Flashcards · 9630 · A Level Topic 6

Circular and periodic motion (A Level) flashcards

12 cards on A Level Topic 6, subtopics 3.6.1 to 3.6.2. Pick a subtopic or study them all.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 12 cards

For a quick read-through or printing. Cover the right-hand side to test yourself.

Radian: The angle at the centre of a circle when the arc length equals the radius: θ = s/r. 360° = 2π rad.
Angular speed: definition and equations: The angle turned per unit time, in rad s⁻¹. ω = v/r = 2πf = 2π/T.
Centripetal acceleration and centripetal force: a = v²/r = ω²r, directed towards the centre. F = mv²/r = mω²r: the resultant force towards the centre.
Why is an object moving in a circle at constant speed accelerating?: Its direction, and so its velocity, is continuously changing. The acceleration is towards the centre, so a resultant force towards the centre is needed.
Why does the centripetal force do no work?: It acts at 90° to the velocity, so it transfers no energy and the speed stays constant.
What provides the centripetal force for a car on a flat bend, a mass on a string, and a satellite?: Car: friction (sideways grip of the tyres). Mass on a string: tension. Satellite: gravitational attraction.

Simple harmonic motion: definition and defining equation: Motion in which the acceleration is proportional to the displacement from equilibrium and always directed towards it: a = −ω²x.
SHM equations for displacement and speed: x = A cos ωt (timing from x = +A, ωt in radians). v = ±ω√(A² − x²).
Maximum speed and maximum acceleration in SHM, and where they occur: Maximum speed ωA, at equilibrium (x = 0). Maximum acceleration ω²A, at the extremes (x = ±A).
Derive the period of a mass-spring system: F = −kx and F = ma give a = −(k/m)x. Comparing with a = −ω²x, ω² = k/m, so T = 2π/ω = 2π√(m/k).
Derive the period of a simple pendulum, and its condition: Restoring force −mg sin θ. For small θ in radians, sin θ ≈ θ = x/l, so a = −(g/l)x, ω² = g/l and T = 2π√(l/g). The mass cancels.
Total energy in SHM, and the period of the Ek and Ep graphs against time: Total energy = ½mω²A² (½kA² for a spring), proportional to A². Ek and Ep each vary with period T/2.