Angular Momentum

Overview

Angular momentum is the rotational analog of linear momentum. It's a conserved quantity in isolated systems and is essential for understanding rotating systems.

Angular Momentum of a Particle

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

Magnitude:

$L = rmv\sin\theta = mvr_\perp$

Where $r_\perp$ is the perpendicular distance from the axis to the line of motion.

Units

• kg·m²/s

Angular Momentum of a Rigid Body

For rotation about a fixed axis:

$L = I\omega$

Where:

• $I$ = moment of inertia about the rotation axis
• $\omega$ = angular velocity

Relationship with Torque

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

This is the rotational analog of $\vec{F} = \frac{d\vec{p}}{dt}$

For a rigid body:

$\tau = I\alpha = I\frac{d\omega}{dt}$

Conservation of Angular Momentum

When no external torque acts on a system:

$L_{\text{initial}} = L_{\text{final}}$

$I_1\omega_1 = I_2\omega_2$

Conditions for Conservation

• No external torques
• Internal forces/torques only
• System is isolated rotationally

Rotational Kinetic Energy

$KE_{\text{rot}} = \frac{1}{2}I\omega^2$

For rolling objects (rotation + translation):

$KE_{\text{total}} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$

With rolling condition $v = r\omega$:

$KE_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{r}\right)^2 = \frac{1}{2}v^2\left(m + \frac{I}{r^2}\right)$

Work-Energy in Rotation

Work done by torque:

$W = \tau\theta = \int\tau \, d\theta$

Power:

$P = \tau\omega$

Examples

Example 1: Figure Skater

A skater with arms extended ($I = 4$ kg·m²) spins at 2 rad/s. She pulls her arms in ($I = 1.5$ kg·m²). Find new angular velocity.

$I_1\omega_1 = I_2\omega_2$

$4 \times 2 = 1.5 \times \omega_2$

$\omega_2 = \frac{8}{1.5} = 5.33 \text{ rad/s}$

Example 2: Merry-Go-Round

A child (30 kg) jumps onto a stationary merry-go-round ($I = 200$ kg·m², $r = 2$ m) at 3 m/s tangentially. Find angular velocity.

Initial angular momentum:

$L_i = mvr = 30 \times 3 \times 2 = 180 \text{ kg·m}^2\text{/s}$

Final moment of inertia:

$I_f = 200 + 30 \times 2^2 = 320 \text{ kg·m}^2$

Final angular velocity:

$\omega_f = \frac{L_i}{I_f} = \frac{180}{320} = 0.56 \text{ rad/s}$

Example 3: Rolling Down Incline

A solid sphere ($I = \frac{2}{5}MR^2$) rolls down a height $h$. Find velocity at bottom.

Using energy conservation:

$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2 + \frac{1}{2}\left(\frac{2}{5}MR^2\right)\left(\frac{v}{R}\right)^2$

$Mgh = \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{7}{10}Mv^2$

$v = \sqrt{\frac{10gh}{7}}$

Compare to sliding (no rotation): $v = \sqrt{2gh}$

Example 4: Kinetic Energy Comparison

Find final KE in Example 1.

$KE_i = \frac{1}{2}I_1\omega_1^2 = \frac{1}{2}(4)(2)^2 = 8 \text{ J}$

$KE_f = \frac{1}{2}I_2\omega_2^2 = \frac{1}{2}(1.5)(5.33)^2 = 21.3 \text{ J}$

KE increased! (Work done by skater pulling arms in)

Example 5: Angular Momentum of Orbiting Object

A satellite (1000 kg) orbits Earth at $r = 7000$ km with $v = 7.5$ km/s.

$L = mvr = 1000 \times 7500 \times 7 \times 10^6 = 5.25 \times 10^{13} \text{ kg·m}^2\text{/s}$

Rolling Objects Comparison

For objects rolling down an incline from height $h$:

Object	$I/MR^2$	$v$ at bottom
Sliding block	0	$\sqrt{2gh}$
Hollow sphere	2/3	$\sqrt{6gh/5}$
Solid sphere	2/5	$\sqrt{10gh/7}$
Hollow cylinder	1	$\sqrt{gh}$
Solid cylinder	1/2	$\sqrt{4gh/3}$

Solid sphere reaches bottom first (lowest $I/MR^2$ ratio)

Vector Nature

• Angular momentum is a vector: $\vec{L} = I\vec{\omega}$
• Direction given by right-hand rule
• For complex systems, components must be considered separately