OxfordAQA International AS and A-level Mathematics: Uniform circular motion and the conical pendulum (9660)

This guide covers section M2.7 (uniform circular motion) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International A-level (A2) Unit M2 (Mechanics), the mechanics option that sits beside Unit S2; it is not examined at International AS. Unit M2 assumes the International AS units, so Newton’s second law and resolving forces are your starting point.

Use it with the revision notes and the practice questions. The course hub is /boards/oxfordaqa/a-level/mathematics/ and the printable checklist is /checklists/oxfordaqa/a-level/mathematics/. To find weak spots first, try the free 10-minute diagnostics. Supporting skills are in the forces and Newton’s laws guide, the vector kinematics guide and the radians and arcs guide.

What this unit covers

Ref	What you must be able to do
M2.7	Describe the motion of a particle in a circle at constant speed
M2.7	Know and use v = rω, a = rω² and a = v²/r
M2.7	Find angular speed in radians per second from revolutions per minute or the time for one revolution
M2.7	Use position, velocity and acceleration vectors in terms of i and j, and show motion is circular
M2.7	Solve problems on the conical pendulum

The specification says problems involve either horizontal circles or situations, such as a satellite in a circular orbit, where the gravitational force is towards the centre of the circle. It also lists these three formulae among those you must learn because they are not in the formulae booklet. Learn them.

Angular speed

When a particle moves round a circle of radius r, the angle it has turned through changes steadily. The rate of change of that angle is its angular speed, ω, measured in radians per second (rad s⁻¹).

One full turn is 2π radians. So if one revolution takes T seconds:

ω = 2π/T

From revolutions per minute (rev/min), multiply by 2π and divide by 60. Angles in this unit are always in radians, so convert before you substitute.

The speed along the circle follows from arc length s = rθ. Divide by time:

v = rω

Speed is constant, but velocity is not: its direction changes all the time. That is why the particle accelerates even at constant speed.

Worked example 1

A wheel of radius 0.35 m turns at 54 rev/min. Find its angular speed, and the speed of a point on its rim.

ω = 54 × 2π/60 = 9π/5 = 5.65 rad s⁻¹ (3 s.f.)
v = rω = 0.35 × 9π/5 = 1.98 m s⁻¹ (3 s.f.)

Acceleration towards the centre

The acceleration of a particle in uniform circular motion has constant magnitude and is directed towards the centre of the circle. Two equivalent forms come from v = rω:

a = v²/r = rω²

Choose the form that uses the data you are given. If you have v and r, use v²/r. If you have ω or the period, use rω².

By Newton’s second law, the resultant force on the particle is mass times this acceleration and also points to the centre. The force is not an extra “centripetal force” added to your diagram. It is the resultant of the real forces: tension, friction, a normal contact force or weight.

Method for a dynamics problem:

1. Draw the real forces only.
2. Mark the direction towards the centre.
3. Resolve towards the centre and write resultant = m × a.
4. Resolve vertically (if the particle does not move vertically, the forces balance).
5. Solve, then give units.

Worked example 2

A particle of mass 0.5 kg moves on a smooth horizontal table, joined by a light inextensible string of length 0.8 m to a fixed point. It completes 10 revolutions in 6.4 s. Find the tension.

T_period = 6.4/10 = 0.64 s
ω = 2π/0.64 = 9.82 rad s⁻¹
Tension = m r ω² = 0.5 × 0.8 × (2π/0.64)² = 38.6 N (3 s.f.)

The tension is the only horizontal force, so it alone supplies the acceleration. Weight and the normal reaction balance vertically.

Worked example 3: friction on a turntable

A small coin lies 0.15 m from the centre of a horizontal turntable. The coefficient of friction between coin and turntable is 0.4. Find the greatest angular speed at which the coin stays in place, in rad s⁻¹ and rev/min.

Vertically, R = mg. Friction provides the acceleration towards the centre. At the greatest speed, friction is limiting:

μR = m r ω²
0.4 × m × 9.8 = m × 0.15 × ω²
ω² = 0.4 × 9.8/0.15 = 26.13
ω = 5.11 rad s⁻¹
In rev/min: 5.112 × 60/(2π) = 48.8 rev/min

The mass cancels. That is a useful check: a heavier coin does not slide sooner.

Gravitationally controlled orbits

For a satellite in a circular orbit, gravity is the only force and it points to the centre. In a problem you will be given a model for the gravitational force, usually inverse square. Then gravity = m a gives an equation for the speed.

Worked example 4

Model the Earth as a sphere of radius R = 6.4 × 10⁶ m, with gravitational acceleration at its surface g = 9.8 m s⁻². A satellite of mass m moves in a circular orbit of radius r = 1.0 × 10⁷ m. The gravitational force on it is mgR²/r². Find its speed and the time for one orbit.

m a = mgR²/r²
a = 9.8 × (6.4×10⁶/10⁷)² = 9.8 × 0.4096 = 4.014 m s⁻²
v²/r = a, so v = √(a r) = √(4.014 × 10⁷) = 6340 m s⁻¹ (3 s.f.)
Period = 2πr/v = 2π × 10⁷/6336 = 9920 s (3 s.f.), about 2.75 hours

The satellite’s mass cancels, so every satellite at this radius has the same speed. State the modelling assumptions if asked: the satellite is a particle, the orbit is circular, and only the Earth’s gravity acts.

Position, velocity and acceleration vectors

Take a particle moving anticlockwise on a circle of radius r, centre at the origin, with angular speed ω. If it starts on the positive x-axis, its position at time t is:

r(t) = r cos(ωt) i + r sin(ωt) j

Differentiate for velocity and again for acceleration:

v = −rω sin(ωt) i + rω cos(ωt) j
a = −rω² cos(ωt) i − rω² sin(ωt) j = −ω² r(t)

Three facts follow:

• The speed is |v| = rω, constant.
• v · r = 0, so the velocity is perpendicular to the radius (tangent to the circle).
• a = −ω² r, so the acceleration points to the centre with magnitude rω².

If the centre is at c instead of the origin, replace r by (r − c): the acceleration is −ω²(r − c). For clockwise motion, swap the roles of sin and cos or change a sign.

Showing that motion is circular. The specification says you may be asked to show this by proving the body stays a constant distance from a fixed point. Find |r − c| for the proposed centre and show it does not depend on t.

Worked example 5

A particle has position vector r = (3 + 2cos 4t)i + (−1 + 2sin 4t)j, with lengths in metres and t in seconds. Show that it moves on a circle, and find its speed and its acceleration when t = π/8.

Subtract the point (3, −1):

r − (3i − j) = 2cos 4t i + 2sin 4t j
|r − (3i − j)|² = 4cos²4t + 4sin²4t = 4

The distance from (3, −1) is always 2 m, so the path is a circle of centre (3, −1) and radius 2 m.

v = −8 sin 4t i + 8 cos 4t j
speed = √(64 sin²4t + 64 cos²4t) = 8 m s⁻¹
a = −32 cos 4t i − 32 sin 4t j

At t = π/8, 4t = π/2, so a = −32j m s⁻². Check: a = rω² = 2 × 4² = 32, directed from the particle at (3, 1) down towards the centre (3, −1). Correct.

If the speed is not constant, or the distance from the supposed centre changes, the motion is not uniform circular motion. For example, 4cos 3t i + 3sin 3t j has distance √(9 + 7cos²3t) from the origin, which varies, so it traces an ellipse.

The conical pendulum

A particle of mass m is attached to a light inextensible string of length L. The other end is fixed. The particle moves in a horizontal circle at constant speed, so the string traces out a cone with its axis vertical.

        pivot O
        /|
     L / |  h = L cos θ
      /θ |
     /   |
    P----+---- centre of circle
      r = L sin θ

Forces on the particle: weight mg downwards and tension T along the string towards O. The angle θ is measured from the vertical. The radius is r = L sin θ.

Vertical (no vertical acceleration):  T cos θ = mg
Horizontal (towards the centre):      T sin θ = m r ω² = m L sin θ ω²

From the horizontal equation, dividing by sin θ (θ is not zero):

T = m L ω²

Substitute into the vertical equation:

m L ω² cos θ = mg, so cos θ = g/(L ω²)

Since cos θ ≤ 1, the particle can only move in a cone if ω² > g/L. The vertical depth of the circle below the pivot is h = L cos θ, which gives ω² = g/h. A faster rotation means a smaller h, so the bob rises.

Worked example 6

A bob of mass 0.4 kg hangs from a string of length 1.2 m and revolves at 5 rad s⁻¹ in a horizontal circle. Find the tension, the angle of the string to the vertical, and the speed of the bob.

T = m L ω² = 0.4 × 1.2 × 25 = 12 N
cos θ = g/(L ω²) = 9.8/(1.2 × 25) = 0.3267
θ = 70.9° (3 s.f.)
r = 1.2 sin 70.93° = 1.134 m
v = rω = 1.134 × 5 = 5.67 m s⁻¹ (3 s.f.)

Check with the vertical equation: T cos θ = 12 × 0.3267 = 3.92, and mg = 0.4 × 9.8 = 3.92. It balances.

Common errors

• Using rev/min or degrees per second in v = rω. Convert to rad s⁻¹ first.
• Adding a “centripetal force” arrow to the force diagram as well as the real forces.
• Using v²/r with ω, or rω² with v. Match the formula to the data.
• In the conical pendulum, using L instead of L sin θ as the radius.
• Writing T = mg in the conical pendulum. Only the vertical component, T cos θ, balances the weight.
• Showing motion is circular by checking only one or two points. Prove the distance is constant for all t.
• Giving the direction of acceleration as along the velocity. It is towards the centre, perpendicular to the velocity.

Next steps

Work through the practice questions, then keep the revision notes for the final weeks. Take the free diagnostics to check where you stand.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, published by OxfordAQA. Section M2.7 of Unit M2 (International A-level only).