AP Physics 1- 5.3 Torque- Study Notes- New Syllabus
AP Physics 1-5.3 Torque – Study Notes
AP Physics 1-5.3 Torque – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Identifying Torques on a Rigid System
- Describing Torques on a Rigid System
Torque
Identifying Torques on a Rigid System
A torque is the rotational equivalent of force. It tells us how effectively a force can cause an object to rotate about an axis.
Torque depends on three factors:
- The magnitude of the applied force \( F \).
- The distance (lever arm) \( r \) from the axis of rotation to the point of application of force.
- The angle \( \theta \) between the force and the position vector.
\( \tau = r F \sin \theta \)
If multiple forces act, each torque is calculated separately, and the net torque is the sum of all torques about the chosen axis.
Describing Torques on a Rigid System
Torque causes angular acceleration in accordance with Newton’s second law for rotation:
\( \tau_{net} = I \alpha \)
where \( I \) = moment of inertia of the rigid system, \( \alpha \) = angular acceleration.
Direction of torque is given by the right-hand rule (curl fingers in the direction of force rotation, thumb points along torque vector).
Torque can do work if applied through an angular displacement:
\( W = \tau \theta \)
If multiple torques act, they may either reinforce or oppose each other depending on their directions (clockwise vs. counterclockwise).
3. Equilibrium Condition
For rotational equilibrium:
\( \sum \tau = 0 \)
This means clockwise torques balance counterclockwise torques.
Example :
A uniform beam of length \( 4 \, \text{m} \) and weight \( 200 \, \text{N} \) is pivoted at its center. A weight of \( 300 \, \text{N} \) is hung at one end. Find the net torque about the pivot.
▶️Answer/Explanation
Torque due to beam’s own weight = 0 (since it acts at the pivot).
Torque due to hanging weight:
\( \tau = r F \sin \theta = (2)(300)(\sin 90^\circ) = 600 \, \text{N·m} \)
Net torque = \( 600 \, \text{N·m} \) (counterclockwise).
Answer: \( \tau_{net} = 600 \, \text{N·m} \)
Example :
A wrench of length \( 0.25 \, \text{m} \) is used to loosen a bolt. If a force of \( 80 \, \text{N} \) is applied perpendicular to the wrench, calculate the torque produced.
▶️Answer/Explanation
\( \tau = r F \sin \theta = (0.25)(80)(\sin 90^\circ) = 20 \, \text{N·m} \)
Answer: \( \tau = 20 \, \text{N·m} \)
Example :
A 50 N force is applied at a distance of 0.6 m from the hinge of a door. The force makes an angle of \( 60^\circ \) with the door surface. Find the effective torque.
▶️Answer/Explanation
\( \tau = r F \sin \theta = (0.6)(50)(\sin 60^\circ) \)
\( \tau = 30 \times 0.866 \approx 25.98 \, \text{N·m} \)
Answer: \( \tau \approx 26 \, \text{N·m} \)