OCR A Level Physics A: Circular motion and centripetal force (H556) – 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 5.2 (Circular motion) of the H556 Physics A specification from OCR (Version 3.0, March 2026): 5.2.1 Kinematics of circular motion and 5.2.2 Centripetal force. Take g = 9.81 m s⁻² where it is needed. Worked explanations sit in the study guide and the revision notes. Also see the subject hub and the checklist to print. If several answers go wrong, take a free 10-minute diagnostic.

Questions

1. (a) Convert 210° to radians. [1] (b) Convert 0.75 rad to degrees. [1] (c) A point on a rotating arm of radius 2.4 m moves through an angle of 1.3 rad. Calculate the length of the arc it travels. [1]

2. A ceiling fan makes 330 complete revolutions in 75 s. Calculate its frequency, its period, its angular velocity, and the speed of a point on a blade 0.46 m from the axis. [4]

3. A Ferris wheel of radius 22 m takes 8.0 minutes for one revolution at a steady rate. Calculate the angular velocity of the wheel and the speed of a rider. [3]

4. A small puck on a frictionless horizontal table is joined by a light string to a fixed pin. It moves in a circle at constant speed.

(a) State the direction of the net force on the puck. [1] (b) Explain why this force does not change the speed of the puck. [2] (c) The string suddenly snaps. Describe the subsequent motion of the puck. [2]

5. A hammer thrower swings a ball of mass 4.0 kg in a horizontal circle of radius 1.8 m at a steady 24 m s⁻¹.

(a) Calculate the centripetal acceleration of the ball. [2] (b) Calculate the tension in the wire, assuming it provides all the centripetal force. [1] (c) Calculate the angular velocity of the ball. [1]

6. A laboratory centrifuge holds a sample tube with its contents 0.090 m from the axis. The contents must have a centripetal acceleration of 1.5 × 10⁴ m s⁻².

(a) Calculate the angular velocity needed. [2] (b) Calculate the frequency in hertz. [1] (c) Express the frequency in revolutions per minute. [1] (d) Compare the acceleration with g. [1]

7. A coin of mass 8.0 g lies on a horizontal turntable that rotates at 45 revolutions per minute. The coin does not slide.

(a) Calculate the angular velocity of the turntable. [1] (b) The coin is 0.12 m from the centre. Calculate the force that friction must supply. [2] (c) The greatest friction force the surface can supply is 0.030 N. Calculate the largest radius at which the coin stays at rest at this rotation rate. [2]

8. A student uses a whirling bung with hanging masses to investigate how the force on the bung depends on its period.

(a) Explain how the hanging masses allow the student to know the centripetal force, and why the string cannot be exactly horizontal. [2] (b) The student times 20 revolutions as 16.2 s, with an uncertainty of 0.3 s. Calculate the percentage uncertainty in this time, and the percentage uncertainty if only one revolution had been timed with the same absolute uncertainty. [2] (c) Suggest two safety precautions. [2]

9. A student keeps the tension at 2.50 N and the mass of the bung at 0.0400 kg, and varies the radius r. The time for 10 revolutions is recorded.

r / m	0.40	0.60	0.80	1.00	1.20
time for 10 revolutions / s	5.0	6.2	7.1	7.9	8.7

(a) Calculate T² for each radius, to 3 significant figures. [2] (b) The graph of T² against r is a straight line through the origin. Use the first and last points to find its gradient. [2] (c) Show that the gradient equals 4π²m/F, and use it to find m. [3] (d) Compare your value with the mass 0.0400 kg and comment. [1]

10. (a) Starting from a = v²/r and v = ωr, show that a = ω²r. [2] (b) Show that the centripetal force on a mass m moving at radius r with period T is F = 4π²mr/T². [1] (c) The period is halved at constant radius. State the effect on the force. [1]

11. A sphere of mass 0.080 kg moves on a smooth horizontal table in a circle of radius 0.60 m, attached to a peg by a light string. Its speed is 3.0 m s⁻¹.

(a) Calculate the tension in the string. [2] (b) Calculate the period of the motion. [2] (c) The string breaks when the tension reaches 12 N. Calculate the greatest speed at which the sphere can move in this circle. [2] (d) The speed is doubled to 6.0 m s⁻¹ at the same radius. Calculate the new tension and compare it with the original. [2] (e) Explain why the string tension does no work on the sphere. [1]

Answers

1. (a) 210 × π/180 = 7π/6 = 3.67 rad [1] (b) 0.75 × 180/π = 43.0° [1] (c) s = rθ = 2.4 × 1.3 = 3.1 m [1] Examiner insight: Answers usually earn credit when the conversion factor is visible and the angle is in radians whenever s = rθ is used.

2. f = 330/75 = 4.4 Hz [1] T = 1/f = 1/4.4 = 0.23 s [1] ω = 2πf = 2π × 4.4 = 28 rad s⁻¹ (27.6) [1] v = ωr = 27.6 × 0.46 = 13 m s⁻¹ (12.7) [1] Examiner insight: Carry the unrounded ω into the speed calculation, because early rounding can change the final digit.

3. T = 8.0 × 60 = 480 s [1] ω = 2π/T = 2π/480 = 0.013 rad s⁻¹ (0.0131) [1] v = ωr = 0.0131 × 22 = 0.29 m s⁻¹ [1] Examiner insight: Write the conversion to seconds as a separate step.

4. (a) Towards the centre of the circle (along the string, towards the pin). [1] (b) At every instant the force is at 90° to the velocity [1], so it has no component along the direction of motion and changes only the direction of the velocity, not its magnitude. [1] (c) There is no longer a net force on the puck [1], so it travels straight on at constant speed, along the tangent at the point where the string snapped. [1] Examiner insight: Say “tangent” and “constant speed”. A description such as “flies outwards” is unlikely to earn credit.

5. (a) a = v²/r = 24²/1.8 [1] = 3.2 × 10² m s⁻² (320) [1] (b) F = ma = 4.0 × 320 = 1.3 × 10³ N [1] (c) ω = v/r = 24/1.8 = 13 rad s⁻¹ [1] Examiner insight: Show the substitution, so that a slip in arithmetic can still be followed.

6. (a) a = ω²r, so ω² = 1.5 × 10⁴/0.090 [1], giving ω = 4.1 × 10² rad s⁻¹ (408) [1] (b) f = ω/2π = 408/2π = 65 Hz [1] (c) 65.0 × 60 = 3.9 × 10³ revolutions per minute [1] (d) 1.5 × 10⁴/9.81 = about 1.5 × 10³ times g [1] Examiner insight: When rearranging for ω, remember the square root. Leaving the answer as ω² is a typical loss.

7. (a) f = 45/60 = 0.75 Hz, so ω = 2π × 0.75 = 4.7 rad s⁻¹ [1] (b) m = 8.0 × 10⁻³ kg, F = mω²r [1] = 8.0 × 10⁻³ × 4.71² × 0.12 = 0.021 N [1] (c) At the limit, 0.030 = mω²r [1], so r = 0.030/(8.0 × 10⁻³ × 4.71²) = 0.17 m [1] Examiner insight: Convert grams to kilograms before substituting. This is a frequent source of an answer that is wrong by a factor of 1000.

8. (a) When the bung moves steadily, the tension in the string equals the weight Mg of the hanging masses, so this is the centripetal force [1]. The bung has weight, which the string must support, so the string must slope slightly upwards from the bung, and the tension is not exactly horizontal. [1] (b) 0.3/16.2 × 100 = 1.9% [1]. One revolution takes 16.2/20 = 0.81 s, giving 0.3/0.81 × 100 = 37%, so timing 20 revolutions greatly reduces the percentage uncertainty [1]. (c) Any two from: wear eye protection; check the string is tied firmly to the bung; keep other people well clear of the circle; use a tray or soft surface under the hanging masses. [2] Examiner insight: For uncertainty questions, give both the numbers and the conclusion drawn from them.

9. (a) T = time/10 then T² [1]: T² = 0.250, 0.384, 0.504, 0.624, 0.757 s² [1] (b) Gradient = (0.757 − 0.250)/(1.20 − 0.40) [1] = 0.63 s² m⁻¹ (0.634) [1] (c) F = 4π²mr/T², so T² = (4π²m/F) × r, and a graph of T² against r has gradient 4π²m/F [1]. Then m = gradient × F/(4π²) [1] = 0.634 × 2.50/39.48 = 0.040 kg (0.0401) [1] (d) The values agree to within about 0.3%, so the data supports F = mω²r (with F constant). [1] Examiner insight: Rearrange into the form y = mx before reading the gradient, so it is clear what the gradient represents.

10. (a) a = v²/r, substituting v = ωr: a = (ωr)²/r [1] = ω²r² /r = ω²r [1] (b) F = mω²r with ω = 2π/T gives F = m(2π/T)²r = 4π²mr/T² [1] (c) F is proportional to 1/T², so halving T multiplies F by 4 [1]. Examiner insight: In a “show that” answer, write out every algebraic line and end on the expression given.

11. (a) F = mv²/r [1] = 0.080 × 3.0²/0.60 = 1.2 N [1] (b) T = 2πr/v [1] = 2π × 0.60/3.0 = 1.3 s (1.26) [1] (c) v² = Fr/m = 12 × 0.60/0.080 = 90 [1], so v = 9.5 m s⁻¹ [1] (d) F = 0.080 × 6.0²/0.60 = 4.8 N [1], which is four times the original tension (doubling v multiplies F by 2²). [1] (e) The tension is perpendicular to the velocity (the displacement) at every instant, so there is no displacement in the direction of the force and no work is done. [1] Examiner insight: A comparison needs a statement, such as “four times as large”, as well as the new value.

Where marks are usually lost

• Using degrees where radians are needed, or the reverse.
• Using revolutions per minute without converting to hertz or radians per second.
• Using a diameter as the radius, or grams as kilograms.
• Forgetting the square root after rearranging v² = Fr/m or ω² = a/r.
• Describing the motion after the string breaks as outwards or radial, rather than along the tangent at constant speed.
• Writing “centripetal force” as an additional force instead of naming the real force that provides it.
• Claiming zero acceleration just because the speed is steady.

Next steps

• Revisit the revision notes and the study guide.
• Tick off progress on the checklist to print and see the subject hub.
• Find gaps with the free 10-minute diagnostics.
• Want a tutor? Book a free trial class.

Official syllabus

Source: the OCR Physics A (H556) specification, Version 3.0 (March 2026), first assessment 2017. Section 5.2 Circular motion; outcomes 5.2.1(a)-(c) and 5.2.2(a)-(d).