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AP Physics C Mechanics- 6.2 Torque and Work- Study Notes- New Syllabus

AP Physics C Mechanics- 6.2 Torque and Work – Study Notes

AP Physics C Mechanics- 6.2 Torque and Work – Study Notes – per latest Syllabus.

Key Concepts:

  • Energy Transfer by Torque
  • Work Done by a Variable Torque
  • Work from the Torque–Angle Graph

AP Physics C Mechanics-Concise Summary Notes- All Topics

Energy Transfer by Torque

A torque can transfer energy into or out of an object or rigid system if it is applied over an angular displacement. This is the rotational equivalent of a force doing work when applied over a linear displacement.

When torque acts through an angular displacement, it can increase or decrease the system’s rotational kinetic energy.

Key Idea: – If torque and angular displacement are in the same direction, the system gains energy (positive work). – If they are in opposite directions, the system loses energy (negative work).

\( \mathrm{W = \tau \,\Delta \theta} \quad (\text{for constant torque}) \)

Example

A constant torque of \( \mathrm{4.0\,N·m} \) acts on a wheel causing it to rotate through an angle of \( \mathrm{2.0\,rad} \). Find the work done by the torque on the wheel.

▶️ Answer / Explanation

Step 1: Use the formula for work done by constant torque:

\( \mathrm{W = \tau \,\Delta \theta} \)

Step 2: Substitute values:

\( \mathrm{W = (4.0)(2.0) = 8.0\,J} \)

Result: The torque does \( \mathrm{8.0\,J} \) of positive work on the wheel, increasing its rotational energy.

Work Done by a Variable Torque

If the torque acting on a rigid system is not constant, the total work done is found by integrating torque over the angular displacement through which it acts.

\( \mathrm{W = \displaystyle \int_{\theta_1}^{\theta_2} \tau\,d\theta} \)

  • \( \mathrm{W} \): work done by torque (in joules, J)
  • \( \mathrm{\tau} \): instantaneous torque (in \( \mathrm{N·m} \))
  • \( \mathrm{d\theta} \): infinitesimal angular displacement (in radians)

Key Idea: This relationship is the rotational analog of \( \mathrm{W = \int F\,dx} \) for linear motion. It connects the changing torque to the total energy transferred to or from a rotating system.

Example

A torque varies with angular displacement as \( \mathrm{\tau = 3\theta} \), where \( \mathrm{\tau} \) is in \( \mathrm{N·m} \) and \( \mathrm{\theta} \) is in radians. Find the work done by this torque as the object rotates from \( \mathrm{0} \) to \( \mathrm{2.0\,rad} \).

▶️ Answer / Explanation

Step 1: Use the general work–torque relation:

\( \mathrm{W = \int_{\theta_1}^{\theta_2} \tau\,d\theta} \)

Step 2: Substitute \( \mathrm{\tau = 3\theta} \):

\( \mathrm{W = \int_0^2 3\theta\,d\theta = 3\int_0^2 \theta\,d\theta} \)

Step 3: Integrate:

\( \mathrm{W = 3\left[\tfrac{1}{2}\theta^2\right]_0^2 = 3(2) = 6\,J} \)

Result: The torque performs \( \mathrm{6\,J} \) of work on the system as it rotates through \( \mathrm{2.0\,rad} \).

 Work from the Torque–Angle Graph

The work done on a rigid system by a given torque can also be determined from a graph of torque as a function of angular position.

Key Principle: The area under the \( \mathrm{\tau} \) vs. \( \mathrm{\theta} \) curve represents the total work done by the torque:

\( \mathrm{W = \text{Area under the } \tau\text{–}\theta \text{ curve}} \)

  • Positive area (above the θ-axis) → positive work (energy gained)
  • Negative area (below the θ-axis) → negative work (energy lost)

Example

The torque on a rotating disk increases linearly from \( \mathrm{0\,N·m} \) at \( \mathrm{\theta = 0} \) to \( \mathrm{4.0\,N·m} \) at \( \mathrm{\theta = 2.0\,rad} \), then remains constant until \( \mathrm{\theta = 4.0\,rad} \). Determine the total work done by the torque.

▶️ Answer / Explanation

Step 1: The work equals the area under the \( \mathrm{\tau\text{–}\theta} \) curve.

The graph consists of:

  • A triangle from \( \mathrm{0} \) to \( \mathrm{2\,rad} \)
  • A rectangle from \( \mathrm{2} \) to \( \mathrm{4\,rad} \)

Step 2: Find each area:

Triangle: \( \mathrm{A_1 = \tfrac{1}{2}(2)(4.0) = 4.0\,J} \)

Rectangle: \( \mathrm{A_2 = (2)(4.0) = 8.0\,J} \)

Step 3: Total work:

\( \mathrm{W = A_1 + A_2 = 4.0 + 8.0 = 12.0\,J} \)

Result: The total work done by the torque is \( \mathrm{12.0\,J} \).

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