1. Angular Velocity and Acceleration
When a particle moves in a circle of radius r, we describe its motion using angular velocity ω (radians per second).
A particle moving in a circle at constant speed is not in equilibrium — its velocity direction is continuously changing, so there is a net acceleration. This acceleration always points towards the centre.
Direction: always towards the centre of the circle. The proof (using calculus or geometry) is not required by Cambridge, but the formula is essential.
The centripetal force is not a separate force. It is the resultant of all real forces acting towards the centre:
This force is provided by tension, gravity, normal reaction, or friction — depending on the context.
2. Horizontal Circles
In horizontal circular motion, the particle moves in a fixed horizontal plane. The vertical forces are in equilibrium; the net horizontal force provides the centripetal acceleration.
The Conical Pendulum
A particle of mass m is attached to a string of length L and moves in a horizontal circle. The string makes angle θ with the vertical.
Conical pendulum: string length L, angle θ from vertical, radius r = L sinθ
Resolving forces for the conical pendulum:
Dividing the equations:
Students often set T sinθ = mg and T cosθ = mrω² — this is wrong. The vertical component balances gravity; the horizontal component provides centripetal force. Always draw a force diagram first.
3. Vertical Circles
In vertical circular motion, the speed of the particle changes as it moves round the circle — gravity does work on it. We use conservation of energy to find the speed at any point, then resolve forces for the tension or normal reaction.
- Step 1 — Energy: Use ½mv² + mgh = constant to find v at any point
- Step 2 — Newton's 2nd Law: Resolve forces towards the centre: net force = mv²/r
Particle on the Inside of a Circle (String / Rod)
Particle at P, height h above bottom A. Angle θ from vertical through centre.
Let the particle have speed vA at the bottom (A) and speed v at point P, height h above A.
Note: h = r − r cosθ = r(1 − cosθ), where θ is measured from the bottom.
Condition for Complete Vertical Circles
For a particle on a string, the string must remain taut at all points. The critical point is the top of the circle, where tension T ≥ 0.
Using energy conservation from bottom to top (height = 2r):
For a particle on the outside of a sphere or rod, the condition changes — normal reaction R ≥ 0 gives a different critical speed.
Particle on the Outside of a Sphere
A particle rests on the outside of a smooth sphere of radius r. It leaves the surface when the normal reaction R = 0.
m = 0.5 kg, L = 0.8 m, θ = 40°. Radius r = L sinθ = 0.8 sin40° = 0.514 m.
Check: 2π√(Lcosθ/g) = 2π√(0.8×0.766/9.8) = 2π√(0.0626) = 2π×0.250 = 1.57 s ✓
At the top, minimum condition is T = 0:
Height from bottom to top = 2r = 1.0 m:
Note: vbot = √(5gr) = √(5×10×0.5) = √25 = 5 m/s ✓
At the bottom, tension T acts upward, weight mg downward. Net force towards centre (upward) = mv²/r:
So tension at bottom = 6 × weight of particle. Notice T = 6mg — a standard result when vbot = √(5gr).
Let θ = angle from vertical, v = speed at that point. Taking v₀ ≈ 0 (small impulse). Drop in height from top = r(1 − cosθ).
Substitute v² from energy into Newton's law with R = 0:
Interactive Circular Motion Simulator
Switch between Horizontal Circle (conical pendulum) and Vertical Circle. Adjust parameters and observe how tension, speed, and reaction forces change.
Practice Questions
⚠ Tension is negative! This means the string would need to push the particle — impossible. The particle leaves the circle before reaching the top. Check: v²min at top = gr = 9.8 × 0.8 = 7.84. But v²top = 4.64 < 7.84. The particle does not complete full circles.
Note for beads on a wire: Unlike a string, the wire can push outward. So a bead can complete circles for any speed — there is no minimum. However the question asks about the condition for completing circles in the context of the bead not losing contact (for a string). Let us proceed with the standard proof.
Proof (string version): At the top, minimum condition T = 0:
Normal reaction at top:
Formula Reference Sheet
Complete reference for circular motion — Cambridge 9231 P3.
- Always draw a force diagram before resolving. Mark forces and the direction towards the centre clearly.
- For vertical circles, apply energy first, then Newton's 2nd Law — never try to skip energy.
- Check whether the particle is on the inside or outside of the circle — this determines which way the normal reaction acts.
- A bead on a wire can have outward normal reaction — it completes circles for all speeds (no minimum).
- State clearly when you use conservation of energy: "no energy loss, so KE + PE = constant" — examiners reward explicit statements.