EASA / CAR Part-66 · Module 2 · 2.2 Mechanics · Cat A

Kinetics in 3D — Linear, Rotational & Periodic Motion

Objective: work through the three kinds of motion the Category A syllabus asks for — uniform and uniformly accelerated motion in a straight line, uniform circular motion with its centripetal force, and periodic (pendular) motion — on one rotatable 3D stage, with the governing equation solved live at every instant.

Uniformly accelerated motion along a trackdrag to rotate

Velocity against time

Velocity v
0m/s
Distance s
0m
Acceleration a
0m/s²
Elapsed time t
0s
v = u + at  ·  s = ut + ½at²  ·  v² = u² + 2as

Uniform acceleration

A constant acceleration adds the same amount of velocity every second, so the distance covered in each successive second keeps growing.

Linear motion. Speed is distance over time; velocity is speed with a direction, and acceleration is the rate at which velocity changes. If the acceleration is constant, the three equations of motion apply: v = u + at, s = ut + ½at² and v² = u² + 2as, where u is the starting velocity, v the velocity after time t, and s the distance covered.

Watch the 1-second markers dropped on the track. Under uniform velocity (a = 0) they are evenly spaced. Under uniform acceleration the gaps grow every second, because the body is faster in each successive second even though the velocity gain per second is the same. That growing gap is what the ½at² term describes.

A negative acceleration is a deceleration: velocity falls, the markers bunch up, and the body eventually stops. The most common aviation case of constant acceleration is free fall, where a = g = 9.81 m/s² downward and mass makes no difference at all.

  • Try this: Set a = 0. Why are the 1-second markers evenly spaced, and what does the velocity graph look like?
  • Try this: Set u = 0, a = 2 m/s². How far has the body gone after 1 s, 2 s and 3 s? Why is it 1 : 4 : 9 and not 1 : 2 : 3?
  • Try this: Set a negative. Read the stopping distance off the readout, then double u. Why does the stopping distance go up four times?

Rotational motion. A body going round a circle at a steady rate has a constant speed but a continually changing velocity, because its direction changes every instant. A changing velocity means an acceleration, so something must be pulling it — that pull is the centripetal force, and it always acts inward, toward the centre.

The angular velocity is ω = 2πn (n in revolutions per second), the tangential speed is v = ωr, and the centripetal acceleration is a₀ = v²/r = ω²r, so the force needed is F = mω²r. Because ω is squared, doubling the rpm needs four times the force. That is why turbine discs, propellers and rotor heads are speed-limited and why an overspeed is treated so seriously.

The outward "centrifugal force" you feel is not a force acting on the mass at all — it is the mass's inertia resisting being pulled off a straight line. Cut the string, and the mass does not fly outward: it flies off along the tangent.

  • Try this: Note the force at 150 rpm, then set 300 rpm. Why does it go up by a factor of four rather than two?
  • Try this: Hold the rpm and halve the radius. What happens to tangential speed, and what happens to the force?
  • Try this: Watch the green tangential arrow. At what point of the circle does it ever point at the centre?

Periodic motion. Motion that repeats itself in equal intervals is periodic; one complete cycle is a period T, and the number of cycles per second is the frequency f = 1/T, in hertz. A pendulum swinging through a small angle is the classic example, and its motion is simple harmonic: the restoring force is proportional to the displacement and always acts back toward the centre.

For a simple pendulum the period is T = 2π√(l/g). Two things stand out. First, the mass does not appear — a heavy bob and a light bob on equal strings keep exactly the same time, which the second bob here demonstrates. Second, the period depends on the square root of the length, so making a pendulum four times longer only doubles its period.

The formula is a small-angle result. Beyond roughly 10–15° the true period runs progressively longer than 2π√(l/g), and the simulation flags the error once the amplitude is large. Every vibrating aircraft component has this same kind of natural frequency; drive it at that frequency and you get resonance, which is why unbalanced rotating parts must be corrected.

  • Try this: Compare the two bobs. Why does the heavier one not lag behind?
  • Try this: Set l = 0.25 m, note T, then set l = 1.00 m. Why does four times the length give only twice the period?
  • Try this: Set g to a lunar 1.6 m/s². Would a pendulum clock taken to the Moon run fast or slow?