Cambridge AS & A Level9702

Motion in a circle

Physics 9702 Chapter Notes

What this chapter covers

Motion in a circle - Kinematics of uniform circular motionMotion in a circle - Centripetal acceleration
ShareWhatsAppPost
Motion in a circle notes

Unable to load PDF

The notes viewer could not load. Please refresh the page.

Read online free. Download a watermarked copy with a free account.

Read the notes

The full Motion in a circle notes as text: skim, search, and jump between subtopics.

~12 min read

In this chapter

  1. 1Describing Circles: Radians and Angular Speed
  2. 2From Rotation to Motion: Linking Speeds
  3. 3Centripetal Force: The 'Center-Seeking' Force
  4. 4The Maths of Circular Motion
  5. 5Applications: The Vertical Circle

1. Describing Circles: Radians and Angular Speed

When an object moves in a circle, we need a way to describe its position and how fast it's rotating. Instead of degrees, physicists use radians to measure angles. One radian is the angle created when the arc length along the circle's edge is equal to the circle's radius. A full circle is 360°, which is equal to 2π radians. Angular speed, symbol ω (omega), is the rate at which the angle changes. It's defined as the angle turned through (in radians) per unit time.

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

$1 \\text{ revolution} = 360^\\circ = 2\\pi \\text{ rad}$

Key term

Angular Speed (ω): The rate of change of angular displacement, measured in radians per second (rad s⁻¹).

Common pitfall

Forgetting to convert revolutions per minute (rpm) into radians per second before using it in physics equations. Remember the two-step conversion: rpm ÷ 60 → rev/s, then rev/s × 2π → rad/s.

Fun fact

The number 2π (about 6.28) appears everywhere in physics, from waves to quantum mechanics, all because the radian is the natural unit for measuring angles in a circle.

Worked example 13 marks

A hard drive platter spins at 7200 rpm (revolutions per minute). What is its angular speed in rad s⁻¹?

  1. 1

    Step 1: Convert revolutions per minute (rpm) to revolutions per second (Hz).

  2. 2

    Frequency, f = 7200 rev min⁻¹ / 60 s min⁻¹ = 120 rev s⁻¹ (or 120 Hz).

  3. 3

    Step 2: Convert revolutions per second to radians per second.

  4. 4

    Each revolution is 2π radians. So, ω = f × 2π.

  5. 5

    ω = 120 × 2π = 240π rad s⁻¹.

  6. 6

    Step 3: Calculate the numerical value.

  7. 7

    ω ≈ 754 rad s⁻¹ (to 3 significant figures).

Recap

  • Angles in circular motion are measured in radians (rad).
  • A full circle contains 2π radians, which is equivalent to 360°.
  • Angular speed (ω) is the angle in radians swept out per second.
  • The units for angular speed are radians per second (rad s⁻¹).

Quick check

  1. Convert 90° to radians, expressing your answer as a fraction of π.1 mark
  2. A wheel turns 10 times in 5 seconds. What is its average angular speed?2 marks

2. From Rotation to Motion: Linking Speeds

An object moving in a circle has both an angular speed (ω) and a linear speed (v). The linear speed (or tangential speed) is its instantaneous speed along the tangent to the circle. These two speeds are directly related. A point further from the centre of a spinning object has to travel a greater distance in the same amount of time, so it moves faster. The relationship is v = rω, where r is the radius of the circular path. We can also relate angular speed to the period (T), the time for one full circle, using ω = 2π/T.

$v = r\\omega$

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

Key term

Linear Speed (v): The instantaneous speed of an object moving along its circular path, directed tangentially to the circle, measured in m s⁻¹.

Examiner insight

Examiners frequently ask questions that require you to switch between linear and angular quantities. Be ready to use v=rω and ω=2π/T as intermediate steps in a larger problem.

Worked example 14 marks

A toy truck travels round a horizontal circular track of radius 0.50 m. It makes three complete revolutions every 10 seconds. Calculate its angular speed and its linear speed.

  1. 1

    Step 1: Find the time for one revolution (the period, T).

  2. 2

    T = Time / Revolutions = 10 s / 3 = 3.33 s.

  3. 3

    Step 2: Calculate the angular speed, ω.

  4. 4

    ω = 2π / T = 2π / 3.33 s = 1.885 rad s⁻¹.

  5. 5

    Step 3: Calculate the linear speed, v.

  6. 6

    v = rω = 0.50 m × 1.885 rad s⁻¹ = 0.9425 m s⁻¹.

  7. 7

    Step 4: State answers to an appropriate number of significant figures (2 s.f. based on the data).

  8. 8

    ω = 1.9 rad s⁻¹, v = 0.94 m s⁻¹.

Recap

  • Linear speed (v) is how fast an object moves along the circular path in m s⁻¹.
  • Angular speed (ω) is how fast it rotates in rad s⁻¹.
  • The two are linked by the formula v = rω.
  • The period (T) is the time for one revolution, and ω = 2π/T.

Quick check

  1. Two children are on a merry-go-round. One is near the centre, one is at the edge. Who has the greater linear speed?1 mark
  2. A bicycle wheel of radius 35 cm has an angular speed of 12 rad s⁻¹. What is the linear speed of a point on the tyre?2 marks

3. Centripetal Force: The 'Center-Seeking' Force

According to Newton's First Law, an object's velocity is constant unless a resultant force acts on it. For an object in circular motion, even at a constant speed, its direction is constantly changing. A change in velocity is acceleration, which means there must be a resultant force. This force is called the centripetal force. It is always directed towards the centre of the circle, and it is always at a right angle (90°) to the object's instantaneous velocity. It's crucial to understand that 'centripetal force' is not a new type of force; it's the label we give to the resultant force that causes circular motion. This force could be tension in a string, gravity from a planet, or friction from the road.

Key term

Centripetal Force: The resultant force acting on an object moving in a circle, which is directed towards the centre of the circle and is responsible for its circular path.

Examiner insight

Marks are consistently awarded for stating that the centripetal force is perpendicular to the velocity and directed towards the centre of the circle. Identifying the real-world origin of the force (e.g., 'friction between tyres and road') is also a key skill.

Common pitfall

Mistaking centripetal force for a new, fundamental force, or mentioning the non-existent 'centrifugal force'. Centripetal force is always provided by a real, identifiable force like tension, gravity, or friction. Never write 'centrifugal force' in an exam.

Worked example 14 marks

A car is travelling round a bend when it hits a patch of oil. The car slides off the road. Explain, using your understanding of circular motion, why the car came off the road.

  1. 1

    To travel in a circular path (the bend), the car requires a resultant force directed towards the centre of the circle. This is the centripetal force.

  2. 2

    This centripetal force is provided by the force of friction between the car's tyres and the road surface.

  3. 3

    When the car hits the oil patch, the friction is significantly reduced, so the required centripetal force can no longer be provided.

  4. 4

    Without sufficient centripetal force, the car can no longer accelerate towards the centre. According to Newton's First Law, it continues to move in a straight line at a tangent to the bend, causing it to slide off the road.

Recap

  • An object moving in a circle is accelerating because its velocity is changing direction.
  • This acceleration is caused by a resultant force called the centripetal force.
  • The centripetal force always acts towards the centre of the circle.
  • The centripetal force is always perpendicular to the object's velocity.
  • Centripetal force is not a fundamental force, but a label for a force like gravity, tension or friction.

Quick check

  1. What provides the centripetal force that keeps the Moon in orbit around the Earth?1 mark
  2. An object moves in a circle at a constant speed. Is its momentum constant? Explain briefly.2 marks

4. The Maths of Circular Motion

Since there is a centripetal force, there must be a centripetal acceleration, a. This acceleration is also directed towards the centre of the circle. Its magnitude depends on the object's linear speed(v) and the radius of the circle (r). The formula is a = v²/r. We can also express this using angular speed (ω) as a = rω². Using Newton's Second Law, F=ma, we can find the equations for centripetal force: F = mv²/r and F = mrω². Choosing the right version of the formula depends on the information given in the question.

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

$a = r\\omega^2$

$F = \\frac{mv^2}{r}$

$F = mr\\omega^2$

Key term

Centripetal Acceleration (a): The acceleration of an object undergoing uniform circular motion, which is directed towards the centre of the circle and has a magnitude of v²/r or rω².

Examiner insight

Students who can confidently choose the correct formula (e.g., using the ω version if given angular speed, or the v version if given linear speed) save time and avoid making conversion errors.

Worked example 15 marks

A rubber bung of mass 50 g is attached to a string and swung in a horizontal circle of radius 80 cm. It completes 2 revolutions per second. Calculate(a) the centripetal acceleration and(b) the tension in the string.

  1. 1

    Step 1: Convert all units to SI units. m = 50 g = 0.050 kg. r = 80 cm = 0.80 m.

  2. 2

    Step 2: Calculate angular speed (ω). Frequency f = 2 rev s⁻¹ = 2 Hz. ω = 2πf = 2π(2) = 4π rad s⁻¹ (≈ 12.57 rad s⁻¹).

  3. 3

    Step 3:(a) Calculate centripetal acceleration (a). Using a = rω² is easiest here.

  4. 4

    a = 0.80 m × (4π rad s⁻¹)² = 0.80 × 157.9 = 126.3 m s⁻².

  5. 5

    To 2 s.f., a = 130 m s⁻².

  6. 6

    Step 4:(b) Calculate centripetal force (F). The tension in the string provides the centripetal force.

  7. 7

    F = ma = 0.050 kg × 126.3 m s⁻² = 6.315 N.

  8. 8

    To 2 s.f., the tension is 6.3 N.

Recap

  • Centripetal acceleration is given by a = v²/r or a = rω².
  • Centripetal force is given by F = mv²/r or F = mrω².
  • Both acceleration and force are vectors directed towards the centre of the circle.
  • Always ensure you use SI units (kg, m, s) in these formulas.

Quick check

  1. If the speed of an object in a circle doubles while the radius stays the same, by what factor does the centripetal force change?1 mark
  2. A 1200 kg car rounds a bend of radius 45 m at a speed of 15 m s⁻¹. Calculate the centripetal force required.2 marks

5. Applications: The Vertical Circle

Analysing forces in a vertical circle is a common exam problem. Unlike a horizontal circle, the force of gravity (weight, mg) plays a changing role. The key is that the resultant of all forces pointing towards the centre must equal the required centripetal force, mv²/r. At the top: Both the tension (T) in the string and the weight (mg) act downwards, towards the centre. So, the resultant force is T + mg. Therefore, T + mg = mv²/r. Tension is at its minimum here. At the bottom: Tension (T) acts upwards (towards the centre), while weight (mg) acts downwards (away from the centre). The resultant force is T - mg. Therefore, T - mg = mv²/r. Tension is at its maximum here.

At top: $T + mg = \\frac{mv^2}{r}$

At bottom: $T - mg = \\frac{mv^2}{r}$

Key term

Resultant Force: The vector sum of all forces acting on an object, which in circular motion must be equal to the required centripetal force.

Examiner insight

A clear free-body diagram for the top and bottom positions is essential. Examiners look for the correct application of Newton's Second Law, where the net force towards the centre is set equal to mv²/r.

Common pitfall

Forgetting to include weight (mg) in the force analysis for vertical circles, or getting the signs wrong (e.g., adding mg at the bottom instead of subtracting). Always draw a free-body diagram.

Worked example 15 marks

A stone of mass 0.20 kg is whirled on the end of a string in a vertical circle of radius 30 cm. The string will break when the tension in it exceeds 8.0 N. Calculate the maximum speed at which the stone can be whirled without the string breaking. (Assume g = 9.81 m s⁻²).

  1. 1

    Step 1: Identify where tension is maximum. Tension is greatest at the bottom of the circle because it must support the stone's weight AND provide the centripetal force.

  2. 2

    Step 2: Write the force equation for the bottom of the circle: Resultant Force = T - mg. This provides the centripetal force. So, T - mg = mv²/r.

  3. 3

    Step 3: The string breaks when T > 8.0 N. The maximum speed occurs at the maximum tension, T = 8.0 N.

  4. 4

    Step 4: Rearrange the equation to solve for v: v² = r(T - mg) / m.

  5. 5

    Step 5: Substitute the values in SI units (r = 0.30 m): v² = (0.30 × (8.0 - (0.20 × 9.81))) / 0.20.

  6. 6

    Step 6: Calculate the result: v² = (0.30 × (8.0 - 1.962)) / 0.20 = (0.30 × 6.038) / 0.20 = 9.057.

  7. 7

    Step 7: Find v: v = √9.057 = 3.009 m s⁻¹. To 2 s.f., the maximum speed is 3.0 m s⁻¹.

Recap

  • In a vertical circle, both tension and weight contribute to the centripetal force.
  • Tension is maximum at the bottom of the circle, where T = mv²/r + mg.
  • Tension is minimum at the top of the circle, where T = mv²/r - mg.
  • The net force towards the centre always equals mv²/r.

Quick check

  1. For a rollercoaster car going over a circular hump, at what point do you feel lightest? Top or bottom?1 mark
  2. A 1.0 kg mass on a string is in a vertical circle of radius 0.5 m at a constant speed of 4.0 m s⁻¹. Calculate the tension at the top. (g = 9.81 m s⁻²)3 marks

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A spinning top completes 5 revolutions in 2 seconds. What is its angular speed in rad s⁻¹?2 marks
  2. A 0.50 kg ball is swung in a horizontal circle of radius 1.2 m at a constant speed of 4.0 m s⁻¹. Calculate the tension in the string.2 marks
  3. The Earth has a radius of 6400 km and rotates once every 24 hours. Calculate the linear speed of a person standing on the equator.4 marks
  4. A 1500 kg car travels at a constant speed of 20 m s⁻¹ around a flat, circular bend of radius 50 m. Calculate the minimum coefficient of static friction between the tyres and the road required for the car to make the turn.4 marks
  5. The International Space Station (mass 4.2 × 10⁵ kg) orbits at an altitude where the gravitational field strength is 8.7 N kg⁻¹. It completes an orbit in 92 minutes. Calculate the radius of its orbit.5 marks
  6. A 200 g mass is attached to a string and swung in a horizontal circle of radius 40 cm. The string makes an angle of 30° to the vertical. Calculate (a) the tension in the string and (b) the speed of the mass.6 marks
  7. A pilot flies an aeroplane at a constant speed of 150 m s⁻¹ in a horizontal circle. The wings are banked at an angle of 25° to the horizontal. Calculate the radius of the circular path.5 marks
  8. A 0.80 kg ball on a 1.0 m string is swung in a vertical circle. The speed of the ball at the highest point is 5.0 m s⁻¹. Calculate (a) the tension in the string at the highest point, and (b) the tension in the string at the lowest point, assuming the speed there is 8.8 m s⁻¹. (Use g = 9.81 m s⁻²).6 marks
  9. Mars orbits the Sun once every 687 days at a distance of 2.3 × 10¹¹ m. The mass of Mars is 6.4 × 10²³ kg. Calculate the gravitational force exerted on Mars by the Sun.5 marks
  10. A small block is placed on a horizontal turntable at a distance of 15 cm from the centre. The turntable is slowly sped up. If the coefficient of static friction between the block and the turntable is 0.40, what is the maximum angular speed (in rad s⁻¹) the turntable can have before the block starts to slip?4 marks

Go deeper

Practise and revise with member-only material for this chapter.

Workbook

Exam-style practice questions with full worked solutions.

Unlock with a plan

Chapter at a Glance

A single-page summary of every key idea for fast revision.

Unlock with a plan

Free notes are just the start.

Unlock every Workbook and Chapter at a Glance, and generate your own worksheets and predicted papers.

Explore plans

Related chapters

Physical quantities and unitsKinematicsDynamicsForces, density and pressureWork, energy and powerDeformation of solids