🎯⭐ INTERACTIVE LESSON

Rotational Dynamics and Angular Momentum

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Rotational Dynamics and Angular Momentum - Complete Interactive Lesson

Part 1: Angular Momentum (L = Iω)

🌀 Angular Momentum

Part 1 of 7 — L=IωL = I\omega

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

Defining Angular Momentum

For a rigid body rotating about a fixed axis:

L=IωL = I\omega

Where:

  • LL = angular momentum (kg·m²/s)
  • II = rotational inertia (kg·m²)
  • ω\omega = angular velocity (rad/s)

For a Point Mass

A particle of mass mm moving in a circle of radius rr:

L=mvrL = mvr

(since I=mr2I = mr^2 and ω=v/r\omega = v/r, so L=mr2v/r=mvrL = mr^2 \cdot v/r = mvr)

Direction and Sign

Like torque, angular momentum follows a sign convention:

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

Units

[L]=kg\cdotpm2/s[L] = \text{kg·m}^2\text{/s}

Linear-Rotational Analogies

LinearRotational
Mass mmRotational inertia II
Velocity vvAngular velocity ω\omega
Momentum p=mvp = mvAngular momentum L=IωL = I\omega
Force FFTorque τ\tau
F=maF = maτ=Iα\tau = I\alpha
F=dp/dtF = dp/dtτ=dL/dt\tau = dL/dt

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 ω\omega)

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

Angular Momentum Quiz 🎯

Angular Momentum Calculations 🧮

  1. A solid disk (M=4M = 4 kg, R=0.5R = 0.5 m) spins at ω=10\omega = 10 rad/s. What is its angular momentum? (in kg·m²/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·m²/s)

  3. A flywheel has L=200L = 200 kg·m²/s and I=25I = 25 kg·m². What is its angular velocity? (in rad/s)

Angular Momentum Concepts 🔍

Exit Quiz — Angular Momentum Basics

Part 2: Conservation of Angular Momentum

🔄 Newton's Second Law for Rotation

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

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

The Rotational Second Law

τnet=Iα\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

τnet=ΔLΔt\tau_{\text{net}} = \frac{\Delta L}{\Delta t}

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

When τnet=0\tau_{\text{net}} = 0:

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

No net torque → angular momentum is conserved!

Applying τ=Iα\tau = I\alpha

Example 1: Spinning a Wheel

A solid disk (M=5M = 5 kg, R=0.4R = 0.4 m) has a tangential force of 2020 N applied at its rim.

  • I=12MR2=12(5)(0.16)=0.4I = \frac{1}{2}MR^2 = \frac{1}{2}(5)(0.16) = 0.4 kg·m²
  • τ=FR=(20)(0.4)=8\tau = FR = (20)(0.4) = 8 N·m
  • α=τ/I=8/0.4=20\alpha = \tau/I = 8/0.4 = 20 rad/s²

Example 2: Pulley Problem

A mass mm hangs from a string wrapped around a pulley (mass MM, radius RR, solid disk). The tension in the string provides the torque:

τ=TR=Iα=12MR2aR\tau = TR = I\alpha = \frac{1}{2}MR^2 \cdot \frac{a}{R}

Combined with mgT=mamg - T = ma, you can solve for both aa and TT.

Rotational Newton's Second Law Quiz 🎯

Rotational Dynamics Calculations 🧮

  1. A solid cylinder (M=8M = 8 kg, R=0.25R = 0.25 m) has a net torque of 5 N·m applied. What is α\alpha? (in rad/s²)

  2. A wheel (I=2I = 2 kg·m²) 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=0.5I = 0.5 kg·m²) 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

Part 3: Ice Skater & Spinning Examples

⚡ Rotational Kinetic Energy

Part 3 of 7 — KErot=12Iω2KE_{\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

KErot=12Iω2KE_{\text{rot}} = \frac{1}{2}I\omega^2

Compare with translational: KEtrans=12mv2KE_{\text{trans}} = \frac{1}{2}mv^2

LinearRotational
12mv2\frac{1}{2}mv^212Iω2\frac{1}{2}I\omega^2

Units

KErotKE_{\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:

KEtotal=12mvcm2+12Iω2KE_{\text{total}} = \frac{1}{2}mv_{\text{cm}}^2 + \frac{1}{2}I\omega^2

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

ShapeKEtotalKE_{\text{total}}
Hoop12mv2+12mv2=mv2\frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2
Disk12mv2+14mv2=34mv2\frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2
Solid sphere12mv2+15mv2=710mv2\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2

Work-Energy Theorem for Rotation

The work done by a torque:

W=τθW = \tau \cdot \theta

The work-energy theorem:

Wnet=ΔKErot=12Iωf212Iωi2W_{\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=τωP = \tau\omega

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

Rotational KE Quiz 🎯

Rotational KE Calculations 🧮

  1. A wheel (I=4I = 4 kg·m²) 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

Part 4: Angular Impulse

🔒 Conservation of Angular Momentum

Part 4 of 7 — No External Torque → LL 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 τnet, ext=0\tau_{\text{net, ext}} = 0, then:

Li=LfL_i = L_f Iiωi=IfωfI_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=0r = 0)
  • External forces are parallel to the axis

Key Consequence

If II decreasesω\omega must increase (and vice versa) to keep LL constant.

ωf=IiIfωi\omega_f = \frac{I_i}{I_f} \omega_i

Important Distinction

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

KEf=12Ifωf2=IiIf×12Iiωi2=IiIfKEiKE_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 II decreases, KEKE 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:

I1ω1+I2ω2=(I1+I2)ωfI_1\omega_1 + I_2\omega_2 = (I_1 + I_2)\omega_f

Example

A disk (I1=2I_1 = 2 kg·m², ω1=10\omega_1 = 10 rad/s) has a ring (I2=3I_2 = 3 kg·m², initially at rest) dropped on top:

2(10)+3(0)=(2+3)ωf2(10) + 3(0) = (2 + 3)\omega_f ωf=20/5=4 rad/s\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=0.5I = 0.5 kg·m²) spins at 8 rad/s. A 2 kg block of clay (I=mr2I = mr^2, r=0.3r = 0.3 m) is dropped on it. What is the final ω\omega? (in rad/s, round to 3 significant figures)

  2. A skater with I=4I = 4 kg·m² and ω=6\omega = 6 rad/s pulls in her arms to I=1.5I = 1.5 kg·m². What is her new ω\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

Part 5: Rotational Kinetic Energy

⭐ 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:

  • Ii=4.0I_i = 4.0 kg·m², ωi=3\omega_i = 3 rad/s
  • L=Iiωi=12L = I_i\omega_i = 12 kg·m²/s

She pulls her arms in:

  • If=1.2I_f = 1.2 kg·m²
  • ωf=L/If=12/1.2=10\omega_f = L/I_f = 12/1.2 = 10 rad/s

Speed increase: ωf/ωi=10/33.3×\omega_f/\omega_i = 10/3 \approx 3.3\times faster!

Energy Analysis

  • KEi=12(4.0)(9)=18KE_i = \frac{1}{2}(4.0)(9) = 18 J
  • KEf=12(1.2)(100)=60KE_f = \frac{1}{2}(1.2)(100) = 60 J
  • Energy increase: 6018=4260 - 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 (R7×108R \sim 7 \times 10^8 m) to a neutron star (R104R \sim 10^4 m).

Before collapse

  • Ri=7×108R_i = 7 \times 10^8 m, rotation period Ti30T_i \approx 30 days

After collapse

  • Rf=104R_f = 10^4 m
  • IMR2I \propto MR^2, so If/Ii=(Rf/Ri)2=(104/7×108)22×1010I_f/I_i = (R_f/R_i)^2 = (10^4/7 \times 10^8)^2 \approx 2 \times 10^{-10}

By conservation: ωf=(Ii/If)ωi\omega_f = (I_i/I_f)\omega_i

ωf5×109×ωi\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 II, increasing spin rate

Real-World Angular Momentum Quiz 🎯

Application Calculations 🧮

  1. A diver (I=14I = 14 kg·m² extended) rotates at 2 rad/s. She tucks to I=3.5I = 3.5 kg·m². 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=800I = 800 kg·m², ω=2\omega = 2 rad/s) has a 40 kg child (r=2r = 2 m from center) jump off tangentially. What is the new ω\omega? (in rad/s, round to 3 significant figures)

Real-World Review 🔍

Exit Quiz — Conservation Examples

Part 6: Problem-Solving Workshop

🛠️ 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 LL: Iiωi=IfωfI_i\omega_i = I_f\omega_f
  3. If external torque exists → use τ=Iα\tau = I\alpha or τ=ΔL/Δt\tau = \Delta L/\Delta t
  4. For energy questions → compute KE=12Iω2KE = \frac{1}{2}I\omega^2 before and after
  5. For rolling problems → remember v=Rωv = R\omega and total KE=12mv2+12Iω2KE = \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=1.2I = 1.2 kg·m², ω=6\omega = 6 rad/s) has a ring (I=0.8I = 0.8 kg·m²) dropped on it from rest. Find the final ω\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

Part 7: Synthesis & AP Review

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

Newton's Second Law (Rotational)

τnet=Iα=ΔLΔt\tau_{\text{net}} = I\alpha = \frac{\Delta L}{\Delta t}

Rotational Kinetic Energy

KErot=12Iω2KE_{\text{rot}} = \frac{1}{2}I\omega^2

Conservation of Angular Momentum

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

Key Relationships

  • LL conserved ↔ no external torque
  • When II decreases → ω\omega increases → KEKE increases (internal work done)
  • Rotational "collisions": I1ω1+I2ω2=(I1+I2)ωfI_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=10M = 10 kg, R=0.2R = 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·m²/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=30m = 30 kg) runs at 4 m/s tangent to the edge of a stationary merry-go-round (uniform disk, M=100M = 100 kg, R=2R = 2 m) and jumps on. What is the final angular velocity? (in rad/s, round to 3 significant figures)

Comprehensive Review 🔍

Final AP Review