OxfordAQA International AS and A-level Mathematics: Uniform circular motion and the conical pendulum (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section M2.7 (uniform circular motion) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International A-level Unit M2 (Mechanics); it is not examined at International AS. Take g = 9.8 m s⁻² and give answers to 3 significant figures unless told otherwise.

Learn the method first in the study guide or the revision notes. Course links: hub, printable checklist and the free 10-minute diagnostics.

Questions

1. A fairground disc turns at 72 revolutions per minute.

(a) Find its angular speed in rad s⁻¹, giving the exact value in terms of π and a decimal. [2] (b) Find the acceleration of a seat 0.6 m from the axis. [2]

2. A car of mass 1200 kg takes a flat circular bend of radius 50 m at a constant 14 m s⁻¹. Model the car as a particle.

(a) Find the sideways friction force needed. [3] (b) Find the least coefficient of friction between the tyres and the road for the car to take the bend at this speed. [2]

3. A satellite of mass m moves in a circular orbit of radius 2.5 × 10⁷ m about a planet. The gravitational force on it is mk/r², where r is its distance from the centre and k = 4.0 × 10¹⁴ m³ s⁻².

(a) Find the speed of the satellite. [3] (b) Find the time for one orbit, in hours. [2] (c) Show that ω²r³ = k for this orbit. [2]

4. A particle moves so that its position vector, in metres, is r = (−2 + 6sin(t/2))i + (4 + 6cos(t/2))j at time t seconds.

(a) Show that the particle moves on a circle, and state its centre and radius. [3] (b) Find the speed of the particle. [2] (c) Find the acceleration when t = π. [2]

5. A particle has position vector r = 4cos(3t)i + 3sin(3t)j. Show that the particle does not move on a circle with centre at the origin. [3]

6. A crate rests on a horizontal platform, 0.8 m from the axis, as the platform rotates at 30 revolutions per minute. Find the least coefficient of friction that stops the crate sliding. [4]

7. A particle P of mass 0.3 kg moves in a horizontal circle of radius 0.5 m on a smooth table. It is attached to a light inextensible string that passes through a small hole in the table at the centre of the circle and carries a particle of mass 0.9 kg hanging at rest below the table.

(a) Find the tension in the string. [1] (b) Find the angular speed of P. [3] (c) Find the speed of P and the time for one revolution. [2]

8. A bob of mass 0.25 kg is attached to a fixed point by a light inextensible string of length 0.8 m. It moves as a conical pendulum, with the string at angle θ to the vertical, taking 1.4 s for one revolution.

(a) Find the angular speed of the bob. [1] (b) Find θ, in degrees. [4] (c) Find the tension in the string. [1] (d) Find the speed of the bob. [2]

9. A bob of mass 0.6 kg is attached to a fixed point by a light string of length 1.5 m. The string breaks if the tension exceeds 20 N. The bob moves as a conical pendulum.

(a) Find the greatest possible angular speed of the bob. [3] (b) Find the angle between the string and the vertical at this angular speed. [3]

10. A particle of mass 0.5 kg has position vector r = (1 + 4cos(2t))i + (−2 + 4sin(2t))j, in metres, at time t seconds.

(a) Show that the particle moves on a circle and find its centre and radius. [2] (b) Find the acceleration and the resultant force on the particle when t = 0. [3] (c) Verify your force using m v²/r. [2]

11. A bob of mass 0.5 kg moves in a horizontal circle of radius 0.6 m, at constant speed. The circle is 0.8 m below the fixed point to which the bob is attached by a light inextensible string.

(a) Find the length of the string and tan θ, where θ is the angle between the string and the vertical. [2] (b) Find the angular speed of the bob. [3] (c) Find the tension in the string. [2] (d) Find the speed of the bob and the time for one revolution. [2]

Answers

1. (a) ω = 72 × 2π/60 [1] = 12π/5 = 7.54 rad s⁻¹ [1] (b) a = rω² [1] = 0.6 × (12π/5)² = 34.1 m s⁻² [1] Examiner insight: Show the conversion factor 2π/60 explicitly, so that a slip in the arithmetic can still be seen to follow a correct method.

2. (a) a = v²/r = 14²/50 = 3.92 m s⁻² [1]. Friction is the only horizontal force, so F = ma [1] = 1200 × 3.92 = 4704 N [1] (b) R = mg = 11 760 N [1], so μ ≥ 4704/11 760 = 0.4 [1] Examiner insight: Say which force supplies the acceleration and resolve vertically for R. The mass cancels, which gives you a check on your answer.

3. (a) Gravity provides the acceleration: m v²/r = mk/r² [1], so v² = k/r = 4.0×10¹⁴/2.5×10⁷ = 1.6×10⁷ [1], giving v = 4000 m s⁻¹ [1] (b) Period = 2πr/v = 2π × 2.5×10⁷/4000 = 39 270 s [1] = 10.9 hours [1] (c) ω = v/r = 4000/2.5×10⁷ = 1.6×10⁻⁴ rad s⁻¹ [1]. Then ω²r³ = 2.56×10⁻⁸ × 1.5625×10²² = 4.0×10¹⁴ = k as required [1] Examiner insight: In “show that”, state the substituted values and the final equality. A bare statement that the two sides agree is a weaker answer.

4. (a) r − (−2i + 4j) = 6sin(t/2)i + 6cos(t/2)j [1]. Its magnitude squared is 36sin²(t/2) + 36cos²(t/2) = 36 for all t [1]. So the particle is always 6 m from (−2, 4): a circle, centre (−2, 4), radius 6 m [1] (b) v = 3cos(t/2)i − 3sin(t/2)j [1], so the speed is 3 m s⁻¹ [1] (c) a = −1.5sin(t/2)i − 1.5cos(t/2)j [1]. At t = π, a = −1.5i m s⁻² [1] Examiner insight: The constant-distance argument needs the identity sin² + cos² = 1 and the phrase “for all t”. Checking one value of t is not a proof.

5. |r|² = 16cos²(3t) + 9sin²(3t) = 9 + 7cos²(3t) [1]. When t = 0, |r| = 4 [1]. When t = π/6, r = 3j and |r| = 3. The distance differs, so it is not constant and the particle is not on a circle centred at the origin [1] Examiner insight: To disprove a property you need only one clear counter-example, but you must say what it shows. Two different distances, stated side by side, make the point.

6. ω = 30 × 2π/60 = π rad s⁻¹ [1]. Friction supplies the acceleration: F = m r ω² [1]. R = mg, and F ≤ μR, so μ ≥ rω²/g [1] = 0.8π²/9.8 = 0.806 [1] Examiner insight: Keep ω as π until the end. Rounding it early to 3.14 changes the last figure of the answer.

7. (a) The hanging particle is in equilibrium, so T = 0.9 × 9.8 = 8.82 N [1] (b) The tension is the only horizontal force on P: T = m r ω² [1]. 8.82 = 0.3 × 0.5 × ω² [1], so ω² = 58.8 and ω = 7.67 rad s⁻¹ [1] (c) v = rω = 3.83 m s⁻¹ [1]. Time for one revolution = 2π/ω = 0.819 s [1] Examiner insight: The tension is the same on both sides of the hole because the string is light and the hole is smooth. Say so, since this is the link between the two particles.

8. (a) ω = 2π/1.4 = 4.49 rad s⁻¹ [1] (b) Vertically, T cos θ = mg [1]. Horizontally, T sin θ = m(L sin θ)ω² [1], so T = mLω² [1]. Then cos θ = g/(Lω²) = 9.8/(0.8 × 20.14) = 0.608, so θ = 52.5° [1] (c) T = mLω² = 0.25 × 0.8 × 20.14 = 4.03 N [1] (d) Radius = 0.8 sin 52.54° = 0.635 m [1], so v = rω = 2.85 m s⁻¹ [1] Examiner insight: Use the unrounded value of ω² (20.14…) in the cosine, because a rounded angle then carried into later parts drifts in the third figure.

9. (a) Horizontally, T sin θ = m(L sin θ)ω² [1], so T = mLω² [1]. With T = 20: ω² = 20/(0.6 × 1.5) = 22.2, so ω = 4.71 rad s⁻¹ [1] (b) Vertically, T cos θ = mg [1]. cos θ = 0.6 × 9.8/20 = 0.294 [1], so θ = 72.9° [1] Examiner insight: Tension grows with ω, so the greatest ω corresponds to the greatest tension. Say this to justify putting T = 20.

10. (a) r − (i − 2j) = 4cos(2t)i + 4sin(2t)j, which has magnitude 4 for all t [1]. So a circle, centre (1, −2), radius 4 m [1] (b) v = −8sin(2t)i + 8cos(2t)j [1]. a = −16cos(2t)i − 16sin(2t)j, so at t = 0, a = −16i m s⁻² [1]. F = ma = −8i N [1] (c) Speed = 8 m s⁻¹, so m v²/r = 0.5 × 64/4 = 8 N [1], which equals |F| = 8 N and the direction is towards the centre (1, −2) from (5, −2) [1] Examiner insight: A vector force needs a direction as well as a size. The negative sign on the i component is the direction here.

11. (a) L = √(0.6² + 0.8²) = 1.0 m [1] and tan θ = 0.6/0.8 = 0.75 [1] (b) T cos θ = mg and T sin θ = m r ω² [1]. Dividing, tan θ = rω²/g [1], so ω² = 9.8 × 0.75/0.6 = 12.25 and ω = 3.5 rad s⁻¹ [1] (c) cos θ = 0.8, so T = mg/cos θ = 4.9/0.8 [1] = 6.125 N [1] (d) v = rω = 0.6 × 3.5 = 2.1 m s⁻¹ [1]. Time = 2π/3.5 = 1.80 s [1] Examiner insight: Dividing the horizontal equation by the vertical one removes both T and m, which is the fastest route when the geometry is given. Check your answer with ω² = g/h = 9.8/0.8.

Where marks are usually lost

• Using rev/min directly in v = rω or a = rω². Convert to rad s⁻¹ first.
• Adding a centripetal force to the diagram. Mark only the real forces and then resolve towards the centre.
• Giving a force or acceleration without direction when the question uses vectors.
• Using L instead of L sin θ as the radius in a conical pendulum.
• Setting the tension equal to mg. Only the vertical component balances the weight.
• Testing one or two values of t when asked to show that motion is circular.
• Rounding ω or an angle early, so that later figures drift.
• Forgetting that the maximum tension gives the maximum angular speed, not the minimum.
• Leaving out the reason that the mass cancels, in friction and orbit questions.

Next steps

• Review the revision notes and study guide.
• Tick off M2.7 on the checklist and visit the course hub.
• Take the free 10-minute diagnostics to find gaps.
• Book a free trial class for help with any question you could not finish.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, published by OxfordAQA. Section M2.7 of Unit M2 (International A-level only).