M5L3 Rolling Motion and Angular Momentum

"In the absence of external torques, the angular momentum of a system remains constant. The universe keeps its books very carefully."
— paraphrase of a conservation law

This lesson is where everything in Module 5 converges. Rolling without slipping is not pure rotation and not pure translation — it is both simultaneously, linked by the constraint v_cm = Rω. To analyze a rolling object on an incline you need M5L1 kinematics, M5L2 dynamics, and M3 energy conservation — all at once. Then the lesson extends the rotational framework to its most powerful conservation law: angular momentum L = Iω is conserved when net external torque is zero. This is why a gyroscope resists tipping, why a diver can control rotation speed in the air, and why neutron stars spin at extraordinary rates. Conservation of angular momentum is one of the deepest symmetries in physics, and it is accessible to you right now.

Two Big Ideas in This Lesson

Part 1: Rolling Without Slipping

The rolling constraint links every kinematic and dynamic quantity:

v_cm = Rω • a_cm = Rα
K_total = ½mv_cm² + ½Iω² (translational + rotational)
Static friction enables rolling — it does no work
Hollow vs solid objects race: shape factor c determines who wins

Part 2: Angular Momentum

The rotational analog of linear momentum:

L = Iω (vector, units: kg·m²/s)
τ_net = dL/dt (most general form of Newton's 2nd)
Conservation: if τ_net = 0, then L = constant
Applications: figure skaters, divers, gyroscopes, neutron stars

Course Competencies and Learning Objectives

A ★ indicates that this page contains content related to that LO.

CC5.3 Analyze rolling motion and apply conservation of angular momentum

★ LO5.3.1 Apply the rolling-without-slipping constraint (v_cm = Rω, a_cm = Rα) to kinematic and dynamic problems

★ LO5.3.2 Calculate the total kinetic energy of a rolling object as the sum of translational and rotational kinetic energy

★ LO5.3.3 Use energy conservation to find the speed of a rolling object at the bottom of an incline

★ LO5.3.4 Predict which of two rolling objects reaches the bottom of an incline first based on their shape factors

★ LO5.3.5 Define angular momentum L = Iω and state its SI units

★ LO5.3.6 Apply the angular impulse-momentum theorem: τ_netΔt = ΔL

★ LO5.3.7 Apply conservation of angular momentum to solve problems involving changing moment of inertia

Required Reading

Section 10.8 covers rolling; Chapter 11 introduces angular momentum. Read all four sections — each one introduces a concept you will need in the computational problems.

10.8 Work and Power for Rotational Motion 11.1 Rolling Motion 11.2 Angular Momentum 11.3 Conservation of Angular Momentum

Optional Reading

Explore More

Physics Classroom: Angular Momentum Khan Academy: Angular Momentum PhET: My Solar System (angular momentum) Khan Academy: Rolling Without Slipping

Media

Video 1 establishes the rolling constraint and total kinetic energy; Video 2 applies energy conservation to the incline race; Video 3 introduces angular momentum and conservation.

Video 1: Rolling Without Slipping — The Constraint

One Object, Two Kinds of Motion at Once

A rolling object is simultaneously translating (its center moves) and rotating (it spins). The rolling-without-slipping condition forces these two motions to be linked.

Why the contact point has zero instantaneous velocity
Deriving v_cm = Rω from the no-slip condition
Total KE = ½mv_cm² + ½I_cmω²
Expressing total KE in terms of v_cm alone: K = ½(m + I/R²)v_cm²
Why static friction enables rolling but does no work

Time: 10:00

Video 2: Rolling Down an Incline — The Race

Why Shape Beats Mass Every Time

Using energy conservation to find the speed at the bottom of an incline, and understanding why the shape factor c determines the outcome regardless of mass or radius.

Energy conservation: mgh = ½mv_cm² + ½Iω² = ½(1 + c)mv_cm²
Solving for v_cm: v = √[2gh/(1+c)]
Shape factor comparison: solid sphere (c=2/5) beats solid disk (c=1/2) beats hollow ring (c=1)
The result is independent of mass and radius — only shape matters
Rolling with slipping: what changes and why friction becomes kinetic

Time: 9:45

Video 3: Angular Momentum and Its Conservation

The Spinning Universe Keeps Its Books

Angular momentum L = Iω is the rotational analog of linear momentum. When no external torque acts, it is conserved — exactly as linear momentum is conserved when no external force acts.

L = Iω — units, direction (right-hand rule), and magnitude
τ_net = dL/dt — the most general form of Newton's 2nd for rotation
Angular impulse: τΔt = ΔL
Conservation: I₁ω₁ = I₂ω₂ when τ_ext = 0
Figure skater, diver, collapsing star, spinning stool demonstrations

Time: 11:30

Deeper Look: Where the Rolling Race Result Comes From, and Angular Momentum as a Vector

Deriving the Incline Race Result

Why Only Shape Factor c Matters

Start from energy conservation for a rolling object descending height h:

                        mgh = ½mvcm² + ½Icmω²
                    

Substitute I_cm = cMR² and ω = v_cm/R:

                        mgh = ½mvcm² + ½(cMR²)(vcm/R)²

                        mgh = ½mvcm²(1 + c)

                        vcm = √[2gh / (1 + c)]

The mass M and radius R cancel entirely. The final speed depends only on g, h, and the shape factor c. Therefore:

Object	c	1 + c	Relative speed (lower c = faster)
Solid sphere	2/5	7/5	Fastest
Solid disk / cylinder	1/2	3/2	2nd
Hollow sphere	2/3	5/3	3rd
Hollow ring / thin cylinder	1	2	Slowest

A sliding block (no rotation, c = 0) would be fastest of all: v = √(2gh). Rolling always arrives later than sliding on a frictionless surface, because some potential energy goes into rotation.

Angular Momentum as a Full Vector

L = r × p — The General Definition

For a single particle: L = r × p = r × mv

For a rigid body rotating about a fixed axis: L = Iω

The direction of L is along the rotation axis, given by the right-hand rule (same as ω). This vector nature matters in three important contexts:

Gyroscope precession: Gravity applies a torque perpendicular to L, so L precesses (rotates) rather than tilting — the gyroscope resists falling over.
Bicycle stability: The spinning wheels have angular momentum along the axle. Any perturbation creates a torque perpendicular to L, causing precession that steers the wheel back upright.
Conservation in 3D: Each component of L is independently conserved if the corresponding torque component is zero. A system can have L_z conserved even if L_x and L_y are not.

Connection to Noether's Theorem

Conservation of angular momentum is not an accident — it follows from the fact that the laws of physics are the same in all directions (rotational symmetry). Emmy Noether proved in 1915 that every continuous symmetry of a physical system corresponds to a conserved quantity. Rotational symmetry → angular momentum. Translational symmetry → linear momentum. Time symmetry → energy.

Rolling Friction: Static vs Kinetic

Why Static Friction Does No Work in Rolling

This is a subtle point that confuses many students. Static friction acts at the contact point. In rolling without slipping, the contact point has zero instantaneous velocity. Since work = F · displacement, and the displacement of the contact point is zero, static friction does zero work.

But friction is essential for rolling:

On an incline, gravity pulls the center of mass forward. Without friction, the contact point would slide — the object would slide, not roll.
Static friction at the contact point provides the torque that causes angular acceleration. It is the mechanism by which the incline "grabs" the object and makes it spin up.
No net work done by static friction, but it does transfer energy between translational and rotational modes.

When rolling slipping occurs: If the torque required exceeds μ_sN, static friction cannot maintain the no-slip condition. The contact point slides, kinetic friction takes over, does negative work, and the object both rolls and slips simultaneously. This is a much more complex problem beyond this course.

Practical Consequence

This is why rolling resistance exists in real tires — the tire deforms at the contact patch, and energy is lost to deformation (internal friction), not to sliding friction. A perfectly rigid wheel rolling on a perfectly rigid surface would have zero rolling resistance.

Practice and Apply — Conceptual

Solving a Rolling Problem

Order the Steps: Rolling Object Down an Incline (Energy Method)

Arrange these steps in the correct order for solving a rolling-on-incline problem using energy conservation.

Identify the object's shape and look up or derive the shape factor c (I_cm = cMR²)
Write the energy conservation equation: mgh = ½mv_cm² + ½I_cmω²
Apply the rolling constraint: substitute ω = v_cm/R into the energy equation
Factor out ½mv_cm² to get: mgh = ½mv_cm²(1 + c)
Solve for v_cm: v_cm = √[2gh/(1 + c)]
Compare results for different shapes: smaller c means larger v_cm (more concentrated mass wins)
If angular velocity is needed, apply ω = v_cm/R using the result from step 5

Angular Momentum: Larger or Smaller?

Sort: Does This Change Increase or Decrease Angular Momentum?

A figure skater is spinning with arms extended. No external torque acts. Sort each action into the correct category.

Actions to Sort

Skater pulls arms in tightly (no external torque)
A second skater pushes on the spinning skater (external torque applied)
Skater extends one leg outward (no external torque)
Ice surface exerts negligible friction (no external torque)
Skater catches a spinning ball thrown to them (adds angular momentum)
Skater drags one blade on the ice to slow down (friction torque applied)

L Stays the Same

L Increases

L Decreases

Conceptual Checkpoints

Predict the Answer — Then Flip

Each card poses a question about rolling motion or angular momentum. Form your answer first, then check.

A solid sphere and a hollow ring of the same mass and radius roll down the same incline from rest. Which reaches the bottom first? Does mass or radius matter?

Solid Sphere Wins — Shape Only

The sphere (c = 2/5) arrives before the ring (c = 1) because it stores less energy as rotation. Mass and radius cancel completely: v = √[2gh/(1+c)]. A 10 kg sphere and a 0.1 kg sphere of any radius would tie each other.

A figure skater pulls their arms in during a spin. Their angular velocity ω increases — does that mean angular momentum increased?

No — L Is Conserved, ω Just Adjusts

With no external torque, L = Iω = constant. Pulling arms in decreases I. Since L is fixed, ω must increase to compensate: ω₂ = I₁ω₁/I₂. Angular momentum is conserved; angular velocity is not. The skater spins faster, but L is unchanged.

A rolling ball at the bottom of a ramp has both translational and rotational KE. Which is larger for a solid sphere?

Translational KE is Always Larger

K_trans = ½mv² and K_rot = ½Iω² = ½(cMR²)(v/R)² = ½cmv². So K_rot/K_trans = c. For a solid sphere (c = 2/5), K_rot = (2/5)K_trans — rotational KE is 40% of translational KE. Total = (7/5)K_trans.

Why do neutron stars spin so fast — some hundreds of times per second — when a normal star rotates only about once per month?

Conservation of Angular Momentum

When a massive star collapses into a neutron star, its radius shrinks from ~10⁶ km to ~10 km — a factor of ~10⁵. Since L = Iω = cMR²ω is conserved and R drops by 10⁵, ω must increase by ~(10⁵)² = 10¹⁰. A one-month rotation becomes milliseconds. The same physics as the figure skater, scaled to stellar dimensions.

Practice and Apply — Computational

Quick Reference: Rolling Motion and Angular Momentum

                        Rolling constraint: vcm = Rω   acm = Rα

                        Contact point velocity = zero

                        Total rolling KE: K = ½(1+c)mvcm²

                        c = Icm/(MR²)

                        Incline speed: v = √[2gh/(1+c)]

                        M and R cancel!

                        Angular momentum: L = Iω

                        Conservation: I₁ω₁ = I₂ω₂ (if τext = 0)

Problem 1 — Rolling Speed at Bottom of Incline

A solid cylinder of mass 3.0 kg and radius 0.10 m rolls without slipping from rest down an incline of height 0.80 m. Find: (a) the speed of the center of mass at the bottom, (b) the angular velocity at the bottom, (c) the translational and rotational kinetic energies.

Given: Solid cylinder → I_cm = ½MR², so c = 1/2. h = 0.80 m, starts from rest.

(a) Speed at bottom:

v_cm = √[2gh/(1+c)] = √[2(9.8)(0.80)/(1 + 0.5)] = √[15.68/1.5] = √10.45 = 3.23 m/s

(b) Angular velocity:

ω = v_cm/R = 3.23/0.10 = 32.3 rad/s

(c) Kinetic energies:

K_trans = ½mv_cm² = ½(3.0)(3.23)² = ½(3.0)(10.43) = 15.65 J

K_rot = ½I_cmω² = ½(½)(3.0)(0.01)(32.3)² = ½(0.015)(1043) = 7.82 J

Total = 23.47 J ≈ mgh = (3.0)(9.8)(0.80) = 23.52 J ✓ (rounding)

Note: K_rot/K_trans = c = 1/2, as predicted. The cylinder puts half as much energy into spinning as into translating.

Problem 2 — The Incline Race

A solid sphere, a solid cylinder, and a hollow ring all have mass 1.5 kg and radius 0.12 m. They are released simultaneously from rest at the top of a 1.2 m high incline. Find the speed of each at the bottom and rank them.

Using v = √[2gh/(1+c)] with g = 9.8 m/s², h = 1.2 m, so 2gh = 23.52 m²/s²:

Solid sphere (c = 2/5):

v = √[23.52/1.4] = √16.80 = 4.10 m/s

Solid cylinder (c = 1/2):

v = √[23.52/1.5] = √15.68 = 3.96 m/s

Hollow ring (c = 1):

v = √[23.52/2.0] = √11.76 = 3.43 m/s

Ranking: Sphere (4.10) > Cylinder (3.96) > Ring (3.43)

Compare to frictionless sliding block (no rotation, c = 0):

v = √[23.52/1.0] = √23.52 = 4.85 m/s — faster than all rolling objects, as expected.

The mass 1.5 kg and radius 0.12 m appeared nowhere in the final answers — they canceled. Replace with any values and the ranking is unchanged.

Problem 3 — Figure Skater: Conservation of Angular Momentum

A figure skater spinning at 2.0 rev/s has a moment of inertia of 4.0 kg·m² with arms extended. She pulls her arms in, reducing her moment of inertia to 1.5 kg·m². Find: (a) her new angular velocity, (b) her initial and final rotational KE, (c) where the extra KE comes from.

Convert: ω₁ = 2.0 rev/s × 2π = 4π rad/s

(a) Conservation of angular momentum (τ_ext = 0):

I₁ω₁ = I₂ω₂

ω₂ = I₁ω₁/I₂ = (4.0)(4π)/(1.5) = 16π/1.5 = 32π/3 ≈ 33.5 rad/s ≈ 5.33 rev/s

(b) Rotational kinetic energies:

K₁ = ½I₁ω₁² = ½(4.0)(4π)² = 2(16π²) = 32π² ≈ 316 J

K₂ = ½I₂ω₂² = ½(1.5)(32π/3)² = ½(1.5)(1024π²/9) = 85.3π² ≈ 842 J

(c) Source of extra kinetic energy:

ΔK = 842 − 316 = 526 J came from the skater's muscles. She did work against the centrifugal tendency of her arms as she pulled them inward. Angular momentum was conserved; energy was not — it was added by biological work.

Problem 4 — Stool + Person: Angular Momentum Transfer

A student (I_student = 3.0 kg·m²) sits on a frictionless rotating stool (I_stool = 0.50 kg·m²) initially at rest. She is handed a spinning bicycle wheel (I_wheel = 0.80 kg·m², ω_wheel = 20 rad/s) with its axle pointing upward. She then flips the wheel so its axle points downward. Find the angular velocity of the student + stool.

Before flipping: Total angular momentum (upward positive):

L_initial = I_wheelω_wheel + 0 = (0.80)(20) = +16 kg·m²/s (student + stool at rest)

After flipping: Wheel now spins at −20 rad/s (same speed, opposite direction):

L_wheel,final = (0.80)(−20) = −16 kg·m²/s

Conservation (no external torque):

L_initial = L_wheel,final + L_{student+stool}

+16 = −16 + (I_student + I_stool)ω

32 = (3.0 + 0.50)ω = 3.5ω

ω = 32/3.5 ≈ 9.14 rad/s (in the original wheel direction)

The student begins to rotate to compensate for the reversal of the wheel's angular momentum. This is the operating principle of reaction wheels used to orient spacecraft without thrusters.

Problem 5 — Rolling with Energy Conservation: Up a Ramp

A hollow sphere (I = ⅔MR²) rolls without slipping on a flat surface at 4.0 m/s and then encounters a ramp. How high does it rise before momentarily stopping? Compare to a solid sphere at the same speed.

Hollow sphere (c = 2/3):

Total KE at bottom: K = ½(1 + c)mv² = ½(1 + 2/3)(m)(4.0)² = ½(5/3)(m)(16) = (40m/3) J

At top: K = 0, PE = mgh

mgh = 40m/3 ⇒ h = 40/(3 × 9.8) = 1.36 m

Solid sphere (c = 2/5):

K = ½(1 + 2/5)(m)(16) = ½(7/5)(m)(16) = (56m/5) J

h = 56/(5 × 9.8) = 1.14 m

Comparison: The hollow sphere rises higher (1.36 m vs 1.14 m) at the same v because it has more rotational KE stored. At the top, both translational and rotational KE must go to zero — but the hollow sphere had more total KE to begin with for the same v_cm.

Ready to Move On?

Self-Check Before M5L4

M5L4 (Static Equilibrium) applies ΣF = 0 and Στ = 0 simultaneously. Make sure you can answer these before continuing:

State the rolling-without-slipping constraint. What are the two equivalent ways to write it?
A solid disk and a hollow ring roll from the same height. Without calculating, which is faster at the bottom? Why?
A star collapses to &frac{1}{10} its original radius. By what factor does its angular velocity change? (Assume I ∝ MR².)
A skater with I = 5.0 kg·m² spinning at 3.0 rad/s tucks in to reduce I to 2.0 kg·m². What is the new ω? Did KE increase, decrease, or stay the same?
Why does a rolling object always arrive later than a sliding (frictionless) object at the bottom of an incline of the same height?

M5L4 Static Equilibrium is the lesson where torque calculation becomes the central computational skill. The pivot-point strategy you will learn there relies on your fluency with τ = rF sinθ from M5L2.