A Level Physics: Circular Motion — 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 available free from your board.

Related: Circular Motion revision notes

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Section A

1. Explain why an object moving in a circle at constant speed is accelerating. [2]

2. Explain why the centripetal force does no work on the object. [2]

3. A student draws a free-body diagram of a car on a flat bend and labels four arrows: weight, normal contact force, friction, and centripetal force. Identify and explain the error. [3]

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Section B

4. A stone of mass 0.25 kg is whirled on a string of length 0.80 m in a vertical circle at a constant speed of 5.0 m s⁻¹.

(a) Calculate the centripetal acceleration. [2]

(b) Calculate the tension in the string at the top of the circle. [3]

(c) Calculate the tension at the bottom. [2]

(d) Calculate the minimum speed at the top for the stone to maintain a circular path. [3]

5. A car of mass 950 kg travels round a flat circular bend of radius 45 m at 18 m s⁻¹.

(a) Calculate the centripetal force required. [2]

(b) State what provides this force. [1]

(c) The maximum frictional force available is 6800 N. Determine whether the car can complete the bend, showing your reasoning. [2]

(d) Suggest two changes that would allow the car to take the bend safely at a higher speed. [2]

6. A wheel of radius 0.15 m completes 3 revolutions in 2.0 s.

(a) Calculate its angular speed. [2]

(b) Calculate the linear speed of a point on the rim. [2]

(c) Calculate the centripetal acceleration of that point using a = ω²r. [2]

7. Define the radian, and show algebraically that a = v²/r and a = ω²r are equivalent expressions for centripetal acceleration. [3]

8. State the equation linking linear speed v to angular speed ω and radius r, and explain what it means physically for two points at different radii on the same rotating disc. [2]

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9. A bicycle wheel of radius 0.35 m rolls along the ground without slipping at a speed of 7.0 m s⁻¹. Calculate the angular velocity of the wheel and its period of rotation. [2]

Answers

1. Velocity is a vector, and its direction changes continuously [1]. A change in velocity is an acceleration, even though the magnitude (speed) is constant [1].

2. The force acts perpendicular to the velocity at every instant [1], and work requires a component of force along the displacement [1].

3. Centripetal force should not be a separate arrow [1]. It is not an additional force but the resultant of the forces already present [1] — here, the friction between tyres and road provides it [1].

4. (a) a = v² ÷ r = 5.0² ÷ 0.80 = 31.25 [1] ≈ 31 m s⁻² [1] (quoted to 2 s.f. to match the data; carry 31.25 into later parts).

(b) At the top, both tension and weight act towards the centre: T + mg = mv²/r [1] T = (0.25 × 31.25) − (0.25 × 9.81) [1] = 7.8125 − 2.4525 = 5.36 ≈ 5.4 N [1].

(c) At the bottom, tension acts towards the centre and weight away: T = mv²/r + mg [1] = 7.8125 + 2.4525 = 10.265 ≈ 10 N [1] (2 s.f.). Note the tensions differ by 2mg = 4.9 N, because the speed is the same at the top and the bottom here. If the stone were moving freely with energy conserved, it would be faster at the bottom and the difference would be 6mg.

(d) At minimum speed T = 0, so weight alone provides the centripetal force [1]: mg = mv²/r, so v = √(gr) [1] = √(9.81 × 0.80) = 2.801 ≈ 2.8 m s⁻¹ [1].

5. (a) F = mv²/r = (950 × 18²) ÷ 45 [1] = 6840 N [1].

(b) Friction between the tyres and the road surface [1].

(c) The required force (6840 N) exceeds the maximum available friction (6800 N) [1], so the car cannot complete the bend — it will skid outwards [1].

(d) Any two, [1] each, maximum [2]: bank the road, so a component of the normal contact force contributes to the centripetal force; increase the radius of the bend, reducing the centripetal force needed at the same speed; improve tyre or road surface condition to increase the maximum friction.

6. (a) Period T = 2.0 ÷ 3 = 0.667 s [1]. ω = 2π/T = 2π ÷ 0.667 = 9.42 rad s⁻¹ [1].

(b) v = ωr = 9.42 × 0.15 [1] = 1.41 m s⁻¹ [1].

(c) a = ω²r = 9.42² × 0.15 [1] = 13.3 m s⁻² [1].

7. One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius [1]. Starting from a = ω²r = ω(ωr) = ωv (since v = ωr) [1], substituting ω = v/r gives a = (v/r)v = v²/r, confirming the two forms are equivalent [1].

8. v = ωr [1]. All points on a rigid rotating disc share the same angular speed ω, but since v = ωr, a point further from the centre has a greater linear speed, even though both points complete a revolution in the same time [1].

9. ω = v ÷ r = 7.0 ÷ 0.35 = 20 rad s⁻¹ [1]. T = 2π ÷ ω = 2π ÷ 20 = 0.31 s [1].

Examiner insight (June 2025): a common mistake was to combine the speed of the wheel with the radius of a different part (a small cog) instead of the radius of the wheel itself. Use the speed and radius of the same rotating object. Try the real question next: Cambridge International AS & A Level Physics 9702, June 2025, Paper 42, Question 1.

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Where marks are usually lost

• Drawing centripetal force as an extra arrow.
• Forgetting that both tension and weight act towards the centre at the top.
• Using degrees instead of radians for angular quantities.
• Saying there is no acceleration because the speed is constant.
• Not comparing required force with available friction explicitly in part (c).
• Mixing up ω = 2π/T with ω = 2π/f — frequency and period are reciprocals, not interchangeable in the formula.
• Forgetting to convert revolutions per second into a period before finding ω.