Unit 6 · Topic 6.2 Beta

Torque and Work

A torque that turns an object through an angle does work W = τΔθ (angle in radians), or the area under a torque-angle graph when the torque varies.

Practice 1: Creating RepresentationsPractice 2: Mathematical Routines

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Question set for this topic

Part 1 · Hook

Why this matters

Wind up a bucket from a well and your arm makes big slow circles while the bucket creeps up. Turning a crank is doing work, just as lifting the bucket directly would be. In rotation, work is torque times angle, and it lands exactly where energy bookkeeping says it should.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A 20 N force pushes a box 3.0 m in the direction of the force. The work done is

  1. 60 J
  2. 6.7 J
  3. 23 J
  4. 0.15 J
Show the answer

W = Fd = (20 N)(3.0 m).

  • Correct: 60 J:
  • 6.7 J:
  • 23 J:
  • 0.15 J:

2. On a force-position graph, the area under the line is

  1. the work done
  2. the power
  3. the impulse
  4. the acceleration
Show the answer

Area = force × distance = work.

  • Correct: the work done:
  • the power:
  • the impulse:
  • the acceleration:

3. A wheel with I = 0.50 kg·m² turns at 4.0 rad/s. Its rotational kinetic energy is

  1. 4.0 J
  2. 1.0 J
  3. 8.0 J
  4. 2.0 J
Show the answer

K = ½Iω² = ½(0.50)(16) = 4.0 J.

  • Correct: 4.0 J:
  • 1.0 J:
  • 8.0 J:
  • 2.0 J:

Part 4 · See it

See it first

Left: a hand pushes up on the handle of a crank wheel, turning it counterclockwise through an angle Δθ; for a constant torque, the work is W = τΔθ, with the angle in radians and work in joules. Right: a graph of torque in N·m against angle in radians. The torque is 4.0 N·m from 0 to 10 rad, then falls in a straight line to zero at 20 rad and stays zero. The shaded area under the line is the work: a rectangle of 4.0 × 10 = 40 J plus a triangle of ½ × 4.0 × 10 = 20 J, 60 J in all.
A crank turned by a torque: the work is τΔθ, the area under the torque-angle graph. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A force at the rim pushes the wheel through a distance s along the rimit does work Fs on the wheel
  2. s = rθ and the torque is rFthe work is Fs = (rF)θ = τΔθ
  3. The torque changes with anglethe work is the area under the torque-angle graph
  4. The net torque does positive work on a wheelits rotational kinetic energy increases by that amount
  5. A torque opposes the turningit does negative work and takes kinetic energy out

Part 6 · Key ideas

Key ideas

  • W = τΔθ for a constant torque, with Δθ in radians: the partner of W = Fd.
  • For a changing torque, W is the area under the τ-θ graph.
  • Positive when the torque turns the object the way it is going; negative when it opposes.
  • The net work done on a turning object equals its change in rotational kinetic energy.
  • Convert revolutions to radians first: 1 rev = 2π rad.

Part 7 · Misconception

A common mistake

The wrong idea: Work done by a torque is τ times the number of revolutions (or degrees).

What actually happens: W = τΔθ only works with the angle in radians, because it comes from W = Fs with s = rθ. Four revolutions is 8π rad, about 25 rad.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

Winding up a bucket

bucketcrank handleDrum radius: 0.0800 mBucket mass: 6.00 kgThe drum turns 25.0 rad,raising the bucket atconstant speed.Ignore friction and themass of the rope. Use g = 9.8 m/s².

1. What torque does the rope exert on the drum while the bucket rises at constant speed?

Type a number in N·m.

Show the answer

At constant speed the rope’s tension equals the bucket’s weight, (6.00 kg)(9.8 m/s²) = 58.8 N. The rope pulls tangent to the drum, so τ = rF = (0.0800 m)(58.8 N) = 4.70 N·m.

  • Answer: 4.70 N·m

2. How much work does the crank do on the drum while it turns through 25.0 rad? Give the value and its unit.

Type a number and its unit.

Show the answer

W = τΔθ = (4.704 N·m)(25.0 rad) = 117.6 J, about 118 J. (No friction and constant speed, so the crank’s torque equals the rope’s.)

  • Answer: 118 J

3. How far does the bucket rise?

Type a number in m.

Show the answer

The rope winds onto the drum without slipping: s = rθ = (0.0800 m)(25.0 rad) = 2.00 m.

  • Answer: 2.00 m

4. How does the work done by the torque on the drum compare with the increase in the bucket’s gravitational potential energy?

  1. They are equal: nothing speeds up and there is no friction
  2. The work is larger, because some goes into spinning the drum
  3. The work is smaller, because the drum’s radius is small
  4. The work is a rotational quantity, so it is not an energy
Show the answer

For the drum-rope-bucket system, the crank’s work all goes into gravitational potential energy: τΔθ = (r · mg)(θ) = mg(rθ) = mgΔy = 117.6 J. Nothing speeds up and there is no friction.

  • Correct: They are equal: nothing speeds up and there is no friction: Right: τΔθ = (rmg)θ = mg(rθ) = mgΔy.
  • The work is larger, because some goes into spinning the drum: The drum turns at a steady rate, so its kinetic energy does not change.
  • The work is smaller, because the drum’s radius is small: A small radius means a small torque but a large angle; the product is the same.
  • The work is a rotational quantity, so it is not an energy: Both are energies in joules; τΔθ = mgΔy here.

Graph

A motor spinning up a wheel

A motor exerts a torque on a wheel (I = 0.50 kg·m²) on a frictionless axle. The graph shows the torque against the angle the wheel has turned since it started from rest.

024680246810121416Angle turned (rad)Torque (N·m)
Data table
Angle turned (rad)τ
06
86
140
160

5. How much work does the motor do on the wheel from 0 to 8.0 rad?

Type a number in J.

Show the answer

Constant torque: W = τΔθ = (6.0 N·m)(8.0 rad) = 48 J, the rectangle under the graph.

  • Answer: 48 J

6. How much work does the motor do on the wheel from 0 to 14.0 rad?

Type a number in J.

Show the answer

Area under the graph: (6.0 N·m)(8.0 rad) + ½(6.0 N·m)(6.0 rad) = 48 + 18 = 66 J.

  • Answer: 66 J

7. The wheel (I = 0.50 kg·m²) starts from rest on a frictionless axle. How fast is it turning at 14.0 rad?

Type a number in rad/s.

Show the answer

The net work equals the change in rotational kinetic energy: 66 J = ½(0.50 kg·m²)ω², so ω = √(2 × 66/0.50) = √264 = 16.2 rad/s, about 16 rad/s.

  • Answer: 16 rad/s

Part 9 · Summary

Summary

A torque that turns an object through an angle does work W = τΔθ (angle in radians), or the area under a torque-angle graph when the torque varies. The net work done on a turning object changes its rotational kinetic energy; a torque that opposes the turning does negative work.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections