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

This study guide covers 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. It sits in Module 5, Newtonian world and astrophysics, and the ideas return in planetary motion (5.4.3). For condensed recall use the revision notes, and check your understanding with the practice questions. Related: the subject hub, the checklist to print, and the free 10-minute diagnostics if you want to find your weak spots first.

Coverage of the outcomes

Outcome	Learners should demonstrate and apply knowledge of
5.2.1(a)	the radian as a measure of angle
5.2.1(b)	period and frequency of an object in circular motion
5.2.1(c)	angular velocity ω, with ω = 2π/T or ω = 2πf
5.2.2(a)	a constant net force perpendicular to the velocity causes a circular path
5.2.2(b)	constant speed in a circle: v = ωr
5.2.2(c)	centripetal acceleration: a = v²/r and a = ω²r
5.2.2(d)(i)	centripetal force: F = mv²/r and F = mω²r
5.2.2(d)(ii)	techniques and procedures for investigating circular motion with a whirling bung

Rearranging equations is the main mathematical skill. Vectors are in the Foundations of Physics study guide; uncertainties are in the practical skills revision notes.

5.2.1(a) The radian

The size of an angle in radians is arc length over radius:

θ = s / r

So 1 radian is the angle at the centre that cuts off an arc as long as the radius. The whole circumference is 2πr, which gives a full turn of 2πr / r = 2π rad. Hence 360° = 2π rad and 180° = π rad.

Conversions: multiply degrees by π/180 to get radians, and multiply radians by 180/π to get degrees. One radian is about 57.3°. Being arc length divided by radius, a radian has no dimensions.

Worked example 1. A robot arm of length 0.60 m swings its tip through 0.85 rad. (a) Find the distance moved by the tip. (b) Express the angle in degrees. (c) Convert 135° to radians.

(a) s = rθ = 0.60 × 0.85 = 0.51 m. (b) 0.85 × 180/π = 48.7°. (c) 135 × π/180 = 3π/4 = 2.36 rad.

Using degrees in a formula that needs radians is a common source of wrong answers in this topic.

5.2.1(b) Period and frequency

The period T is the time taken by a single revolution, in seconds. The frequency f is the number of revolutions per second, in hertz (Hz). They are linked by

f = 1 / T

Divide revolutions per minute by 60 to get hertz.

5.2.1(c) Angular velocity

Angular velocity ω is the angle swept out per unit time, measured in rad s⁻¹:

ω = Δθ / Δt

One revolution is 2π rad in a time T, so

ω = 2π / T = 2πf

Worked example 2. A turntable rotates at 45 revolutions per minute. Find f, T and ω.

f = 45/60 = 0.75 Hz T = 1/f = 1.33 s ω = 2πf = 2π × 0.75 = 4.71 rad s⁻¹

Worked example 3. The minute hand of a tower clock is 1.8 m long. Find the angular velocity of the hand and the speed of its tip.

T = 3600 s, so ω = 2π/3600 = 1.75 × 10⁻³ rad s⁻¹. v = ωr = 1.75 × 10⁻³ × 1.8 = 3.1 × 10⁻³ m s⁻¹.

5.2.2(a) Why a force is needed

Velocity is a vector. An object going round a circle at constant speed changes direction all the time, so it accelerates, and by Newton’s second law a net force must act.

The force that produces a circular path has to be perpendicular to the velocity at every instant. A force at right angles to the motion turns the velocity without changing its size, so the speed stays constant. Because the force is at 90° to the displacement it does no work, and the kinetic energy of the object is unchanged. A component of force along the velocity would change the speed.

The force and the acceleration point towards the centre. If the force vanishes, Newton’s first law applies and the object moves off in a straight line along the tangent.

The centripetal force is not an extra force. It is the name for the net force towards the centre, and some real force or combination of forces supplies it:

Situation	Force supplying the centripetal force
Bung whirled on a string	tension in the string
Coin on a rotating turntable	friction between coin and surface
Rider pressed against the wall of a spinning drum	normal contact force from the wall
Charged particle in a magnetic field	magnetic force
Planet orbiting a star	gravitational force

On a free-body diagram, draw the real forces only.

5.2.2(b) Speed in a circle: v = ωr

In one period the object travels the circumference 2πr, so

v = 2πr / T

With ω = 2π/T this gives

v = ωr

The speed is constant but the velocity, along the tangent, keeps changing direction. Every point on a rigid spinning disc has the same ω and T, but a point at a larger radius travels a longer circumference in the same time, so it has a greater speed.

5.2.2(c) Centripetal acceleration

Let the velocity turn through a small angle Δθ in a time Δt. Its magnitude v is unchanged, so the change in velocity is the short side of an isosceles triangle, Δv ≈ vΔθ, directed towards the centre. The object moves an arc rΔθ = vΔt, so Δθ = vΔt/r. Then

a = Δv/Δt = v × (v/r) = v² / r

Using v = ωr gives the second form:

a = ω²r

Use v²/r when you have speed and radius, and ω²r when you have period, frequency or ω.

Worked example 4. A fairground ride is a drum of radius 2.2 m turning at 30 revolutions per minute. Find the acceleration of a rider against the wall, and the force on a 60 kg rider.

f = 30/60 = 0.50 Hz, so ω = 2π × 0.50 = 3.14 rad s⁻¹. a = ω²r = 3.14² × 2.2 = 21.7 m s⁻² (about 2.2 times the acceleration of free fall, using g = 9.81 m s⁻²). F = ma = 60 × 21.7 = 1.3 × 10³ N, provided by the wall.

5.2.2(d)(i) Centripetal force

Combining F = ma with the two forms of acceleration:

F = mv² / r = mω²r

For fixed m and r the force is proportional to v²: double the speed and you need four times the force. For fixed m and v it is inversely proportional to r.

Worked example 5. A rubber bung of mass 0.060 kg moves round a horizontal circle of radius 0.75 m, held by a string that breaks when the tension reaches 18 N. Find the greatest speed, and the corresponding ω and period.

F = mv²/r, so v² = Fr/m = (18 × 0.75)/0.060 = 225, giving v = 15 m s⁻¹. ω = v/r = 15/0.75 = 20 rad s⁻¹. T = 2π/ω = 0.31 s.

Worked example 6. A block of mass 0.20 kg moves on a smooth horizontal table in a circle of radius 0.40 m. It is joined by a string, passing through a hole at the centre, to a hanging mass of 0.50 kg. The block moves steadily at the radius stated. Find its speed and period (g = 9.81 m s⁻²).

The tension equals the weight of the hanging mass: F = 0.50 × 9.81 = 4.905 N. v² = Fr/m = (4.905 × 0.40)/0.20 = 9.81, so v = 3.13 m s⁻¹. ω = v/r = 7.83 rad s⁻¹, so T = 2π/ω = 0.80 s.

5.2.2(d)(ii) The whirling bung investigation

Aim. Test how the centripetal force depends on period, radius and mass.

Apparatus. A rubber bung on a string passed through a smooth glass or plastic tube held in the hand, with a weight holder and slotted masses hanging from the lower end. A piece of tape on the string marks the radius. You also need a stopwatch, metre rule, balance and goggles.

How it works. When the bung moves at a steady rate in a near-horizontal circle, the tension in the string equals the weight Mg of the hanging masses, so the centripetal force is F ≈ Mg. You choose the marker position to set the radius r.

Method for F against T (r and m constant).

1. Measure the mass of the bung and set the radius by placing the marker r from the bung.
2. Whirl the bung so that the marker stays just below the tube. Time 20 revolutions, then divide by 20 to get T.
3. Repeat with different hanging masses, and repeat each reading to find a mean.
4. Calculate f² = 1/T² for each F.

Since F = m(2πf)²r = 4π²mr f², plotting F against f² should give a straight line through the origin whose gradient is 4π²mr.

Worked data analysis. Here m = 0.0250 kg and r = 0.80 m. The expected gradient is 4π² × 0.0250 × 0.80 = 0.79 N s².

Hanging mass / kg	F = Mg / N	time for 20 revolutions / s	T / s	f² / s⁻²
0.200	1.96	12.7	0.635	2.48
0.300	2.94	10.4	0.520	3.70
0.400	3.92	9.0	0.450	4.94
0.500	4.91	8.1	0.405	6.10
0.600	5.89	7.4	0.370	7.30

A line through the first and last points gives a gradient (5.89 − 1.96)/(7.30 − 2.48) = 0.81 N s². Then m = gradient/(4π²r) = 0.81/(4π² × 0.80) = 0.026 kg, within about 3% of the balance reading. That agreement supports F = mω²r, and points scattering evenly about a straight line through the origin show F is proportional to f².

Other variation. Keep F and m constant and change r: T² = 4π²mr/F, so T² against r is a straight line through the origin.

Improving the reliability and the uncertainty.

• Timing 20 revolutions reduces the percentage uncertainty in T. With a timing uncertainty of 0.2 s, one revolution of 0.635 s has about 31% uncertainty, but 20 revolutions (12.7 s) has about 1.6%.
• Repeat and average each timing.
• Keep the marker just below the tube so the radius stays constant.

Sources of systematic error. The string is not exactly horizontal, because the bung must be supported against its weight. The string tension therefore acts at a small angle above the horizontal and is greater than the horizontal force needed. Friction between string and tube also makes the tension at the bung differ slightly from Mg.

Safety. Wear eye protection, knot the string firmly to the bung, keep other people well clear and place a tray under the hanging masses, because the bung flies off if the string breaks.

Common errors

• Writing “centripetal force” on a free-body diagram as well as the real force that supplies it.
• Saying that an outward “centrifugal force” pulls the object outwards. In the frame of the ground there is no outward force on the object.
• Using revolutions per minute or degrees directly in v = ωr.
• Using the diameter in place of the radius.
• Forgetting that doubling the speed multiplies the force by four, not two.

Next steps

Consolidate with the revision notes, then apply it in the practice questions.

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).