Centripetal acceleration and force — the formulae — Physics

Before this

This leads to

The number the explorer was settling on

In the last lecture, a 0.3 kg ball moved round a circle of radius 1.2 m at about 5.03 m/s. The explorer measured its average acceleration over smaller and smaller pieces of the circle: 13.4 m/s² over half a turn, 19.0 over a quarter turn, 20.1, 20.8, then 21.0 m/s². The numbers were settling on one value, always pointing towards the centre. That lecture promised a formula for it.

Here it is, and it is short: $a = \frac{v ^{2}}{r}$ . For the ball, $\frac{5.0 3 ^{2}}{1.2} \approx 21.05$ m/s², exactly the value the explorer was approaching. This lecture shows where the formula comes from, why the speed is squared, and how to use it on roundabouts, washing machines and satellites.

Where v²/r comes from

This is the hard part of the lecture. It uses one idea from geometry: similar triangles. Take it one step at a time.

Let an object move at speed $v$ round a circle of radius $r$ . Take two points, A and B, a short time $Δ t$ apart. In that time, the object travels a distance $v Δ t$ . When A and B are very close, the straight line from A to B is almost the same as the curved path, so the chord AB has length $v Δ t$ .

Now look at two triangles. The first is made by the two radii, OA and OB, and the chord. It has two sides of length $r$ and an angle $θ$ at the centre. The second is made by the two velocities, drawn from one point, as in the last lecture. Each velocity is at right angles to its radius. So when the radius turns through $θ$ , the velocity turns through $θ$ too. This second triangle has two sides of length $v$ , the same angle $θ$ between them, and the third side $Δ v$ .

Left: the position triangle, with two radii r, the angle θ at the centre, and the chord AB of length vΔt. Right: the velocity triangle, with two velocities of size v, the same angle θ between them, and the change Δv. Both triangles are isosceles with the same angle, so they have the same shape.

Two isosceles triangles with the same angle between their equal sides are similar: one is a scaled copy of the other. So the ratio of the third side to the equal sides is the same in both:

\frac{Δ v}{v} = \frac{v Δ t}{r}

Multiply both sides by $v$ , then divide both sides by $Δ t$ :

a = \frac{Δ v}{Δ t} = \frac{v ^{2}}{r}

The only approximation was treating the chord as the arc. As $Δ t$ gets smaller, that approximation becomes exact. That is why the explorer's averages settled on one value. Step through the argument once more below.

Step through — deriving a = v²/r

Press Next to add one step. Read the reason before you go on.

Each step of the derivation, with its reason.

In a short time Δt, the object moves a distance vΔt, so the chord AB ≈ vΔt.
The radii OA and OB and the chord form an isosceles triangle: sides r, r, vΔt, with angle θ at the centre.
Each velocity is at right angles to its radius, so the velocities are also θ apart.
Drawn tail to tail, the velocities form an isosceles triangle: sides v, v, Δv, with angle θ.
The two triangles are similar, so Δv ÷ v = vΔt ÷ r.
Multiply by v: Δv = v²Δt ÷ r.
Divide by Δt: a = Δv ÷ Δt = v² ÷ r, towards the centre.

Why the speed is squared

The $v^{2}$ is not an accident. Speed matters twice. First, a faster object moves round the circle sooner, so its direction changes more often each second. Second, each change of direction is a change of a bigger velocity, so each $Δ v$ is bigger. Double the speed, and both effects double: $2 \times 2 = 4$ times the acceleration. The radius works the other way. On a bigger circle, the path bends more gently, so the direction changes more slowly.

Three checks on the formula

A formula you derived yourself still deserves testing. Physicists use three quick checks, and they are worth learning as habits.

Units. $\frac{v ^{2}}{r}$ has units of $\frac{( m/s ) ^{2}}{m} = \frac{m ^{2} / s ^{2}}{m} = m/s^{2}$ . Those are the units of acceleration, as they must be. A formula with the wrong units is always wrong.

Extreme cases. If the radius becomes huge, the circle is almost a straight line. Then $\frac{v ^{2}}{r}$ becomes tiny, and an object moving in a straight line at constant speed should have no acceleration. If the speed is zero, nothing moves, and again $a = 0$ . The formula passes both tests.

A measured value. The explorer in the last lecture never used the formula. It only subtracted velocity arrows and divided by time. Its averages settled on about 21 m/s², and $\frac{v ^{2}}{r}$ gives 21.05 m/s². Two completely different routes give the same number.

Centripetal acceleration and force

An object moving at speed $v$ on a circle of radius $r$ has acceleration $a = \frac{v ^{2}}{r}$ , towards the centre.

By Newton's second law, the resultant force needed is $F = \frac{m v ^{2}}{r}$ , towards the centre.

Remember the rule from the last lecture: $F = \frac{m v ^{2}}{r}$ tells you how big the resultant force must be. It does not add a new force. A real force, such as tension, friction or gravity, has to supply it.

Angles, turns and angular velocity

Many problems give the time for one turn, or the number of turns per second, instead of the speed. For these, it helps to measure how fast the angle changes.

Physicists measure angles in radians. One radian is the angle at the centre when the arc is exactly as long as the radius. A full circle has a circumference of $2 π r$ , so a full turn is $2 π$ radians, about 6.28 rad. With angles in radians, an arc of angle $θ$ has length $r θ$ .

The angular velocity $ω$ (the Greek letter omega) is the angle turned per second, in rad/s. The period $T$ is the time for one turn, and the frequency $f$ is the number of turns per second, in hertz (Hz). One turn is $2 π$ radians in time $T$ , so:

ω = \frac{2 π}{T} = 2 π f

In one second, the object turns through $ω$ radians, so it travels an arc of $r ω$ . That arc is its speed: $v = ω r$ . Put this into $a = \frac{v ^{2}}{r}$ :

a = \frac{( ω r ) ^{2}}{r} = ω^{2} r, F = m ω^{2} r

Use $\frac{v ^{2}}{r}$ when you know the speed, and $ω^{2} r$ when you know the period or frequency. They are the same formula in two forms. Check with the ball: $T = 1.5$ s, so $ω = \frac{2 π}{1.5} \approx 4.19$ rad/s, and $ω^{2} r \approx 4.1 9^{2} \times 1.2 \approx 21.05$ m/s². The same answer.

Stop and think

Two children sit on a turning roundabout, one near the centre and one at the edge. They go round together, once every 6 s. Which one has the larger speed? Which has the larger acceleration?

Show the answer

They have the same period, so the same angular velocity $ω$ . The child at the edge has the larger $r$ , so a larger speed, $v = ω r$ , and a larger acceleration, $a = ω^{2} r$ . With the same $ω$ , acceleration grows with the radius. That may seem to contradict "bigger circle, smaller acceleration". It does not: that rule holds at the same speed. Here the speeds are different.

Worked examples

Worked Example 1 — A car on a roundabout

A 1200 kg car goes round a roundabout of radius 30 m at a steady 12 m/s. Find its acceleration and the centripetal force. What provides it?

$a = \frac{v ^{2}}{r} = \frac{1 2 ^{2}}{30} = \frac{144}{30} = 4.8$ m/s², towards the centre.

$F = ma = 1200 \times 4.8 = 5760$ N. On a flat road, the only horizontal force is friction between the tyres and the road, so friction must supply 5760 N towards the centre. That is about half the car's weight ( $1200 \times 9.8 = 11760$ N). Check the units: kg × m/s² = N.

Now make a prediction about the same car.

Predict, then reveal — twice as fast round the same roundabout

The same car goes round the same roundabout at 24 m/s instead of 12 m/s. At 12 m/s, friction had to supply 5760 N. How much friction is needed now?

Four times as much: 23 040 N. Doubling v multiplies v² by 4, so F = 1200 × 24² ÷ 30 = 23 040 N. That is about twice the car's weight. Friction between tyres and a road cannot usually supply that much, so the car would skid.

This is why speed on bends is so dangerous. The grip a tyre can give is limited, but the grip needed grows with the square of the speed. Going a little faster on a bend asks for a lot more friction.

Worked Example 2 — A washing machine on spin

A washing-machine drum of radius 0.25 m spins at 1200 revolutions per minute (rpm). Find its angular velocity and the acceleration of the clothes at the edge.

First the frequency: $1200 \div 60 = 20$ turns per second, so $f = 20$ Hz. Then $ω = 2 π f = 2 π \times 20 \approx 125.7$ rad/s.

$a = ω^{2} r \approx 125. 7^{2} \times 0.25 \approx 3950$ m/s². Compare with $g = 9.8$ m/s²: $3950 \div 9.8 \approx 400$ . The clothes accelerate at about 400 times $g$ .

The drum wall pushes the clothes inwards with a huge force, so they go round. Water held loosely in the fabric needs the same huge inward push. Where there is a hole in the drum, there is no wall to push. So the water there stops following the circle and carries on along the tangent, out through the hole. As the last lecture showed, it is not thrown out. It is simply no longer pushed in.

Worked Example 3 — The ball, finished

The last lecture estimated the tension in the ball's string as "about 6 N". Find it exactly.

$F = \frac{m v ^{2}}{r} = \frac{0.3 \times 5.0 3 ^{2}}{1.2} \approx 6.32$ N. The table is smooth and the string is horizontal, so the tension is the only horizontal force. It must be 6.32 N, pulling towards the hand. The estimate from a quarter turn, 0.3 × 19.0 ≈ 5.7 N, was about 10% too low.

Worked Example 4 — The International Space Station

The International Space Station orbits about 400 km above the Earth's surface. Take the Earth's radius as $6.4 \times 1 0^{6}$ m and the gravitational field there as about 8.7 m/s². Find its orbital speed and the time for one orbit.

Gravity is the only force on the station, so gravity supplies the centripetal acceleration: $\frac{v ^{2}}{r} = 8.7$ . The radius is measured from the Earth's centre: $r = 6.4 \times 1 0^{6} + 0.4 \times 1 0^{6} = 6.8 \times 1 0^{6}$ m.

v^{2} v = 8.7 \times 6.8 \times 1 0^{6} \approx 5.92 \times 1 0^{7} \approx 7690 m/s

That is about 7.7 km/s. One orbit is $2 π r \approx 4.27 \times 1 0^{7}$ m, so it takes about $4.27 \times 1 0^{7} \div 7690 \approx 5550$ s, about 93 minutes. NASA describes the station as travelling at about 8 km/s at an average height of about 400 km, so the estimate is close. The gravitation lectures return to orbits in detail.

Three classic mistakes

Trap 1 — Using the diameter, or the height, as r

$r$ is the radius of the circle the object moves on, measured from the centre. For a satellite, that is the Earth's radius plus the height, not the height alone.

Trap 2 — Putting rpm straight into ω²r

Convert first: rpm ÷ 60 gives $f$ in Hz, and then $ω = 2 π f$ in rad/s. Using 1200 for $ω$ in the washing machine would give an acceleration about 90 times too large.

Trap 3 — Forgetting the square

$F = \frac{m v ^{2}}{r}$ , not $\frac{m v}{r}$ . A quick check: doubling the speed must give four times the force.

Try it yourself

Use $g = 9.8$ m/s² where needed.

A 0.5 kg ball on a string moves in a horizontal circle of radius 0.8 m on a smooth table, at 6 m/s. Find its acceleration and the tension in the string.

Hint 1
$a = \frac{v ^{2}}{r}$ .

Hint 2
The tension is the only horizontal force, so it equals $ma$ .

Full answer
$a = \frac{36}{0.8} = 45$ m/s², towards the centre. Tension $= ma = 0.5 \times 45 = 22.5$ N, pulling towards the centre. Check: $\frac{m v ^{2}}{r} = \frac{0.5 \times 36}{0.8} = 22.5$ N.
A 30 kg child sits 3 m from the centre of a playground roundabout that turns once every 6 s. Find the angular velocity, the child's speed, the acceleration and the resultant force on the child.

Hint 1
$ω = \frac{2 π}{T}$ .

Hint 2
Then $v = ω r$ and $a = ω^{2} r$ .

Full answer
$ω = \frac{2 π}{6} \approx 1.05$ rad/s. $v = ω r \approx 1.047 \times 3 \approx 3.14$ m/s. $a = ω^{2} r \approx 1.04 7^{2} \times 3 \approx 3.29$ m/s². $F = ma \approx 30 \times 3.29 \approx 98.7$ N, towards the centre, provided by friction from the seat and the child's grip. Check: $\frac{v ^{2}}{r} \approx \frac{3.1 4 ^{2}}{3} \approx 3.29$ m/s².
A cyclist rounds a bend. By what factor does the friction needed change in each case? (a) The speed is halved. (b) The radius of the bend is halved, at the same speed. (c) The speed is tripled and the radius is also tripled.

Hint 1
$F$ is proportional to $v^{2}$ and to $\frac{1}{r}$ .

Hint 2
In (c), work out the effect of each change and multiply.

Full answer
(a) $(\frac{1}{2})^{2} = \frac{1}{4}$ : a quarter of the friction. (b) Halving $r$ doubles $F$ : twice the friction. (c) Tripling $v$ multiplies $F$ by 9; tripling $r$ divides it by 3. Overall $9 \div 3 = 3$ : three times the friction. A wider bend helps, but speed counts more.
Newton's Moon test, with numbers. The Moon's orbit has a radius of about $3.85 \times 1 0^{8}$ m, and it takes about 27.3 days to go round once. Find its centripetal acceleration. Compare it with $g \div 3600$ , the value Newton predicted from the inverse-square law (the Moon is about 60 Earth radii away).

Hint 1
Convert the period to seconds: $27.3 \times 24 \times 3600$ .

Hint 2
$a = ω^{2} r = \frac{4 π ^{2} r}{T ^{2}}$ .

Full answer
$T = 27.3 \times 86400 \approx 2.36 \times 1 0^{6}$ s. $a = \frac{4 π ^{2} r}{T ^{2}} \approx \frac{4 π ^{2} \times 3.85 \times 1 0 ^{8}}{( 2.36 \times 1 0 ^{6} ) ^{2}} \approx 0.0027$ m/s². And $g \div 3600 = 9.8 \div 3600 \approx 0.0027$ m/s². They agree to within about 1%. In one second, an acceleration of 0.0027 m/s² makes the Moon fall about $\frac{1}{2} \times 0.0027 \times 1^{2} \approx 0.0014$ m, the 1.3–1.4 mm of the last lecture.

The bigger idea

The formula $a = \frac{v ^{2}}{r}$ came from nothing but geometry and the meaning of acceleration. It does not care what the object is or what pulls on it. That is why one formula serves a ball on a string, a car, a washing machine, a space station and the Moon. The physics of each case is only the question "which real force supplies $\frac{m v ^{2}}{r}$ ?"

The $v^{2}$ is the reason speed limits on bends matter so much. The next lecture applies the formulas to real designs: cars on banked roads, rides that loop, and satellites in orbit.

Key concepts in this lesson

Centripetal acceleration is a = v²/r, towards the centre, derived from two similar isosceles triangles.
The resultant force needed is F = mv²/r, and it must be supplied by real forces.
Angular velocity ω = 2π/T = 2πf in rad/s, with v = ωr, so a = ω²r and F = mω²r.
Doubling the speed on the same circle needs four times the force; doubling the radius at the same speed needs half.
Always use the radius from the centre, and convert rpm to rad/s before using ω.