Circular Motion

Why an object moving at constant speed around a circle is accelerating, which real force supplies that acceleration in each situation the syllabus names, and how torque sets things turning.

60 min · Module 5: Advanced Mechanics

Assumes you have read Projectile Motion.

The full teaching. Start here.

Syllabus requirementsModule 5: Advanced Mechanics

conduct investigations to explain and evaluate, for objects executing uniform circular motion, the relationships that exist between:
centripetal force
mass
speed
radius
analyse the forces acting on an object executing uniform circular motion in a variety of situations, for example:
cars moving around horizontal circular bends
a mass on a string
objects on banked tracks (ACSPH100)ACSPH100
solve problems, model and make quantitative predictions about objects executing uniform circular motion in a variety of situations, using the following relationships:
$a_{c} = \frac{v ^{2}}{r}$
$v = \frac{2 π r}{T}$
$F_{c} = \frac{m v ^{2}}{r}$
$ω = \frac{Δ θ}{t}$
investigate the relationship between the total energy and work done on an object executing uniform circular motion
investigate the relationship between the rotation of mechanical systems and the applied torque
$τ = r_{⊥} F = r F sin θ$

What this actually asks for

Five dot points, and they ask for different kinds of work.

The first is a practical: you need to be able to show, with data, how the force needed depends on mass, speed and radius, and to judge how good that data is.

The second says analyse the forces. For each named situation you must say which real force points at the centre, and write the equation that follows. This is where most of the marks in the topic sit.

The third is calculation, using four relationships that are all on the formulae sheet. The fourth asks why the force does no work, and the fifth brings in torque, which is the turning effect of a force and a separate idea from centripetal force despite sitting in the same section.

The short version

An object moving around a circle at constant speed is accelerating, because its direction is changing, and a change of direction is a change of velocity. The acceleration points at the centre of the circle and has size

$a_{c} = \frac{v ^{2}}{r}$

By Newton's second law, something must be pulling it towards the centre with a net force

$F_{c} = \frac{m v ^{2}}{r}$

That net force is not a new kind of force. It is always a real force you can name: tension in a string, friction from a road, the normal force from a banked track, gravity for a satellite. Every question in this topic starts by naming it.

The speed comes from the period, $v = 2 π r / T$ , and the angular velocity is $ω = Δ θ / t$ , with the angle in radians.

Two results to have ready:

On a flat bend, friction supplies the force, so the fastest safe speed is the one where $m v^{2} / r$ equals the greatest friction the tyres can give.
On a banked bend with no friction, $tan θ = v^{2} / r g$ . Mass cancels.

The centripetal force is always perpendicular to the motion, so it does no work, and the kinetic energy stays constant. Torque is $τ = r F sin θ$ , largest when the force is perpendicular to the lever arm.

The fastest way to lose marks: drawing an outward centrifugal force, or adding a force labelled centripetal on top of the real forces.

Constant speed, changing velocity

Projectile motion had a constant acceleration and a changing speed. Circular motion reverses both: the speed is constant, and the acceleration is never the same twice, because it keeps turning to point at the centre.

Why an object at constant speed can be accelerating

AnswersIf the speedometer reading never changes, how can the car be accelerating?

Acceleration is the rate of change of velocity, not of speed. Velocity is a vector, so it changes when its size changes, when its direction changes, or both.

An object going around a circle has a direction of travel that is always along the tangent, and the tangent keeps turning. So the velocity is changing at every instant even though its size, the speed, is not. A changing velocity is an acceleration.

Which way does it point? Look at the velocity a moment ago and the velocity now. The change between them, the arrow you would have to add to the first to get the second, points in towards the centre. The object is continuously being bent inwards from the straight line it would otherwise follow.

And Newton's second law then says there must be a net force pointing inwards too. Without one, the object would carry on in a straight line along the tangent. That is exactly what happens when a string breaks or a car hits oil on a bend.

Speed

Radius2.0 m

Speed4.0 m s⁻¹

Mass1.5 kg

Force required

The same situation drawn as a graph, shown because this page has no JavaScript to animate it.

The numbers behind this graph

Speed (m s⁻¹)	Force (N)	Acceleration (m s⁻²)
8	48	32.00
16	192	128.00
24	432	288.00
32	768	512.00
40	1200	800.00

The speed never changes and the velocity changes constantly. The arrow drawn from the centre keeps its length and sweeps around, which is the whole distinction in one picture.

Watch the green arrow drawn from the centre. It is a copy of the velocity, and it keeps a constant length while its direction sweeps round. That is uniform circular motion in one picture: a fixed speed and a velocity that is never the same twice. The red arrow, the net force, always points at the centre and is always at right angles to the velocity.

CheckpointAnswer before reading on.

A satellite moves around the Earth in a circular orbit at a constant speed.

Which statement about the satellite is correct?

AIts velocity is changing and it is accelerating towards the centre of the orbit.
BIts velocity is constant, so its acceleration is zero.
CIt is accelerating in its direction of motion, which is why it keeps moving.
DIt is accelerating away from the Earth, balanced by gravity pulling it inwards.

Hint 1Speed and velocity are different quantities. Ask which one the question says is constant.

Hint 2Velocity has a direction. If the direction changes, the velocity changes, even when the speed does not.

Hint 3Acceleration is the rate of change of velocity. Is the velocity changing? Which way is the change pointing?

CheckpointAnswer before reading on.

A ball on a string is swung in a horizontal circle on a smooth, frictionless table. At the instant shown the ball is at the most northerly point of the circle and moving east. The string suddenly breaks.

Which describes the path of the ball after the string breaks?

AA straight line due east, along the tangent to the circle at that point.
BA straight line due north, directly away from the centre.
CA curve that gradually straightens out, continuing to bend a little towards the south.
DA straight line to the north east, between the tangent and the outward direction.

Hint 1Once the string breaks, what horizontal forces act on the ball?

Hint 2With no net force, Newton's first law says the velocity cannot change.

Hint 3At the moment of breaking, the ball's velocity is along the tangent to the circle. Which direction is that here?

Where $v^{2} / r$ comes from

The formulae sheet gives $a_{c} = v^{2} / r$ without explanation. The argument behind it is short, and knowing it makes the formula hard to misremember.

Derivation

Centripetal acceleration

Starts from

the geometry of the position and velocity vectors over a short time interval

Ends at

a_{c} = \frac{v ^{2}}{r}

Holds only if

The speed is constant, so the motion is uniform circular motion
The time interval is taken to be very short, so a short arc and its chord are the same length

Centripetal acceleration

4 steps

1
Follow the object for a short time
Acceleration is a rate of change, so we need to compare the velocity at two instants close together.
In a short time $Δ t$ the object moves a distance $v Δ t$ around the circle, and the radius to it turns through a small angle $Δ θ$ .
2
Notice the velocity turns through the same angle
This is the key observation. The velocity is always perpendicular to the radius, so when the radius turns by an angle, the velocity turns by the same angle.
Draw the two velocity arrows tail to tail. Both have length $v$ , and the angle between them is $Δ θ$ . The arrow joining their tips is the change in velocity, $Δ v$ , and it points towards the centre.
3
Compare two similar triangles
The position triangle and the velocity triangle are both isosceles with the same apex angle, so their sides are in proportion.
The position triangle has two sides $r$ and a short third side $v Δ t$ . The velocity triangle has two sides $v$ and a short third side $Δ v$ . Similar triangles give
$\frac{Δ v}{v} = \frac{v Δ t}{r}$
4
Divide by the time
Change in velocity over time is the acceleration, which is what we wanted.
$a_{c} = \frac{Δ v}{Δ t} = \frac{v ^{2}}{r}$
One factor of $v$ comes from how fast the object moves around the circle, and the other from how large the velocity is that is being turned. That is why speed enters squared.

Centripetal accelerationon the NESA formulae sheet

a_{c} = \frac{v ^{2}}{r}

Symbols: $a_{c}$ acceleration directed at the centre of the circlem s⁻²
$v$ speed around the circlem s⁻¹
$r$ radius of the circular pathm
Valid when: The object moves on a circular path at an instant when its speed is v. For uniform circular motion it holds everywhere on the path and the acceleration is entirely towards the centre.
Not valid when: The path is not a circle of the stated radius. When the speed is also changing, this gives only the part of the acceleration pointing at the centre, and a second, tangential part exists alongside it.
Rearranged: for v: $v = a_{c} r$ for r: $r = \frac{v ^{2}}{a _{c}}$
Where it turns up: Finding how hard an object is being pulled towards the centre, as a multiple of g, before any mass is known
Showing that an object moving at constant speed is nonetheless accelerating, which is the conceptual core of the topic
The first half of a centripetal force calculation, since multiplying by the mass gives the force
Where marks go missing: Forgetting to square the speed
Describing the acceleration as outwards, or as along the direction of motion
Concluding that an object at constant speed has zero acceleration

Where this comes from

Centripetal forceon the NESA formulae sheet

F_{c} = \frac{m v ^{2}}{r}

Symbols: $F_{c}$ net force towards the centre needed to keep the object on the circleN
$m$ mass of the objectkg
$v$ speed around the circlem s⁻¹
$r$ radius of the circular pathm
Valid when: The object is moving in a circle of radius r at speed v. The result is the size of the net force towards the centre, whatever real forces happen to supply it.
Not valid when: It is treated as an extra force to add to a free body diagram. It is never a separate force: it is the name for the resultant of the real forces, such as tension, friction, gravity or a normal force, along the radius.
Rearranged: for v: $v = \frac{F _{c} r}{m}$ for r: $r = \frac{m v ^{2}}{F _{c}}$ for m: $m = \frac{F _{c} r}{v ^{2}}$
Where it turns up: Finding the friction a car's tyres must supply on a flat bend, and from that the fastest safe speed
Finding the tension in a string holding a mass in a horizontal circle
Predicting how the required force changes when the speed, mass or radius is changed, which is the investigation in the first dot point
Where marks go missing: Drawing a force labelled centripetal force on a free body diagram alongside the tension or friction that is actually providing it
Saying that the force required doubles when the speed doubles, when it quadruples
Using a speed in km per hour without converting to metres per second

How the force depends on mass, speed and radius

The first dot point asks you to investigate the relationships, so it is worth reading $F_{c} = m v^{2} / r$ one variable at a time.

Mass. Double the mass and you double the force needed. A loaded truck needs twice the grip of the same truck empty on the same bend at the same speed.

Speed. Double the speed and you need four times the force. This is the single most examined feature of the relationship, and the reason speed limits on bends are so much lower than on the straight.

Radius. Halve the radius and you double the force. A tighter turn at the same speed is a harder turn, but the dependence is gentler than on speed.

Force required

The shape is the answer to why cars come off bends. Speed enters squared, so a bend taken 40 per cent faster needs twice the grip.

The numbers behind this graph

Speed (m s⁻¹)	Force (N)	Acceleration (m s⁻²)
8	1536	1.28
16	6144	5.12
24	13824	11.52
32	24576	20.48
40	38400	32.00

Mass1200.0 kg

Radius50 m

Speed marked15.0 m s⁻¹

Force at the marked speed: 5400 N
Centripetal acceleration there: 4.50 m s⁻²
As a fraction of the weight: 0.46 g
Period of one revolution: 20.9 s

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

Force required

Radius enters to the first power and speed enters squared, which is why halving the radius is a milder change than doubling the speed.

Mass1200.0 kg

Speed15.0 m s⁻¹

Radius marked50 m

Force at the marked radius: 5400 N
Centripetal acceleration there: 4.50 m s⁻²
Period of one revolution: 20.9 s

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

CheckpointAnswer before reading on.

A car rounds a bend at a certain speed. It then rounds a second bend with half the radius, travelling at twice the speed.

By what factor has the centripetal force on the car changed?

AIt is 8 times as large.
BIt is 4 times as large.
CIt is unchanged.
DIt is 2 times as large.

Hint 1Write down the centripetal force relationship and note the power of each variable.

Hint 2Speed is squared in $F_{c} = m v^{2} / r$ ; radius is not.

Hint 3Doubling $v$ multiplies the force by $2^{2}$ . Halving $r$ divides by $\frac{1}{2}$ , which multiplies by $2$ .

The practical: a whirling stopper

The usual school investigation whirls a rubber stopper on a string that passes through a vertical tube, with masses hanging from the lower end. The tension in the string pulls the stopper towards the centre, and the hanging weight $M g$ sets that tension. A clip on the string holds the radius fixed.

For each hanging mass, time a number of revolutions, find the period, and get the speed from $v = 2 π r / T$ . If $F_{c} = m v^{2} / r$ holds, a graph of $M g$ against $v^{2}$ is a straight line through the origin with gradient $m / r$ , where $m$ is the stopper's mass. Changing the radius and keeping the force fixed tests the other variable.

Two things decide whether the result is worth anything. First, time ten or twenty revolutions, not one, because reaction time is a large fraction of a single period. Second, check the gradient against $m / r$ calculated from the measured stopper mass and radius. A straight line with the wrong gradient does not support the relationship.

The standard limitation is that the string cannot be exactly horizontal. The stopper's weight pulls it down, so only the horizontal part of the tension points at the centre, and the true radius is a little less than the length of string. Friction at the lip of the tube also takes some of the hanging weight. Both make $M g$ larger than the true centripetal force.

Exam question

Harder · about 9 min

5 marks

A student whirls a rubber stopper of mass $0.015$ kg in a horizontal circle on a string that passes through a vertical glass tube. Masses hung from the bottom of the string keep it taut, and a marker keeps the radius at $0.75$ m. The student times ten revolutions for each hanging mass:

$M = 0.050$ kg: $9.6$ s
$M = 0.100$ kg: $6.8$ s
$M = 0.150$ kg: $5.5$ s
$M = 0.200$ kg: $4.8$ s

Explain how this data can be processed and graphed to evaluate the relationship $F_{c} = \frac{m v ^{2}}{r}$ , what the graph should show if the relationship holds, and identify one limitation of the method and its effect on the results.

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1What provides the centripetal force on the stopper? How big is it for each hanging mass?

Hint 2The speed comes from the period: $v = 2 π r / T$ , with $T$ one tenth of the time recorded.

Hint 3If $F = m v^{2} / r$ , then $F$ against $v^{2}$ is a straight line through the origin with gradient $m / r$ .

Where students lose marks on this one

Plotting $F$ against $v$ and describing the curve as confirming the relationship.
Why it happens: The speed is the quantity calculated, so it is plotted as it comes.
A curve can fit many relationships. Plotting against $v^{2}$ gives a specific prediction, a straight line through the origin with a known gradient, which the data can fail.
Using the mass of the stopper to calculate the centripetal force.
Why it happens: The stopper is the object in circular motion, so its mass seems to be the one that matters.
The stopper's mass sets how much force is needed. The hanging mass sets how much force is supplied. The experiment tests whether the two agree.

Written for this site.

Period, frequency and angular velocity

Orbital speed from the periodon the NESA formulae sheet

v = \frac{2 π r}{T}

Symbols: $v$ speed around the circle, constant in uniform circular motionm s⁻¹
$r$ radius of the circular path, measured from its centrem
$T$ period, the time for one complete revolutions
Valid when: The object moves around a circle at constant speed, so one circumference is covered in exactly one period.
Not valid when: The speed changes around the path, as it does for a mass swung in a vertical circle, or the radius is taken as a diameter or as the length of a string that is not horizontal.
Rearranged: for T: $T = \frac{2 π r}{v}$ for r: $r = \frac{v T}{2 π}$
Where it turns up: Turning a count of revolutions in a measured time into a speed, the first step in almost every circular motion calculation
Finding the speed of a satellite or a planet from its orbital period
Checking the period predicted by a centripetal force calculation against a stopwatch reading in the practical
Where marks go missing: Using the frequency in revolutions per second where the period is needed, which gives a speed that is too small by a factor of the frequency squared
Using the diameter in place of the radius
Timing a single revolution in a practical, where the reaction time of the timer is a large fraction of the period

Angular velocityon the NESA formulae sheet

ω = \frac{Δ θ}{t}

Symbols: $ω$ angular velocity, the rate at which the angle swept out growsrad s⁻¹
$Δ θ$ angle turned through, in radiansrad
$t$ time taken to turn through that angles
Valid when: The angle is measured in radians. For uniform circular motion the angular velocity is constant, one revolution is 2π radians, and so ω = 2π/T and v = ωr.
Not valid when: The angle is left in degrees or in revolutions, in which case the number produced is not in rad s⁻¹ and the relationship v = ωr gives nonsense.
Rearranged: for \omega: $ω = \frac{2 π}{T}$ for v: $v = ω r$
Where it turns up: Describing a rotating object where every point shares one angular velocity but has its own speed
Converting a rotation rate in revolutions per minute into rad s⁻¹
Linking the rotation rate to the linear speed at a given radius through v = ωr
Where marks go missing: Leaving the angle in degrees
Using revolutions per second as if it were rad s⁻¹, which is out by a factor of 2π
Assuming points at different radii on a rotating disc share a speed, when they share only an angular velocity

Angular velocity describes how fast the angle grows, rather than how fast the object moves along its path. On a rotating disc every point shares one angular velocity, but a point on the rim travels further per revolution than a point near the hub, so it has a larger speed. The link between them is $v = ω r$ , and that only works with the angle in radians.

CheckpointAnswer before reading on.

A child sits $2.5$ m from the centre of a merry go round, which completes one revolution every $4.0$ s.

Calculate the child's speed.

Your answerm s⁻¹

Give it to 2 significant figures.

Hint 1In one period the child travels once around a circle. How far is that?

Hint 2The circumference of a circle of radius $r$ is $2 π r$ .

Hint 3Use $v = 2 π r / T$ with $r = 2.5$ m and $T = 4.0$ s.

CheckpointAnswer before reading on.

A bicycle wheel turns through $3.0$ complete revolutions in $1.5$ s at a steady rate.

Calculate its angular velocity in rad s⁻¹.

Your answerrad s⁻¹

Give it to 3 significant figures.

Hint 1Angular velocity needs the angle in radians, not in revolutions.

Hint 2One revolution is $2 π$ radians.

Hint 3The angle turned is $3.0 \times 2 π$ rad. Divide by the time.

Naming the force: a mass on a string

Every situation in the second dot point is solved the same way. Draw the real forces. Decide which way the centre is. Set the net force in that direction equal to $m v^{2} / r$ .

Worked example5 marks

A puck on a string

A $0.25$ kg puck is attached to a string and moves in a horizontal circle of radius $0.80$ m on a frictionless air table. It completes $3.0$ revolutions in $2.0$ s. Find its period, speed, angular velocity, centripetal acceleration and the tension in the string.

Name the forces and find the centre

The whole topic comes down to identifying which real force points at the centre. Doing it first stops a made up force appearing later.

Vertically, the weight is balanced by the upward push of the air cushion, so nothing happens vertically. Horizontally, only the tension acts, pointing along the string to the centre. So the tension is the centripetal force: $T_{string} = m v^{2} / r$ .

Get the period and the speed

Most circular motion data arrives as a count of revolutions in a time. Converting it to a period comes before any other formula.

$T = \frac{2.0}{3.0} = 0.667 s v = \frac{2 π r}{T} = \frac{2 π \times 0.80}{0.667} = 7.54 m s^{- 1}$

Angular velocity, with the angle in radians

Three revolutions is not three radians. The conversion is where this quantity usually goes wrong.

$ω = \frac{Δ θ}{t} = \frac{3.0 \times 2 π}{2.0} = 9.42 rad s^{- 1}$

A check: $v = ω r = 9.42 \times 0.80 = 7.54$ m s⁻¹, which agrees.

Acceleration, then the force

Acceleration first means the answer to a question that gives no mass is already on the page.

$a_{c} = \frac{v ^{2}}{r} = \frac{7.5 4 ^{2}}{0.80} = 71 m s^{- 2}$

$T_{string} = m a_{c} = 0.25 \times 71 = 18 N$

Answer

Period $0.67$ s, speed $7.5$ m s⁻¹, angular velocity $9.4$ rad s⁻¹, centripetal acceleration $71$ m s⁻² towards the centre, and tension $18$ N.

Is that answer sensible?

An acceleration of $71$ m s⁻² is over seven times $g$ , which sounds huge until you notice the puck goes round one and a half times every second on a short string. Swing anything that fast by hand and you feel the string pulling hard. The two routes to the speed agreeing is the internal check.

Cars on flat bends

On a flat road the weight and the normal force are both vertical. They balance, and neither of them can turn a car. The only horizontal force available is friction between the tyres and the road, acting sideways, towards the centre of the bend.

It is static friction, not sliding friction, because the tyres are not skidding sideways. And static friction has a maximum. If the bend asks for more than the tyres can supply, the car cannot follow the bend and it runs wide.

Worked example5 marks

How fast can the car take the corner?

A $1200$ kg car rounds a flat bend of radius $45$ m at $54$ km h⁻¹. Find the friction force needed, and the greatest speed at which it could take the bend if the tyres can supply at most $7.5$ kN sideways.

Convert the speed

Every speed on a road is quoted in km per hour and every formula needs metres per second. Because the speed is squared, forgetting this changes the answer by a factor of thirteen.

$54 \div 3.6 = 15 m s^{- 1}$

Name the force, then calculate it

The question says friction, but in an exam you will often have to say it yourself, and that identification is a mark.

The friction from the road is the centripetal force.

$a_{c} = \frac{1 5 ^{2}}{45} = 5.0 m s^{- 2} F = m a_{c} = 1200 \times 5.0 = 6.0 \times 1 0^{3} N$

Set the friction to its maximum and solve for speed

A limiting case question always works the same way: set the force to its largest possible value and ask what speed that allows.

$\frac{m v ^{2}}{r} = 7500 \Rightarrow v = \frac{7500 \times 45}{1200} = 16.8 m s^{- 1}$

Answer

The tyres must supply $6.0$ kN of friction towards the centre. The greatest safe speed is $16.8$ m s⁻¹, about $60$ km h⁻¹.

Is that answer sensible?

The limit is only a little above the actual speed, which fits: the car was already using $6.0$ of the available $7.5$ kN. A $12$ per cent increase in speed needs $25$ per cent more force, because force goes with speed squared.

Common mistake

Drawing a force labelled centripetal force on the free body diagram, alongside friction or tension.

Why it happens: The formula has a named force in it, so it feels like a force that should be drawn.

The centripetal force is the net force towards the centre. It is what the real forces add up to, not an extra one. Drawing both counts the same push twice, and the diagram then implies an acceleration twice as big as the real one.

A free body diagram shows forces that some other object exerts: the Earth, a string, the road. Label them with what they are. Then write, beside the diagram, that their resultant towards the centre equals $m v^{2} / r$ .

CheckpointAnswer before reading on.

A car turns sharply left at constant speed. A passenger on a slippery leather seat slides towards the right hand door until the door stops them.

Which statement best explains the passenger's motion?

AWith almost no sideways force acting, the passenger keeps moving in a straight line while the car turns left, until the door pushes them left, towards the centre of the bend.
BA centrifugal force pushes the passenger outwards, to the right, away from the centre of the bend.
CThe centripetal force on the passenger acts to the right, so they accelerate to the right.
DThe car's turning transfers momentum to the passenger in the opposite direction, by Newton's third law.

Hint 1Describe the motion from outside the car, as someone standing on the footpath would see it.

Hint 2Before the door touches them, what horizontal forces act on the passenger, given the seat is slippery?

Hint 3With almost no horizontal force, the passenger keeps moving in a straight line while the car turns left underneath them.

Exam question

Harder · about 8 min

5 marks

A car of mass $1500$ kg travels around a flat, horizontal bend of radius $60$ m at a constant $72$ km h⁻¹.

(a) Identify the force that provides the centripetal force on the car. (1 mark)

(b) Calculate the size of this force. (2 marks)

(c) On a wet day the greatest sideways friction the tyres can supply on this bend is $12$ kN. Calculate the greatest speed at which the car can safely round the bend. (2 marks)

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1On a flat road, the weight and the normal force are both vertical. Which force is horizontal and points towards the centre?

Hint 2Convert the speed to m s⁻¹ before substituting. Divide km h⁻¹ by $3.6$ .

Hint 3For (c), set the centripetal force equal to the maximum friction and solve $F = m v^{2} / r$ for $v$ .

Where students lose marks on this one

Substituting $72$ directly as the speed and obtaining $1.3\times 10^{5}$ N.
Why it happens: The speed is given in the units used on road signs and is substituted as printed.
Every quantity goes into $F = m v^{2} / r$ in SI units. Convert km h⁻¹ to m s⁻¹ by dividing by $3.6$ , and because the speed is squared, an unconverted speed puts the answer out by a factor of about thirteen.
Naming the centripetal force itself as the answer to (a).
Why it happens: The centripetal force is treated as a force in its own right rather than a name for the net inward force.
The question asks which real force does the job. Name the interaction: friction from the road on the tyres.

Written for this site.

Banked tracks

Relying on friction is a problem when the road is wet. A banked bend tilts the road so the outside edge is higher, and that lets the normal force do some or all of the turning.

The normal force always acts perpendicular to the surface. On a tilted surface it tilts inwards, towards the centre of the bend, so it gains a horizontal component. At exactly the right speed, that component is the whole centripetal force and no friction is needed at all.

The forces on a car on a banked bend, at the design speed

1The normal force, perpendicular to the track
The road pushes perpendicular to its own surface. Because the surface is tilted by $θ$ , the push is tilted by $θ$ from the vertical, inwards. This single tilt is what banking achieves: it gives a force that already existed a sideways part.
2Vertical part: balances the weight
The car does not rise or sink, so the vertical forces balance: $N cos θ = m g$ . Notice this makes the normal force larger than the weight, by a factor of $1/ cos θ$ . The inclined plane result $N = m g cos θ$ does not apply here, because the car is not accelerating along the slope.
3Horizontal part: the centripetal force
The only horizontal force is this component of the normal force, and it points at the centre. So it is the centripetal force: $N sin θ = m v^{2} / r$ .
4The weight, straight down
The weight is vertical, so it has no horizontal part and does not help turn the car. Its only role is to fix how large the normal force must be.
5The banking angle
The angle between the track and the horizontal is the same as the angle between the normal force and the vertical. That equality is what lets the normal force be resolved with $sin θ$ horizontally and $cos θ$ vertically.
6Away from the design speed: friction
Slower than the design speed, the horizontal part of $N$ is more than is needed, and the car tends to slide down the bank, so friction acts up the slope. Faster, it is not enough, the car tends to slide up and out, and friction acts down the slope, adding to the inward force.

The centre of the bend is to the left. The track rises towards the outside. With no friction, two forces act, and the normal force is resolved horizontally and vertically, not along and across the slope.

Derivation

The design speed of a banked track

Starts from

the two force equations for a car on a banked track with no friction

Ends at

tan θ = \frac{v ^{2}}{r g}

Holds only if

The car moves in a horizontal circle of radius r
No friction acts, so the weight and the normal force are the only forces
The speed is constant

The design speed of a banked track

3 steps

1
Resolve vertically
The car's acceleration is horizontal, so there is no vertical acceleration and the vertical forces must balance.
$N cos θ = m g$
2
Resolve horizontally
The net horizontal force points at the centre and must equal the centripetal force.
$N sin θ = \frac{m v ^{2}}{r}$
3
Divide to eliminate the normal force
N is unknown and not asked for. Dividing one equation by the other removes it in a single step, and removes the mass with it.
$\frac{N s i n θ}{N c o s θ} = \frac{m v ^{2} / r}{m g} \Rightarrow tan θ = \frac{v ^{2}}{r g}$
The mass has cancelled. A banked bend is designed for a speed, not for a vehicle, and a truck and a motorbike share the same design speed.

Design speed of a banked tracknot on the formulae sheet, learn it

tan θ = \frac{v ^{2}}{r g}

Symbols: $θ$ banking angle of the track, measured from the horizontal°
$v$ the one speed at which no sideways friction is neededm s⁻¹
$r$ radius of the bendm
$g$ gravitational field strength, 9.8 near the Earth's surfacem s⁻²
Valid when: The vehicle moves in a horizontal circle on a surface banked at θ, and the only forces are its weight and the normal force. It gives the single speed for which the horizontal part of the normal force is exactly the centripetal force.
Not valid when: The vehicle travels faster or slower than the design speed, where friction must act along the slope and the two force equations each gain a friction term. It also assumes the circle is horizontal, not the slope.
Rearranged: for v: $v = r g tan θ$ for \theta: $θ = tan^{- 1} (\frac{v ^{2}}{r g})$ for r: $r = \frac{v ^{2}}{g t a n θ}$
Where it turns up: Designing the banking angle of a road or velodrome for a chosen speed
Explaining why a banked bend is safer than a flat one at the speed it was built for
Showing that the design speed does not depend on the vehicle's mass
Where marks go missing: Resolving the weight into components along and across the slope, as for an inclined plane, rather than resolving the normal force horizontally and vertically
Writing N = mg cos θ, which is the inclined plane result and is wrong here because the acceleration is horizontal, not along the slope
Quoting the equation without the derivation when a question asks for it to be shown

Where this comes from

Worked example4 marks

A banked highway bend

A highway bend of radius $120$ m is banked at $15°$ . Find the speed at which a car needs no friction to take the bend. For a $1200$ kg car, find the normal force at that speed, and explain what happens at $25$ m s⁻¹.

Design speed from the derived result

The derivation has already been done, so it can be applied directly. In an exam that says show, the derivation itself would be the answer.

$v = r g tan θ = 120 \times 9.8 \times tan 15° = 315 = 17.7 m s^{- 1}$

That is about $64$ km h⁻¹.

Normal force from the vertical balance

The vertical equation contains only N and known quantities, so it gives N directly.

$N = \frac{m g}{c o s θ} = \frac{1200 \times 9.8}{c o s 15°} = 1.22 \times 1 0^{4} N$

Its horizontal part is $N sin 15° = 3.15 \times 1 0^{3}$ N, and $m v^{2} / r = 1200 \times 315/120 = 3.15 \times 1 0^{3}$ N. They match, as they must at the design speed.

Compare the force needed at the higher speed

Above the design speed the normal force alone cannot supply enough. The question is how much is missing and what supplies it.

$\frac{m v ^{2}}{r} = \frac{1200 \times 2 5 ^{2}}{120} = 6.25 \times 1 0^{3} N$

The normal force supplies only about half of this. The car tends to slide up and out, so friction acts down the slope to provide the rest.

Answer

The design speed is $17.7$ m s⁻¹ (about $64$ km h⁻¹), where the normal force is $12$ kN. At $25$ m s⁻¹ roughly $6.3$ kN of inward force is needed and friction down the slope must provide the part that the normal force does not.

Is that answer sensible?

A gentle $15°$ bank on a highway bend designed for around sixty kilometres per hour is the right scale for a real road. The normal force being slightly larger than the weight, by about three and a half per cent, fits a small banking angle.

Speed needing no friction

A banking angle does not make a bend safe at all speeds. It picks one speed, and friction covers the difference either side of it.

Radius of the bend120 m

Banking angle marked15 °

Design speed: 17.75 m s⁻¹
The same in km per hour: 64 km h⁻¹
Centripetal acceleration there: 2.63 m s⁻²
Mass appears nowhere: the design speed is the same for a car and a truck

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

CheckpointAnswer before reading on.

A curve on a racing circuit has a radius of $80$ m and is banked at $20°$ to the horizontal.

Calculate the speed at which a car can round this curve with no sideways friction from the track. Take $g = 9.8$ m s⁻².

Your answerm s⁻¹

Give it to 2 significant figures.

Hint 1At the design speed, only two forces act: the weight and the normal force.

Hint 2The vertical part of the normal force balances the weight. The horizontal part is the centripetal force.

Hint 3Dividing those two equations gives $tan θ = v^{2} / r g$ . Rearrange for $v$ .

Analysetypically 4 to 7 marks

Demands: Break the situation into its parts and show how they relate and affect each other.
Shape: Identify the components, then the relationships, then what follows from them.
Loses marks: Summarising the stimulus rather than taking it apart.

Through a marker’s eyes

3 marks

Analyse the forces acting on a car travelling around a banked curve at its design speed. (3 marks)

The attempt

1 out of 3

The car has its weight acting down and the normal force acting up. There is also a centripetal force acting towards the centre, which keeps the car on the curve. The banking helps the car go around the corner.

What the marker sees

The response earns a mark for naming the weight and the normal force. It then loses the other two.

Saying the normal force acts up is wrong: it acts perpendicular to the banked surface, which is what gives it a horizontal part. And the centripetal force is listed as a third force, when it is really the horizontal part of the normal force. The response never says how the forces relate to each other, which is what analyse asks for. The banking helps is a description with no mechanism.

The same answer, fixed

3 out of 3

Two forces act on the car: its weight $m g$ vertically down, and the normal force $N$ perpendicular to the banked surface, tilted towards the centre of the curve at the banking angle $θ$ from the vertical.
Vertically the car does not accelerate, so the vertical component of the normal force balances the weight: $N cos θ = m g$ .
The horizontal component of the normal force, $N sin θ$ , is the only horizontal force, and it points towards the centre. It therefore provides the whole centripetal force, $N sin θ = m v^{2} / r$ , so no friction is required at the design speed. Combining the two gives $tan θ = v^{2} / r g$ .

Exam question

Harder · about 6 min

4 marks

A car travels around a bend of radius $r$ that is banked at an angle $θ$ to the horizontal. At one particular speed $v$ the car needs no friction from the road to follow the bend.

Using an analysis of the forces on the car, show that $tan θ = \frac{v ^{2}}{r g}$ .

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1With no friction, only two forces act. Name them and give their directions.

Hint 2The car moves in a horizontal circle, so resolve horizontally and vertically, not along and across the slope.

Hint 3Vertically the forces balance. Horizontally the net force is $m v^{2} / r$ . Divide one equation by the other.

Where students lose marks on this one

Writing $N=mg\cos\theta$.
Why it happens: It is the result for a block on an inclined plane and is remembered as a general fact about slopes.
On an inclined plane the acceleration is along the slope, so the forces across the slope balance. On a banked bend the acceleration is horizontal, so it is the vertical forces that balance, giving $N cos θ = m g$ and a normal force larger than the weight.

Written for this site.

Work and energy

The fourth dot point asks about the relationship between the total energy and the work done on an object in uniform circular motion. The answer is short and it catches people because they expect a force to do something to the energy.

Why the centripetal force does no work

AnswersA force acts the whole time. Why does it never change the kinetic energy?

Work is $W = F s cos θ$ , where $θ$ is the angle between the force and the displacement. Only the part of a force along the direction of motion does work.

In uniform circular motion the net force points at the centre and the motion is along the tangent. Those are at right angles at every instant, so $cos θ = 0$ and no work is done, over any part of the path.

No work means no change in kinetic energy, which is the same statement as the speed staying constant. The force changes the direction of the velocity and never its size. The total energy of the object is constant: its kinetic energy does not change, and in a horizontal circle, or a circular orbit, neither does its potential energy.

CheckpointAnswer before reading on.

A $0.50$ kg mass on a string moves in a horizontal circle of radius $1.2$ m at a constant speed on a frictionless table. The tension in the string is $15$ N.

How much work does the tension do on the mass during half a revolution?

A $0$ J
B $57$ J
C $36$ J
DIt cannot be found without the speed.

Hint 1Work is $W = F s cos θ$ , where $θ$ is the angle between the force and the displacement.

Hint 2At every instant the tension points at the centre and the mass moves along the tangent.

Hint 3What is $cos 90°$ ? And does the mass's kinetic energy change?

Exam question

Standard · about 4 min

3 marks

A net force acts on an object in uniform circular motion at every instant. Explain why the object's kinetic energy nevertheless stays constant.

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1Kinetic energy changes only when net work is done.

Hint 2Work depends on the angle between the force and the displacement.

Hint 3The net force points along the radius. Which way is the object moving?

Where students lose marks on this one

Saying the force does no work because the object returns to where it started.
Why it happens: Zero displacement over a full revolution looks like zero work.
That argument fails for half a revolution, where the displacement is a diameter, and the kinetic energy is still unchanged. The real reason works for any part of the path: the force is perpendicular to the motion at every instant.

Written for this site.

Torque

The final dot point is about what sets things rotating in the first place. A force applied to a body that can turn about a pivot produces a turning effect, the torque.

Torqueτ, N m: The turning effect of a force about a pivot, equal to the distance from the pivot to where the force acts, multiplied by the component of the force perpendicular to that distance.

Torqueon the NESA formulae sheet

τ = r_{⊥} F = r F sin θ

Symbols: $τ$ torque, the turning effect of the force about the pivotN m
$r$ distance from the pivot to where the force is appliedm
$F$ size of the applied forceN
$θ$ angle between the force and the line from the pivot to its point of application°
$r_{⊥}$ perpendicular distance from the pivot to the line of action of the forcem
Valid when: A single force acts on a body that can rotate about a fixed pivot or axis. The angle is the one between the force and the lever arm, not between the force and the horizontal.
Not valid when: The angle is measured from something other than the lever arm, or several forces act and only one torque is calculated. Where several forces act, their torques are combined, with clockwise and anticlockwise given opposite signs.
Rearranged: for F: $F = \frac{τ}{r s i n θ}$ for r: $r = \frac{τ}{F s i n θ}$
Where it turns up: Explaining why a door handle is placed far from the hinge and why a longer spanner loosens a tight nut
Finding the force needed at a given point to produce a required torque
The torque on a current carrying coil in a magnetic field in Module 6, which is this same relationship applied to the motor
Where marks go missing: Using cos θ where the angle given is between the force and the lever arm
Measuring the distance from the wrong point, such as from the end of the object rather than the pivot
Giving the unit as joules; torque and work share N m dimensionally but torque is written N m

The two forms are the same thing seen two ways. $r F sin θ$ keeps the full distance and uses only the perpendicular part of the force. $r_{⊥} F$ keeps the full force and uses only the perpendicular distance to its line of action. Both give the same number.

Torque produced

Pushing a spanner along its own length achieves nothing. Only the perpendicular part of a force turns anything.

Distance from the pivot0.30 m

Size of the force150 N

Angle marked60 °

Torque at the marked angle: 38.97 N m
Greatest possible torque: 45.00 N m at 90 degrees
Fraction of the best: 87 per cent
Perpendicular part of the force: 129.9 N

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

This is why a door handle sits at the edge furthest from the hinges, why a long spanner shifts a stuck nut, and why pushing a door towards its hinges does nothing at all.

Worked example3 marks

Tightening a wheel nut

A wheel nut must be tightened to a torque of $40$ N m using a spanner $0.20$ m long. Find the smallest force that will do this, and the force needed if it is applied at $60°$ to the spanner.

The smallest force acts perpendicular

sin θ is at most 1, at 90°, so for a fixed torque the force is least when it is perpendicular to the lever arm.

$F = \frac{τ}{r} = \frac{40}{0.20} = 200 N$

At an angle, only the perpendicular part counts

The other part of the force pushes along the spanner into the nut and turns nothing, so more total force is needed.

$F = \frac{τ}{r s i n θ} = \frac{40}{0.20 \times s i n 60°} = 231 N$

Answer

At least $200$ N, applied perpendicular to the spanner at its end. At $60°$ the force must rise to $231$ N.

Is that answer sensible?

Two hundred newtons is about the weight of a $20$ kg mass, which is a firm but achievable push for an adult. And the angled force being only about $15$ per cent larger fits: $sin 60°$ is close to one.

CheckpointAnswer before reading on.

A door is $0.90$ m wide and hinged along one edge. Four people each push on it with a force of $40$ N.

Which push produces the greatest torque about the hinges?

AAt the outer edge, perpendicular to the door.
BHalfway across the door, perpendicular to it.
CAt the outer edge, at $30°$ to the surface of the door.
DAt the outer edge, pushing along the door towards the hinge.

Hint 1Torque depends on the distance from the pivot and on the angle.

Hint 2The angle that gives the most torque is the one where $sin θ$ is largest.

Hint 3Calculate $r F sin θ$ for each option and compare.

CheckpointAnswer before reading on.

A mechanic pulls on the end of a spanner $0.25$ m long with a force of $120$ N. The force makes an angle of $70°$ with the spanner.

Calculate the torque applied to the nut.

Your answerN m

Give it to 2 significant figures.

Hint 1Only the part of the force perpendicular to the spanner turns the nut.

Hint 2The angle in $τ = r F sin θ$ is the angle between the force and the lever arm, which is what the question gives.

Hint 3 $τ = 0.25 \times 120 \times sin 70°$ .

Vertical circles

The syllabus is about uniform circular motion, but a mass on a string swung in a vertical circle appears in examinations as a force analysis, and the method is the one used all through this lesson.

The difference is that the weight now lies along the radius at the top and the bottom. At the bottom the centre is above, so the tension must exceed the weight:

$T - m g = \frac{m v ^{2}}{r}$

At the top the centre is below, and the weight helps:

$T + m g = \frac{m v ^{2}}{r}$

This is why a string is most likely to break at the bottom of a vertical circle, and why there is a minimum speed at the top. A string can pull but cannot push, so the smallest possible tension is zero, and at that point the weight alone must be the centripetal force: $v_{m i n} = g r$ . Any slower and gravity pulls the object inside the circle.

Strictly, the speed in a vertical circle is not constant, because gravity does work on the way down and against the way up. The equations above still hold at each instant, using the speed at that point.

CheckpointAnswer before reading on.

A bucket of water is swung in a vertical circle of radius $0.90$ m, so that at the top of the circle the bucket is upside down.

Calculate the minimum speed the bucket must have at the top of the circle if the water is to stay in it. Take $g = 9.8$ m s⁻².

Your answerm s⁻¹

Give it to 2 significant figures.

Hint 1At the top of the circle, which way is the centre? Which forces on the water point that way?

Hint 2The water needs a net downward force of $m v^{2} / r$ at the top. Its weight supplies some or all of that; the base of the bucket can push down and supply the rest.

Hint 3At the minimum speed the bucket's base only just stops pushing, so the weight alone is the centripetal force: $m g = m v^{2} / r$ .

Centrifugal force, properly

Students are told centrifugal force is not real, and then feel it every time a car turns. Both are right, and the resolution is worth understanding.

Seen from the footpath, a frame that is not accelerating, there is no outward force on a passenger. They tend to go straight, the car turns under them, and the door pushes them inwards. Every force has an object that exerts it.

Seen from inside the turning car, the passenger appears to be pushed outwards by nothing. Newton's laws only hold in non accelerating frames, so to use them in a rotating frame you have to add a fictitious outward force, $m v^{2} / r$ , to make the accounting work. That is the centrifugal force. It is a correction for doing physics from a turning viewpoint, not an interaction between two objects.

The HSC course works in the footpath frame throughout. In an examination, an outward force on a free body diagram is marked wrong.

Where this reappears

Circular motion is the first topic where the same relationship keeps coming back with a different force in the role of centripetal force.

Later in this module, gravity holds a satellite in orbit, and setting $GM m / r^{2} = m v^{2} / r$ gives the orbital speed and, with $v = 2 π r / T$ , Kepler's third law. In Module 6, a charged particle moving across a uniform magnetic field feels a force $q v B$ that is always perpendicular to its velocity, so it moves in a circle with $q v B = m v^{2} / r$ . The torque relationship returns there too, as the turning effect on a current carrying coil in a motor.

The habit that carries across all of them is the one built here: find the real force that points at the centre, and set it equal to $m v^{2} / r$ .

The short version

Constant speed, changing velocity

Why an object at constant speed can be accelerating

Where v2/r comes from

Centripetal acceleration

How the force depends on mass, speed and radius

The practical: a whirling stopper

Period, frequency and angular velocity

Naming the force: a mass on a string

A puck on a string

Cars on flat bends

How fast can the car take the corner?

Banked tracks

The design speed of a banked track

A banked highway bend

Work and energy

Why the centripetal force does no work

Torque

Tightening a wheel nut

Vertical circles

Centrifugal force, properly

Where this reappears

Where $v^{2} / r$ comes from