Physics · Kinematics

Angular velocity and angular frequency

Q: How are angular velocity and orbital speed related?

For pure rotation about a fixed axis, $\vec v=\vec\omega\times\vec r$. Tangential speed is $v=|\omega_z|r_\perp$, with r⊥ the perpendicular distance from the axis. Points of a rigid body share angular velocity but have different linear velocities. Uniform rotation with $T=4 s$ gives angular-speed magnitude $\frac{\frac{\pi{}}{2} rad}{s}$; a point at 3 m moves at approximately $\frac{4.71 m}{s}$ and one at 1 m at $\frac{1.57 m}{s}$. Negative ωz denotes turning sense and must not be reported as negative speed. Additional body translation requires adding reference velocity. A flexible or twisting body need not have one shared ω throughout.

Q: Why must you calculate in radians?

Radian measure defines $\theta=\frac sr$, making $s=r\theta$, $v=|\omega|r$ and standard sine/cosine derivatives work without extra conversion factors. Convert degree angular rates to $\frac{rad}{s}$ by multiplying by $\frac\pi{180}$. The radian is a dimensionless SI unit whose name preserves the angular meaning. Calculator mode must nevertheless match the actual input: degree mode for $\sin30^\circ$ and radian mode for $\sin\frac\pi6$. A force-geometry formula explicitly using a degree angle does not therefore require RAD automatically. What matters is angle unit and the function’s convention. One complete revolution is 360° or 2π rad; consistent conversion makes the two representations equivalent.

Angular velocity is angle change per time. The shorthand $\omega=2\pi f=\frac{2\pi}T$ gives magnitude for uniform revolution; tangential speed additionally depends on distance from the axis.

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Formula

Calculate

ω = 2 π f = \frac{2 π}{T} v = ω \cdot r

ω in rad/s · f in Hz · T in s · v in m/s · r in m

Variables & units

Variables & units
Symbol	Meaning	Unit
$ω$	Angular-speed magnitude in this uniform shorthand; ωz may carry a sign	rad/s
$f$	Positive cycle frequency for uniform rotation	Hz
$T$	Positive duration of one uniform revolution	s
$v$	Tangential-speed magnitude	m/s
$r$	Perpendicular distance from the fixed rotation axis	m

LaTeX

\omega = 2\pi f = \frac{2\pi}{T} \qquad v = \omega \cdot r

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Diagram: a rotating disc with a rotation arrow ω, below it a dimension arrow r from the centre to the rim. At the rim a velocity arrow v, at half the radius an arrow v/2 half as long, both tips on a dashed straight line through the centre. — All points on the rigid disk share angular velocity. Tangential speed is v = |ωz|r and grows in proportion to distance from the fixed axis.

Angular velocity calculator

Find angular velocity, frequency, rpm, period or tangential speed for uniform rotation about a fixed axis.

Calculation model

Find

Revolution frequency f

Revolutions per second in Hz or per minute in rpm, e.g. 120 rpm, which is 2 Hz.

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

Working · rearrange for Ω, substitute, evaluate

1. Rearrange for Ω
$\Omega=2\pi f$
2. Substitute values (in SI units)
- f = 120 U/min = 2 Hz
$2\cdot 3.14159\cdot 2=12.5664$
3. Result
Ω = 12.57 rad/s

Result: Angular speed Ω (magnitude of ω)

Ω = 12.57 rad/s

With f = 120 U/min, Ω = 12.57 rad/s.

Quick tips

One revolution is 2π rad, so ω = 2π · f. Estimate: 1 revolution per second is just over 6 rad/s, and 60 rpm is exactly 1 Hz.
The further out a point sits, the faster it moves: v = ω · r. All points of a rigid disc share the same ω, but the outermost has the largest v.

Assumptions and limits

Uniform nonzero revolution without reversals, or a well-defined harmonic phase. Positive frequency and complete period; Ω is magnitude. This does not determine turning sense or instantaneous values in accelerating rotation.

Derivation & background

One revolution is 2π rad. Uniform rotation therefore gives $|\omega_z|=2\pi f=\frac{2\pi}T$ ; $\omega_z=\frac{\mathrm d\theta}{\mathrm dt}$ includes turning sense. Rigid-body points share angular velocity while their tangential speeds grow with axis distance. For a harmonic oscillator, Ω is phase rate; the body need not follow a circular path. Divide rpm by 60 before multiplying by 2π.

Exam blueprint

Validity range

Fixed axis or well-defined phase. $\omega=2\pi f=\frac{2\pi}T$ describes magnitude for uniform revolution; v is speed. Nonuniform rotation requires differentiation/integration and, where relevant, turning sense.

Derivation steps

Radians connect angle and arc length; multiplying cycle rate by 2π gives angular speed.

1Choose a fixed axis and positive turning sense; track θ as a continuous angle in radians.
2Define $\omega_z=\frac{\mathrm d\theta}{\mathrm dt}$ and $\Omega=|\omega_z|$ . Signed component and magnitude are distinct quantities.
3Uniform revolution covers $|\Delta\theta|=2\pi$ in time T, hence $\Omega=\frac{2\pi}T$ .
4Positive cycle frequency $f=\frac1T$ gives $\Omega=2\pi f$ . Negative turning sense gives $\omega_z=-\Omega$ .
5For a fixed axis, differentiating $s=r_\perp\theta$ gives tangential speed $v=\Omega r_\perp$ .
6For varying ωz, obtain instantaneous values from θ(t) and angle change from $\int\omega_z(t)\,\mathrm dt$ ; one revolution’s $\frac{2\pi}T$ is not generally an instantaneous value.

Rearrangements

Frequency

f=\frac{|\omega_z|}{2\pi}

For uniform revolution f is positive; use the signed axial component’s magnitude.

Radius

r_\perp=\frac v{|\omega_z|}

Pure rotation about a fixed axis, nonzero angular velocity, and tangential speed v.

Period

T=\frac{2\pi}{|\omega_z|}

Uniform nonzero revolution; T is a positive complete period.

Task variant

A drill runs at 3000 rpm. Compute f and ω.

Uniform revolution: $f=\frac{3000}{60}\,\mathrm{Hz}=50\,\mathrm{Hz}$ , $\Omega=2\pi f=100\pi\,\mathrm{rad/s}\approx314.159\,\mathrm{rad/s}$ .
Without turning sense only magnitude is determined.

In a simplified spherical Earth model, radius is $6.371\times{}10^{6} m$ and uniform full rotation period is 86400 s. Find the equatorial point’s speed in the model’s nonrotating reference.

Approach: $\Omega=\frac{2\pi}T$ , $v=\Omega R$ .
The supplied model value gives $\Omega=\frac{2\pi}{86400}\,\mathrm{rad/s}\approx7.27221\times10^{-5}\,\mathrm{rad/s}$ and $v\approx463.312\,\mathrm{m/s}$ .
Actual rotation relative to distant stars has a slightly shorter period.

Common mistakes

Using rpm values directly as frequency.

Divide by 60 first: $3000\,\mathrm{min^{-1}}=50\,\mathrm{Hz}$ .

Equating ω and f.

$\omega=2\pi f$ ; the factor 2π separates angular rate from revolution rate.

Calculating in degrees instead of radians.

$v=\omega r$ only holds with ω in $\frac{rad}{s}$ ; $360^\circ=2\pi\,\mathrm{rad}$ .

Assigning the same orbital speed to all points of a wheel.

ω is the same everywhere, but $v=\omega r$ grows with radius.

Exam context

A building block for centripetal force tasks, satellite orbits and harmonic oscillations; converting rpm to $\frac{rad}{s}$ is often the first step.

Formula cluster

Rotation

Connects rotational quantities with orbital ones and leads to the centripetal force.

Fz = m·v²/r

Centripetal Force (Circular Motion)

T²/a³ = 4π²/[G(M+m)]

Kepler Third Law

M = F·r·sin(θ)

Torque

Worked example

A carousel turns once in $T=4\,\mathrm s$ : $\omega=\frac{2\pi}{4}\,\mathrm{rad/s}\approx1.57\,\mathrm{rad/s}$ . A seat at $r=3\,\mathrm m$ has the orbital speed $v=\omega r\approx4.7\,\mathrm{m/s}$ .

Applications

Engine speeds and gearboxes, centrifuges, wind turbines (blade tip speed), Earth rotation and satellites, oscillation theory

How do you calculate with Angular Velocity and Angular Frequency?

Here is how to work through a typical Angular Velocity and Angular Frequency (ω = 2πf; v = ω·r) task step by step:

1
Task
A drill runs at 3000 rpm. Compute f and ω.
Solution path
1. Uniform revolution: $f=\frac{3000}{60}\,\mathrm{Hz}=50\,\mathrm{Hz}$ , $\Omega=2\pi f=100\pi\,\mathrm{rad/s}\approx314.159\,\mathrm{rad/s}$ .
2. Without turning sense only magnitude is determined.
2
Task
In a simplified spherical Earth model, radius is $6.371\times{}10^{6} m$ and uniform full rotation period is 86400 s. Find the equatorial point’s speed in the model’s nonrotating reference.
Solution path
1. Approach: $\Omega=\frac{2\pi}T$ , $v=\Omega R$ .
2. The supplied model value gives $\Omega=\frac{2\pi}{86400}\,\mathrm{rad/s}\approx7.27221\times10^{-5}\,\mathrm{rad/s}$ and $v\approx463.312\,\mathrm{m/s}$ .
3. Actual rotation relative to distant stars has a slightly shorter period.

Understand and apply the equation

Angular velocity describes angular change per time and has a sign for a specified axis. In uniform rotation it connects rotation rate and revolution period. A disk, accelerated motion, and a reversal separate magnitude from direction.

How much angle changes per unit time?

Two points on the same rigid rotating disk go around the axis together. The outer point travels a longer distance. Their angular velocities are equal, but their tangential speeds differ. This page explains how rotation rate, revolution period, and an unwrapped angle fit together.[1][4]

You need time intervals, circumference, and a positive rotation direction. A full revolution contains two pi radians. Frequency counts complete cycles per second; angular frequency measures radians per second. The same relation describes oscillation phase even when no body actually moves around a circle.[1][4]

Angle change rather than cycle counts alone

\begin{aligned}\omega_z(t)&=\frac{\mathrm d\theta}{\mathrm dt}\\\overline\omega_z&=\frac{\theta_2-\theta_1}{t_2-t_1}\end{aligned}

A fixed positive axis determines positive turning sense by the right-hand rule. Track θ as a continuous angle in radians.[1]

For rotation in the xy plane with positive z out of the page, counterclockwise rotation is positive. Clockwise rotation has $\omega_z<0$ . Magnitude $\Omega=|\omega_z|$ is nonnegative. A speed v therefore cannot be equated blindly to a negative ωz.

Angle displays often reset from 360° to 0°. Unwrap these resets for changes spanning several revolutions: two full positive revolutions give $\Delta\theta=4\pi$ even though initial and final orientation look identical. During back-and-forth rotation, net angle change can be smaller than total swept angle.

Why radians are more than an arbitrary label

\begin{aligned}\theta&=\frac sr\\\Delta\theta_{\mathrm{rev}}&=2\pi\,\mathrm{rad}\\\theta_{\mathrm{rad}}&=\frac\pi{180}\theta_{\mathrm{deg}}\end{aligned}

s is the corresponding signed arc length at fixed radius r > 0.[1]

The radian is a dimensionless derived SI unit because $\frac{\mathrm m}{\mathrm m}=1$ . Writing rad still identifies an angle rather than an arbitrary number. Angular velocity has dimension $\mathrm{s^{-1}}$ , with the informative unit $\mathrm{rad/s}$ . Hertz instead counts complete cycles per second. Equal dimensions do not remove the conversion factor $2\pi$ between the two.[5]

The simple relation $s=r\theta$ and standard sine/cosine derivatives use radians. Convert an angular rate in degrees per second by multiplying by $\frac{\pi}{180}$ first. To evaluate $\sin30^\circ$ , degree mode is appropriate; the input must match the chosen trigonometric convention.

How do you convert frequency, period and rpm to angular speed?

\begin{aligned}\Omega&=2\pi f=\frac{2\pi}T\\f&=\frac1T\end{aligned}

For uniform rotation, f and T are positive cycle rate and period; Ω is angular-speed magnitude.[1]

A uniform positive revolution in T seconds covers $\Delta\theta=+2\pi\,\mathrm{rad}$ , giving $\omega_z=+\frac{2\pi}{T}$ . Negative turning sense gives component $-\frac{2\pi}{T}$ . The shorthand $\omega=2\pi f$ often means only magnitude. Define that meaning before calculating, especially for an axial component or reversal. For a positive rotational-speed magnitude n in rpm, first obtain $f=\frac n{60}$ in hertz, then $\Omega=2\pi f$ in rad/s.

For varying rotation, $\frac{2\pi}{T}$ over a completed revolution without reversal gives only that revolution’s mean angular-speed magnitude. It is not automatically the instantaneous value at its start or end. A continuous angle record or smaller measurement windows are needed to describe acceleration in more detail.

What points share and what depends on radius

\begin{aligned}\vec v&=\vec\omega\times\vec r\\v&=\Omega r_\perp\end{aligned}

For pure rotation about a fixed axis, r⊥ is perpendicular distance to the axis and v is speed.[2]

A rigid body has one shared angular velocity at each instant. A point on its axis has zero tangential speed, while twice the distance gives twice the speed. For a planar disc, r⊥ is circle radius. If the body also translates, add reference velocity to rotational velocity.[2]

\begin{aligned}a_{\mathrm t}&=\alpha_zr_\perp\\a_{\mathrm r}&=\Omega^2r_\perp\\\alpha_z&=\frac{\mathrm d\omega_z}{\mathrm dt}\end{aligned}

Tangential acceleration has a sign; radial acceleration points inward with the stated magnitude.[2]

Even constant Ω produces radial acceleration because velocity direction changes. Varying ωz adds tangential acceleration. Zero angular velocity, zero angular acceleration and zero linear acceleration are therefore different statements. The required net radial force is described by centripetal force.

Two original rotation examples

Rotation rate with negative turning sense

Given: Rigid rotor about fixed z: 120 rpm clockwise, viewed from positive z. A point is at r⊥ = 0.18 m.
Find: Calculate f, T, ωz and tangential speed.

Approach: Uniform revolution: first obtain cycles per second, then apply turning sense and radius.
$\begin{aligned}f&=\frac{N_{\mathrm{rev}}}{\Delta t}\\\omega_z&=-2\pi f\\v&=|\omega_z|r\end{aligned}$
Substitute the given values with matching units.
$\begin{aligned}f&=\frac{120}{60}=2\,\mathrm{Hz}\\{} T&=\frac{1}{f}=0.50\,\mathrm s\end{aligned}$
Use factor 2π and negative turning sense.
$\omega_z=-2\pi f=-4\pi\,\mathrm{rad/s}\approx-12.5664\,\mathrm{rad/s}$
Use Ω rather than the signed component for linear speed.
$v=|\omega_z|r_\perp=4\pi\cdot0.18\approx2.26195\,\mathrm{m/s}$

Result: f = 2 Hz, T = 0.50 s, ωz ≈ −12.57 rad/s. The point’s tangential speed is 2.262 m/s.

Check: One full revolution covers $2\pi\cdot0.18\,\mathrm m$ . Dividing by $0.50\,\mathrm s$ again gives $0.72\pi\,\mathrm{m/s}$ . The negative axial component denotes turning sense, not negative distance.

Time-dependent angular velocity

Given: θ(t) = (1.2 rad/s²)t² + (0.50 rad/s³)t³, with t in seconds; fixed axis, t from 0 to 0.40 s.
Find: Compare mean angular velocity, final angular velocity and final angular acceleration.

Approach: Derivative and interval mean are different quantities.
$\begin{aligned}\overline\omega_z&=\frac{\theta(t_1)-\theta(0)}{t_1}\\\omega_z(t)&=\frac{\mathrm d\theta}{\mathrm dt}\\\alpha_z(t)&=\frac{\mathrm d\omega_z}{\mathrm dt}\end{aligned}$
Substitute the given values with matching units.
$\Delta\theta=1.2\cdot0.40^2+0.50\cdot0.40^3=0.224\,\mathrm{rad}$
Divide by duration for the mean.
$\overline\omega_z=\frac{0.224}{0.40}=0.560\,\mathrm{rad/s}$
Differentiate for the instantaneous value.
$\begin{aligned}\omega_z(t)&=2.4t+1.5t^2\\{} \omega_z(0.40)&=1.20\,\mathrm{rad/s}\end{aligned}$
Differentiate once more; coefficient units follow from the given function.
$\begin{aligned}\alpha_z(t)&=2.4+3t\\{} \alpha_z(0.40)&=3.60\,\mathrm{rad/s^2}\end{aligned}$

Result: Mean angular velocity is 0.560 rad/s, final angular velocity 1.20 rad/s and final angular acceleration 3.60 rad/s².

Check: Integrating ωz(t) from 0 to 0.40 s recovers 0.224 rad. Because α varies with time, assuming constant α without justification would be invalid.

Uniform, constantly accelerated or general

\theta(t)=\theta_0+\int_0^t\omega_z(\tau)\,\mathrm d\tau

Angle change is signed area under ωz(t).[1]

Constant ωz gives $\theta=\theta_0+\omega_zt$ . Constant αz gives $\omega_z=\omega_{z,0}+\alpha_zt$ and $\theta=\theta_0+\omega_{z,0}t+\frac12\alpha_zt^2$ . Each simplification requires its stated constant quantity. Negative initial ω with positive α can first slow rotation and later reverse it. The constant angular acceleration model tracks these initial values and the subsequent turning sense together.[3]

With $\omega_{z,0}=-2\,\mathrm{rad/s}$ and $\alpha_z=+1\,\mathrm{rad/s^2}$ , rotation stops after $2\,\mathrm s$ and reverses. After $4\,\mathrm s$ the unwrapped angle equals its initial value again. Angular travel is still $4\,\mathrm{rad}$ , because both parts count positively. The third example evaluates the following diagram’s two areas step by step.

Signed areas show the reversal

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For the third example,

\omega_z(t)=-2\,\mathrm{rad/s}+(1\,\mathrm{rad/s^2})t

. Negative and positive triangular areas cancel in net angle; their magnitudes add to angular travel

4\,\mathrm{rad}

.[1][3]

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Oscillation frequency and practical limits

In $x(t)=A\cos(\Omega t+\varphi_0)$ , $\Omega$ denotes oscillation phase rate. Harmonic phase evolution obeys $\Omega=2\pi f$ . The body need not move around a circle, so $\Omega r$ does not automatically represent its velocity. Differentiation instead gives $v_x(t)=-A\Omega\sin(\Omega t+\varphi_0)$ .[4]

Different parts of a flexible or twisting rotor need not share angular velocity. In spatial motion, ω⃗ is a vector; Euler-angle derivatives are not generally its components directly. This page’s simple scalar relationships use a fixed axis or a well-defined oscillation phase.

A rotational-speed display over a finite window gives an average. Frequency or period alone cannot describe starting transients, acceleration and reversals. Also specify measurement interval, turning sense, radius and the frame used for linear velocity.

A reversal has angular travel but zero net angle

A reversal sharply separates net angle from angular travel. Integrate the following profile once with its sign and once as a magnitude.[1][3]

A reversal has angular travel but zero net angle

Given: Motion about a fixed z-axis starts at $\omega_{z,0}=-2.0\,\mathrm{rad/s}$ and $\theta_0=0$ . Angular acceleration is constant at +1.0 rad/s² for 4.0 s. Signed angles are recorded continuously.
Find: Find reversal time, net angle, total angular travel, and both averages.

Approach: integrate signed angular velocity for net angle and its magnitude for angular travel.
$\begin{aligned}\omega_z(t)&=\omega_{z,0}+\alpha_zt\\\Delta\theta&=\int_0^{4\,\mathrm s}\omega_z(t)\,\mathrm dt\\\Theta_{\mathrm{travel}}&=\int_0^{4\,\mathrm s}|\omega_z(t)|\,\mathrm dt\end{aligned}$
The two triangular areas have equal magnitudes but opposite signs for net angle.
$\begin{aligned}t_{\mathrm{turn}}&=-\frac{\omega_{z,0}}{\alpha_z}=2.0\,\mathrm s\\\Delta\theta_1&=-\frac12(2\,\mathrm s)(2\,\mathrm{rad/s})=-2\,\mathrm{rad}\\\Delta\theta_2&=+2\,\mathrm{rad}\\\Delta\theta&=0,&\Theta_{\mathrm{travel}}&=4\,\mathrm{rad}\\\overline\omega_z&=0,&\overline{|\omega_z|}&=1\,\mathrm{rad/s}\end{aligned}$

Result: The body reverses after 2 s and returns to its initial unwrapped angle after 4 s, yet has travelled a total angular distance of 4 rad.

Check: The independent angle relation gives $\theta(4\,\mathrm s)=(-2)(4)+\frac12(1)(4)^2=0\,\mathrm{rad}$ ; at reversal $\theta=-2\,\mathrm{rad}$ . A zero mean therefore does not mean the body remained still.

Five problems with solutions

Calculate · Foundation

A disc rotates uniformly at 75 rpm in positive z sense. Find ωz and speed at 0.20 m.

Show hint

First divide rpm by 60.

Learning goal: Relate rotational frequency and point speed.

Show solution and calculation

Approach: positive uniform rotation gives $f=1.25\,\mathrm{Hz}$ and $\omega_z=2.5\pi\,\mathrm{rad/s}\approx7.854\,\mathrm{rad/s}$ .
Radius then gives $v=0.5\pi\,\mathrm{m/s}\approx1.571\,\mathrm{m/s}$ .

Check: circumference $2\pi r$ divided by period $\frac1f$ gives the same speed.

\begin{aligned}f&=\frac{75}{60}\,\mathrm{Hz}=1.25\,\mathrm{Hz}\\\omega_z&=+2\pi f=2.5\pi\,\mathrm{rad/s}\\v&=|\omega_z|r=0.5\pi\,\mathrm{m/s}\end{aligned}

Calculate · Application

A continuous angle changes from 10° to 250° in 2 s without reversal. Find mean angular velocity in rad/s.

Show hint

Find angle change and convert degrees to radians.

Learning goal: Average continuous angle change over time.

Show solution and calculation

Approach: the angle is already unwrapped, so the change is uniquely $240^\circ=\frac{4\pi}3\,\mathrm{rad}$ .
Dividing by the interval gives $\overline\omega_z=\frac{2\pi}3\,\mathrm{rad/s}\approx2.094\,\mathrm{rad/s}$ .
Without a detailed time profile this does not determine instantaneous final velocity.

\begin{aligned}\Delta\theta&=(250-10)\frac\pi{180}\,\mathrm{rad}=\frac{4\pi}3\,\mathrm{rad}\\\overline\omega_z&=\frac{\Delta\theta}{\Delta t}=\frac{2\pi}3\,\mathrm{rad/s}\\&\approx2.094\,\mathrm{rad/s}\end{aligned}

Calculate · Application

$\omega_z(t)=(2\,\mathrm{rad/s^2})t$ for $0\le t\le3\,\mathrm s$ . Find final angular velocity and angle change.

Show hint

Substitute final time; integrate triangular area for angle.

Learning goal: Integrate an angular-velocity history.

Show solution and calculation

Approach: substitute directly for the final rate but integrate for angular change.
Thus $\omega_z(3\,\mathrm s)=6\,\mathrm{rad/s}$ and $\Delta\theta=9\,\mathrm{rad}$ .
The triangle under the straight line checks the integral.
Final rate times three seconds would be $18\,\mathrm{rad}$ , twice too large.

\begin{aligned}\omega_z(3\,\mathrm s)&=6\,\mathrm{rad/s}\\\Delta\theta&=\int_0^{3\,\mathrm s}\omega_z(t)\,\mathrm dt\\&=\frac12(2\,\mathrm{rad/s^2})(3\,\mathrm s)^2=9\,\mathrm{rad}\end{aligned}

Interpret · Application

Two points on one rigid uniformly rotating disk have radii 5 cm and 15 cm. Both have revolution period 2 s. Compare f, |ω| and speeds; what shape does a v-versus-r plot have?

Show hint

All points share rotation, not orbital circumference.

Learning goal: Distinguish shared angular velocity and different linear speed.

Show solution and calculation

Approach: Both frequencies are 0.5 Hz and both angular-speed magnitudes are π rad/s.
Speeds are approximately 0.157080 and 0.471239 m/s.
The outer point is three times faster.
At fixed angular speed, v against r is a line through the origin with slope equal to that angular speed.

\begin{aligned}f&=\frac1T=0.5\,\mathrm{Hz},&|\omega|&=2\pi f=\pi\,\mathrm{rad\,s^{-1}}\\{}v_1&=0.05\pi\,\mathrm{m\,s^{-1}},&v_2&=0.15\pi\,\mathrm{m\,s^{-1}}\end{aligned}

[1]

Error analysis · Application

An encoder changes from 350° to 10° in 0.20 s. Motion is known positive, without reversal or an additional full revolution. Someone uses −340° and reports negative mean angular velocity. Correct by unwrapping the angle.

Show hint

The second displayed angle corresponds to 370° continuously.

Learning goal: Avoid using a wrapped angle as continuous angular change.

Show solution and calculation

Approach: Positive continuous change is 20°, not −340°.
Conversion gives π divided by nine radians.
Over 0.20 s this yields mean angular velocity about +1.74533 rad/s.
Without direction and revolution information unwrapping would be ambiguous; the display gives only angle modulo one turn.

\begin{aligned}\Delta\theta&=370^\circ-350^\circ=20^\circ=\frac\pi9\,\mathrm{rad}\\{}\overline\omega&=\frac{\pi\,\mathrm{rad}}{9\cdot0.20\,\mathrm s}\approx1.74533\,\mathrm{rad\,s^{-1}}\end{aligned}

[1]

In brief

Distinguish signed angular velocity, its magnitude, and frequency in complete cycles per second. Convert angles to radians and account for extra revolutions. Integrating signed velocity gives net angle; integrating its magnitude gives angular travel. Radius is additionally needed to obtain tangential speed.[1][3]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 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.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 2, 2026Refined question headings and added scientifically relevant paragraph links; retained existing derivations, examples, diagrams, exercises and sources.
October 1, 2026Clarified angular velocity, magnitude, and phase; added a complete reversal example and quantitative signed areas. Preserved previous rotation-rate and derivative examples and all five exercises.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

Cite this reference page

Quanta. Angular Velocity and Angular Frequency. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/winkelgeschwindigkeit-kreisfrequenz

When citing a specific scientific statement, also cite the specialist source linked to that statement.

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Frequently asked questions about Angular Velocity and Angular Frequency

How do you convert revolutions per minute to

\frac{rad}{s}

?+

Divide positive rotational-speed magnitude n in rpm by 60 to obtain $f=\frac n{60}$ in revolutions per second. Then multiply by 2π radians per revolution: $\Omega=\frac{2\pi n}{60}$ . For 3000 rpm this gives 50 Hz and approximately $\frac{314.16 rad}{s}$ . Turning sense is additional information: about a positively defined axis, ωz may be +Ω or −Ω. A display averaged over a measuring window also gives an average. If rotation varies, it does not automatically give instantaneous final angular velocity. Use radians for the simple relationships to tangential speed and acceleration. The factor 2π remains necessary even though radians are dimensionless in SI.

What is the difference between frequency and angular frequency?+

Frequency f counts complete cycles per second and is expressed in hertz. Uniform revolution or harmonic phase evolution has angular frequency $\Omega=2\pi f$ in $\frac{rad}{s}$ . A cycle corresponds to 2π rad, so numerical values differ. Signed rotation also requires turning sense; f is usually counted positively. For an oscillator, Ω describes phase rather than necessarily a body rotating on a circle. Radians are dimensionless in SI, but writing rad distinguishes angular rate from cycle counting. For varying rotation, $\frac{2\pi}T$ over a full revolution is only its mean, not generally instantaneous angular velocity. Account for this when calculating speed or radial acceleration.

How are angular velocity and orbital speed related?+

For pure rotation about a fixed axis, $\vec v=\vec\omega\times\vec r$ . Tangential speed is $v=|\omega_z|r_\perp$ , with r⊥ the perpendicular distance from the axis. Points of a rigid body share angular velocity but have different linear velocities. Uniform rotation with $T=4 s$ gives angular-speed magnitude $\frac{\frac{\pi{}}{2} rad}{s}$ ; a point at 3 m moves at approximately $\frac{4.71 m}{s}$ and one at 1 m at $\frac{1.57 m}{s}$ . Negative ωz denotes turning sense and must not be reported as negative speed. Additional body translation requires adding reference velocity. A flexible or twisting body need not have one shared ω throughout.

Why must you calculate in radians?+

Radian measure defines $\theta=\frac sr$ , making $s=r\theta$ , $v=|\omega|r$ and standard sine/cosine derivatives work without extra conversion factors. Convert degree angular rates to $\frac{rad}{s}$ by multiplying by $\frac\pi{180}$ . The radian is a dimensionless SI unit whose name preserves the angular meaning. Calculator mode must nevertheless match the actual input: degree mode for $\sin30^\circ$ and radian mode for $\sin\frac\pi6$ . A force-geometry formula explicitly using a degree angle does not therefore require RAD automatically. What matters is angle unit and the function’s convention. One complete revolution is 360° or 2π rad; consistent conversion makes the two representations equivalent.

How fast does the Earth rotate and what follows from it?+

Rotation relative to distant stars takes approximately one sidereal day, $86164\,\mathrm s$ , shorter than the mean solar day of $86400\,\mathrm s$ . This gives $\Omega=\frac{2\pi}T\approx7.29\times10^{-5}\,\mathrm{rad/s}$ . In a simplified spherical model with radius $R=6.371\times10^6\,\mathrm m$ , an equatorial point moves at $v=\Omega R\approx465\,\mathrm{m/s}$ , exceeding $\frac{1600 km}{h}$ . At geographic latitude the perpendicular axis distance is approximately $R\cos\varphi$ , vanishing at the poles. The value $\frac{300 m}{s}$ is only a rough central-European scale. A sign requires a chosen axis. Rotation contributes speed to launches and matters for centrifugal and Coriolis effects; their full calculation requires additional geometry and reference-frame information.

Quanta exam set

Curated exam set for "Angular Velocity and Angular Frequency":

Question (front)

Which formula describes Angular Velocity and Angular Frequency?

Answer in your set

Question (front)

How do you rearrange Angular Velocity and Angular Frequency for Frequency?

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Which common mistake happens with Angular Velocity and Angular Frequency?

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Retain Angular Velocity and Angular Frequency for exams

Create a curated FSRS exam set for ω = 2πf; v = ω·r: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

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Angular velocity and angular frequency

Variables & units

Angular velocity calculator

Quick tips

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Rearrangements

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Rotation

Worked example

Applications

How do you calculate with Angular Velocity and Angular Frequency?

How much angle changes per unit time?

Angle change rather than cycle counts alone

Why radians are more than an arbitrary label

How do you convert frequency, period and rpm to angular speed?

What points share and what depends on radius

Two original rotation examples

Rotation rate with negative turning sense

Time-dependent angular velocity

Uniform, constantly accelerated or general

Signed areas show the reversal

Oscillation frequency and practical limits

A reversal has angular travel but zero net angle

A reversal has angular travel but zero net angle

Five problems with solutions

A disc rotates uniformly at 75 rpm in positive z sense. Find ωz and speed at 0.20 m.

A continuous angle changes from 10° to 250° in 2 s without reversal. Find mean angular velocity in rad/s.

ωz(t)=(2 rad/s2)t\omega_z(t)=(2\,\mathrm{rad/s^2})tωz​(t)=(2rad/s2)t for 0≤t≤3 s0\le t\le3\,\mathrm s0≤t≤3s. Find final angular velocity and angle change.

Two points on one rigid uniformly rotating disk have radii 5 cm and 15 cm. Both have revolution period 2 s. Compare f, |ω| and speeds; what shape does a v-versus-r plot have?

An encoder changes from 350° to 10° in 0.20 s. Motion is known positive, without reversal or an additional full revolution. Someone uses −340° and reports negative mean angular velocity. Correct by unwrapping the angle.

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More formulas: Mechanics

Retain Angular Velocity and Angular Frequency for exams

$\omega_z(t)=(2\,\mathrm{rad/s^2})t$ for $0\le t\le3\,\mathrm s$ . Find final angular velocity and angle change.