Angular Momentum

Angular Momentum of a Particle — $\vec{L} = \vec{r} \times \vec{p}$

Part 1 of 7

The angular momentum of a particle about a point $O$ is:

$\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v}$

where $\vec{r}$ is the position vector from $O$ to the particle.

Magnitude

$|\vec{L}| = rmv\sin\theta = mv r_{\perp}$

where:

• $\theta$ = angle between $\vec{r}$ and $\vec{v}$
• $r_{\perp} = r\sin\theta$ = perpendicular distance from $O$ to the line of motion (the moment arm)

Direction

Use the right-hand rule: curl fingers from $\vec{r}$ toward $\vec{p}$, thumb points in the direction of $\vec{L}$.

For 2D motion in the $xy$-plane, $\vec{L}$ points along $\hat{z}$:

• Counterclockwise → $\vec{L} = +L\hat{z}$
• Clockwise → $\vec{L} = -L\hat{z}$

Angular Momentum of a Particle Moving in a Straight Line

Even a particle in straight-line motion has angular momentum about any point not on its path!

For a particle moving at constant velocity $v$ passing at closest distance $d$ from point $O$:

$L = mvd = \text{constant}$

This is because $r\sin\theta = d$ remains constant as the particle moves.

Worked Example

A 0.5 kg ball moves at $10$ m/s along a line that passes $3$ m from the origin.

$L = mvd = 0.5 \times 10 \times 3 = 15 \text{ kg}\cdot\text{m}^2/\text{s}$

This remains constant because $\vec{\tau} = \vec{r} \times \vec{F} = 0$ (no force acts).

Angular Momentum in Circular Motion

For a particle of mass $m$ moving in a circle of radius $r$ at speed $v$:

$L = mvr$

Since $v = r\omega$:

$L = mr^2\omega$

For uniform circular motion, $\vec{v} \perp \vec{r}$, so $\sin\theta = 1$.

Cross Product in Components

For a particle at $(x, y, 0)$ with velocity $(v_x, v_y, 0)$:

$\vec{L} = m(\vec{r} \times \vec{v}) = m(xv_y - yv_x)\hat{z}$

This is the $z$-component: $L_z = m(xv_y - yv_x)$.

Summary

Concept	Expression
Definition	$\vec{L} = \vec{r} \times m\vec{v}$
Magnitude	$L = mvr\sin\theta = mvr_{\perp}$
Circular motion	$L = mvr = mr^2\omega$
Straight line	$L = mvd$ (constant)
Direction	Right-hand rule ($\vec{r}$ to $\vec{p}$)
Units	$kg\cdot m^{2}/s$

Next: Part 2 — Angular momentum of rigid bodies ($L = I\omega$).

Angular Momentum of Rigid Bodies — $L = I\omega$

Part 2 of 7

For a rigid body rotating about a fixed axis with angular velocity $\omega$:

$L = I\omega$

where $I$ is the moment of inertia about that axis:

$I = \sum m_i r_i^2 \quad \text{(discrete)} \qquad I = \int r^2\,dm \quad \text{(continuous)}$

Common Moments of Inertia

Object	Axis	$I$
Thin rod (center)	Perpendicular, center	$\frac{1}{12}ML^2$
Thin rod (end)	Perpendicular, end	$\frac{1}{3}ML^2$
Solid disk/cylinder	Central axis	$\frac{1}{2}MR^2$
Thin ring/hoop	Central axis	$MR^2$
Solid sphere	Through center	$\frac{2}{5}MR^2$
Hollow sphere	Through center	$\frac{2}{3}MR^2$

Parallel Axis Theorem

If $I_{\text{cm}}$ is known about the center of mass, the moment of inertia about any parallel axis at distance $d$ is:

$I = I_{\text{cm}} + Md^2$

Worked Example

A uniform rod of mass $M$ and length $L$ — find $I$ about one end.

$I_{\text{end}} = I_{\text{cm}} + M(L/2)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{3}ML^2 \quad \checkmark$

Angular Momentum About a Point vs. an Axis

For rotation about a fixed axis, $L$ along that axis is simply $I\omega$.

But the total angular momentum vector may not be parallel to $\omega$ unless the object is symmetric about the rotation axis. This leads to:

$\vec{L} = I\vec{\omega}$

only when $\vec{\omega}$ is along a principal axis of inertia. Otherwise, $\vec{L}$ precesses — we'll explore this in Part 5.

Combining Rotation and Translation

For a rigid body that translates and rotates (e.g., rolling), the total angular momentum about a fixed point $O$ has two contributions:

$\vec{L}_O = \underbrace{\vec{r}_{\text{cm}} \times M\vec{v}_{\text{cm}}}_{\text{orbital}} + \underbrace{I_{\text{cm}}\vec{\omega}}_{\text{spin}}$

Rolling Without Slipping

For a disk rolling without slipping at speed $v_{\text{cm}}$ ($v = R\omega$):

$L = Mv_{\text{cm}}R + \frac{1}{2}MR^2 \cdot \frac{v_{\text{cm}}}{R} = Mv_{\text{cm}}R + \frac{1}{2}Mv_{\text{cm}}R = \frac{3}{2}Mv_{\text{cm}}R$

about the contact point. (This also equals $I_{\text{contact}}\omega = \frac{3}{2}MR^2 \cdot \omega$.)

Using the Contact Point

For rolling without slipping, the instantaneous axis of rotation is the contact point. Taking torques about this point eliminates the friction force from the equation (since its moment arm is zero).

Summary

Concept	Expression
Rigid body $L$	$L = I\omega$
Parallel axis	$I = I_{\text{cm}} + Md^2$
Perpendicular axis (planar)	$I_z = I_x + I_y$
Rolling body $L$	$L = Mv_{\text{cm}}R + I_{\text{cm}}\omega$
Sphere rolling $L$	$\frac{7}{5}MvR$ (about contact)

Next: Part 3 — Torque and $d\vec{L}/dt$.

Torque and $d\vec{L}/dt$

Part 3 of 7

The rotational analog of Newton's second law:

$\vec{\tau}_{\text{net}} = \frac{d\vec{L}}{dt}$

This is the most general form — it holds even when $I$ changes with time.

Derivation

$\frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p}) = \underbrace{\dot{\vec{r}} \times \vec{p}}_{= \vec{v} \times m\vec{v} = 0} + \vec{r} \times \dot{\vec{p}} = \vec{r} \times \vec{F} = \vec{\tau}$

Special Cases

Condition	Result
Fixed axis, constant $I$	$\tau = I\alpha$
Fixed axis, varying $I$	$\tau = \frac{d}{dt}(I\omega)$
$\tau = 0$	$\vec{L} = \text{const}$ (conservation)

Angular Impulse

The angular analog of the impulse-momentum theorem:

$\vec{J}_{\text{angular}} = \int_{t_1}^{t_2} \vec{\tau}\,dt = \Delta\vec{L} = \vec{L}_f - \vec{L}_i$

For a constant torque: $\vec{J} = \vec{\tau}\,\Delta t$

Worked Example

A figure skater extends her arms (initial $I_1 = 4$ $kg\cdot m^{2}$, $\omega_1 = 2$ rad/s). She brings her arms in, changing her moment of inertia to $I_2 = 1.5$ $kg\cdot m^{2}$ over $0.5$ seconds.

If no external torque acts:

$L_1 = L_2 \implies I_1\omega_1 = I_2\omega_2$

$\omega_2 = \frac{4 \times 2}{1.5} = \frac{16}{3} \approx 5.33 \text{ rad/s}$

The average internal torque she exerts during the transition:

$\tau_{\text{avg}} = \frac{\Delta L}{\Delta t}$

But $\Delta L = 0$ for the system! The internal torque changes $I$ and $\omega$ while keeping $L$ constant.

Varying Moment of Inertia

When $I$ changes with time (e.g., a rod extending while spinning):

$\tau = \frac{d}{dt}(I\omega) = I\frac{d\omega}{dt} + \omega\frac{dI}{dt}$

This is not the same as $I\alpha$! The $\omega\,dI/dt$ term accounts for the redistribution of mass.

Example: Wrapping Rope

A disk ($I = \frac{1}{2}MR^2$) has a rope wound around it. The rope unwinds under a hanging mass $m$.

Torque: $\tau = mgR$ (ignoring rope mass)

$mgR = I\alpha = \frac{1}{2}MR^2 \alpha$

$\alpha = \frac{2mg}{MR}$

Linear acceleration of the hanging mass: $a = R\alpha = \frac{2mg}{M}$

Note: if $m \ll M$, then $a \approx 0$ (the heavy disk barely accelerates). If $M \ll m$, then $a \to \infty$ — but we'd need to account for the mass falling at $a = g$.

Correct treatment with tension $T$:

For hanging mass: $mg - T = ma$

For disk: $TR = \frac{1}{2}MR^2 \alpha$, $T = \frac{1}{2}Ma$

$mg - \frac{1}{2}Ma = ma \implies a = \frac{mg}{m + M/2}$

Summary

Concept	Expression
Newton's 2nd (rotation)	$\vec{\tau} = d\vec{L}/dt$
Constant $I$	$\tau = I\alpha$
Variable $I$	$\tau = I(d\omega/dt) + \omega(dI/dt)$
Angular impulse	$\int \tau\,dt = \Delta L$
Cross product	$\vec{\tau} = \vec{r} \times \vec{F}$

Next: Part 4 — Conservation of angular momentum.

Conservation of Angular Momentum

Part 4 of 7

If the net external torque on a system is zero:

$\vec{\tau}_{\text{ext}} = 0 \implies \frac{d\vec{L}}{dt} = 0 \implies \vec{L} = \text{constant}$

This is one of the most powerful conservation laws in physics!

When Is Angular Momentum Conserved?

Scenario	$\tau_{\text{ext}} = 0$?	$L$ conserved?
Central force ($\vec{F} \parallel \vec{r}$)	Yes	Yes
Gravity about Earth's center	Yes (for orbital $L$)	Yes
Ice skater pulling arms in	Yes (no external torque)	Yes
Collision with fixed pivot	About pivot, yes	$L$ about pivot conserved
Rolling down incline	No ($\tau_{\text{gravity}} \neq 0$)	No

Kinetic Energy Changes

When a skater pulls her arms in, $L$ is conserved but $K$ is NOT:

$K_1 = \frac{1}{2}I_1\omega_1^2, \quad K_2 = \frac{1}{2}I_2\omega_2^2$

Since $\omega_2 = (I_1/I_2)\omega_1$:

$K_2 = \frac{1}{2}I_2\left(\frac{I_1}{I_2}\omega_1\right)^2 = \frac{I_1^2\omega_1^2}{2I_2} = K_1 \cdot \frac{I_1}{I_2}$

Since $I_2 < I_1$: $K_2 > K_1$. Kinetic energy increases!

Where does this energy come from? From the internal work done by the skater's muscles as she pulls her arms inward against the centrifugal tendency.

Worked Example

$I_1 = 5$ $kg\cdot m^{2}$, $\omega_1 = 4$ rad/s, $I_2 = 2$ $kg\cdot m^{2}$:

$K_1 = \frac{1}{2}(5)(16) = 40 \text{ J}$

$K_2 = \frac{1}{2}(2)(100) = 100 \text{ J}$

The skater does $60$ J of internal work.

Angular Momentum in Collisions

Bullet-Rod Problem

A bullet of mass $m$ and speed $v$ hits the end of a rod (mass $M$, length $L$) that is pivoted at the other end. The bullet embeds in the rod.

Conserve angular momentum about the pivot (forces at the pivot exert zero torque about the pivot):

$L_i = mvL$

$L_f = \left(\frac{1}{3}ML^2 + mL^2\right)\omega$

$\omega = \frac{mvL}{\frac{1}{3}ML^2 + mL^2} = \frac{mv}{(\frac{M}{3}+m)L}$

Note: Linear momentum is NOT conserved (the pivot exerts an impulse). But angular momentum about the pivot IS conserved.

Why Choose the Pivot Point?

Forces at the pivot have zero moment arm → zero torque about the pivot. This makes angular momentum conserved about that specific point, even though transient impulsive forces act.

Kepler's Second Law from Angular Momentum

For a planet in orbit, gravity is a central force ($\vec{F} \parallel \vec{r}$), so $\vec{\tau} = \vec{r} \times \vec{F} = 0$.

$L = mr v_{\perp} = \text{const}$

The area swept per unit time:

$\frac{dA}{dt} = \frac{1}{2}r v_{\perp} = \frac{L}{2m} = \text{const}$

This is Kepler's second law: a planet sweeps out equal areas in equal times — a direct consequence of angular momentum conservation under a central force.

Summary

Concept	Key Result
Conservation condition	$\tau_{\text{ext}} = 0 \implies L = \text{const}$
Skater pulls arms in	$\omega$ increases, $K$ increases
$K$ change	$K = L^2/(2I)$ — inversely proportional to $I$
Collision with pivot	Conserve $L$ about pivot
Central forces	$\vec{\tau} = 0$ → Kepler's second law

Next: Part 5 — Precession and gyroscopes.

Precession and Gyroscopes

Part 5 of 7

When $\vec{\tau}$ is perpendicular to $\vec{L}$, the torque doesn't change the magnitude of $L$ — it changes its direction. This causes precession.

$\vec{\tau} = \frac{d\vec{L}}{dt}$

If $\vec{\tau} \perp \vec{L}$, then $d\vec{L} \perp \vec{L}$, meaning $\vec{L}$ rotates without changing magnitude.

Gyroscope Precession

A spinning gyroscope tilted at angle $\theta$ from vertical, with spin angular momentum $L = I\omega$. Gravity creates a torque:

$\tau = Mgr\sin\theta$

where $r$ is the distance from the pivot to the center of mass.

The precession angular velocity:

$\boxed{\Omega = \frac{\tau}{L\sin\theta} = \frac{Mgr}{I\omega}}$

Key Features

• Precession is slower when the spin is faster ($\Omega \propto 1/\omega$)
• The spin axis traces a cone around the vertical
• $\vec{L}$ traces a horizontal circle

Vector Analysis of Precession

Consider $\vec{L}$ making angle $\theta$ with the vertical. The horizontal component:

$L_{\text{horiz}} = L\sin\theta$

In time $dt$, the torque causes $\vec{L}$ to sweep through angle $d\phi$:

$|d\vec{L}| = L\sin\theta \cdot d\phi = \tau \cdot dt$

$\frac{d\phi}{dt} = \Omega = \frac{\tau}{L\sin\theta}$

For the gravitational torque $\tau = Mgr\sin\theta$:

$\Omega = \frac{Mgr\sin\theta}{I\omega\sin\theta} = \frac{Mgr}{I\omega}$

The $\sin\theta$ cancels — the precession rate is independent of the tilt angle!

Nutation

In reality, a released gyroscope also exhibits nutation — a rapid bobbing superimposed on the precession. This is a higher-order effect that damps out due to friction, leaving steady precession.

Applications of Precession

1. Bicycle Wheel Gyroscope

Hold a spinning bicycle wheel by one end of its axle. The wheel doesn't fall — it precesses around the vertical axis.

$\Omega = \frac{Mgr}{I\omega}$

2. Earth's Axial Precession

The Earth's rotation axis precesses due to the gravitational torque from the Sun and Moon on Earth's equatorial bulge.

• Period: ~26,000 years
• Current pole star: Polaris
• In ~13,000 years: Vega will be near the pole

3. Spinning Top

A toy top exhibits precession while spinning fast. As $\omega$ decreases due to friction:

• $\Omega$ increases (precesses faster)
• Eventually $\omega$ becomes too small to sustain gyroscopic stability
• The top wobbles and falls over

Worked Example

A disk of $M = 0.5$ kg, $R = 0.1$ m spins at $\omega = 100$ rad/s. It's mounted $r = 0.15$ m from the pivot.

$I = \frac{1}{2}MR^2 = \frac{1}{2}(0.5)(0.01) = 0.0025 \text{ kg}\cdot\text{m}^2$

$\Omega = \frac{(0.5)(9.8)(0.15)}{0.0025 \times 100} = \frac{0.735}{0.25} = 2.94 \text{ rad/s}$

Summary

Concept	Expression
Precession condition	$\vec{\tau} \perp \vec{L}$
Precession rate	$\Omega = Mgr/(I\omega)$
$\Omega$ vs $\omega$	Inversely proportional
$\Omega$ vs $\theta$	Independent
Nutation	Rapid bobbing; damps out
Gyroscopic stability	Large $L$ resists direction change

Next: Part 6 — Problem-solving workshop.

Angular Momentum — Problem-Solving Workshop

Part 6 of 7

Strategy Guide

Step	Action
1	Choose the reference point wisely (pivot eliminates unknown forces)
2	Determine if $\vec{\tau}_{\text{ext}} = 0$ about that point → conservation
3	For collisions: conserve $L$ about the impact point or pivot
4	For orbits: use $L = mrv_{\perp}$ and central force → $\tau = 0$
5	For rolling: combine orbital + spin angular momentum
6	Check: is KE also conserved? (elastic) or not? (inelastic)

Problem 1: Ball Hits Rod

A ball of mass $m$ moving at speed $v$ hits the end of a stationary rod of mass $M$ and length $L$ that is free to rotate about its center. The collision is perfectly elastic.

Conservation of angular momentum about the center:

$mvL/2 = \frac{1}{12}ML^2 \omega + mv'(L/2)$

Conservation of kinetic energy:

$\frac{1}{2}mv^2 = \frac{1}{2}\cdot\frac{1}{12}ML^2 \omega^2 + \frac{1}{2}mv'^2$

These two equations solve for $\omega$ and $v'$.

For the special case $m = M/3$:

$v' = \frac{m - M/3}{m + M/3}\cdot v \cdot \frac{L/2 \text{ factors}}{\cdots}$

The algebra is involved but the approach is systematic:

1. Write $L$-conservation about the pivot
2. Write $K$-conservation
3. Solve the two equations for two unknowns ($\omega$, $v'$)

Problem 2: Atwood + Pulley

An Atwood machine has masses $m_1 > m_2$ connected by a string over a pulley of mass $M$ and radius $R$ (solid disk). The string doesn't slip on the pulley.

Angular momentum approach (about pulley center):

$\tau_{\text{net}} = (m_1 - m_2)gR$

$I_{\text{total}} = \frac{1}{2}MR^2 + m_1 R^2 + m_2 R^2 = \left(\frac{M}{2} + m_1 + m_2\right)R^2$

$\alpha = \frac{\tau}{I} = \frac{(m_1 - m_2)g}{(M/2 + m_1 + m_2)R}$

$a = R\alpha = \frac{(m_1-m_2)g}{M/2 + m_1 + m_2}$

Compare to the massless pulley result: $a = (m_1-m_2)g/(m_1+m_2)$. The massive pulley adds $M/2$ to the effective inertia.

Workshop Takeaways

Problem Type	Key Approach
Object dropped on spinner	$L$ conserved; solve for $\omega_f$
Collision with pivot	Conserve $L$ about pivot
Massive pulley Atwood	Pulley adds $I/R^2$ to effective mass
Rod released from horizontal	Energy conservation with $I_{\text{end}} = ML^2/3$
Choosing reference point	Pick where unknown forces act

Next: Part 7 — Comprehensive review & applications.

Angular Momentum — Review & Applications

Part 7 of 7 — Comprehensive Assessment

Formula Reference

Concept	Expression
Particle	$\vec{L} = \vec{r} \times m\vec{v}$
Rigid body	$L = I\omega$
Newton's 2nd	$\vec{\tau} = d\vec{L}/dt$
Conservation	$\tau_{\text{ext}} = 0 \implies L = \text{const}$
Precession	$\Omega = Mgr/(I\omega)$
Rolling ($L$ about contact)	$L = (I_{\text{cm}} + MR^2)\omega$
Angular impulse	$\int \tau\,dt = \Delta L$
Kinetic energy	$K_{\text{rot}} = L^2/(2I)$

Common Moments of Inertia

Object	$I_{\text{cm}}$
Rod (center)	$ML^2/12$
Rod (end)	$ML^2/3$
Disk/cylinder	$MR^2/2$
Ring/hoop	$MR^2$
Solid sphere	$2MR^2/5$
Spherical shell	$2MR^2/3$

AP Free Response — Ballistic Pendulum with Rotation

A bullet ($m = 10$ g, $v_0 = 400$ m/s) strikes the bottom of a vertical rod ($M = 2$ kg, $L = 1$ m) pivoted at the top. The bullet embeds.

(a) Find $\omega$ just after impact.

Conserve $L$ about the pivot:

$mv_0 L = \left(\frac{1}{3}ML^2 + mL^2\right)\omega$

$0.01(400)(1) = \left(\frac{1}{3}(2)(1) + 0.01(1)\right)\omega$

$4 = (0.667 + 0.01)\omega = 0.677\omega$

$\omega = 5.91 \text{ rad/s}$

(b) Find the maximum angle the rod swings upward.

Energy conservation after impact:

$\frac{1}{2}I\omega^2 = (M + m)g \cdot d_{\text{cm}} \cdot (1 - \cos\theta)$

COM rises by $d_{\text{cm}}(1-\cos\theta)$ where $d_{\text{cm}}$ is the distance from pivot to the system's COM.

$d_{\text{cm}} = \frac{M(L/2) + mL}{M + m} = \frac{2(0.5) + 0.01(1)}{2.01} = \frac{1.01}{2.01} \approx 0.502 \text{ m}$

$\frac{1}{2}(0.677)(5.91)^2 = 2.01(9.8)(0.502)(1-\cos\theta)$

$11.82 = 9.89(1-\cos\theta) \implies \cos\theta = -0.195 \implies \theta \approx 101°$

The rod swings past horizontal!

🎉 Topic Complete — Angular Momentum

You've mastered:

Part	Topic	Status
1	$\vec{L} = \vec{r} \times \vec{p}$ (particles)	✅
2	$L = I\omega$ (rigid bodies)	✅
3	Torque and $d\vec{L}/dt$	✅
4	Conservation of angular momentum	✅
5	Precession and gyroscopes	✅
6	Problem-solving workshop	✅
7	Review & applications	✅

Key Insight: Angular momentum conservation is the rotational analog of linear momentum conservation, but with a crucial addition: you must choose your reference point wisely. Forces at the pivot have zero moment arm, making angular momentum conserved about that point even during violent collisions.