Water Resources HubFE section 7 of 16 · free theory
Dynamics
Kinematics, Newton's second law, work-energy, impulse-momentum, and vibration — the whole FE dynamics block in one place.
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Rectilinear kinematics: the three equations that matter
Position s, velocity v, and acceleration a are linked by differentiation and integration. When acceleration is constant, three relations cover almost every exam question:
v = ds/dt a = dv/dt v dv = a ds
| s | position along the line of motion |
| v | velocity, ds/dt |
| a | acceleration, dv/dt |
v = v0 + at
s = s0 + v0t + ½at²
v² = v0² + 2a(s − s0)
| v0 | initial velocity (at t = 0) |
| s0 | initial position (often taken as zero) |
Pick the equation that mentions the quantity you want and the quantities you know. If time is not given and not asked, use v² = v0² + 2aΔs — it eliminates t directly.
Curvilinear kinematics: projectiles and path coordinates
Projectile motion is just two independent rectilinear motions: constant velocity horizontally, constant acceleration (−g) vertically. Normal–tangential coordinates describe acceleration along any curved path:
x = v0xt y = v0yt − ½gt²
| v0x = v0 cos θ | horizontal launch component |
| v0y = v0 sin θ | vertical launch component |
R = v0² sin 2θ / g hmax = v0² sin²θ / (2g) T = 2v0 sin θ / g
| R | horizontal range (launch and landing at the same elevation) |
| hmax | maximum height above the launch level |
| T | total time of flight |
at = dv/dt an = v²/ρ
| at | tangential acceleration — rate of change of speed, along the path |
| an | normal (centripetal) acceleration — toward the centre of curvature, ρ is the radius of curvature |
| a = √(at² + an²) | magnitude of total acceleration |
The range formula needs sin 2θ, not sin θ — a favourite trap. And an always points toward the centre of curvature, even when the particle is slowing down; the slowing is carried entirely by at.
Kinetics: Newton's second law and friction
Draw the free-body diagram first, then write ΣF = ma along each coordinate direction. On an incline, friction opposes the relative motion:
ΣF = ma
| m | mass — if the problem gives weight W, convert first: m = W/g |
| a | acceleration of the mass centre in the chosen direction |
Ffriction = μN
| μ | coefficient of friction (μs for impending slip, μk for sliding) |
| N | normal force on the surface (on an incline, N = mg cos θ) |
Weight is a force; mass is mass. In US units, a 32.2-lb block has a mass of 1 slug, not 32.2. In SI the same slip is rarer because mass is usually given in kg — but watch for problems that hand you newtons of weight instead.
Work–energy: when you know distances, not times
The work–energy principle is the kinetics twin of v² = v0² + 2aΔs. If the question gives you speeds and distances but no time, this is almost always the intended route:
T1 + ΣU1–2 = T2 T = ½mv²
| T | kinetic energy (always non-negative) |
| U | work of each force over the displacement |
Uweight = ±mgΔh Uspring = ½k(s1² − s2²) Ufriction = −Ffd
| Δh | vertical drop is positive work for gravity; a rise is negative |
| s1, s2 | spring stretch/compression at positions 1 and 2 |
Normal forces and forces perpendicular to the displacement do no work — only components along the path count. Friction always does negative work when sliding occurs.
Impulse–momentum: when you know times, not distances
The mirror image of work–energy: use it when the problem hands you a duration, an impact, or a stream of fluid. Momentum is a vector — directions matter:
mv1 + Σ∫F dt = mv2
| ∫F dt | linear impulse of a force; for a constant force over Δt it is simply F Δt |
e = (vB2 − vA2) / (vA1 − vB1)
| e | coefficient of restitution: 1 for elastic (no kinetic-energy loss), 0 for perfectly plastic (bodies stick together) |
Impacts are solved with two equations: conservation of linear momentum of the system, plus the restitution relation. One equation alone is not enough — the exam counts on you forgetting the second.
Vibration: natural frequency, damping, and resonance
FE vibration questions are usually single-degree-of-freedom and ask for the natural frequency, the period, or what happens when the forcing frequency matches it:
ωn = √(k/m) fn = ωn/(2π) τ = 2π√(m/k)
| ωn | undamped natural circular frequency (rad/s) |
| fn | natural frequency (Hz, cycles/s) |
| τ | period of one oscillation (s) |
ζ = c/cc cc = 2√(km) fd = fn√(1 − ζ²)
| ζ | damping ratio (ζ < 1 underdamped, ζ = 1 critically damped, ζ > 1 overdamped) |
| fd | damped natural frequency — always slightly below fn |
Resonance: when the forcing frequency Ω approaches ωn, the steady-state amplitude grows large (bounded only by damping). A machine running near its natural frequency will rattle itself apart — that is the physical idea behind every resonance question.
Worked example Projectile: range, height, and flight time
Given:
- Launch speed v0 = 25 m/s at θ = 35° above the horizontal, from ground level.
- g = 9.81 m/s². Find R, hmax, and T.
Solution:
- Resolve the velocity: v0x = 25 cos 35° = 20.5 m/s, v0y = 25 sin 35° = 14.3 m/s.
- Time of flight: T = 2 × 14.3 / 9.81 = 2.92 s. (Landing when y returns to zero.)
- Range: R = 20.5 × 2.92 = 59.9 m. Cross-check with R = 25² sin 70°/9.81 = 59.9 m. ✓
- Max height: hmax = 14.3²/(2 × 9.81) = 10.5 m.
Answer: R ≈ 59.9 m, hmax ≈ 10.5 m, T ≈ 2.92 s.
Worked example Block sliding down an incline with friction
Given:
- 10-kg block released from rest on a 30° incline, slides d = 5 m down it.
- Kinetic friction μk = 0.20. Find the speed at the bottom.
Solution:
- Normal force: N = mg cos 30° = 10 × 9.81 × 0.866 = 85.0 N. Friction: Ff = 0.20 × 85.0 = 17.0 N, acting up the incline.
- Work of gravity: Ug = mg sin 30° × 5 = 245 J (positive — moving downhill). Work of friction: Uf = −17.0 × 5 = −85.0 J.
- Work–energy: 0 + 245 − 85.0 = ½(10)v² → v = √(2 × 160/10) = 5.66 m/s.
Answer: v ≈ 5.66 m/s at the bottom.
Worked example Natural and damped frequency of a mass-spring system
Given:
- Mass m = 20 kg on a spring with k = 5000 N/m.
- Damping ratio ζ = 0.15. Find fn, the period τ, and fd.
Solution:
- ωn = √(5000/20) = 15.8 rad/s. Convert to cycles: fn = 15.8/(2π) = 2.52 Hz.
- Period: τ = 1/2.52 = 0.397 s.
- Damped frequency: fd = 2.52 × √(1 − 0.15²) = 2.52 × 0.989 = 2.49 Hz — slightly below the undamped value, as expected.
Answer: fn ≈ 2.52 Hz, τ ≈ 0.397 s, fd ≈ 2.49 Hz.