Why circular motion needs a centripetal force — Physics

Before this

Newton's first law
Newton's second law
Speed and velocity · coming soon

This leads to

A car that accelerates without speeding up

A car drives round a roundabout. Its speedometer shows a steady 30 km/h the whole way round. Is the car accelerating?

Most people say no. The speed is not changing, so how could it be? But that answer uses an idea of acceleration that is too small. The car is accelerating. There must be a force on it to make it do so, and that force must point in a very particular direction. This lecture shows why, using only Newton's laws and one careful look at what velocity means.

Velocity has a direction

Speed tells you how fast something moves. Velocity tells you how fast and in which direction. So velocity can change in two ways: its size can change (speeding up or slowing down), or its direction can change. Either way, the velocity is different, so it has changed.

Acceleration is the rate of change of velocity, not of speed:

a = \frac{Δ v}{Δ t}

Here $Δ v$ is the change in velocity, the new velocity minus the old one, and it has a direction too. This is the point that most people miss. A change of direction is a change of velocity, even if the speed stays the same. So it needs an acceleration.

An object moving clockwise round a circle at constant speed. At every point the velocity arrow has the same length (the same speed), but it points along the circle in a different direction. Because the direction keeps changing, the velocity keeps changing.

Now apply Newton's first law. An object with no resultant force on it keeps moving in a straight line at a steady speed. A circle is not a straight line. So an object moving in a circle cannot have zero resultant force on it. Something must be pushing or pulling it off the straight path, all the time.

Stop and think

A car goes round a bend at a steady 40 km/h. A student says: "The speed is constant, so the forces on the car are balanced." Is the student right?

Show the answer

No. Balanced forces mean zero resultant force, and then the first law says the car would move in a straight line. The car is turning, so its velocity is changing, so it is accelerating. By the second law, there must be a resultant force. Constant speed is not the same as constant velocity.

Which way does the change point?

The force must point the same way as the acceleration, and the acceleration points the same way as $Δ v$ . So the question is: when an object moves round a circle, which way does its change in velocity point? This is the hard part of the lecture. Take it slowly.

Take two points on the circle, A and B. At A the velocity is $v_{A}$ . A moment later, at B, it is $v_{B}$ . Both have the same length, because the speed is constant, but they point in different directions. The change is

Δ v = v_{B} - v_{A}

To subtract two arrows, draw them from the same starting point. Then $Δ v$ is the arrow from the tip of $v_{A}$ to the tip of $v_{B}$ . It is the arrow you must add to $v_{A}$ to turn it into $v_{B}$ .

Left: two points A and B, 60° apart on the circle, with their velocities. Right: the same two velocity arrows drawn from one point. The change, Δv, goes from the tip of vA to the tip of vB. It points straight down, parallel to the dashed line from the point halfway between A and B to the centre.

Look at the right-hand picture. The two velocity arrows have the same length, so they form two equal sides of a triangle. The third side, $Δ v$ , points straight down. On the circle, "straight down" from the point halfway between A and B is exactly towards the centre. This is no accident. The picture is symmetric about the line through the halfway point and the centre, so $Δ v$ must lie along that line.

Now bring A and B closer together, so that the halfway point and the actual position become the same point. The change in velocity still points towards the centre. So at every instant, the acceleration of an object moving round a circle at constant speed points towards the centre of the circle. Use the explorer below to watch this happen.

Explore — the change in velocity on a circle

A 0.3 kg ball moves round a circle of radius 1.2 m on a smooth table, making one turn every 1.5 s. Choose the angle between A and B. Watch which way Δv points, and how the average acceleration changes as the angle gets smaller. Try 180°, 90°, 60° and 10°.

Angle from A to B: 60°

The ball moves at about 5.03 m/s round a circle of radius 1.2 m, one turn every 1.5 s. Δv always points towards the centre, from the point halfway between A and B. Its size, and the average acceleration Δv ÷ Δt, depend on the angle:

Angle A to B	Time Δt	Size of Δv	Average acceleration
180°	0.75 s	≈ 10.05 m/s	≈ 13.4 m/s²
90°	0.375 s	≈ 7.11 m/s	≈ 19.0 m/s²
60°	0.25 s	≈ 5.03 m/s	≈ 20.1 m/s²
30°	0.125 s	≈ 2.60 m/s	≈ 20.8 m/s²
10°	≈ 0.042 s	≈ 0.88 m/s	≈ 21.0 m/s²

Two things happened as the angle shrank. First, $Δ v$ always pointed at the centre. Second, the average acceleration rose and then settled near one value. The settled value, about 21 m/s² here, is the acceleration of the ball at each instant. It is more than twice $g$ . The next lecture finds a formula for it. For now, the direction is the point.

Centripetal acceleration and force

An object moving in a circle at constant speed accelerates towards the centre of the circle. This is called centripetal acceleration.

By Newton's second law, the resultant force on it also points towards the centre. This is called the centripetal force.

"Centripetal" comes from two Latin words: centrum, centre, and petere, to seek. It means "centre-seeking".

Why the speed stays the same

Here is a deeper question. If a force acts on the ball all the time, why does the ball not speed up? The answer is the direction of the force. Any force can be split into two parts: one along the direction of motion, and one at right angles to it. The part along the motion changes the speed. The part at right angles changes the direction. In steady circular motion, the force points at the centre, and the velocity points along the circle. These are always at right angles. So the force has no part along the motion. It can only turn the ball, never speed it up or slow it down.

That is it. That is the whole idea. A force at right angles to the motion bends the path without changing the speed. Keep it at right angles all the time, and the path bends into a circle.

Not a new kind of force

This is where many students go wrong. "Centripetal force" is not a new kind of force, like gravity or friction. It is a job description. It names whatever real force, or combination of forces, is pulling the object towards the centre. In every problem, ask: which real force is doing the centripetal job here?

Situation	What provides the centripetal force
A car going round a bend on a flat road	Friction between the tyres and the road, acting sideways into the bend
A ball whirled on a string on a smooth table	The tension in the string
A satellite or the Moon orbiting the Earth	The Earth's gravity
A ball rolling round the inside of a bowl	The part of the bowl's normal force that points towards the centre
A rider at the top of a loop on a ride	Weight and the normal force from the seat, both pointing down

On a free-body diagram, draw only the real forces: weight, tension, friction, normal force. Never add an extra arrow labelled "centripetal force". The centripetal force is the resultant of the real forces, not one more force on top of them.

What happens when the force stops?

Suppose the string holding the ball breaks. The ball has nothing pulling it towards the centre any more. Which way does it go? Decide before you look.

Predict, then reveal — the string breaks

A ball is whirled round in a horizontal circle on a smooth table. Seen from above, it is moving clockwise. At the moment it reaches the top of the circle, the string is cut. Which path does the ball take?

The ball moves off in a straight line along the tangent: the direction it was moving at the moment the string was cut.

The ball goes along the tangent. That is Newton's first law at work: with no force, keep going in a straight line.

So what about the feeling of being thrown outwards when a car turns sharply? Watch what really happens. The car turns. Your body, by the first law, tries to keep going straight. So the car turns underneath you, and the side of the car moves towards you until the door presses on you. Then the door pushes you inwards, towards the centre of the bend, and you turn with the car. From the road, there is no force pushing you outwards. There is only your tendency to go straight, and an inward push that stops you. Some books call the outward feeling "centrifugal force". From the ground, it is not a force at all.

Stop and think

In a spin dryer, water comes out of wet clothes through the holes in the drum. If no force pushes the water outwards, how does it leave?

Show the answer

The drum wall pushes the clothes inwards, so the clothes move in a circle. Water held loosely in the clothes needs an inward force too. At a hole, there is no wall to push it inwards, so the water stops turning. It carries on in a straight line along the tangent, which takes it through the hole. The water is not thrown out. It is simply no longer pushed in.

Worked examples

Worked Example 1 — A car on a bend

A car goes round a flat, circular bend of radius 50 m at a steady 15 m/s. What provides the centripetal force, and which way does it act? What happens if the road is icy?

The forces on the car are its weight (down), the normal force from the road (up) and friction from the road. Weight and the normal force are vertical, and they balance. So the only horizontal force is friction. Friction must provide the centripetal force. It acts horizontally, sideways on the tyres, towards the centre of the bend.

On ice, friction is very small. It cannot provide enough inward force, so the car cannot follow the bend. It does not slide straight outwards. It carries on more nearly in a straight line, along the direction it was travelling, and so it leaves the curve on the outside of the bend.

Worked Example 2 — A ball on a string, without the formula

A 0.3 kg ball on a 1.2 m string moves in a horizontal circle on a smooth table, making one turn every 1.5 s. Find its speed, and estimate its acceleration from the change in velocity over a quarter turn.

One turn is the circumference: $2 π \times 1.2 \approx 7.54$ m. So the speed is $7.54 \div 1.5 \approx 5.03$ m/s.

Over a quarter turn, the velocity turns through 90°. Two arrows of length 5.03 at right angles differ by an arrow of length $5.0 3^{2} + 5.0 3^{2} \approx 7.11$ m/s. A quarter turn takes $1.5 \div 4 = 0.375$ s. So the average acceleration is about $7.11 \div 0.375 \approx 19.0$ m/s², towards the centre.

This is an average over a long stretch of the circle. The explorer showed that shorter stretches give a slightly larger value, settling near 21 m/s². So the tension must be about $0.3 \times 21 \approx 6$ N, pulling the ball towards the hand. Check the direction: the string can only pull, and it pulls towards the centre, which is just what the ball needs.

Worked Example 3 — A bucket of water upside down

A student swings a bucket of water fast in a vertical circle. At the top, the bucket is upside down, yet the water stays in. Why?

At the top, the centre of the circle is below the water. So the water needs a resultant force downwards, towards the centre. Two forces can provide it: the water's weight, and the push of the bucket's base on the water, which also acts downwards at the top.

If the bucket moves fast enough, the water needs more downward force than its weight alone. The bucket's base supplies the rest, so the water stays pressed against the base. In a sense, the water is falling. But the bucket is curving downwards even faster, so the base keeps pushing on the water. If the bucket moves too slowly, the water needs less downward force than its weight. Then gravity pulls it down faster than the circle requires, and it falls out.

Worked Example 4 — The Moon is falling

What provides the centripetal force on the Moon, and how far does the Moon "fall" each second?

The Moon moves in a nearly circular orbit around the Earth. The only significant force on it is the Earth's gravity, so gravity provides the centripetal force. Without gravity, the Moon would move off along a tangent.

In one second, the Moon falls below the straight tangent line by only about 1.3 mm. An apple near the ground falls 4.9 m in its first second. The ratio is roughly 1 to 3600. Newton noticed that the Moon is about 60 times farther from the Earth's centre than the apple is, and $6 0^{2} = 3600$ . Gravity gets weaker with the square of the distance. The same gravity that drops the apple keeps the Moon falling round the Earth. It moves sideways fast enough that it keeps missing.

Three classic mistakes

Trap 1 — "Constant speed means no acceleration"

Acceleration is the rate of change of velocity. In a circle, the direction changes all the time, so the velocity changes, and there is an acceleration towards the centre.

Trap 2 — Drawing the centripetal force as an extra force

The centripetal force is the resultant of the real forces (tension, friction, gravity, normal force). Adding a separate "centripetal force" arrow counts the same force twice.

Trap 3 — "It flies outwards"

When the inward force stops, the object moves along the tangent, in the direction it was already moving. It does not fly out along the radius. Nothing pushes it outwards.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Name the real force that provides the centripetal force in each case. (a) A runner going round a curved track. (b) The Earth orbiting the Sun. (c) Clothes going round in a spinning washing-machine drum. (d) An electron moving round a nucleus, in a simple model of the atom.

Hint 1
For each, ask: what is touching the object, or pulling on it, towards the centre?

Hint 2
In (d), the nucleus is positive and the electron is negative.

Full answer
(a) Friction between the runner's shoes and the track, pointing into the bend. (b) The Sun's gravity on the Earth. (c) The normal force from the drum wall, pushing the clothes inwards. (d) The electric attraction between the positive nucleus and the negative electron. In each case the centripetal force is a real force you already know, doing the centripetal job.
A 0.5 kg stone is whirled on a string in a horizontal circle of radius 0.8 m on a smooth table, making 2 turns every second. (a) Find its speed. (b) Is its velocity constant? (c) Which way does the resultant force on it point? (d) Does the tension make the stone speed up? Explain.

Hint 1
2 turns every second means each turn takes 0.5 s.

Hint 2
For (d), compare the direction of the tension with the direction of motion.

Full answer
(a) One turn is $2 π \times 0.8 \approx 5.03$ m, and it takes 0.5 s, so the speed is about $5.03 \div 0.5 \approx 10.1$ m/s. (b) No. The speed is constant, but the direction changes all the time, so the velocity changes. (c) Towards the centre, along the string. (d) No. The tension is always at right angles to the motion, so it has no part along the direction of motion. It changes the direction only, not the speed.
An object moves round a circle at 5 m/s, taking 2 s for each turn. Find the size and direction of its change in velocity over half a turn, and its average acceleration over that half turn.

Hint 1
After half a turn, the object is moving in exactly the opposite direction.

Hint 2
Take the starting direction as positive: $Δ v = (- 5) - (+ 5)$ .

Full answer
Take the starting direction as positive. Then $Δ v = (- 5) - (+ 5) = - 10$ m/s: a change of 10 m/s, opposite to the starting direction. That direction points towards the centre from the point a quarter of a turn along, halfway between the start and the end. Half a turn takes 1 s, so the average acceleration is $10 \div 1 = 10$ m/s². As the explorer showed, the instantaneous acceleration is larger than this rough average.
Newton's test. An apple near the Earth's surface falls 4.9 m in its first second. The Moon is about 60 times as far from the Earth's centre. If gravity gets weaker with the square of the distance, how far should the Moon fall towards the Earth in one second? Compare your answer with the measured value, about 1.3 mm.

Hint 1
60 times the distance means gravity is $6 0^{2}$ times weaker.

Hint 2
In the same time, the fall is in the same ratio as the acceleration: divide 4.9 m by 3600.

Full answer
$6 0^{2} = 3600$ , so the Moon's fall should be $4.9 \div 3600 \approx 0.00136$ m, about 1.4 mm. The measured value is about 1.3 mm. The two agree to within a few per cent, which is the agreement that convinced Newton that one law of gravity governs both the apple and the Moon. The small difference comes from rounded values, such as "about 60 times".

The bigger idea

For a long time, many thinkers treated circular motion as natural, especially the motion of the heavens, needing no explanation. Newton turned this round. Straight-line motion is natural; a circle needs a cause. Once you ask "what is pulling this towards the centre?", the Moon, a car on a bend and a ball on a string become the same problem. Newton's answer for the Moon, that the force which drops an apple also bends the Moon's path, joined the physics of the Earth and the physics of the sky into one subject.

The next lecture turns the direction into numbers: how large the centripetal acceleration and force are, for a given speed and radius.

Key concepts in this lesson

Velocity has a direction, so an object moving in a circle at constant speed has a changing velocity and is accelerating.
The change in velocity, and so the acceleration, points towards the centre of the circle: centripetal acceleration.
By Newton's second law, the resultant force also points towards the centre; this resultant is called the centripetal force.
The centripetal force is always provided by real forces, such as tension, friction, gravity or a normal force, and is never drawn as an extra force.
If the centripetal force stops, the object moves along the tangent in a straight line; from the ground, no outward force acts on it.