Rotational Dynamics and Angular Momentum

🌀 Angular Momentum

Part 1 of 7 — $L = I\omega$

Just as linear momentum $p = mv$ describes the "quantity of motion" in a straight line, angular momentum $L = I\omega$ describes the "quantity of rotational motion."

Defining Angular Momentum

For a rigid body rotating about a fixed axis:

$L = I\omega$

Where:

• $L$ = angular momentum $(kg\cdot m^{2}/s)$
• $I$ = rotational inertia $(kg\cdot m^{2})$
• $\omega$ = angular velocity (rad/s)

For a Point Mass

A particle of mass $m$ moving in a circle of radius $r$:

$L = mvr$

(since $I = mr^2$ and $\omega = v/r$, so $L = mr^2 \cdot v/r = mvr$)

Direction and Sign

Like torque, angular momentum follows a sign convention:

• CCW rotation → $L > 0$ (positive)
• CW rotation → $L < 0$ (negative)

Units

$[L] = \text{kg}\cdot\text{m}^2\text{/s}$

Linear-Rotational Analogies

Key Insight

Angular momentum is large when:

• The object has a large rotational inertia (lots of mass far from axis)
• The object spins fast (large $\omega$)

A massive, slowly spinning flywheel can have the same $L$ as a tiny, rapidly spinning top.

Angular Momentum Quiz 🎯

Angular Momentum Calculations 🧮

1. A solid disk ($M = 4$ kg, $R = 0.5$ m) spins at $\omega = 10$ rad/s. What is its angular momentum? $(in kg\cdot m^{2}/s)$

2. A particle of mass 0.5 kg moves at 8 m/s in a circle of radius 2 m. What is its angular momentum? $(in kg\cdot m^{2}/s)$

3. A flywheel has $L = 200$ $kg\cdot m^{2}/s$ and $I = 25$ $kg\cdot m^{2}$. What is its angular velocity? (in rad/s)

Angular Momentum Concepts 🔍

Exit Quiz — Angular Momentum Basics ✅

🔄 Newton's Second Law for Rotation

Part 2 of 7 — $\tau_{\text{net}} = I\alpha$

Newton's Second Law $F = ma$ has a rotational analogue: net torque equals rotational inertia times angular acceleration.

The Rotational Second Law

$\tau_{\text{net}} = I\alpha$

This is the most important equation in rotational dynamics. It tells us:

• A net torque causes angular acceleration
• More rotational inertia means less angular acceleration for the same torque
• The angular acceleration is in the same direction as the net torque

Equivalent Form

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

Net torque equals the rate of change of angular momentum — the direct analogue of $F = dp/dt$.

When $\tau_{\text{net}} = 0$:

$\frac{\Delta L}{\Delta t} = 0 \Rightarrow L = \text{constant}$

No net torque → angular momentum is conserved!

Applying $\tau = I\alpha$

Example 1: Spinning a Wheel

A solid disk ($M = 5$ kg, $R = 0.4$ m) has a tangential force of $20$ N applied at its rim.

• $I = \frac{1}{2}MR^2 = \frac{1}{2}(5)(0.16) = 0.4$ $kg\cdot m^{2}$
• $\tau = FR = (20)(0.4) = 8$ N·m
• $\alpha = \tau/I = 8/0.4 = 20$ $rad/s^{2}$

Example 2: Pulley Problem

A mass $m$ hangs from a string wrapped around a pulley (mass $M$, radius $R$, solid disk). The tension in the string provides the torque:

$\tau = TR = I\alpha = \frac{1}{2}MR^2 \cdot \frac{a}{R}$

Combined with $mg - T = ma$, you can solve for both $a$ and $T$.

Rotational Newton's Second Law Quiz 🎯

Rotational Dynamics Calculations 🧮

1. A solid cylinder ($M = 8$ kg, $R = 0.25$ m) has a net torque of 5 N·m applied. What is $\alpha$? $(in rad/s^{2})$

2. A wheel ($I = 2$ $kg\cdot m^{2}$) starts from rest and a constant torque of 6 N·m is applied for 4 seconds. What is the final angular velocity? (in rad/s)

3. A disk ($I = 0.5$ $kg\cdot m^{2}$) decelerates from 40 rad/s to rest in 8 seconds. What is the magnitude of the braking torque? (in N·m)

Round all answers to 3 significant figures.

Rotational Dynamics Review 🔍

Exit Quiz — Rotational Second Law ✅

⚡ Rotational Kinetic Energy

Part 3 of 7 — $KE_{\text{rot}} = \frac{1}{2}I\omega^2$

A spinning object has kinetic energy due to its rotation — even if its center of mass isn't moving. This rotational kinetic energy follows the same pattern as translational KE.

Rotational Kinetic Energy

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

Compare with translational: $KE_{\text{trans}} = \frac{1}{2}mv^2$

Units

$KE_{\text{rot}}$ is measured in joules (J), just like any other form of energy.

Total Kinetic Energy for Rolling Objects

An object that both translates and rotates has:

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

For rolling without slipping ($v = R\omega$):

Work-Energy Theorem for Rotation

The work done by a torque:

$W = \tau \cdot \theta$

The work-energy theorem:

$W_{\text{net}} = \Delta KE_{\text{rot}} = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$

Power

The rotational power (rate of doing work):

$P = \tau\omega$

This is analogous to $P = Fv$ in linear motion.

Rotational KE Quiz 🎯

Rotational KE Calculations 🧮

1. A wheel ($I = 4$ $kg\cdot m^{2}$) spins at 10 rad/s. What is its rotational KE? (in J)

2. A solid sphere (mass 3 kg, radius 0.1 m) spins at 20 rad/s (not translating). What is its rotational KE? (in J, round to 3 significant figures)

3. A torque of 8 N·m acts through an angle of 25 rad on a wheel. How much work is done? (in J)

Energy Concepts 🔍

Exit Quiz — Rotational KE ✅

🔒 Conservation of Angular Momentum

Part 4 of 7 — No External Torque → $L$ is Conserved

Just as linear momentum is conserved when there is no external force, angular momentum is conserved when there is no external torque.

The Conservation Law

If $\tau_{\text{net, ext}} = 0$, then:

$L_i = L_f$ $I_i \omega_i = I_f \omega_f$

What Counts as "No External Torque"?

External torque is zero when:

• No external forces act on the system
• External forces act at the axis of rotation ($r = 0$)
• External forces are parallel to the axis

Key Consequence

If $I$ decreases → $\omega$ must increase (and vice versa) to keep $L$ constant.

$\omega_f = \frac{I_i}{I_f} \omega_i$

Important Distinction

Angular momentum is conserved, but rotational kinetic energy is generally NOT conserved when $I$ changes:

$KE_f = \frac{1}{2}I_f\omega_f^2 = \frac{I_i}{I_f} \times \frac{1}{2}I_i\omega_i^2 = \frac{I_i}{I_f} KE_i$

If $I$ decreases, $KE$ increases — the energy comes from internal work (muscles, etc.).

Rotational Collisions

When two rotating objects interact (e.g., a disk drops onto a turntable), angular momentum is conserved:

$I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega_f$

Example

A disk ($I_1 = 2$ $kg\cdot m^{2}$, $\omega_1 = 10$ rad/s) has a ring ($I_2 = 3$ $kg\cdot m^{2}$, initially at rest) dropped on top:

$2(10) + 3(0) = (2 + 3)\omega_f$ $\omega_f = 20/5 = 4 \text{ rad/s}$

Note: KE is NOT conserved (this is an inelastic rotational "collision").

Conservation Quiz 🎯

Conservation Calculations 🧮

1. A turntable ($I = 0.5$ $kg\cdot m^{2}$) spins at 8 rad/s. A 2 kg block of clay ($I = mr^2$, $r = 0.3$ m) is dropped on it. What is the final $\omega$? (in rad/s, round to 3 significant figures)

2. A skater with $I = 4$ $kg\cdot m^{2}$ and $\omega = 6$ rad/s pulls in her arms to $I = 1.5$ $kg\cdot m^{2}$. What is her new $\omega$? (in rad/s)

3. In problem 2, by what factor does her KE increase? (round to 3 significant figures)

Conservation Concepts 🔍

Exit Quiz — Conservation of Angular Momentum ✅

⭐ Figure Skater & Collapsing Star Examples

Part 5 of 7 — Conservation in Action

The conservation of angular momentum produces some of nature's most dramatic phenomena — from figure skaters spinning faster to neutron stars rotating hundreds of times per second.

The Figure Skater

A figure skater begins a spin with arms extended:

• $I_i = 4.0$ $kg\cdot m^{2}$, $\omega_i = 3$ rad/s
• $L = I_i\omega_i = 12$ $kg\cdot m^{2}/s$

She pulls her arms in:

• $I_f = 1.2$ $kg\cdot m^{2}$
• $\omega_f = L/I_f = 12/1.2 = 10$ rad/s

Speed increase: $\omega_f/\omega_i = 10/3 \approx 3.3\times$ faster!

Energy Analysis

• $KE_i = \frac{1}{2}(4.0)(9) = 18$ J
• $KE_f = \frac{1}{2}(1.2)(100) = 60$ J
• Energy increase: $60 - 18 = 42$ J

Where does the extra 42 J come from? Internal work by the skater's muscles pulling her arms inward against the centripetal acceleration.

The Collapsing Star

When a massive star runs out of fuel, its core collapses from roughly the size of the Sun ($R \sim 7 \times 10^8$ m) to a neutron star ($R \sim 10^4$ m).

Before collapse

• $R_i = 7 \times 10^8$ m, rotation period $T_i \approx 30$ days

After collapse

• $R_f = 10^4$ m
• $I \propto MR^2$, so $I_f/I_i = (R_f/R_i)^2 = (10^4/7 \times 10^8)^2 \approx 2 \times 10^{-10}$

By conservation: $\omega_f = (I_i/I_f)\omega_i$

$\omega_f \approx 5 \times 10^9 \times \omega_i$

The period goes from ~30 days to milliseconds! This explains why pulsars (rotating neutron stars) spin incredibly fast.

Other Examples

• Helicopter tail rotor: prevents the body from spinning (reaction to main rotor torque)
• Cat righting reflex: cats change their body shape mid-air to reorient
• Diver's tuck: pulling into a tuck position reduces $I$, increasing spin rate

Real-World Angular Momentum Quiz 🎯

Application Calculations 🧮

1. A diver ($I = 14$ $kg\cdot m^{2}$ extended) rotates at 2 rad/s. She tucks to $I = 3.5$ $kg\cdot m^{2}$. What is her angular velocity while tucked? (in rad/s)

2. What is the ratio of her tucked KE to her extended KE?

3. A merry-go-round ($I = 800$ $kg\cdot m^{2}$, $\omega = 2$ rad/s) has a 40 kg child ($r = 2$ m from center) jump off tangentially. What is the new $\omega$? (in rad/s, round to 3 significant figures)

Real-World Review 🔍

Exit Quiz — Conservation Examples ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Angular Momentum Practice

Time to work through challenging problems involving angular momentum, rotational dynamics, and energy.

Problem-Solving Strategy

1. Identify the system and check for external torques
2. If no external torque → use conservation of $L$: $I_i\omega_i = I_f\omega_f$
3. If external torque exists → use $\tau = I\alpha$ or $\tau = \Delta L/\Delta t$
4. For energy questions → compute $KE = \frac{1}{2}I\omega^2$ before and after
5. For rolling problems → remember $v = R\omega$ and total $KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Common AP Scenarios

• Object drops onto rotating platform (inelastic collision)
• Person walks on turntable
• Object changes shape while spinning
• Atwood machine with massive pulley

Workshop Problems — Set 1 🎯

Workshop Calculations 🧮

1. A turntable ($I = 1.2$ $kg\cdot m^{2}$, $\omega = 6$ rad/s) has a ring ($I = 0.8$ $kg\cdot m^{2}$) dropped on it from rest. Find the final $\omega$. (in rad/s, round to 3 significant figures)

2. How much kinetic energy is lost in the collision above? (in J, round to 3 significant figures)

3. A 60 kg person stands on the edge of a 200 kg, 3 m radius turntable (uniform disk) initially at rest. The person begins walking at 1.5 m/s tangentially (relative to the ground). What is the turntable's angular velocity? (in rad/s, round to 3 significant figures)

Strategy Check 🔍

Exit Quiz — Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Angular Momentum

Let's synthesize everything about angular momentum and tackle AP-level questions.

Complete Summary

Angular Momentum

$L = I\omega \quad \text{(rigid body)} \qquad L = mvr \quad \text{(point mass)}$

Newton's Second Law (Rotational)

$\tau_{\text{net}} = I\alpha = \frac{\Delta L}{\Delta t}$

Rotational Kinetic Energy

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

Conservation of Angular Momentum

$\text{If } \tau_{\text{net, ext}} = 0: \quad I_i\omega_i = I_f\omega_f$

Key Relationships

• $L$ conserved ↔ no external torque
• When $I$ decreases → $\omega$ increases → $KE$ increases (internal work done)
• Rotational "collisions": $I_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega_f$

AP-Style Questions — Set 1 🎯

AP Calculation Practice 🧮

1. A solid cylinder ($M = 10$ kg, $R = 0.2$ m) starts from rest and a constant torque of 4 N·m is applied. What is its angular momentum after 5 seconds? $(in kg\cdot m^{2}/s)$

2. A hoop (mass 2 kg, radius 0.5 m) rolls without slipping at 3 m/s. What is its total kinetic energy? (in J)

3. A child ($m = 30$ kg) runs at 4 m/s tangent to the edge of a stationary merry-go-round (uniform disk, $M = 100$ kg, $R = 2$ m) and jumps on. What is the final angular velocity? (in rad/s, round to 3 significant figures)

Comprehensive Review 🔍

Final AP Review ✅