6.2 Torque and Work

Story: six twisting pushes

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. Ben opens a heavy door

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A playground merry-go-round sits still. You grab a handle at the edge and run beside it, pushing as hard as you can.

After half a turn you let go. It is now spinning. Your push gave it kinetic energy.

If you had pushed closer to the center, with the same force, it would spin more slowly. If you had pushed for a whole turn, it would spin faster.

How much energy does a twisting push give a spinning object, and how fast will it end up turning?

2. Priya spins the merry-go-round

Quick check: Priya pushes the merry-go-round with the same torque, but keeps pushing for a whole turn instead of half a turn. How much work does she do?

3. Sam pushes the wrong way

4. Lucia stops the merry-go-round

Quick check: a wheel spins with 40 J of kinetic energy. A brake twists against the spin and does −40 J of work on it. What happens?

5. Theo winds up the well bucket

6. Jamal pedals slow, then fast

Quick check: Jamal keeps the same torque on his pedals but turns them three times as fast. What happens to his power?

7. Check yourself

Think of your answer first, then tap to see it.

a) Where should you push on a heavy door to open it most easily, and in which direction?

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As far from the hinges as you can (by the handle), and at right angles to the door. That gives the biggest torque τ = rF sin θ.

b) A brake pad rubs on a spinning wheel's rim, against the motion. Is the work done by the brake positive or negative? What happens to the wheel's kinetic energy?

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Negative. The brake's torque points against the spin, so W = τΔθ is negative and the wheel's kinetic energy goes down.

c) A constant torque of 15 N·m turns a wheel through 2.0 rad. How much work does it do? If the wheel started at rest with no friction, what is its kinetic energy now?

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W = τΔθ = 15 × 2.0 = 30 J. All of it becomes rotational kinetic energy: K = 30 J.

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Three quick questions. Get all three right on the first try and you can skip ahead.

1. Which push gives a door the biggest torque about its hinges?

2. A steady torque of 12 N·m turns a wheel, starting from rest, through 3.0 rad. There is no friction. What is the wheel's kinetic energy now?

3. Ella twists a wheel with 20 N·m for half a turn. Noah twists it with 10 N·m for a whole turn. Both act in the direction of the spin. Which does more work?

The idea

Picture yourself pushing a merry-go-round. You walk beside it with your hand on the rail. You keep pushing. It turns a little, then more, then halfway round.

The harder you twist, the more energy you give it. The farther it turns while you push, the more energy too. A twist that turns nothing gives no energy at all. Think of pushing on a locked door.

axis start Δθ r push F push F at right angles to the radius r torque τ = r F angle turned Δθ work W = τ Δθ

That energy is the work done by your twist. In Unit 3, work was push times distance: W = Fd. For spinning things, the push is a torque τ. The distance is the angle turned Δθ. In words: work equals torque times the angle turned:

W = τ Δθ

Trap: using the whole force F in τ = rF when the push is at an angle. Instead: use only the perpendicular part: τ = rF sinθ, where θ is the angle between the radius and the force. A push along the radius gives zero torque.
Trap: putting the number of revolutions (or degrees) into W = τΔθ. Instead: change to radians first: 1 revolution = 2π rad, so 3 turns = 6π ≈ 18.8 rad.

The work-energy theorem for spinning objects says the net work done by torques equals the change in rotational kinetic energy:

τnet Δθ = ΔKrot = ½ I ωf² − ½ I ωi²

Trap: calling a torque of 50 N·m "50 joules", because N·m looks like J. Instead: torque is N·m and is not energy. Only torque × angle (in rad) is work in joules: 50 N·m × 2.0 rad = 100 J.

When the torque changes as the object turns, the work is the area under the torque against angle graph (τ on the vertical axis, θ on the horizontal axis).

Power is how fast the work is done. For a steady torque on something turning at angular speed ω:

P = τ ω

area = work W θ (rad) τ (N·m)

Worked: the merry-go-round

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The merry-go-round has rotational inertia I = 500 kg·m² and starts at rest. You push with F = 100 N, always perpendicular to the radius, at r = 2.0 m from the axis, for half a turn. Ignore friction. How much work do you do, and how fast is it spinning when you let go?

  1. Torque: τ = rF = 2.0 m × 100 N = 200 N·m. the force is perpendicular to the radius, so all of it twists.
  2. Angle: half a turn is Δθ = π rad ≈ 3.14 rad. one turn is 2π rad, and W = τΔθ needs radians.
  3. Work: W = τΔθ = 200 N·m × 3.14 rad = 628 J. a steady torque through an angle does work τΔθ (like Fd).
  4. Energy: it started at rest, so Krot = 0 + 628 J = 628 J. work-energy theorem. With no friction all your work becomes rotational KE.
  5. Speed: ½Iω² = 628 J, so ω = √(2 × 628 / 500) = √2.51 = 1.59 rad/s. solve K = ½Iω² for ω. That is about one turn every 4 seconds.

Lab: twist the disk

A handle on the disk is pushed with force F, always perpendicular to the radius, at distance r from the axis. An optional friction torque at the axle fights the spin. The bars show the work you do, how much became kinetic energy (ΔK), and how much became thermal energy in the axle. Check that Wyou = ΔK + thermal at every moment.

Try this: double F, then instead double r. Both double the torque and give the same result. Then give it a start speed ω₀ and add friction equal to your torque (F·r): the disk keeps a steady speed and all your work turns into heat.

Examples

Example 1: tightening a bolt basic

You push a wrench with 40 N, perpendicular to the handle, 0.25 m from the bolt. The bolt turns a quarter turn. How much work do you do?

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Read the steps as text
  1. τ = rF = 0.25 m × 40 N = 10 N·m. Force is perpendicular to the wrench.
  2. Δθ = ¼ × 2π = π/2 = 1.57 rad. Convert turns to radians.
  3. W = τΔθ = 10 × 1.57 = 15.7 J ≈ 16 J. Steady torque through an angle.

Example 2: winding up a well bucket medium

A well crank has a handle 0.30 m from the axle. You push the handle with 50 N, perpendicular to the crank, for 10 full turns. The rope wraps on a drum of radius 0.080 m and lifts a 12 kg bucket of water at a steady speed. (a) How much work do you do? (b) How much goes into the bucket's gravitational energy? (c) Where does the rest go?

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Read the steps as text
  1. τ = 0.30 × 50 = 15 N·m; Δθ = 10 × 2π = 62.8 rad. Torque from your push; turns to radians.
  2. (a) W = 15 × 62.8 = 942 J. W = τΔθ.
  3. Rope wound in: 10 turns × 2π × 0.080 m = 5.03 m, so the bucket rises 5.03 m. Each turn wraps one drum circumference.
  4. (b) ΔU = mgh = 12 × 9.8 × 5.03 = 591 J. Gravity energy of the bucket-Earth system.
  5. (c) 942 − 591 = 351 J goes to thermal energy (friction in the axle and rope). Steady speed means no change in K, so the rest of the work must be lost to friction.

Example 3: a grinding wheel AP

A grinding wheel (I = 0.020 kg·m²) starts at rest. A motor gives it a steady torque of 0.50 N·m while it turns through 40 rad. (a) Find its final angular speed. (b) Later, at a working speed of 300 rad/s, the motor supplies 15 N·m to keep it turning against the tool. What power does the motor deliver?

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Read the steps as text
  1. W = τΔθ = 0.50 × 40 = 20 J. Work by the motor torque.
  2. (a) ½Iω² = 20 J, ω = √(2 × 20 / 0.020) = √2000 = 44.7 rad/s. From rest, so ΔK = K.
  3. (b) P = τω = 15 × 300 = 4500 W = 4.5 kW. Power of a torque at angular speed ω.

Example 4: reading a torque-angle graph AP

A disk (I = 0.25 kg·m²) spins at 4.0 rad/s. A torque in the direction of spin then grows steadily from 0 to 8.0 N·m while the disk turns 2.0 rad, and stays at 8.0 N·m for the next 3.0 rad. Find the final angular speed.

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Read the steps as text
  1. Area of the triangle (0 to 2 rad): ½ × 2.0 × 8.0 = 8.0 J. Work is the area under τ-θ.
  2. Area of the rectangle (2 to 5 rad): 3.0 × 8.0 = 24 J. Total W = 32 J. Add the pieces.
  3. Ki = ½ × 0.25 × 4.0² = 2.0 J, so Kf = 2.0 + 32 = 34 J. Work-energy theorem.
  4. ωf = √(2 × 34 / 0.25) = √272 = 16.5 rad/s. Solve ½Iω² = 34 J.

Practice (AP style)

1. A steady torque of 6.0 N·m turns a wheel through 3.0 revolutions. How much work does it do?

Show answer

C. 3.0 rev = 3.0 × 2π = 18.8 rad. W = 6.0 × 18.8 = 113 J. Choice A forgets to convert revolutions to radians.

2. A student pulls on the rim of a wheel with a force that points straight toward the axle. The wheel turns because of a motor. How much work does the student's force do on the wheel's rotation?

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A. A force along the radius has no perpendicular part, so τ = 0 and W = τΔθ = 0, however hard she pulls.

3. A spinning wheel has 50 J of rotational kinetic energy. A steady friction torque of 2.0 N·m acts at its axle. Through what angle does it turn before stopping?

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C. Friction does −50 J of work: −2.0 × Δθ = −50 J, so Δθ = 25 rad (about 4 turns).

4. A motor exerts a torque of 20 N·m on a shaft spinning at a steady 50 rad/s. What power does it deliver?

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C. P = τω = 20 × 50 = 1000 W.

5. Disk A has rotational inertia I, disk B has 2I. Both start at rest. The same torque acts on each through the same angle. Which is true afterward?

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B. Same τ and Δθ means the same work, so the same K. K = ½Iω², so the disk with smaller I spins faster: ωA = √2 ωB.

6. A graph shows the torque on a wheel (vertical axis) against the angle it has turned (horizontal axis). What does the area under the graph represent?

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C. Area = τ × Δθ = work. (Torque against time would give angular impulse, topic 6.3.)

7. (Short free response, qualitative to quantitative) A student says: "If I push the merry-go-round from rest with twice the force through the same angle, it will end up spinning twice as fast." (a) Is she right? Explain in words. (b) Check with numbers: I = 500 kg·m², r = 2.0 m, Δθ = π rad, F = 100 N and then 200 N.

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(a) No. Twice the force gives twice the torque, so twice the work and twice the kinetic energy. But K = ½Iω², so ω grows only by √2 ≈ 1.41, not 2.

(b) F = 100 N: W = 200 × π = 628 J, ω = √(2 × 628 / 500) = 1.59 rad/s. F = 200 N: W = 1257 J, ω = √(2 × 1257 / 500) = 2.24 rad/s. Ratio 2.24 / 1.59 = 1.41 = √2.

8. (Short free response, experimental design) You want to find the rotational inertia of a bicycle wheel on a low-friction axle. You have a spring scale, string, a meter stick and a sensor that reads angle and angular speed. Describe a procedure and how you would use a graph to find I.

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Wrap string around the hub (radius r, measured with the meter stick). Pull the string with the spring scale so it reads a steady force F, starting from rest. The torque is τ = rF. Use the sensor to record ω after several different angles Δθ.

From the work-energy theorem, τΔθ = ½Iω², so ω² = (2τ/I) Δθ. Plot ω² (vertical) against Δθ (horizontal): it should be a straight line through the origin with slope 2τ/I. Then I = 2τ / slope. Repeat each run a few times and average to reduce error; keep the pull perpendicular to the radius (string leaves the hub tangentially).

Free response (Mathematical routines). A grinding wheel with rotational inertia I = 0.080 kg·m² starts from rest. A motor applies a steady torque τ = 2.0 N·m while the wheel turns 5.0 revolutions. Ignore friction.

(a) Derive an expression for the final angular speed in terms of τ, the number of turns N and I.

(b) Calculate the final angular speed.

(c) Calculate the power the motor delivers at the end.

(d) If the torque were doubled for the same 5.0 turns, by what factor would the final ω change?

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(a) Δθ = 2πN. Work-energy: τ(2πN) = ½Iω², so ω = √(4πNτ / I).

(b) Δθ = 2π × 5.0 = 31.4 rad, W = 2.0 × 31.4 = 62.8 J, ω = √(2 × 62.8 / 0.080) = √1571 ≈ 39.6 rad/s.

(c) P = τω = 2.0 × 39.6 ≈ 79 W.

(d) ω ∝ √τ, so ω grows by √2 ≈ 1.41 (not 2): twice the work gives twice the energy, and K grows as ω².

Point guide (5 points): 1 for converting turns to radians; 1 for τΔθ = ½Iω²; 1 for ω ≈ 40 rad/s; 1 for P = τω ≈ 79 W; 1 for the factor √2 with reason.

Free response (Translation between representations). A wheel (I = 0.50 kg·m²) starts from rest. A graph of the torque on it against the angle turned looks like this: the line rises in a straight line from 0 at θ = 0 to 6.0 N·m at θ = 2.0 rad, then stays flat at 6.0 N·m until θ = 5.0 rad.

(a) Use the graph to find the total work done on the wheel.

(b) Find the wheel's angular speed at θ = 5.0 rad.

(c) Which describes the graph of the wheel's rotational kinetic energy against θ? (A) a straight line all the way; (B) a curve that gets steeper from 0 to 2.0 rad, then a straight rising line; (C) a rising line, then flat; (D) a curve that gets less steep all the way. Explain.

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(a) Work = area under τ against θ: triangle ½ × 2.0 × 6.0 = 6.0 J plus rectangle 3.0 × 6.0 = 18 J, total 24 J.

(b) ½Iω² = 24 J, so ω = √(2 × 24 / 0.50) = √96 ≈ 9.8 rad/s.

(c) B. The slope of K against θ equals the torque. From 0 to 2.0 rad the torque grows, so the K graph gets steeper (a curve, K reaches 6.0 J). After that the torque is constant, so K rises in a straight line from 6.0 J to 24 J.

Point guide (4 points): 1 for using the area; 1 for 24 J; 1 for ω ≈ 9.8 rad/s; 1 for choice B with "slope of K-θ = torque".

Common mistakes

Using the whole force when it is at an angle

The mistake: using τ = rF when the force is not perpendicular to the radius.

Why it is wrong: only the perpendicular part twists: τ = rF sin θ, where θ is the angle between r and F.

How to spot it: draw the radius line to the point where the force acts. If the force is not at 90° to it, take the perpendicular part.

Angle in turns or degrees

The mistake: W = 6 N·m × 3 rev = 18 J.

Why it is wrong: W = τΔθ only gives joules when Δθ is in radians. 3 rev = 6π rad = 18.8 rad.

How to spot it: if the angle came from "turns", "rev" or degrees, convert before you multiply.

Thinking work changes ω in proportion

The mistake: "Twice the work, twice the spin rate."

Why it is wrong: work changes kinetic energy, and K goes as ω². Twice the energy from rest gives √2 times ω.

How to spot it: always go through K: W → ΔK → ω = √(2K/I).

Forgetting friction's (negative) work

The mistake: setting your work equal to the gain in K when there is friction at the axle.

Why it is wrong: the friction torque does negative work, which turns into thermal energy. Use the net torque: (τyou − τfriction)Δθ = ΔK.

How to spot it: if the problem says "steady speed" while you push, ΔK = 0, so friction took all your work.