Physics › Periodic motion › Circular motion

Circular motion

Going round a corner at a steady speed counts as accelerating, because velocity is a vector and its direction never stops changing. One angle-based toolkit, the radian and omega, turns every circular problem into three short formulas.

Year 13AQA 3.6.1.1CIE 12.1.1, 12.1.2, 12.1.3, 12.2.1, 12.2.2, 12.2.3, 12.2.4OCR A 5.2.1(a), 5.2.1(b), 5.2.1(c), 5.2.2(a), 5.2.2(b), 5.2.2(c), 5.2.2(d)(i)EDEXCEL 103, 104, 105, 106, 107

Builds on Newton's laws and the resultant force and Scalars and vectors.

The maths behind it: Radians, arcs and small angles on InkMaths.

IN THIS TOPIC

  • Explain why constant-speed circular motion is accelerated motion, and why it needs a centripetal force.
  • Work in radians, with ω = v/r = 2πf.
  • Derive a = v²/r from a vector triangle, and see where its inward direction comes from.
  • Apply a = v²/r = ω²r and F = mv²/r = mω²r, and name the real force supplying it each time.

COMMON MISCONCEPTION

You're flung outwards by centrifugal force.

Steady speed, changing velocity

An object circling at constant speed is accelerating, continuously, because velocity is a vector and its direction changes at every instant. The acceleration points to the centre of the circle, always at right angles to the velocity; that perpendicularity is how the direction can change while the speed does not.

Circular motion: the turning velocity (animated figure)constant speed is not constant velocitywatch the arrows:velocity: length pinnedacceleration: always inwardso a force must aim at the centre:gravity, for the satellite;the magnetic force, for the charge
FIG. 1The dot goes round at constant speed, and the cyan velocity arrow never changes length; but it never stops changing direction, so the velocity is changing and there must be an acceleration. The coral arrow shows it: aimed at the centre at every instant. Restyled, the same motion is a satellite held by gravity or a charge held by a magnetic field.

By Newton's second law, a centre-pointing acceleration demands a centre-pointing resultant force, the centripetal force. Nothing flings you outward. Take a corner in a car and your body continues in a straight line, Newton's first law, while the car turns underneath you; the sensation of being thrown outward is the car door arriving to push you inward. Remove the centripetal force and the object does not fly radially away. It departs along the tangent.

The angular toolkit

Circles are easier in angle language. One radian is the angle whose arc length equals the radius, which gives θ = s/r and makes a full circle 2π radians.

The radian: the angle whose arc length equals the radius, and the natural unit for angular speed1 radarc s = rrθ = s / rω = Δθ / Δta full circle is 2π radians
FIG. 2One radian: the arc equals the radius. From it, θ = s/r, and ω is the angle swept per second.

The angular speed ω is the angle swept per second, in rad s−1, and it links to everything else through

ω=vr=2πf\omega = \frac{v}{r} = 2\pi fON THE AQA DATA SHEET

Direction of angular velocity is not considered at this level, so ω here means a magnitude. Keep the calculator in radian mode for the whole unit. The moment trig arrives next lesson, degree mode quietly poisons every answer.

WORKED EXAMPLE

One ride, the full toolkit

A fairground ride carries riders in a circle of radius 6.0 m, completing each revolution in 4.0 s. Find the angular speed, the linear speed and the centripetal acceleration.

ω = 2π/T = 2π/4.0 = 1.6 rad s−1.

v = ωr = 1.57 × 6.0 = 9.4 m s−1.

a = ω2r = 1.572 × 6.0 = 15 m s−2, about 1.5g, aimed at the centre. Work ω, then v, then a. That order gets you through almost every circular-motion calculation.

The vector triangle behind the acceleration

Where does the centripetal acceleration come from? From the velocity vector alone, and the argument is short enough to reproduce under exam conditions. Follow the object from one point on the circle to another a small angle Δθ later. The speed never changes, so the two velocity vectors have the same length. Each is tangent to the circle, so swinging the radius through Δθ swings the velocity through the same Δθ.

Deriving the centripetal acceleration: two velocity vectors a small angle apart, and the triangle that closes themABvvvvΔvΔθsame speed, an angle Δθ apartthe two velocities, tail to tailΔv = vΔθ, and it aims at the centre
FIG. 3Left, two instants a small angle apart, with equal-length velocity arrows tangent at each. Right, the same two velocities redrawn from a common tail; the coral side closing their tips is the change in velocity, and it aims at the centre.

Redraw the two velocities from a common tail and the change in velocity is the third side of the triangle, running from the tip of the first to the tip of the second. Two sides of length v with Δθ between them make an isosceles triangle, so for a small angle the closing side is a short arc of a circle of radius v, and its length is Δv = vΔθ.

The direction is the part worth pausing on. The two velocities sit symmetrically about the middle of the arc, so their difference lies exactly along the radius there, pointing inwards. Shrink Δθ and that midpoint closes on the object itself, which is the reason the acceleration at any instant is directed at the centre. The coral arrow on the left of the figure is that same Δv carried back to the circle, and the dashed line shows it running through the centre.

The time to sweep Δθ is Δt = Δθ/ω, straight from the definition of angular speed, so

a=ΔvΔt=vΔθΔθ/ω=vωa = \frac{\Delta v}{\Delta t} = \frac{v\Delta\theta}{\Delta\theta/\omega} = v\omega

and Δθ cancels, as it had to, since the answer cannot depend on how small a step you chose to take.

Acceleration and force

One substitution each way turns a = vω into the pair the questions use. Put ω = v/r into it for the first form and v = ωr for the second:

a=v2r=ω2ra = \frac{v^{2}}{r} = \omega^{2}rON THE AQA DATA SHEET

Multiply by mass for the force:

F=mv2r=mω2rF = \frac{mv^{2}}{r} = m\omega^{2}rON THE AQA DATA SHEET
Centripetal force is never a new force: something real, tension, gravity or friction, must point to the centreball on a stringtensionmoon in orbitgravitycar on a bendfrictionno force to the centre, no circle:the object flies off along the tangent
FIG. 4The centripetal force is always provided by something real: tension for a ball on a string, gravity for the Moon, friction for a car on a bend.

Here is the central idea of the topic. Centripetal force is never a new force. The phrase names a job, and some real force fills the post, tension or gravity or friction or a normal reaction, sometimes several together. “What provides the centripetal force?” wants one of those by name. Answering “centripetal force” restates the question instead of answering it.

GUIDED PRACTICE

The spin cycle's violence

A washing machine drum of radius 0.25 m spins at 1200 revolutions per minute. Convert to an angular speed, then find the centripetal acceleration at the drum wall.

Show the working

1200 rpm means 20 revolutions per second, so ω = 20 × 2π = 126 rad s−1.

a = ω2r = 1262 × 0.25 = 3.9 × 103 m s−2, about 400g. The drum wall provides that force on the clothes; the water, with nothing to push on it at the holes, leaves. Spin-drying works because the water is not made to follow the circular path.

INDEPENDENT PRACTICE

The slowest loop

A ball on a string is whirled in a vertical circle of radius 0.80 m. Find the minimum speed at the top for the string to stay taut.

Show the working

At the top, weight and tension both point downwards and together supply the centripetal force. The minimum case has the tension at zero and the weight supplying all of it, so mg = mv2/r.

v = the square root of gr = the square root of 9.81 × 0.80 = 2.8 m s−1. The mass cancelled; every ball on every 0.80 m string shares the same threshold.

ASSESSMENT FOCUS

  • For “explain why the object accelerates at constant speed”, three short sentences do it. Velocity is a vector. Its direction changes. A changing velocity is an acceleration.
  • Name the provider. Tension, gravity, friction or a normal reaction supplies the centripetal force; writing “centripetal force” as the provider names the job rather than the force.
  • Radian mode. The formulas assume it, the data sheet assumes it, and any question evaluating x = A cos ωt in degree mode fails by the same silent factor.
  • Lose the centripetal force and the object leaves along the tangent, never radially outward. Sketch questions test exactly this.
  • Choose the convenient form. Use v²/r when the speed is given, ω²r when the period or frequency is, via ω = 2πf = 2π/T.
  • Edexcel asks you to derive a = v²/r from a vector diagram, and the three steps are the marks. Equal-length velocity vectors because the speed is constant, Δv = vΔθ from the isosceles triangle with Δv aimed at the centre, then divide by Δt = Δθ/ω. The other boards quote the result and examine its use, but that same triangle is still the cleanest answer to “why does the acceleration point inwards?”, which every board asks.

CHECK YOURSELF

A 900 kg car takes a bend of radius 50 m at a steady 15 m s−1. Find the centripetal force required, state what provides it, and explain what happens if the road cannot supply that much.

Show a hint

One formula, then think about which real force supplies it on a flat road.

Show the answer

F=mv2/rF = mv^{2}/r = 900 × 152 / 50 = 4050 N, directed towards the bend's centre.

On a flat road the provider is friction between the tyres and the surface.

If friction cannot reach 4050 N, on ice say, the car cannot hold that circle. It follows a straighter path and drifts wide along a tangent, which is a skid described politely.

Circular motion is accelerated motion, aimed at the centre.

Something real has to supply the force.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

Question workbook

PDF · 41 KB · print edition · white background

Question workbook

PDF · 41 KB · screen edition · navy

Answer book

PDF · 57 KB · fully worked · navy
19 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the circular motion questions page.

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  • Explain why constant-speed circular motion is accelerated motion, and why it needs a centripetal force.
  • Work in radians, with ω = v/r = 2πf.
  • Derive a = v²/r from a vector triangle, and see where its inward direction comes from.
  • Apply a = v²/r = ω²r and F = mv²/r = mω²r, and name the real force supplying it each time.

Open the full revision checklist to track your progress across the whole unit.