#### Share this article:

### Circular motion and centripetal force

- Carbon dioxide puck on a glass plate
- The motion of the Moon around the Earth
- Fine beam tube
- Whirling a rubber bung on a string
- Whirling a rubber bung and letting go
- Penny on a turntable
- Looping the loop
- Sketching a satellite orbit and predicting its period
- Experimental test of F = mv²/R
- Further test of mv²/R
- Test of mv²/R on a turntable
- Estimating the Moon’s orbit time
- Orbits of satellites and moons
- Proof of F = mv²/R
- Centrifugal motion at a fun fair
- What pushes planets along?

## Circular motion and centripetal force

for 14-16

Common sense suggests that any object in circular motion strives constantly to recede from the centre. This observation troubled great thinkers like Descartes and Newton. It took Newton some 20 years to incorporate circular motion into his thinking about forces and motion, as published in *Principia Mathematica*.

Newton’s laws of motion are now commonly taught in school science. It should come as no surprise that students find the idea of a centripetal force difficult.

At the introductory stage, some teachers prefer to use the term 'centrifugal force' because it draws on students’ common belief. Others hold that this start will lead to difficulties. One thing is certain: a mixture of the two is very confusing.

In this collection, the idea of a centripetal force is consistently used. We begin with qualitative experiments that focus students’ attention on the direction of the force causing a circular motion; later experiments are quantitative. Acceleration, *a*, needs a force, *F*, in the same direction, generally expressed as *F* = *ma*. Acceleration in a circle, v2/*R*, is centripetal, and so the force required to cause it is also centripetal.

**Demonstration**

To show the link between constant speed in a straight line and constant speed in circular motion.

Apparatus and Materials

- Carbon dioxide puck (ring with cardboard lid)
- Glass plate
- Wedges, 4
- Dry ice attachment
- String, length about 50cm
- Suction sticker, for central anchor point
- Frame to support multiflash equipment
- Xenon strobe and camera
- Weak spring (OPTIONAL)
- Glass cleaner

Health & Safety and Technical Notes

Xenon strobe: avoid flash frequencies in the range 15-20 Hz, and avoid flickering light, which makes some people can feel unwell. Rarely, some people experience photosensitive epilepsy.

Dry ice is very cold. Handle with thermal gloves, and wear safety spectacles.

Take care when handling the glass plate.

Ethanol (industrial denatured alcohol) is very flammable and harmful.

Read our standard health & safety guidance

Clean the glass plate carefully with methylated spirit or window cleaning fluid. Polish it with a duster and carefully level it with the wedges.

The carbon dioxide cylinder has a pipe inside like a soda-water siphon. Keep it the right way up to release the gas into the dry-ice attachment. Pack solid CO_{2} under the puck, on the glass table.

The procedure below is carried out using multiflash photographs and examples are provided (see below).

Place the xenon strobe (or substitute lighting} so that its light strikes the glass plate at a glancing angle and does not reflect into the camera. In step 4, use a stopwatch to find the orbital time of the orbiting puck from a ‘standard push’. Decide how many images of the puck you want so that the images are clearly spaced; hence decide the frequency of the strobe.

The photographs should be projected from the negative so that the pucks are actual size and then no scaling factors are needed.

A digital camera attached to a computer or a video camera coupled with a VTR which has frame by frame projection can also be used.

Possible alternatives to the xenon strobe:

- Mechanical strobe near to camera and (1 or 2) bright lamps (up to 500 W)
- Home-made flashing light stuck to puck and camera

Procedure

- Make a small quantity of dry ice in the usual way using the CO
_{2}cylinder and dry ice attachment. Fill the puck with dry ice. - Give the puck a gentle push.
- Push it gently again, in the opposite direction, so that students see it travelling both ways and are convinced it is not gliding downhill.
- Attach the puck to a string at one end and to a centrally placed suction cap at the other end. Give the puck a gentle push; it will follow a circular path around the central point. Point out that its speed remains constant, but its direction constantly changes.
- Cut the string while the puck is moving in a circle: it will go off on a tangent to its circular path.
OPTIONAL

Insert the spring into the string connecting the puck to the central point. Give the puck a slightly harder push; it will follow an elliptical path. Get students to observe how the speed of the puck changes along the orbit.

Teaching Notes

- In steps 2 and 3, the puck travels in a straight line at constant velocity. Ask: Is this natural motion? Are any forces necessary?
- The weight of the puck is balanced by the upward contact force of the glass plate and there are no resultant forces on the puck in its direction of motion. It just continues to glide at constant velocity.
- This is an illustration of Newton’s first law. The puck is in equilibrium. This does not always mean staying at rest but rather continuing to move with uniform motion.
- In step 4, ask:
*"What makes the puck change direction? What forces act on it after the initial push?"*The vertical forces are balanced, as in steps 2 and 3, but tension in the string indicates that there is now an unbalanced horizontal force. That force is always at right angles to the velocity of the puck. Point out that this is typical of circular motion. - Before doing step 5, you might want to ask students to predict how the motion of the puck will change when the string is broken. In any case, link the puck’s change in motion back to what they saw in steps 2 and 3.
- Attached below are some example photos:

**Puck A**

- Illuminated with a lamp near to the glass plate and a mechanical strobe in front of the camera.
- Note the constant speed of the puck around its orbit. Any deviation from constancy may be due to residual friction or a sloping glass plate.
- Note also that the position of the string indicate constant angular velocity, which can be calculated from the angle and the time between two flashes.

**Puck B**

- Illuminated with a flashing light on the puck.
- Note the constant speed around the orbit.

**Puck C**

- Illuminated with a lamp near to the glass plate and a mechanical strobe in front of the camera. The string is burnt through using a taper.
- Note the circular orbit before the string burns and the tangential motion afterwards.

**Puck D**

- The puck is connected to the suction cap with a spring, so an elliptic orbit ensues.
- Note the change in speed along the orbit.
- Note also the different angles swept out in the same time intervals. Equal areas are swept out in equal times. (Kepler’s law II)

### Resources

Download the support sheet / student worksheet for this practical.

### Up next

### The motion of the Moon around the Earth

**Class practical**

This shows an everyday example of circular motion, the Moon going round the Earth.

Apparatus and Materials

Chart showing the star pattern.

Health & Safety and Technical Notes

Read our standard health & safety guidance

Procedure

- Locate the Moon against the background pattern of stars.
- Follow the Moon’s motion by observing its position relative to the stars at different times on the same night, and also at the same time on consecutive nights.

Teaching Notes

- Questions to ask:
*"Is the motion of the Moon around the Earth a natural motion? Is the Moon in equilibrium? Is any force needed to keep it going? Does the Earth attract or repel the Moon?"* - The speed of the Moon is constant but its direction of motion changes constantly, so the velocity is changing. The Moon is not in equilibrium; there must be a force to create this acceleration. The Earth gravitionally pulls the Moon inwards, and keeps it in a stable orbit.
- Newton’s third law says that the Moon must pull the Earth towards it. This could be linked to the tides, if this has been covered already.

### Up next

### Fine beam tube

**Demonstration**

This shows an electron beam being deflected by a magnetic field into a circular path.

Apparatus and Materials

- Fine beam tube and stand
- HT Power supply, 0-350 V and 6.3 V for heater
- Rheostat (10-15 ohms)
- Power supply, low voltage, variable for field coils
- Magnadur magnets, 2
- Helmholtz coils, 2
- Leads, 4mm shrouded, 6

Health & Safety and Technical Notes

Take care when using the HT supply. Make all connections, using shrouded connectors, with the HT turned off. Once switched on, do not make changes to the connections. An electric shock from a HT supply can be severe, possibly fatal.

Take care when handling the fine beam tube. Use the purpose-made holder and stand to avoid damage.

Read our standard health & safety guidance

Set up the fine beam tube as in the Leybold or Teltron manufacturer’s instructions. No voltage should be connected to the deflecting plates. These should both be connected to the anode.

Procedure

- Use a single Magnadur magnet to deflect the beam. A pair of magnets will then give a bigger and more symmetrical deflection, as shown.
- Connect the low voltage power supply to the Helmholtz coils (combined resistance 4 ohms) and set to about 8 V, giving currents of 0.5 - 2.0 amps. Adjust the supply voltage so that the beam becomes circular.
- Increase the current, pointing out that this decreases the radius of the electron beam path.
- Turn the tube slightly so that the beam moves in a spiral. This shows that the circular motion produced by this field does not stop after one revolution.

Teaching Notes

- These demonstrations show that the beam is bent where the magnetic field is strongest - and that the force always acts at right angles to the motion of the beam. For steady bending a large uniform field is needed and the Helmholtz coils are used for this.
- The track left by the electrons shows that they move in a straight line until the magnetic field is applied in a plane at right angles. The beam then moves in a circular path, with a radius dependent on the strength of the magnetic field.

### Up next

### Whirling a rubber bung on a string

**Class practical**

To begin to explore circular motion in more detail.

Apparatus and Materials

- Rubber bung
- String, length about 50cm

Health & Safety and Technical Notes

Ensure there is sufficient area around each student when whirling. Keep away from windows, etc

Securely attach the rubber bungs to the string. Ensure that the string is held tightly as it is whirled around the head.

Observers should wear safety spectacles.

Read our standard health & safety guidance

A rubber bung is convenient for this experiment and safer than whirling stones or steel washers.

Procedure

- Whirl the bung fairly gently in a horizontal plane above the head.
- Allow the string to wrap round a finger. If the bung is whirling slowly at first, the consequent increase in speed will be apparent.

Teaching Notes

- Ask:
*"Is the force on the bung a push or a pull?"*(Strings never push.) The same string which pulls the bung inwards also pulls the hand holding it outwards. - Step 2 is a model of a satellite about 200 km from the Earth where there is still some air resistance and so energy is transferred to the atmosphere and the satellite descends. As that happens, its time to orbit the Earth grows smaller and the satellite speeds up.
- In this experiment a vertical gravitational force acts on the bung. A vertical component of the string's tension balances this force: note that the horizontal plane in which the bung orbits is slightly below the height at which your finger holds the string.
- (Some students will maintain that there is a centrifugal force acting on the bung. They would certainly infer the presence of such an outward force if they were in a rotating frame of reference riding on the bung. See guidance note:
Centrifugal motion at a fun fair

- Offer another example, a thought experiment:
*"Sit on a slippery bus seat, with eyes closed. When the bus turns round a sharp left corner, in which direction do you appear to move?"*This may be a difficult question to answer because of our intuition.*"There is nothing to stop your motion (friction on the bus seat or bracing yourself not to move) continuing straight on (Newton’s first law). However, the bus turns underneath you and you hit the wall (or person) on your right."*

### Up next

### Whirling a rubber bung and letting go

**Demonstration**

To show the tangential motion of an object when it is released from circular motion.

Apparatus and Materials

- Rubber bung or tennis ball
- String, length 1 m approx

Health & Safety and Technical Notes

Securely attach the rubber bung or ball to the string. Hold the string tightly as you whirl it around. At the point when the bung is released, the tangent to the direction of motion should point in a direction clear of obstacles.

Read our standard health & safety guidance

Procedure

- Swing the bung round in a circle, with a radius of about 50 cm, horizontally above the head.
- At the moment when it is nearest to the class, release the string so that it travels tangentially towards the side of the room.

Teaching Notes

- It is useful to say that the bung will be released when it is nearest to the class and count it down by saying NOW...NOW.... then release it. The class will no doubt duck, convinced that centrifugal force will drive it out to them.. It will travel tangentially out (to a clear space) and not radially out (into the class).
- Ask:
*"How does it travel before release?"*(In a circle.)*"How does it travel after release?"*(It continues its motion in a straight line along the tangent.)

### Up next

### Introducing circular motion

### Up next

### Penny on a turntable

**Demonstration**

A further example of the motion of an object travelling in a circle.

Apparatus and Materials

- Turntable
- Penny

Health & Safety and Technical Notes

Read our standard health & safety guidance

A record turntable would be suitable for this demonstration.

Procedure

- Fix smooth paper on the turntable. Place a penny about 7 cm from the centre.
- Gradually increase the speed to something over 1 revolution per second - to, say, 78 rev / min - so that the penny slides off.
- Replace the paper with a polystyrene ceiling tile fixed to the turntable. Stick a drawing pin into the tile near to the penny to stop the penny sliding, and increase the turntable speed [the pin should not be used to hold down the penny and prevent it from moving].

Teaching Notes

- In step 2, once limiting friction is exceeded then the penny moves. A little ink under the penny leaves a track.
- Ask students to describe what happens from the point of view of an outside observer at rest (the penny moves along a tangent) and then to speculate what that would look like from the point of view of an observer standing on the turntable (the initial motion of the penny is radial).
- When the penny slides, initially it does so at a tangent but friction with the rotating table makes its motion turn into a spiral.
- After step 2, ask:
*"Where should the pin be placed, in order to prevent the coin sliding? (At A, B, C, D or elsewhere?)"*The force of the pin on the penny plus the frictional force between the penny and the tile provide a centripetal force pulling the penny into an orbit. (Correct answer is B.) - With the extra force from the pin, the table can rotate faster without the penny sliding.
- Students may find it helpful to imagine a toy truck moving along in a straight line. To get it to turn left, then it could be nudged sideways to the left; in other words towards the centre, of which the left turning arc is a part. To enable it to continue along that arc, the truck will need periodic nudges. And the nudges must change direction so that they are always at right angles to the truck.

### Up next

### Looping the loop

**Demonstration**

A fun demonstration of the relationship between energy stored gravitationally, and energy stored kinetically; it can also be used to consider the forces involved during a complete loop through a vertical circle.

Apparatus and Materials

- Flexible curtain rail, 3m to 4m long
- Ball bearing
- Toy
hot

wheels kit (OPTIONAL)

Health & Safety and Technical Notes

Be very careful over working at heights when setting up this apparatus and when using it. No one should climb on stools or benches.

Read our standard health & safety guidance

Toys such as 'Hot Wheels' allow these demonstrations to be set up easily at floor level.

It is rewarding to construct a version of this apparatus, very well mounted with the top section detachable. A convenient method of mounting is to glue blocks of wood 3 cm x 3 cm x 3 cm at 30 cm intervals around the curtain rail. The blocks are screwed to the track with counter-sunk screws (so that the steel ball does not hit the screw-heads as it goes round the rail). The blocks could be drilled so that the holes fit the ends of clamps attached to retort stands with bosses (see illustration). Alternatively, 10 cm nails can be put through the holes in the blocks.

Bend the curtain rail so that the ball can 'loop the loop'. This is more effective with a 3.5 m length or even a 4 m one. The initial fall should be as steep as possible and the loop needs to be tight.

If you plan to discuss circular motion: the size of the gap should be such that while the ball is in the gap the parabolic motion it gains is approximately equal to the circular motion it would have had if the track was there.

A "Roller Coaster Physics" kit is now (October 2011) available from the supplier: Data Harvest. This allows the user to build a variety of models of roller coaster and investigate several related concepts. It includes a 62-page teachers' guide, with lesson plans and design briefs. Order number KX78880.

Procedure

- Use a tray of sand, or a good wicket keeper, to catch the ball.
- Use the apparatus to explore what happens when the ball is released from different positions on the track. (See teaching notes.)
- Extension for circular motion discussion: remove the top piece and release the ball from different heights until it completes the circle.

Teaching Notes

- It is great fun to challenge students to set the steel ball off at the right height,
*h*(assuming that no energy is dissipated, the ball bearing just goes round the loop when*h = 5R/2*). You might ask them why you need to store more energy gravitationally at the start than is needed to lift the ball to the height of the top of the loop. - Ask: "What makes the ball go round a circle? What pushes or pulls the ball with a real force to make it do that?"
- Point out that there must be some inward force towards the centre of the loop
- Ask: "What provides the force at the top of the loop B?" (the track
and

gravity). After removing a bit of the track in step 3, the ball will still go round the loop, provided the speed is right. Too fast and it flies upwards along a projectile (parabolic) path and too slow and it falls downwards along a projectile (parabolic) path. - Ask: "What provides the force at the sides halfway up the loop (A and C)?" (The track.) The effect of gravity pulls vertically and only slows the ball down a bit.
- Ask: "What provides the force at the bottom of the track (D)?" (The resultant between the inward push of the track and the downward force of gravity.)

### Up next

### Sketching a satellite orbit and predicting its period

## Sketching a satellite orbit and predicting its period

Practical Activity for 14-16

**Demonstration**

Using a scale drawing to predict the time for a satellite close to the Earth. The results can later be compared with mathematical calculations based on *v ^{2}/R*.

Apparatus and Materials

- Brown paper sheet about 1.5 m long, 15 to 20 cm wide - or a roll of long paper
- Thin wire, strong, about 3.5 m
- Large mass or hook to anchor one end of the wire on the floor or near it
- Metre rule with cm and mm markings

Health & Safety and Technical Notes

Be careful with the thin wire, and avoid getting cuts to the hand.

Read our standard health & safety guidance

This works best on the floor, or the side bench of a laboratory, with small groups of students.

The arc drawn is twice as long as needed for 2 minutes' travel, but the double length enables a symmetrical drawing which will yield a good estimate more easily.

Procedure

- Using a thin wire anchored on the floor at one end and held taut with a pencil at the other end, draw an arc about 1.3 m long and with a radius of 3.3 m. The centre of the circle is not on the paper but arrange the paper so that the arc is. This represents part of a circular orbit for a satellite at an altitude of 200 km. (A scale drawing with 0.5 mm = 1 km - see Teaching Notes.)
- On that arc, XAY, mark the mid point A and part of the radius to A. Drop down along that radius the calculated fall 72 km (to scale) AM. (On the same scale 0.5 x 72 = 36 mm.)
- Draw the tangent at A, symmetry helping. And draw a chord XMY parallel to the tangent, with mid point at M.
- Transfer the fall AM out to the place where it should be shown as a fall from the tangent to the orbit, a fall NY. Measure the travel distance AY, which is covered in their chosen time 120 seconds (973 km).
- Calculate the time for a complete orbit, knowing the total travel distance 2π
*R*, which is 2π x 6,600 metres.

Teaching Notes

- A scale of 0.5 mm to 1 km is best for the drawing - a smaller drawing would be difficult to measure. On this scale,
*R =*6,400 + 200 km is represented by a radius of 0.5 x 6,600 = 3,300 mm = 3.3 m. - Assume the satellite is always falling inward with acceleration
*g*(≈ 10 m/s/s). First calculate how far the satellite will fall from the tangent in 1 second (5 m). It is clearly difficult to draw a scale diagram with 5 m drawn on it and the radius of the orbit, 6,600 km, and so a longer time needs to be chosen. the satellite's free fall from the tangent to its orbit in a time of 120 seconds will be*s = 1/2 gt*= 72 km.^{2} - This will give a travel distance AY of about 973 km. The time for a complete orbit,
*T,*will then be 120/*T*=973 / 2π x 6,600, giving a value of*T*of about 85 minutes. Students are generally very surprised to find that near-Earth satellites orbit with such short periods. Such satellites are used for astronomy, earth-observation (mapping and spying) and weather-forecasting. (Note that communications satellites must remain geostationary, and so orbit at much greater distances from the Earth.) - It helps to provide illustrative date about a variety of near-Earth satellites. For example, the first artificial satellite, Sputnik 1, had an orbit period of 96.2 minutes. NASA has a website that offers real-time tracking of satellites, including the Hubble Space Telescope and the International Space Station.
- Likewise the European Space Agency has a satellites in orbit webpage, where you can find orbit information for its Earth Observation missions. Note that ESA Space Science missions do not have near-earth orbits.
- You could go on to calculate the Moon's orbital time, using the same scale. See the guidance note:

Estimating the Moon's orbit time

### Up next

### Experimental test of F = mv²/R

**Class practical**

Measuring the variables involved in circular motion, to find out whether the formula for centripetal acceleration, *F* = *m**v*^{ 2}*R*, is reasonable.

Apparatus and Materials

*For each student pair*

- Stopwatch or stopclock
- Centripetal force kit
- Metre rule

Health & Safety and Technical Notes

Ensure the bungs are securely attached and that there is sufficient space around each student when whirling the bung.

Read our standard health & safety guidance

To make a centripetal force kit, as shown in the diagram below, you will need:

- Glass tube with smooth fire-polished rim, 15-20 cm long and 3-4 mm bore. To make this easier to grasp, it can be placed inside a sleeve of tightly-fitting rubber tubing.
- Rubber bung 2-3 cm diameter, 2.5 cm long with a dowel in one of its own holes
- Wire hook
- Small, square piece of card with two holes, used as an indicator
- Metal washers, up to 15 (200 g)
- Thin string or nylon cord, about 1.5 m long

The indicator will be adjusted so that it is just below the glass tube when the bung is at the required radius. You can use a paper clip in place of the square indicator, but this is less convenient when rotating.

Pass one end of a piece of cord 1.5 m long first through the holes in a square indicator, then through the glass tube and finally through one hole of the bung. Pass the end back through the other hole of the bung and anchor it by plugging the hole with a short wooden rod (dowelling). Finally, hitch the string round the wooden rod or tie it off. At the other end of the cord, slide the washers which provide the accelerating force, above the wire hook.

Procedure

- Hold the glass tube, making sure there is no one too near, and whirl the bung around your head, keeping the indicator just below the glass tube. The indicator will probably rotate slightly when it is not touching the tube; this can be helpful in deciding whether it is clear of the tube or not. While you continue doing this your partner should measure the time for 50 complete orbits. This is best done by counting '3,2,1,0,1,2,3..' with the timer started at 0.
- Repeat step 1, with your partner doing the whirling and you the timing. Average both results and work out the orbital period,
*T*, in seconds. - Measure the radius,
*R*, of the orbit in metres. From*T*and*R*, calculate the orbital speed of the bung,*v*, in metres per second. - Find the mass of the bung in kg and then calculate the value of the predicted centripetal force in newtons, using
*m**v*^{ 2}*R*. - Ignoring friction at the top of the glass tube, the centripetal force is equal to the weight of the washers. Is
*m**v*^{ 2}*R*, roughly equal to the weight of the washers? Take g = 10 N/kg. - The force F may be varied by adding more washers.

Teaching Notes

- This experiment assumes that students have already been taught that centripetal acceleration,
*a*=*v*^{ 2}*R*. They are now finding out whether the formula for centripetal acceleration,*F*=*m**a*=*m**v*^{ 2}*R*, is reasonable.*F*is the real force (the weight of the washers acting vertically downwards) and*m**v*^{ 2}*R*is the predicted centripetal force for a given orbit. - Students can produce a table of results and in each case the force creating the circular motion,
*F*, the tension in the string, is the weight of the washers. They compare this force with the calculated value of the force which is needed to perform a given orbit,*m**v*^{ 2}*R*. - There are two difficulties which students commonly have with this experiment:
- Using
*F*=*m**a*in the derivation, when the centripetal force is at right angles to the motion and the speed of the bung is constant - Understanding that the weight of the washers provides the centripetal force
- Students may not be able to whirl the bung around keeping the string horizontal and so the actual radius of the orbit is less. However this does not affect the results because the tension in the string has to be resolved too and both cosine factors cancel out. In any case the cosine is very close to 1.
- If there is time, get students to explore different variables:
- Vary the force
*F*by adding more washers (up to 20). Look for a relationship between*F*and*v*. (They should find that the force is proportional to*v*^{ 2}.) - Vary the mass for a constant radius.
- Changing to a different radius complicates the discussion because the radius is involved in calculating the velocity. It is much easier to demonstrate empirically that force is proportional to
*v*^{ 2}than to show that*F*is actually equal to*m**v*^{ 2}*R*because*R*is involved in calculating*v*.

### Up next

### Further test of mv²/R

**Class practical**

This is a modification of the experiment:

Experimental test of F = mv²/R

Apparatus and Materials

*For each pair of students*

- Metal ball, 50-100 g (or rubber bung)
- Thin string or nylon cord, length about 1.5 m
- Rod, length 20 cm, diameter 1 cm
- Curtain ring
- Long spring
- Short, weak spring, made from think Manganin wire wound onto a pencil
- Forcemeter, 0-10 N or masses
- Balance to find mass or metal ball (or rubber bung)
- Thread

Health & Safety and Technical Notes

Ensure there is sufficient area around each student when whirling. Keep away from windows, etc.

Securely attach rubber bungs to the string. Ensure the string is held tightly as it is whirled around the head.

Observers should wear safety spectacles.

Read our standard health & safety guidance

The curtain ring is held in place on the wooden handle with two washers and nails. It is free to rotate around the handle. Note: File off the sharp point of the nail.

**Attaching the long spring:** Using a light string, tie one end of the long spring to the curtain ring, at a distance of 10 cm. Tie the other end of the long spring to a metal ball, at a distance of about 30 cm.

**Attaching the short spring:** Using similar string, fix one end of the small spring to the curtain ring, at a distance of 2.5 cm. Feed string tied to the other end of the small spring through the long spring to prevent the thread from getting tangled and tie it to the mass end of the long spring. Finally, fix a short piece of string to either end of the small spring so that it cannot stretch to more than 7.5 cm.

Procedure

- Gently swing the mass in a horizontal circle above your head, gradually increasing the rotation speed so that the long spring stretches and the small spring is stretched almost to its full extent. Your partner now times 50 full rotations.
- Measure the radius of the orbit (the length from the ring to the ball) by placing the device flat on a bench with the mass pulled outwards so that the springs are stretched by the same amount as in step 1.
- Use the data collected in steps 1 and 2 to calculate the orbital speed,
*v*. - Hang the spring vertically and load it with masses (or pull it with the forcemeter) so that the springs are stretched by the same amount as in step 1. Record the force
*F*needed to do this. - Measure the mass, and calculate the theoretical centripetal force
*mv*. Compare with the actual force^{2}/R*F*.

Teaching Notes

- The long spring serves both to produce a variable force - so that the orbit is stable - and to measure that force. The tension in the long string is measured by inserting a piece of string through it which is connected to a small spring - to show when the spring is stretched a standard amount, and to prevent over stretching of the spring. The small spring is only used as a signal and does not contribute significantly to the central force.
- When the device is whirled faster and faster the long spring stretches more and more until it begins to pull its main safe-guarding thread taut. The weak spring remains unstretched with its thread loose, until the safeguarding thread is just taut. Then, at that speed, the weak spring acts as a signal.
- The tension in the long spring is the centripetal force,
*F*. The predicted size of the force required for an orbit is*mv*. The point of the experiment is to compare the predicted force to the spring tension providing the centripetal force.^{ 2}/R

### Up next

### Test of mv²/R on a turntable

**Demonstration**

An alternative method of investigating centripetal force quantitatively.

Apparatus and Materials

- Turntable
- Truck attachments (see Technical notes and the diagram below)
- Stopwatch or stopclock
- Fractional horsepower motor (with gear box)
- LV variable power supply
- chemical balance with a sensitivity of no less than 0.1 g
- Forcemeter, reading up to 10 N
- Lead plates

Health & Safety and Technical Notes

Do not increase the rotational speed of the turntable too much; otherwise there is a risk of objects flying off

and causing injury.

Read our standard health & safety guidance

Fix the truck attachment securely to the top of the turntable, as illustrated below. The section of model railway track extends radially and the toy truck is pulled towards the centre by a spiral spring.

The turntable has to be rotated steadily. This can be done by hand, but it is more convenient to drive the turntable using the fractional horsepower motor and the gear box which can be attached to it. The speed of rotation is then conveniently controlled by adjusting the power supplied to the motor.

The position of the truck for various orbital speeds can be marked with a flag, or an electric contact can be arranged to light a lamp when the truck makes a circuit

at a particular radius. In some models, the end plate is moveable so that different radii can be selected and the truck always hits the end plate.

The experiment is sensitive to the levelling of the turntable.

It may be necessary to increase the mass of the truck for satisfactory results.

The supplier Fracmo sells fractional horsepower motors:

Procedure

- Gradually increase the track's speed of rotation until the truck
just

reaches the stop at the end of the track. Time 50 revolutions of the turntable and calculate the time for one revolution. - Measure the radius of the circle described by the rotating truck,
*R*, and calculate the speed of the truck in its orbit,*v*. - Find the mass of the truck,
*m*, using the balance. - With the turntable at rest, use a forcemeter to measure the force,
*F*, required to extend the spring by the same amount as it does when the truck hits the end stop. - Compare this actual force with the theoretical value of the centripetal force,
*mv²/R*.

Teaching Notes

- This experiment assumes that students have already been taught that centripetal acceleration,
*a = v²/R*. They are now finding out whether the formula for centripetal acceleration,*F = ma = mv²/R*, is reasonable.*F*is the real force (the force of the spring pulling on the rotating truck) and*mv²/R*is the predicted centripetal force. - You could increase the mass of the truck by loading it with one or more lead plates, repeating the demonstration and calculation for each mass.

### Up next

### Estimating the Moon’s orbit time

The Moon’s orbital period can be estimated by considering the Moon as a satellite of the Earth and using a scale drawing, e.g. the drawing produced in the experiment:

Sketching a satellite orbit and predicting its period

The Moon's distance from the Earth is about 60 Earth-radii. Whereas a scale of 0.5 mm to a kilometre can be used for a near-Earth satellite, it is necessary to use a scale of 0.5 mm to 60 km for the Moon's orbit.

Here’s a possible script for class discussion which might follow immediately after the Sketching a satellite orbit and predicting its period experiment.

Imagine that gravity extends undiminished out to the Moon. Calculate the fall in 2 minutes: 72 km as before.

On a scale drawing with 0.5 mm to 60 km that fall would be only 72/120 mm; too small to work with. Let the Moon travel the same arc on the diagram as for the Earth satellite falling for 2 minutes. (That was somewhere between 40 and 80 cm.)

But now the fall of 36 mm from the tangent to that place on the diagram no longer represents 72 km. It represents a fall of 72 x 60 km. How much time would a falling body need for that under full gravity? Look at *s* = ½*a**t*^{ 2}. With *s* 60 times as big, with the same *a*, *t*^{ 2} must be 60 times as big. Then*t* must be √60 times as big: that is: 7¾ times as big. [To show that √60 ≈ 7 ¾ you could do this calculation:

(7¾)^{2} = (31/4)^{2} = 96116… within 0.06 % of 96016 = 60

Then the Moon would travel the arc in 7¾ x 2 minutes; and it would travel the whole circle in 7¾ x the orbital period obtained for the near-Earth satellite; that is, between 11 and 12 hours. [Many students will not be worried about 11-12 hours; after all it is half a day and explains why the Moon disappears (sometimes) during the day and only appears at night!]

Even if answers for the near-Earth were far from 90 minutes, the new answer is clearly wrong for the Moon, which takes a month to orbit the Earth. Therefore, if the Moon is constantly pulled from the tangent to its orbit by the Earth's gravity, the force of gravity must be weaker out at the Moon.

[You could then take the Moon's month of 27.3 days for granted and calculate the amount of dilution. But the answer would not look clear and simple to beginners. It is probably better to suggest an inverse-square dilution of gravity and try it in the calculation from the drawing. If it is the pull of gravity that holds the Moon in its orbit, making it fall from the tangent to the orbit again and again; it must be a much weaker gravity. The acceleration must be much less than 10 m/s/s out at the Moon.]

When 17th century astronomers started puzzling about this, several people suggested that gravity may 'thin out’ according to an inverse-square law. According to that, if gravity is so much at a certain distance, it is 1/4 at double distance; 1/9 as much at treble distance; 1/100 as much at 10 times as far away from the attracting body.

An apple near the Earth is pulled so strongly that it falls with acceleration 10 m/s/s .The Earth attracts an apple as if all the Earth were concentrated at the centre 6,400 km below the surface, a whole Earth-radius from the apple. But we know that the Moon is about 60 Earth-radii away from us, 60 times as far from the Earth's centre as an apple. So, if gravity does follow an inverse-square law; it must thin out by a factor (1/60)^{2} when we change from apple to Moon. If so, free fall under gravity at the Moon would not have an acceleration l0 m/s/s; but it would have acceleration (10/60)^{2} or 10/3,600 m/s/s.

How would that change affect the calculation of the Moon’s orbit time?

Look at*s* = ½*a**t*^{ 2}. With *s*. If *a* is 3,600 times smaller, then, for the same *s*, *t*^{ 2} must be 3,600 times bigger: and t must be 60 times longer. As a result of diluting gravity; you might expect the Moon to take 60 times the previous estimate for its whole orbit.

That is 60 x (7¾ x 90) minutes, or 60 x (7¾ x 90) / (60 x 24) days, or very close to 29 days.

It looks as if the Moon may be ‘falling’, to keep its orbit, with inverse-square-diluted gravity.

[It is amazing that the change of scale from the Earth orbit satellite to the Moon gives such a good result. It is easier to do than it is to read the instructions so have a go!]

Newton is credited with doing this calculation and so students who attempt it are following in the footsteps of a great scientist.

### Up next

### Orbits of satellites and moons

This sequence of ideas can help students understand orbital motion.

Throw a ball out horizontally. It falls to the ground some distance away. A rifle bullet, fired faster but also horizontally, reaches the ground after a kilometre or so.

A thought experiment: fire a bullet so fast that it covers an appreciable part of the Earth’s circumference before it reaches the ground. What effect does the Earth’s curvature have on the bullet’s fate? Fire it even faster and it ‘falls over the edge of the Earth’. The Earth falls away from the bullet’s original direction at exactly the same rate as the bullet falls.

To the bullet, all parts of the Earth are the same. It soon forgets where it started from. With just the right speed, it will always be falling over the edge and so it will go on round the Earth (keeping just above the ground) until it arrives back at the starting point and hits us from behind.

The bullet started off as a projectile with a constant horizontal velocity and a vertical acceleration due to gravity. Therefore at every point in its orbit the Earth satellite must have a constant acceleration towards the centre of the Earth.

Newton imagined that if he threw the stone fast enough it would orbit the Earth because it is always falling towards the Earth at the same rate as the Earth "falls away" from it. The projectile becomes a satellite.

This thought experiment is sometimes known as *Newton’s cannon* and is available as a computer simulation or video.

In practice air resistance absorbs energy and down comes the bullet, so you must start outside the atmosphere. When a rocket is used to launch a satellite, the motions of rocket and satellite can be analyzed like this:

- The rocket starts off nearly vertically. The exhaust gases exert an upward push greater than the rocket’s weight, so that the rocket accelerates upwards.
- Fuel exhausted; motor cuts out; first stage jettisoned.
- Parabolic (free-fall trajectory until the path is horizontal at maximum altitude.

Final stage ignites and accelerates its relatively small mass to high velocity. Satellite unlatched and left in orbit. Final stage is also in orbit but at a slower speed so it gets left behind.

### Up next

### Proof of F = mv2/R

#### Method A

This follows directly from the experiment

Sketching a satellite orbit and predicting its period

The mathematics follows directly from the sketch produced in that experiment, reproduced below. This is Newton’s method.

It relies crucially on the crossed chords theorem for a circle, which should be given.

The circle represents the orbit of a satellite of radius *R*, moving with speed *v* . The satellite moves from A to B in a time *t*. Without a force the satellite would have moved to K at constant speed.

Now 'switch-on' gravity and the satellite will fall a distance *h* in the same time from the tangent from A to the point B. It doesn’t matter whether you let it fall from A first, and then continue in the tangential direction, or vice versa. Anyone who objects that the fall from K to B is not along the radius should look at their scale diagram again: it is almost impossible to see the difference between h and the radial drop. [You may need to talk about in the limit

.]

From the crossed chords property, *h*(2R-*h*) = *x*^{ 2}

but 2*R*>> *h* therefore 2v = *x*^{ 2} and so *h* = *x*^{ 2}2*R* **(equation 1)**

now *x* = AK which is almost the *arcAB* = *v**t* **(equation 2)**

Combining 1 and 2, *h* = (*v**t*)^{2}2*R* **(equation 3)**

*h* is the vertical fall and so using *s* = ½*a**t*^{ 2} = *h* **(equation 4)**

Then from **(equation 3)** and **(equation 4)**

½*a**t*^{ 2} = (*v**t*)^{2}2*R*

leading to *a* = *v*^{ 2}*R*

Using *F* = *m**a* then *F* = *m**v*^{ 2}R

The same holds for the motion at all places round the circle. The vertical is always taken to mean the direction from the satellite to the centre of the attracting body.

#### Method B

This method relies on an understanding of vectors.

The circle represents the orbit of a satellite of radius *R* , moving with speed *v* .
The satellite moves from A to B in a time *t* .

Draw vector AP to represent the initial velocity of the satellite at A, which is along a tangent at A. Draw a second vector of the same length, BQ, to represent the later velocity at B.

Redraw the initial and later vectors, both starting from the same point D. Both have magnitude equal to *v* . FG, representing the change in velocity, must be added to the old velocity to make the new velocity.

AOB and FDG are similar triangles.

change in velocity*v* = AB*R*

acceleration = change in velocitytime taken A to B = AB x *v**R* x time to A to B = *v*^{ 2}*R*

Using *F* = *m**a* then *F* = *m**v*^{ 2}R

The equation, *F* = *m**v*^{ 2}R, illustrates these relationships:

- the higher the speed
*v*, the bigger the force needed to hold objects in orbit, so the bigger the central acceleration - for the same speed, the smaller the radius, or the sharper the curve, the bigger the force and therefore the bigger the acceleration must be.

The acceleration goes up with orbital speed, *v*, but decreases as the radius, *R*, increases.

### Up next

### Centrifugal motion at a fun fair

Consider using this thought experiment

about a common example of circular motion to stimulate a fruitful classroom discussion.

Suppose that two boys, A and B, visit one of those amusements at a fair in which people sit on a floor that rotates. A and B enter the room while the floor is at rest, and sit on the polished floor. Knowing the trick of the performance, A glues himself to the floor. When the floor begins to spin A notes that a mysterious force seems to pull him outward. But for the glue, it would make him slide out to the wall.

B, without glue; slides out to the wall if A does not hold on to him, exerting an inward pull on him. Each boy feels he is struggling against centrifugal force

.

But now let a stationary observer take a bird's eye view from above. Seen from outside the spinning room, A and B are each moving in a circular orbit; and each needs a real inward force to keep him in orbit. For B, the force is the inward pull which A provides: for A it is the pull of the sticky floor on him. The boy A merely imagines an outward force on B because he has to apply a real inward force to him. As the outsider sees, these inward forces are not neutralizing a mysterious outward force; they are making an inward acceleration; they are making A and B move in a curve.

The outside observer offers a further comment. He sees that when A lets go; B continues along a tangent (if there is no friction). B's successive positions along that tangent are farther and farther out from the centre of the circle; so, as seen by A (revolving with the floor), B seems to be sliding out along a radius. But really B is just continuing a straight (tangent) path, a simple example of Newton's first law.

Centrifugal force is a delusion due to living in the rotating system and trying to forget it.

The rotating-floor discussion leads straight to this view. To people sitting on the floor in a concealing fog – and ignoring its motion – there is an outward field of force, endowing every mass *m* with an outward force *m**v*^{ 2}*R*.

Unless some real agent applies an inward force to balance this, any object left alone will seem to slide outward with acceleration *v*^{ 2}*R*. A sober view from outside says that both the outward field of force and the outward sliding are delusions due to living in a rotating framework and not allowing for its motion.

### Up next

### What pushes planets along?

By the 17th century, people like Descartes and Newton were questioning the Greek view that the circular motions of celestial objects were natural.

For Aristotle, the answer to the question *"why does an object go on moving?"* had been *"Because a force continues to push it along"*. Galileo suggested that no force is needed to keep an object moving with constant velocity. Newton took this as his first law of motion.

Newton's answer to *"What force pushes a planet along?"* was *"No force is necessary, the motion simply continues"*. At the time, this was a revolutionary idea. Newton's explanation: an inward force is needed for a curved orbit, continually pulling the planet away from simple straight line motion. Any satellite must fall inward from the tangent to its circular orbit, again and again and again. That falling constitutes an inward acceleration.

Gravity, Newton argued, provides the inward pull acting on every satellite. The acceleration due to gravity is *v*^{ 2}*R*, where *v* is the satellite’s orbital speed and *R* is the radius of its circular orbit.