What is torque (moment of force)?
Torque is how strongly a force turns a body about an axis. It is also called the moment of force.
τ = r × F, size τ = rF sinθ, where r is the distance from the axis to where the force acts, and θ is the angle between r and F.
- Unit: N m (not written as joule, because it is not energy).
- Dimensions: [M L² T⁻²].
- r sinθ is the moment arm: the perpendicular distance from the axis to the line of the force.
- Direction: by the right-hand rule, along the axis. Anticlockwise is usually taken as positive.
A couple is two equal and opposite forces not on the same line. It turns a body without moving it forward, like turning a tap or a steering wheel. Its torque = F × distance between the lines.
Angular momentum
Linear momentum p = mv says how hard it is to stop a moving body. Angular momentum says how hard it is to stop a turning one.
For a particle: L = r × p, size L = r p sinθ = m v r (when v is at right angles to r). Unit: kg m²/s (or J s).
For a rigid body spinning about a fixed axis: L = Iω, where I is the moment of inertia.
Link with torque: just as F = dp/dt, we have τ = dL/dt. Proof in one line: L = r × p, so dL/dt = (dr/dt × p) + (r × dp/dt) = (v × mv) + (r × F) = 0 + τ.
Conservation of angular momentum
If the total external torque on a system is zero, then dL/dt = 0, so L stays constant. For a fixed axis: I₁ω₁ = I₂ω₂.
- Skater or dancer pulls arms in → I falls → ω rises.
- Diver tucks the body → spins more times before entering water.
- A planet moves faster when it is nearer the Sun (Kepler's second law comes from this).
Note: kinetic energy of rotation ½Iω² is not conserved here. When arms are pulled in, the person does work, so KE rises.
Equilibrium of rigid bodies
A rigid body is in mechanical equilibrium when it has no linear acceleration and no angular acceleration. Two conditions:
- Translational equilibrium: ΣF = 0 (no sliding).
- Rotational equilibrium: Στ = 0 about any point (no turning).
Principle of moments (lever): load × load arm = effort × effort arm. Clockwise torque = anticlockwise torque.
A couple has ΣF = 0 but Στ ≠ 0, so the body is not in equilibrium: it turns. Centre of gravity is the point about which the total gravitational torque on the body is zero; a body hung from its CG stays in any position.
Types (for understanding): stable (a ball in a bowl), unstable (a pencil on its tip), neutral (a ball on a flat floor).
Try it at home
Sit on a swivel chair with two filled water bottles, arms stretched. Ask a friend to spin you gently, then pull the bottles to your chest. Feel the chair speed up. Next, balance a 30 cm scale on a pencil and put one coin at 5 cm from the centre. Where must two coins go on the other side? Predict (2.5 cm), then check.
Key formulas and definitions
- τ = r × F, τ = rF sinθ (N m)
- Moment arm = r sinθ
- Couple: τ = F × d
- L = r × p, L = mvr (v ⟂ r)
- L = Iω (fixed axis)
- τ = dL/dt
- τ_ext = 0 ⇒ L constant, I₁ω₁ = I₂ω₂
- Equilibrium: ΣF = 0 and Στ = 0
Worked examples
1. A 20 N force is applied at right angles to a door, 0.8 m from the hinge. Find the torque.
τ = rF sin90° = 0.8 × 20 × 1 = 16 N m.
2. The same 20 N force acts at 0.8 m but at 30° to the door. Find τ.
τ = 0.8 × 20 × sin30° = 16 × 0.5 = 8 N m. Half the turning effect.
3. Find the torque of F = (2î + 3ĵ) N acting at r = (4î + 1ĵ) m about the origin.
τ = r × F = (x F_y − y F_x) k̂ = (4×3 − 1×2) k̂ = 10 k̂ N m.
4. A 0.2 kg ball moves in a circle of radius 0.5 m at 4 m/s. Find its angular momentum about the centre.
L = m v r = 0.2 × 4 × 0.5 = 0.4 kg m²/s.
5. A wheel's angular momentum grows from 2 to 8 kg m²/s in 3 s. Find the average torque.
τ = ΔL/Δt = (8 − 2)/3 = 2 N m.
6. A dancer spins at 2 rev/s with I = 6 kg m². She pulls in her arms and I becomes 2 kg m². New speed?
I₁ω₁ = I₂ω₂ → 6 × 2 = 2 × ω₂ → ω₂ = 6 rev/s. Three times faster.
7. On a see-saw pivoted at the centre, a 40 kg child sits 2 m from the pivot. Where should a 20 kg child sit to balance? Also find the pivot force (g = 10 m/s²).
Στ = 0: 400 × 2 = 200 × x → x = 4 m on the other side. ΣF = 0: N = 400 + 200 = 600 N (plank weight ignored).
8. A 2 m uniform plank of 10 kg rests on two supports at its ends. A 30 kg box sits 0.5 m from the left end. Find both support forces (g = 10).
Weights: plank 100 N at 1 m, box 300 N at 0.5 m. Torques about left end: N_R × 2 = 100×1 + 300×0.5 = 250 → N_R = 125 N. ΣF: N_L = 400 − 125 = 275 N.
Common mistakes
- Using the full distance r when the force is slanted. Use r sinθ (the moment arm).
- Writing torque in joules. Torque is N m; it is not work or energy.
- Thinking equilibrium needs only ΣF = 0. A couple has ΣF = 0 but still turns the body; Στ must be 0 too.
- Thinking kinetic energy is also conserved when a skater pulls arms in. L is conserved; KE increases.