Conservation of Angular Momentum - AP Physics C Mechanics

SECTION 1

Historical Context & Motivation

The idea that something about rotational motion is "conserved" emerged gradually over several centuries, evolving from Kepler's empirical observation that planets sweep out equal areas in equal times to the fully formalized vector treatment in Newtonian and Lagrangian mechanics. Unlike linear momentum, which is relatively intuitive—a heavier or faster object is harder to stop—angular momentum requires careful attention to the axis of rotation, the distribution of mass, and the coupling between spinning and orbiting motions. The story of angular momentum conservation is therefore a story about physicists learning to see rotational symmetry as a deep feature of nature, not merely a convenient calculation trick.

1609 • Kepler's Second Law

Johannes Kepler published his law of areas: a line joining a planet to the Sun sweeps out equal areas in equal intervals of time. This was, in retrospect, the first implicit statement of angular momentum conservation for a central-force orbit.

1687 • Newton's Principia

Isaac Newton derived Kepler's area law from his laws of motion and the inverse-square gravitational force, showing that any central force—one directed along the radial line—produces zero torque about the center, thereby conserving angular momentum.

1736 • Euler's Rigid-Body Mechanics

Leonhard Euler formalized the equations of rotational motion for rigid bodies, introducing the moment of inertia tensor and establishing the mathematical machinery that would underpin the modern treatment of angular momentum.

1788 • Lagrange & Analytical Mechanics

Joseph-Louis Lagrange showed in his Mécanique analytique that every continuous symmetry of a mechanical system yields a conserved quantity. Rotational symmetry of the Lagrangian directly implies conservation of angular momentum—a precursor to Noether's theorem.

1918 • Noether's Theorem

Emmy Noether proved rigorously that every differentiable symmetry of the action of a physical system has a corresponding conservation law. Rotational invariance of the Lagrangian guarantees conservation of angular momentum in full generality.

The central question that conservation of angular momentum answers is deceptively simple: what stays the same when a system spins, and under what conditions? From collapsing gas clouds that form rapidly spinning neutron stars to the figure skater who pulls in her arms and accelerates her spin, the principle provides a single quantitative framework. In this lesson, we will develop the formal statement of the law, connect it to the rotational analogue of Newton's second law, and apply it to the kinds of problems that appear on the AP Physics C: Mechanics exam.

SECTION 2

Core Principles & Definitions

Before tackling conservation, we need a precise definition of angular momentum and the conditions under which it is conserved. Angular momentum is the rotational analogue of linear momentum; just as linear momentum p = mv measures the tendency of a translating object to keep translating, angular momentum L measures the tendency of a rotating or orbiting object to keep rotating. The vector nature of L is critical: it has a direction (given by the right-hand rule) as well as a magnitude, and conservation applies to the full vector.

Angular Momentum of a Particle

For a point particle, L = r × p, where r is the position vector from the chosen reference point to the particle and p = mv is its linear momentum. Because L is defined via a cross product, it is always perpendicular to the plane containing r and v.

Angular Momentum of a Rigid Body

For a rigid body rotating about a fixed axis, L = Iω, where I is the moment of inertia about that axis and ω is the angular velocity. When the axis is also a principal axis, L is parallel to ω; otherwise the full inertia tensor is needed.

Torque as the Agent of Change

The rotational analogue of Newton's second law states τ_net = dL/dt. If the net external torque on a system is zero, the time derivative of L vanishes and angular momentum is conserved.

Conservation Statement

When τ_ext = 0, the total angular momentum of the system is constant:

L_i

L_f

. Internal torques (e.g., friction between components) can redistribute angular momentum among parts but cannot change the system total.

System Definition Matters

Conservation applies to a chosen system about a chosen axis. A torque that is external to one subsystem may be internal to a larger system. Carefully defining the system boundary is often the key first step in solving angular momentum problems.

✦

KEY TAKEAWAY

Think of angular momentum conservation like a bank account with no deposits or withdrawals. The system can shuffle funds between internal accounts—one part spins faster while another slows—but the total balance never changes as long as no external torque (no outside transaction) acts. In engineering, this principle lets designers predict how a satellite will reorient itself simply by spinning an internal flywheel: the flywheel gains angular momentum in one direction, so the spacecraft must gain an equal and opposite angular momentum, keeping the net at zero.

SECTION 3

Visual Explanation

The following diagram illustrates the classic demonstration of angular momentum conservation: a figure skater (modeled as a cylinder with extendable arms) changes her moment of inertia by pulling her arms inward, causing her angular velocity to increase while the product Iω remains constant. The diagram shows the before and after states side by side, with the key quantities labeled.

Left panel: the skater has her arms extended, giving a large moment of inertia I₁ and a correspondingly slow angular velocity ω₁. Right panel: arms are pulled in, reducing I₂, and ω₂ increases so that L = Iω remains constant. The pink arrow between the panels indicates the transition.

Notice that the ice exerts negligible torque on the skater about the vertical axis (friction is very small), so the system satisfies the condition τ_ext ≈ 0. The internal muscular forces that pull the arms inward are internal forces and produce internal torques that redistribute angular momentum among body parts but cannot change the system total. The result is a dramatic increase in spin rate, which is easily observed in competitive figure skating.

SECTION 4

Mathematical Framework

The mathematical structure of angular momentum conservation follows directly from Newton's second law in its rotational form. We start with the definition for a single particle, generalize to a system, and derive the conservation law by requiring zero net external torque.

ANGULAR MOMENTUM OF A PARTICLE

L⃗ = r⃗ × p⃗ = r⃗ × (mv⃗)

where r⃗ is the position vector from the reference point to the particle, p⃗ = mv⃗ is the linear momentum, and × denotes the cross product. The magnitude is |L| = rp sin θ = rmv sin θ, where θ is the angle between r⃗ and v⃗.

ROTATIONAL NEWTON'S SECOND LAW

τ⃗_net = dL⃗/dt

The net external torque on a system equals the time rate of change of its total angular momentum. This is the rotational analogue of F⃗_net = dp⃗/dt. Note: internal torques cancel in pairs (Newton's third law), so only external torques contribute.

CONSERVATION LAW

If τ⃗_ext = 0, then L⃗_total = constant

When the net external torque vanishes, dL⃗/dt = 0, so L⃗ does not change. This holds for each component independently: if τ_z,ext = 0, then L_z is conserved even if the other components are not.

FIXED-AXIS SPECIAL CASE

I₁ω₁ = I₂ω₂

For rotation about a fixed axis with no external torque about that axis, the product of moment of inertia and angular velocity remains constant. This is the form most commonly used in AP Physics C problems involving collapsing or expanding systems.

💡

Component-wise Conservation

Angular momentum is a vector, so conservation applies component by component. On the AP exam, you may encounter problems where torque is zero about one axis but not another. In such cases, the component of L along the torque-free axis is conserved while the other components change. Always identify which axis has zero net external torque before writing the conservation equation.

It is worth noting the derivation from first principles. Taking the time derivative of L⃗ = r⃗ × p⃗ and applying the product rule: dL⃗/dt = (dr⃗/dt) × p⃗ + r⃗ × (dp⃗/dt). The first term is v⃗ × (mv⃗) = 0 because the cross product of any vector with itself is zero. The second term is r⃗ × F⃗_net = τ⃗_net. Thus τ⃗_net = dL⃗/dt is derived directly from Newton's second law, confirming that angular momentum conservation is not an independent postulate but a consequence of the underlying force laws when net external torque is absent.

SECTION 5

Detailed Breakdown — Common Scenarios

Angular momentum conservation manifests in a wide variety of physical situations. On the AP Physics C exam, the most frequently tested scenarios fall into several categories: systems where the moment of inertia changes (like the spinning skater), collisions involving rotation (such as a projectile embedding in a rotating rod), and orbital motion under central forces. The diagram below classifies these scenarios and identifies what stays constant in each case.

Three major categories of angular momentum conservation problems seen on the AP Physics C exam. The cyan box covers variable-inertia problems, the pink box covers rotational collision problems, and the amber box covers central-force orbital problems. The green banner at the bottom summarizes the universal problem-solving strategy.

Scenario Details

In changing-inertia problems, the system's mass distribution changes while no external torque acts. The classic example is a point mass sliding radially inward on a frictionless turntable: as the mass moves closer to the axis, I decreases and ω increases. The kinetic energy actually increases in this process because the internal force does positive work. This is a common source of confusion: angular momentum conservation does not imply kinetic energy conservation. The work done by the internal forces accounts for the energy change.

In rotational collision problems, two objects interact over a short time interval. If the collision is about a pivot where the pivot force exerts no torque (because its moment arm is zero), then angular momentum about the pivot is conserved through the collision. A common example is a ball of putty striking the end of a rod that is free to rotate about its other end. The ball's initial angular momentum mvd (where d is the distance from the pivot to the point of impact) equals the system's final angular momentum I_totalω_f. Note that linear momentum is generally not conserved in such problems because the pivot exerts an impulsive force.

In central-force orbit problems, the force is always directed toward (or away from) a fixed center, so the torque about that center is zero. For a particle of mass m, the angular momentum L = mr²(dθ/dt) = mrvsinθ is constant throughout the orbit. At the closest approach (perihelion) the velocity is tangential and sin θ = 1, so L = mr_minv_max; at the farthest point (aphelion), L = mr_maxv_min. This recovers Kepler's equal-area law.

A close spacecraft fires a localized plume above a sunlit planet, with a smaller coasting craft state in the distance. Equal time intervals during ideal gravitational coasting sweep equal areas about the planet at the ellipse’s focus. Angular speed changes along the orbit while angular momentum remains constant.

Equal areas during coasting. Equal changes of mean anomaly, ΔM = 0.6 rad, correspond to equal coast times and equal swept areas about the occupied focus. The angular span is larger near periapsis. Equal changes of mean anomaly, ΔM = 0.6 rad, correspond to equal coast times and equal swept areas about the occupied focus. The angular span is larger near periapsis.

SECTION 6

Worked Example — Rotational Collision

A uniform thin rod of mass M = 2.0 kg and length L = 1.2 m is pivoted at one end and hangs vertically at rest. A small ball of mass m = 0.50 kg is launched horizontally with speed v₀ = 8.0 m/s and strikes the rod at its free end, embedding itself. Find the angular velocity of the rod-ball system immediately after the collision.

Rotational Collision: Ball Embeds in Pivoted Rod

1
Step 1 — Identify the System and Conservation Principle

The system consists of the rod and the ball. The pivot exerts a force, but that force acts at the pivot point (r = 0), so it produces zero torque about the pivot. Gravity also acts, but during the brief collision interval the angular impulse from gravity is negligible. Therefore angular momentum about the pivot is conserved through the collision.
2
Step 2 — Calculate Initial Angular Momentum

Before the collision, the rod is at rest so its angular momentum is zero. The ball moves horizontally with speed v₀ at a perpendicular distance L from the pivot (it strikes the free end). Its angular momentum about the pivot is: L_i = mv₀L = (0.50 kg)(8.0 m/s)(1.2 m).

L_i = 4.80 kg·m²/s
3
Step 3 — Calculate Final Moment of Inertia

After the collision, the ball is embedded at the end of the rod. The total moment of inertia about the pivot is: I_total = I_rod + I_ball = (1/3)ML² + mL² = (1/3)(2.0)(1.2)² + (0.50)(1.2)² = 0.960 + 0.720.

I_total = 1.68 kg·m²
4
Step 4 — Apply Conservation and Solve

Setting L_i = L_f: mv₀L = I_totalω_f. Therefore ω_f = L_i / I_total = 4.80 / 1.68.

ω_f ≈ 2.86 rad/s
5
Step 5 — Check: Is Kinetic Energy Conserved?

KE_i = ½mv₀² = ½(0.50)(8.0)² = 16.0 J. KE_f = ½I_totalω² = ½(1.68)(2.86)² ≈ 6.87 J. The collision is perfectly inelastic (the ball embeds), so kinetic energy is not conserved; about 9.1 J is lost to deformation and heat. This is expected: angular momentum is conserved but mechanical energy is not.

ΔKE ≈ −9.1 J (energy lost to inelastic deformation)

A vertical pivoted rod with a captured ball stands beside a separate horizontal disk with a mass at its rim. During a short impact, choose the pivot as the angular-momentum reference. Its support force has zero lever arm about that pivot; angular momentum can be conserved even though linear momentum and kinetic energy are not.

An impact caught by a pivot. During a short impact, choose the pivot as the angular-momentum reference. Its support force has zero lever arm about that pivot; angular momentum can be conserved even though linear momentum and kinetic energy are not. During a short impact, choose the pivot as the angular-momentum reference. Its support force has zero lever arm about that pivot; angular momentum can be conserved even though linear momentum and kinetic energy are not.

SECTION 7

Common Pitfalls & Comparisons

Students frequently make predictable errors when applying angular momentum conservation. The table below compares common correct and incorrect reasoning patterns, as well as the relationship between angular momentum conservation and the conservation of linear momentum and energy.

Situation / Question	Common Mistake	Correct Reasoning
Ball strikes pivoted rod: is linear momentum conserved?	Yes—momentum is always conserved in collisions.	No. The pivot exerts an external impulsive force on the system, so linear momentum is not conserved. Angular momentum about the pivot is conserved because the pivot force has zero moment arm.
Skater pulls arms in: is kinetic energy conserved?	Yes—no external forces do work.	No. Internal muscular forces do positive work, increasing KE. L is conserved, but KE increases (since KE = L²/2I and I decreases).
Choosing the reference point for torque	Any reference point gives the same L.	Angular momentum depends on the chosen reference point. Pick the point where the external forces produce zero torque (typically a pivot or center of mass) to simplify the problem.
Disk dropped onto spinning disk	Use I₁ω₁ = I₂ω₂ with I₂ = I₁ (same disk).	I₂ = I₁ + I_dropped because both disks rotate together afterward. The system's total I increases, so ω decreases.
Object in orbit: can L change direction?	L is just a number (scalar).	L is a vector. Conservation means both magnitude and direction are fixed. The orbital plane does not precess under a central force.

Common errors and correct approaches in angular momentum problems

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WHEN IS WHAT CONSERVED?

Angular momentum, linear momentum, and kinetic energy are three distinct quantities with independent conservation conditions. A perfectly inelastic rotational collision conserves angular momentum (if no external torque) but loses kinetic energy and may not conserve linear momentum. Always check each conservation law separately: ask whether the relevant external agent (force, torque, or non-conservative work) is absent before declaring a quantity conserved.

SECTION 8

Connection to Advanced Theory

The conservation of angular momentum as presented in AP Physics C is the tip of a much deeper mathematical iceberg. In Lagrangian and Hamiltonian mechanics, angular momentum conservation is a direct consequence of Noether's theorem: any system whose Lagrangian is invariant under continuous rotations possesses a conserved angular momentum. The table below draws explicit parallels between the AP-level treatment and the more advanced formulations you may encounter in upper-division physics.

Aspect	AP Physics C Treatment	Advanced / Analytical Mechanics
Why is L conserved?	Because τ_ext = 0 → dL/dt = 0	Because the Lagrangian is rotationally symmetric (Noether's theorem)
Mathematical object	Scalar L = Iω (about a fixed axis) or vector L = r × p	Component of generalized momentum p_θ = ∂ℒ/∂θ̇ conjugate to cyclic angle θ; tensor $L_i$ = I_ij ω_j for general 3D rotation
Precession / gyroscopic motion	Qualitative: torque changes direction of L, causing precession	Euler's equations for rigid body: I₁ω̇₁ − (I₂ − I₃)ω₂ω₃ = τ₁, etc. Precession rate Ω = τ/(Iω)
Quantum mechanics	Not addressed	Angular momentum is quantized: L² = ℓ(ℓ+1)ħ², $L_z$ = mℓħ. Rotational symmetry of the Hamiltonian yields conserved quantum numbers.

AP Physics C angular momentum vs. advanced treatments

For the AP exam, the key takeaway from these advanced connections is conceptual: angular momentum conservation is not merely a useful computational shortcut but reflects a fundamental symmetry of nature. Whenever a physical law looks the same regardless of how you orient your coordinate system, angular momentum is conserved. This deep link between symmetry and conservation is one of the most powerful organizing principles in all of physics, and mastering the AP-level version prepares you to engage with it in greater mathematical generality.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A figure skater is spinning with arms extended. She then pulls her arms close to her body. Which of the following correctly describes the changes in her angular velocity ω, moment of inertia I, and rotational kinetic energy KE?

A. ω increases, I decreases, KE stays the same
B. ω increases, I decreases, KE increases
C. ω increases, I decreases, KE decreases
D. ω stays the same, I decreases, KE decreases

Show answer (B)

The correct answer is B. By pulling her arms in, the skater reduces her moment of inertia I. Since angular momentum L = Iω is conserved (no external torque from the ice about the vertical axis), ω must increase. The rotational kinetic energy KE = ½Iω² = L²/(2I) increases because I decreases while L is constant—the internal muscular work done to pull the arms inward is the source of this additional kinetic energy. Choice A is wrong because KE does not stay the same; the relationship KE = L²/(2I) shows that decreasing I increases KE. Choice C is wrong because KE increases, not decreases. Choice D is wrong because ω must increase to keep L constant when I decreases. Illustrative arms-in rotation: I decreases from 4 to 2 kg·m² while ω increases from 2 to 4 rad/s, keeping L = 8 kg·m²/s. K increases from 8 to 16 J through internal work.

Illustrative arms-in rotation: I decreases from 4 to 2 kg·m² while ω increases from 2 to 4 rad/s, keeping L = 8 kg·m²/s. K increases from 8 to 16 J through internal work.

Question 2: BASIC CALCULATION

A uniform disk of mass 4.0 kg and radius 0.30 m spins freely at 12 rad/s about its central axis. A ring of mass 2.0 kg and the same radius is dropped concentrically onto the disk. After friction brings them to a common angular velocity, what is that final angular velocity? (I_disk = ½MR², I_ring = mR²)

A. 4.0 rad/s
B. 6.0 rad/s
C. 8.0 rad/s
D. 12 rad/s

Show answer (B)

The correct answer is B. I_disk = ½(4.0)(0.30)² = 0.18 kg·m². I_ring = (2.0)(0.30)² = 0.18 kg·m². Initial angular momentum:

L_i

= I_disk × ω_i = (0.18)(12) = 2.16 kg·m²/s. The ring is initially at rest, so it contributes zero angular momentum. After coupling:

L_f

= (I_disk + I_ring)ω_f = (0.36)ω_f. Setting

L_i

L_f

: 2.16 = 0.36ω_f → ω_f = 6.0 rad/s. Choice A (4.0 rad/s) would result from incorrectly tripling the inertia. Choice C (8.0 rad/s) would result from using I_ring = ½mR² instead of mR². Choice D (12 rad/s) ignores the added ring entirely.

Question 3: INTERMEDIATE

A small puck of mass m = 0.20 kg slides at v = 5.0 m/s across a frictionless horizontal surface and strikes the end of a thin, uniform rod of mass M = 1.0 kg and length L = 0.80 m. The rod is free to rotate about a frictionless pivot at its center. The puck bounces back with a speed of 1.0 m/s. What is the angular velocity of the rod immediately after the collision?

A. 9.0 rad/s
B. 12 rad/s
C. 18 rad/s
D. 22.5 rad/s

Show answer (A)

The correct answer is A. The pivot is at the center of the rod, so the puck strikes at distance L/2 = 0.40 m from the pivot. Angular momentum about the pivot is conserved because the pivot force has zero moment arm. Taking the initial direction of the puck as positive:

L_i

= mv(L/2) = (0.20)(5.0)(0.40) = 0.40 kg·m²/s. After the collision, the puck moves in the opposite direction at 1.0 m/s: L_puck,f = m(−1.0)(L/2) = (0.20)(−1.0)(0.40) = −0.080 kg·m²/s. The rod has I = (1/12)ML² = (1/12)(1.0)(0.80)² = 0.0533 kg·m². Setting

L_i

= L_rod,f + L_puck,f: 0.40 = (0.0533)ω + (−0.080). So 0.0533ω = 0.480, giving ω = 0.480/0.0533 ≈ 9.0 rad/s. Choice B (12 rad/s) results from neglecting the puck's rebound angular momentum, giving 0.0533ω = 0.40 and ω ≈ 7.5, then rounding incorrectly, or from using an incorrect moment of inertia. Choice C (18 rad/s) results from incorrectly using the full length L instead of L/2 for the moment arm of the puck while keeping the same moment of inertia, effectively doubling the numerator. Choice D (22.5 rad/s) results from omitting the puck's rebound angular momentum and using an incorrect moment of inertia such as I = ML²/12 with L/2 substituted for L.

Question 4: APPLIED

7 pts

A satellite in a circular orbit of radius r₁ = 7.0 × 10⁶ m around Earth fires its thrusters briefly to increase its speed tangentially, entering an elliptical orbit. At the apogee of the new orbit, its distance from Earth's center is r₂ = 1.4 × 10⁷ m. Assuming the thrust impulse is effectively instantaneous: (a) Using conservation of angular momentum, express the satellite's speed v₂ at apogee in terms of v₁ (the speed just after the thruster burn at r₁), r₁, and r₂. (b) If v₁ = 8.5 × 10³ m/s, calculate v₂. (c) Explain why angular momentum is conserved during the elliptical orbit but kinetic energy is not.

Show answer

Rubric (7 points): (a) [2 points] • (1 pt) Recognizing that at both the point of closest approach (just after the burn, where v is tangential) and at apogee (where v is again tangential), L = mrv. So mr₁v₁ = mr₂v₂. • (1 pt) Solving: v₂ = v₁(r₁/r₂). (b) [2 points] • (1 pt) Substituting: v₂ = (8.5 × 10³)(7.0 × 10⁶)/(1.4 × 10⁷) = (8.5 × 10³)(0.50). • (1 pt) Correct answer: v₂ = 4.25 × 10³ m/s (or 4250 m/s). (c) [3 points] • (1 pt) Gravity is a central force directed toward Earth's center, so the torque τ = r × F about Earth's center is zero at all points in the orbit; hence angular momentum is conserved throughout. • (1 pt) Gravity does work on the satellite because it has a component along (or opposite to) the direction of motion when the orbit is elliptical. As the satellite moves from perigee to apogee, it moves against the gravitational field component along its velocity, so gravity does negative work and kinetic energy decreases. • (1 pt) Correctly noting that total mechanical energy (KE + PE) is conserved during the orbit, but KE alone is not; KE converts to gravitational PE and vice versa.

Question 5: CRITICAL THINKING

7 pts

A horizontal turntable (a uniform disk of mass M and radius R) is spinning freely at angular velocity ω₀. A person of mass m, initially standing at the center of the turntable, walks slowly outward along a radial line until reaching the edge. (a) Derive an expression for the angular velocity ω_f of the system when the person reaches the edge. Treat the person as a point mass. (b) Determine the ratio of the final rotational kinetic energy to the initial rotational kinetic energy. Is energy lost, gained, or constant? Explain physically what accounts for the energy change. (c) Now suppose the person throws a ball of mass

m_b

tangentially (relative to the ground) with speed

v_b

while standing at the edge. Find the new angular velocity of the turntable-person system after the throw, in terms of ω_f, I_system,

m_b

v_b

, and R.

Show answer

Rubric (7 points): (a) [2 points] • (1 pt) I_initial = I_turntable = ½MR² (person at center contributes nothing).

L_i

= (½MR²)ω₀. • (1 pt) I_final = ½MR² + mR². By conservation (no external torque): (½MR²)ω₀ = (½MR² + mR²)ω_f → ω_f = ω₀ × (½M)/(½M + m) = ω₀M/(M + 2m). (b) [3 points] • (1 pt) KE_i = ½(½MR²)ω₀² = ¼MR²ω₀². KE_f = ½(½MR² + mR²)ω_f². • (1 pt) Using ω_f from (a): KE_f/KE_i = [(½M + m)(ω₀M/(M+2m))²] / [(½M)ω₀²] = M/(M + 2m). Since M/(M + 2m) < 1, the final KE is less than the initial KE. • (1 pt) Energy is lost. The person must exert a friction force on the turntable as they walk outward. In the rotating frame, the person does negative work against the Coriolis force (or equivalently, from the lab frame, the person's muscles do negative work in maintaining the radial path, converting kinetic energy into internal/metabolic energy). Alternatively: the constraint forces between feet and turntable do negative net work on the system. (c) [2 points] • (1 pt) Before the throw the system (turntable + person) has angular momentum L = I_system × ω_f. After the throw, the ball carries angular momentum L_ball =

m_b

v_b

× R (taking the tangential throw direction as the positive rotation direction). By conservation: I_system × ω_f = I_system' × ω_new +

m_b

v_b

R, where I_system' = I_system −

m_b

R² (the ball is no longer part of the system). If the ball was not originally part of the system (it was carried by the person), then I_system already includes the person but not

m_b

separately. Properly: define the system as turntable + person + ball initially. I_total,initial = (½MR² + mR² +

m_b

R²), all at angular velocity ω_f. After throw: ω_new = [I_total × ω_f −

m_b

v_b

R] / (½MR² + mR²), where I_total = ½MR² + mR² +

m_b

R². • (1 pt) Simplifying: ω_new = ω_f +

m_b

R(ω_f R −

v_b

)/(½MR² + mR²). If

v_b

> ω_f R, the turntable-person system slows down; if

v_b

< ω_f R, it speeds up.

SUMMARY

Summary

Angular momentum is the rotational analogue of linear momentum, defined for a particle as L⃗ = r⃗ × p⃗ and for a rigid body rotating about a fixed axis as L = Iω. The rotational form of Newton's second law, τ⃗_net = dL⃗/dt, tells us that angular momentum changes only when a net external torque acts. When that torque is zero—whether because all forces pass through the axis, or because the system is isolated—the total angular momentum is conserved: $L_i$ = $L_f$ .

Key application categories include variable-inertia systems (I₁ω₁ = I₂ω₂), rotational collisions (conserve L about the pivot, but KE and linear momentum may not be conserved), and central-force orbits (r₁v₁ = r₂v₂ at turning points). Remember that angular momentum is a vector: conservation applies component by component, and the choice of reference point matters. At its deepest level, this conservation law reflects the rotational symmetry of the physical laws governing the system, a connection formalized by Noether's theorem.