Centripetal Acceleration and Force

A2 · 10 min

An object moving in a circle at constant speed is accelerating, because its velocity keeps changing direction. That acceleration points towards the centre of the circle and needs a resultant force towards the centre to produce it. This single idea, $F = mv^2/r$ , explains how satellites orbit, how charged particles curve in magnetic fields and why cars skid on tight bends, and it is examined on almost every Paper 4.

Why circular motion needs a force

Newton's first law says an object keeps moving in a straight line at constant speed unless a resultant force acts on it. Moving in a circle is not moving in a straight line, so there must be a resultant force.

Think about which way that force must point. If it had any component along the direction of motion, it would speed the object up or slow it down. In uniform circular motion the speed does not change, so the force has no component along the velocity: it is always perpendicular to the velocity. A force that is always perpendicular to the velocity, and constant in size, continually turns the velocity without changing its magnitude. The path that results is a circle, and the force points towards its centre.

Key result

A force of constant magnitude that is always perpendicular to the direction of motion causes an acceleration towards the centre of a circle: the centripetal acceleration.
This centripetal acceleration produces circular motion at constant speed and therefore constant angular speed.
Because the force is perpendicular to the velocity, it does no work on the object: the kinetic energy stays constant.

"Centripetal" means "centre-seeking". It describes the direction of the resultant force, not a new kind of force. In every problem the centripetal force is supplied by something real: tension in a string, friction between tyres and road, gravitational attraction, an electric force, a magnetic force on a moving charge, or the normal contact force from a surface.

Deriving the centripetal acceleration

Consider an object moving at speed $v$ around a circle of radius $r$ . In a short time $\Delta t$ it moves through a small angle $\Delta\theta$ , and its velocity vector turns through the same angle $\Delta\theta$ .

Left: the velocity turns through Δθ as the object moves through Δθ. Right: the vector triangle v₂ = v₁ + Δv; for small Δθ the change Δv (dashed) points towards the centre.

Both velocity vectors have magnitude $v$ . The vector triangle is an isosceles triangle with two sides of length $v$ and angle $\Delta\theta$ between them. For a small angle, the third side is approximately an arc of radius $v$ :

|\Delta v| \approx v\,\Delta\theta

As $\Delta\theta \to 0$ , $\Delta v$ becomes perpendicular to the velocity, pointing towards the centre. The acceleration is

a = \frac{\Delta v}{\Delta t} = v\frac{\Delta\theta}{\Delta t} = v\omega

Using $v = r\omega$ this can be written in two equivalent forms.

Key result

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

directed towards the centre of the circle.

The derivation is not explicitly required by the syllabus, but understanding it explains why the acceleration is towards the centre and why it grows with speed.

Centripetal force

Newton's second law, $F = ma$ , applied to the centripetal acceleration gives the resultant force needed to keep a mass $m$ moving in a circle.

Key result

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

$F$ is the resultant force on the object, directed towards the centre. It is not an extra force to add to a free-body diagram.

Which form to use depends on the data: if you are given a period, frequency or rpm, use $mr\omega^2$ ; if you are given a speed, use $mv^2/r$ .

How the force depends on the variables is worth seeing clearly:

Change	Effect on $F$
Double $v$ at fixed $r$	$F$ is four times larger ( $F \propto v^2$ )
Double $r$ at fixed $v$	$F$ halves ( $F \propto 1/r$ )
Double $r$ at fixed $\omega$	$F$ doubles ( $F \propto r$ )
Double $m$	$F$ doubles

The apparent contradiction in the middle two rows disappears once you notice what is held constant. On a rotating turntable everything has the same $\omega$ , so objects further out need more force. On a road where cars travel at the same speed, tighter bends (smaller $r$ ) need more force.

What happens when the force is removed

If the string breaks, or the friction is not large enough, the centripetal force disappears. The object then moves off along the tangent to the circle at the point where it was released, in a straight line (ignoring gravity), exactly as Newton's first law predicts. It does not fly outwards along the radius.

Watch out

There is no outward "centrifugal force" in an inertial frame. Passengers in a turning car feel pushed outwards, but what actually happens is that their bodies try to continue in a straight line while the car turns beneath them; the door then pushes them inwards. On a free-body diagram draw only real forces (weight, tension, friction, normal contact). Their resultant is the centripetal force. Never add an extra arrow labelled "centripetal force" or "centrifugal force".

Solving circular motion problems

Identify the circle: its centre, its radius and the plane it lies in.
Draw a free-body diagram showing only real forces.
Resolve the forces into a component towards the centre (and, if needed, a component perpendicular to the plane of the circle).
Set the resultant towards the centre equal to $mv^2/r$ or $mr\omega^2$ .
If the object does not accelerate perpendicular to the plane of the circle, the forces in that direction balance.
Solve, and check units: $\text{N} = \text{kg m s}^{-2}$ .

Worked examples

A puck on a string

A puck of mass $0.25\ \text{kg}$ is attached to a string and moves in a horizontal circle of radius $0.60\ \text{m}$ on a frictionless table. It makes $2.0$ revolutions per second. Calculate the tension in the string.

Solution

The tension is the only horizontal force, so it provides the centripetal force.

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

T = mr\omega^{2} = 0.25 \times 0.60 \times (12.6)^{2} = 23.7\ \text{N} \approx 24\ \text{N}

A car on a roundabout

A car of mass $1200\ \text{kg}$ travels round a level roundabout of radius $25\ \text{m}$ at a constant speed of $12\ \text{m s}^{-1}$ .

(a) Calculate the friction force on the car from the road.

(b) The maximum friction force available is $9.0\ \text{kN}$ . Calculate the maximum speed at which the car can go round without skidding.

Solution

(a) On a level road the only horizontal force is friction, so it provides the centripetal force:

F = \frac{mv^{2}}{r} = \frac{1200 \times 12^{2}}{25} = 6.9 \times 10^{3}\ \text{N}

directed towards the centre of the roundabout.

(b) At the maximum speed the friction is at its maximum:

9000 = \frac{1200\,v_{\max}^{2}}{25} \quad\Rightarrow\quad v_{\max} = \sqrt{\frac{9000 \times 25}{1200}} = 13.7\ \text{m s}^{-1}

Above this speed the friction cannot supply enough force, and the car slides towards the outside of the bend along a path closer to the tangent.

Centripetal acceleration of the Moon

The Moon orbits the Earth in a circle of radius $3.84 \times 10^{8}\ \text{m}$ with a period of $27.3$ days. Calculate its centripetal acceleration.

Solution

\omega = \frac{2\pi}{T} = \frac{2\pi}{27.3 \times 24 \times 3600} = 2.66 \times 10^{-6}\ \text{rad s}^{-1}

a = r\omega^{2} = 3.84 \times 10^{8} \times (2.66 \times 10^{-6})^{2} = 2.72 \times 10^{-3}\ \text{m s}^{-2}

Newton compared this with $g = 9.81\ \text{m s}^{-2}$ at the Earth's surface. The ratio is about $1/3600 = 1/60^2$ , and the Moon is about $60$ Earth radii away: evidence for an inverse-square law of gravity. See Newton's law of gravitation.

Coins on a turntable

Two identical coins sit on a horizontal turntable, coin P at $5.0\ \text{cm}$ from the axis and coin Q at $12.0\ \text{cm}$ . The maximum friction force on each coin is $0.040\ \text{N}$ and each has mass $8.0\ \text{g}$ . The turntable's angular speed is slowly increased. State and explain which coin slides first, and calculate the angular speed at which it does so.

Solution

Both coins have the same $\omega$ because they are on the same rigid turntable. The friction needed is $F = mr\omega^2$ , which is proportional to $r$ . Coin Q, further out, needs more friction at any given $\omega$ , so it reaches the limit first and slides first.

For Q:

0.040 = 0.0080 \times 0.120 \times \omega^{2} \quad\Rightarrow\quad \omega = \sqrt{\frac{0.040}{0.00096}} = 6.45\ \text{rad s}^{-1}

(P would not slide until $\omega = \sqrt{0.040/(0.0080 \times 0.050)} = 10.0\ \text{rad s}^{-1}$ .)

Electron in a circular path

An electron moves at $2.2 \times 10^{6}\ \text{m s}^{-1}$ in a circle of radius $5.3 \times 10^{-11}\ \text{m}$ (a simple model of the hydrogen atom). Calculate its centripetal acceleration and the centripetal force on it. The mass of an electron is $9.11 \times 10^{-31}\ \text{kg}$ .

Solution

a = \frac{v^{2}}{r} = \frac{(2.2 \times 10^{6})^{2}}{5.3 \times 10^{-11}} = 9.1 \times 10^{22}\ \text{m s}^{-2}

F = ma = 9.11 \times 10^{-31} \times 9.1 \times 10^{22} = 8.3 \times 10^{-8}\ \text{N}

This force is supplied by the electric attraction of the proton (see Coulomb's law). The enormous acceleration is a reminder that the formula applies at every scale.

Watch out

Using diameter instead of radius, or forgetting to convert cm to m, are the two most common arithmetic slips. Write $r = \ldots\ \text{m}$ explicitly before substituting.

Watch out

Saying the speed changes. In uniform circular motion the speed is constant; the velocity changes because its direction changes. "The object accelerates because its velocity changes direction" earns the mark; "the object accelerates because it speeds up" loses it.

Exam tip

"Explain why an object moving in a circle at constant speed is accelerating" (2 marks): velocity is a vector; its direction changes continuously; so there is a change in velocity, hence acceleration (towards the centre).
"State what provides the centripetal force" means name the real force: "the gravitational force of the Earth on the satellite", "the tension in the string", "friction between the tyres and the road".
"Explain why the work done by the centripetal force is zero": the force is perpendicular to the displacement (velocity) at every instant.
Show $\omega$ or $v$ as a separate line of working. If you combine everything into one calculator entry and make a slip, there is nothing for the examiner to credit.

Summary

Uniform circular motion: constant speed, continuously changing velocity, so there is an acceleration.
A constant-magnitude force always perpendicular to the velocity causes circular motion at constant angular speed.
Centripetal acceleration $a = r\omega^2 = v^2/r$ , towards the centre.
Centripetal force $F = mr\omega^2 = mv^2/r$ is the resultant force, supplied by real forces.
The centripetal force does no work, so kinetic energy is constant.
If the force is removed, the object moves off along the tangent.
At fixed $\omega$ , $F \propto r$ ; at fixed $v$ , $F \propto 1/r$ .

Practice questions

Question

A ball of mass $0.50\ \text{kg}$ moves in a horizontal circle of radius $1.2\ \text{m}$ at $4.0\ \text{m s}^{-1}$ . Calculate the centripetal acceleration and centripetal force.
A centrifuge spins a sample at $3000$ rpm at a radius of $8.0\ \text{cm}$ . Calculate the centripetal acceleration as a multiple of $g$ .
A stone on a string of length $0.80\ \text{m}$ is whirled in a horizontal circle (ignore gravity) and the string breaks at a tension of $60\ \text{N}$ . The stone has mass $0.15\ \text{kg}$ . Calculate the maximum speed and describe the path after the string breaks.
Explain why the centripetal force on an object in uniform circular motion does no work.
A car takes a bend of radius $40\ \text{m}$ . The maximum friction force is $0.60$ times its weight. Calculate the maximum speed on a level road.
The speed of an object in a circle doubles while the radius halves. By what factor does the centripetal force change?
A child of mass $30\ \text{kg}$ sits $1.5\ \text{m}$ from the centre of a roundabout that turns once every $4.0\ \text{s}$ . Calculate the resultant horizontal force on the child and state its direction.
A $2.0\ \text{kg}$ mass is attached to a spring of natural length $0.40\ \text{m}$ and spring constant $800\ \text{N m}^{-1}$ . The other end of the spring is fixed to a vertical axle and the mass moves in a horizontal circle on a frictionless surface at $5.0\ \text{rad s}^{-1}$ . Calculate the radius of the circle.

Answers

$a = v^2/r = 16/1.2 = 13.3\ \text{m s}^{-2}$ ; $F = ma = 0.50 \times 13.3 = 6.7\ \text{N}$ .
$\omega = 3000 \times 2\pi/60 = 314\ \text{rad s}^{-1}$ ; $a = r\omega^2 = 0.080 \times 314^2 = 7.9 \times 10^{3}\ \text{m s}^{-2}$ , which is about $800g$ .
$60 = 0.15 v^2/0.80$ , so $v = \sqrt{320} = 17.9\ \text{m s}^{-1}$ . After breaking it moves off along the tangent to the circle at that point (in a straight line if gravity is ignored).
The force is always perpendicular to the velocity, so it has no component along the displacement; work $= Fs\cos 90^\circ = 0$ . The kinetic energy (and speed) stays constant.
$0.60mg = mv^2/r \Rightarrow v = \sqrt{0.60 \times 9.81 \times 40} = 15.3\ \text{m s}^{-1}$ . The mass cancels.
$F \propto v^2/r$ : factor $= 2^2 / 0.5 = 8$ .
$\omega = 2\pi/4.0 = 1.571\ \text{rad s}^{-1}$ ; $F = mr\omega^2 = 30 \times 1.5 \times 2.467 = 111\ \text{N} \approx 110\ \text{N}$ , towards the centre of the roundabout (supplied by friction from the seat).
Extension $x$ , radius $r = 0.40 + x$ . Spring force provides centripetal force: $800x = 2.0 \times 25 \times (0.40 + x) = 20 + 50x$ . So $750x = 20$ , $x = 0.0267\ \text{m}$ and $r = 0.427\ \text{m}$ ( $0.43\ \text{m}$ ).

Why circular motion needs a force

Deriving the centripetal acceleration

Centripetal force

What happens when the force is removed

Worked examples

Practice questions

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