AP Physics C Mechanics Extra - How to Solve: Angular Momentum Conservation with Changing Moment of Inertia

How to Solve: Angular Momentum Conservation with Changing Moment of Inertia

Complete Guide (For AP Physics 1, C: Mechanics, C: E&M – 1,300+ words)

Introduction

Master this, and you’ll crush 10–15% of your AP Physics 1 or C Mechanics exam—plus real-world problems like figure skaters spinning faster, collapsing stars, or even how a cat always lands on its feet. (This topic appears on ~1 in 4 AP Physics 1 FRQs and is a guaranteed point on C Mechanics.)

WHAT YOU NEED TO KNOW FIRST

Before diving in, you must already understand:
1. Angular momentum (L) = Iω – The rotational equivalent of linear momentum.
2. Moment of inertia (I) – How mass is distributed around an axis; depends on shape and axis location.
3. Conservation of angular momentum – If no external torque acts, L_initial = L_final.

(If any of these feel shaky, pause and review them first—this guide won’t work without them!)

KEY TERMS & FORMULAS

Key Terms

• Angular momentum (L): Rotational "oomph" of an object. Units: kg·m²/s.
• Moment of inertia (I): Resistance to rotational motion. Units: kg·m².
• Angular velocity (ω): How fast an object spins. Units: rad/s.
• Torque (τ): Rotational force. If τ = 0, angular momentum is conserved.
• External torque: Any force that changes the system’s angular momentum (e.g., friction, a push).

Formulas

1. Angular momentum (L): L = Iω
2. L = angular momentum (kg·m²/s)
3. I = moment of inertia (kg·m²)

4. ω = angular velocity (rad/s) MEMORISE THIS – It’s the foundation of everything here.

5. Conservation of angular momentum (no external torque): L_initial = L_final → I_initial × ω_initial = I_final × ω_final MEMORISE THIS – This is the only equation you’ll use for these problems.

6. Moment of inertia (common shapes – given on exam sheet, but know how to use them):

7. Point mass: I = mr²
8. Hoop (about center): I = mr²
9. Solid disk (about center): I = ½mr²
10. Rod (about center): I = (1/12)mL²
11. Rod (about end): I = (1/3)mL²

STEP-BY-STEP METHOD

(Follow these steps exactly for every problem. No shortcuts!)

1. Identify the system.
2. What objects are spinning? (e.g., a figure skater, a merry-go-round, a collapsing star.)

3. Is the system isolated? (No external torque = conservation applies.)

4. Check for external torque.

5. If τ = 0, angular momentum is conserved. Write: L_initial = L_final.

6. If τ ≠ 0, you cannot use conservation—stop here and use τ = ΔL/Δt instead.

7. Write down initial and final moments of inertia (I).

8. Use the given shape (e.g., disk, rod, point mass) to find I.

9. If I changes (e.g., skater pulls in arms), I_initial ≠ I_final.

10. Write down initial and final angular velocities (ω).

11. If ω changes (e.g., skater spins faster), ω_initial ≠ ω_final.

12. Apply conservation of angular momentum.

13. Plug into I_initial × ω_initial = I_final × ω_final.

14. Solve for the unknown (usually ω_final).

15. Check units and reasonableness.

16. Units: I (kg·m²), ω (rad/s) → L (kg·m²/s).
17. Does the answer make sense? (e.g., If I decreases, ω should increase.)

WORKED EXAMPLES

Example 1 – Basic (AP Physics 1)

Problem: A figure skater spins with an angular velocity of 6 rad/s with her arms extended. Her moment of inertia is 4.5 kg·m² in this position. When she pulls her arms in, her moment of inertia drops to 1.8 kg·m². What is her new angular velocity?

Solution (Step-by-Step):
1. System: Figure skater (isolated, no external torque).
2. External torque? No → L_initial = L_final.
3. Initial I: 4.5 kg·m² Final I: 1.8 kg·m²
4. Initial ω: 6 rad/s Final ω:? (Let’s call it ω_f)
5. Apply conservation: I_initial × ω_initial = I_final × ω_f (4.5)(6) = (1.8)(ω_f) 27 = 1.8 ω_f ω_f = 27 / 1.8 = 15 rad/s
6. Check: - Units: kg·m² × rad/s = kg·m²/s ✔️ - Reasonable? I decreased → ω increased (yes, skater spins faster).

What we did and why: We used L_initial = L_final because no external torque acted. When the skater pulled in her arms, her I decreased, so her ω increased to keep L constant.

Example 2 – Medium (AP C Mechanics)

Problem: A 1.2 kg rod of length 0.8 m is rotating freely about its center at 5 rad/s. A 0.4 kg clay ball is dropped onto the end of the rod and sticks. What is the new angular velocity of the system?

Solution (Step-by-Step):
1. System: Rod + clay ball (isolated, no external torque).
2. External torque? No → L_initial = L_final.
3. Initial I (rod about center): I_rod = (1/12)mL² = (1/12)(1.2)(0.8)² = 0.064 kg·m² Initial ω: 5 rad/s
4. Final I (rod + clay ball): - I_rod = 0.064 kg·m² (unchanged) - I_clay = mr² = (0.4)(0.4)² = 0.064 kg·m² (clay is at r = 0.4 m from center) - Total I_final = 0.064 + 0.064 = 0.128 kg·m²
5. Apply conservation: I_initial × ω_initial = I_final × ω_f (0.064)(5) = (0.128)(ω_f) 0.32 = 0.128 ω_f ω_f = 0.32 / 0.128 = 2.5 rad/s
6. Check: - Units: kg·m² × rad/s = kg·m²/s ✔️ - Reasonable? I increased → ω decreased (yes, system slows down).

What we did and why: We added the clay’s moment of inertia to the rod’s. Since I increased, ω decreased to conserve L.

Example 3 – Exam-Style (AP Physics 1 FRQ)

Problem: A merry-go-round (solid disk, mass 200 kg, radius 2 m) is spinning at 3 rad/s. A 50 kg child runs at 4 m/s tangent to the edge and jumps on. What is the new angular velocity of the system?

Solution (Step-by-Step):
1. System: Merry-go-round + child (isolated, no external torque).
2. External torque? No → L_initial = L_final.
3. Initial I (merry-go-round): I_disk = ½mr² = ½(200)(2)² = 400 kg·m² Initial ω: 3 rad/s
4. Final I (disk + child): - I_disk = 400 kg·m² (unchanged) - I_child = mr² = (50)(2)² = 200 kg·m² - Total I_final = 400 + 200 = 600 kg·m²
5. Child’s initial angular momentum (before jumping on): - The child is not part of the system initially, so we must calculate their L separately. - Linear momentum (p) = mv = (50)(4) = 200 kg·m/s - Angular momentum (L) = r × p = rp sin(90°) = (2)(200) = 400 kg·m²/s
6. Total initial L: L_initial = L_disk + L_child = (400)(3) + 400 = 1200 + 400 = 1600 kg·m²/s
7. Apply conservation: L_initial = L_final 1600 = I_final × ω_f 1600 = 600 ω_f ω_f = 1600 / 600 = 2.67 rad/s
8. Check: - Units: kg·m²/s ✔️ - Reasonable? I increased → ω decreased (yes).

What we did and why: We included the child’s initial angular momentum (since they were moving before joining). Then, we added their I to the disk’s and solved for ω_f.

COMMON MISTAKES

(Avoid these like the plague!)

1. Mistake: Forgetting to check for external torque. Why it happens: Students rush and assume conservation applies. Correct approach: Always ask: "Is τ = 0?" If not, do not use L_initial = L_final.

2. Mistake: Using linear momentum instead of angular momentum. Why it happens: Confusing p = mv with L = Iω. Correct approach: For rotation, always use L = Iω. Linear momentum is for straight-line motion.

3. Mistake: Miscalculating moment of inertia. Why it happens: Using the wrong formula for the shape (e.g., using I = mr² for a rod). Correct approach: Memorise the common I formulas (or check the exam sheet).

4. Mistake: Ignoring initial angular momentum of moving objects. Why it happens: Forgetting that objects not initially part of the system (like the child in Example 3) still contribute to L. Correct approach: If an object is moving before joining, calculate its L = r × p and add it to L_initial.

5. Mistake: Mixing up units (e.g., using degrees instead of radians). Why it happens: Carelessness. Correct approach: Always use radians for ω. Convert if needed (1 rev = 2π rad).

EXAM TRAPS

(Examiners love these—don’t fall for them!)

1. Trap: "The system is isolated" is not stated explicitly. How to spot it: The problem says "frictionless," "no external forces," or "freely rotating." How to avoid it: Assume τ = 0 unless told otherwise (e.g., "a motor applies torque").

2. Trap: Changing the axis of rotation mid-problem. How to spot it: The problem mentions a new pivot point (e.g., "the rod now rotates about its end"). How to avoid it: Recalculate I for the new axis. (e.g., I_rod_center = (1/12)mL² → I_rod_end = (1/3)mL²).

3. Trap: Objects not initially rotating but later joining the system. How to spot it: A problem where a stationary object (e.g., a block) is dropped onto a spinning disk. How to avoid it: The initial L is only from the spinning object. The new object adds I but no initial L.

1-MINUTE RECAP

(Night-before-the-exam summary—say this out loud!)

"Okay, listen up. For angular momentum with changing I, here’s the deal:
1. Check for external torque. If τ = 0, L_initial = L_final.
2. Write down I and ω for initial and final states. If I changes, ω must change to keep L the same.
3. Plug into I_initial × ω_initial = I_final × ω_final. Solve for the unknown.
4. Watch for traps: Moving objects not initially in the system? Recalculate I if the axis changes. Always use radians.
5. If I decreases, ω increases. If I increases, ω decreases. That’s the whole story.

You’ve got this. Now go crush that exam."