Torque and Work: Study Guide (Physics 1 6.2)

This guide is for the algebra-based Physics 1 course. In Unit 3 a force did work by acting through a distance. Here a torque does work by acting through an angle. No calculus is needed.

Torque transfers energy only when something turns

Push hard on a spanner gripping a bolt that will not move. You exert a large torque, but the bolt does not turn. No energy goes into the bolt. Now push the same spanner on a loose bolt, and it turns. Energy is transferred.

The rule is the same as for forces. A force does work only if its point of application moves. A torque does work only if the object turns while the torque acts. A torque with zero angular displacement does zero work, however large it is.

Deriving W = τΔθ

Take a force F acting at the rim of a wheel, at distance r from the axle, always perpendicular to the radius (tangent to the circle). Let the wheel turn through angle Δθ.

1. The point where the force acts moves along an arc. The arc length is s = rΔθ, with Δθ in radians.
2. The force points along the arc the whole time, so the work it does is W = Fs.
3. Substitute s: W = F(rΔθ) = (rF)Δθ.
4. For a perpendicular force, rF is the torque τ. So:

Work done by a constant torque: W = τΔθ

If the force is at an angle to the radius, only the perpendicular component F⊥ moves along the arc, and the parallel component does nothing. Since τ = rF⊥, the result is the same. A force pointing straight at the axle has F⊥ = 0, so it exerts no torque and does no work on the rotation.

Units. N·m × rad = J, because a radian is a ratio of two lengths and has no dimension. Always put Δθ in radians. One revolution is 2π rad; 90° is π/2 rad. With degrees, a torque of 8.0 N·m turned through 90° would seem to do 720 J; the correct value is 8.0 × π/2 = 12.6 J.

Positive, negative and zero work

Choose a positive sense of rotation, for example counterclockwise. Then:

Torque compared with rotation	Sign of work	Energy
Same sense (a motor driving a fan)	positive	into the system; it speeds up if nothing else acts
Opposite sense (friction at an axle, a brake)	negative	out of the system; it slows down if nothing else acts
No rotation while it acts (a stuck bolt)	zero	no transfer
Force through the axle (radial)	zero	no transfer

When several torques act, add their works with signs. The total is the net work, which also equals τ_net × Δθ when every torque is constant.

Torque, work and rotational kinetic energy

In Unit 3 you met the work–energy theorem: net work on an object equals its change in kinetic energy. The rotational version is the same idea. For a rigid object turning about a fixed axis:

W_net = ΔK = ½Iω² − ½Iω₀²

Positive net work speeds the object up; negative net work slows it down. This is often the fastest route to a final angular velocity, because it skips time and angular acceleration completely.

You can check the link with Topic 5.6. For a constant net torque, α = τ_net / I, and rotational kinematics gives ω² = ω₀² + 2αΔθ. Multiply by ½I: ½Iω² − ½Iω₀² = IαΔθ = τ_net Δθ. The two approaches agree.

When the torque changes: area under the graph

Many torques change as an object turns: a wind-up spring loses torque as it unwinds, and a motor’s torque depends on how it is driven. For a changing torque, plot torque (vertical) against angular position (horizontal). The work done is the area between the graph and the angle axis.

• Split the area into rectangles (constant torque) and triangles (torque changing steadily).
• Area below the axis (negative torque while θ increases) counts as negative work.
• A rectangle of height τ and width Δθ gives back W = τΔθ, so the area rule includes the constant-torque case.

Worked example 1: pushing a roundabout

Question. A playground roundabout has rotational inertia 180 kg·m² about its axle and starts at rest. A parent pushes tangentially at the rim, 1.5 m from the axle, with a constant force of 60 N while it turns through a quarter turn (Figure 1). Friction at the axle exerts a constant opposing torque of 12 N·m. Take counterclockwise (the direction of the push) as positive. Find (a) the work done by the push, (b) the work done by friction, and (c) the angular velocity after the quarter turn.

1. Angular displacement: a quarter turn is Δθ = 2π ÷ 4 = π/2 = 1.57 rad.
2. Torque of the push: τ = rF = 1.5 m × 60 N = 90 N·m.
3. (a) W_push = τΔθ = 90 × 1.571 = +141 J. Check: arc length s = 1.5 × 1.571 = 2.36 m, and Fs = 60 × 2.36 = 141 J. ✓
4. (b) W_friction = −12 × 1.571 = −18.8 J. It is negative because the friction torque opposes the rotation.
5. Net work: 141.4 − 18.8 = 122.5 J. (Or τ_net Δθ = (90 − 12) × 1.571 = 122.5 J.)
6. (c) Starting from rest, W_net = ½Iω², so ω = √(2W_net / I) = √(2 × 122.5 ÷ 180) = √1.361 = 1.17 rad/s.

Check. The rim then moves at rω = 1.5 × 1.17 = 1.75 m/s, a sensible walking-pace speed for a gentle push. Without friction, ω would be √(2 × 141.4 ÷ 180) = 1.25 rad/s, a little larger, as expected.

Worked example 2: a changing torque from a graph

Question. A motor spins a wheel up from rest. The wheel’s rotational inertia is 0.15 kg·m², and friction is negligible. The motor’s torque varies with angular position as in Figure 2. Find the work done by the motor and the wheel’s final angular velocity.

1. Split the area under the graph.
   • 0 to 2.0 rad, triangle: ½ × 2.0 rad × 6.0 N·m = 6.0 J.
   • 2.0 to 5.0 rad, rectangle: 3.0 rad × 6.0 N·m = 18 J.
   • 5.0 to 7.0 rad, triangle: ½ × 2.0 rad × 6.0 N·m = 6.0 J.
2. Total work: 6.0 + 18 + 6.0 = 30 J.
3. All of it becomes rotational kinetic energy: ½Iω² = 30 J, so ω = √(2 × 30 ÷ 0.15) = √400 = 20 rad/s.

Interpretation. The wheel keeps speeding up all the way to 7.0 rad, even while the torque is falling, because the torque is still positive and still doing positive work. Speeding up only stops when the torque reaches zero. Using one torque value (6.0 N·m × 7.0 rad = 42 J) would overestimate the work, because the torque is smaller than 6.0 N·m for part of the turn.

Worked example 3: deriving a coasting angle

Question. A ceiling fan with rotational inertia I is spinning at ω₀ when it is switched off. A constant friction torque of size τ_f slows it to rest. (a) Derive an expression for the angle it turns through before stopping. (b) Evaluate it for I = 0.20 kg·m², ω₀ = 25 rad/s and τ_f = 0.050 N·m. (c) If the fan had been spinning twice as fast, how would the stopping angle change?

1. (a) Only friction does work, and it opposes the motion: W_net = −τ_f Δθ.
2. Work–energy: −τ_f Δθ = 0 − ½Iω₀².
3. So Δθ = Iω₀² / (2τ_f).
4. (b) Δθ = 0.20 × 25² ÷ (2 × 0.050) = 125 ÷ 0.10 = 1250 rad, which is 1250 ÷ 2π ≈ 199 revolutions. The initial kinetic energy was ½ × 0.20 × 625 = 62.5 J.
5. (c) Δθ depends on ω₀², so doubling ω₀ makes the stopping angle 4 times larger: 5000 rad, or about 796 revolutions.

Check the expression. Larger I or larger ω₀ means more stored energy, so more turning before it is used up. A larger friction torque removes energy faster per radian, so the fan stops sooner. The units are kg·m² × (rad/s)² ÷ N·m = J ÷ N·m = rad. ✓

Planning an experiment

To measure the work done by a torque in the lab, you need the torque at each angle and the angle turned:

• Wrap a string around an axle of known radius r and pull it with a force sensor, keeping the string tangent to the axle.
• The torque at each moment is τ = rF. The angle turned equals the length of string unwound divided by r (Δθ = d / r), or you can read it with a rotary motion sensor.
• Plot τ against θ and find the area. Compare it with the change in ½Iω², measuring ω at the end with a photogate or the rotary sensor.

Repeat runs and use a low-friction axle, or measure the friction torque separately by timing a free spin-down, so you can account for negative work done by friction.

Common misconceptions

• “A large torque always does a lot of work.” Without rotation, the work is zero.
• “Torque and work are the same because both are in N·m.” Torque is a turning effect; work is energy transferred, and a torque does work only through an angle.
• Using degrees or revolutions in W = τΔθ. Convert to radians first.
• “Friction torque does positive work because it is a torque.” It opposes the rotation, so its work is negative.
• Multiplying the largest torque by the whole angle when the torque changes. Use the area under the torque–angle graph (Worked example 2).
• “The wheel slows down when the torque starts to fall.” It keeps speeding up while the net torque is in the direction of rotation.
• Using the whole force instead of its perpendicular part. Only F⊥ = F sin θ, where θ is the angle between the radius and the force, contributes to the torque and the work.

Where this leads

Topic 6.3 looks at the same torque from a different angle: a torque acting for a time gives an angular impulse and changes angular momentum, just as a torque acting through an angle does work and changes kinetic energy. Topic 6.5 uses work and energy with rolling objects. Try the practice questions now, then use the revision notes and the checklist to consolidate. Next topic: Angular Momentum and Angular Impulse. You can also go back to Topic 6.1 or the course roadmap.