about a specified axis, and correctly combine center-of-mass translation with rotation during rolling. Drive losses and mass redistribution need separate contributions.

How do you calculate with Rotational kinetic energy of a rigid body?

Here is how to work through a typical Rotational kinetic energy of a rigid body (Erot = ½Iω²) task step by step:

1
Task
A disk has $I=0.25\,\mathrm{kg\,m^2}$ and $\omega=8\,\mathrm{rad\,s^{-1}}$ . Find Erot.
Solution path
Erot=½·0.25·8²=8.0 J.

Understand and apply the equation

A flywheel stores kinetic energy even when its center of mass is stationary. Mass distribution about the chosen axis and squared angular speed determine that energy. Derive it from particle motion, balance a slowing rotor, then separate translation and rotation in a rolling body.

Why a stationary centre can still store energy

A flywheel can turn rapidly while its centre stays fixed. Its mass elements still move and carry kinetic energy. Their energies add; opposite velocities do not cancel energy. First study a fixed axis, then include centre-of-mass translation as a separate term for a rolling body.[1]

Prior knowledge: kinetic energy and conversion of rotation rates. Axis and mass distribution together determine moment of inertia.

How do you calculate rotational kinetic energy about a fixed axis?

First choose an inertial frame and fixed rotation axis. The body rotates rigidly about it: particles share angular velocity, but tangential speed depends on perpendicular axis distance. I must describe that exact axis and mass distribution. This is not a motor’s entire energy budget; it initially counts the specified mechanical motion.[1]

E_{\mathrm{rot}}=\frac12 I\omega_z^2

$\omega_z$ is the signed axial component. Squaring gives the same energy for either rotation direction.

For $I=0.40\,\mathrm{kg\,m^2}$ and $\omega_z=10\,\mathrm{rad\,s^{-1}}$ , the energy is $E=20\,\mathrm J$ . At $-10\,\mathrm{rad\,s^{-1}}$ , the energy is also 20 J, whereas axial angular momentum has the opposite sign. Doubling angular speed quadruples energy at unchanged inertia. Doubling inertia at unchanged angular speed doubles energy.

Equal energy in either direction

QUANTA STUDYquanta-study.de

I=0.250\,\mathrm{kg\,m^2}

as in exercise 1. At

\omega_z

=−12 or +12

\mathrm{rad\,s^{-1}}

, energy is 18 J in either case. Energy does not identify direction; doubling angular speed quadruples it.[1]

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Add the particle energies

A particle of mass $m_i$ at perpendicular axial distance $r_{\perp i}$ has $v_i=r_{\perp i}|\omega_z|$ . Insert this into its kinetic energy. Rigid shared rotation allows $\omega_z^2$ to leave the sum. The remaining sum defines moment of inertia; a mass integral replaces the sum for continuous matter.[1]

\begin{aligned}E&=\sum_i\frac12m_i(r_{\perp i}\omega_z)^2=\frac12\left(\sum_i m_ir_{\perp i}^2\right)\omega_z^2\\I&=\int r_\perp^2\,\mathrm dm\end{aligned}

The derivation does not assume all particles have equal tangential speed.

A thin ring and uniform disk of equal mass and outer radius have $I_{\mathrm{ring}}=MR^2$ and $I_{\mathrm{disk}}=\frac12MR^2$ about their symmetry axes. At equal rotation rate the ring stores twice the energy. This comparison does not establish general engineering superiority: strength, dimensions and losses are not assessed.[1]

How do you convert rpm and what does energy tell you about rotation direction?

\omega_z=\frac{2\pi N}{60}\quad(N\text{ in min}^{-1}),\qquad |\omega_z|=\sqrt{\frac{2E}{I}},\qquad I=\frac{2E}{\omega_z^2}

N is signed revolutions per minute here; rearrangements require I > 0 and $\omega_z$ ≠ 0 when dividing.

A minute has 60 seconds and one revolution is 2π rad. Inserting 1800 rather than 188.50 into the energy equation is therefore not a small rounding error. kg·m²·s⁻² is joule; rad keeps angular meaning visible without adding a dimension. Energy alone determines only a speed magnitude, not direction.

For the corresponding component $L_z=I\omega_z$ , $E=\frac{L_z^2}{2I}$ . This does not permit arbitrary substitution of the full angular-momentum magnitude: non-principal-axis rotation can give transverse components. General rigid rotation uses $E=\frac12\vec\omega\cdot\mathsf I\vec\omega$ with the inertia tensor; one scalar does not describe every spatial motion.[4]

Example: energy, braking work and power

A rotor slows down

Given: Rigid fixed-axis rotor, $I=0.180\,\mathrm{kg\,m^2}$ . Signed rate decreases from +1800 to +600 rpm. Constant external braking torque $\tau_z=-3.60\,\mathrm{N\,m}$ ; I remains unchanged.
Find: Find energy decrease and independently check with work and average power.

Approach: A fixed axis and constant moment of inertia permit the rotational-energy expression. Constant signed braking torque additionally gives angular acceleration and work for an independent balance.
$\begin{aligned}E&=\frac12I\omega^2\\{}\alpha_z&=\frac{\tau_z}I\\{}W&=\tau_z\Delta\theta\end{aligned}$
Convert both rotation rates to angular velocities.
$\begin{aligned}\omega_i&=60\pi\ \mathrm{rad/s}\\{} \omega_f&=20\pi\ \mathrm{rad/s}\end{aligned}$
Calculate energies separately, then subtract.
$\begin{aligned}E_i&=324\pi^2\ \mathrm J\\{} E_f&=36\pi^2\ \mathrm J\\{} \Delta E&=-288\pi^2\ \mathrm J\end{aligned}$
$\alpha_z=\frac{\tau_z}{I}=-20\,\frac{\mathrm{rad}}{\mathrm{s^2}}$ is constant. Time to the positive final speed is $2\pi\,\mathrm s$ .
$\Delta t=\frac{\omega_f-\omega_i}{\alpha_z}=2\pi\ \mathrm s$
Mean angular velocity times duration gives angular displacement; $\tau_z$ Δ $\theta$ is braking work.
$\begin{aligned}\Delta\theta&=80\pi^2\ \mathrm{rad}\\{} W&=\tau_z\Delta\theta=-288\pi^2\ \mathrm J\end{aligned}$

Result: Rotational energy decreases from about 3197.75 J to 355.31 J, releasing 2842.45 J. The rotor still turns positively afterward.

Check: Over $\Delta t\approx6.283\,\mathrm s$ , average $P=\frac{\Delta E}{\Delta t}\approx-452.39\,\mathrm W$ . Magnitude of $P=\tau_z\omega_z$ decreases from 678.58 to 226.19 W; its average agrees here for constant braking torque. Determining where energy appears as heat or electrical energy requires another balance.

Example: count center-of-mass motion too

In the chosen inertial frame, split particle velocity into center-of-mass velocity $V_{\mathrm{CM}}$ and relative velocity. The summed cross term vanishes because $\sum_i m_i\vec r_i^{\prime}=\vec0$ . Total energy is therefore center-of-mass translation plus rotation relative to it. $I_{\mathrm{CM}}$ and $\omega$ describe the relative motion. Pure rolling links the terms by $v_{\mathrm{CM}}=R|\omega|$ . The no-slip rolling condition is an additional kinematic assumption, not an extra source of energy.[2]

E_{\mathrm{kin}}=\frac12Mv_{\mathrm{CM}}^2+\frac12I_{\mathrm{CM}}\omega^2

Both terms must use the same frame.

Uniform disk on an ideal incline

Given: Uniform disk $M=3.00\,\mathrm{kg}$ , $R=0.200\,\mathrm{m}$ starts at rest. Center drops $h=0.450\,\mathrm{m}$ . Pure rolling without slipping on a fixed incline, negligible losses; g = 9.81 $\mathrm{m\,s^{-2}}$ .
Find: Find $v_{\mathrm{CM}}$ , $|\omega|$ and both kinetic-energy parts.

Approach: The uniform disk rolls without slipping on a fixed incline and starts at rest. Without losses, lost gravitational energy becomes centre-of-mass translation plus rotation, without double counting.
$\begin{aligned}Mgh&=\frac12Mv_{\mathrm{CM}}^2+\frac12I_{\mathrm{CM}}\omega^2\\{}I_{\mathrm{CM}}&=\frac12MR^2\\{}|\omega|&=\frac{v_{\mathrm{CM}}}R\end{aligned}$
Substitution: About the symmetry axis, $I_{\mathrm{CM}}=\frac12MR^2=0.0600\,\mathrm{kg\,m^2}$ .
$Mgh=\frac12Mv^2+\frac14MR^2\frac{v^2}{R^2}=\frac34Mv^2$
Solve for v, then use the rolling constraint for $\omega$ .
$v=\sqrt{\frac{4gh}{3}}=\sqrt{5.886}\ \mathrm{m/s}\approx2.42611\ \mathrm{m/s}$
Evaluate translation and rotation separately.
$\begin{aligned}E_{\mathrm{trans}}&=8.829\ \mathrm J\\{} E_{\mathrm{rot}}&=4.4145\ \mathrm J\end{aligned}$

Result: $v_{\mathrm{CM}}\approx2.43\,\mathrm{m\,s^{-1}}$ and $|\omega|\approx12.13\,\mathrm{rad\,s^{-1}}$ . One third of kinetic energy is rotational, two thirds translational.

Check: The sum 13.2435 J equals $Mgh$ . Alternatively the instantaneous contact axis has $I_{\mathrm{contact}}=I_{\mathrm{CM}}+MR^2=0.180\,\mathrm{kg\,m^2}$ ; $\frac12I_{\mathrm{contact}}\omega^2$ gives the same total energy, so do not add translation again. At a 20 degree incline, the ideal disk model requires $\mu_s\ge\frac{\tan(20^\circ)}3\approx0.1213$ for static friction.

Energy transfer and balance limits

\begin{aligned}W&=\int_{\theta_i}^{\theta_f}\tau_z\,\mathrm d\theta\\P&=\tau_z\omega_z\end{aligned}

For rigid rotation about the same fixed axis, work is not simply torque times time.[3]

With constant I and this axis motion, resultant external work equals rotational-energy change. Braking torque has negative power when $\tau_z$ and $\omega_z$ have opposite signs. Torque can exist at $\omega$ = 0, where instantaneous mechanical power is zero. Electrical input power, bearing heating and additional moving components must not be equated without checking.[3]

When weights move inward, I changes and motion during adjustment is not purely rigid. Even with conserved $L_z$ , rotational energy can increase: $E=\frac{L_z^2}{2I}$ grows as I decreases. Adjustment forces do work; conservation of angular momentum alone does not imply conserved rotational energy.

When one inertia number is insufficient

Three-dimensional angular velocity can have several principal-axis components. In that frame rotational energy is $\frac12(I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2)$ . Pure motion about one fixed axis still has energy $\frac12I_{zz}\omega_z^2$ even for a non-principal axis with transverse angular momentum. The principal-axis restriction concerns scalar vector relation $\vec L=I\vec\omega$ , not every use of the energy expression.[4]

Deformation, internal relative motion and relativistic speeds require more than this classical rigid model. An area second moment in m⁴ from beam theory cannot replace mass inertia in $\mathrm{kg\,m^2}$ . Specify axis, frame and included energy terms before calculating a check value.

Five problems with complete checks

Calculate · Foundation

$I=0.250\,\mathrm{kg\,m^2}$ and $\omega_z=-12.0\,\mathrm{rad\,s^{-1}}$ . Find E and $L_z$ .

Show hint

Energy uses $\omega_z^2$ , while axial $L_z$ uses $\omega_z$ .

Learning goal: Calculate energy and signed angular momentum.

Show solution and calculation

Approach: Use the same axis in $E=\frac12I\omega_z^2$ and $L_z=I\omega_z$ .
Substitution gives $E=18\,\mathrm J$ and $L_z=-3\,\mathrm{kg\,m^2\,s^{-1}}$ .

Check: $\frac{L_z^2}{2I}=18\,\mathrm J$ . Squaring makes energy positive, while angular momentum retains rotation direction.

\begin{aligned}E&=18.0\ \mathrm J\\{} L_z&=-3.00\ \mathrm{kg\,m^2/s}\end{aligned}

Calculate · Application

A rigid rotor has $I=0.160\,\mathrm{kg\,m^2}$ . What energy increase takes it from +10 to +25 $\mathrm{rad\,s^{-1}}$ ?

Show hint

Take the difference of squared speeds, not the square of their difference.

Learning goal: Find rotor energy change.

Show solution and calculation

Approach: With unchanged inertia, $\Delta E=\frac12I(\omega_f^2-\omega_i^2)$ .
Initial energy is 8.00 J and final energy 50.0 J, hence $\Delta E=42.0\,\mathrm J$ .

Check: a speed ratio of 2.5 gives an energy ratio of 6.25, and $\frac{50}{8}=6.25$ . The state change determines net mechanical work rather than electrical consumption.

\Delta E=42.0\ \mathrm J

Interpret · Application

A thin ring and a uniform disk of equal mass roll without slipping at the same center-of-mass speed. Compare total energies.

Show hint

Insert $I=MR^2$ or $I=\frac{MR^2}2$ and $\omega^2=\frac{v^2}{R^2}$ .

Learning goal: Compare rolling total energies.

Show solution and calculation

Approach: Combine $E=\frac12Mv^2+\frac12I_{\mathrm{CM}}\omega^2$ with $|\omega|=\frac vR$ .
The ring has $I_{\mathrm{CM}}=MR^2$ , the disk $I_{\mathrm{CM}}=\frac12MR^2$ .
At the same mass and centre-of-mass speed, $E_{\mathrm{ring}}=Mv^2$ and $E_{\mathrm{disk}}=\frac34Mv^2$ , giving the ratio $\frac43$ .

Check: translational contributions agree; only rotational contributions differ.

\frac{E_{\mathrm{ring}}}{E_{\mathrm{disk}}}=\frac43

Interpret · Application

Two rigid rotors about fixed principal axes have I1=0.10 and I2=0.40 $\mathrm{kg\,m^2}$ . Both have 5 J rotational energy and rotate positively. Compare angular velocities and momenta.

Show hint

Solve energy for $\omega$ , then use L=Iω.

Learning goal: Distinguish equal rotational energy from equal angular momentum.

Show solution and calculation

Approach: At equal energy, angular velocity decreases with the square root of inertia whereas angular momentum increases with it.
The rotors turn at 10 and 5 $\mathrm{rad\,s^{-1}}$ with momenta 1 and 2 $\mathrm{kg\,m^2\,s^{-1}}$ .
Substitution yields 5 J for both.
Equal energy means neither equal angular velocity nor equal momentum.

\begin{aligned}\omega_i&=\sqrt{\frac{2E}{I_i}}\\{}\omega_1&=10\,\mathrm{rad\,s^{-1}},&\omega_2&=5\,\mathrm{rad\,s^{-1}}\\{}L_1&=1\,\mathrm{kg\,m^2\,s^{-1}},&L_2&=2\,\mathrm{kg\,m^2\,s^{-1}}\end{aligned}

[1]

Error analysis · Application

A rigid rotor with $I=0.20\,\mathrm{kg\,m^2}$ changes from +8 to −8 $\mathrm{rad\,s^{-1}}$ . Someone claims final energy −6.4 J and hence $\Delta E=-12.8\,\mathrm{J}$ . Find initial and final energy and net change; does that change determine motor consumption?

Show hint

Energy contains angular velocity squared.

Learning goal: Do not transfer rotation sign to quadratic energy.

Show solution and calculation

Approach: Both squares are 64, so both energies are positive 6.4 J and net change is zero.
Angular momentum, in contrast, reverses sign.
Braking and accelerating again may remove, recover or dissipate energy.
Zero net change alone does not determine electrical motor consumption without a process and loss model.

\begin{aligned}E_i=E_f&=\frac12(0.20)(8^2)\,\mathrm J=6.4\,\mathrm J\\{}\Delta E&=\frac12I(\omega_f^2-\omega_i^2)=0\end{aligned}

[1]

What to remember

Rotational energy sums the kinetic energies of all mass elements. It is positive for either direction and grows with angular speed squared. Use the correct axis and add translation when the centre moves. Work and power check state change; equal initial and final energies do not determine electrical consumption.[1]

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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Change history

October 2, 2026Refined question headings for the fixed energy axis, rotation rate and direction; added paragraph connections to the rolling condition and mechanical power.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an individual introduction and conclusion, checked existing calculations, clarified notation and model references, and added an original learning figure.

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Quanta. Rotational kinetic energy of a rigid body. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/rotationsenergie-starrer-koerper

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Frequently asked questions about Rotational kinetic energy of a rigid body

What does rotational energy describe physically?+

Rotational energy counts particle kinetic energies about a specified axis. In rigid rotation all particles share angular velocity, but different perpendicular distances give different tangential speeds. Adding their energies yields $E_{\mathrm{rot}}=\frac12I\omega^2$ . Thus I includes mass distribution and the chosen axis, rather than total mass alone. For $I=0.40\,\mathrm{kg\,m^2}$ and ω = 10 rad/s, energy is 20 J; $-10\,\mathrm{rad\,s^{-1}}$ also gives 20 J. Energy therefore cannot reveal direction. Doubling angular speed at unchanged I gives 80 J. If the center of mass also moves, that motion must be included using the appropriate center-of-mass decomposition.

How can I check a calculation involving rotational energy?+

First check that I belongs to the specified axis and rotation rate has been converted to rad/s. With $I=0.40\,\mathrm{kg\,m^2}$ and ω = 10 rad/s, energy is 20 J. Back-substitution into $|\omega|=\sqrt{\frac{2E}I}$ returns 10 rad/s but not its sign. Then use an independent relation: a rotor with I = 0.180 kg·m² slows from 1800 to 600 rpm, giving $\Delta E=-288\pi^2\,\mathrm J\approx-2842.45\,\mathrm J$ . Constant braking torque −3.60 N·m acting through $80\pi^2\,\mathrm{rad}$ produces the same work. Energy change can be negative even though each kinetic energy remains nonnegative.

What assumptions and limits apply to rotational energy?+

The simple derivation assumes classical rigid shared rotation about a specified axis. I must describe the actual mass distribution about that exact axis. Pure z rotation has Erot = $\frac12I_{zz}\omega_z^2$ even if z is not principal; the full-vector relation $\vec L=I\vec\omega$ using scalar I needs the corresponding principal-axis condition. General spatial rotation uses an inertia tensor. Rolling combines center-of-mass translation with rotation using ICM. The instantaneous contact-axis form already includes both energy terms. Moving weights, deformation or internal relative motion need additional terms. Pure rolling also requires sufficient static friction; the ideal disk on a 20° incline needs μs at least about 0.1213.

How do I check units and scale for rotational energy?+

I has units kg·m² and ω has rad/s; radians are dimensionless. Thus $I\omega^2$ gives joules. Convert rotation rate in rpm with $\omega=\frac{2\pi N_{\mathrm{rpm}}}{60\,\mathrm{s\,min^{-1}}}$ before substituting. A rate of 1800 rpm means $60\pi\,\mathrm{rad\,s^{-1}}$ , not 1800 rad/s. With I = 0.180 kg·m², the example rotor has about 3197.75 J, and at 600 rpm about 355.31 J. Round after evaluating the energy difference. An area second moment in m⁴ cannot replace mass inertia here. At equal angular speed, a thin ring rotating about its symmetry axis stores twice the rotational energy of a uniform disk of equal mass and radius.

When should I combine rotational energy with another relation?+

Combine rotational energy with work W = ∫τz dθ and power P = τzωz about the same fixed axis. Braking work equals the rotor’s energy change, and average power is −452.39 W. Rolling instead requires center-of-mass accounting: a uniform 3.00 kg disk descending 0.450 m gains 13.2435 J in total. Translation contributes 8.829 J and rotation 4.4145 J. The rolling constraint $v_{\mathrm{CM}}=R|\omega|$ connects them. Alternatively, Icontact = ICM + MR² makes $\frac12I_{\mathrm{contact}}\omega^2$ already equal to total kinetic energy. Adding translation again would double-count it. Conservation of angular momentum alone does not guarantee conserved energy when I changes.

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Related formulas

v_{\mathrm{CM}}=R\omega

Rolling without slipping

More formulas: Mechanics

\sum \vec F_{\mathrm{ext}} = \vec F_{\mathrm{res}} = m \vec a

Newton's Second Law

F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}

Newton's Law of Universal Gravitation

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

F = k \cdot x

Hooke's Law (Spring Force)

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Rotational kinetic energy of a rigid body

Variables & units

Rotational kinetic energy calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Rotation and energy

Worked example

Applications

How do you calculate with Rotational kinetic energy of a rigid body?

Why a stationary centre can still store energy

How do you calculate rotational kinetic energy about a fixed axis?

Equal energy in either direction

Add the particle energies

How do you convert rpm and what does energy tell you about rotation direction?

Example: energy, braking work and power

A rotor slows down

Example: count center-of-mass motion too

Uniform disk on an ideal incline

Energy transfer and balance limits

When one inertia number is insufficient

Five problems with complete checks

I=0.250 kg m2I=0.250\,\mathrm{kg\,m^2}I=0.250kgm2 and ωz=−12.0 rad s−1\omega_z=-12.0\,\mathrm{rad\,s^{-1}}ωz​=−12.0rads−1. Find E and LzL_zLz​.

A rigid rotor has I=0.160 kg m2I=0.160\,\mathrm{kg\,m^2}I=0.160kgm2. What energy increase takes it from +10 to +25 rad s−1\mathrm{rad\,s^{-1}}rads−1?

A thin ring and a uniform disk of equal mass roll without slipping at the same center-of-mass speed. Compare total energies.

Two rigid rotors about fixed principal axes have I1=0.10 and I2=0.40 kg m2\mathrm{kg\,m^2}kgm2. Both have 5 J rotational energy and rotate positively. Compare angular velocities and momenta.

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Retain Rotational kinetic energy of a rigid body for exams

$I=0.250\,\mathrm{kg\,m^2}$ and $\omega_z=-12.0\,\mathrm{rad\,s^{-1}}$ . Find E and $L_z$ .

A rigid rotor has $I=0.160\,\mathrm{kg\,m^2}$ . What energy increase takes it from +10 to +25 $\mathrm{rad\,s^{-1}}$ ?

Two rigid rotors about fixed principal axes have I1=0.10 and I2=0.40 $\mathrm{kg\,m^2}$ . Both have 5 J rotational energy and rotate positively. Compare angular velocities and momenta.