A Level Physics: Circular Motion — Revision Notes

Condensed for the full weeks. For the full explanation, use the Motion in a Circle study guide, and test yourself with the practice questions.

Angular quantities

angular velocity   omega = theta / t = 2 pi / T = 2 pi f     rad s^-1
linear speed       v = r omega

Angles must be in radians: 2π rad = 360°, so 1 rad ≈ 57.3° — every formula on this page assumes radians, and switching a calculator to degree mode by mistake is the most common source of a wrong answer.

Centripetal acceleration and force

a = v^2 / r  =  r omega^2
F = m v^2 / r  =  m r omega^2

Both always directed towards the centre.

The single most important idea: centripetal force is not a new force. It is the resultant of the real forces already acting — tension, gravity, friction, normal contact, magnetic force. Naming it as an extra arrow on a free-body diagram is wrong; identify which real force (or combination) is actually providing it instead.

Situation	What provides the centripetal force
Ball on a string	Tension
Car on a flat bend	Friction between tyres and road
Planet in orbit	Gravitational attraction
Electron in a magnetic field	Magnetic force BQv
Car on a banked track	Horizontal component of the normal contact force

Banked tracks (beyond the syllabus — background only, not examinable)

Banked-track problems do not appear in the Cambridge 9702 Motion in a Circle syllabus content, but the idea extends the centripetal-force relationships above in a way that helps consolidate them.

On a frictionless banked track at angle θ, only two forces act: weight (down) and the normal contact force (perpendicular to the track surface). Resolving:

Vertical:    N cos(theta) = mg
Horizontal:  N sin(theta) = m v^2 / r

Dividing:    tan(theta) = v^2 / (r g)

Worked example. A track is banked at 20° and designed for cars travelling at 25 m/s. Find the radius it should be built to.

tan(20) = v^2 / (r g)
r = v^2 / (g tan(20)) = 25^2 / (9.81 x 0.364) = 625 / 3.57 = 175 m

At exactly this speed and radius, no friction is needed at all — friction only becomes necessary if the car’s actual speed differs from the design speed.

Why it accelerates at constant speed

Velocity is a vector, with both magnitude and direction. In circular motion the direction changes continuously, so velocity changes, so there is acceleration — even though the speed (the magnitude of velocity) stays constant throughout.

The force is perpendicular to velocity at every instant, therefore no work is done and kinetic energy is constant.

Vertical circles — the two critical points

TOP:     T + mg = m v^2 / r      ->   T = m v^2 / r - mg
BOTTOM:  T - mg = m v^2 / r      ->   T = m v^2 / r + mg

Tension is greatest at the bottom and least at the top. Subtracting the two equations gives a difference of 2mg only if the speed is the same at both points (e.g. a stone whirled at constant speed). If the object moves freely and energy is conserved, it is faster at the bottom: ½mv_bottom² = ½mv_top² + mg(2r), so v_bottom² = v_top² + 4gr, and the tensions then differ by 6mg.

Minimum speed at the top occurs when T = 0 and gravity alone supplies the centripetal force:

mg = m v^2 / r      ->      v_min = sqrt(g r)

Below that speed the object leaves the circular path — the string would go slack, or the bucket of water would spill — because gravity alone is more than enough to provide the (now smaller) required centripetal force, so the path curves away from the intended circle.

Exam traps

• Drawing centripetal force as an extra arrow instead of identifying the real force providing it.
• Using degrees where radians are required.
• Saying there is no acceleration because the speed is constant.
• Claiming the centripetal force does work — it never does.
• At the top of a vertical circle, forgetting that both tension and weight act downwards, towards the centre.
• Mixing v and ω in the same expression without converting.
• Forgetting that a banked-track derivation requires resolving the normal contact force into horizontal and vertical components, not just quoting the final tan θ formula.
• Assuming friction is always needed on a banked track — at exactly the design speed for a given angle and radius, none is required at all.

Self-test

1. Give the two expressions for centripetal acceleration.
2. Why is an object in uniform circular motion accelerating?
3. What provides the centripetal force for a car on a flat bend?
4. Derive the minimum speed at the top of a vertical circle.
5. Where is the tension greatest in a vertical circle, and by how much does it differ from the minimum?
6. (Extension, beyond the syllabus) Derive the relationship between banking angle, speed and radius for a frictionless banked track.
7. (Extension, beyond the syllabus) A track is banked at 15° for cars travelling at 20 m/s. Find the design radius.

Answers: 1. a = v²/r and a = rω². 2. Velocity is a vector and its direction changes continuously, so the velocity changes even though the speed does not. 3. Friction between the tyres and the road surface. 4. At minimum speed the tension is zero, so weight alone supplies the centripetal force: mg = mv²/r, giving v = √(gr). 5. At the bottom; if the speed is the same at the top and bottom it exceeds the tension at the top by 2mg (if energy is conserved instead, so the object is faster at the bottom, by 6mg). 6. Resolving the normal contact force: N cos θ = mg (vertical) and N sin θ = mv²/r (horizontal); dividing gives tan θ = v²/(rg). 7. r = v² ÷ (g tan θ) = 20² ÷ (9.81 × tan 15°) = 400 ÷ (9.81 × 0.268) = 152 m.