Circular Motion

Understanding angular speed, centripetal acceleration and centripetal force

SPEC 3.6.1.1 • FURTHER MECHANICS

Motion in a Circular Path

An object moving in a circle at constant speed has a constantly changing velocity. Velocity is a vector with both magnitude and direction - when direction changes, velocity changes.

Key Principle

Motion in a circular path at constant speed implies there is an acceleration (centripetal acceleration) and requires a centripetal force.

Newton's First Law

An object continues in uniform motion unless acted on by a resultant force. Since direction changes, there must be a resultant force - the centripetal force.

Radians

Angles in circular motion are measured in radians. There are 2π radians in a full circle (360°).

Definition

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

θ_{r a d} = θ_{d e g} \times \frac{π}{180}

s = r θ

Δ θ = \frac{Δ s}{r}

Exam Tip

The equation s = rθ only works when θ is in radians. Always check your calculator is in radian mode!

Angular Speed

Definition

Angular speed (ω) is the angle rotated through per unit time. Unit: rad s⁻¹

ω = \frac{Δ θ}{Δ t}

ω = \frac{2 π}{T}

ω = 2 π f

Where T is the period (time for one revolution) and f is the frequency. Since T = 1/f, these are equivalent.

Linking Linear and Angular Speed

From Δs = rΔθ, dividing by Δt gives:

v = r ω

Linear speed = radius × angular speed. Points further from the centre move faster!

QuantitySymbolUnitKey Formulas

Angular speedωrad s⁻¹

ω = v / r = 2 π f = 2 π / T

Linear speedvm s⁻¹

v = r ω = 2 π r f

PeriodTs

T = 1/ f = 2 π / ω

01Worked Example - Angular Speed

Calculate the angular velocity of an object that makes:

(a) 25 revolutions per second

(b) 33.3 revolutions per minute

a)

f = 25 Hz, T = 1/f = 0.04 s

•

ω = 2 π f = 2 π \times 25 = 50 π =

157 rad s⁻¹

b)

f = 33.3/60 = 0.556 Hz, T = 1.80 s

•

ω = 2 π / T = 2 π /1.80 =

3.49 rad s⁻¹

Centripetal Acceleration

Even when an object moves at constant speed in a circle, it is accelerating because:

Velocity is a vector - it has both magnitude (speed) and direction

The direction of velocity is constantly changing as the object moves around the circle

A change in velocity means there must be an acceleration (a = Δv/Δt)

Key Result

Using vector analysis, we can show that the change in velocity Δv always points towards the centre of the circle. Since a = Δv/Δt, the acceleration also points towards the centre. This is called centripetal acceleration (from Latin: "centre-seeking").

Centripetal Acceleration Formulas

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

a = ω^{2} r

02Worked Example - Period and Acceleration

An object moves at 5 m s⁻¹ in a circle of radius 3 m. Find:

(a) the period (b) the frequency (c) the angular velocity (d) the centripetal acceleration

a)

Circumference = 2πr = 2π × 3 = 6π m; T = 6π/5 = 3.77 s

b)

f = 1/T = 1/3.77 = 0.265 Hz

c)

ω = v/r = 5/3 = 1.67 rad s⁻¹

d)

a = v²/r = 25/3 = 8.33 m s⁻²

Centripetal Force

By Newton's second law (F = ma), centripetal acceleration requires a centripetal force.

Definition

Centripetal force is the resultant force acting towards the centre, causing circular motion. It is always perpendicular to velocity.

F = \frac{m v ^{2}}{r}

F = m ω^{2} r

What Provides Centripetal Force?

It's NOT a new force - it's provided by: Tension (string), Friction (car on road), Gravity (orbits), Normal force (wall of death)

03Worked Example - Stone on String

A 4 kg stone is swung in a horizontal circle of radius 2 m at 4 m s⁻¹.

(a) What is the centripetal force? (b) How many revolutions per second? (c) What happens to F if r halves (v constant)?

a)

F = m v^{2} / r = 4 \times 16/2 =

32 N

b)

T = 2πr/v = 2π×2/4 = π s; f = 1/π = 0.318 rev/s

c)

If r = 1 m: F = 4×16/1 = 64 N (force doubles)

Worked Example - Car on Bend04

The friction force between a 1000 kg car and road is 10000 N. What is the maximum speed for a corner of radius 40 m?

1

Friction provides centripetal force: F = 10000 N

2

F = m v^{2} / r \Rightarrow 10000 = 1000 \times v^{2} /40

3

v^{2} = 10000 \times 40/1000 = 400

4

v = 400 =

20 m s⁻¹

Note: Use mass (1000 kg), NOT weight. Above 20 m s⁻¹ the car skids.

Artificial Gravity

A rotating space station creates artificial gravity. The "floor" is the outer edge; the normal force provides centripetal force.

Key Result

For artificial gravity equal to g, the rotation rate needed is:

ω = \frac{g}{r}

How It Works

Astronauts stand on the outer rim. The normal force from the "floor" provides centripetal force. If ω²r = g, they feel normal Earth weight.

Typical Mistake

Using diameter instead of radius in circular motion calculations!

Worked Example - Centrifuge05

An astronaut is spun in a centrifuge of diameter 6 m. What velocity gives acceleration of 8g?

1

a = 8g = 8 × 9.81 = 78.48 m s⁻²

2

Radius = diameter/2 = 3 m (common mistake: using diameter!)

3

a = v²/r → v = √(ar) = √(78.48 × 3) = 15.3 m s⁻¹

Carousel / Conical Pendulum

On carousel rides, chairs swing out at an angle. Tension resolves into vertical (balances weight) and horizontal (centripetal force) components.

Force Resolution

Vertical: T cos θ = mg

Horizontal: T sin θ = mω²r

Key Result

Dividing horizontal by vertical:

tan θ = \frac{r ω ^{2}}{g}

Angle depends on ω and r, but NOT on mass!

06Worked Example - Conical Pendulum

A 0.5 kg mass on a 0.8 m string rotates as a conical pendulum. If the string makes an angle of 30° with the vertical, calculate:

(a) the tension in the string

(b) the angular velocity

a)

Vertical equilibrium: T cos θ = mg

•

T = mg/cos θ = (0.5 × 9.81)/cos 30° = 4.905/0.866 = 5.66 N

b)

Horizontal: T sin θ = mω²r, where r = L sin θ = 0.8 × sin 30° = 0.4 m

•

From tan θ = rω²/g: ω² = g tan θ/r = (9.81 × tan 30°)/0.4 = 14.16

•

ω = 3.76 rad s⁻¹

Motion in a Vertical Circle

In vertical circles (bucket of water, loop-the-loop), gravity affects the motion differently at different points.

Bucket of Water

Water stays in if centripetal acceleration > g. At the top, minimum speed when T = 0: mg = mv²/r, so v_min = √(gr)

Important

At TOP: both T and mg point to centre (add). At BOTTOM: T points to centre, mg points away (subtract).

07Worked Example - Vertical Circle

A 0.2 kg ball on a 0.5 m string is swung in a vertical circle at constant speed 4 m s⁻¹. Calculate the tension in the string:

(a) at the top of the circle

(b) at the bottom of the circle

•

First find centripetal force needed: F = mv²/r = (0.2 × 4²)/0.5 = 6.4 N

a)

At TOP: T + mg = mv²/r (both forces toward centre)

•

T = 6.4 − (0.2 × 9.81) = 6.4 − 1.96 = 4.44 N

b)

At BOTTOM: T − mg = mv²/r (T toward centre, mg away)

•

T = 6.4 + 1.96 = 8.36 N

Tension is greater at the bottom than at the top!

Practice

Question 1: A wheel of radius 0.3 m rotates at 1200 rpm. Calculate (a) the angular speed in rad s⁻¹ and (b) the linear speed of a point on the rim.

a)

f = 1200/60 = 20 Hz; ω = 2πf = 40π = 125.7 rad s⁻¹

b)

v = rω = 0.3 × 125.7 = 37.7 m s⁻¹

Practice

Question 2: A satellite orbits Earth at 42000 km from the centre with period 1 day (86400 s). Calculate its centripetal acceleration.

1

r = 4.2 × 10⁷ m

2

ω = 2π/T = 2π/86400 = 7.27 × 10⁻⁵ rad s⁻¹

3

a = ω²r = (7.27×10⁻⁵)² × 4.2×10⁷ = 0.22 m s⁻²

This is a geostationary satellite!

Practice

Question 3: A bucket of water is swung in a vertical circle of radius 0.8 m. What is the minimum speed at the top for the water not to fall out?

1

At minimum speed, T = 0

2

Only weight provides centripetal force: mg = mv²/r

3

v = √(gr) = √(9.81 × 0.8) = 2.8 m s⁻¹

Summary

Key Formulas

QuantityFormula(s)

Radians

θ_{r a d} = θ_{d e g} \times π /180

;

s = r θ

Angular speed

ω = v / r = 2 π f = 2 π / T

Linear speed

v = r ω

Centripetal acceleration

a = v^{2} / r = ω^{2} r

Centripetal force

F = m v^{2} / r = m ω^{2} r

Conical pendulum

tan θ = r ω^{2} / g

Vertical circle (min at top)

v_{min} = g r

Exam Tips

Always use radius, not diameter

Use mass (kg), not weight (N)

Identify what provides centripetal force

For vertical circles: add forces at top, subtract at bottom