Angular Motion Equations Calculator
Angular Motion Equations: Any Three Give the Rest
The rotational equations of motion are the linear ones with every symbol swapped for its angular twin: θ for s, ω for v, α for a. Tell the page which three of θ, ω₀, ω, α and t you were given and it returns the other two, says which equation it used, and counts the complete turns — because an angle of 39 radians is six and a bit revolutions, and marks are lost on that conversion more often than on the algebra.
Rotational SUVAT: give any three, get the other two
A flywheel is turning at 2.0 rad/s when a motor applies a constant angular acceleration of 1.5 rad/s² for 6.0 s. Find the angle turned through, the final angular velocity, and how many complete revolutions that is.
The four equations, and the linear ones they came from
- θ
- angular displacement, in radians. One revolution is 2π rad, so 39 rad is 6.21 revolutions. The arc length a rim point travels is rθ, and that formula is true in radians and in nothing else
- ω₀, ω
- angular velocity at the start and the end, in rad/s. A revolutions-per-minute figure is not an ω until it has been multiplied by 2π/60. Both may be negative, which just names the other sense of rotation
- α
- angular acceleration, in rad/s², assumed constant. The tangential acceleration of a point at radius r is αr; its centripetal acceleration is ω²r, and those are two different accelerations at right angles to each other
- t
- the length of the interval, in seconds
- the swap
- s→θ, u→ω₀, v→ω, a→α, t→t. That is the entire trick. If you can do the linear equations of motion you can do these, because they are the same four equations with different letters — and they are derived the same way, by integrating a constant acceleration twice
- what is NOT in this set
- anything that needs the radius. These four equations describe the angle and never the body. Mass, moment of inertia, torque and radius enter only when you ask about forces and energy — see rotational kinetic energy for that half
Worked example
A flywheel is turning at 2.0 rad/s when a motor applies a constant angular acceleration of 1.5 rad/s² for 6.0 s. Find the angle turned through, the final angular velocity, and how many complete revolutions that is.
Identify the mode. You have ω₀ = 2.0 rad/s, α = 1.5 rad/s² and t = 6.0 s — the second option, the standard forward problem. The equation that takes exactly those three and returns θ is θ = ω₀t + ½αt².
Substitute. θ = (2.0)(6.0) + ½(1.5)(6.0)² = 12 + ½(1.5)(36) = 12 + 27 = 39 rad. Keep the two terms separate until the end; collapsing them early is where arithmetic slips hide.
Final angular velocity. ω = ω₀ + αt = 2.0 + (1.5)(6.0) = 11 rad/s. That is 105 revolutions per minute, from 11 × 60/2π.
Check it another way. The mean angular velocity over a constant-α interval is (ω₀ + ω)/2 = (2.0 + 11)/2 = 6.5 rad/s, and 6.5 × 6.0 = 39 rad. Two different equations agreeing is the cheapest check there is, and all four of this page’s modes return 39 rad on these numbers precisely because the five quantities are consistent.
Convert to revolutions. 39 / 2π = 6.21 revolutions — six complete turns and 0.207 of another, which is 74.5°. So a spoke that started pointing along the reference line finishes 74.5° past it, not 39 radians past it. The diagram above draws that last part-turn as the arc, with the six complete turns counted beside it.
And the distance a rim point travelled. If the flywheel has a radius of 0.25 m, that point covered rθ = 0.25 × 39 = 9.75 m. Using 6.21 instead of 39 here gives 1.55 m and is wrong: rθ needs the whole angle in radians, not the part-turn and not the revolution count.
The rotational equations beside the linear ones they are copied from
| Linear | Rotational | What it leaves out | Use it when |
|---|---|---|---|
| v = u + at | ω = ω₀ + αt | θ | You want a rate and have a time. |
| s = ut + ½at² | θ = ω₀t + ½αt² | ω | The standard forward problem: start, push, how far round? |
| v² = u² + 2as | ω² = ω₀² + 2αθ | t | No time in the question and none wanted in the answer. |
| s = (u + v)t/2 | θ = (ω₀ + ω)t/2 | α | You know both rates and the time but not the acceleration. |
| F = ma | τ = Iα | everything kinematic | A different page: see torque and I. |
Converting the units that actually appear in questions
| Given as | To rad/s | Worked example |
|---|---|---|
| revolutions per minute (rpm) | × 2π/60, i.e. ÷ 9.5493 | 3000 rpm = 314.16 rad/s |
| revolutions per second (Hz) | × 2π | 50 Hz = 314.16 rad/s |
| degrees per second | × π/180 | 540°/s = 9.4248 rad/s |
| period of one turn T, in s | 2π/T | T = 0.80 s gives 7.854 rad/s |
| rim speed v and radius r | v/r | 3.9 m/s at r = 0.50 m gives 7.85 rad/s |
Why the same four equations work for rotation, and the two places it stops working
Nothing new is being learned here. If a wheel turns with a constant angular acceleration, then θ, ω and α stand in exactly the same differential relationship as s, v and a: ω is dθ/dt and α is dω/dt. Integrate a constant twice and you get the same four equations you already know, with different letters on them. That is the whole content of the topic, and it is why a student who is fluent in linear kinematics can pick up rotational kinematics in an afternoon.
Radians are compulsory, not conventional. The radian is defined as arc length divided by radius, which is what makes the arc swept through angle θ equal to rθ and the rim speed equal to rω. Put degrees into v = rω and you are out by a factor of 57.3. The SI Brochure classifies the radian as having dimension one, so rad/s and s−1 survive the same unit check — convenient for dimensional analysis, and a trap when you are trying to work out from the units alone whether a number is an angular velocity or a frequency. They are not the same thing: 50 Hz and 50 rad/s differ by 2π.
The revolution count is where the marks go. θ = 39 rad is a perfectly good answer and it is also six complete turns plus 74.5°. Which of those two the question wants depends on what it asks. A question about where a mark on the rim ends up wants the part-turn. A question about how far a point on the rim has travelled wants rθ with the full 39 rad, because the point really did go round six times. A question about the number of revolutions wants θ/2π. The page gives all three so the only decision left is which one the question asked for.
The first place it stops working: a changing α. These equations are not a general description of rotation, they are the special case of uniform angular acceleration. A real motor’s torque usually falls as it speeds up, so α falls too and θ = ω₀t + ½αt² overestimates the angle. Boards handle this by either saying “constant angular acceleration” in the question or by giving you an ω–t graph and asking for the area under it. If you are given a graph, do not try to fit these equations to it — read the area, which is θ whatever shape the curve is.
The second: a reversal inside the interval. If α opposes ω₀, the wheel stops at t = −ω₀/α and then spins up the other way. The equations keep working, but θ becomes the net angle and no longer the total angle swept. The page detects this and says so. A wheel that turns 10 rad forward and 10 rad back has θ = 0 and has still turned 20 rad of arc, and a question that asks “through what angle does it turn” is ambiguous enough that the safe answer gives both numbers with one sentence saying which is which.
Where this goes wrong in practice. Four slips account for almost everything. One: an rpm figure typed in as an ω. Two: degrees typed in as radians, which the arithmetic cannot detect because both are just numbers. Three: the sign of α on a spin-down — braking is a negative α when the rotation is positive, and the third mode’s minus sign in front of the ½αt² term catches people out even when the sign of α is right. Four: answering in revolutions when the question wanted radians, or the reverse. None of those four is a physics error, and all four cost the same mark as one.
Frequently asked questions
Can I really just swap the letters?
Yes, for the kinematics. θ, ω and α are related by differentiation exactly as s, v and a are, so integrating a constant α twice gives the same four equations. The swap stops being safe when you move to dynamics: τ = Iα looks like F = ma but I depends on the shape and on which axis you chose, in a way that mass never does. Two bodies of the same mass can have moments of inertia differing by a factor of eight.
My answer is 39 radians. Is that allowed? It is bigger than 2π.
Perfectly allowed. An angle is not confined to one turn any more than a distance is confined to one lap. 39 rad means the wheel went round 6.21 times. Only when the question asks about a position — which way is the spoke pointing — do you reduce it to the part-turn, which is 0.207 rev or 74.5°.
When do I need the radius?
Never, for these four equations. The radius appears when you convert an angular quantity to a linear one for a point on the body: arc s = rθ, rim speed v = rω, tangential acceleration at = rα. It also appears, squared, in the moment of inertia once you move on to torque and energy. A rigid body has one ω and one α everywhere, and every point on it has a different v and a different a — which is the reason for working in angles in the first place.
What is the difference between angular velocity and frequency?
A factor of 2π, and they are measured in units that pass the same dimensional check, so you cannot catch the mistake from the units. ω = 2πf. A wheel at 50 Hz has ω = 314 rad/s. Writing ω = 50 there understates the rate by 6.28 times and every subsequent answer inherits it. If a question gives you hertz or rpm, convert before you do anything else and write the converted value down.
The page says the rotation reverses. What should I write in the exam?
Both numbers, with a sentence. Give θ, the net angle, and say that the body stopped at t = −ω₀/α and reversed, so the total angle swept is larger. Markers are looking for the recognition that the two are different; a bare net angle on a question that meant total arc loses the mark even when the arithmetic is flawless.
Which of the four equations will my board let me use?
All four are standard, but formula booklets differ on which are printed and therefore on which you are expected to have memorised. The CBSE curriculum puts this material in Unit V and explicitly says the moments of inertia come with “no derivation”; Cambridge 9702 prints its constants and a short formula list and expects the kinematic relations to be known. Check your own booklet before the paper: if ω² = ω₀² + 2αθ is not on it, it is the one to memorise, because it is the only one that eliminates t.
Does a negative θ mean anything physical?
It means the body finished on the other side of where it started, measured in the sense you chose as positive. There is nothing wrong with it. It happens whenever ω₀ and α have opposite signs and the interval is long enough for the reversal to carry the body back past the start. The sign is bookkeeping, and the only mistake is to drop it.
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References
- PHYSICS (042) Curriculum, Class XI–XII, Central Board of Secondary Education, session 2025–26. Unit V, Motion of System of Particles and Rigid Body, reads in part: “Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications” and “Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation)”. Unit III, Laws of Motion, reads in part: “Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road)”. The parenthesis “no derivation” is why this page gives you the standard results rather than the integrals behind them. Quoted to establish what is in scope, never to claim any endorsement by the board.
- NCERT, Physics: Textbook for Class XI, Part I, Chapter 7 “Systems of Particles and Rotational Motion”. Read 11 November 2026. Table 7.1 gives eight bodies and no more: thin circular ring about the perpendicular axis through its centre (MR²) and about a diameter (MR²/2); thin rod perpendicular to the rod at its midpoint (ML²/12); circular disc about the perpendicular axis through its centre (MR²/2) and about a diameter (MR²/4); hollow cylinder about its own axis (MR²); solid cylinder about its own axis (MR²/2); solid sphere about a diameter (2MR²/5). The theorem of perpendicular axes is §7.10 and the theorem of parallel axes is §7.10.1, stated there without proof. The thin spherical shell is not in Table 7.1, which is why this page derives 2MR²/3 for you instead of quoting it.
- BIPM, The International System of Units (SI), 9th edition (2019). The radian is the coherent SI unit of plane angle and has dimension one — which is why rad/s and s−1 pass the same unit check, why the radian can appear and vanish from a dimensional analysis without anything being wrong, and why every rotational formula on this site is written in radians and not degrees. Table 4 gives the newton as kg m s−2.
- Cambridge International AS & A Level Physics 9702, syllabus for examination in 2025, 2026 and 2027. Its data sheet gives the acceleration of free fall as 9.81 m s−2. Worth knowing what this syllabus leaves out, because it explains why a question from one board can look unanswerable to a student taught on another: topic 3.2 asks only for “a qualitative understanding of frictional forces and viscous/drag forces including air resistance” and states that “no treatment of the coefficients of friction and viscosity is required”. Quoted by outcome number so the difference is checkable rather than asserted.