The complete lecture — every idea comes alive in the live panel on the right as you read. Scroll down; the spanner turns, the see-saw balances, the cone topples, and in the equilibrium section you can pull the strings of a real 3D force table yourself.
1 — Rigid body & rotational motion
So far our objects were particles that only slide. Real objects — a ceiling fan, a car wheel, a bottle cap — also spin.
- Rigid body — all its particles keep fixed distances from one another; it never bends or stretches (an ideal model).
- Rotational motion — every particle moves in a circle about a fixed line, the axis of rotation.
Watch the disc: the dot near the rim races while the dot near the hub crawls, yet both sweep the same angle in the same time. That shared angle is why rotation needs its own quantities — θ, ω, α.
Exam point: a car wheel translates with the car and rotates about its axle — combined motion = translation of the centre + rotation about the centre.
2 — Torque: the turning effect of a force
A force applied off-axis makes a body turn. The turning effect is torque (moment of force), τ:
Torqueτ = r F sin θ = (moment arm) × F
moment arm l = r sin θ = ⊥ distance from axis to the line of action
vector: τ = r × F · unit: N·m · anticlockwise positive
- θ = 90° — τ = rF, maximum: push at right angles to the spanner.
- θ = 0° — τ = 0: a force whose line of action passes through the axis cannot turn the body.
- Bigger r — bigger τ: the mechanic slips a pipe over the spanner instead of pushing harder.
worked — spanner
150 N at 90° on a 0.30 m spanner vs 120 N at 60° on a 0.60 m pipe-extended spanner?
τ₁ = 0.30×150×1 = 45 N·m · τ₂ = 0.60×120×0.866 ≈ 62.4 N·m — the longer arm wins
3 — Couple: two forces, pure rotation
Turn a steering wheel: two hands, equal forces, opposite directions, different lines of action. That pair is a couple.
Moment of a coupleτ = F × d (one force × ⊥ distance between the two forces)
net force = F − F = 0 → no translation, pure rotation
same moment about every point — a couple has no unique pivot
worked — steering wheel
Two 25 N forces on opposite rims of a 40 cm wheel?
τ = F×d = 25 × 0.40 = 10 N·m
Everyday couples: turning a tap, winding a clock key, a cyclist's two pedals, wringing a wet cloth — zero net force, pure turning.
4 — Centre of mass & centre of gravity
- Centre of mass — the point where the whole mass may be considered concentrated; a force through it gives pure translation.
- Centre of gravity (CG) — the point through which the total weight acts. In a uniform field (any school lab) CG = CM.
Regular bodies: CG at the geometric centre — midpoint of a metre rule, centre of a disc, intersection of a triangle's medians. It can lie outside the material: centre of a ring.
Board practical — irregular lamina1) hang lamina freely from hole A → CG settles vertically below the pin
2) hang a plumb line from the same pin, draw the line
3) repeat from holes B and C → the lines intersect at the CG
4) check: lamina balances on a pencil tip at that point
Why it works: a freely hung body is in equilibrium only when its weight and the pin's reaction share the same vertical line — so the CG lies on every plumb line.
5 — First condition of equilibrium: ΣF = 0
A body is in translational equilibrium when the vector sum of all forces on it is zero:
First conditionΣF = 0 → ΣFx = 0 and ΣFy = 0
static equilibrium: at rest (book on a table)
dynamic equilibrium: uniform velocity (paratrooper at terminal speed)
The 3D bench on the right is the lab's force table: three strings pull a central ring over pulleys. Adjust the angles and weights — when the three tensions add to zero, the ring floats exactly over the centre pin. That is ΣF = 0 made visible.
But ΣF = 0 is not enough — a couple has zero net force and still spins the body. Equilibrium needs a second condition.
6 — Second condition: Στ = 0 (principle of moments)
For rotational equilibrium the torques must also cancel:
Second conditionΣτ = 0 about any axis
Σ(clockwise torques) = Σ(anticlockwise torques)
On the right, a 40 kg child sits 1.5 m left of the pivot. The see-saw tips — until the 30 kg child slides out to 2.0 m on the other side and the live Στ readout hits zero. Lighter child, longer arm: w₁d₁ = w₂d₂.
worked — beam on two supports
Uniform 6 m beam (200 N) on end supports; 450 N load 2 m from A. Reactions?
R_A+R_B = 650 N · about A: R_B×6 = 450×2 + 200×3 = 1500 → R_B = 250 N, R_A = 400 N
Technique: take torques about the point where an unknown force acts — it vanishes from the equation (zero moment arm).
7 — States of equilibrium
Tilt a body slightly and let go. What its CG height does decides everything:
| State | On tilting, CG… | Behaviour | Example |
|---|
| Stable | rises | returns | cone on base, book on table |
| Unstable | falls | topples | cone on tip, pencil on point |
| Neutral | stays level | rests where left | ball or cylinder on the floor |
A body topples once the vertical line through its CG leaves its base. Hence the two design rules: lower the CG (racing cars; heavy luggage in a bus's lower hold) and widen the base (tripod legs, a wrestler's stance).
Truck loading (board favourite): stack heavy crates at the bottom — on a slope or sharp turn the weight line must stay inside the wheels' base, or the truck rolls.
8 — Angular displacement, velocity & acceleration
- Angular displacement θ — angle swept, in radians: θ = s/r. 1 rev = 2π rad = 360°; 1 rad ≈ 57.3°.
- Angular velocity ω — ω = Δθ/Δt (rad/s); for f rev/s, ω = 2πf. Direction by the right-hand rule, along the axis.
- Angular acceleration α — α = Δω/Δt (rad/s²).
Equations of angular motion (constant α) — twins of the linear setωf = ωi + αt · θ = ωi t + ½αt² · 2αθ = ωf² − ωi²
worked — fan spin-up
Fan reaches 300 rpm from rest in 10 s. α? revolutions?
ωf = 2π(300/60) ≈ 31.4 rad/s → α = 3.14 rad/s² · θ = ½×3.14×100 = 157 rad = 25 rev
9 — Linking linear & angular: v = rω, a = rα
All particles of a rotating body share the same ω, but linear speed grows with radius. From s = rθ:
Linear ⟷ angulars = rθ · v = rω · a = rα (θ in radians only)
That is why a long fan blade's tip whooshes while the hub barely moves, why the rim of a CD outruns its centre, and why a wheel of radius r on a car at speed v spins at ω = v/r.
worked — motorcycle wheel
r = 0.35 m at 8 rev/s — rim speed?
ω = 2π×8 ≈ 50.3 rad/s → v = rω = 17.6 m/s ≈ 63 km/h
10 — Worked numericals & exam recap
numerical — door torque
100 N at 30° to a 1.2 m door, about the hinges?
τ = 1.2×100×sin30° = 60 N·m (90° would give 120 N·m)
numerical — braking wheel
60 rad/s to rest in 12 s. α and θ?
α = −5 rad/s² · θ = (0−3600)/(2×−5) = 360 rad ≈ 57.3 rev
- τ = rF sinθ = moment arm × force; max at 90°, zero through the axis; anticlockwise positive.
- Couple: τ = F·d; zero net force, pure rotation, same about every point.
- CG: where the weight acts; irregular lamina → three plumb lines intersect at it.
- Equilibrium = ΣF = 0 and Στ = 0; static or dynamic.
- Stable / unstable / neutral ↔ CG rises / falls / stays level; lower CG, wider base.
- θ = s/r, ω = 2πf, α; angular equations mirror the linear set; v = rω, a = rα.