A Level Physics 12.1 Kinematics of Uniform Circular Motion MCQs – 30 Practice Questions with Answers (Cambridge 9702)

These 30 original practice questions cover Cambridge AS & A Level Physics 9702 Section 12.1, Kinematics of Uniform Circular Motion, an A Level topic that builds on the AS mechanics of Topics 2 to 5. They practise measuring angles in radians, angular velocity, and the link between angular and linear speed for objects moving in circles.

Formula recap: θ = s / r (angle in radians)  |  360° = 2π rad  |  ω = Δθ / Δt = 2π / T = 2πf  |  v = ωr. See the full A Level Physics Formula Sheet for every equation used in this topic.

Angles in Radians

Q1. One radian is the angle subtended at the centre of a circle by an arc whose length is

  • (A) equal to the diameter of the circle
  • (B) equal to the circumference of the circle
  • (C) equal to the radius of the circle
  • (D) one quarter of the circumference of the circle
Show Answer

Answer: C – By definition, θ = s / r. When the arc length s equals the radius r, θ = r / r = 1 rad.

Q2. One complete revolution is equal to an angle of

  • (A) π rad
  • (B) 2π rad
  • (C) π/2 rad
  • (D) 360π rad
Show Answer

Answer: B – A complete revolution is 360°, and 360° = 2π rad ≈ 6.28 rad.

Q3. An angle of 90° expressed in radians is

  • (A) π/4 rad
  • (B) π/3 rad
  • (C) 2π/3 rad
  • (D) π/2 rad
Show Answer

Answer: D – 90° is one quarter of a revolution, so the angle is (90/360) × 2π = π/2 rad.

Q4. One radian is approximately equal to

  • (A) 57.3°
  • (B) 0.0175°
  • (C) 28.6°
  • (D) 180°
Show Answer

Answer: A – Since π rad = 180°, one radian = 180° / π ≈ 57.3°.

Q5. A point on the rim of a rotating wheel moves along an arc of length 0.30 m. The radius of the wheel is 0.50 m. The angle through which the wheel has turned is

  • (A) 0.17 rad
  • (B) 1.7 rad
  • (C) 0.60 rad
  • (D) 6.0 rad
Show Answer

Answer: C – θ = s / r = 0.30 / 0.50 = 0.60 rad.

Q6. A point moves through an angle of 2.0 rad on a circle of radius 0.25 m. The distance travelled along the arc is

  • (A) 8.0 m
  • (B) 0.50 m
  • (C) 0.13 m
  • (D) 2.5 m
Show Answer

Answer: B – s = rθ = 0.25 × 2.0 = 0.50 m.

Q7. A turntable of radius 0.50 m rotates so that a point on its rim moves along an arc of 1.57 m. The angle through which the turntable has turned is

  • (A) 0.79 rad
  • (B) 1.57 rad
  • (C) 6.28 rad
  • (D) 3.14 rad
Show Answer

Answer: D – θ = s / r = 1.57 / 0.50 = 3.14 rad, which is π rad, i.e. half a complete revolution.

Q8. The equation v = ωr gives the correct speed only when the angle is measured in radians because

  • (A) it is derived from s = rθ, which holds only when θ is in radians
  • (B) degrees cannot be used for rotating objects
  • (C) a radian is a larger unit than a degree
  • (D) angular velocity is always a whole number in radians
Show Answer

Answer: A – The radian is defined by θ = s / r, so s = rθ is true by definition. Dividing by time gives v = ωr, and both steps need θ in radians.

Q9. For a small angle θ = 0.10 rad, the values of sin θ and tan θ are both approximately

  • (A) 5.7
  • (B) 0.90
  • (C) 0.10
  • (D) 0.0010
Show Answer

Answer: C – For small angles measured in radians, sin θ ≈ tan θ ≈ θ, so both are approximately 0.10.

Q10. Three-quarters of a complete revolution is equal to an angle of

  • (A) π/2 rad
  • (B) 3π/4 rad
  • (C) π rad
  • (D) 3π/2 rad
Show Answer

Answer: D – One revolution is 2π rad, so three-quarters of a revolution is (3/4) × 2π = 3π/2 rad (= 270°).

Angular Velocity

Q11. The angular velocity ω of an object moving in a circle is defined as

  • (A) ω = v × r
  • (B) ω = Δθ / Δt
  • (C) ω = 2πr
  • (D) ω = θ × t
Show Answer

Answer: B – Angular velocity is the rate of change of angle, ω = Δθ / Δt, measured in rad s−1.

Q12. A motor rotates at 600 revolutions per minute. Its angular velocity is

  • (A) 10 rad s−1
  • (B) 600π rad s−1
  • (C) 20π rad s−1
  • (D) 2π rad s−1
Show Answer

Answer: C – 600 rpm = 600 / 60 = 10 revolutions per second, so f = 10 Hz and ω = 2πf = 20π rad s−1 (≈ 62.8 rad s−1).

Q13. An object moving in a circle has a period of 0.50 s. Its angular velocity is

  • (A) 4π rad s−1
  • (B) π/4 rad s−1
  • (C) 2π rad s−1
  • (D) 0.50π rad s−1
Show Answer

Answer: A – ω = 2π / T = 2π / 0.50 = 4π rad s−1 (≈ 12.6 rad s−1).

Q14. The Earth rotates once on its axis every 24 hours. The angular velocity of the Earth is approximately

  • (A) 7.3 × 10−3 rad s−1
  • (B) 2.6 × 10−1 rad s−1
  • (C) 7.3 × 10−5 rad s−1
  • (D) 2π rad s−1
Show Answer

Answer: C – T = 24 × 3600 = 86 400 s, so ω = 2π / 86 400 ≈ 7.3 × 10−5 rad s−1.

Q15. The angular velocity of the minute hand of a clock is approximately

  • (A) 1.7 × 10−3 rad s−1
  • (B) 0.105 rad s−1
  • (C) 1.5 × 10−4 rad s−1
  • (D) 6.28 rad s−1
Show Answer

Answer: A – The minute hand completes one revolution in 3600 s, so ω = 2π / 3600 ≈ 1.7 × 10−3 rad s−1. (0.105 rad s−1 is the second hand.)

Q16. A wheel rotates with a frequency of 50 Hz. Its angular velocity is

  • (A) 50 rad s−1
  • (B) 100 rad s−1
  • (C) 25π rad s−1
  • (D) 100π rad s−1
Show Answer

Answer: D – ω = 2πf = 2π × 50 = 100π rad s−1 (≈ 314 rad s−1).

Q17. Two coins are fixed at different distances from the centre of the same rotating disc. Which statement is correct?

  • (A) Both coins have the same linear speed but different angular velocities.
  • (B) Both coins have the same angular velocity but different linear speeds.
  • (C) Both coins have the same angular velocity and the same linear speed.
  • (D) The coin further from the centre has the larger angular velocity.
Show Answer

Answer: B – The disc is rigid, so every point sweeps out the same angle in the same time: ω is the same. But v = ωr, so the coin at the larger radius has the larger linear speed.

Q18. A flywheel rotates with a constant angular velocity of 5.0 rad s−1. The angle it sweeps out in 4.0 s is

  • (A) 1.25 rad
  • (B) 9.0 rad
  • (C) 20 rad
  • (D) 45 rad
Show Answer

Answer: C – Δθ = ωΔt = 5.0 × 4.0 = 20 rad.

Q19. A fan rotates at 120 revolutions per minute. Its angular velocity is

  • (A) 4π rad s−1
  • (B) 120π rad s−1
  • (C) 2π rad s−1
  • (D) 240 rad s−1
Show Answer

Answer: A – 120 rpm = 120 / 60 = 2.0 revolutions per second, so ω = 2π × 2.0 = 4π rad s−1 (≈ 12.6 rad s−1).

Q20. Which object has the greatest angular velocity?

  • (A) An object with a period of 4.0 s
  • (B) An object rotating at a frequency of 0.20 Hz
  • (C) An object rotating at 20 revolutions per minute
  • (D) An object with a period of 2.0 s
Show Answer

Answer: D – A: ω = 2π/4.0 ≈ 1.6 rad s−1; B: ω = 2π × 0.20 ≈ 1.3 rad s−1; C: 20 rpm = 1/3 rev s−1, ω ≈ 2.1 rad s−1; D: ω = 2π/2.0 = π ≈ 3.1 rad s−1, the greatest.


Linear Speed on a Circle

Q21. A point on a rotating disc is 0.30 m from the centre. The disc has an angular velocity of 4.0 rad s−1. The linear speed of the point is

  • (A) 1.2 m s−1
  • (B) 13 m s−1
  • (C) 0.075 m s−1
  • (D) 4.3 m s−1
Show Answer

Answer: A – v = ωr = 4.0 × 0.30 = 1.2 m s−1.

Q22. On a turntable, one mark is 0.10 m from the centre and a second mark is 0.20 m from the centre. The speed of the second mark compared with the speed of the first mark is

  • (A) the same
  • (B) twice as large
  • (C) half as large
  • (D) four times as large
Show Answer

Answer: B – Both marks have the same ω, and v = ωr, so doubling the radius doubles the speed.

Q23. A satellite orbits the Earth in a circular orbit of radius 7.0 × 106 m with a period of 100 minutes. The speed of the satellite is approximately

  • (A) 7.3 m s−1
  • (B) 7.3 × 102 m s−1
  • (C) 7.3 × 103 m s−1
  • (D) 4.4 × 104 m s−1
Show Answer

Answer: C – T = 100 × 60 = 6000 s. v = 2πr / T = (2π × 7.0 × 106) / 6000 ≈ 7.3 × 103 m s−1.

Q24. On a fairground ride, passengers move in a horizontal circle of radius 6.0 m, completing one revolution every 8.0 s. The speed of a passenger is

  • (A) 0.75 m s−1
  • (B) 1.3 m s−1
  • (C) 48 m s−1
  • (D) 4.7 m s−1
Show Answer

Answer: D – v = 2πr / T = (2π × 6.0) / 8.0 ≈ 4.7 m s−1.

Q25. A disc rotates with a constant angular velocity of 2.5 rad s−1. The time taken for the disc to complete 15 revolutions is approximately

  • (A) 6.0 s
  • (B) 38 s
  • (C) 94 s
  • (D) 2.4 s
Show Answer

Answer: B – 15 revolutions = 15 × 2π = 30π ≈ 94 rad. t = Δθ / ω = 94 / 2.5 ≈ 38 s.

Q26. A vinyl record rotates at 33 revolutions per minute. The speed of a point 0.15 m from the centre of the record is approximately

  • (A) 0.52 m s−1
  • (B) 5.0 m s−1
  • (C) 0.083 m s−1
  • (D) 3.5 m s−1
Show Answer

Answer: A – f = 33 / 60 = 0.55 Hz, so ω = 2πf ≈ 3.5 rad s−1. Then v = ωr = 3.5 × 0.15 ≈ 0.52 m s−1.

Q27. An object moves in a circle at a constant speed. Which statement about its velocity is correct?

  • (A) The velocity increases steadily as the object goes round.
  • (B) The velocity is constant because the speed is constant.
  • (C) The velocity changes because the direction of motion changes continuously.
  • (D) The velocity is zero once every revolution.
Show Answer

Answer: C – Velocity is a vector. Even at constant speed, the direction changes continuously, so the velocity changes and the object accelerates towards the centre; this centripetal acceleration is covered fully in Section 12.2.

Q28. A car travels at 20 m s−1. Its wheels have a radius of 0.30 m and roll without slipping. The angular velocity of a wheel is approximately

  • (A) 6.0 rad s−1
  • (B) 0.015 rad s−1
  • (C) 600 rad s−1
  • (D) 67 rad s−1
Show Answer

Answer: D – The rim of the wheel moves at the speed of the car, so ω = v / r = 20 / 0.30 ≈ 67 rad s−1.

Q29. A point on a circle of radius 0.20 m sweeps out an angle of π rad in a time of 0.50 s. The speed of the point is approximately

  • (A) 1.3 m s−1
  • (B) 0.20 m s−1
  • (C) 6.3 m s−1
  • (D) 0.63 m s−1
Show Answer

Answer: A – ω = Δθ / Δt = π / 0.50 = 2π ≈ 6.3 rad s−1. Then v = ωr = 6.3 × 0.20 ≈ 1.3 m s−1.

Q30. The Moon orbits the Earth in an approximately circular orbit of radius 3.84 × 108 m with a period of 27.3 days. The speed of the Moon is approximately

  • (A) 27 m s−1
  • (B) 1.0 × 103 m s−1
  • (C) 1.0 × 105 m s−1
  • (D) 2.4 × 109 m s−1
Show Answer

Answer: B – T = 27.3 × 24 × 3600 ≈ 2.36 × 106 s. v = 2πr / T = (2π × 3.84 × 108) / (2.36 × 106) ≈ 1.0 × 103 m s−1.

Keep practising: work through more free physics questions on the Physics MCQs hub page, and keep the A Level Physics Formula Sheet open while you revise circular motion, gravitation and the rest of the A Level course.

These are original practice questions written for revision purposes. They are not Cambridge past-paper questions, and this page is not affiliated with Cambridge Assessment International Education.