OCR A Level Physics A: Circular motion and centripetal force (H556) – Revision Notes

Condensed recall for section 5.2 (Circular motion) of the H556 Physics A specification from OCR (Version 3.0, March 2026), covering 5.2.1 Kinematics of circular motion and 5.2.2 Centripetal force. For full explanations and worked derivations use the study guide. Next, attempt the practice questions. Also useful: the subject hub, the checklist to print and the free 10-minute diagnostics.

Definitions and key equations

Reference	Quantity or idea	Equation or statement
5.2.1(a)	Radian	angle at the centre that cuts off an arc as long as the radius; θ = s/r
5.2.1(a)	Full turn	2π rad = 360°, so 1 rad = 57.3°
5.2.1(b)	Period T	time for one revolution, in s
5.2.1(b)	Frequency f	revolutions per second, in Hz; f = 1/T
5.2.1(c)	Angular velocity ω	ω = Δθ/Δt = 2π/T = 2πf, in rad s⁻¹
5.2.2(a)	Cause of circular path	a constant net force perpendicular to the velocity
5.2.2(b)	Speed in a circle	v = ωr = 2πr/T
5.2.2(c)	Centripetal acceleration	a = v²/r = ω²r, towards the centre
5.2.2(d)(i)	Centripetal force	F = mv²/r = mω²r = 4π²mr/T²

Radians and conversions

• Going from degrees to radians, multiply by π/180; going back, multiply by 180/π.
• Arc length s = rθ only works when θ is in radians.
• Being arc length divided by radius, a radian has no dimensions.

Angle	90°	180°	270°	360°
In radians	π/2	π	3π/2	2π

Method: solving a circular motion problem

1. Convert everything to SI units: seconds, metres, kilograms, and hertz or rad s⁻¹ rather than revolutions per minute.
2. List what you are given: any two of r, v, ω, T, f, m, a, F.
3. Find ω from T or f if you need it: ω = 2π/T = 2πf.
4. Find v = ωr, or choose the acceleration form that matches your data.
5. Use F = ma with the acceleration towards the centre.
6. Identify the real force that supplies it: tension, friction, normal contact, gravity or a magnetic force.
7. Check units and give the answer to a suitable number of significant figures.

Worked reminder. A mass of 0.90 kg is swung at a steady 2.0 revolutions per second on a rope of radius 0.35 m.

ω = 2πf = 2π × 2.0 = 12.57 rad s⁻¹ v = ωr = 12.57 × 0.35 = 4.4 m s⁻¹ a = ω²r = 12.57² × 0.35 = 55.3 m s⁻² F = ma = 0.90 × 55.3 = 50 N (two significant figures)

Why there must be a net force

• Velocity has a direction, so turning counts as changing velocity: the object accelerates even at constant speed.
• The net force and the acceleration both point to the centre of the circle.
• The force is perpendicular to the velocity, so it changes the direction but not the speed, and does no work. The kinetic energy stays constant.
• If the force stops acting, the object moves in a straight line along the tangent at constant speed.
• Centripetal force is the name for the net force towards the centre. It is supplied by a real force. Do not draw it as an extra arrow beside the real force on a free-body diagram.

Situation	Real force providing the centripetal force
Bung on a string	tension
Coin on a turntable	friction
Rider against the wall of a spinning drum	normal contact force
Charged particle in a magnetic field	magnetic force
Planet in orbit	gravity

Model explanations

• Why does a net force act? The velocity changes direction, so the object accelerates. By F = ma there must be a net force, in the direction of the acceleration, which is towards the centre.
• Why does the speed stay constant? The force is perpendicular to the velocity, so it has no component along the motion. It does no work and cannot change the kinetic energy.
• What happens if the string breaks? No net force acts, so Newton’s first law says the object keeps going in a straight line, at constant velocity, on the tangent where it was released.

Proportionality rules

For F = mv²/r:

• Double the speed, same r and m: force is four times as large.
• Double the radius, same v and m: force is halved.
• Double the mass, same v and r: force is doubled.

For F = mω²r:

• Double ω, same r and m: force is four times as large.
• Double the radius, same ω and m: force is doubled.

Notice that radius appears in the denominator in the first form but in the numerator in the second. Choose the form that matches what is held constant.

Whirling bung investigation (5.2.2(d)(ii))

Aspect	Notes
Set-up	bung on a string through a smooth tube; weight holder with masses below; tape marker sets the radius
Centripetal force	the weight Mg of the hanging masses, because the tension in the string equals it for steady motion
Independent variable	hanging mass (force), radius or bung mass, one at a time
Dependent variable	period T from the time for 20 revolutions divided by 20
Controls	keep the other two of F, r and m constant
Graph for F	F against f² (or 1/T²): straight line through the origin, gradient 4π²mr
Graph for r	T² against r: straight line through the origin, gradient 4π²m/F
Reducing uncertainty	time many revolutions, repeat and average, keep the marker at a constant position
Systematic effect	the string cannot be exactly horizontal, and friction at the tube changes the tension slightly
Safety	eye protection, firm knot, clear space around the circle, tray under the masses

Must-know distinctions

• Frequency or angular velocity. f counts turns per second; ω counts radians per second. ω = 2πf.
• Speed or angular velocity. v is distance per second along the circle; ω is angle per second. They are related by v = ωr.
• Constant speed or constant velocity. The speed is constant; the velocity is not, because its direction changes.
• Centripetal or centrifugal. Centripetal means towards the centre and describes the net force on the object. There is no outward force acting on the object itself.
• Period or time for several turns. Divide the total time by the number of revolutions.

Quick self-test

1. Define the radian.
2. Convert 2.5 rad to degrees.
3. Convert 72° to radians, as a multiple of π.
4. A wheel has a period of 0.80 s. Find its frequency and angular velocity.
5. A point 0.30 m from the axis of a disc rotates with ω = 6.0 rad s⁻¹. Find its speed.
6. A particle moves at 12 m s⁻¹ in a circle of radius 0.50 m. Find its centripetal acceleration.
7. A mass of 0.50 kg moves with ω = 8.0 rad s⁻¹ at a radius of 0.25 m. Find the centripetal force.
8. By what factor does the force change if the speed is tripled at constant radius and mass?
9. Explain why a body moving round a circle at constant speed is accelerating.
10. A whirling bung of mass 0.040 kg is used at a radius of 0.50 m. What is the expected gradient of a graph of F against f², with units?
11. A mass of 0.15 kg moves in a circle of radius 0.40 m with period 0.50 s. Find the force needed.

Answers

1. The angle at the centre of a circle that cuts off an arc whose length is the same as the radius.
2. 2.5 × 180/π = 143° (143.2°).
3. 72 × π/180 = 2π/5 rad (1.26 rad).
4. f = 1/0.80 = 1.25 Hz; ω = 2π/0.80 = 7.9 rad s⁻¹ (7.85).
5. v = ωr = 6.0 × 0.30 = 1.8 m s⁻¹.
6. a = v²/r = 144/0.50 = 288 m s⁻² (2.9 × 10² m s⁻²).
7. F = mω²r = 0.50 × 64 × 0.25 = 8.0 N.
8. 3² = 9, so the force is nine times as large.
9. Its direction of travel keeps changing, so its velocity keeps changing even though the speed does not. That rate of change is an acceleration, directed towards the centre.
10. Gradient = 4π²mr = 4π² × 0.040 × 0.50 = 0.79 N s² (equivalently kg m).
11. F = 4π²mr/T² = 4π² × 0.15 × 0.40/0.50² = 9.5 N.

Where marks are usually lost

• Using degrees for s = rθ, or forgetting to convert revolutions per minute.
• Writing “centripetal force” as a separate force on a diagram.
• Saying the object has no acceleration because the speed is constant.
• Using the diameter as the radius.
• Forgetting to square, or to take the square root, when rearranging v² or ω².
• Saying that a broken string sends the object radially outwards. The motion is along the tangent at constant speed.
• Not stating that the hanging weight equals the tension, which is why it gives the centripetal force in the bung investigation.
• Timing one revolution only, instead of many revolutions, in the investigation.
• Missing the units of a gradient, which are N s² for F against f².

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).