Uniform Circular Motion — Centripetal Acceleration a = v²/r (NCERT 3.10)
1. What Is Uniform Circular Motion?
Uniform circular motion (UCM) is motion in a circle at constant speed. The word "uniform" refers to the speed only — and this is the conceptual pivot of the whole topic: the velocity is tangent to the circle at every instant, so its direction changes continuously even while its magnitude refuses to. A changing velocity vector is, by definition, acceleration. UCM is therefore accelerated motion, and the accelerating agent is whatever force bends the path inward — tension in a string, friction on a car tyre, gravity for a satellite. Every assertion–reason item on this page is built on that speed-versus-velocity distinction, which extends the vector logic of Topic 1 into dynamics.
The chapter closes here because UCM is the doorway to Laws of Motion (what force supplies a = v²/r?), Gravitation (satellites: gravity is the centripetal force) and Rotational Motion. The kinematics you build now — ω, T, v = ωr, a = v²/r — is reused verbatim in all three.
2. Complete Theory: Angular Variables and the v²/r Derivation
Angular description. Let the radius to the particle sweep an angle Δθ (in radians). The angular velocity is ω = Δθ/Δt; for one full revolution 2π radians in period T, ω = 2π/T. The arc-length relation Δs = rΔθ immediately ties the two descriptions: v = ωr — every point of a rotating rigid body shares the same ω, but outer points move faster linearly. Frequency ν = 1/T and rpm convert through ω = 2πν = 2π(rpm)/60.
Deriving a = v²/r (the NCERT geometric argument — learn the steps, they are asked). Take the velocity vectors at two nearby points separated by central angle Δθ. Both have magnitude v, and the angle between the two velocity vectors is also Δθ (tangents rotate with the radius). The vector change Δv closes the isosceles velocity triangle: |Δv| = 2v sin(Δθ/2) ≈ vΔθ for small Δθ. Dividing by Δt and using Δθ = vΔt/r: a = |Δv|/Δt = v·(vΔt/r)/Δt = v²/r. For small Δθ the direction of Δv points along the bisector — i.e., radially inward, toward the centre; hence the name centripetal (centre-seeking). Substituting v = ωr gives the twin form a = ω²r.
Properties worth internalising. The acceleration is perpendicular to the velocity at every instant — a perpendicular force does zero work (W = F·d cos 90° = 0), which is exactly why the speed stays constant. The magnitude a = v²/r is constant, but the vector a rotates (always pointing to the centre), so UCM's acceleration is itself a changing vector — a favourite subtlety in statement questions. Nothing in UCM "throws the body outward": the inward force is real; the "centrifugal force" is a frame artefact, treated properly in later chapters.
3. Visualising the Kinematics of the Circle
Figure 3.9 — Velocity is tangent; acceleration points to the centre
Orange arrows: velocity (always tangent, fixed length v). Green arrows: centripetal acceleration (always inward, fixed length v²/r). Both rotate as the particle moves.
Exam read-out: v ⊥ a at every point — that perpendicularity is why the force does no work and the speed never changes. Note both arrows rotate together; their magnitudes are individually constant, their directions are not.
4. Solved Examples
Example 1 — Car on a circular track (r = 50 m, v = 10 m/s)
A car rounds a circular track of radius 50 m at a steady 10 m/s. Find its centripetal acceleration, angular velocity and period.
Solution (step by step): a = v²/r = (10)²/50 = 2 m s⁻², directed toward the centre — note the units: m s⁻², an honest acceleration even though the speedometer never moves. ω = v/r = 10/50 = 0.2 rad s⁻¹. Period T = 2π/ω = 2π/0.2 = 10π ≈ 31.4 s per lap. Dimensional cross-check: [v²/r] = (LT⁻¹)²/L = LT⁻² ✓ — the check habit of Chapter 1 applies to every derived formula.
Example 2 — Fan blade at 300 rpm (r = 0.3 m)
A fan blade tip rotates at 300 rpm on a circle of radius 0.3 m. Find ω, the tip's speed, and its acceleration.
Solution: ω = 2π(300)/60 = 10π ≈ 31.4 rad s⁻¹. Tip speed v = ωr = 10π × 0.3 = 3π ≈ 9.42 m/s — comparable to a sprinter, at the tip of a desk fan. Acceleration a = ω²r = (10π)² × 0.3 = 30π² ≈ 30 × 9.87 ≈ 296 m s⁻² — about 30 g! The lesson: fast spin, small circle → enormous centripetal acceleration, which is why unbalanced fan blades tear loose.
5. Practice Questions
Q1. A particle moves in a circle of radius 2 m with period 4 s. Find ω, v and a.
ANSWER: ω = 2π/T = π/2 ≈ 1.57 rad/s. v = ωr = π ≈ 3.14 m/s. a = v²/r = ω²r = (π/2)² × 2 = π²/2 ≈ 4.93 m/s² toward the centre.
Q2 (MCQ). A stone is whirled in a horizontal circle at constant speed. If the string slips to half the radius while v stays the same, the centripetal acceleration becomes: (a) half (b) double (c) four times (d) unchanged
ANSWER: (b) double — a = v²/r with v fixed gives a ∝ 1/r; halving r doubles a. (Watch the variant: if ω is held constant instead, a = ω²r halves with r.)
Q3. A stone on a 0.8 m string spins at 30 rpm. Find ω, v and a.
ANSWER: ω = 2π(30)/60 = π ≈ 3.14 rad/s. v = ωr = 0.8π ≈ 2.51 m/s. a = ω²r = π² × 0.8 ≈ 7.90 m/s², directed along the string toward the hand.
6. Key Formulas & Takeaways
| Relation | Condition / remark |
|---|---|
| ω = Δθ/Δt = 2π/T = 2πν | θ in radians; rad is dimensionless |
| v = ωr | Tangential speed; ω common to the whole rigid body |
| a = v²/r = ω²r | Centripetal, always toward the centre, ⊥ to v |
| T = 2πr/v = 2π/ω | Period of revolution; ν = 1/T |
| rpm → ω: × 2π/60 | The standard conversion; 60 rpm = 2π rad/s |
| W = F·d cos 90° = 0 | Perpendicular centripetal force cannot change speed |
| [v²/r] = L T⁻² | Dimensional check of every UCM formula |
What supplies the inward force is Laws of Motion business (tension, friction, gravity, normal). Here we only describe the motion; there, we explain it — same equation, new job.
7. Frequently Asked Questions
Why is uniform circular motion called accelerated motion even though the speed is constant?
Acceleration is the rate of change of velocity, and velocity is a vector — it has magnitude and direction. In uniform circular motion the speed (magnitude) is constant, but the direction of the velocity changes continuously as the body moves around the circle, always remaining tangent to the path. A changing velocity vector means a nonzero acceleration, of magnitude v²/r, directed toward the centre.
Derive the expression for centripetal acceleration in uniform circular motion.
Take velocity vectors v and v′ at two nearby points separated by angle Δθ. Their magnitudes are equal (v), and the angle between the vectors is Δθ. The change Δv closes the velocity triangle, so |Δv| = 2v sin(Δθ/2) ≈ v Δθ for small Δθ. Since the arc length Δs = v Δt = r Δθ, we get Δθ = v Δt/r. Hence a = |Δv|/Δt = v Δθ/Δt = v²/r, directed toward the centre along the bisector of the two velocity vectors. Equivalently a = ω²r using v = ωr.
What is the relation between linear speed and angular velocity?
v = ωr. The angular velocity ω = Δθ/Δt (in rad/s, with 2π rad per revolution) is the same for every point of a rotating rigid body, but the linear speed v grows with distance r from the axis — the tip of a fan blade moves faster than a point halfway along it. The period follows from T = 2πr/v = 2π/ω, and an rpm figure converts by ω = 2π(rpm)/60.
Does the centripetal acceleration change the speed of the body?
No. The centripetal acceleration is always perpendicular to the velocity (radially inward versus tangential motion), and a perpendicular force does zero work (W = F·d with cos 90° = 0). It changes only the direction of the velocity, keeping the speed constant. Speed changes only if there is an additional tangential acceleration — but then the motion is no longer uniform circular motion.
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