Horizontal Circular Motion

The most-examined applications of circular motion happen in a horizontal plane with vertical equilibrium: the conical pendulum, the banked track, the flat roundabout and the rotating turntable. The shared method is always the same two-line resolution — vertically the forces balance (no vertical acceleration), horizontally they supply the centripetal force — and almost every result falls out by dividing those two equations. Master that template here and these problems become routine.

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1 Where this sits in AQA 7367

This is the central applications block of the Mechanics option, Paper 3 (7367/3M), sitting between circular-motion basics and vertical circles. With Paper 3 weighted AO1 40% / AO2 25% / AO3 35%, banked-track and conical-pendulum questions are favourite AO3 carriers because they bundle resolution, friction limits and the centripetal condition into one multi-stage problem. The prerequisites are $a = v^2/r = r\omega^2$a=v2/r=rω2 (previous lesson), resolving forces, and the friction model $F \le \mu N$F≤μN from A-Level Maths.

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2 Core theory

The universal template

For any horizontal-circle problem with the circle in a horizontal plane:

$$\begin{aligned} \text{Vertically (equilibrium):}\quad & \sum F_{\text{vertical}} = 0,\\ \text{Horizontally (centripetal):}\quad & \sum F_{\text{inward}} = \frac{mv^2}{r} = mr\omega^2. \end{aligned}$$Vertically (equilibrium):Horizontally (centripetal):​∑Fvertical​=0,∑Finward​=rmv2​=mrω2.​

Identify the real forces, resolve them into these two equations, and solve — usually by dividing the horizontal equation by the vertical one to eliminate an unknown reaction or tension.

The conical pendulum

A particle of mass $m$m on a light string of length $l$l moves in a horizontal circle while the string sweeps out a cone at angle $\theta$θ to the vertical. The radius of the circle is

$r = l\sin\theta.$r=lsinθ.

The forces are the weight $mg$mg (down) and the tension $T$T (along the string, towards the support). Resolving:

$$\begin{aligned} \text{Vertically:}\quad & T\cos\theta = mg \quad (1)\\ \text{Horizontally:}\quad & T\sin\theta = \frac{mv^2}{r} = mr\omega^2 \quad (2) \end{aligned}$$Vertically:Horizontally:​Tcosθ=mg(1)Tsinθ=rmv2​=mrω2(2)​

Dividing $(2)$(2) by $(1)$(1) eliminates $T$T:

$\tan\theta = \frac{v^2}{rg} = \frac{r\omega^2}{g}.$tanθ=rgv2​=grω2​.

Substituting $r = l\sin\theta$r=lsinθ into the $\omega$ω-form, $\tan\theta = \dfrac{l\sin\theta\,\omega^2}{g}$tanθ=glsinθω2​, and dividing through by $\sin\theta$sinθ (using $\tan\theta/\sin\theta = 1/\cos\theta$tanθ/sinθ=1/cosθ),

$\boxed{\;\omega^2 = \frac{g}{l\cos\theta}\;}\qquad\Longrightarrow\qquad T_{\text{period}} = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{l\cos\theta}{g}}.$ω2=lcosθg​​⟹Tperiod​=ω2π​=2πglcosθ​​.

Two consequences worth noting: the bob can never reach $\theta = 90^\circ$θ=90∘ (that needs infinite $\omega$ω, since $\cos\theta \to 0$cosθ→0); and faster rotation means a larger angle and a shallower cone.

Banked tracks — the no-friction "design speed"

On a road banked at angle $\theta$θ, the normal reaction $N$N is perpendicular to the surface and so has an inward horizontal component. With no friction the forces are $N$N and the weight $mg$mg:

$$\begin{aligned} \text{Vertically:}\quad & N\cos\theta = mg,\\ \text{Horizontally:}\quad & N\sin\theta = \frac{mv^2}{r}. \end{aligned}$$Vertically:Horizontally:​Ncosθ=mg,Nsinθ=rmv2​.​

Dividing,

$\boxed{\;\tan\theta = \frac{v^2}{rg}\;}$tanθ=rgv2​​

This single speed, $v = \sqrt{rg\tan\theta}$v=rgtanθ​, is the design speed: travel at it and the banking alone supplies the centripetal force, with no sideways friction needed — the smoothest, lowest-wear ride.

Banked tracks — with friction

Off the design speed, friction $F \le \mu N$F≤μN makes up the difference. At the maximum speed the car is on the point of sliding up and outward, so limiting friction $\mu N$μN acts down the slope:

$$\begin{aligned} \text{Vertically:}\quad & N\cos\theta - \mu N\sin\theta = mg,\\ \text{Horizontally:}\quad & N\sin\theta + \mu N\cos\theta = \frac{mv^2}{r}. \end{aligned}$$Vertically:Horizontally:​Ncosθ−μNsinθ=mg,Nsinθ+μNcosθ=rmv2​.​

Dividing eliminates $N$N and $m$m:

$\frac{v_{\max}^2}{rg} = \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta} = \frac{\tan\theta + \mu}{1 - \mu\tan\theta}.$rgvmax2​​=cosθ−μsinθsinθ+μcosθ​=1−μtanθtanθ+μ​.

For the minimum speed (about to slide down and inward), friction acts up the slope — flip the sign of $\mu$μ in both equations, giving $\dfrac{v_{\min}^2}{rg} = \dfrac{\tan\theta - \mu}{1 + \mu\tan\theta}$rgvmin2​​=1+μtanθtanθ−μ​ (and if $\tan\theta \le \mu$tanθ≤μ the car cannot slide inward at any speed, so $v_{\min} = 0$vmin​=0).

Flat road and rotating turntable

On a flat (unbanked) road friction alone supplies the inward force, so the limiting case $\mu mg = \dfrac{mv^2}{r}$μmg=rmv2​ gives

$v_{\max} = \sqrt{\mu g r}.$vmax​=μgr​.

A particle resting on a rotating turntable at radius $r$r is held in its circle by friction; on the point of sliding outward, $\mu mg = mr\omega^2$μmg=mrω2, so

$\omega_{\max} = \sqrt{\frac{\mu g}{r}}.$ωmax​=rμg​​.

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3 Worked examples with mark scheme

Worked Example 1 — conical pendulum, find $\omega$ω and the period

A conical pendulum has string length $0.8\ \mathrm{m}$0.8 m; the string makes $30^\circ$30∘ with the vertical. Find the angular speed and the period. (Take $g = 9.8\ \mathrm{m\,s^{-2}}$g=9.8 ms−2.)

$$\begin{aligned} \omega^2 &= \frac{g}{l\cos\theta} = \frac{9.8}{0.8\cos 30^\circ} = \frac{9.8}{0.8 \times \tfrac{\sqrt3}{2}} = \frac{9.8}{0.6928} = 14.14,\\ \omega &= 3.76\ \mathrm{rad\,s^{-1}},\\ T_{\text{period}} &= \frac{2\pi}{\omega} = \frac{2\pi}{3.76} = 1.67\ \mathrm{s}. \end{aligned}$$ω2ωTperiod​​=lcosθg​=0.8cos30∘9.8​=0.8×23​​9.8​=0.69289.8​=14.14,=3.76 rads−1,=ω2π​=3.762π​=1.67 s.​

Mark scheme: M1 $\omega^2 = g/(l\cos\theta)$ω2=g/(lcosθ); A1 $\omega = 3.76\ \mathrm{rad\,s^{-1}}$ω=3.76 rads−1; M1 $T = 2\pi/\omega$T=2π/ω; A1 $1.67\ \mathrm{s}$1.67 s. (4 marks.)

Worked Example 2 — conical pendulum, find the angle and tension

A conical pendulum has string length $1.2\ \mathrm{m}$1.2 m and bob mass $0.5\ \mathrm{kg}$0.5 kg; it revolves at $2$2 revolutions per second. Find the angle to the vertical and the tension.

$\omega = 2\pi \times 2 = 4\pi\ \mathrm{rad\,s^{-1}}.$ω=2π×2=4π rads−1.

From $\omega^2 = \dfrac{g}{l\cos\theta}$ω2=lcosθg​,

$$\cos\theta = \frac{g}{l\omega^2} = \frac{9.8}{1.2(4\pi)^2} = \frac{9.8}{189.5} = 0.0517 \;\Rightarrow\; \theta = \arccos(0.0517) = 87.0^\circ.$$cosθ=lω2g​=1.2(4π)29.8​=189.59.8​=0.0517⇒θ=arccos(0.0517)=87.0∘.

The cone is almost flat — at high spin rates the bob flies out nearly horizontal. The tension from $(1)$(1):

$T = \frac{mg}{\cos\theta} = \frac{0.5 \times 9.8}{0.0517} = 94.8\ \mathrm{N}.$T=cosθmg​=0.05170.5×9.8​=94.8 N.

Mark scheme: M1 $\omega = 2\pi f$ω=2πf; M1 $\cos\theta = g/(l\omega^2)$cosθ=g/(lω2); A1 $\theta = 87.0^\circ$θ=87.0∘; M1 $T = mg/\cos\theta$T=mg/cosθ; A1 $94.8\ \mathrm{N}$94.8 N. (5 marks.) The very small $\cos\theta$cosθ makes the tension large — a good physical sanity check.

Worked Example 3 — banked track, design speed

A road is banked at $15^\circ$15∘ for a bend of radius $200\ \mathrm{m}$200 m. Find the design speed (the speed needing no friction).

$$\tan 15^\circ = \frac{v^2}{rg} \;\Rightarrow\; v^2 = rg\tan 15^\circ = 200 \times 9.8 \times 0.2679 = 525.1 \;\Rightarrow\; v = 22.9\ \mathrm{m\,s^{-1}}.$$tan15∘=rgv2​⇒v2=rgtan15∘=200×9.8×0.2679=525.1⇒v=22.9 ms−1.

Mark scheme: M1 $\tan\theta = v^2/(rg)$tanθ=v2/(rg); M1 rearrange for $v$v; A1 $22.9\ \mathrm{m\,s^{-1}}$22.9 ms−1 (about $82\ \mathrm{km\,h^{-1}}$82 kmh−1). At this speed the resultant of $N$N and $mg$mg is exactly horizontal and inward.

Worked Example 4 — banked track with friction, maximum speed

A bend of radius $100\ \mathrm{m}$100 m is banked at $20^\circ$20∘; the coefficient of friction between tyres and road is $\mu = 0.3$μ=0.3. Find the maximum speed for no slipping.

At maximum speed limiting friction acts down the slope. Dividing the resolved equations,

$$\begin{aligned} \frac{v_{\max}^2}{rg} &= \frac{\sin 20^\circ + 0.3\cos 20^\circ}{\cos 20^\circ - 0.3\sin 20^\circ} = \frac{0.3420 + 0.2819}{0.9397 - 0.1026} = \frac{0.6239}{0.8371} = 0.7452,\\ v_{\max}^2 &= 0.7452 \times 100 \times 9.8 = 730.3 \;\Rightarrow\; v_{\max} = 27.0\ \mathrm{m\,s^{-1}}. \end{aligned}$$rgvmax2​​vmax2​​=cos20∘−0.3sin20∘sin20∘+0.3cos20∘​=0.9397−0.10260.3420+0.2819​=0.83710.6239​=0.7452,=0.7452×100×9.8=730.3⇒vmax​=27.0 ms−1.​

Mark scheme: M1 both resolved equations with friction down the slope; M1 divide to eliminate $N$N; A1 ratio $0.745$0.745; A1 $v_{\max} = 27.0\ \mathrm{m\,s^{-1}}$vmax​=27.0 ms−1. (4 marks.) The hardest mark is the direction of friction — at top speed the car tends to slide outward/up, so friction opposes that and acts down the slope.

Worked Example 5 — flat roundabout

A car rounds a flat roundabout of radius $30\ \mathrm{m}$30 m; the coefficient of friction is $0.5$0.5. Find the maximum speed.

$v_{\max} = \sqrt{\mu g r} = \sqrt{0.5 \times 9.8 \times 30} = \sqrt{147} = 12.1\ \mathrm{m\,s^{-1}}.$vmax​=μgr​=0.5×9.8×30​=147​=12.1 ms−1.

Mark scheme: M1 $\mu mg = mv^2/r$μmg=mv2/r (friction is the only inward force); A1 $v_{\max} = 12.1\ \mathrm{m\,s^{-1}}$vmax​=12.1 ms−1 (about $43.6\ \mathrm{km\,h^{-1}}$43.6 kmh−1). The mass cancels — the limiting speed is independent of how heavy the car is.

Worked Example 6 — normal reaction on a banked track

A car of mass $1200\ \mathrm{kg}$1200 kg rounds a bend of radius $40\ \mathrm{m}$40 m banked at $18^\circ$18∘ at exactly the design speed. Find (a) the design speed and (b) the normal reaction from the road, and compare it with the car's weight. (Take $g = 9.8$g=9.8.)

(a) $v = \sqrt{rg\tan\theta} = \sqrt{40 \times 9.8 \times \tan 18^\circ} = \sqrt{40 \times 9.8 \times 0.3249} = \sqrt{127.4} = 11.3\ \mathrm{m\,s^{-1}}.$v=rgtanθ​=40×9.8×tan18∘​=40×9.8×0.3249​=127.4​=11.3 ms−1.

(b) At the design speed there is no friction, so resolving vertically $N\cos\theta = mg$Ncosθ=mg:

$N = \frac{mg}{\cos\theta} = \frac{1200 \times 9.8}{\cos 18^\circ} = \frac{11760}{0.9511} = 12365\ \mathrm{N}.$N=cosθmg​=cos18∘1200×9.8​=0.951111760​=12365 N.

The weight is $mg = 11760\ \mathrm{N}$mg=11760 N, so the reaction is about $1.05$1.05 times the weight. Mark scheme: M1 design speed from $\tan\theta = v^2/(rg)$tanθ=v2/(rg); A1 $11.3\ \mathrm{m\,s^{-1}}$11.3 ms−1; M1 $N = mg/\cos\theta$N=mg/cosθ; A1 $12365\ \mathrm{N}$12365 N. The normal reaction on a bank is always larger than the weight, because $N$N must also supply the inward force — taking $N = mg$N=mg is a common error.

Worked Example 7 — particle in a smooth hemispherical bowl

A particle of mass $0.3\ \mathrm{kg}$0.3 kg moves in a horizontal circle on the smooth inner surface of a hemispherical bowl of radius $0.5\ \mathrm{m}$0.5 m. The radius drawn from the centre of the sphere to the particle makes $40^\circ$40∘ with the downward vertical. Find the angular speed and the normal reaction. (Take $g = 9.8$g=9.8.)

The only forces are the weight $mg$mg and the normal reaction $N$N, which (the surface being smooth) acts along the radius towards the centre $O$O of the sphere. The radius of the horizontal circle is $r = a\sin\theta$r=asinθ. Resolving:

$$\begin{aligned} \text{Vertically:}\quad & N\cos\theta = mg,\\ \text{Horizontally:}\quad & N\sin\theta = mr\omega^2 = m(a\sin\theta)\omega^2. \end{aligned}$$Vertically:Horizontally:​Ncosθ=mg,Nsinθ=mrω2=m(asinθ)ω2.​

The horizontal equation simplifies (cancel $\sin\theta$sinθ) to $N = ma\omega^2$N=maω2; substituting into the vertical equation gives $ma\omega^2\cos\theta = mg$maω2cosθ=mg, i.e.

$\omega^2 = \frac{g}{a\cos\theta} = \frac{9.8}{0.5\cos 40^\circ} = \frac{9.8}{0.3830} = 25.59 \;\Rightarrow\; \omega = 5.06\ \mathrm{rad\,s^{-1}}.$ω2=acosθg​=0.5cos40∘9.8​=0.38309.8​=25.59⇒ω=5.06 rads−1.

Then $N = ma\omega^2 = 0.3(0.5)(25.59) = 3.84\ \mathrm{N}$N=maω2=0.3(0.5)(25.59)=3.84 N. Mark scheme: M1 both resolved equations with $N$N along the radius; M1 use $r = a\sin\theta$r=asinθ and simplify; A1 $\omega^2 = g/(a\cos\theta)$ω2=g/(acosθ); A1 $\omega = 5.06\ \mathrm{rad\,s^{-1}}$ω=5.06 rads−1; A1 $N = 3.84\ \mathrm{N}$N=3.84 N. Notice the result is identical to a conical pendulum of string length $a$a: the bowl's normal reaction plays exactly the role of the string's tension. Spotting that equivalence is a genuine mark-scheme insight.

Worked Example 8 — turntable, maximum revolutions per minute

A coin rests $0.15\ \mathrm{m}$0.15 m from the centre of a horizontal turntable; the coefficient of friction between coin and turntable is $0.4$0.4. Find the greatest number of revolutions per minute the turntable can make before the coin slides. (Take $g = 9.8$g=9.8.)

Friction is the only horizontal (inward) force; on the point of sliding it is limiting, $\mu mg = mr\omega^2$μmg=mrω2:

$\omega_{\max} = \sqrt{\frac{\mu g}{r}} = \sqrt{\frac{0.4 \times 9.8}{0.15}} = \sqrt{26.13} = 5.11\ \mathrm{rad\,s^{-1}}.$ωmax​=rμg​​=0.150.4×9.8​​=26.13​=5.11 rads−1.

Converting to revolutions per minute,

$N = \frac{\omega_{\max}}{2\pi} \times 60 = \frac{5.11}{2\pi} \times 60 \approx 48.8\ \mathrm{rev\,min^{-1}}.$N=2πωmax​​×60=2π5.11​×60≈48.8 revmin−1.

Mark scheme: M1 $\mu mg = mr\omega^2$μmg=mrω2; A1 $\omega_{\max} = 5.11\ \mathrm{rad\,s^{-1}}$ωmax​=5.11 rads−1; M1 convert $\mathrm{rad\,s^{-1}}$rads−1 to $\mathrm{rev\,min^{-1}}$revmin−1 $(\div 2\pi,\ \times 60)$(÷2π, ×60); A1 $48.8\ \mathrm{rev\,min^{-1}}$48.8 revmin−1. The mass cancels (so the answer is the same for a coin or a brick), and the unit conversion at the end is where marks are most often dropped.

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4 Specimen-style exam question