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โš™๏ธAP Physics C: Mechanics Unit 6 Review

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6.2 Torque and Work

โš™๏ธAP Physics C: Mechanics
Unit 6 Review

6.2 Torque and Work

Written by the Fiveable Content Team โ€ข Last updated September 2025
Verified for the 2026 exam
Verified for the 2026 examโ€ขWritten by the Fiveable Content Team โ€ข Last updated September 2025
โš™๏ธAP Physics C: Mechanics
Unit & Topic Study Guides
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Energy Transfer by Torque

Torque, the rotational equivalent of force, can transfer energy into or out of a rigid system when applied over an angular displacement. This energy transfer is fundamental to understanding how rotating objects gain or lose energy.

  • When torque acts in the same direction as rotation, positive work is done (energy is added to the system)
  • When torque opposes the direction of rotation, negative work is done (energy is removed from the system)
  • The magnitude of energy transfer depends on both the torque applied and how far the object rotates

For example, when you apply torque to a bicycle pedal, you transfer energy into the bicycle's drivetrain, causing it to accelerate. Conversely, when brakes apply a torque that opposes wheel rotation, they remove energy from the system, causing deceleration.

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Work-Torque Relationship

The mathematical relationship between work and torque allows us to quantify the energy transfer in rotating systems. This relationship is analogous to the work-force relationship in linear systems.

To calculate the work done by a torque, we integrate the torque with respect to angular displacement:

W=โˆซฮธ1ฮธ2ฯ„dฮธW=\int_{\theta_{1}}^{\theta_{2}} \tau d \theta

Where:

  • WW is the work done by the torque (measured in joules)
  • ฯ„\tau is the torque as a function of angular position (measured in Nยทm)
  • ฮธ1\theta_{1} and ฮธ2\theta_{2} are the initial and final angular positions (measured in radians)

For the special case of constant torque, the calculation simplifies to:

W=ฯ„ฮ”ฮธW = \tau \Delta \theta

This equation tells us that the work equals the product of the torque and the angular displacement through which it acts.

Graphical Work Analysis

Visualizing the relationship between torque and angular position provides valuable insights into the work done in rotational systems.

When we plot torque versus angular position:

  • The area under the curve represents the work done by the torque
  • Areas above the x-axis indicate positive work (energy added to the system)
  • Areas below the x-axis indicate negative work (energy removed from the system)

This graphical approach is particularly useful for analyzing situations with variable torque, where the torque changes as the object rotates. By finding the area under the curve, we can determine the total work without needing to solve complex integrals analytically.

For example, in a car engine, the torque varies throughout the rotation cycle. By analyzing the torque-angle graph, engineers can calculate the total work output per cycle and optimize engine performance.

Practice Problem 1: Work Done by Constant Torque

A student applies a constant torque of 15 Nยทm to a wheel, causing it to rotate through an angle of 2.5 radians. How much work does the student do on the wheel?

Solution

Since the torque is constant, we can use the simplified equation for work:

W=ฯ„ฮ”ฮธW = \tau \Delta \theta

Substituting the given values: W=15ย N\cdotpmร—2.5ย radW = 15 \text{ Nยทm} \times 2.5 \text{ rad} W=37.5ย JW = 37.5 \text{ J}

Therefore, the student does 37.5 joules of work on the wheel.

Practice Problem 2: Work from Torque-Angle Graph

A variable torque is applied to a rotating shaft according to the function ฯ„=20โˆ’5ฮธ\tau = 20 - 5\theta Nยทm, where ฮธ\theta is in radians. Calculate the work done by this torque as the shaft rotates from ฮธ=0\theta = 0 to ฮธ=3\theta = 3 radians.

Solution

To find the work, we need to integrate the torque function with respect to angular displacement:

W=โˆซ03(20โˆ’5ฮธ)dฮธW = \int_{0}^{3} (20 - 5\theta) d\theta

W=[20ฮธโˆ’5ฮธ22]03W = [20\theta - \frac{5\theta^2}{2}]_{0}^{3}

W=(20ร—3โˆ’5ร—322)โˆ’(20ร—0โˆ’5ร—022)W = (20 \times 3 - \frac{5 \times 3^2}{2}) - (20 \times 0 - \frac{5 \times 0^2}{2})

W=60โˆ’22.5โˆ’0W = 60 - 22.5 - 0

W=37.5ย JW = 37.5 \text{ J}

Therefore, the work done by the variable torque is 37.5 joules.

Vocabulary

The following words are mentioned explicitly in the College Board Course and Exam Description for this topic.

TermDefinition
angular displacementThe change in angular position of a rotating object, measured in radians.
energy transferThe process by which energy moves into or out of a system through the action of forces or torques.
rigid systemA collection of objects or particles that maintain fixed distances from each other and rotate as a single unit.
torqueA measure of the rotational effect of a force on a rigid body, calculated as the product of the force component perpendicular to the position vector and the distance from the axis of rotation.
workEnergy transferred to or from a system by forces or torques acting on it.