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AP®PhysicsCollege BoardPhysics 1: Algebra-BasedStudy GuidesUnit 6: Energy & Momentum of Rotating SystemsRotational Kinetic Energy, Torque & WorkTorque & Work

Torque & Work (College Board AP® Physics 1: Algebra-Based)

Study Guide

Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

Work done by torque

  • When a torque is exerted over an angular displacement, energy is transferred into or out of an object or rigid system

    • Some examples of this include rotating cranes and fairground rides

  • The work done on a rigid system by a torque is given by:

W space equals space tau increment theta

  • Where:

    • W = work done, in straight J

    • tau = torque, in straight N times straight m

    • increment theta = angular displacement, in rad

  • Equivalent amounts of work can be done on a rigid system by

    • exerting a smaller torque over a greater angle of rotation

    • exerting a larger torque over a smaller angle of rotation

  • This is analogous to systems with linear acceleration, where work done is the product of the force and the linear displacement

Comparing work done on translational and rotational systems

Comparing work done by a force and by a torque. Top: work is done by a force F moving an object over a displacement Δx increasing linear velocity v. Bottom: work is done by a torque τ rotating a wrench over an angular displacement Δθ, increasing angular velocity ω.
Work done by a torque is analogous to the work done by a force, but instead leads to an increase in angular velocity

Worked Example

A horizontally mounted wheel with rotational inertia I spins about a frictionless axle. At time t = 0, the angular speed of the wheel is omega. A constant torque tau is applied to the wheel which causes it to come to rest in time t.

Derive an expression for the average power required to dissipate the wheel's energy in terms of I, omega and t.

Answer:

Step 1: Analyze the scenario

  • Initially, the wheel spins with an initial angular speed of omega subscript 0 space equals space omega

  • After time increment t space equals space t, the wheel has a final angular speed of omega space equals space 0

  • The torque tau applied to the wheel is constant, therefore the angular acceleration alpha is also constant

  • Hence, we can use the rotational form of Newton's second law as well as the equations of constant angular acceleration in our derivation

A rotating wheel with angular velocity ω is slowed by a torque τ = Iα, shown by an arrow pointing to the wheel's rim and a black curved arrow showing rotation direction.

Step 2: Write an expression for the average power supplied to the wheel

  • The work done on the wheel is:

W space equals space tau increment theta

  • The torque applied to the wheel is:

tau space equals space I alpha

  • Therefore, the average power supplied to the wheel is:

P subscript a v g end subscript space equals space fraction numerator W over denominator increment t end fraction space equals space fraction numerator tau increment theta over denominator increment t end fraction space equals space fraction numerator I alpha increment theta over denominator increment t end fraction

Step 3: Write an expression for the angular acceleration of the wheel

  • The rotational kinematic equation relating omega, omega subscript 0, alpha and t is:

omega space equals space omega subscript 0 space plus space alpha t

  • Where omega subscript 0 space equals space omega and omega space equals space 0:

0 space equals space omega space plus space alpha t space space space space space rightwards double arrow space space space space space alpha space equals space minus omega over t

Step 4: Write an expression for the angular displacement of the wheel

  • The rotational kinematic equation relating omega, omega subscript 0, theta and t is:

increment theta space equals space 1 half open parentheses omega subscript 0 space plus space omega close parentheses t

  • Where omega subscript 0 space equals space omega and omega space equals space 0:

increment theta space equals space 1 half open parentheses omega space plus space 0 close parentheses t space space space space space rightwards double arrow space space space space space increment theta space equals space 1 half omega t

Step 5: Combine the equations and simplify

  • Substituting these equations into the expression for power:

P subscript a v g end subscript space equals space fraction numerator I alpha increment theta over denominator increment t end fraction

P subscript a v g end subscript space equals space fraction numerator I open parentheses negative omega over t close parentheses open parentheses 1 half omega t close parentheses over denominator t end fraction

P subscript a v g end subscript space equals space minus fraction numerator I omega squared over denominator 2 t end fraction

Torque-angular position graph

  • A changing torque can be plotted as a function of angular displacement

  • The work done on a rigid system is equal to the product of torque and angular displacement

  • Therefore, the area under the torque-angular position graph is equal to

    • work done

    • change in kinetic energy

area space equals space tau increment theta space equals space W space equals space increment K

Determining work done from a torque-angular displacement graph

The relationship between torque (Nm) and angular displacement (radians), illustrating that the area under the curve represents the work done.
The work done is the area under the torque-angular displacement graph

  • The graph may show positive or negative values of torque

    • Positive areas represent positive work i.e. energy added to the system

    • Negative areas represent negative work i.e. energy removed from the system

  • This is analogous to linear work done being equal to the area under a force-displacement graph

Worked Example

A cylinder of mass 5.0 kg and radius 0.20 m is mounted on an axle parallel to its axis and through its center of mass. The rotational inertia of the cylinder about this axis is 1 half M R squared. The cylinder is initially at rest before a force is applied to a point on its edge which causes it to rotate. The graph shows the net torque exerted on the cylinder as a function of angular position.

Plot of torque (τ) in N·m versus angular position (Δθ) in radians. The graph features stepped changes at π/2, π, and 3π/2 with values of 10, 0, and -5 N·m.

Determine the angular velocity of the cylinder after it completes one full revolution.

Answer:

Step 1: Analyze the scenario

  • One full revolution is equivalent to 2π rad

  • A constant positive torque of 10 N·m is applied until the cylinder rotates through π rad

  • At theta = π rad, the direction of the applied torque changes

  • A constant negative torque of −5 N·m is then applied as the cylinder rotates through another π rad

  • Overall, the work done on the cylinder:

    • is positive from theta = 0 to theta = π, so the angular velocity will increase

    • is negative from theta = π to theta = 2π, so the angular velocity will decrease

Step 2: Determine the work done over the 2π rotation

  • The area under a torque-angular position graph is equal to the work done, or the change in kinetic energy

Graph showing torque (τ in N·m) vs angular displacement (Δθ in radians). Green areas represent positive work (10π) from 0 to π rad and negative work (−5π) from π to 2π radians.

  • Area under the positive section = 10 cross times straight pi space equals space 10 straight pi space straight J

  • Area under the negative section = negative 5 cross times straight pi space equals space minus 5 straight pi space straight J

  • Therefore, the work done, or net change in kinetic energy is:

W space equals space increment K space equals space 10 space minus space 5 space equals space 5 straight pi space straight J

Step 3: Determine the final angular velocity of the cylinder

  • The change in kinetic energy is equal to

increment K space equals space 1 half I omega squared space equals space 1 half open parentheses 1 half M R squared close parentheses omega squared space equals space 1 fourth M R squared omega squared

  • Where M = 5.0 kg and R = 0.20 m (and omega subscript 0 = 0)

  • Therefore, after a full revolution open parentheses increment theta space equals space 2 straight pi close parentheses, the cylinder's angular velocity is:

omega space equals space square root of fraction numerator 4 increment K over denominator M R squared end fraction end root

omega space equals space square root of fraction numerator 4 cross times 5 straight pi over denominator 5.0 cross times 0.20 squared end fraction end root

omega space equals space 17.8 space rad divided by straight s

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    Author
    Reviewer
    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

    Caroline Carroll

    Author: Caroline Carroll

    Expertise: Physics Subject Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.