Linear and Angular Speed – Uniform Circular Motion (Senior 3 Physics) | Ganzaa

Uniform Circular Motion

Lessons

Senior 3 · Physics

Uniform Circular Motion

4 lessons

Lesson 2 of 4

Linear and Angular Speed

25 min

Sparks along a straight line

At a market stall, a knife sharpener presses a blade against a spinning grinding wheel. A shower of sparks shoots out. The sparks are tiny, glowing pieces of metal and stone torn off at the edge of the wheel, and they do not curl round with the wheel: they fly off in straight lines. On a wet murram road, mud from a bicycle tyre does the same. It leaves the tyre and flies off in a straight line, which is why bicycles and motos have mudguards.

Each spark and each lump of mud leaves in the direction it was moving at the moment it let go. That direction tells us which way the velocity of a body in circular motion points.

Experiment 2: where does a marble go when the wall ends?

You need: a marble, a strip of stiff card about 4 cm wide, tape, and a smooth table.

Bend the card into a ring about 25 cm across and tape it to the table, standing on its edge. Cut out a gap about 5 cm wide in the ring.
Roll the marble fast round the inside of the ring, starting well away from the gap, so that it runs along the wall.
Before it reaches the gap, predict its path after the gap: will it keep curving, go straight outwards from the centre, or go straight on? Mark your prediction with a pencil.
Watch where it goes. Repeat three times.

Figure 2.1: While the wall pushes on it, the marble moves in a circle. At the gap the push stops, and the marble goes straight on along the tangent to the circle at the point where it left.

Observation: the marble leaves the gap in a straight line that just touches the circle at the gap. It does not keep curving, and it does not move straight away from the centre.

Important

Safety. Work on a table with a raised edge or a book at the side, so that the marble does not roll onto the floor. Pick up any marble that falls straight away: it can make someone slip.

Definition

A tangent to a circle is a straight line that touches the circle at one point without crossing it. At that point it is at right angles (90°) to the radius.

The linear velocity (tangential velocity) $\vec{v}$ of a body in circular motion is its velocity along the path. Its size is the linear speed $v = \frac{\text{distance along the arc}}{\text{time}}$ , in $\mathrm{m\,s^{-1}}$ , and its direction is along the tangent at the body's position.

Explanation: the velocity keeps changing direction

Figure 2.2: In uniform circular motion the speed is constant, so the velocity arrows all have the same length. Each arrow lies along the tangent, so the direction keeps changing as the body goes round.

Velocity is a vector: it has a size and a direction. At P the body moves up the page, at Q to the left, at S down the page and at U to the right. Even though the speed never changes, the velocity is different at every point. A changing velocity means an acceleration, so a body in uniform circular motion is always accelerating. You will find out in the next lesson which way this acceleration points and what causes it.

Linking linear speed and angular velocity

In a time $t$ , a body on a circle of radius $r$ turns through an angle $\theta$ and travels along an arc of length $s$ . From the definition of the radian,

s = r\theta

Divide both sides by the time $t$ :

\frac{s}{t} = r\,\frac{\theta}{t} \qquad\Rightarrow\qquad v = r\omega

For one complete turn the distance is the circumference $2\pi r$ and the time is the period $T$ , which gives two more useful forms:

v = \frac{2\pi r}{T} = 2\pi r f

Straight-line quantity	Circular (angular) quantity	Link
distance along the path, $s$ (m)	angular displacement, $\theta$ (rad)	$s = r\theta$
linear speed, $v$ ( $\mathrm{m\,s^{-1}}$ )	angular velocity, $\omega$ ( $\mathrm{rad\,s^{-1}}$ )	$v = r\omega$
time for one lap	period, $T$ (s)	$v = \frac{2\pi r}{T}$

Figure 2.3: All points of a turning body share the same angular velocity $\omega$ , but their linear speeds are proportional to their distance from the centre, $v = r\omega$ .

Worked example 1: a bicycle wheel

A bicycle wheel of radius 0.35 m makes 3.0 turns per second. Find (a) its angular velocity and (b) the speed of a point on the tyre.

(a) $\omega = 2\pi f = 2\pi \times 3.0\ \text{Hz} = 18.8\ \mathrm{rad\,s^{-1}}$

(b) $v = r\omega = 0.35\ \text{m} \times 18.85\ \mathrm{rad\,s^{-1}} = 6.6\ \mathrm{m\,s^{-1}}$

That is $6.6 \times 3.6 = 24\ \mathrm{km\,h^{-1}}$ . If the tyre does not slip, this is also the speed of the bicycle along the road.

Worked example 2: riding on the turning Earth

The Earth's radius is $6.38 \times 10^{6}\ \text{m}$ and it turns once in about $86\,400\ \text{s}$ . Find the speed of a point on the equator.

v = \frac{2\pi r}{T} = \frac{2\pi \times 6.38 \times 10^{6}\ \text{m}}{86\,400\ \text{s}} = 464\ \mathrm{m\,s^{-1}}

That is about $1670\ \mathrm{km\,h^{-1}}$ , faster than a passenger jet. Rwanda lies just south of the equator, so the ground under your school moves at almost this speed. You do not feel it because you, the air and the ground all move together.

Worked example 3: from road speed to wheel turns

A car travels at $72\ \mathrm{km\,h^{-1}}$ on tyres of radius 0.30 m. How many revolutions per minute do the wheels make?

Speed: $v = \frac{72}{3.6} = 20\ \mathrm{m\,s^{-1}}$ .

\omega = \frac{v}{r} = \frac{20\ \mathrm{m\,s^{-1}}}{0.30\ \text{m}} = 66.7\ \mathrm{rad\,s^{-1}}

f = \frac{\omega}{2\pi} = 10.6\ \text{Hz}

In one minute: $10.6 \times 60 = 640\ \text{rev/min}$ (to 2 significant figures).

Warning

Put $\omega$ in $\mathrm{rad\,s^{-1}}$ before using $v = r\omega$ . A wheel of radius 0.50 m turning at 2 rev/s does NOT have $v = 0.50 \times 2 = 1\ \mathrm{m\,s^{-1}}$ . First change to radians: $\omega = 2\pi \times 2 = 12.6\ \mathrm{rad\,s^{-1}}$ , so $v = 0.50 \times 12.6 = 6.3\ \mathrm{m\,s^{-1}}$ .

Tip

Same body, same $\omega$ . Everything fixed to one turning object (spokes, rim, a child on a merry-go-round) turns through the same angle in the same time. Only the speed $v = r\omega$ grows as you move outwards: the outer horses on a roundabout ride go fastest.

Relationship between angular velocity and speed (Khan Academy). Watch how the arc length s = rθ leads to v = rω.

✦

Practice

What is a tangent to a circle? In which direction does the velocity of a body in circular motion point?
A point moves round a circle of radius 0.50 m with an angular velocity of $6.0\ \mathrm{rad\,s^{-1}}$ . Find its linear speed.
The minute hand of a wall clock is 12 cm long. Find the speed of its tip.
A wheel of radius 0.40 m makes 120 turns each minute. Find the speed of a point on its rim.
A car moves at $15\ \mathrm{m\,s^{-1}}$ on tyres of radius 0.30 m. Find the angular velocity of the wheels and the number of revolutions per minute.
Two children sit on a merry-go-round, one 1.0 m and the other 2.5 m from the centre. It turns once every 5.0 s. Find the angular velocity and the speed of each child.
Explain why mud flies off a turning bicycle tyre, and why a mudguard is curved to follow the top of the wheel.
Seen from above, a pupil whirls a stone anticlockwise in a horizontal circle. She lets go when the stone is due north of her hand. In which direction does the stone fly off? Explain.

Show answersHide answers

A straight line that touches the circle at one point, at right angles to the radius there. The velocity points along the tangent, in the direction of motion.
$v = r\omega = 0.50 \times 6.0 = 3.0\ \mathrm{m\,s^{-1}}$
$\omega = \frac{2\pi}{3600} = 1.75 \times 10^{-3}\ \mathrm{rad\,s^{-1}}$ , so $v = 0.12 \times 1.75 \times 10^{-3} = 2.1 \times 10^{-4}\ \mathrm{m\,s^{-1}}$ .
$f = \frac{120}{60} = 2.0\ \text{Hz}$ ; $v = 2\pi r f = 2\pi \times 0.40 \times 2.0 = 5.0\ \mathrm{m\,s^{-1}}$ .
$\omega = \frac{15}{0.30} = 50\ \mathrm{rad\,s^{-1}}$ ; $f = \frac{50}{2\pi} = 7.96\ \text{Hz}$ , about 480 rev/min.
Both: $\omega = \frac{2\pi}{5.0} = 1.26\ \mathrm{rad\,s^{-1}}$ . Speeds: $1.0 \times 1.26 = 1.3\ \mathrm{m\,s^{-1}}$ and $2.5 \times 1.26 = 3.1\ \mathrm{m\,s^{-1}}$ .
Mud that loses its grip on the tyre keeps the velocity it had, which is along the tangent, so it flies off in a straight line. A curved mudguard lies close to the tyre all round the top, so it catches mud leaving along any of these tangents.
Due west. Moving anticlockwise, the stone at the north point is travelling westwards along the tangent, and when released it keeps that velocity.

Key points

The velocity of a body in circular motion points along the tangent, at 90° to the radius.
In uniform circular motion the speed is constant but the direction of the velocity changes, so the body is accelerating.
$s = r\theta$ and $v = r\omega = \frac{2\pi r}{T} = 2\pi r f$ , with $\omega$ in $\mathrm{rad\,s^{-1}}$ .
All points of a rigid turning body share the same $\omega$ ; points farther out move faster.

Sources and credits

6.1: Rotation Angle and Angular Velocity, 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: Metal sparks.jpg by Przemek P, Wikimedia Commons, licensed under CC BY-SA 3.0.
Video: Relationship between angular velocity and speed | Physics | Khan Academy, Khan Academy, embedded from YouTube.

Written by Ganzaa: the lesson text, the marble experiment, the diagrams, worked examples and practice exercises.

1Sparks along a straight line

2Experiment 2: where does a marble go when the wall ends?

3Explanation: the velocity keeps changing direction

4Linking linear speed and angular velocity

Sparks along a straight line

Experiment 2: where does a marble go when the wall ends?

Explanation: the velocity keeps changing direction

Linking linear speed and angular velocity