Lesson 13: Torque and Rotational KinematicsAP

Rotation is translation in a new costume (CED Unit 7): every linear quantity has an angular twin, and every linear law survives the translation. Angle θ (in radians — the only unit in which the equations below work) replaces position; angular velocity ω = Δθ/Δt replaces velocity; angular acceleration α = Δω/Δt replaces acceleration. The bridge equations connect the two worlds at a point a distance r from the axis: s = rθ, v = rω, a_tan = rα, alongside the familiar centripetal a_c = rω². The rotational kinematic equations mirror the linear set exactly (ω = ω_0 + αt, θ = ω_0 t + ½αt², …) under constant α.

Torque: the rotational force

A force changes rotation only through torque: τ = rF sinφ, where φ is the angle between the lever arm and the force. Equivalently, torque is the force times the moment arm — the perpendicular distance from the axis to the force's line of action. Pushing a door at the handle, perpendicular to the door, gives maximum torque; pushing near the hinge, or along the door's face, gives nearly none — which is why door handles live where they do. Torque's unit is the newton-metre; although dimensionally identical to the joule, torque is not energy and is never measured in joules.

Moment of inertia and Newton's second law for rotation

Mass resists linear acceleration; moment of inertia I resists angular acceleration. For a point mass, I = mr²; for extended objects I depends on how the mass is distributed relative to the axis — the shape factors below (in units of MR²) are provided on problems when needed, but their ordering is worth internalising:

The second law translates as τ_net = Iα. As with F = ma, it is the net torque about a chosen axis that matters, and the axis must be stated — torque and I are meaningless without one.

Worked example 1 — torque on a wrench

A mechanic applies 40 N to a wrench 0.25 m long, pulling at 60° to the handle. Find the torque about the bolt. Given: r = 0.25 m, F = 40 N, φ = 60° between handle and force.

τ = rF sinφ = 0.25 × 40 × sin60° = 10 × 0.866 ≈ 8.7 N·m

Check: pushing perpendicular (φ = 90°) would give 10 N·m — the maximum available; pulling along the handle (φ = 0) gives zero, as anyone who has tried it knows. Units N·m, reported as torque, not joules ✓.

Worked example 2 — angular acceleration of a disk

A disk with I = 2.0 kg·m² is acted on by a net torque of 6.0 N·m. Starting from rest, find (a) the angular acceleration, (b) the angular velocity after 4.0 s, (c) the angle turned through. Given: τ_net = 6.0 N·m, I = 2.0 kg·m², ω_0 = 0.

(a) α = τ_net/I = 6.0/2.0 = 3.0 rad/s²
(b) ω = ω_0 + αt = 0 + 3.0 × 4.0 = 12 rad/s
(c) θ = ω_0 t + ½αt² = ½ × 3.0 × 16 = 24 rad ≈ 3.8 revolutions

Check: the calculation is the linear car problem of Lesson 1 with symbols swapped — by design. A point on the rim (r = 0.30 m) would end with tangential speed v = rω = 3.6 m/s, a sensible scale.

Real-world application: gears, levers and flywheels

A bicycle's gear system is a torque-exchange market: a low gear applies the rider's force at a large chainring-to-cog ratio, trading angular speed for torque at the wheel to climb hills; a high gear does the reverse on the flat. Engine flywheels exploit I directly — a heavy rim concentrates mass at large r, storing rotational energy that smooths the pulses between piston strokes. Even the placement of a door handle and the length of a breaker bar are τ = rF sinφ in daily dress.

Experimental design note

Measure an unknown I with a falling mass: wrap a string around the object's axle (radius r), hang mass m, release, and time the fall through height h. Energy or dynamics (mg − T = ma with T·r = Iα and a = rα) yields I from the measured acceleration. The experiment's credibility rests on measuring a precisely (photogate timing, several repeats) because I depends on 1/a and errors amplify.

Common mistakes (from College Board scoring commentary)

• Using the full distance r when the force is not perpendicular — the sinφ (or moment arm) is essential.
• Quoting torque in joules, or energy in newton-metres interchanged carelessly.
• Applying τ = Iα with I computed about the wrong axis, or mixing equations for different axes.
• Working in degrees inside ω = Δθ/Δt or s = rθ; radians are compulsory there.

AP Exam Experience

MCQ 1. A force of 10 N is applied tangent to the rim of a wheel of radius 0.40 m. The torque about the axle is:

• (A) 0.40 N·m
• (B) 2.5 N·m
• (C) 4.0 N·m
• (D) 25 N·m

Answer: (C). Tangential means φ = 90°, so τ = rF = 0.40 × 10 = 4.0 N·m.

MCQ 2. Two identical forces act on a door: one at the handle (0.80 m from the hinge), one halfway (0.40 m), both perpendicular to the door. The ratio of their torques is:

• (A) 1 : 1
• (B) 2 : 1
• (C) 4 : 1
• (D) 1 : 2

Answer: (B). Torque is proportional to lever arm at fixed force and angle, so doubling r doubles τ.

MCQ 3. A wheel with I = 3.0 kg·m² experiences a net torque of 12 N·m. Its angular acceleration is:

• (A) 0.25 rad/s²
• (B) 4.0 rad/s²
• (C) 9.0 rad/s²
• (D) 36 rad/s²

Answer: (B). α = τ_net/I = 12/3.0 = 4.0 rad/s².

FRQ. A uniform disk of mass 2.0 kg and radius 0.25 m is mounted on a fixed axle. A string wrapped around its rim is pulled with a constant tension of 8.0 N. The disk starts from rest (I_disk = ½MR²). (a) Calculate the disk's moment of inertia. (b) Calculate the torque applied to the disk. (c) Calculate the disk's angular velocity after 5.0 s. (d) Calculate the tangential speed of a point on the rim at that time, and the number of revolutions completed. Scoring guideline. (a) 1 point: I = ½(2.0)(0.25²) = 0.0625 kg·m². (b) 1 point: τ = rF = 0.25 × 8.0 = 2.0 N·m (tangential pull, sinφ = 1). (c) 1 point: α = 2.0/0.0625 = 32 rad/s²; ω = αt = 160 rad/s. (d) 2 points: v = rω = 0.25 × 160 = 40 m/s (1 point); θ = ½αt² = ½ × 32 × 25 = 400 rad ≈ 64 revolutions (1 point).

Worked example 3 — measuring a moment of inertia

A spool of axle radius 0.050 m has a string wrapped around it attached to a 0.50 kg hanging mass. Released, the mass falls with acceleration 0.40 m/s² (g = 10 m/s²). Find the spool's moment of inertia about its axle. Method: the tension is found from the falling mass's own equation first — it is not mg, because the mass accelerates.

For the mass: mg − T = ma → T = 0.50 × (10 − 0.40) = 4.8 N
Torque on spool: τ = Tr = 4.8 × 0.050 = 0.24 N·m
Angular acceleration: α = a/r = 0.40/0.050 = 8.0 rad/s²
I = τ/α = 0.24/8.0 = 0.030 kg·m²

Check: had you lazily used T = mg = 5.0 N, you would get I = 0.031 — close here because a ≪ g, but the error grows with a; the rigorous habit costs one extra line. Note how the experiment of the design note is now fully computational.

Rotational kinetic energy: the bridge to Lesson 14

A rotating object stores K_rot = ½Iω² — for the disk of Worked example 2 (I = 2.0 kg·m², ω = 12 rad/s) that is ½ × 2.0 × 144 = 144 J, delivered by the 6.0 N·m torque acting through 24 rad of turning (W = τθ = 6.0 × 24 = 144 J ✓ — the rotational work–energy theorem, W = τΔθ, mirroring W = FΔx). A point mass m at radius r has I = mr² and K = ½(mr²)ω² = ½m(rω)² = ½mv² — the rotational formula contains the linear one, as a consistent translation must.

Torque equilibrium: the statics hiding inside dynamics

Set τ_net = 0 and α = 0 and rotation's second law becomes the law of the seesaw. A 30 kg child sits 1.5 m from a playground pivot; where must a 60 kg adult sit to balance? Equal opposing torques: 30 × 10 × 1.5 = 60 × 10 × d → d = 0.75 m. The g cancels — balancing is about mass × distance, not weight — which is why the same trick works on the Moon. Static equilibrium (ΣF = 0 and Στ = 0) is the whole of bridge and crane design, and it appears in AP questions disguised as ladders, signs on brackets, and arms holding weights.

Worked example 4 — a beam set rotating

A uniform beam 4.0 m long, mass 10 kg, is pivoted at its centre. A 5.0 kg mass is hung 1.6 m from the centre on one side. Find the initial angular acceleration (the beam's own I about its centre is (1/12)mL²).

Torque (the beam's weight acts at the pivot → no torque): τ = 5.0 × 10 × 1.6 = 80 N·m
Total I = (1/12)(10)(4.0²) + (5.0)(1.6²) = 13.3 + 12.8 = 26.1 kg·m²
α = τ/I = 80/26.1 ≈ 3.1 rad/s²

Check: notice the hung mass contributes almost half the total inertia despite being the lighter object — distance squared is a loud voice in I. The beam's own weight contributes nothing to the torque (it acts at the axis) but plenty to I: two different roles for one mass, separated cleanly by the equations.