Circular Motion Basics

When a particle moves in a circle it is always changing direction, so its velocity vector is always changing — which means it is accelerating even when its speed is constant. That acceleration points towards the centre of the circle (it is centripetal, "centre-seeking"), and by Newton's second law it demands an inward resultant force. Everything in the next four lessons — conical pendulums, banked tracks, vertical loops, satellites — is built on the two relations $v = r\omega$v=rω and $a = v^2/r = r\omega^2$a=v2/r=rω2 established here.

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1 Where this sits in AQA 7367

Circular motion opens the kinematics-of-rotation strand of the Mechanics option on Paper 3 (7367/3M). Paper 3 is weighted AO1 40% / AO2 25% / AO3 35%, so beyond the standard formulae expect modelling questions that pick the right inward force and resolve it. The prerequisites are radian measure, $\vec F = m\vec a$F=ma, and the resolution of forces from A-Level Maths. The single conceptual leap is that a changing direction is an acceleration — we derive the magnitude $v^2/r$v2/r from first principles so it is understood, not merely quoted.

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2 Core theory

Angular speed

For a particle on a circle of radius $r$r, let $\theta$θ be the angle (in radians) swept from a fixed reference. The angular speed is the rate of change of that angle,

$\omega = \frac{\mathrm{d}\theta}{\mathrm{d}t}, \qquad [\omega] = \mathrm{rad\,s^{-1}}.$ω=dtdθ​,[ω]=rads−1.

In one complete revolution $\theta$θ increases by $2\pi$2π in one period $T$T, so

$\omega = \frac{2\pi}{T} = 2\pi f,$ω=T2π​=2πf,

where $f = 1/T$f=1/T is the frequency in revolutions per second (hertz). For uniform motion $\omega$ω is constant and $\theta = \omega t$θ=ωt.

Linear and angular speed

A point at radius $r$r travels an arc $s = r\theta$s=rθ (radian measure). Differentiating, the linear (tangential) speed is

$v = \frac{\mathrm{d}s}{\mathrm{d}t} = r\frac{\mathrm{d}\theta}{\mathrm{d}t} = r\omega.$v=dtds​=rdtdθ​=rω.

The velocity is tangent to the circle; only its direction changes in uniform motion, never its magnitude.

Centripetal acceleration — a derivation

Write the position of a particle moving anticlockwise at constant $\omega$ω as

$\vec r = r\cos(\omega t)\,\vec i + r\sin(\omega t)\,\vec j.$r=rcos(ωt)i+rsin(ωt)j​.

Differentiating twice with respect to $t$t,

$$\begin{aligned} \vec v &= \dot{\vec r} = -r\omega\sin(\omega t)\,\vec i + r\omega\cos(\omega t)\,\vec j,\\ \vec a &= \ddot{\vec r} = -r\omega^2\cos(\omega t)\,\vec i - r\omega^2\sin(\omega t)\,\vec j = -\omega^2 \vec r. \end{aligned}$$va​=r˙=−rωsin(ωt)i+rωcos(ωt)j​,=r¨=−rω2cos(ωt)i−rω2sin(ωt)j​=−ω2r.​

So the acceleration is $\vec a = -\omega^2\vec r$a=−ω2r: it points opposite to $\vec r$r, i.e. straight at the centre, with magnitude

$\boxed{\;a = r\omega^2 = \frac{v^2}{r}\;}\qquad(\text{using } v = r\omega).$a=rω2=rv2​​(using v=rω).

A quick check: $\vec v \cdot \vec a = (-r\omega\sin)( -r\omega^2\cos) + (r\omega\cos)(-r\omega^2\sin) = 0$v⋅a=(−rωsin)(−rω2cos)+(rωcos)(−rω2sin)=0, confirming the acceleration is perpendicular to the velocity — it turns the velocity without changing its speed.

A second derivation — the velocity triangle

It is worth seeing the same result geometrically, because it makes why the acceleration points inward vivid. Over a small time $\delta t$δt the particle turns through an angle $\delta\theta = \omega\,\delta t$δθ=ωδt. The velocity has constant magnitude $v$v but rotates through $\delta\theta$δθ; the change in velocity $\delta\vec v$δv is the third side of an isosceles triangle whose two equal sides have length $v$v and whose apex angle is $\delta\theta$δθ. For a small angle the base of that triangle is

$|\delta\vec v| \approx v\,\delta\theta.$∣δv∣≈vδθ.

The magnitude of the acceleration is then

$a = \lim_{\delta t \to 0}\frac{|\delta\vec v|}{\delta t} = \lim_{\delta t \to 0}\frac{v\,\delta\theta}{\delta t} = v\,\frac{\mathrm{d}\theta}{\mathrm{d}t} = v\omega = \frac{v^2}{r},$a=limδt→0​δt∣δv∣​=limδt→0​δtvδθ​=vdtdθ​=vω=rv2​,

agreeing with the calculus result. Crucially, as $\delta t \to 0$δt→0 the vector $\delta\vec v$δv becomes perpendicular to $\vec v$v and points towards the centre — the velocity is always being "bent inward". This isosceles-triangle argument is the one most often asked for in a "show that" derivation and needs no calculus.

Centripetal force

By Newton's second law the inward acceleration requires an inward resultant force,

$F = ma = \frac{mv^2}{r} = mr\omega^2,$F=ma=rmv2​=mrω2,

directed towards the centre. This is not a new kind of force. It is the name for the resultant of the real forces — tension, gravity, friction, normal reaction or a combination — resolved along the inward radius. The correct exam habit is: draw the real forces, then write "resultant inward force $= \dfrac{mv^2}{r}$=rmv2​". Never add "centripetal force" as an extra arrow.

Period and frequency

$T = \frac{2\pi}{\omega} = \frac{2\pi r}{v}, \qquad f = \frac{1}{T} = \frac{\omega}{2\pi}.$T=ω2π​=v2πr​,f=T1​=2πω​.

Uniform versus non-uniform motion

If the speed is constant the only acceleration is the centripetal $v^2/r$v2/r towards the centre. If the speed is changing (a bead sliding round a rough loop, a vertical circle under gravity), there is also a tangential acceleration along the direction of motion,

$a_{\text{tangential}} = \frac{\mathrm{d}v}{\mathrm{d}t}, \qquad a_{\text{centripetal}} = \frac{v^2}{r},$atangential​=dtdv​,acentripetal​=rv2​,

and these two components are perpendicular, so the total acceleration has magnitude

$a = \sqrt{\left(\frac{\mathrm{d}v}{\mathrm{d}t}\right)^2 + \left(\frac{v^2}{r}\right)^2}.$a=(dtdv​)2+(rv2​)2​.

The centripetal component still uses the instantaneous speed $v$v, even when $v$v is varying — a subtlety that matters in the vertical-circle lesson, where gravity continually changes the speed but the radial equation $T - mg\cos\theta = mv^2/r$T−mgcosθ=mv2/r holds at every instant with the current $v$v.

Why radians are non-negotiable

The relations $s = r\theta$s=rθ, $v = r\omega$v=rω and $a = r\omega^2$a=rω2 are all consequences of the arc-length definition $s = r\theta$s=rθ, which is only true when $\theta$θ is in radians (it is the definition of radian measure). In degrees, $s = \dfrac{\pi}{180}r\theta$s=180π​rθ, and every formula would carry an ugly $\pi/180$π/180. This is why $\omega$ω must always be converted to $\mathrm{rad\,s^{-1}}$rads−1 — revolutions per second become $\omega = 2\pi f$ω=2πf, revolutions per minute become $\omega = 2\pi N/60$ω=2πN/60, and a calculator left in degree mode will silently wreck a circular-motion answer.

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3 Worked examples with mark scheme

Worked Example 1 — centripetal force from $\omega$ω

A particle of mass $0.5\ \mathrm{kg}$0.5 kg moves in a horizontal circle of radius $2\ \mathrm{m}$2 m at angular speed $3\ \mathrm{rad\,s^{-1}}$3 rads−1. Find the centripetal force.

$F = mr\omega^2 = 0.5 \times 2 \times 3^2 = 0.5 \times 2 \times 9 = 9\ \mathrm{N}.$F=mrω2=0.5×2×32=0.5×2×9=9 N.

Mark scheme: M1 $F = mr\omega^2$F=mrω2; A1 $9\ \mathrm{N}$9 N, directed towards the centre. (2 marks.) Choose the $r\omega^2$rω2 form when $\omega$ω is given and the $v^2/r$v2/r form when $v$v is given — converting unnecessarily invites arithmetic slips.

Worked Example 2 — centripetal acceleration from $v$v

A car travels round a circular bend of radius $50\ \mathrm{m}$50 m at $20\ \mathrm{m\,s^{-1}}$20 ms−1. Find its centripetal acceleration.

$a = \frac{v^2}{r} = \frac{20^2}{50} = \frac{400}{50} = 8\ \mathrm{m\,s^{-2}}.$a=rv2​=50202​=50400​=8 ms−2.

Mark scheme: M1 $a = v^2/r$a=v2/r; A1 $8\ \mathrm{m\,s^{-2}}$8 ms−2 towards the centre. Note this is about $0.8g$0.8g — a sharp bend at speed makes serious demands on the available friction, which is exactly why bends are banked (next lesson).

Worked Example 3 — "show that" with gravity as the inward force

A satellite orbits the Earth (radius $R$R) at height $h$h above the surface; gravity provides the centripetal force. Show that its speed is $v = \sqrt{\dfrac{gR^2}{R+h}}$v=R+hgR2​​, where $g$g is the surface gravitational acceleration.

The gravitational force at the surface is $mg = \dfrac{GMm}{R^2}$mg=R2GMm​, so $GM = gR^2$GM=gR2. At orbital radius $R+h$R+h this force provides the centripetal force:

$$\begin{aligned} \frac{GMm}{(R+h)^2} &= \frac{mv^2}{R+h}\\ \frac{gR^2}{(R+h)^2} &= \frac{v^2}{R+h} \qquad(\text{using } GM = gR^2)\\ v^2 &= \frac{gR^2}{R+h}\\ v &= \sqrt{\frac{gR^2}{R+h}}. \qquad\blacksquare \end{aligned}$$(R+h)2GMm​(R+h)2gR2​v2v​=R+hmv2​=R+hv2​(using GM=gR2)=R+hgR2​=R+hgR2​​.■​

Mark scheme: M1 equate gravitational and centripetal forces; M1 substitute $GM = gR^2$GM=gR2 (the key insight); A1 reach the printed result with no algebraic gaps. In a "show that" every line must be present — the answer is given, so the marks are entirely for the reasoning.

Worked Example 4 — period to $\omega$ω and $v$v

A particle completes one revolution every $0.5\ \mathrm{s}$0.5 s in a circle of radius $0.4\ \mathrm{m}$0.4 m. Find its angular speed and linear speed.

$$\begin{aligned} \omega &= \frac{2\pi}{T} = \frac{2\pi}{0.5} = 4\pi \approx 12.57\ \mathrm{rad\,s^{-1}},\\ v &= r\omega = 0.4 \times 4\pi = 1.6\pi \approx 5.03\ \mathrm{m\,s^{-1}}. \end{aligned}$$ωv​=T2π​=0.52π​=4π≈12.57 rads−1,=rω=0.4×4π=1.6π≈5.03 ms−1.​

Mark scheme: M1 $\omega = 2\pi/T$ω=2π/T; A1 $4\pi$4π; M1 $v = r\omega$v=rω; A1 $1.6\pi$1.6π. Leaving the answer as an exact multiple of $\pi$π and then giving the decimal is good practice — examiners accept either, but the exact form shows control.

Worked Example 5 — converting revolutions per minute

A centrifuge rotor spins at $1200\ \mathrm{rev\,min^{-1}}$1200 revmin−1. A sample sits $0.08\ \mathrm{m}$0.08 m from the axis. Find (a) the angular speed in $\mathrm{rad\,s^{-1}}$rads−1, (b) the centripetal acceleration, and (c) that acceleration as a multiple of $g$g. (Take $g = 9.8$g=9.8.)

(a) Each revolution is $2\pi$2π radians and there are $60$60 seconds in a minute:

$\omega = 1200 \times \frac{2\pi}{60} = 40\pi \approx 125.7\ \mathrm{rad\,s^{-1}}.$ω=1200×602π​=40π≈125.7 rads−1.

(b) $a = r\omega^2 = 0.08 \times (40\pi)^2 = 0.08 \times 1600\pi^2 = 128\pi^2 \approx 1263\ \mathrm{m\,s^{-2}}.$a=rω2=0.08×(40π)2=0.08×1600π2=128π2≈1263 ms−2.

(c) $\frac{a}{g} = \frac{1263}{9.8} \approx 129.$ga​=9.81263​≈129.

Mark scheme: M1 convert $\mathrm{rev\,min^{-1}}$revmin−1 to $\mathrm{rad\,s^{-1}}$rads−1 $(\times 2\pi/60)$(×2π/60); A1 $40\pi$40π; M1 $a = r\omega^2$a=rω2; A1 $\approx 1263\ \mathrm{m\,s^{-2}}$≈1263 ms−2; A1 $\approx 129g$≈129g. The conversion is the trap — forgetting either the $2\pi$2π (per revolution) or the $/60$/60 (per minute) is the single commonest error in this style of question, and the huge "$129g$129g" is exactly why centrifuges can separate materials by density.

Worked Example 6 — two points on the same rotating body

A rigid disc rotates about its centre. Point $A$A is $0.1\ \mathrm{m}$0.1 m from the centre and point $B$B is $0.4\ \mathrm{m}$0.4 m from the centre. Compare (a) their angular speeds, (b) their linear speeds, and (c) their centripetal accelerations.

Both points are on the same rigid body, so they sweep the same angle in the same time — they share one angular speed $\omega$ω.

(a) $\omega_A = \omega_B = \omega$ωA​=ωB​=ω (equal).

(b) $v = r\omega$v=rω, so $\dfrac{v_B}{v_A} = \dfrac{r_B}{r_A} = \dfrac{0.4}{0.1} = 4$vA​vB​​=rA​rB​​=0.10.4​=4: point $B$B moves four times faster.

(c) $a = r\omega^2$a=rω2, so $\dfrac{a_B}{a_A} = \dfrac{r_B}{r_A} = 4$aA​aB​​=rA​rB​​=4: point $B$B also has four times the acceleration.

Mark scheme: B1 equal $\omega$ω (same rigid body); M1 use $v = r\omega$v=rω and $a = r\omega^2$a=rω2 to form ratios; A1 both ratios $= 4$=4. The key insight — the same body shares one $\omega$ω, but $v$v and $a$a grow with the radius — explains everything from why the rim of a CD reads faster than the centre to why a discus thrower releases at arm's length.

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4 Specimen-style exam question

(Specimen-style.) A particle $P$P of mass $0.3\ \mathrm{kg}$0.3 kg is attached to one end of a light inextensible string of length $0.6\ \mathrm{m}$0.6 m; the other end is fixed at the centre $O$O of a smooth horizontal table. $P$P moves in a horizontal circle on the table, completing $2$2 revolutions per second. (a) Find the angular speed and the linear speed of $P$P. (b) Find the tension in the string.

(a) Two revolutions per second means $f = 2\ \mathrm{Hz}$f=2 Hz:

$\omega = 2\pi f = 2\pi(2) = 4\pi \approx 12.57\ \mathrm{rad\,s^{-1}}, \qquad v = r\omega = 0.6 \times 4\pi = 2.4\pi \approx 7.54\ \mathrm{m\,s^{-1}}.$ω=2πf=2π(2)=4π≈12.57 rads−1,v=rω=0.6×4π=2.4π≈7.54 ms−1.

(b) On the smooth horizontal table the only horizontal force is the string tension, which therefore is the centripetal force:

$T = mr\omega^2 = 0.3 \times 0.6 \times (4\pi)^2 = 0.18 \times 16\pi^2 = 2.88\pi^2 \approx 28.4\ \mathrm{N}.$T=mrω2=0.3×0.6×(4π)2=0.18×16π2=2.88π2≈28.4 N.