A Level Physics: Circular Motion — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — 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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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 [1] = 31.25 m s⁻² [1].

(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] = 7.81 − 2.45 = 5.36 N [1].

(c) At the bottom, tension acts towards the centre and weight away: T = mv²/r + mg [1] = 7.81 + 2.45 = 10.3 N [1]. Note the tensions differ by 2mg = 4.9 N.

(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) = 2.80 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: bank the road, so a component of the normal contact force contributes to the centripetal force [1]; increase the radius of the bend; improve tyre or road surface condition to increase the maximum friction [1].

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].

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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.