AP Physics C Mechanics- 6.4 Conservation of Angular Momentum- Study Notes- New Syllabus
AP Physics C Mechanics- 6.4 Conservation of Angular Momentum – Study Notes
AP Physics C Mechanics- 6.4 Conservation of Angular Momentum – Study Notes – per latest Syllabus.
Key Concepts:
- Total Angular Momentum of a System
- Cause of Change in Angular Momentum
- Applications and Conditions of Angular Momentum Conservation
Total Angular Momentum of a System
The total angular momentum of a system about a given rotational axis is the vector sum of the angular momenta of all individual parts or particles that make up the system:
\( \mathrm{L_{total} = \sum_i L_i = \sum_i I_i\omega_i} \)
- \( \mathrm{L_i} \): angular momentum of each part of the system
- \( \mathrm{I_i} \): moment of inertia of part \( \mathrm{i} \)
- \( \mathrm{\omega_i} \): angular velocity of part \( \mathrm{i} \)
Key Idea: Angular momentum is additive. When analyzing complex systems (e.g., planets, rotating machinery, multiple connected objects), the total angular momentum about a common axis equals the sum of all individual contributions.
Example
Two disks are mounted on a common frictionless axle. Disk A has \( \mathrm{I_A = 0.30\,kg·m^2} \) and \( \mathrm{\omega_A = 6.0\,rad/s} \); disk B has \( \mathrm{I_B = 0.20\,kg·m^2} \) and \( \mathrm{\omega_B = 3.0\,rad/s} \). Find the total angular momentum of the system about the axle.
▶️ Answer / Explanation
Step 1: Compute each disk’s angular momentum:
\( \mathrm{L_A = I_A\omega_A = (0.30)(6.0) = 1.8} \)
\( \mathrm{L_B = I_B\omega_B = (0.20)(3.0) = 0.6} \)
Step 2: Add them for total angular momentum:
\( \mathrm{L_{total} = L_A + L_B = 1.8 + 0.6 = 2.4\,kg·m^2/s} \)
Result: The total angular momentum about the axle is \( \mathrm{2.4\,kg·m^2/s} \).
Cause of Change in Angular Momentum
Any change in the total angular momentum of a system must result from an interaction with the surroundings (an external torque). If no external torques act, the system’s total angular momentum remains constant it is conserved.
\( \mathrm{\sum \tau_{ext} = \dfrac{dL_{total}}{dt}} \)
Angular momentum can only change if there is a net external torque acting on the system. Internal forces and torques cannot change the system’s total angular momentum.
Newton’s Third Law and Equal Angular Impulses
When two objects interact, the angular impulse that one exerts on the other is equal in magnitude and opposite in direction to the impulse the second exerts on the first:
\( \mathrm{J_{\tau,1\rightarrow2} = -J_{\tau,2\rightarrow1}} \)
Key Idea: This is the rotational analog of Newton’s Third Law. In an isolated system, internal torques between components cancel, ensuring total angular momentum conservation.
Example
Two ice skaters initially at rest push off from each other. Skater A has moment of inertia \( \mathrm{I_A = 3.0\,kg·m^2} \), and skater B has \( \mathrm{I_B = 4.0\,kg·m^2} \). After pushing off, skater A spins at \( \mathrm{2.0\,rad/s} \). Find skater B’s angular velocity.
▶️ Answer / Explanation
Step 1: Total initial angular momentum = 0.
Step 2: After pushing, \( \mathrm{L_A + L_B = 0} \).
\( \mathrm{I_A\omega_A + I_B\omega_B = 0} \)
Step 3: Solve for \( \mathrm{\omega_B} \):
\( \mathrm{\omega_B = -\dfrac{I_A\omega_A}{I_B} = -\dfrac{(3.0)(2.0)}{4.0} = -1.5\,rad/s} \)
Result: Skater B spins at \( \mathrm{1.5\,rad/s} \) in the opposite direction, consistent with Newton’s third law and conservation of angular momentum.
Selecting a System for Angular Momentum Conservation
A system may be chosen such that the total angular momentum is constant when the net external torque is zero:
\( \mathrm{\sum \tau_{ext} = 0 \Rightarrow L_{total} = \text{constant}} \)
Key Idea: When analyzing interactions (e.g., collisions, explosions, or separations), defining the system to include all interacting parts allows internal torques to cancel and total angular momentum to remain conserved.
Change in Angular Speed of Nonrigid Systems
For a nonrigid system, angular speed can change without altering total angular momentum if the system changes its shape that is, if mass moves closer to or farther from the rotational axis.
\( \mathrm{L = I\omega = \text{constant}} \Rightarrow \omega \propto \dfrac{1}{I}} \)
Key Idea: When \( \mathrm{I} \) decreases (mass moves inward), \( \mathrm{\omega} \) increases, and vice versa. This principle explains why a spinning figure skater speeds up when pulling in her arms.
Example
A figure skater spins with an initial angular velocity of \( \mathrm{3.0\,rad/s} \) and moment of inertia \( \mathrm{4.0\,kg·m^2} \). When she pulls her arms inward, her moment of inertia decreases to \( \mathrm{2.0\,kg·m^2} \). Find her new angular velocity, assuming no external torque.
▶️ Answer / Explanation
Step 1: By conservation of angular momentum:
\( \mathrm{I_i\omega_i = I_f\omega_f} \)
Step 2: Substitute known values:
\( \mathrm{(4.0)(3.0) = (2.0)\omega_f} \)
Step 3: Solve for \( \mathrm{\omega_f} \):
\( \mathrm{\omega_f = 6.0\,rad/s} \)
Result: Her angular speed doubles because her moment of inertia was halved, while angular momentum remained constant.
Change in Angular Momentum Equals Angular Impulse
If a system’s total angular momentum changes, that change equals the angular impulse exerted on the system:
\( \mathrm{\Delta L = \int \tau_{ext}\,dt = J_\tau} \)
Key Idea: The angular impulse represents the rotational “push” that alters the angular momentum. If \( \mathrm{\tau_{ext} = 0} \), then \( \mathrm{\Delta L = 0} \) and angular momentum is conserved.
Example
A net external torque of \( \mathrm{2.0\,N·m} \) acts on a flywheel for \( \mathrm{5.0\,s} \). If the flywheel’s initial angular momentum was \( \mathrm{4.0\,N·m·s} \), find its final angular momentum.
▶️ Answer / Explanation
Step 1: Find angular impulse:
\( \mathrm{J_\tau = \tau_{ext}\,\Delta t = (2.0)(5.0) = 10.0\,N·m·s} \)
Step 2: Apply \( \mathrm{\Delta L = J_\tau} \):
\( \mathrm{L_f = L_i + \Delta L = 4.0 + 10.0 = 14.0\,N·m·s} \)
Result: The flywheel’s final angular momentum is \( \mathrm{14.0\,N·m·s} \).
Angular Momentum is Conserved in All Interactions
In every type of interaction — including collisions, explosions, and internal rearrangements — the total angular momentum of a system is conserved, provided that there are no net external torques acting on it.
\( \mathrm{L_{i,\,total} = L_{f,\,total}} \)
- Conservation of angular momentum is a universal law, analogous to linear momentum conservation.
- It holds true regardless of the type of interaction, as long as the system is isolated from external torques.
Example
Two disks on a frictionless axle collide and stick together. Disk A has \( \mathrm{I_A = 0.20\,kg·m^2} \) and rotates at \( \mathrm{10\,rad/s} \); disk B has \( \mathrm{I_B = 0.10\,kg·m^2} \) and is initially at rest. Find their common angular velocity after collision.
▶️ Answer / Explanation
Step 1: Since the system is isolated, \( \mathrm{L_i = L_f} \).
\( \mathrm{I_A\omega_A = (I_A + I_B)\omega_f} \)
Step 2: Solve for \( \mathrm{\omega_f} \):
\( \mathrm{\omega_f = \dfrac{I_A\omega_A}{I_A + I_B} = \dfrac{(0.20)(10)}{0.30} = 6.67\,rad/s} \)
Result: The combined disks rotate together at \( \mathrm{6.7\,rad/s} \), conserving total angular momentum.
Zero Net External Torque: Constant Total Angular Momentum
If the net external torque acting on a selected object or system is zero, then the system’s total angular momentum remains constant over time.
\( \mathrm{\sum \tau_{ext} = 0 \Rightarrow \dfrac{dL_{total}}{dt} = 0 \Rightarrow L_{total} = \text{constant}} \)
Key Idea: This is the mathematical statement of angular momentum conservation. Internal torques within the system can redistribute angular momentum between components, but cannot change the total value for the system as a whole.
Example
A rotating platform (moment of inertia \( \mathrm{2.5\,kg·m^2} \)) spins at \( \mathrm{2.0\,rad/s} \). A student of mass \( \mathrm{60\,kg} \) moves inward, reducing the system’s total moment of inertia to \( \mathrm{1.0\,kg·m^2} \). No external torque acts. Find the new angular velocity of the system.
▶️ Answer / Explanation
Step 1: Apply conservation of angular momentum:
\( \mathrm{I_i\omega_i = I_f\omega_f} \)
Step 2: Substitute values:
\( \mathrm{(2.5)(2.0) = (1.0)\omega_f} \)
\( \mathrm{\omega_f = 5.0\,rad/s} \)
Result: The platform and student spin faster (\( \mathrm{\omega = 5.0\,rad/s} \)) as the moment of inertia decreases — total angular momentum remains constant.
Nonzero Net External Torque: Angular Momentum Transfer
If a system experiences a nonzero net external torque, then angular momentum is transferred between the system and its surroundings. The system’s angular momentum changes according to the applied torque and time interval.
\( \mathrm{\sum \tau_{ext} = \dfrac{dL}{dt}} \Rightarrow \Delta L = \int \tau_{ext}\,dt} \)
Key Idea: An external torque acts as a mechanism for exchanging angular momentum — for example, when brakes slow a wheel, or when a motor applies torque to spin up a disk.
Example
A wheel of moment of inertia \( \mathrm{0.60\,kg·m^2} \) initially rotates at \( \mathrm{8.0\,rad/s} \). A braking torque of \( \mathrm{4.0\,N·m} \) is applied for \( \mathrm{3.0\,s} \). Find (a) the change in angular momentum and (b) the final angular velocity.
▶️ Answer / Explanation
Step 1: Find change in angular momentum:
\( \mathrm{\Delta L = \tau_{ext}\,\Delta t = (−4.0)(3.0) = −12.0\,N·m·s} \)
(negative sign indicates torque opposes rotation)
Step 2: Initial angular momentum:
\( \mathrm{L_i = I\omega_i = (0.60)(8.0) = 4.8\,N·m·s} \)
Step 3: Final angular momentum:
\( \mathrm{L_f = L_i + \Delta L = 4.8 − 12.0 = −7.2\,N·m·s} \)
The negative result means the wheel reversed direction.
Step 4: Find final angular velocity:
\( \mathrm{\omega_f = \dfrac{L_f}{I} = \dfrac{−7.2}{0.60} = −12.0\,rad/s} \)
Result: The wheel’s angular momentum decreased by \( \mathrm{12.0\,N·m·s} \), reversing its direction due to the external braking torque.