Free NEET Physics resource

Circular Motion: Kinematics

Concept & Practice Guide · 4-page PDF · Approximately 30 minutes · Updated 13 October 2026

A particle can move at constant speed and still accelerate because its velocity changes direction. Understand the picture first; use the equations after that.

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1. Start with the picture

For a circle of radius r, the arc length covered through an angle θ (in radians) is s = rθ. Displacement is the straight-line chord, not the arc. After one complete revolution, distance is 2πr but displacement is zero.

ω = 2π/T = 2πf · f = 1/T · v = rω

Teacher's checkpoint: On the same rigid rotating disc, all points have the same angular speed. Points farther from the axis have greater linear speed.

At every instant, velocity is tangent to the circle; centripetal acceleration points towards the centre. The direction of the velocity changes even when its magnitude stays constant.

2. Acceleration: radial and tangential

In uniform circular motion (UCM), the acceleration is radially inward. After a turn through angle Δθ at constant speed v, the magnitude of the change in velocity is:

a_c = v²/r = ω²r · |Δv| = 2v sin(Δθ/2)

Quarter turn: |Δv| = √2v. Half turn: |Δv| = 2v. Full turn: |Δv| = 0.

Average is not instantaneous: Over a full revolution, average velocity and average acceleration are both zero, although instantaneous acceleration is nonzero.

When speed changes in non-uniform circular motion, radial acceleration changes the direction of velocity, while tangential acceleration changes its magnitude.

a_t = dv/dt = rα · |a| = √(a_c² + a_t²)

Common mistake: Centripetal acceleration is the inward component of acceleration. The net inward force comes from real forces such as friction, tension or gravity; centripetal force is not an additional force to add.

3. Five worked examples

01 · Frequency to acceleration

Radius 0.50 m, frequency 2 Hz. Find ω, v and a_c.

Reasoning: ω = 2πf = 4π rad/s; v = rω = 2π m/s; a_c = ω²r = 8π² m/s².

02 · Changing direction

At a constant speed of 10 m/s, find |Δv| after a 90° turn.

Reasoning: |Δv| = 2(10)sin45° = 10√2 m/s. Constant speed does not mean constant velocity.

03 · Average versus instantaneous

Half a revolution at speed π m/s around a circle of radius 2 m.

Reasoning: Time = 2 s; displacement magnitude = 4 m. Average velocity magnitude = 2 m/s. |Δv| = 2π m/s, so average acceleration magnitude = π m/s². Compare instantaneous a_c = π²/2 ≈ 4.93 m/s².

04 · Two perpendicular components

At an instant, r = 2 m, v = 6 m/s, a_t = 4 m/s². Find net acceleration.

Reasoning: a_c = 36/2 = 18 m/s². Net magnitude = √(18² + 4²) = √340 ≈ 18.44 m/s².

05 · Points on the same disc

One point at 25 cm from the axis moves at 5 m/s. Find the speed at 10 cm.

Reasoning: ω = 5/0.25 = 20 rad/s. The second speed is 20 × 0.10 = 2 m/s.

4. Practice: 10 NEET-style questions

Choose one answer for each question. The final two use statement-based and assertion-reason formats.

After a complete revolution in UCM, which quantity is zero?A. Average velocity · B. Distance travelled · C. Instantaneous acceleration · D. Average speed
A particle moves in a circle of radius 0.50 m with a period of 2 s. Its speed is:A. π/2 m/s · B. π m/s · C. 2π m/s · D. 1/π m/s
At constant speed v, |Δv| after a quarter-turn is:A. v · B. √2v · C. 2v · D. 0
Two points are at radii 2r and 5r on one rigid rotating disc. Their linear speed ratio is:A. 5:2 · B. 1:1 · C. 2:5 · D. 4:25
At fixed radius, doubling the speed multiplies centripetal acceleration by:A. 2 · B. 4 · C. 1/2 · D. 1/4
Over a complete UCM revolution, average acceleration magnitude is:A. 0 · B. v²/r · C. 2v/T · D. v/T
If a_c = 18 m/s² and a_t = 4 m/s², net acceleration magnitude is:A. √20 · B. √340 · C. 22 · D. 9 m/s²
Viewed face-on, a wheel spins anticlockwise. Its angular-velocity vector points:A. In the wheel plane, inward · B. Perpendicular, away from viewer · C. Tangent to rim · D. Perpendicular, towards viewer
Statements: For UCM, I: speed is constant. II: acceleration vector is constant.A. Both true · B. I true, II false · C. I false, II true · D. Both false
Assertion: UCM has nonzero acceleration despite constant speed. Reason: Velocity direction changes continuously.A. Both true; reason explains assertion · B. Both true; reason does not explain assertion · C. Assertion true, reason false · D. Assertion false, reason true

Reveal answers with reasons

A — Net displacement is zero.
A — v = 2πr/T.
B — Vector change after 90° is √2v.
C — At common ω, v ∝ r.
B — a_c ∝ v² at fixed r.
A — Final velocity equals initial velocity.
B — Radial and tangential components are perpendicular.
D — Right-hand rule for a face-on view.
B — In UCM, inward acceleration changes direction.
A — Changing velocity direction causes acceleration.

What should you study next?

Move from kinematics to the forces behind circular motion: centripetal force, banking of roads, and vertical circles.

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