
Physics I course sequence · Interactive lesson
Angular Velocity, Period & Frequency
Every point of a spinning body moves at its own speed, yet the body turns at one rate. This lesson makes that rate precise: read it as the slope of θ(t), point it along the axle with your right hand, and translate it into the periods, frequencies and rpm that instruments and machines actually report.
UPI-09.02 · Lesson 2 of 7 in this unit. Position is not a completion score.Build the model
A rotation rate is a slope, and its direction is the axle.
The key idea
Angular velocity is the rate at which a rotating body's angle changes, ω = dθ/dt in rad s⁻¹; its sign, or its direction along the axle by the right-hand rule, records which way the body turns.
Example. A ship's radar antenna making 24 turns per minute has f = 0.40 Hz, T = 2.5 s and ω = 2π × 0.40 = 2.51 rad s⁻¹; turning clockwise as seen from above, its ω⃗ points straight down.
The complete physical model and assumptions
Angular velocity exists because a rigid body turning about a fixed axle needs only one coordinate. Every point keeps its distance from the axle and every radius line keeps its angle to every other, so a single angle θ(t), in radians from a fixed reference line, locates the whole body, and a single rate says how fast it turns. Two readings give the average rate Δθ/Δt, the slope of a chord on the θ–t graph; letting the interval shrink gives the instantaneous rate ω = dθ/dt, the slope of the tangent. The sign of dθ/dt records the sense of turning, and the right-hand rule promotes it to a vector along the axle, the one direction every point of the body shares. When ω holds constant the motion repeats every T = 2π/|ω|, at frequency f = 1/T. The model has edges. A period computed from an instantaneous ω is wrong whenever the rate varies within a turn; only the average over one whole turn fixes T. A body that is not rigid has no single ω: the Sun's equator rotates in about 25 days and its polar regions in about 35. And although angular velocities combine as vectors, finite rotations about different axes do not, so θ is not a vector even though ω is.
1One angle, one rate, for the whole body
Fix a rigid body to an axle and follow any point on it. Its angle θ, measured in radians from a fixed reference line in the plane perpendicular to the axle, changes by the same Δθ as every other point's, because a rigid body cannot bend: the angle between any two of its radius lines is frozen. So one function θ(t) describes the entire rotation, and its rate belongs to the body rather than to any one point. A vinyl record at 33⅓ rpm shows it. A speck on the label and a speck on the outer groove both sweep 2π rad every 1.80 s, so both turn at 2π/1.80 s = 3.49 rad s⁻¹, even though the outer speck travels much farther each turn. Distance travelled depends on radius; angle turned does not. Keep θ in radians: only then is arc length simply radius times angle, a simplicity every later relation built on ω inherits.
2Average is a chord, instantaneous is a tangent
The average angular velocity over an interval, ωavg = Δθ/Δt, is the slope of the straight chord joining two points on the θ–t graph. It uses two readings and ignores everything between them. Shrink the interval and the chord pivots towards the tangent; its limiting slope is the instantaneous angular velocity ω = dθ/dt. Take a wind-turbine rotor starting from rest with θ(t) = 0.050t³ rad. At t = 4.0 s the derivative gives ω = 0.15t² = 2.40 rad s⁻¹. Averages that start at 4.0 s close in on that value: 3.05 rad s⁻¹ over the next 1.0 s, 2.46 rad s⁻¹ over 0.10 s, 2.41 rad s⁻¹ over 0.010 s. The average over the first 4.0 s is only θ/t = 3.2 rad ÷ 4.0 s = 0.80 rad s⁻¹, a third of the final rate, because the rotor spent most of that time turning slowly. Current angle divided by elapsed time is always an average from t = 0, never the rate now.
3The sign becomes a vector along the axle
Choose a direction along the axle and call it +z. Seen from the +z end looking back at the body, counterclockwise turning makes θ grow, so ωz = dθ/dt is positive; clockwise turning makes it negative. The right-hand rule turns that sign into a direction in space: curl the fingers of your right hand the way the body turns, and your thumb points along ω⃗. The arrow lies on the axis because the axis is the only line the whole body agrees on; its points are moving in every direction around the circle. Two cases fix the convention. A car driving north has wheels whose tops move north, so each wheel's ω⃗ points west, to the driver's left, and swings round to east when the car reverses. The Earth turns eastward, so its ω⃗ runs along the axis from the South Pole out through the North Pole, towards Polaris.
4Period, frequency and the missing 2π
When ω is constant the body repeats itself, and three numbers describe the same motion. The period T is the time for one revolution, Δθ = 2π, so T = 2π/|ω|. The frequency f = 1/T counts revolutions per second, in hertz, and ω = 2πf. Machines quote a fourth form, revolutions per minute. A car engine idling at 750 rpm makes 750/60 = 12.5 revolutions each second, so f = 12.5 Hz, T = 1/12.5 s = 80 ms and ω = 2π × 12.5 = 78.5 rad s⁻¹. The trap is that the radian is dimensionless: rad s⁻¹, Hz and rev s⁻¹ all reduce to s⁻¹, so a dimensional check cannot catch a missing or doubled 2π. Carry the unit words anyway, rad s⁻¹ for ω and Hz or rev s⁻¹ for f, and let them tell you when a 2π is owed.
5An uneven turn still has a period
T = 2π/|ω| assumes ω never changes, yet plenty of rotations repeat without being uniform. A playground roundabout that gets a shove each time its handle comes past speeds up, then coasts more slowly until the next push. Model it as ω(t) = ω̄[1 + k cos(2πt/T)] with T = 4.0 s, k = 0.50 and ω̄ = 2π/T = 1.57 rad s⁻¹. At its fastest ω = 2.36 rad s⁻¹ and at its slowest 0.785 rad s⁻¹, so 2π/ω(t) would swing between 2.7 s and 8.0 s. Yet every revolution takes exactly 4.0 s: integrating ω over any 4.0 s window adds precisely 2π to θ, because the cosine term integrates to zero over a whole cycle. The period of an uneven rotation is fixed by the average angular velocity over one full turn, T = 2π/ωavg. Shorter windows depend on where they start: from t = 0.5 s to 2.5 s the average is only 1.22 rad s⁻¹.
6Rates about one axle add; periods do not
Angular velocities about one shared axle combine like one-dimensional velocities: take components along the same +z and add or subtract them with their signs. Periods do not combine that way, so convert to ω first. A clock's hands share an axle and both turn clockwise, so take +z into the wall and both components are positive: ωₘᵢₙᵤₜₑ = 2π/3600 s = 1.745 × 10⁻³ rad s⁻¹ and ωₕₒᵤᵣ = 2π/43 200 s = 1.454 × 10⁻⁴ rad s⁻¹. The minute hand gains on the hour hand at the difference, 1.600 × 10⁻³ rad s⁻¹, and must gain a full 2π between one overlap and the next, which takes 2π/(1.600 × 10⁻³ rad s⁻¹) = 3927 s, or 65.45 minutes, not the 60 minutes that comparing periods suggests. Had the hour hand turned the other way, the rates would add and the hands would meet every 55.38 minutes.
Key equations
Read the units, assumptions and limits before substituting.
Average angular velocity
ωavg = Δθ/Δt = (θ₂ − θ₁)/(t₂ − t₁)
The slope of the chord on a θ–t graph. Two readings are enough, and whatever the body did between them stays invisible.
θ in rad, t in s, ωavg in rad s⁻¹; Δθ is the net signed change, not the total angle sweptInstantaneous angular velocity
ωz = dθ/dt = lim Δθ/Δt as Δt → 0
The slope of the tangent. Differentiate θ(t) and evaluate at the instant: one number that holds for every point of a rigid body.
ωz in rad s⁻¹; θ counted positive counterclockwise as seen from the +z end of the axleDirection by the right-hand rule
ω⃗ = ωz ẑ
Curl the right hand's fingers with the rotation and the thumb gives ω⃗. It lies along the axle, the only line every point of the body shares.
ẑ is the unit vector along the axle; the component ωz carries the sign; |ω⃗| in rad s⁻¹Period and frequency of uniform rotation
T = 2π/|ω| · f = 1/T · ω = 2πf
Turns a rate in rad s⁻¹ into the time per revolution and the revolutions per second that a strobe or tachometer reports.
T in s; f in Hz, meaning revolutions per second; valid only while ω is constantFrom rpm to rad s⁻¹
ω = 2πN/60 ≈ 0.1047 N
Machines are rated in rpm, equations need rad s⁻¹: 1000 rpm is 104.7 rad s⁻¹ and 16.7 Hz, never 1000 of either.
N in revolutions per minute; 2π rad per revolution and 60 s per minutePeriod of an uneven, repeating rotation
T = 2π/ωavg, averaged over one full turn
When ω rises and falls within each turn, 2π/ω at a single instant misjudges the period; only the whole-turn average gives the time per revolution.
ωavg in rad s⁻¹ over an interval in which Δθ = 2π; T in sInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Shrink Δt from 2.0 s to 0.1 s: the dashed secant swings onto the tangent and ωavg climbs from 1.22 to 2.08 rad s⁻¹, closing on 2.13. Stretch Δt to 4.0 s, one full turn, and ωavg = 1.57 rad s⁻¹ = 2π/T from any t. Set k = 0 and every secant is the tangent.
Live results
- ω AT t (TANGENT)
- 2.13 rad s⁻¹
- ωavg OVER Δt (SECANT)
- 1.22 rad s⁻¹
- NET Δθ OVER Δt
- 2.43 rad
- IF STEADY: 2π/ωavg
- 5.16 s
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Instant t: 0.5 s; Averaging interval Δt: 2.0 s; Push unevenness k: 0.5
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Many students draw the angular velocity arrow tangent to the rim, pointing the way the edge moves. That arrow cannot be right, because at any instant the rim moves in every direction at once: relative to the axle, the top of a bicycle wheel moves forward, the bottom backward, the front edge down and the back edge up. No single tangent speaks for the body, and choosing a different rim point would give a different arrow for the same rotation. The one direction the rotation itself singles out is the axle, so ω⃗ lies along it, with its sense set by the right-hand rule. Nothing travels along that arrow. It packs the axis, the sense and the rate into one vector, and it is exactly the form that lets v⃗ = ω⃗ × r⃗ give the velocity of every point from a single quantity.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A cordless drill drives a screw into a wall. Seen from behind the drill, the chuck turns clockwise at a steady 1650 rpm. Find the frequency, the period and the angular velocity, and state which way ω⃗ points.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Frequency counts revolutions per second: f = 1650 rev ÷ 60 s = 27.5 Hz.
- Period is the time for one revolution: T = 1/f = 1/27.5 s = 0.0364 s, or 36.4 ms.
- Each revolution is 2π rad, so ω = 2πf = 2π × 27.5 = 172.8 rad s⁻¹. Check with the period: 2π/0.03636 s = 172.8 rad s⁻¹.
- Direction: curl your right-hand fingers clockwise as you see the chuck from behind, and your thumb points away from you, along the bit and into the wall, the same way a standard screw advances as it tightens.
Answer. f = 27.5 Hz, T = 36.4 ms and ω = 173 rad s⁻¹, with ω⃗ along the bit, pointing into the wall.
MediumExample 2
A front-loading washing machine drum starts from rest on a gentle cycle. Seen through the door, with θ positive counterclockwise and +z pointing out of the door towards you, θ(t) = 0.45t² − 0.030t³ rad for 0 ≤ t ≤ 12 s. Find ωz at 5.0 s and at 11 s with the direction of ω⃗ each time, the average angular velocity over the 12 s, and the instants when the drum actually turns at that average rate.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Differentiate term by term: ωz(t) = dθ/dt = 0.90t − 0.090t² rad s⁻¹, which factorises as 0.090t(10 − t).
- At 5.0 s: ωz = 4.5 − 2.25 = +2.25 rad s⁻¹. Positive means counterclockwise as you watch, so ω⃗ points out of the door towards you.
- At 11 s: ωz = 9.9 − 10.89 = −0.99 rad s⁻¹. The factor (10 − t) changed sign at t = 10 s, so the drum has reversed: it turns clockwise as seen, and ω⃗ points into the machine.
- Average over the cycle: θ(12) = 0.45 × 144 − 0.030 × 1728 = 64.8 − 51.84 = 12.96 rad, so ωavg = 12.96 rad ÷ 12 s = 1.08 rad s⁻¹.
- Set the derivative equal to the average: 0.90t − 0.090t² = 1.08, or t² − 10t + 12 = 0, so t = 5 ± √13 = 1.39 s and 8.61 s. This chord has two parallel tangents.
- The average hides the reversal: the drum reached θ(10) = 15.0 rad and came back 2.04 rad, sweeping 17.04 rad in all, but ωavg counts only the net 12.96 rad.
Answer. ωz(5.0 s) = +2.25 rad s⁻¹, ω⃗ out of the door; ωz(11 s) = −0.99 rad s⁻¹, ω⃗ into the machine; ωavg = 1.08 rad s⁻¹, matched by the instantaneous rate at t = 1.39 s and 8.61 s.
HardExample 3
A coaxial drone carries two rotors on one vertical shaft. Seen from above, the upper two-blade rotor turns counterclockwise at 3300 rpm and the lower three-blade rotor turns clockwise at 3000 rpm. With +z upward, find each rotor's ωz and the direction of its ω⃗, the angular velocity of the lower rotor relative to the upper one, and how often a blade of one rotor passes directly over a blade of the other.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Convert the speeds: 3300 rpm = 55.0 rev s⁻¹ and 3000 rpm = 50.0 rev s⁻¹, so the magnitudes are 2π × 55.0 = 345.6 rad s⁻¹ and 2π × 50.0 = 314.2 rad s⁻¹.
- Signs from the right-hand rule with +z up. The upper rotor turns counterclockwise seen from above, so ω₁ = +345.6 rad s⁻¹ and its ω⃗ points up. The lower rotor turns clockwise, so ω₂ = −314.2 rad s⁻¹ and its ω⃗ points down.
- Components along a shared axis subtract like one-dimensional velocities, and working in rev s⁻¹ keeps the numbers exact: −50.0 − 55.0 = −105.0 rev s⁻¹, so ω₂ − ω₁ = 2π × (−105.0) = −659.7 rad s⁻¹. Riding on the upper rotor you would see the lower one turn clockwise at 105.0 rev s⁻¹; counter-rotation adds the rates.
- Count alignments per relative turn. The upper blades are π rad apart and the lower blades 2π/3 rad apart, so a blade of one lies over a blade of the other whenever the relative angle is a multiple of π/3: six alignments per relative revolution.
- Pass rate = 6 × 105.0 = 630 passes per second, so the time between passes is 1/630 s = 1.59 ms. Had both rotors turned counterclockwise, the relative rate would be only 55.0 − 50.0 = 5.0 rev s⁻¹, and the blades would pass just 30 times per second.
Answer. ω₁ = +346 rad s⁻¹ (up) and ω₂ = −314 rad s⁻¹ (down); relative to the upper rotor the lower turns at −660 rad s⁻¹, 105 rev s⁻¹ clockwise from above; blades pass 630 times per second, 1.59 ms apart.
Quick check
Select an answer and explain it before checking.
Review & continue
Angular velocity is the rate at which a rotating body's angle changes, ω = dθ/dt in rad s⁻¹; its sign, or its direction along the axle by the right-hand rule, records which way the body turns.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics I completion tasks.