4.2 Circular motion
⤓ Presentation⤓ PowerPointPresentation: opens offline in any browser, simulations live. PowerPoint: editable.
A2 content for Unit 4 (WPH14), specification points 87–91. Orbits, where gravity is the centripetal force, are on the Unit 5 gravitational fields page (5.7).
What this page covers
7 points
- Angular displacement in radians and degrees, and converting between them
- Angular velocity ω, period and frequency; \(v = \omega r\) and \(T = 2\pi/\omega\)
- Deriving the centripetal acceleration \(a = v^2/r = r\omega^2\) from a vector diagram
- Why a resultant (centripetal) force is needed, and \(F = mv^2/r = mr\omega^2\)
- Cars on flat bends and banked tracks (qualitatively)
- Vertical loops, the conical pendulum and fairground rides
- Testing \(F = mr\omega^2\) with a graph, including combining percentage uncertainties
In one minute
6 points
- A body moving in a circle is always changing direction, so it is always accelerating towards the centre, and a real resultant force must supply that acceleration: \(F = mv^2/r = mr\omega^2\).
- An angle in radians is arc length ÷ radius: θ = s/r. One full turn = 2π rad = 360°.
- Angular velocity ω is the angle turned per second (rad s⁻¹): v = ωr, T = 2π/ω, ω = 2πf.
- The centripetal acceleration is \(a = v^2/r = r\omega^2\). The force is provided by real forces: tension, friction, a normal reaction, weight, lift.
- Remove the force and the object moves off along the tangent, not outwards.
- In a vertical loop the minimum speed at the top is \(\sqrt{gr}\), when weight alone supplies the centripetal force.
Contents
14 sections
- What is angular displacement, and why use radians?
- What are angular velocity, period and frequency?
- Why is a body moving at constant speed in a circle accelerating?
- What force keeps a body moving in a circle?
- How do vehicles get round bends?
- What happens in a vertical circle?
- How do you analyse a conical pendulum?
- How can you test \(F = mr\omega^2\) by experiment?
- Exam tips for 4.2
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is angular displacement, and why use radians?
From the definition, the angle in radians is the arc length divided by the radius:
\[\theta = \frac{s}{r} \qquad\text{so}\qquad s = r\theta\]with θ in rad, s the arc length (m) and r the radius (m). This is not in the formulae list; learn it, because it is the definition of the radian. A whole circle has arc length 2πr, so 360° = 2π rad. To convert: radians = degrees × π/180 and degrees = radians × 180/π; 1 rad ≈ 57.3°.
The radian is a ratio of two lengths, so it has no dimensions; "rad" is written to remind you which angle measure is in use. Useful values: 90° = π/2 rad, 180° = π rad, one revolution = 2π rad. Every equation on this page that contains θ or ω needs the angle in radians.
The minute hand of a station clock is 0.85 m long. Calculate (a) the angle, in radians, through which it turns in 25 minutes and (b) the distance moved by its tip in this time.
(a) The hand turns 360° in 60 minutes: angle = (25 ÷ 60) × 360° = 150° = 150 × π/180 = 2.62 rad
(b) Arc length = radius × angle in radians: s = 0.85 × 2.618 = 2.23 m ≈ 2.2 m
Using 150 instead of 2.62 would give 128 m. An angle in degrees in s = rθ is the most common slip in this topic.
What are angular velocity, period and frequency?
Here v is the linear speed (m s⁻¹), ω the angular velocity (rad s⁻¹), r the radius (m) and T the period (s). Both equations are in the formulae list. Why v = ωr: in time Δt the arc travelled is rΔθ, so the speed is rΔθ/Δt = rω. One revolution is 2π rad, so T = 2π/ω and therefore ω = 2πf.
The rotor of a wind turbine makes 18 revolutions per minute. Each blade is 42 m long. Calculate (a) the period, (b) the angular velocity and (c) the speed of a blade tip.
(a) f = 18 ÷ 60 = 0.30 Hz, so T = 1 ÷ 0.30 = 3.33 s ≈ 3.3 s
(b) ω = 2π ÷ T = 2π × 0.30 = 1.885 rad s⁻¹ ≈ 1.9 rad s⁻¹
(c) v = ωr = 1.885 × 42 = 79.2 m s⁻¹ ≈ 79 m s⁻¹
A point halfway along the blade has the same ω but only half the speed, about 40 m s⁻¹.
Rotation rates are often given in rev min⁻¹. Convert in one line: ω = 2π × (rev min⁻¹) ÷ 60, and write the conversion down so that a slip is easy to spot.
Why is a body moving at constant speed in a circle accelerating?
Velocity is a vector. In a circle the direction of the velocity changes continuously, even if its magnitude (the speed) does not. A change of velocity in a time interval is an acceleration, so uniform circular motion is accelerated motion. The acceleration is directed towards the centre of the circle and is called the centripetal acceleration.
Deriving \(a = v^2/r\)
- The body moves from A to B at constant speed v while the radius turns through a small angle Δθ.
- The velocities \(v_A\) and \(v_B\) are each perpendicular to their radius, so the angle between them is also Δθ.
- Drawn tail to tail, they form an isosceles triangle. The change in velocity \(\Delta v = v_B - v_A\) joins their tips and points towards the centre.
- For small Δθ the short side is almost an arc of radius v, so \(|\Delta v| = v\Delta\theta\).
- The time taken is Δt = arc AB ÷ speed = rΔθ/v.
- So \(a = \dfrac{|\Delta v|}{\Delta t} = v\Delta\theta \times \dfrac{v}{r\Delta\theta} = \dfrac{v^2}{r}\); with v = ωr this is also rω².
The centripetal acceleration (m s⁻²) is always directed towards the centre. The equation is in the formulae list, but you may be asked to derive it. In uniform circular motion its magnitude is constant while its direction rotates with the body.
A sample tube in a centrifuge is 8.5 cm from the axis. The rotor turns at 7200 revolutions per minute. Calculate the centripetal acceleration of the sample and express it as a multiple of g.
\(\omega = 2\pi \times 7200 \div 60 = 2\pi \times 120 = 754\ \text{rad s}^{-1}\).
With r in metres: \(a = r\omega^2 = 0.085 \times 754.0^2 = 4.83\times10^{4}\ \text{m s}^{-2}\) ≈ \(4.8 \times 10^{4}\ \text{m s}^{-2}\)
a/g = 4.83 × 10⁴ ÷ 9.81 ≈ 4900 g
Check with the other form: v = ωr = 64.1 m s⁻¹, and v²/r gives the same 4.83 × 10⁴ m s⁻².
Constant speed and constant velocity are not the same thing. In a circle only the speed is constant.
What force keeps a body moving in a circle?
By Newton's first law, a body with no resultant force moves in a straight line at constant speed. To make it follow a circle, a resultant force must act, and by Newton's second law it is in the direction of the acceleration: towards the centre. It is called the centripetal force. The force is perpendicular to the velocity, so it does no work: it changes the direction of motion but not the speed or the kinetic energy.
\[F = ma = \frac{mv^2}{r} = mr\omega^2\]F is the resultant force towards the centre (N) and m the mass (kg); the equation is in the formulae list. Use the v form when you know the speed and the ω form when you know the rotation rate or period. "Centripetal force" is a job description, not a new kind of force: some real force, or the resultant of several, does the job.
| Situation | What provides the centripetal force |
|---|---|
| ball whirled on a string in a horizontal circle | tension in the string |
| car on a flat bend | sideways friction between the tyres and the road |
| car or bicycle on a banked track | horizontal component of the normal reaction (plus any friction) |
| aircraft banking to turn | horizontal component of the lift |
| passenger on a rotor ride, back against the wall | normal reaction from the wall |
| satellite or moon in orbit | gravitational attraction (Unit 5) |
Drawing an arrow labelled "centripetal force" on a free-body diagram, or an outward "centrifugal force". Draw only the real forces; their resultant towards the centre is the centripetal force.
How do vehicles get round bends?
A flat bend
On a level road the only horizontal force on the car is friction between the tyres and the road, acting sideways towards the centre of the bend. Friction has a maximum value, while the force needed, mv²/r, rises with the square of the speed, so there is a maximum safe speed for each bend. If the speed is too high, friction cannot supply mv²/r; the car follows a path of larger radius and slides towards the outside of the bend.
A car of mass 1250 kg travels round a level bend of radius 45 m. (a) Calculate the sideways friction needed at 14 m s⁻¹. (b) The greatest sideways friction the tyres can provide on this road is 9.6 kN. Calculate the maximum speed at which the car can take the bend.
(a) Friction is the only horizontal force, so it must equal mv²/r: \(F = \dfrac{1250 \times 14^2}{45} = 5444\ \text{N}\) ≈ 5.4 kN
(b) Set mv²/r equal to the maximum friction: \(v_{max} = \sqrt{\dfrac{Fr}{m}} = \sqrt{\dfrac{9600 \times 45}{1250}} = 18.6\ \text{m s}^{-1}\) ≈ 19 m s⁻¹
Doubling the speed needs four times the friction: F ∝ v².
A banked track: the qualitative idea
Velodromes, race tracks and some motorway slip roads are banked: the outer edge is higher than the inner edge. The normal reaction acts perpendicular to the surface, so it is tilted towards the centre of the bend, and its horizontal component provides some or all of the centripetal force. At one particular speed the horizontal component of the normal reaction alone is exactly mv²/r, and no sideways friction is needed at all. Faster than this, friction acts down the slope as well; slower, it acts up the slope. Either way, banking reduces the reliance on friction, so the bend can be taken more safely at higher speeds. A banking aircraft or a cyclist leaning into a bend uses the same idea: a force that would otherwise be vertical is tilted so that it gains a horizontal component.
What happens in a vertical circle?
In a vertical circle the weight acts down at every point, so its role changes around the loop. Take towards the centre as positive at each point and write: resultant force towards centre = mv²/r.
- At the top (rider on the inside of a loop, or a ball on a string), the weight and the reaction or tension both act down, towards the centre: \(N + mg = mv^2/r\).
- At the bottom, the reaction or tension acts up, towards the centre, and the weight acts away from it: \(N - mg = mv^2/r\).
- The speed is not constant: the body speeds up on the way down and slows on the way up, so use conservation of energy to link the speeds at different heights.
At the minimum speed at the top, the reaction (or tension) is zero, so weight alone provides the centripetal force: mg = mv²/r, giving
\[v_{min} = \sqrt{gr}\]This is not in the formulae list, so derive it. Slower than this, the weight is more than enough to provide the centripetal force; the body falls inside the circle and the string goes slack or the rider leaves the track.
A rider of mass 65 kg goes round a vertical loop of radius 7.5 m on a fairground ride, sitting on the inside of the track. At the top of the loop the rider's speed is 11 m s⁻¹. Friction and air resistance are negligible. (a) Calculate the force exerted on the rider by the seat at the top. (b) Calculate the minimum speed at the top for the rider to stay in contact with the seat without a harness. (c) Calculate the force from the seat at the bottom of the loop.
(a) At the top both forces act towards the centre: \(N + mg = mv^2/r\), so \(N = \dfrac{65 \times 11^2}{7.5} - 65 \times 9.81 = 1049 - 638 = 411\ \text{N}\) ≈ 410 N
(b) Contact is just lost when N = 0: mg = mv²/r, so \(v = \sqrt{9.81 \times 7.5} = 8.58\ \text{m s}^{-1}\) ≈ 8.6 m s⁻¹
(c) Speed at the bottom from energy, as the rider falls 2r = 15 m: \(v^2 = 11^2 + 2 \times 9.81 \times 15 = 415.3\ \text{m}^2\ \text{s}^{-2}\) (v = 20.4 m s⁻¹).
At the bottom the reaction acts towards the centre and the weight away from it: \(N = mg + \dfrac{mv^2}{r} = 638 + \dfrac{65 \times 415.3}{7.5} = 4237\ \text{N}\) ≈ 4.2 kN
Before you substitute, write the equation with the force towards the centre first: "N + mg = mv²/r at the top", "T − mg = mv²/r at the bottom". A sign error in this line spoils everything that follows.
How do you analyse a conical pendulum?
A conical pendulum is a mass on a string (or a chair on a chain) moving in a horizontal circle, with the string sweeping out a cone. Only two forces act: the tension T along the string and the weight mg.
- Vertically there is no acceleration: T cos θ = mg.
- Horizontally the component of tension towards the centre provides the centripetal force: T sin θ = mv²/r.
- Dividing the two equations removes T and m: \(\tan\theta = \dfrac{v^2}{rg}\). Faster rotation gives a larger angle, and the mass of the rider does not matter.
- The radius is the radius of the horizontal circle, not the length of the string.
On a fairground swing-chair ride, each chair hangs from a chain 5.0 m long attached to the end of an arm 3.0 m from the central axis. When the ride turns steadily, each chain makes 35° with the vertical. A chair and rider have a combined mass of 80 kg. Calculate (a) the radius of the chair's circle, (b) the speed of the chair, (c) the time for one revolution and (d) the tension in the chain.
(a) Add the horizontal reach of the chain to the arm: r = 3.0 + 5.0 sin 35° = 3.0 + 2.87 = 5.87 m
(b) tan θ = v²/(rg): v² = 5.868 × 9.81 × tan 35° = 40.3, so v = 6.35 m s⁻¹ ≈ 6.3 m s⁻¹
(c) One revolution is a distance 2πr: period = 2π × 5.868 ÷ 6.349 = 5.81 s ≈ 5.8 s
(d) Vertical equilibrium: T = mg/cos 35° = 80 × 9.81 ÷ 0.8192 = 958 N ≈ 960 N
Check: T sin 35° = 550 N, and mv²/r = 80 × 40.3 ÷ 5.868 = 550 N, so the horizontal component is exactly the centripetal force.
Using the chain length (5.0 m) as the radius. The body moves in a horizontal circle; its radius is the horizontal distance from the axis of rotation to the body.
How can you test \(F = mr\omega^2\) by experiment?
Keep m and r fixed, vary ω, and measure the force needed. Since F = (mr)ω², a graph of F against ω² should be a straight line through the origin with gradient mr.
- One method: a mass fixed to a force sensor on a motor-driven arm; a light gate counts revolutions so that the period, and hence ω = 2π/T, is found from the time for many turns.
- Simpler version: a rubber bung whirled on a string that passes through a glass tube, with hanging masses providing the tension. The radius is kept constant with a small clip fixed to the string, and the period is timed over 10–20 revolutions.
- Safety: safety glasses, a clear space, and a string checked for wear before whirling anything.
Use the graph above, for which r = 0.200 ± 0.002 m, to determine the mass on the arm. The uncertainty in the gradient is estimated as ±3%. Give the mass with its absolute uncertainty.
Gradient from a large triangle on the line: \(\dfrac{5.39 - 0.61}{180 - 20} = 0.0299\ \text{kg m}\).
Gradient = mr, so m = 0.0299 ÷ 0.200 = 0.149 kg.
For a quotient, add percentage uncertainties: r has 0.002 ÷ 0.200 × 100 = 1.0%, so m has 3% + 1.0% = 4%. m = 0.149 ± 0.006 kg
The known mass, 0.150 kg, lies inside this range, so the results support F = mrω².
Timing many revolutions and dividing reduces the percentage uncertainty in the period. Because ω = 2π/T is squared, the percentage uncertainty in T counts twice in the percentage uncertainty in ω², and so in a calculated value of mrω².
Exam tips for 4.2
Every formula with θ or ω assumes radians. Convert degrees first, and check that your calculator is in the right mode before any trigonometry on a conical pendulum or banked track.
"What provides the centripetal force?" wants a real force (tension, friction, a normal reaction, a component of lift or weight) and its direction: towards the centre of the circle.
For each position on a vertical circle, take towards the centre as positive, write resultant force = mv²/r, and only then substitute. Find speeds at different heights with conservation of energy.
For a conical pendulum, banked track or banking aircraft, resolve vertically (equilibrium) and horizontally (mv²/r). Dividing the two equations gives tan θ = v²/(rg) and removes the unknown force.
A complete explanation links three steps: the velocity changes direction → so there is an acceleration towards the centre → so a resultant force towards the centre is needed (Newton's second law). Then name what provides it.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A body moving in a circle at constant speed has no acceleration. | Its velocity changes direction all the time. | It accelerates towards the centre at v²/r. |
| An outward "centrifugal force" pushes you to the outside of a bend. | No body exerts such a force on you; your body simply tends to keep moving in a straight line. | The only resultant force is inwards; the seat or door pushes you towards the centre. |
| Centripetal force is an extra force to add to a free-body diagram. | It is the name for the resultant of the real forces. | Draw tension, friction, reaction, weight; their resultant towards the centre equals mv²/r. |
| If the string breaks, the ball flies outwards along the radius. | With no force there is no change of velocity. | It moves off along the tangent at the speed it had. |
| Points further from the axis of a turning wheel have a larger angular velocity. | Every radius line of a rigid body turns through the same angle in the same time. | Same ω everywhere; the speed v = ωr is larger further out. |
| At the top of a loop at the minimum speed no forces act on the rider. | The weight never disappears. | The reaction is zero and the weight alone provides the centripetal force. |
How this topic connects
1.2 Forces and Newton's laws
Circular motion is F = ma with the acceleration pointing to the centre; the first law explains why a force is needed at all, and resolving forces handles the conical pendulum and banked track.
4.5 Magnetic fields and 4.7 Particle accelerators
A charged particle moving at right angles to a uniform magnetic field moves in a circle because the magnetic force BQv provides mv²/r; this gives r = p/BQ, the basis of cyclotrons and particle detectors.
5.7 Gravitational fields and orbits, and 5.5 Simple harmonic motion
For a satellite or planet, gravity is the centripetal force: GMm/r² = mv²/r. The same ω appears in SHM: the shadow of a point moving round a circle at constant ω moves with simple harmonic motion, and T = 2π/ω carries over unchanged.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| θ = s/r; 2π rad = 360° | angle in radians; converting angles | No — learn it |
| \(\omega = \Delta\theta/\Delta t\) | definition of angular velocity | No — learn it |
| v = ωr | linear speed from angular velocity | Yes |
| T = 2π/ω | period from angular velocity | Yes |
| f = 1/T, ω = 2πf | frequency and rotation rate | No — learn it / derive it |
| \(a = v^2/r = r\omega^2\) | centripetal acceleration | Yes (derivation may be asked) |
| \(F = mv^2/r = mr\omega^2\) | resultant force towards the centre | Yes |
| \(v_{min} = \sqrt{gr}\) | top of a vertical circle, reaction or tension zero | No — derive it |
| \(\tan\theta = v^2/(rg)\) | conical pendulum (angle to the vertical) | No — derive it |
Check your understanding
1. Convert an angle of 2.5 rad to degrees.
2.5 × 180/π = 143°.
2. A computer hard disk spins at 5400 revolutions per minute. Calculate its angular velocity.
ω = 2π × 5400 ÷ 60 = 565 rad s⁻¹.
3. A car turns sharply left. Explain why a passenger feels pressed against the right-hand door.
The passenger's body tends to continue in a straight line. The car turns left underneath them until the right-hand door reaches them; the door then pushes them to the left, towards the centre of the turn, providing their centripetal force. There is no outward force on the passenger.
4. A 0.060 kg ball on a string 0.50 m long moves in a horizontal circle on a smooth table at 4.0 m s⁻¹. Calculate the tension.
\(T = \dfrac{mv^2}{r} = \dfrac{0.060 \times 4.0^2}{0.50} = 1.9\ \text{N}\).
5. A toy car runs round the inside of a vertical loop of track of radius 0.25 m. Calculate the minimum speed at the top of the loop for the car to stay on the track.
\(v = \sqrt{gr} = \sqrt{9.81 \times 0.25} = 1.57\ \text{m s}^{-1}\) ≈ 1.6 m s⁻¹. At this speed the track exerts no force on the car and its weight alone provides the centripetal force.
Specification checklist
- (87) I can give an angular displacement in radians or in degrees and convert between the two.
- (88) I can define angular velocity and use v = ωr, T = 2π/ω and f = 1/T.
- (89) I can derive \(a = v^2/r = r\omega^2\) from a vector diagram of two velocities.
- (90) I can explain why a resultant force towards the centre is needed to start and keep a body moving in a circle, and say what provides it in a given situation.
- (91) I can use \(F = ma = mv^2/r = mr\omega^2\) for bends, banked tracks, vertical circles, conical pendulums and rides.