OxfordAQA International AS and A-level Physics: Circular motion and angular speed (9630) – Revision Notes

These notes condense OxfordAQA International AS and A-level Physics (9630), Version 4.4, section 3.6.1 (Circular motion). The specification marks section 3.6 as International A-level only, so none of it is International AS content. For full explanations, read the study guide. Then test yourself with the practice questions. The course hub and printable checklist show where this unit sits in the course.

The big idea

• Constant speed round a circle still means the velocity is changing, because its direction changes.
• A changing velocity is an acceleration. It points to the centre.
• Newton’s second law then needs a resultant force towards the centre: the centripetal force.
• You will not be asked to derive a = v²/r. Learn the result and use it.

Key definitions

Term	Meaning	Unit
Radian	Angle at the centre when the arc is as long as the radius; θ = s/r	rad
Period T	Time for one complete revolution	s
Frequency f	Revolutions per second; f = 1/T	Hz
Angular speed ω	Angle turned per unit time	rad s⁻¹
Centripetal acceleration	Acceleration directed towards the centre	m s⁻²
Centripetal force	Resultant force towards the centre	N

Formulas

Quantity	Equation	Use it when
Angular speed	ω = v/r = 2πf = 2π/T	Linking speed, radius, frequency and period
Linear speed	v = ωr	Finding rim speed from ω
Centripetal acceleration	a = v²/r = ω²r	v is given (first form) or f, T, ω is given (second form)
Centripetal force	F = mv²/r = mω²r	F is the resultant towards the centre
Full turn	360° = 2π rad	Converting angles

Method in steps

1. Draw the real forces on the object. Do not draw a centripetal or centrifugal arrow.
2. Pick the direction towards the centre. Add the forces along it (towards the centre positive).
3. Convert revolutions per minute to revolutions per second, then to ω with ω = 2πf. Convert any angles to radians.
4. Write resultant force = mv²/r or mω²r, whichever fits the data.
5. Rearrange before substituting. Give the answer with units and sensible significant figures.
6. Check by finding the other form of a, or the force from the other equation. They should agree.

What supplies the force

Situation	Provider of centripetal force
Mass on a string, smooth surface	Tension
Object on a spinning disc	Friction
Car on a flat bend	Sideways grip of the tyres (friction)
Orbiting satellite	Gravitational attraction
Rider held against a rotating cylinder wall	Push of the wall (normal contact force)

Worked reminders

Convert and find speed. A disc turns at 33 revolutions per minute and a point lies 0.15 m from the axis. Then f = 33/60 = 0.55 Hz, ω = 2π × 0.55 = 3.46 rad s⁻¹ and v = 3.456 × 0.15 = 0.52 m s⁻¹.

Force from speed. A 0.30 kg object moves at 12 m s⁻¹ on a circle of radius 0.75 m. Then F = 0.30 × 12²/0.75 = 58 N.

Rearranging for speed. The same string can supply at most 24 N to a 1.5 kg mass at radius 0.50 m. The greatest speed is v = √(Fr/m) = √(24 × 0.50/1.5) = √8.0 = 2.8 m s⁻¹.

Period to force. A 0.30 kg mass has a period of 0.80 s on a circle of radius 0.55 m. Then ω = 2π/0.80 = 7.85 rad s⁻¹ and F = mω²r = 0.30 × 7.854² × 0.55 = 10 N.

Hanging-mass rig. A mass on a smooth table is held on a string over a pulley by a hanging mass of 0.25 kg. At steady motion the tension is Mg = 0.25 × 9.81 = 2.45 N, and that tension is the centripetal force on the table mass.

Orbit. A 600 kg satellite has ω = 1.2 × 10⁻³ rad s⁻¹ at a radius of 7.0 × 10⁶ m. Then a = ω²r = (1.2 × 10⁻³)² × 7.0 × 10⁶ = 10.08, so a ≈ 10 m s⁻², and F = ma ≈ 6.0 × 10³ N.

Scaling. If the speed doubles at a fixed radius, the force required rises by a factor of four. If the radius doubles at a fixed speed, the force required halves.

Must-know distinctions

• Speed and velocity. Speed is constant in this unit. Velocity is not.
• Linear speed and angular speed. v is in m s⁻¹ and changes with radius. ω is in rad s⁻¹ and is the same for every point on a rigid rotating object.
• Frequency and angular speed. f counts revolutions per second. ω = 2πf counts radians per second.
• Centripetal force and the real forces. Centripetal force is the resultant of the real forces, taken towards the centre. It does not add to them.
• Circular path and straight tangent. If the force stops, the object moves off at constant velocity along the tangent. It does not spiral out radially.
• Force and work. The centripetal force acts at 90° to the velocity, so it transfers no energy and the speed stays constant.
• v²/r and ω²r. Same quantity, different data. Use whichever you were given.

Graph reminders

• F against v² for a fixed m and r: straight line through the origin, gradient m/r.
• F against ω² for a fixed m and r: straight line through the origin, gradient mr.
• F against 1/T² for a fixed m and r: straight line through the origin, gradient 4π²mr.
• To reduce the percentage uncertainty in a period, time a large number of revolutions and divide.

Quick self-test

1. Convert 240° to radians. Give the answer in terms of π and as a decimal to 3 significant figures.
2. Convert 1.8 rad to degrees, to 3 significant figures.
3. A turntable rotates at a frequency of 1.25 Hz. Find its period and angular speed.
4. A point moves at 6.0 m s⁻¹ in a circle of radius 1.5 m. Find its centripetal acceleration.
5. A 0.30 kg ball is on a string of length 0.75 m, travelling at 12 m s⁻¹ in a horizontal circle on a smooth surface. Find the tension.
6. A wheel turns at 10 rad s⁻¹ and a point on the rim is 0.25 m from the axis. Find the frequency, and the speed of the point.
7. A 2.0 kg object travels in a circle of radius 0.30 m at 5.0 rad s⁻¹. Find the resultant force on it.
8. State the direction of the acceleration of an object moving at constant speed in a circle.
9. Explain why the centripetal force transfers no energy to the object.
10. A satellite moves in a circular orbit. What provides its centripetal force?
11. The speed of an object on a circle is doubled and the radius is unchanged. By what factor does the force needed change?
12. A student plots F against ω² for a bung of fixed mass on a string of fixed length and finds the gradient is 0.027 kg m. State what the gradient represents.

Answers

1. 240 × π/180 = 4π/3 = 4.19 rad
2. 1.8 × 180/π = 103°
3. T = 1/1.25 = 0.80 s; ω = 2π × 1.25 = 7.85 rad s⁻¹
4. a = 6.0²/1.5 = 24 m s⁻²
5. T = 0.30 × 12²/0.75 = 58 N (57.6 N)
6. f = ω/2π = 10/2π = 1.59 Hz; v = ωr = 10 × 0.25 = 2.5 m s⁻¹
7. F = mω²r = 2.0 × 5.0² × 0.30 = 15 N
8. Towards the centre of the circle, perpendicular to the velocity.
9. The force is always at right angles to the displacement along the path, so the work done is zero and the kinetic energy and speed are unchanged.
10. The gravitational attraction towards the body it orbits.
11. F is proportional to v², so it increases by a factor of 4.
12. The gradient equals mr, the product of the mass and the radius. Dividing by the radius gives the mass.

Where marks are usually lost

• Writing the unit of ω as Hz or s⁻¹ with no mention of radians. Say rad s⁻¹.
• Forgetting the 2π when converting frequency to angular speed.
• Using revolutions per minute directly in a formula without converting.
• Using the diameter as the radius.
• Giving “centrifugal force” as the explanation of why an object moves outward. In an exam answer, use the object’s own inertia and the real inward force.
• Writing “the speed changes” when explaining why there is an acceleration. The velocity changes; the speed does not.
• Naming the centripetal force as the provider of the motion, instead of naming tension, friction, a normal force or gravity.
• Forgetting to take the square root when rearranging F = mv²/r for v.
• Quoting an unrounded value, or too many significant figures, when the data have two.

Take one of the free 10-minute diagnostics when you want a quick check of weaker areas.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams from May/June 2020 onwards, published by OxfordAQA. Section 3.6.1 Circular motion (International A-level only).