§9.5 — Angular Momentum

Until now we have studied linear motion of objects. But many objects rotate. Car wheels, soccer balls, planets, and even electrons have rotational motion. Angular momentum is the rotational analog of linear momentum, and its conservation law is equally powerful.

Definition of angular momentum

For a particle at position \( \mathbf{r} \) with linear momentum \( \mathbf{p} \), the angular momentum about a point (e.g., the origin) is:

\[ \mathbf{L} = \mathbf{r} \times \mathbf{p} \]

or:

\[ \mathbf{L} = \mathbf{r} \times m\mathbf{v} \]

This is a cross product with both magnitude and direction.

Magnitude:

\[ L = r p \sin\theta = mrv\sin\theta \]

where \( \theta \) is the angle between \( \mathbf{r} \) and \( \mathbf{v} \).

Unit: kilogram-meter-squared-per-second (kg·m²/s).

Linear ⟷ Rotational analogs

The striking parallels between linear and rotational motion:

Linear	Rotational
Position \( x \)	Angle \( \theta \)
Velocity \( v \)	Angular velocity \( \omega \)
Momentum \( p = mv \)	Angular momentum \( L = I\omega \)
Force \( F \)	Torque \( \tau \)
\( F = \frac{dp}{dt} \)	\( \tau = \frac{dL}{dt} \)
Conservation of momentum	Conservation of angular momentum

Torque and angular momentum

Torque is the rotational analog of force:

\[ \boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} \]

Magnitude:

\[ \tau = rF\sin\phi \]

where \( \phi \) is the angle between the lever arm \( \mathbf{r} \) and force \( \mathbf{F} \).

Importance of lever arm: The longer the lever arm, the greater the rotational effect!

Relation to angular momentum:

\[ \boldsymbol{\tau} = \frac{d\mathbf{L}}{dt} \]

This is Newton's second law for rotation: torque = rate of change of angular momentum.

Example 1: Stone on a string

A stone (point mass) tied to a string spinning in a circle:

• Mass: \( m = 0.1 \) kg
• Circle radius: \( r = 2 \) m
• Linear speed: \( v = 3 \) m/s

Question: Angular momentum about the center?

Solution:

Velocity and position are perpendicular (\( \sin\theta = 1 \)):

\[ L = mrv = 0.1 \times 2 \times 3 = 0.6 \text{ kg·m}^2\text{/s} \]

Example 2: Door and hinges

A door attached to hinges (rotation axis) on one side.

You push perpendicular to the door at distance \( r = 1 \) m from the hinge.

• Force: \( F = 50 \) N

You push the same force at distance \( r = 0.1 \) m from the hinge. How differently does the door respond?

Solution:

Case 1 (hand at far end): \[ \tau_1 = rF = 1 \times 50 = 50 \text{ N·m} \]

Case 2 (hand near hinge): \[ \tau_2 = 0.1 \times 50 = 5 \text{ N·m} \]

In the first case, the torque is 10 times larger! This is why doors have handles at the far end.

Conservation of angular momentum

If external torque is zero (\( \boldsymbol{\tau}_{\text{ext}} = 0 \)), then:

\[ \frac{d\mathbf{L}}{dt} = 0 \implies \mathbf{L} = \text{constant} \]

Angular momentum is conserved.

Example 3: Figure skater

A figure skater spinning on ice:

• Arms outstretched: angular momentum \( L_1 = I_1 \omega_1 \)
• Arms tucked: angular momentum \( L_2 = I_2 \omega_2 \)

No external torque (frictionless ice), so \( L \) is constant:

\[ I_1 \omega_1 = I_2 \omega_2 \]

When you tuck your arms:

• Moment of inertia decreases: \( I_2 < I_1 \)
• Angular velocity increases: \( \omega_2 > \omega_1 \)

Numerical example: If \( I_2 = 0.5 I_1 \), then \( \omega_2 = 2 \omega_1 \) — spins twice as fast!

Example 4: Neutron star

During stellar collapse:

• Radius decreases: \( R_i = 7 \times 10^8 \) m → \( R_f = 20 \) km = \( 2 \times 10^4 \) m
• Initial angular velocity: \( \omega_i = 1 \) rev/sec

Question: Final angular velocity?

Solution:

Assume mass and spherical shape are conserved. Moment of inertia for a sphere:

\[ I = \frac{2}{5}MR^2 \]

From conservation of \( L \):

\[ I_i \omega_i = I_f \omega_f \]

\[ \frac{2}{5}M R_i^2 \omega_i = \frac{2}{5}M R_f^2 \omega_f \]

\[ R_i^2 \omega_i = R_f^2 \omega_f \]

\[ \omega_f = \omega_i \left(\frac{R_i}{R_f}\right)^2 = 1 \times \left(\frac{7 \times 10^8}{2 \times 10^4}\right)^2 \]

\[ \omega_f = (3.5 \times 10^4)^2 \approx 1.2 \times 10^9 \text{ rev/s} \]

1.2 billion revolutions per second! Neutron stars spin incredibly fast.

Torque in everyday life

1. Doors and knives: Long lever arm = easy rotation.

2. Bicycles: When riding fast, small torques preserve rotational motion through angular momentum.

3. Acrobatics: When airborne (jumping/diving), zero external torque allows controlled body rotation.

4. Binary stars: Two stars orbit each other, but total angular momentum is conserved.

What you should be able to do

• Calculate angular momentum: \( \mathbf{L} = \mathbf{r} \times m\mathbf{v} \)
• Calculate torque: \( \boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} \)
• Apply the rotational second law: \( \boldsymbol{\tau} = \frac{d\mathbf{L}}{dt} \)
• Recognize that zero external torque means conserved angular momentum
• Draw parallels between linear and rotational quantities

Preview of §9.6

So far we have looked at angular momentum of a particle. But what about rigid bodies? For a wheel or cylinder, we must sum the angular momentum of all particles — this is the moment of inertia.

📚 See also: Halliday Vol 1, Ch 9, §9.5.

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