- Define and explain Circular Motion in your own words
- Use key terms such as simple harmonic motion accurately
- Apply what you have learned to new examples and questions
- Avoid the common mistakes learners make with this topic
GCSE mechanics was the warm-up. Now you will handle motion that curves, oscillates and collides — the mathematics behind satellites, springs and car crashes. Precision matters here: every sign and every square counts.
This lesson focuses on Circular Motion: deriving a = v²/r and applying F = mv²/r to orbits and roundabouts.
Deriving a = v²/r and applying F = mv²/r to orbits and roundabouts.
Key ideas
Circular motion needs a constant centre-seeking force
An object circling at steady speed is accelerating because its velocity direction keeps changing. The centripetal acceleration is v²/r towards the centre, so the required force is F = mv²/r. Remove the force — cut the string — and the object flies off along a tangent, which is why mud flings off a spinning wheel.
SHM describes smooth back-and-forth oscillation
In simple harmonic motion the restoring force grows with displacement: a = −ω²x, the minus sign showing acceleration always points back to equilibrium. A mass on a spring has period T = 2π√(m/k); a pendulum has T = 2π√(l/g). Maximum speed occurs at the centre of the swing, maximum acceleration at the extremes.
Key term — simple harmonic motion: Oscillation where acceleration is proportional to displacement and always towards equilibrium: a = −ω²x.
A 0.5 kg ball is whirled on a 2.0 m string at a steady speed of 4.0 m/s. Calculate the tension in the string.
The tension provides the centripetal force: F = mv²/r. Substitute: F = 0.5 × 4.0² ÷ 2.0. 4.0² = 16, so F = 0.5 × 16 ÷ 2.0 = 8 ÷ 2.0 = 4.0.
Answer: 4.0 N.
- Inventing an outward 'centrifugal force' In an inertial frame there is only the inward centripetal force — the outward feeling is just inertia resisting the turn.
- Treating momentum as a scalar Momentum has direction; opposite momenta subtract, which is why head-on collisions need a sign convention.
Practice
(1200 × 400) ÷ 50 = 480,000 ÷ 50 = 9600 N.
√(0.2 ÷ 80) = √0.0025 = 0.05, so T = 2π × 0.05 ≈ 0.31 s.
Δp = 0.5 × 6 = 3 kg m/s; F = 3 ÷ 0.2 = 15 N.
Before: 2 × 3 = 6 kg m/s. After: 3 kg × v = 6, so v = 2 m/s.
Quick check
Which of these best defines "simple harmonic motion"?
In SHM, where are speed and acceleration greatest?
Why do airbags reduce injury in a crash?
- Circular Motion: deriving a = v²/r and applying F = mv²/r to orbits and roundabouts.
- Circular motion needs a constant centre-seeking force: An object circling at steady speed is accelerating because its velocity direction keeps changing.
- centripetal force: The resultant force towards the centre that keeps an object moving in a circle.
- Watch out for: inventing an outward 'centrifugal force'