Angular Kinematics Equations
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Direct answer
Every linear kinematics equation has a rotational twin obtained by swapping x → θ, v → ω, a → α: ω = ω0 + αt, θ = ω0t + ½αt², ω² = ω0² + 2αθ, and θ = ½(ω0 + ω)t, all valid for constant angular acceleration α and all angles measured strictly in radians. The bridge to linear motion is s = rθ, v = ωr and a_t = αr along the tangent, joined by the centripetal component a_c = ω²r = v²/r toward the axis. Angular quantities are vectors along the rotation axis by the right-hand rule, so a disc spinning up has α parallel to ω while a disc slowing has them antiparallel — which decides signs in every rolling numerical.
What you must remember
- The three equations: ω = ω0 + αt; θ = ω0t + ½αt²; ω² = ω0² + 2αθ — valid only for constant α, the exact analogues of v = u + at, s = ut + ½at², v² = u² + 2as.
- Average angular speed: θ = ½(ω0 + ω)t for constant α; average of ω over time equals the arithmetic mean of initial and final values.
- Radian bridge: s = rθ, v = rω, a_t = rα; every rim quantity is radius times the angular quantity, which is why radians (dimensionless) are compulsory.
- Two acceleration components: tangential a_t = rα changes the speed, centripetal a_c = ω²r changes the direction; total acceleration is √(a_t² + a_c²) at angle tan⁻¹(a_c/a_t) from the tangent.
- Vector direction: ω and α point along the axis by the right-hand rule; anticlockwise in the plane of view is "out of the page".
- Unit conversions: 1 revolution = 2π rad; a disc at rpm → multiply by 2π/60 for rad/s — 1200 rpm is 125.7 rad/s.
- Rolling link: a wheel rolling without slipping obeys v = ωR and a = αR.
From equation to graph
A grinding wheel spins at 1200 rpm and is brought uniformly to rest in 40 s. Convert first: ω0 = 1200 × 2π/60 = 40π ≈ 125.7 rad/s. Then α = (0 − 125.7)/40 = −3.14 rad/s², a number every student should reach without hesitation because 40π/40 = π. The angle swept: θ = ½(ω0 + ω)t = ½ × 125.7 × 40 = 2514 rad, and in revolutions θ/2π = 2514/6.283 = 400 turns. The full solution used two formulas and one conversion; the 400-revolution answer is the kind examiners design to be clean when the method is right and ugly when it is not.
The graphs mirror linear kinematics. With constant α, the ω-t plot is a straight line of slope α and the θ-t plot is a parabola; the area under the ω-t curve equals the angle turned, just as area under v-t gives displacement. When α itself varies, you integrate: θ = ∫ω dt and ω = ∫α dt, and the average ω is the time-averaged value, not (ω_min + ω_max)/2 unless α is constant. A JEE Advanced favourite shows a curved ω-t graph and asks for the instant of maximum angular acceleration — the answer is where the graph is steepest, not where ω is largest.
Where students slip
Revolutions versus radians sink more marks here than any concept: substituting "revolutions" into ω² = ω0² + 2αθ without multiplying by 2π produces answers wrong by a factor of 6.28, and the options include exactly that value. Second, the two accelerations get conflated — a point on a uniformly spinning disc has a_t = 0 but a_c = ω²r fully nonzero, so "is the point accelerating?" is a yes, despite constant speed; assertion-reason questions are built on precisely this. Third, sign discipline: a wheel rotating clockwise and slowing has ω negative and α positive (if out-of-page is positive); drawing the rotation sense and the axis direction before writing equations prevents the sign cascade that ruins the second half of any coupled pulley problem. Main tests the plug-in numericals; Advanced couples these equations to moment of inertia or rolling within one question.
Frequently asked questions
What are the angular equivalents of the three linear kinematics equations?
ω = ω0 + αt, θ = ω0t + ½αt² and ω² = ω0² + 2αθ, obtained from the linear set by replacing s, v, u, a with θ, ω, ω0, α, valid for constant angular acceleration.
Why must angles be in radians in these equations?
The bridge formulas s = rθ and v = ωr hold only for radians; the radian is dimensionless, keeping both sides dimensionally consistent.
How do rpm convert to rad/s?
Multiply by 2π/60, since one revolution is 2π radians and one minute is 60 seconds; 1200 rpm becomes about 125.7 rad/s.
Can a point on a rotating disc have zero tangential acceleration yet nonzero total acceleration?
Yes — at constant ω the tangential part αr vanishes but the centripetal part ω²r persists, so the point accelerates inward while its speed stays constant.
How do you find angular displacement from an ω-t graph?
By area: the region under the ω-t curve between two times equals the angle turned, exactly as area under v-t gives displacement.