Circular motion: horizontal, vertical and conical

AshPhys · Revision cheat sheet · Sheet 1 of 6

9702 A Level Ch.16Chapter 16: Circular motionLesson 16.1Sections 16.1–16.7A Level only (not AS)

Goal: Analyse any circular motion with ω, v = rω and F = mv²/r: strings, bends, banked tracks, vertical circles, conical pendulums.

forcevelocityacceleration

01The big picture

An object moving in a circle at constant speed has a velocity whose direction changes all the time, so it is accelerating. The acceleration points to the centre: the centripetal acceleration $a = v^2/r = r\omega^2$ .

By Newton’s second law that needs a resultant force towards the centre, $F = mv^2/r = mr\omega^2$ . It is perpendicular to the velocity, so it changes the direction of motion but not the speed, and it does no work.

The centripetal force is not an extra force: it is the resultant of the real forces (tension, friction, gravity, contact force, lift). The recipe never changes: draw the real forces, resolve along the radius with “towards the centre” positive, and set the resultant equal to $mv^2/r$ . In a horizontal circle the vertical components also balance.

Angles are in radians. One revolution is $2\pi$ rad in one period $T$ , so the angular speed is $\omega = 2\pi/T = 2\pi f$ and the speed is $v = r\omega$ .

Fig. 1 · Velocity, force and acceleration

Velocity is along the tangent; the resultant force and the acceleration both point to the centre, at 90° to it.

Uniform circular motion (syllabus) [syllabus wording]: Understand that a force of constant magnitude that is always perpendicular to the direction of motion causes centripetal acceleration.
Radian [paraphrase]: The angle subtended at the centre of a circle by an arc equal in length to the radius. So $\theta = s/r$ and $2\pi$ rad $= 360°$ .
Angular speed $\omega$ [paraphrase]: The rate of change of angular displacement, $\omega = \Delta\theta/\Delta t$ , in rad s⁻¹.
Centripetal force [paraphrase]: The resultant force on an object in circular motion, directed towards the centre of the circle.

02Deriving the equations

D1 Centripetal acceleration and force

In a short time $\Delta t$ the object turns through a small angle $\Delta\theta$ . Its speed $v$ stays the same; only the direction of its velocity changes.

Fig. 2 · Why Δv points to the centre

Redraw v₁ and v₂ from one point: the change Δv joins their tips and is parallel to the radius, pointing to the centre.

$v_1$ and $v_2$ have the same length $v$ and differ in direction by $\Delta\theta$ . For a small angle, $\Delta v$ is an arc of radius $v$ :
$\Delta v = v\,\Delta\theta$
$\Delta v$ points towards the centre. Divide by $\Delta t$ , using $\omega = \Delta\theta/\Delta t$ :
$a = \frac{\Delta v}{\Delta t} = v\,\frac{\Delta\theta}{\Delta t} = v\omega$
Substitute $\omega = v/r$ (or $v = r\omega$ ). Newton’s second law, $F = ma$ , then gives the resultant force needed.
$a = \frac{v^{2}}{r} = r\omega^{2}$

Result

a = \frac{v^{2}}{r} = r\omega^{2} \qquad F = \frac{mv^{2}}{r} = mr\omega^{2}

Circular motion: horizontal, vertical and conical · 9702 A Level Ch.16Sheet 2 of 6

02Deriving the equations (continued)

D2 Horizontal circles: string, flat bend, banked track

The centre is level with the object, so the resultant force is horizontal and the vertical forces balance.

Fig. 3 · Banked track, rear view

N\cos\theta

balances the weight;

N\sin\theta

points to the centre of the bend and provides

mv^2/r

.

Ball on a string on a smooth table: the tension alone points to the centre.
$F_\mathrm{T} = \frac{mv^{2}}{r}$
Car on a flat bend: friction from the road points to the centre. If $mv^2/r$ needs more friction than the road can give, the car skids to a wider path.
$F_\mathrm{friction} = \frac{mv^{2}}{r}$
Banked track with no friction (or a banking aircraft, with lift in place of $N$ ). Resolve the normal contact force $N$ :
$N\cos\theta = mg \qquad N\sin\theta = \frac{mv^{2}}{r}$
Divide the second equation by the first to eliminate $N$ :
$\tan\theta = \frac{v^{2}}{rg}$

Result

\tan\theta = \frac{v^{2}}{rg}

D3 Vertical circles

A mass on a string, or a car in a loop. Take “towards the centre” as positive: down at the top, up at the bottom. The speed changes round the circle.

Fig. 4 · Vertical circle on a string

At the top, tension and weight both point to the centre. At the bottom, tension points to the centre and weight away from it.

Top: tension and weight both point down, towards the centre.
$F_\mathrm{T} + mg = \frac{mv_\mathrm{top}^{2}}{r}$
Bottom: tension points up (towards the centre), weight down.
$F_\mathrm{T} - mg = \frac{mv_\mathrm{bottom}^{2}}{r}$
The string just stays taut at the top when $F_\mathrm{T} = 0$ : gravity alone provides the centripetal force.
$mg = \frac{mv^{2}}{r} \;\Longrightarrow\; v_\mathrm{min} = \sqrt{gr}$
With no energy losses, kinetic energy at the bottom = kinetic energy at the top + $mg(2r)$ :
$v_\mathrm{bottom}^{2} = v_\mathrm{top}^{2} + 4gr$
Car over a hump (top of a circle, contact force $N$ up): $mg - N = mv^2/r$ , so it leaves the road when $N = 0$ , at $v = \sqrt{gr}$ .

Result

\text{top: } F_\mathrm{T} + mg = \frac{mv^{2}}{r} \qquad \text{bottom: } F_\mathrm{T} - mg = \frac{mv^{2}}{r}

Circular motion: horizontal, vertical and conical · 9702 A Level Ch.16Sheet 3 of 6

02Deriving the equations (continued)

D4 Conical pendulum

A bob on a string of length $L$ moves in a horizontal circle of radius $r$ ; the string sweeps out a cone at angle $\theta$ to the vertical.

Fig. 5 · Conical pendulum

F_\mathrm{T}\cos\theta

balances the weight;

F_\mathrm{T}\sin\theta

points to the centre of the horizontal circle, of radius

r = L\sin\theta

.

Vertically the bob is in equilibrium:
$F_\mathrm{T}\cos\theta = mg$
The horizontal component points to the centre and provides the centripetal force:
$F_\mathrm{T}\sin\theta = \frac{mv^{2}}{r}$
Divide to eliminate $F_\mathrm{T}$ (the same result as the banked track):
$\tan\theta = \frac{v^{2}}{rg}$
With $r = L\sin\theta$ and $v = r\omega$ , the horizontal equation gives $F_\mathrm{T} = mL\omega^2$ . With the vertical one, $\cos\theta = g/(L\omega^2)$ , so the period is
$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L\cos\theta}{g}}$

Result

\tan\theta = \frac{v^{2}}{rg} \qquad T = 2\pi\sqrt{\frac{L\cos\theta}{g}}

03Key equations

Angular speed

\omega = \frac{\Delta\theta}{\Delta t} = \frac{2\pi}{T} = 2\pi f

Speed and angular speed

v = r\omega

Centripetal acceleration

a = \frac{v^{2}}{r} = r\omega^{2}

towards the centre

Centripetal force

F = \frac{mv^{2}}{r} = mr\omega^{2}

resultant of the real forces

Vertical circle: top

F_\mathrm{T} + mg = \frac{mv^{2}}{r}

Vertical circle: bottom

F_\mathrm{T} - mg = \frac{mv^{2}}{r}

Least speed at the top

v_\mathrm{min} = \sqrt{gr}

string slack, loop, hump

Banked track, cone

\tan\theta = \frac{v^{2}}{rg}

Conical pendulum period

T = 2\pi\sqrt{\frac{L\cos\theta}{g}}

Symbols and units

$\theta$ angle turned, bank or cone angle $\mathrm{rad}$
$\omega$ angular speed $\mathrm{rad\,s^{-1}}$
$T$ period (one revolution) $\mathrm{s}$
$f$ frequency (rev per second) $\mathrm{Hz}$
$r$ radius of the circle $\mathrm{m}$
$v$ speed $\mathrm{m\,s^{-1}}$
$a$ centripetal acceleration $\mathrm{m\,s^{-2}}$
$F$ centripetal (resultant) force $\mathrm{N}$
$F_\mathrm{T}$ tension $\mathrm{N}$
$N$ normal contact force $\mathrm{N}$
$m$ mass $\mathrm{kg}$
$L$ length of the string $\mathrm{m}$
$g$ free-fall acceleration, 9.81 $\mathrm{m\,s^{-2}}$

Circular motion: horizontal, vertical and conical · 9702 A Level Ch.16Sheet 4 of 6

04Sample problems with solutions

Sample 1 · Horizontal circle on a string
A 0.20 kg puck on a smooth horizontal table is tied to a string and moves in a circle of radius 0.50 m, making 3.0 revolutions per second. Calculate its angular speed, its speed and the tension in the string.

Revolutions per second is the frequency $f$ :
$\omega = 2\pi f = 2\pi \times 3.0\ \mathrm{s^{-1}} = 18.85\ \mathrm{rad\,s^{-1}}$
Speed:
$v = r\omega = 0.50\ \mathrm{m} \times 18.85\ \mathrm{rad\,s^{-1}} = 9.42\ \mathrm{m\,s^{-1}}$
The tension is the only horizontal force, so it is the centripetal force:
$F_\mathrm{T} = mr\omega^{2} = 0.20 \times 0.50 \times 18.85^{2} = 35.5\ \mathrm{N}$

Answer:

\omega = 19\ \mathrm{rad\,s^{-1}},\quad v = 9.4\ \mathrm{m\,s^{-1}},\quad F_\mathrm{T} = 36\ \mathrm{N}

Unit check:

\mathrm{kg} \times \mathrm{m} \times (\mathrm{rad\,s^{-1}})^{2} = \mathrm{kg\,m\,s^{-2}} = \mathrm{N}\ \text{(the radian has no unit)}\ \checkmark

Sample 2 · Vertical circle
A 0.50 kg ball on a string 0.80 m long is whirled in a vertical circle. At the top its speed is 4.0 m s⁻¹. Calculate the tension at the top, the speed at the bottom and the tension at the bottom. Ignore air resistance.

Top: tension and weight both point to the centre. Rearrange for the tension:
$F_\mathrm{T} = \frac{mv^{2}}{r} - mg = \frac{0.50 \times 4.0^{2}}{0.80} - 0.50 \times 9.81 = 10.0 - 4.91 = 5.1\ \mathrm{N}$
Energy from top to bottom, a drop of $2r = 1.60$ m:
$v_\mathrm{bottom}^{2} = 4.0^{2} + 4 \times 9.81 \times 0.80 = 47.4 \;\Longrightarrow\; v_\mathrm{bottom} = 6.88\ \mathrm{m\,s^{-1}}$
Bottom: tension up, weight down:
$F_\mathrm{T} = mg + \frac{mv^{2}}{r} = 4.91 + \frac{0.50 \times 47.4}{0.80} = 34.5\ \mathrm{N}$
Sense check: the bottom tension exceeds the top tension by $6mg = 29.4$ N, as the derivation predicts.

Answer:

F_\mathrm{T,top} = 5.1\ \mathrm{N},\quad v_\mathrm{bottom} = 6.9\ \mathrm{m\,s^{-1}},\quad F_\mathrm{T,bottom} = 35\ \mathrm{N}

Unit check:

\frac{\mathrm{kg}\,(\mathrm{m\,s^{-1}})^{2}}{\mathrm{m}} = \mathrm{kg\,m\,s^{-2}} = \mathrm{N}\ \checkmark

Circular motion: horizontal, vertical and conical · 9702 A Level Ch.16Sheet 5 of 6

04Sample problems with solutions (continued)

Sample 3 · Conical pendulum
A 0.30 kg bob on a string 1.2 m long moves in a horizontal circle with the string at 25° to the vertical. Calculate the tension, the speed of the bob and the period.

Vertical equilibrium:
$F_\mathrm{T} = \frac{mg}{\cos\theta} = \frac{0.30 \times 9.81}{\cos 25^\circ} = 3.25\ \mathrm{N}$
Radius of the circle (not the string length):
$r = L\sin\theta = 1.2 \times \sin 25^\circ = 0.507\ \mathrm{m}$
From $\tan\theta = v^2/(rg)$ :
$v = \sqrt{rg\tan\theta} = \sqrt{0.507 \times 9.81 \times \tan 25^\circ} = 1.52\ \mathrm{m\,s^{-1}}$
Period, checked against $2\pi\sqrt{L\cos\theta/g}$ :
$T = \frac{2\pi r}{v} = \frac{2\pi \times 0.507}{1.52} = 2.09\ \mathrm{s}$

Answer:

F_\mathrm{T} = 3.2\ \mathrm{N},\quad v = 1.5\ \mathrm{m\,s^{-1}},\quad T = 2.1\ \mathrm{s}

Unit check:

\sqrt{\mathrm{m} \times \mathrm{m\,s^{-2}}} = \mathrm{m\,s^{-1}}\ \checkmark

Sample 4 · Banked track
A track is banked at 12° to the horizontal. Calculate the radius of the bend for which a car at 25 m s⁻¹ needs no friction.

With no friction only $N$ and $mg$ act, so the banked-track result applies:
$\tan\theta = \frac{v^{2}}{rg}$
$r$ is in the denominator and $\tan\theta$ is the lone term on the other side: swap them in one step (cross-multiplication).
$\tan\theta = \frac{v^{2}}{rg} \;\Longrightarrow\; r = \frac{v^{2}}{g\tan\theta}$
Substitute:
$r = \frac{25^{2}}{9.81 \times \tan 12^\circ} = \frac{625}{2.085} = 299.7\ \mathrm{m}$

Answer:

r = 3.0 \times 10^{2}\ \mathrm{m}

Unit check:

\frac{(\mathrm{m\,s^{-1}})^{2}}{\mathrm{m\,s^{-2}}} = \mathrm{m}\ \checkmark

Circular motion: horizontal, vertical and conical · 9702 A Level Ch.16Sheet 6 of 6

05Practice problems

P1
A wheel turns at 2.5 revolutions per second. Calculate its angular speed.
Final answer
16 rad s⁻¹ (15.7)
Answer: 16 rad s⁻¹ (15.7)
P2
Express an angle of 150° in radians.
Final answer
2.6 rad (5π/6)
Answer: 2.6 rad (5π/6)
P3
A disc turns at 500 revolutions per minute. Calculate the speed of a point 5.5 cm from its centre.
Final answer
2.9 m s⁻¹
Answer: 2.9 m s⁻¹
P4
A 1200 kg car takes a flat bend of radius 50 m at 15 m s⁻¹. Calculate the friction force needed.
Final answer
5.4 × 10³ N
Answer: 5.4 × 10³ N
P5
Calculate the least speed at the top of a vertical loop of radius 6.0 m for a roller-coaster car to stay on the track.
Final answer
7.7 m s⁻¹
Answer: 7.7 m s⁻¹
P6
A car drives over a hump-backed bridge whose top is part of a circle of radius 20 m. Calculate the greatest speed at which it stays in contact with the road.
Final answer
14 m s⁻¹
Answer: 14 m s⁻¹
P7
2.0 kg of water in a bucket is swung in a vertical circle of radius 0.90 m. At the bottom its speed is 6.0 m s⁻¹. Calculate the force the bucket exerts on the water there.
Final answer
1.0 × 10² N (99.6)
Answer: 1.0 × 10² N (99.6)
P8
A conical pendulum has a string 0.75 m long at 30° to the vertical. Calculate its period.
Final answer
1.6 s
Answer: 1.6 s
P9
An aircraft makes a level turn at 120 m s⁻¹, banked at 35°. Calculate the radius of the turn.
Final answer
2.1 × 10³ m
Answer: 2.1 × 10³ m
P10
A 0.050 kg mass on a string is whirled in a vertical circle with no energy losses. Calculate how much larger the tension is at the bottom than at the top.
Final answer
2.9 N (6mg)
Answer: 2.9 N (6mg)

06Other useful information

Common mistakes

✗ Wrong: Use ω in degrees per second or revolutions per minute.
✓ Right: ω must be in rad s⁻¹: multiply rev s⁻¹ by 2π; divide rpm by 60 first.
✗ Wrong: Draw “centripetal force” as an extra arrow on the free-body diagram.
✓ Right: Draw only the real forces. Their resultant towards the centre equals mv²/r.
✗ Wrong: At the top of a vertical circle, $F_\mathrm{T} - mg = mv^2/r$ .
✓ Right: At the top both forces point to the centre: $F_\mathrm{T} + mg = mv^2/r$ .
✗ Wrong: Use the string length as the radius of a conical pendulum.
✓ Right: r = L sin θ.
✗ Wrong: Constant speed, so the object is in equilibrium.
✓ Right: Its velocity changes, so there is a resultant force. In a horizontal circle only the vertical components balance.

Exam tips

Draw the free-body diagram first and mark the centre: “towards the centre” is positive.
Horizontal circle: write two equations (vertical balance, horizontal = mv²/r) and divide them to eliminate the unknown force.
Vertical circle: the speed changes, so link the top and bottom with energy conservation.
The centripetal force is perpendicular to the motion, so it does no work and never changes the kinetic energy.

Explore

What the syllabus asks

9702 topic 12, Motion in a circle: the radian, angular displacement, ω = 2π/T, v = rω, a = rω² = v²/r and F = mrω² = mv²/r. Coursebook Chapter 16.
A Level only: not part of AS.
Banked tracks, vertical circles and conical pendulums are not listed separately: exam questions use them as applications of F = mv²/r, so learn the method rather than the special formulas.
Orbits (gravity as the centripetal force) are in Chapter 17.