Torque and Work: Study Guide (Physics C: Mechanics 6.2)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 6.2 called Torque and Work. This guide is the calculus-based one. It derives W = ∫τ dθ, integrates torques that change with angle, and builds the rotational work–energy theorem from τ = Iα. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

A torque transfers energy only through an angle

In Topic 3.2, a force does work on an object only if the point where it acts has a displacement with a component along the force. The rotational version is similar: a torque transfers energy into or out of a rigid body only if the body turns through an angle while the torque acts.

• A mechanic leans on a spanner but the rusted nut does not move. There is a large torque, but Δθ = 0, so no work is done and no energy is transferred to the nut.
• An electric drill turns a bit through many revolutions while exerting a torque in the direction of rotation. Energy flows from the motor into the bit: positive work.
• Friction in a bearing exerts a torque against the rotation. Energy leaves the rotating body: negative work.

From force work to torque work

Take a rigid body turning about a fixed axis through O. A force F acts at point P, a distance r from the axis, at angle φ between the position vector r and F (Figure 1).

1. When the body turns through a small angle dθ, point P moves along a circular arc of length ds = r dθ (θ in radians). The displacement is tangent to the circle.
2. Only the tangential component of the force, F_t = F sin φ, lies along that displacement. The radial component, F cos φ, is perpendicular to it and does no work.
3. So the work done is dW = F_t ds = (F sin φ)(r dθ) = (rF sin φ) dθ.
4. The bracket is the torque about the axis from Topic 5.3: τ = rF sin φ. So dW = τ dθ.
5. Add up over the whole rotation from θ₁ to θ₂:

W = ∫ τ dθ (from θ₁ to θ₂)

If the torque is constant, it comes outside the integral: W = τΔθ.

Units. τ in N·m times θ in rad gives N·m = J, because the radian is dimensionless. Torque and energy share base units, but they are different quantities: always write torques in N·m and work in J.

Positive, negative and zero work

Choose a positive sense of rotation (for example, counterclockwise positive). Then τ and dθ carry signs, and τ dθ gives the sign of the work automatically:

Situation	Sign of τ dθ	Work
Torque in the same sense as the rotation	+	positive: energy in
Torque opposite to the rotation (friction, brakes)	−	negative: energy out
Force through the axis, or along the radius	τ = 0	zero
Body does not turn (Δθ = 0)	dθ = 0	zero

A collection of torques. Each torque does its own work, W_i = ∫τ_i dθ. Because every point of a rigid body turns through the same dθ, the total is W_net = ∫(τ₁ + τ₂ + …) dθ = ∫τ_net dθ.

The rotational work–energy theorem

Start from Newton’s second law in rotational form (Topic 5.6), for a rigid body on a fixed axis: τ_net = Iα.

1. Use the chain rule to write α in terms of angle: α = dω/dt = (dω/dθ)(dθ/dt) = ω dω/dθ.
2. So τ_net = Iω dω/dθ, which rearranges to τ_net dθ = Iω dω.
3. Integrate from (θ₀, ω₀) to (θ, ω): ∫τ_net dθ = I∫ω dω = ½Iω² − ½Iω₀².

W_net = ΔK_rot

This is the rotational partner of the translational work–energy theorem from Topic 3.2, and it uses the same chain-rule step as the derivation of v² = v₀² + 2a(x − x₀) in Topic 1.2. It needs a rigid body (so I is constant) and a fixed axis. It also explains the straight K–θ graph from Topic 6.1: with a constant net torque, K grows by τ_net for every radian.

Area under a torque–angle graph

Since W = ∫τ dθ, the work done by a torque between two angles is the signed area between its τ–θ graph and the θ-axis. Area above the axis is positive work; area below is negative work. When a torque changes with angle, the area is often easier than a formula: count squares, use trapezia, or integrate the equation of the line. Figure 2 shows this for Worked example 1.

Comparing scenarios. The theorem lets you compare without solving everything:

• Doubling a constant torque, or doubling the angle it acts through, doubles the work and so doubles the change in rotational kinetic energy.
• The same work gives the same change in kinetic energy whatever the rotational inertia. A body with larger I ends up with a smaller ω, because K = ½Iω².
• A torque that changes with angle does not do equal work in equal angles. Look at where the graph is high.

Worked example 1: a motor whose torque falls with angle

Question. Take counterclockwise as positive. A grinding wheel with rotational inertia I = 0.40 kg·m² starts from rest. A motor drives it with a torque that falls as the wheel turns: τ_m(θ) = 12 − 1.5θ (τ in N·m, θ in rad) for 0 ≤ θ ≤ 8.0 rad. A constant friction torque of 1.0 N·m opposes the rotation throughout. (a) Find the work done by the motor and by friction over the 8.0 rad. (b) Find the wheel’s angular velocity at θ = 8.0 rad. (c) Find the greatest angular velocity, and the angle where it happens.

(a) Work by each torque.

1. Motor: W_m = ∫₀^8.0 (12 − 1.5θ) dθ = [12θ − 0.75θ²]₀^8.0 = 96 − 48 = +48 J. On Figure 2 this is the triangle under the motor line: ½ × 8.0 rad × 12 N·m = 48 J. ✓
2. Friction: τ_f = −1.0 N·m (it opposes counterclockwise rotation). W_f = ∫₀^8.0 (−1.0) dθ = −8.0 J.

(b) Final angular velocity.

1. W_net = 48 − 8.0 = 40 J = ½Iω² − 0.
2. ω = √(2 × 40/0.40) = √200 = 14.1 rad/s ≈ 14 rad/s.

(c) Greatest angular velocity.

1. ω is greatest where dω/dt = α = 0, that is where τ_net = 0: 12 − 1.5θ − 1.0 = 0, so θ = 7.33 rad.
2. W_net up to there: ∫₀^7.33 (11 − 1.5θ) dθ = 11 × 7.33 − 0.75 × 7.33² = 40.3 J.
3. ω_max = √(2 × 40.3/0.40) = 14.2 rad/s, at θ = 7.3 rad. After that, friction exceeds the motor torque, net work is negative, and the wheel slows slightly before θ = 8.0 rad.

Check. A student who treats the motor torque as a constant 12 N·m gets W_m = 96 J and ω ≈ 21 rad/s, far too high. The torque falls with angle, so you must integrate (or use the area).

Worked example 2: a gate with a return spring

Question. A rotating gate turns about a vertical axis and has rotational inertia I about it. A spring pulls it back towards the closed position with a torque −κθ, where θ is the angle from closed (opening is positive). Starting from rest at θ = 0, a person pushes it open with a constant torque τ_p. Friction is negligible. (a) Derive ω(θ). (b) Find the angle at which the gate turns fastest, and the angle at which it stops. (c) Evaluate both, and the greatest ω, for I = 25 kg·m², κ = 40 N·m/rad and τ_p = 30 N·m.

(a) Derive ω(θ).

1. Work by the person: W_p = ∫₀^θ τ_p dθ = τ_p θ (positive).
2. Work by the spring: W_s = ∫₀^θ (−κθ) dθ = −½κθ² (negative: the spring torque opposes the opening).
3. Work–energy theorem: τ_p θ − ½κθ² = ½Iω² − 0.
4. So ω(θ) = √((2τ_p θ − κθ²)/I).

(b) Fastest and stopping angles.

1. ω is greatest where the net work stops increasing: d(τ_p θ − ½κθ²)/dθ = τ_p − κθ = 0, so θ = τ_p/κ. This is where the net torque is zero, as in Worked example 1.
2. The gate stops when ω = 0 again: 2τ_p θ − κθ² = 0, so θ = 2τ_p/κ. There, the positive work by the person exactly cancels the negative work by the spring.

(c) Numbers.

1. Fastest at θ = 30/40 = 0.75 rad. Net work there: 30 × 0.75 − 20 × 0.75² = 22.5 − 11.25 = 11.25 J. ω_max = √(2 × 11.25/25) = 0.95 rad/s.
2. Stops at θ = 2 × 30/40 = 1.5 rad (about 86°).

Check the units. κθ² has units (N·m/rad)(rad²) = N·m·rad = J. ✓ Interpretation. Doubling the push τ_p doubles both angles. The net work at the fastest point is τ_p²/(2κ), so it grows four times and the greatest ω doubles.

Common misconceptions

• “A large torque always does a lot of work.” No rotation, no work. A torque on a stuck nut transfers no energy.
• Using revolutions or degrees in W = τΔθ. Convert to radians: 1 rev = 2π rad.
• W = τΔθ for a torque that changes with angle. Integrate τ(θ), or use the area under the graph.
• Forgetting negative work. Friction and brake torques take energy out; include them in W_net.
• Calling torque an energy because both can be written in N·m. Write torque in N·m and work in J.
• “The radial component of a force does some work.” It is perpendicular to the motion of the point and does none.
• Confusing ∫τ dθ with ∫τ dt. The area under τ against θ is work; the area under τ against t is something else (Topic 6.3).
• Thinking the fastest moment is the end of the push. ω is greatest where the net torque falls to zero.

Where this leads

Topic 6.3, Angular Momentum and Angular Impulse, asks what a torque does over a time interval instead of an angle: ∫τ dt changes angular momentum, just as ∫τ dθ changes rotational kinetic energy. Topic 6.5, Rolling, combines both kinds of kinetic energy with the work done by forces and torques. Next, try the practice questions, then use the revision notes and the checklist. When you are ready, move on to Topic 6.3, Angular Momentum and Angular Impulse, or return to the course roadmap.