Uniform Circular Motion

Lessons

Senior 3 · Physics

Uniform Circular Motion

4 lessons

Lesson 3 of 4

Centripetal Acceleration and Centripetal Force

25 min

The hand that keeps pulling

Tie a small stone to a string and whirl it round your head. You can feel the string pulling your hand outwards, and you must keep pulling it inwards the whole time. If the string snaps, the stone does not keep going round: as you saw with the marble in Lesson 2, it flies off along a tangent.

So an inward pull is needed to keep a body moving in a circle. How big must it be, and why does it point towards the centre? This lesson answers both questions, with an experiment and then with vectors.

Experiment 3: measuring the pull needed for a circle

You need: a rubber bung (mass $m$ about 20 g), about 1.5 m of strong thin string, a smooth plastic or glass tube about 15 cm long with its ends covered in tape, slotted masses with a hanger, a paper clip, a metre rule and a stopwatch.

Tie the bung firmly to one end of the string. Pass the string through the tube and tie the hanger with masses $M$ to the other end.
Measure 0.50 m of string from the bung to the top of the tube and fix the paper clip on the string just below the tube.
Hold the tube upright and whirl the bung in a horizontal circle above your head. Adjust the speed until the paper clip stays just below the tube without touching it: the radius $r$ is then 0.50 m.
A partner times 20 turns. Repeat with larger hanging masses.

Figure 3.1: The weight $W = M g$ of the hanging masses provides the tension in the string, and the tension is the inward pull on the bung.

$M$ (kg)	$W = M g$ (N)	Time for 20 turns (s)	$T$ (s)	$v = \frac{2\pi r}{T}$ ( $\mathrm{m\,s^{-1}}$ )	$\frac{mv^2}{r}$ (N)
0.050	0.49	18.3	0.915	3.43	0.47
0.100	0.98	12.8	0.640	4.91	0.96
0.150	1.47	10.5	0.525	5.98	1.43
0.200	1.96	9.1	0.455	6.90	1.91

These are illustrative results for $m = 0.020\ \text{kg}$ and $r = 0.50\ \text{m}$ , with $g = 9.8\ \mathrm{N\,kg^{-1}}$ . The larger the pull, the faster the bung must go round. In the last column, $\frac{mv^2}{r}$ is almost equal to $W$ every time; it is slightly smaller because of friction where the string rubs on the tube. The explanation below shows why the pull must equal $\frac{mv^2}{r}$ .

Important

Safety. Wear eye protection and whirl the bung outdoors or in an open hall, with everyone else at least 3 m away. Check the knot on the bung before every run. Whirl above head height, never towards faces, and stop at once if the string frays.

Explanation: the acceleration points to the centre

Figure 3.2 shows a body moving from A to B round a circle of radius $r$ at constant speed $v$ , while the radius turns through a small angle $\theta$ . The velocities $\vec{v}_A$ and $\vec{v}_B$ have the same size $v$ but different directions.

Figure 3.2: (b) Drawing $\vec{v}_A$ and $\vec{v}_B$ from one point shows the change in velocity $\Delta\vec{v}$ . For a small angle it points towards the centre of the circle.

The angle between the two velocity arrows equals the angle $\theta$ turned by the radius, because each velocity is at 90° to its radius. For a small angle the arrow $\Delta v$ is almost an arc of a circle of radius $v$ , so

\Delta v \approx v\,\theta

The time to go from A to B is the arc length divided by the speed: $\Delta t = \frac{r\theta}{v}$ . The acceleration is therefore

a = \frac{\Delta v}{\Delta t} = \frac{v\,\theta}{\left(\frac{r\theta}{v}\right)} = \frac{v^2}{r}

and, using $v = r\omega$ ,

a = \frac{(r\omega)^2}{r} = r\omega^2 = v\omega

Because $\Delta\vec{v}$ points towards the centre, so does the acceleration. It is called the centripetal acceleration ("centripetal" means "centre-seeking"). By Newton's second law, $F = m a$ , there must be a resultant force in the same direction:

F = \frac{mv^2}{r} = mr\omega^2

Definition

Centripetal acceleration ( $a$ ): the acceleration of a body moving in a circle, directed towards the centre, of size $a = \frac{v^2}{r} = r\omega^2$ . SI unit: $\mathrm{m\,s^{-2}}$ .

Centripetal force ( $F$ ): the resultant force towards the centre that keeps a body moving in a circle, $F = \frac{mv^2}{r} = mr\omega^2$ . SI unit: the newton (N). Example: a 0.020 kg bung moving at $4.9\ \mathrm{m\,s^{-1}}$ on a 0.50 m string needs $F = \frac{0.020 \times 4.9^2}{0.50} = 0.96\ \text{N}$ .

Figure 3.3: At every point of the circle the centripetal force is at right angles to the velocity. It changes the direction of motion, never the speed.

What supplies the centripetal force?

Centripetal force is not a new kind of force. It is the name for whichever real force, or resultant of real forces, points to the centre.

Body moving in a circle	Force that points to the centre
a stone or bung whirled on a string	tension in the string
a moto or car going round a flat roundabout	friction from the road on the tyres
the Moon going round the Earth	the Earth's gravitational pull
clothes in the drum of a spin dryer	the push (normal reaction) of the drum wall
a chair on a swing ride	the horizontal part of the tension in its chains

Worked example 1: the bung in Experiment 3

The 0.020 kg bung moves in a circle of radius 0.50 m with a period of 0.64 s. Find its speed, its centripetal acceleration and the centripetal force.

v = \frac{2\pi r}{T} = \frac{2\pi \times 0.50\ \text{m}}{0.64\ \text{s}} = 4.91\ \mathrm{m\,s^{-1}}

a = \frac{v^2}{r} = \frac{(4.91\ \mathrm{m\,s^{-1}})^2}{0.50\ \text{m}} = 48\ \mathrm{m\,s^{-2}}

F = ma = 0.020\ \text{kg} \times 48.2\ \mathrm{m\,s^{-2}} = 0.96\ \text{N}

This matches the 0.98 N weight of the 0.100 kg hanging mass, within the experiment's uncertainty.

Worked example 2: a car on a roundabout

A car of mass 1200 kg goes round a roundabout of radius 20 m at $10\ \mathrm{m\,s^{-1}}$ ( $36\ \mathrm{km\,h^{-1}}$ ). (a) Find the centripetal force. (b) What force supplies it? (c) What force would be needed at $20\ \mathrm{m\,s^{-1}}$ ?

(a) $F = \frac{mv^2}{r} = \frac{1200 \times 10^2}{20} = 6000\ \text{N}$ , towards the centre of the roundabout.

(b) Sideways friction between the tyres and the road.

(c) $F = \frac{1200 \times 20^2}{20} = 24\,000\ \text{N}$ . Doubling the speed makes the force $2^{2} = 4$ times bigger. If the road cannot supply that much friction, the car slides off the curve.

Worked example 3: using angular velocity

A 0.050 kg coin sits 0.10 m from the centre of a turntable that makes 2.0 turns per second. Find the friction force needed to keep the coin in place.

\omega = 2\pi f = 2\pi \times 2.0 = 12.6\ \mathrm{rad\,s^{-1}}

F = mr\omega^2 = 0.050 \times 0.10 \times 12.57^2 = 0.79\ \text{N}

Warning

Do not add "centripetal force" as an extra arrow. On a free-body diagram of the whirled bung, draw only the real forces: the tension and the weight. The centripetal force is their resultant towards the centre, not a third force. And there is no outward force on the bung: the string pulls it inwards only.

Tip

Feel the square. $F = \frac{mv^2}{r}$ : double the mass, double the force; double the speed, FOUR times the force; halve the radius, double the force. That is why a sharp bend is dangerous at speed.

Uniform Circular Motion: Crash Course Physics #7 (CrashCourse). Watch for why the acceleration points to the centre and for the formula a = v²/r.

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Practice

Define centripetal acceleration and state its direction.
A car moves at $12\ \mathrm{m\,s^{-1}}$ round a bend of radius 48 m. Find its centripetal acceleration.
A 0.50 kg ball on a string moves in a horizontal circle of radius 0.80 m at $4.0\ \mathrm{m\,s^{-1}}$ . Find the tension in the string (ignore gravity).
A 0.20 kg mass turns at $10\ \mathrm{rad\,s^{-1}}$ on a circle of radius 0.25 m. Find the centripetal force.
A 60 kg child sits 2.0 m from the centre of a merry-go-round that turns once every 4.0 s. Find the centripetal force on the child. Which force provides it?
The centripetal force on a whirled bung is 8.0 N. Find the force needed (a) at three times the speed, same radius; (b) at the same speed on half the radius.
In Experiment 3 the hanging mass is doubled and the radius kept the same. By what factor must the speed of the bung change?
No string holds the Moon, yet it moves in a near-circle round the Earth. What provides its centripetal force? Why does this force not speed the Moon up?

Show answersHide answers

The acceleration of a body in circular motion, $a = \frac{v^2}{r} = r\omega^2$ , directed towards the centre of the circle.
$a = \frac{12^2}{48} = 3.0\ \mathrm{m\,s^{-2}}$
$T = \frac{mv^2}{r} = \frac{0.50 \times 4.0^2}{0.80} = 10\ \text{N}$
$F = mr\omega^2 = 0.20 \times 0.25 \times 10^2 = 5.0\ \text{N}$
$\omega = \frac{2\pi}{4.0} = 1.57\ \mathrm{rad\,s^{-1}}$ , $F = 60 \times 2.0 \times 1.57^2 = 300\ \text{N}$ (to 2 s.f.), provided by friction on the seat and the child's grip on the bar.
(a) $8.0 \times 3^2 = 72\ \text{N}$ ; (b) $8.0 \times 2 = 16\ \text{N}$ .
$F = \frac{mv^2}{r}$ with $m$ and $r$ fixed, so $v^{2}$ doubles and $v$ increases by a factor $\sqrt{2} = 1.41$ .
The Earth's gravitational pull on the Moon. It always acts at right angles to the Moon's velocity, so it changes only the direction of motion, not the speed.

Key points

In uniform circular motion the change in velocity, and so the acceleration, points to the centre.
Centripetal acceleration $a = \frac{v^2}{r} = r\omega^2$ ; centripetal force $F = \frac{mv^2}{r} = mr\omega^2$ .
The centripetal force is always supplied by a real force: tension, friction, gravity or a normal reaction.
It is at right angles to the velocity, so it changes the direction, not the speed.

Sources and credits

6.2: Centripetal Acceleration and 6.3: Centripetal Force, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Consulted for the physics; no text was copied.
Photo: Swing ride.jpg by Tiia Monto, Wikimedia Commons, licensed under CC BY 4.0.
Video: Uniform Circular Motion: Crash Course Physics #7, CrashCourse, embedded from YouTube.

Written by Ganzaa: the lesson text, the whirling-bung experiment and its illustrative results, the derivation, diagrams, worked examples and practice exercises.

1The hand that keeps pulling

2Experiment 3: measuring the pull needed for a circle

3Explanation: the acceleration points to the centre

4What supplies the centripetal force?

The hand that keeps pulling

Experiment 3: measuring the pull needed for a circle

Explanation: the acceleration points to the centre

What supplies the centripetal force?