In this episode, you will introduce the importance of circular motion and explain the need for a centripetal force to keep an object moving along a circular path.

Lesson Summary

• Discussion: Observing circular motion (10 minutes)
• Demonstration: Whirling bucket and centripetal force (15 minutes)
• Demonstration: Whirling coin (5 minutes)
• Discussion: Centripetal forces (15 minutes)
• Demonstration: Further demos (10 minutes)

Motion in a circle is an everyday occurrence and the students should be given time to discuss their experiences of such motion. It is important that they should actually feel the force exerted when an object on a string is whirled round their head.

Discussion: Observing circular motion

Invite suggestions of objects which move in a circular path:

• The hammer swung by a hammer thrower
• Clothes being dried in a spin drier
• Chemicals being separated in a centrifuge
• Cornering in a car or on a bike
• A stone being whirled round on a string
• A plane looping the loop
• A DVD, CD or record spinning on its turntable
• Satellites moving in orbits around the Earth
• A planet orbiting the sun (almost circular orbit for many)
• Many fairground rides
• An electron in orbit about a nucleus

Remember that motion in a circle is only a special case of motion in a curve. So why do we study it? It is fairly common and the maths is easier!

Demonstration: Whirling bucket and centripetal force

The whirling bucket is the classic centripetal force experiment. Put a little water in a bucket – tie a string firmly to the bucket handle and then swing the bucket in a vertical circle. As long as the rate of rotation is great enough the water stays in the bucket! Slowing the rate of rotation can get the water to almost fall out at the top of the path and you can usually hear it slopping around at this critical point.

Make sure that the handles of the bucket do not come off and that the bucket does not hit the floor at the lowest point of the circle, or the ceiling. Many extensions of this are possible such as swinging a tray loaded with beakers by four strings!

This is best done outside; in any case avoid the temptation to stand on anything.

Point out that you are pulling on the string to make the bucket go round. The force with which you are pulling on the string provides the centripetal force needed to keep the bucket orbiting.

Check that your students understand the direction in which this force acts – towards your hand. You can only pull string along its length. Explain the name: centripetal means centre seeking. Centripetal is an adjective describing the force; it is not the name of a special type of force, such as tension, gravity, magnetic force etc.

(At the same time, you can feel a force pulling on your hand. This force is the equal and opposite reaction to your pull. Since it acts outwards from the centre of the circle, it can be described as centrifugal , or centre-fleeing. However, you may wish to avoid this term, and ask your students to stick to centripetal to describe an inwards force.)

Demonstration: Whirling coin

Pull open a wire coat hanger so that it forms a square. File the end of the hook flat and then bend the hook until it points towards the opposite corner of the square. Balance a 1p coin on the hook, put one finger in the corner of the square opposite the hook and then spin the coat hanger in a vertical circle – the coin stays in place! This is a very simple but excellent demonstration of centripetal force.

The force of the hook on the penny always acts towards the centre of rotation. Can you beat the record (five 1p pieces stacked on top of each other)? With only one penny balanced and with great care you may be able to bring the coat hanger to rest without the penny falling off.

Discussion: Centripetal forces

What produces the force to keep the object in a circular path?

The actual way the force is produced depends on the particular example:

Planetary orbits (almost!)  →  gravitation

Electron orbits  →  electrostatic force on electron

Centrifuge  →  contact force (reaction) at the walls

Gramophone needle  →  the walls of the groove in the record

Car cornering  →  friction between road and tyres

Car cornering on banked track  →  component of gravity

Aircraft banking  →  horizontal component of lift on the wings

So a centripetal force may be a contact force or electrostatic, magnetic, gravitational, etc.

Why must there be an unbalanced force if an object is to follow a circular path? Emphasise Newton’s First law. If an object is to move in a circle there must be a force pushing or pulling it out of the straight line path. This force must act towards the centre of the circle and it is this that we call the centripetal force. It is the unbalanced force on the orbiting object.

Imagine whirling a stone on the end of a string, so that it follows a horizontal orbit. If you remove the centripetal force by cutting the string, the stone will move off along a straight line along the tangent to the circle (ignoring gravity for a moment) and not along a radius.

Talk to the students about the following two examples of circular motion in practice:

Sitting in the back seat of a car as it corners: If the car turns to the left, you feel as if you are being thrown to the right. In fact, your bum is in contact with the seat, and gets pulled round to the left (providing there is sufficient friction). The upper half of your body tries to carry on in a straight line. Viewed from a point above the car, your upper half will be seen to be trying to follow a tangential path while the car turns to the left.

Watching a marble roll on the surface of a table in a train as the train corners: again, if the train turns to the left, the marble will appear to drift off to the right. It is following a straight line path, tangential to the curve. There is no friction to pull it to the left, so no centripetal force.

An interesting example is a helium-filled balloon inside a cornering car. The balloon leans in towards the centre of the circle. The air in the car tries to continue in a straight line, so it is slewing to the right inside the car. The balloon is lighter than the air, so it gets pushed towards the lower pressure at the centre of the circle.

Demonstration: Further demonstrations

Try out a number of other experiments on circular motion.

Episode 224-1: Demonstrations involving circular motion (Word, 26 KB)

This episode discusses: linear and angular velocity; degrees and radians; and angular acceleration. And is accompanied by worked examples, student question sets, a student experiment and a demonstration.

Lesson Summary

• Discussion: Linear and angular velocity (10 minutes)
• Worked example: Calculating ω (10 minutes)
• Discussion: Degrees and radians (5 minutes)
• Student questions: Calculating v and ω (20 minutes)
• Discussion: Angular acceleration (10 minutes)
• Worked example: Centripetal force (5 minutes)
• Student questions: Calculations on centripetal force (10 minutes)
• Student experiment: Verification of the equation for centripetal force (40 minutes)
• Demonstration: Alternative method of verifying the equation for centripetal force (40 minutes)

Discussion: Linear and angular velocity

Explain the difference between linear and angular velocity.

The instantaneous linear velocity at a point in the circle is usually given the letter v and measured in metres per second (m s^-1).

The angular velocity is the angle through which the radius to this point on the circle turns in one second. This is usually given the letter ω  (Greek omega) and is measured in radians per second (rad s^-1) (See below)

Time period for one rotation:

T = distancevelocity

T = 2 π rv

T = 2 π ω

Therefore linear and angular velocity are related by the formula:

Linear velocity = radius of circle × angular velocity,

v = r ×  ω

Worked examples: Calculating ω

A stone on a string: the stone moves round at a constant speed of 3 m s^-1 on a string of length 0.75 m.

Linear velocity of stone at any point on the circle is 3 m s^-1 directed along a tangent to the point.

Note that although the magnitude of the linear velocity (i.e. the speed) is constant its direction is constantly changing as the stone moves round the circle.

Angular velocity of stone at any point on the circle = 3 m s^-10.75 m

ω  = 4 rad s^-1

Discussion: Degrees and radians

You will have to explain the relationship between degrees and radians. The radian is a more natural unit for measuring angles.

One radian (or rad for short) is defined as the angle subtended at the centre of a circle radius r by an arc of length r.

Thus the complete circumference 2 π r subtends an angle of 2 π rr  radians

Thus in a complete circle of 360 degrees there are 2 π   radians.

Therefore 1 radian = 360 ° 2 π

1 radian = 57.3 °

Student questions: Calculating v and ω

Some radian ideas and practice calculations of v, ω .

Episode 225-1: Radians and angular speed (Word, 45 KB)

Discussion: Angular acceleration

If an object is moving in a circle at a constant speed, its direction of motion is constantly changing. This means that its linear velocity is changing and so it has a linear acceleration. The existence of an acceleration means that there must also be an unbalanced force acting on the rotating object.

Derive the formula for centripetal acceleration ( α = v²r, α = r ω , α = r ω ²):

Consider an object of mass m moving with constant angular velocity ( ω ) and constant speed (v ) in a circle of radius r with centre O.

It moves from P to Q in a time t .

The change in velocity Δ v is parallel to PO and Δ v = v sin( θ )

When θ becomes small (that is when Q is very close to P) sin( θ ) is close to θ in radians.

So Δ v = v θ

Dividing both sides by t gives:

Δ vt = v θ t

Since Δ vt = acceleration and θ t =  ω , we have

α  = v ×  ω

Since we also have

v =  ω  × r ,

this can be written as

α  = v²r

α  = v ×  ω

α  =  ω ² × r

Applying Newton's Second Law ( F = m × a) gives:

F = mv²r

F = mr ω ²

This is the equation for centripetal force; students should learn to identify the appropriate form for use in any given situation.

Worked examples: Centripetal force

A stone of mass 0.5 kg is swung round in a horizontal circle (on a frictionless surface) of radius 0.75 m with a steady speed of 4 m s^-1 .

Calculate:

(a) the centripetal acceleration of the stone

acceleration = v²r

acceleration = (4 m s^-1)²0.75 m

acceleration = 21.4 m s^-2

(b) the centripetal force acting on the stone.

F = m × a

F = 0.5 kg × 21.4 m s^-2

F = 10.7 N

Notice that this is a linear acceleration and not an angular acceleration. The angular velocity of the stone is constant and so there is no angular acceleration.

Student questions: Calculations on centripetal force

Episode 225-2: Centripetal force calculations (Word, 26 KB)

Student experiment: Verification of the equation for centripetal force using the whirling bung

Episode 225-3: Verification of the equation for centripetal force (Word, 28 KB)

A Java applet version of this experiment is available on the National Taiwan Normal University website.

Demonstration: Alternative method of verifying the equation for centripetal force

This demonstration is an alternative method of verifying the equation for centripetal force.

Episode 225-4: Verifying the equation for centripetal force (Word, 44 KB)