Angular Displacement and Angular Speed

A2 · 9 min

Anything that goes round in a circle (a wheel, a satellite, a point on the spinning Earth, an electron in a magnetic field) is best described not by how far it travels but by how much angle it sweeps out. This note sets up the language of circular motion: the radian, angular displacement, angular speed, and the link $v = r\omega$ between angular and linear speed. Every later topic in A Level physics that involves rotation or oscillation (orbits, simple harmonic motion, alternating current) uses these ideas, and Paper 4 questions routinely start with a one-mark conversion between them.

Why measure angles in radians

Degrees are an arbitrary human choice: 360 was picked because it divides nicely. Physics needs an angle unit that is tied to the geometry of the circle itself, so that formulas such as arc length come out without awkward conversion factors. That unit is the radian.

Definition

One radian is the angle subtended at the centre of a circle by an arc of length equal to the radius of the circle.

An arc of length s at radius r subtends an angle θ = s / r radians at the centre.

If the arc length is $s$ and the radius is $r$ , the angle in radians is simply how many radii fit along the arc:

Key result

\theta = \frac{s}{r} \qquad\Longleftrightarrow\qquad s = r\theta \quad (\theta \text{ in radians})

A full circle has circumference $2\pi r$ , so one complete revolution is $2\pi$ radians:

360^\circ = 2\pi\ \text{rad}, \qquad 1\ \text{rad} = \frac{180^\circ}{\pi} \approx 57.3^\circ

Because $\theta$ is a ratio of two lengths, the radian is dimensionless. It is still written as "rad" in units (for example $\text{rad s}^{-1}$ ) to remind the reader that an angle is involved.

Useful conversions to know without thinking:

Degrees	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$	$180^\circ$	$270^\circ$	$360^\circ$
Radians	$\pi/6$	$\pi/4$	$\pi/3$	$\pi/2$	$\pi$	$3\pi/2$	$2\pi$

To convert degrees to radians multiply by $\pi/180$ ; to convert radians to degrees multiply by $180/\pi$ .

Angular displacement

When an object moves around a circle, its position can be described by the angle between the radius to the object and some fixed reference radius. The angular displacement is the change in that angle: the angle swept out by the radius as the object moves. It is measured in radians (or degrees, but A Level formulas assume radians).

A point on a wheel that turns through three complete revolutions has an angular displacement of $3 \times 2\pi = 6\pi$ rad, regardless of the radius at which the point sits. Two points at different radii on the same rigid wheel always share the same angular displacement, even though the point further out travels a greater distance. This is exactly why angular quantities are so convenient for rotation: one number describes the whole rigid body.

Angular speed

Definition

Angular speed $\omega$ is the angle swept out per unit time by the radius joining the object to the centre of the circle, that is, the rate of change of angular displacement:

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

Its unit is the radian per second, $\text{rad s}^{-1}$ .

For uniform circular motion the angular speed is constant. In one period $T$ (the time for one complete revolution) the angle swept is $2\pi$ , so

Key result

\omega = \frac{2\pi}{T} = 2\pi f

where $f = 1/T$ is the frequency (revolutions per second, in Hz).

Engineers often quote rotation rates in revolutions per minute (rpm). To convert to $\text{rad s}^{-1}$ , multiply by $2\pi$ (radians per revolution) and divide by $60$ (seconds per minute).

Linking angular speed and linear speed

In a time $\Delta t$ the radius sweeps an angle $\Delta\theta$ , and the object moves along an arc of length $\Delta s = r\Delta\theta$ . Dividing by $\Delta t$ :

\frac{\Delta s}{\Delta t} = r\frac{\Delta\theta}{\Delta t} \quad\Rightarrow\quad v = r\omega

Key result

v = r\omega

$v$ is the linear speed along the circle (tangential speed), $r$ the radius, and $\omega$ the angular speed in $\text{rad s}^{-1}$ . The formula only works with $\omega$ in radians per second.

The direction of the velocity is always along the tangent to the circle, at right angles to the radius. Even when $v$ is constant in magnitude, the velocity is continually changing direction, which is why circular motion needs a resultant force. That idea is developed in centripetal acceleration and force.

For a rigid rotating body every point has the same $\omega$ , so $v \propto r$ : points twice as far from the axis move twice as fast.

Solving angular-speed problems

Convert any angles to radians and any times to seconds.
If you are given a period, frequency or rpm, find $\omega$ first using $\omega = 2\pi/T = 2\pi f$ .
Use $v = r\omega$ to move between angular and linear speed, with $r$ in metres.
For distances along the circle, use $s = r\theta$ .
Quote the unit: $\text{rad s}^{-1}$ for $\omega$ , $\text{m s}^{-1}$ for $v$ .

Worked examples

Arc length from an angle in degrees

A pendulum bob on a string of length $0.40\ \text{m}$ swings through an angle of $75^\circ$ . Calculate the angle in radians and the length of the arc travelled by the bob.

Solution

Convert to radians:

\theta = 75 \times \frac{\pi}{180} = 1.309\ \text{rad} \approx 1.31\ \text{rad}

Arc length:

s = r\theta = 0.40 \times 1.309 = 0.524\ \text{m} \approx 0.52\ \text{m}

The answer is given to 2 significant figures to match the data.

The rotating Earth

The Earth rotates once on its axis every $24$ hours. The radius of the Earth is $6.37 \times 10^{6}\ \text{m}$ .

(a) Calculate the angular speed of the Earth.

(b) Calculate the linear speed of a point on the Equator.

Solution

(a)

\omega = \frac{2\pi}{T} = \frac{2\pi}{24 \times 3600} = 7.27 \times 10^{-5}\ \text{rad s}^{-1}

(b) On the Equator the radius of the circle is the radius of the Earth:

v = r\omega = 6.37 \times 10^{6} \times 7.27 \times 10^{-5} = 463\ \text{m s}^{-1}

(c) At latitude $50^\circ$ the point moves in a smaller circle whose radius is the distance from the axis, $r = R\cos 50^\circ = 6.37 \times 10^{6} \times 0.643 = 4.09 \times 10^{6}\ \text{m}$ . The angular speed is the same, so

v = 4.09 \times 10^{6} \times 7.27 \times 10^{-5} = 298\ \text{m s}^{-1}

The key physics: every point on the Earth has the same $\omega$ , but $v$ depends on the radius of the circle actually traced out.

A car wheel

A car travels at a constant $25\ \text{m s}^{-1}$ . Its wheels have diameter $0.62\ \text{m}$ and do not slip. Calculate the angular speed of a wheel in $\text{rad s}^{-1}$ and in revolutions per minute.

Solution

If the wheel does not slip, the speed of the rim relative to the axle equals the speed of the car. Radius $r = 0.31\ \text{m}$ .

\omega = \frac{v}{r} = \frac{25}{0.31} = 80.6\ \text{rad s}^{-1} \approx 81\ \text{rad s}^{-1}

Revolutions per second: $f = \omega / 2\pi = 12.8\ \text{Hz}$ . Revolutions per minute: $12.8 \times 60 = 770$ rpm.

Hard drive platter

A computer hard-disc platter spins at $7200$ revolutions per minute. A data track lies $4.0\ \text{cm}$ from the axis. Calculate the speed of the track relative to the read head.

Solution

\omega = \frac{7200 \times 2\pi}{60} = 754\ \text{rad s}^{-1}

v = r\omega = 0.040 \times 754 = 30\ \text{m s}^{-1}

Remember to convert $4.0\ \text{cm}$ to $0.040\ \text{m}$ before multiplying.

When do the hands of a clock next coincide?

At 12

the hour hand and minute hand of a clock point in the same direction. Calculate the angular speed of each hand and hence the time after 12

at which they next point in the same direction.

Solution

The minute hand has period $3600\ \text{s}$ and the hour hand $12 \times 3600 = 43\,200\ \text{s}$ :

\omega_m = \frac{2\pi}{3600} = 1.745 \times 10^{-3}\ \text{rad s}^{-1}, \qquad \omega_h = \frac{2\pi}{43\,200} = 1.454 \times 10^{-4}\ \text{rad s}^{-1}

The hands next coincide when the minute hand has gained one full revolution ( $2\pi$ rad) on the hour hand:

(\omega_m - \omega_h)\,t = 2\pi \quad\Rightarrow\quad t = \frac{2\pi}{1.745 \times 10^{-3} - 1.454 \times 10^{-4}} = 3930\ \text{s}

That is $65.5$ minutes, so the hands coincide at about 13

(exactly

65\tfrac{5}{11}

minutes after 12

Watch out

Calculator in the wrong mode. Formulas such as $s = r\theta$ and $v = r\omega$ assume radians. If you use $\theta$ in degrees you will be wrong by a factor of $57.3$ . Before any trigonometry in a circular motion or oscillation question, check your calculator mode.

Watch out

Confusing $f$ and $\omega$ . Frequency $f$ is in revolutions per second (Hz); angular speed $\omega$ is in radians per second. They differ by a factor of $2\pi$ . A wheel turning at $10\ \text{Hz}$ has $\omega = 62.8\ \text{rad s}^{-1}$ , not $10\ \text{rad s}^{-1}$ .

Watch out

Using the wrong radius. For a point on the surface of a rotating sphere at latitude $\phi$ , the circle it traces has radius $R\cos\phi$ , not $R$ . Always ask: what is the radius of the circle this particular object moves in?

Exam tip

"Define the radian" is a common one- or two-mark question. The mark scheme wants: angle subtended at the centre of a circle by an arc equal in length to the radius. Missing "at the centre" or "arc length equal to radius" loses the mark.
When asked to "show that" $\omega$ has a certain value, write the formula, substitute with units, and give the answer to one more significant figure than the value shown.
The unit of angular speed is $\text{rad s}^{-1}$ . Writing $\text{s}^{-1}$ alone is usually accepted, but "rpm" or "Hz" is not.

Summary

One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius; $2\pi$ rad $= 360^\circ$ .
Arc length $s = r\theta$ with $\theta$ in radians.
Angular displacement is the angle swept out by the radius.
Angular speed $\omega = \Delta\theta/\Delta t$ in $\text{rad s}^{-1}$ ; for uniform motion $\omega = 2\pi/T = 2\pi f$ .
Linear (tangential) speed $v = r\omega$ ; velocity is along the tangent.
All points on a rigid rotating body share the same $\omega$ ; their speeds are proportional to their distance from the axis.

Practice questions

Question

Convert $2.5\ \text{rad}$ to degrees, and $135^\circ$ to radians in terms of $\pi$ .
A fan blade rotates at $1500$ rpm. Calculate its angular speed in $\text{rad s}^{-1}$ .
A satellite completes one orbit of radius $7.0 \times 10^{6}\ \text{m}$ in $98$ minutes. Calculate its angular speed and its orbital speed.
The tip of a helicopter rotor blade of length $5.2\ \text{m}$ moves at $210\ \text{m s}^{-1}$ . Calculate the angular speed of the rotor and the speed of a point $2.0\ \text{m}$ from the axis.
A vinyl record rotates at $33\tfrac{1}{3}$ rpm. Calculate the time for one revolution and the angle in degrees turned through in $0.10\ \text{s}$ .
A bicycle wheel of radius $0.34\ \text{m}$ rolls without slipping through a distance of $50\ \text{m}$ . Calculate the angular displacement of the wheel in radians and the number of revolutions.
Explain why two children sitting at different distances from the centre of a rotating roundabout have the same angular speed but different linear speeds.
Two runners on a circular track start together and run in the same direction. Runner A completes a lap in $72\ \text{s}$ and runner B in $80\ \text{s}$ . Using angular speeds, calculate the time taken for A to lap B for the first time.

Answers

$2.5 \times 180/\pi = 143^\circ$ . $135 \times \pi/180 = 3\pi/4$ rad.
$\omega = 1500 \times 2\pi / 60 = 157\ \text{rad s}^{-1}$ ( $50\pi$ ).
$T = 98 \times 60 = 5880\ \text{s}$ ; $\omega = 2\pi/5880 = 1.07 \times 10^{-3}\ \text{rad s}^{-1}$ ; $v = r\omega = 7.0 \times 10^{6} \times 1.07 \times 10^{-3} = 7.5 \times 10^{3}\ \text{m s}^{-1}$ .
$\omega = v/r = 210/5.2 = 40.4\ \text{rad s}^{-1}$ . At $2.0\ \text{m}$ : $v = 2.0 \times 40.4 = 81\ \text{m s}^{-1}$ .
$f = 33.33/60 = 0.556\ \text{Hz}$ , so $T = 1.80\ \text{s}$ . $\omega = 2\pi f = 3.49\ \text{rad s}^{-1}$ ; in $0.10\ \text{s}$ , $\theta = 0.349\ \text{rad} = 20^\circ$ .
$\theta = s/r = 50/0.34 = 147\ \text{rad}$ . Revolutions $= 147/2\pi = 23.4$ .
The roundabout is rigid, so every radius sweeps the same angle in the same time: same $\omega$ . Linear speed $v = r\omega$ , so the child at the larger radius travels a longer arc in the same time and has the greater speed.
$\omega_A = 2\pi/72 = 0.08727\ \text{rad s}^{-1}$ , $\omega_B = 2\pi/80 = 0.07854\ \text{rad s}^{-1}$ . A laps B when it has gained $2\pi$ : $t = 2\pi/(0.08727 - 0.07854) = 720\ \text{s}$ . (Check: in $720\ \text{s}$ A runs $10$ laps and B runs $9$ .)

Why measure angles in radians

Angular displacement

Angular speed

Linking angular speed and linear speed

Worked examples

Practice questions

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