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

These notes condense section M2.7 of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The content belongs to International A-level (A2) Unit M2 (Mechanics) and is not examined at International AS. For full explanations, use the study guide; to test yourself, use the practice questions.

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Formulae to learn

The specification lists these among the formulae you must learn, because they are not in the formulae booklet.

Result	Meaning
v = rω	speed from radius and angular speed
a = rω²	acceleration towards the centre, from ω
a = v²/r	acceleration towards the centre, from v
ω = 2π/T	angular speed from the time T for one revolution
ω (rad s⁻¹) = n × 2π/60	angular speed from n rev/min
F = ma	resultant force towards the centre

Units: ω in rad s⁻¹, v in m s⁻¹, a in m s⁻². Angular speed is the rate of turning in radians per second.

Key ideas

• Speed is constant. Velocity changes direction, so there is an acceleration.
• The acceleration is towards the centre and perpendicular to the velocity.
• There is no extra “centripetal force”. The resultant of the real forces (tension, friction, normal reaction, weight, gravity) points to the centre and equals ma.
• Problems involve horizontal circles, or orbits where gravity points to the centre.

Method for a force problem

1. Draw the real forces on the particle.
2. Mark the direction to the centre of the circle.
3. Resolve towards the centre: resultant = m × a, with a = v²/r or rω².
4. Resolve vertically: forces balance if there is no vertical motion.
5. Solve and give units and the right accuracy.

Setting up common situations

Situation	Force towards the centre	Vertical balance
Particle on smooth table, string to a fixed point	Tension	R = mg
Object on a turntable or flat bend	Friction (at most μR)	R = mg
Satellite in circular orbit	Gravitational force	none (all force is towards the centre)
Conical pendulum	Horizontal part of tension, T sin θ	T cos θ = mg

For limiting friction, use F = μR and note that the mass cancels with R = mg.

Vector form

Anticlockwise, centre at the origin, radius r, angular speed ω, starting on the positive x-axis:

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

• |v| = rω, constant.
• v · r = 0.
• Centre at c: the acceleration is −ω²(r − c).
• For other starting points or clockwise motion, check the signs by finding v at t = 0.

To show motion is circular: pick the centre c, find |r − c|, and show it is a constant for every t. Use cos² + sin² = 1.

Reminder example. For r = (−2 + 6 sin(t/2))i + (4 + 6cos(t/2))j, the vector r − (−2i + 4j) has length 6 for all t, so the path is a circle of radius 6. Here ω = 1/2 and the speed is 6 × 1/2 = 3 m s⁻¹.

Conical pendulum

Bob of mass m, string length L, angle θ to the vertical, radius r = L sin θ, depth below the pivot h = L cos θ.

T cos θ = mg
T sin θ = m r ω² = m L sin θ ω²

Results (divide by sin θ):

Result	Use
T = m L ω²	tension from ω and L
cos θ = g/(L ω²)	angle from ω and L
ω² = g/h	angular speed from the depth h
tan θ = r ω²/g	angle from radius and ω
ω² > g/L	needed for the string to leave the vertical

Period of one revolution: T_period = 2π/ω = 2π√(L cos θ/g).

Reminder example. L = 0.75 m and θ = 40°. Then h = 0.75 cos 40° = 0.5745 m, and the period is 2π√(0.5745/9.8) = 1.52 s (3 s.f.). Then ω = 4.13 rad s⁻¹.

Quick conversions

Given	Angular speed ω
30 rev/min	π rad s⁻¹
60 rev/min	2π rad s⁻¹
120 rev/min	4π rad s⁻¹
Period 2 s	π rad s⁻¹
Period 0.5 s	4π rad s⁻¹
f revolutions per second	2πf rad s⁻¹

Going back: n rev/min = ω × 60/(2π). Check that your answer is sensible. A rate of 5 rad s⁻¹ is a little under 1 revolution per second, which is about 48 rev/min.

Orbit reminder

When the problem gives the gravitational force on a satellite as mk/r², the mass cancels:

m v²/r = m k/r²
v² = k/r

Then the period is 2πr/v. Equivalently, ω²r³ = k, so the square of the period is proportional to r³.

Reminder example: k = 3.6 × 10¹⁴ m³ s⁻² and r = 9.0 × 10⁶ m. Then v = √(3.6×10¹⁴/9.0×10⁶) = √(4.0×10⁷) = 6320 m s⁻¹ (3 s.f.), and the period is 2π × 9.0×10⁶/6325 = 8940 s (3 s.f.).

Reminder: finding the angle from a period

If a conical pendulum has string length L and period P:

1. ω = 2π/P.
2. cos θ = g/(L ω²).
3. If the right side is greater than 1, the data are impossible, because the bob is turning too slowly to leave the vertical.

Must-know distinctions

• Speed v and angular speed ω. v is in m s⁻¹, ω in rad s⁻¹. They are linked by v = rω.
• Tension T and period T. Use different symbols in your working, such as T for tension and P for period.
• The circle’s radius and the string length. In a conical pendulum, r = L sin θ, not L.
• Real forces and the resultant. The resultant towards the centre is what equals ma.
• Uniform circular motion and any circular path. The formulae here need constant speed.

Quick self-test

Take g = 9.8 m s⁻² and give 3 significant figures unless exact.

1. Convert 90 rev/min to rad s⁻¹ in terms of π.
2. A particle moves in a circle of radius 0.6 m with period 0.8 s. Find ω.
3. A point moves at 12 m s⁻¹ round a circle of radius 4 m. Find its acceleration.
4. A particle of mass 2 kg moves at constant speed on a circle of radius 1.5 m with ω = 4 rad s⁻¹. Find the resultant force.
5. Show that r = (5 + 3cos 2t)i + (1 + 3sin 2t)j describes motion on a circle, stating the centre and radius.
6. For the motion in question 5, find the speed.
7. Write down the acceleration vector when t = 0 for the motion in question 5.
8. A conical pendulum has string length 2 m and rotates at 3 rad s⁻¹. Find cos θ.
9. For the pendulum in question 8 with bob mass 0.3 kg, find the tension.
10. A bob moves in a horizontal circle 0.49 m below its pivot. Find ω.
11. State the condition on ω for a conical pendulum of length L to move in a cone.
12. A coin on a turntable stays put at 0.2 m from the axis when ω = 4 rad s⁻¹ but not at 5 rad s⁻¹. What can you say about μ?

Answers

1. 90 × 2π/60 = 3π rad s⁻¹
2. ω = 2π/0.8 = 7.85 rad s⁻¹
3. a = v²/r = 144/4 = 36 m s⁻²
4. F = m r ω² = 2 × 1.5 × 16 = 48 N, towards the centre
5. r − (5i + j) = 3cos 2t i + 3sin 2t j, with length √(9cos² 2t + 9sin² 2t) = 3. So the centre is (5, 1) and the radius 3.
6. v = −6 sin 2t i + 6cos 2t j, so the speed is 6 (units of length per unit time), which is rω = 3 × 2.
7. a = −12cos 2t i − 12sin 2t j, so at t = 0, a = −12i
8. cos θ = 9.8/(2 × 9) = 9.8/18 = 0.544 (θ = 57.0°)
9. T = m L ω² = 0.3 × 2 × 9 = 5.4 N
10. ω² = g/h = 9.8/0.49 = 20, so ω = 4.47 rad s⁻¹ (exactly 2√5)
11. ω² > g/L, equivalently ω > √(g/L)
12. At the limit, μg = rω². So 16 × 0.2/9.8 ≤ μ < 25 × 0.2/9.8, giving 0.327 ≤ μ < 0.510 (3 s.f.)

Where marks are usually lost

• Leaving ω in rev/min or using degrees. Work in radians per second.
• Choosing a = v²/r when only ω is known, then rounding early. Keep full values until the end.
• Forgetting that the horizontal radius of a conical pendulum is L sin θ.
• Writing T = mg for the string tension.
• Drawing a centripetal force in addition to the real forces.
• Showing that two points are equidistant from a centre and calling that a proof. The distance must be constant for all t.
• Quoting the acceleration of a vector problem without a direction or sign. In vector form the sign carries the direction.
• Omitting units, or giving a bare decimal where the question asked for an answer in terms of π.

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).