OxfordAQA International AS and A-level Physics: Circular motion and angular speed (9630) – 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 OxfordAQA International AS and A-level Physics (9630), Version 4.4, section 3.6.1 (Circular motion), which the specification marks as International A-level only. Read the study guide first. Where g is needed, use 9.81 N kg⁻¹. Give answers to a sensible number of significant figures.

Questions

1. (a) Convert 150° into radians, as a multiple of π. [1] (b) Convert 2.4 rad into degrees. [1]

2. A fan blade turns at a frequency of 3.5 Hz. (a) Calculate the period of the rotation. [1] (b) Calculate the angular speed of the blade. [2]

3. The drum of a washing machine has a radius of 0.22 m and spins at 1200 revolutions per minute. For a point on the inside surface of the drum, calculate: (a) the frequency in hertz [1] (b) the angular speed [1] (c) the linear speed [1] (d) the centripetal acceleration [1]

4. A ball on a light string moves at constant speed in a horizontal circle on a smooth table. Explain why the ball accelerates even though its speed is constant, state the direction of the acceleration, and name the force that causes it. [4]

5. A 0.25 kg bob travels round a horizontal circle of radius 0.90 m on a smooth table, completing 2.2 revolutions every second. (a) Calculate the angular speed of the bob. [2] (b) Calculate the tension in the string. [2] (c) State which force provides the centripetal force. [1]

6. A 12 g token sits on a flat tray that spins at 60 revolutions per minute. The most friction the tray can exert on the token is 0.060 N. (a) Calculate the angular speed of the tray. [2] (b) The token is 0.10 m from the axis. Calculate the friction force needed to keep it moving in a circle, and state whether it stays in place. [3] (c) The token is moved to 0.14 m from the axis. Show whether it stays in place. [2]

7. A car of mass 900 kg travels at 11 m s⁻¹ round a flat circular bend of radius 35 m. (a) Calculate the sideways force the road must exert on the car. [2] (b) The tyres can supply at most 4.8 kN. Calculate the greatest speed the car can have on this bend. [2] (c) Calculate the greatest speed on a bend of radius 60 m, assuming the same limit. [1] (d) Explain, using F = mv²/r, why a larger radius allows a higher greatest speed. [1]

8. A rubber bung with a mass of 0.080 kg circles at a radius of 0.50 m on a frictionless table. A light string from the bung passes through a smooth hole in the table to a hanging mass M. For each M, a student times 10 steady revolutions (t).

M / kg	t for 10 revolutions / s	1/T² / s⁻²
0.10	12.69	0.621
0.15	10.36	0.932
0.20	8.97	1.243
0.25	8.02	1.555
0.30	7.33	1.861

(a) Explain why the tension in the string equals Mg and why this tension is the centripetal force on the bung. [2] (b) For M = 0.20 kg, show that the data are consistent with F = mω²r. [2] (c) Use the first and last rows to calculate the gradient of a graph of Mg against 1/T². [2] (d) Use your gradient to find the mass of the bung. [2]

9. A satellite of mass 1200 kg is in a circular orbit of radius 1.06 × 10⁷ m with a period of 1.00 × 10⁴ s (supplied data). (a) Calculate the angular speed of the satellite. [1] (b) Calculate its orbital speed. [1] (c) Calculate its centripetal acceleration. [2] (d) Calculate the resultant force on the satellite. [1] (e) State what provides this force and its direction. [1]

10. A mass of 0.18 kg moves in a horizontal circle of radius 0.65 m on a smooth table, held by a string that breaks if the tension exceeds 14 N. (a) Calculate the greatest speed at which the mass can move without breaking the string. [2] (b) Calculate the corresponding angular speed. [1] (c) Calculate the corresponding frequency. [1] (d) The string is replaced by one of length 1.3 m and the mass moves at the speed found in (a). Calculate the tension needed and explain the change. [2]

11. A fairground ride carries a rider of mass 60 kg in a horizontal circle of radius 2.8 m. One revolution takes 3.4 s. (a) Calculate the angular speed. [2] (b) Calculate the speed of the rider. [1] (c) Calculate the resultant force on the rider. [2] (d) Express the rider’s acceleration as a multiple of g. [1] (e) The ride speeds up until one revolution takes 2.4 s. Calculate the new resultant force and the factor by which it has increased. [3]

Answers

1. (a) 150 × π/180 = 5π/6 rad [1] (b) 2.4 × 180/π = 138° (137.5°) [1] [2] Examiner insight: Show the conversion factor, π/180 or 180/π, so the working is visible.

2. (a) T = 1/f = 1/3.5 = 0.29 s [1] (b) ω = 2πf = 2π × 3.5 [1] = 22 rad s⁻¹ (22.0) [1] [3] Examiner insight: Give the unit rad s⁻¹ for angular speed. Hz or “per second” hides the radian.

3. (a) f = 1200/60 = 20 Hz [1] (b) ω = 2πf = 2π × 20 = 126 rad s⁻¹ [1] (c) v = ωr = 125.7 × 0.22 = 27.6 m s⁻¹ [1] (d) a = ω²r = 125.7² × 0.22 = 3.5 × 10³ m s⁻² (3474) [1] [4] Examiner insight: Convert rpm to revolutions per second before multiplying by 2π, and carry the unrounded ω forward.

4. The ball’s velocity changes because its direction is always changing [1]. A change of velocity is an acceleration, even at constant speed [1]. The acceleration is directed towards the centre of the circle, at right angles to the velocity [1]. By Newton’s second law this needs a resultant force towards the centre, which here is the tension in the string [1]. [4] Examiner insight: “It accelerates because it goes round a circle” restates the question. Say that velocity is a vector and its direction changes.

5. (a) ω = 2πf = 2π × 2.2 [1] = 14 rad s⁻¹ (13.8) [1] (b) T = mω²r = 0.25 × 13.82² × 0.90 [1] = 43 N (43.0) [1] (c) The tension in the string. [1] [5] Examiner insight: In part (c) name a real force. “Centripetal force” does not say what pulls the bob inwards.

6. (a) f = 60/60 = 1.0 Hz [1]; ω = 2π × 1.0 = 6.3 rad s⁻¹ (6.28) [1] (b) m = 12 × 10⁻³ kg and F = mω²r [1]; F = 0.012 × 6.283² × 0.10 = 0.047 N [1]; this is below the 0.060 N maximum, so the token stays in place [1] (c) F = 0.012 × 6.283² × 0.14 = 0.0663 N [1]; this is greater than 0.060 N, so the token slips [1] [7] Examiner insight: Convert grams to kilograms. End with a comparison and a conclusion in words; a force without a statement about staying or slipping leaves the question unanswered.

7. (a) F = mv²/r = 900 × 11²/35 [1] = 3.1 × 10³ N (3111 N) [1] (b) v = √(Fr/m) = √(4800 × 35/900) [1] = 13.7 m s⁻¹ [1] (c) v = √(4800 × 60/900) = 17.9 m s⁻¹ [1] (d) For a given speed, the force needed F = mv²/r is smaller when r is larger, so the friction limit is reached at a higher speed. [1] [6] Examiner insight: Rearrange for v first and take the square root. In part (d), refer to the equation.

8. (a) The hanging mass is in equilibrium, so the tension equals Mg [1]. The tension is the only horizontal force on the bung and points to the centre, so it is the centripetal force [1]. (b) For M = 0.20 kg, T = 8.97/10 = 0.897 s, so ω = 2π/0.897 = 7.00 rad s⁻¹ [1]. Then mω²r = 0.080 × 7.005² × 0.50 = 1.96 N, which equals Mg = 0.20 × 9.81 = 1.96 N [1]. (c) ΔF = (0.30 − 0.10) × 9.81 = 1.962 N [1]; gradient = 1.962/(1.861 − 0.621) = 1.6 N s² (1.58) [1] (d) F = 4π²mr/T², so gradient = 4π²mr, giving m = gradient/(4π²r) [1] = 1.582/(4π² × 0.50) = 0.080 kg (0.0801) [1] [8] Examiner insight: In a “consistent with” question, state both values and say they agree.

9. (a) ω = 2π/T = 2π/(1.00 × 10⁴) = 6.28 × 10⁻⁴ rad s⁻¹ [1] (b) v = ωr = 6.283 × 10⁻⁴ × 1.06 × 10⁷ = 6.66 × 10³ m s⁻¹ [1] (c) a = ω²r [1] = (6.283 × 10⁻⁴)² × 1.06 × 10⁷ = 4.18 m s⁻² [1] (d) F = ma = 1200 × 4.185 = 5.02 × 10³ N [1] (e) Gravitational attraction of the planet, directed towards the centre of the planet (the centre of the orbit). [1] [6] Examiner insight: Keep powers of ten visible in the working and match the data’s three significant figures.

10. (a) v = √(Fr/m) = √(14 × 0.65/0.18) [1] = 7.1 m s⁻¹ (7.11) [1] (b) ω = v/r = 7.110/0.65 = 11 rad s⁻¹ (10.9) [1] (c) f = ω/2π = 10.94/2π = 1.7 Hz (1.74) [1] (d) At fixed speed, F ∝ 1/r, so doubling the radius halves the tension [1]; T = 14/2 = 7.0 N [1] [6] Examiner insight: In part (d), give the reason for the change as well as the number.

11. (a) ω = 2π/T = 2π/3.4 [1] = 1.85 rad s⁻¹ [1] (b) v = ωr = 1.848 × 2.8 = 5.2 m s⁻¹ (5.17) [1] (c) F = mω²r = 60 × 1.848² × 2.8 [1] = 5.7 × 10² N (574 N) [1] (d) a = ω²r = 9.56 m s⁻²; a/g = 9.56/9.81 = 0.97 (about 1g) [1] (e) ω = 2π/2.4 = 2.618 rad s⁻¹ [1]; F = 60 × 2.618² × 2.8 = 1.15 × 10³ N [1]; the factor is 1151/574 = 2.0, equal to (3.4/2.4)² because F ∝ 1/T² [1] [9] Examiner insight: F proportional to 1/T² gives a quick check on the factor in the last part.

Where marks are usually lost

• Writing m s⁻¹ for angular speed. It is rad s⁻¹.
• Using revolutions per minute directly in the formulas, or the diameter as the radius.
• Squaring v/r instead of squaring v and then dividing by r.
• Writing “centrifugal force” in an explanation, or adding a centripetal arrow to a free-body diagram.
• Saying a circling object has no acceleration because its speed is constant, or not converting grams to kilograms.
• Giving an answer to a “does it stay or slip?” question without a comparison in words.

Next steps

• Use the revision notes.
• Re-read the study guide for missed questions.
• Visit the OxfordAQA A-level Physics hub and tick off the checklist.
• Try one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams from May/June 2020 onwards, published by OxfordAQA. Section 3.6.1 Circular motion (International A-level only).