AP Physics 1- 6.2 Torque and Work- Study Notes- New Syllabus
AP Physics 1-6.2 Torque and Work – Study Notes
AP Physics 1-6.2 Torque and Work – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Work by Torque
Work by Torque
A torque can transfer energy into or out of an object or rigid system if the torque is exerted over an angular displacement.
Key Points:
- The work done on a rigid system by a torque depends on both the magnitude of the torque and the angular displacement during which it acts.
- Work is done only by the component of torque in the direction of angular displacement.
- Work by torque is a scalar quantity; it may be positive, negative, or zero depending on the relative sense of torque and displacement.
- Positive work: torque and angular displacement act in the same rotational sense.
- Negative work: torque and angular displacement act in opposite rotational senses.
Relevant Equation:
\( W = \tau \, \theta \)
where:
- \( W \) = work done by torque (J)
- \( \tau \) = constant torque applied (N·m)
- \( \theta \) = angular displacement (radians)
Graphical Representation:
- The work done on a rigid system by a given torque can also be found from the area under the curve of a graph of torque as a function of angular displacement.
- This is analogous to the way linear work can be found from the area under the force–displacement curve.
Example :
A constant torque of 12 N·m is applied to a wheel, causing it to rotate through an angular displacement of 5 radians. Find the work done.
▶️Answer/Explanation
Using the formula: \( W = \tau \theta \)
\( W = 12 \cdot 5 = 60 \, \text{J} \)
Final Answer: The work done is 60 J.
Example :
A variable torque is applied to a rigid body, and the torque–angular displacement graph is a straight line from \( \tau = 0 \) at \( \theta = 0 \) to \( \tau = 10 \, \text{N·m} \) at \( \theta = 4 \, \text{rad} \). Find the work done.
▶️Answer/Explanation
Step 1: Work = Area under the torque–displacement curve.
The graph is a triangle: Area = \( \tfrac{1}{2} \times \text{base} \times \text{height} \)
= \( \tfrac{1}{2} \times 4 \times 10 = 20 \, \text{J} \)
Final Answer: The work done is 20 J.