Uniform circular motion

What is uniform circular motion?

When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion.

NCERT Physics Part I (Class 11), Section 3.10, page 40

In plain words: “uniform” refers only to the speed. A car going round a roundabout at a steady 36 km/h is accelerating, because the direction of its velocity keeps turning, and something must keep pulling it inwards to make it turn.

The direction of the acceleration

Let the object be at P, with position vector r from the centre C and velocity v, and a short time Δt later at P′, with r′ and v′ (NCERT Fig. 3.18, page 41). The velocity is always along the tangent, so v is perpendicular to r and v′ to r′. Both have the same length v.

Draw v and v′ from one point: the change Δv = v′ − v (a vector subtraction) joins the head of v to the head of v′. Each velocity is its radius vector turned through 90° (in the sense of motion) and scaled by v/R, so Δv is Δr turned through 90°: it is perpendicular to the chord PP′. So Δv lies along the line that bisects the angle between r and r′, pointing towards C (NCERT page 41). As Δt → 0, P′ closes in on P, this bisector becomes the radius at P, and the acceleration, the limit of Δv/Δt, points to the centre at every point of the circle.

The magnitude: a = v^²/R

The angle between v and v′ equals the angle Δθ between r and r′, because each velocity is at right angles to its radius. So the triangle of position vectors (sides R, R, |Δr|) and the triangle of velocities (sides v, v, |Δv|) are similar (page 41):

So |Δv| = (v/R)|Δr|. Divide both sides by Δt and let Δt → 0. The left side becomes a_c, the magnitude of the acceleration. On the right, for a small Δθ the chord |Δr| is almost the arc PP′ = vΔt, so |Δr|/Δt → v (NCERT page 41):

(NCERT Eq. 3.43, page 42.) The name “centripetal” means “centre-seeking”; NCERT notes that the term was proposed by Newton, and that Christiaan Huygens first published a thorough analysis of centripetal acceleration in 1673 (page 42).

Therefore, a centripetal acceleration is not a constant vector.

NCERT Physics Part I (Class 11), Section 3.10, page 42

Its magnitude stays the same but its direction keeps turning, which is why the equations for constant acceleration do not apply here. The inward pull needed to produce this acceleration is the centripetal force, mv^²/R.

Angular speed

As the object moves from P to P′, the radius turns through the angular distance Δθ. The angular speed is the angle turned per unit time (NCERT Eq. 3.44):

Strictly, Δθ/Δt is the average angular speed over Δt; the instantaneous value is its limit as Δt → 0. Here θ is accumulated angular distance, so ω = dθ/dt ≥ 0. For a signed angular coordinate φ, angular speed is |dφ/dt|. In uniform circular motion the two are equal. With θ in radians, ω is in rad/s. The arc length is Δs = RΔθ; dividing by Δt gives v = Rω, and putting this into v^²/R gives a_c = ω^²R (NCERT Eqs. 3.45–3.46):

Time period and frequency

The time period T is the time for one revolution, and the frequency ν = 1/T is the number of revolutions per second (NCERT page 42). In one revolution the object covers 2πR and the radius turns through 2π rad, so (NCERT Eqs. 3.47–3.48):

The projection of uniform circular motion on a diameter is simple harmonic motion, the link used in SHM and uniform circular motion.

Solved examples

Common mistakes to avoid

Quick revision summary

• Uniform circular motion: constant speed on a circle; the velocity is along the tangent and changes direction.
• The acceleration points to the centre: a_c = v^²/R = ω^²R = 4π^²ν^²R.
• Angular speed ω = Δθ/Δt (rad/s); v = ωR.
• T = time per revolution, ν = 1/T, ω = 2π/T = 2πν, v = 2πR/T.
• The centripetal acceleration has constant size but changing direction: it is not a constant vector.

Questions students ask

Is centripetal acceleration the same as centripetal force?

No, centripetal acceleration and centripetal force are different quantities linked by Newton’s second law: the acceleration, a_c = v^²/R towards the centre, describes how the velocity changes, while the centripetal force, F = ma_c = mv^²/R, is the net inward force that produces it. The centripetal force is not a new kind of force; real forces such as the tension in a string, gravity or friction supply it. A 2 kg stone moving at 3 m s^⁻¹ on a circle of radius 1.5 m has a_c = 6 m s^⁻² and needs a net inward force of 12 N.

Does uniform circular motion have angular acceleration?

No, uniform circular motion has zero angular acceleration, because the angular velocity ω stays constant in size and, for a fixed axis, in direction. It still has a centripetal acceleration of ω^²R, which is a linear acceleration, not an angular one. For a fixed axis, an angular acceleration α = dω/dt appears when the rotation speeds up or slows down, and the particle then also has a tangential acceleration αR.

What is non-uniform circular motion?

Non-uniform circular motion is motion on a circle with a changing speed, such as a stone whirled on a string in a vertical circle. Its acceleration then has a tangential part, dv/dt, along or against the velocity, as well as the centripetal part v^²/R: when v = 6 m s^⁻¹ on a circle of radius 3 m and the speed is rising at 9 m s^⁻², the parts are 12 and 9 m s^⁻², giving √(12^² + 9^²) = 15 m s^⁻², not towards the centre. Neither the JEE Main 2027 nor the NEET (UG) 2026 syllabus uses the term, but both list motion in a vertical circle, the standard non-uniform case, in Unit 4 (Work, Energy and Power).

Is uniform circular motion simple harmonic motion?

No, uniform circular motion is periodic but not simple harmonic: the object goes round a circle at constant speed, while simple harmonic motion is a to-and-fro motion along a line, with an acceleration −ω^²x proportional to the displacement from the mean position. The two are closely linked, because the projection (shadow) of uniform circular motion on any diameter is simple harmonic motion with the same ω, as shown in NCERT Section 13.4 (pages 264–265).

What are the SI units and dimensions of angular speed and centripetal acceleration?

Angular speed is measured in radians per second, rad s^⁻¹, with dimensions [T^⁻¹], because the radian is the ratio of an arc to a radius and has no dimensions; for the same reason ω is often written simply in s^⁻¹. Centripetal acceleration, like any acceleration, is measured in m s^⁻², with dimensions [L T^⁻²]. Both forms of the formula agree: v^²/R gives (m s^⁻¹)^²/m = m s^⁻², and ω^²R gives s^⁻² × m = m s^⁻².

Is angular speed the same as angular velocity?

No, angular speed is only the magnitude of angular velocity: the angular speed is the non-negative magnitude of the rate of change of a signed angle, ω = |dφ/dt|, in rad s^⁻¹, while the angular velocity is a vector of that size pointing along the axis of rotation, by the right-hand rule. In uniform circular motion about a fixed axis both are constant, which is why NCERT’s Section 3.10 needs only the angular speed; the vector form comes with rotational motion.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. A car goes round a circular track of radius 50 m at 10 m/s. Find its centripetal acceleration.

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   a = v^²/R = 100/50 = 2 m s^⁻², towards the centre of the track.

2. Q2. A wheel turns at 120 revolutions per minute. Find its angular speed in rad/s.

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   ν = 120/60 = 2 Hz, so ω = 2π × 2 ≈ 12.6 rad/s.

3. Q3. If the speed of an object in uniform circular motion is doubled, keeping the radius the same, what happens to its centripetal acceleration?

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   a = v^²/R, so it becomes four times as large.