🔄 Rotational Dynamics and Angular Momentum

Moment of Inertia

Moment of inertia $I$ is the rotational equivalent of mass. It measures resistance to rotational acceleration.

$I = \sum m_i r_i^2$

For continuous objects: $I = \int r^2 \, dm$

where:

• $I$ = moment of inertia (kg·m²)
• $m_i$ = mass of particle $i$
• $r_i$ = distance of particle $i$ from axis of rotation

💡 Key Idea: Moment of inertia depends on both mass AND how that mass is distributed relative to the axis. Same object can have different moments of inertia for different axes!

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Common Moments of Inertia

| Object | Axis | Moment of Inertia | |--------|------|-------------------| | Point mass | Distance $r$ from axis | $I = mr^2$ | | Thin rod | Through center, perpendicular | $I = \frac{1}{12}mL^2$ | | Thin rod | Through end, perpendicular | $I = \frac{1}{3}mL^2$ | | Solid disk/cylinder | Through center, along axis | $I = \frac{1}{2}mr^2$ | | Hollow cylinder (thin) | Through center, along axis | $I = mr^2$ | | Solid sphere | Through center | $I = \frac{2}{5}mr^2$ | | Hollow sphere (thin) | Through center | $I = \frac{2}{3}mr^2$ |

Note: These formulas are given on AP Physics 1 formula sheet!

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Properties of Moment of Inertia

Depends on Axis

Same object, different axis → different $I$

Example: Rod

• Center axis: $I = \frac{1}{12}mL^2$
• End axis: $I = \frac{1}{3}mL^2$ (larger!)

Depends on Mass Distribution

• Mass farther from axis → larger $I$
• Mass closer to axis → smaller $I$

Why figure skaters spin faster with arms in: Pulling arms in decreases $I$, so $\omega$ increases to conserve angular momentum!

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Newton's Second Law for Rotation

$\tau_{net} = I\alpha$

where:

• $\tau_{net}$ = net torque (N·m)
• $I$ = moment of inertia (kg·m²)
• $\alpha$ = angular acceleration (rad/s²)

Analogy: $F = ma$ for linear motion

Interpretation:

• Larger torque → larger angular acceleration
• Larger moment of inertia → harder to accelerate rotationally

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Rotational Kinetic Energy

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

Analogy: $KE = \frac{1}{2}mv^2$ for linear motion

Total kinetic energy of rolling object: $KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2$

For rolling without slipping ($v_{cm} = r\omega$): $KE_{total} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\left(\frac{v_{cm}}{r}\right)^2$

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Angular Momentum

Angular momentum $L$ is the rotational equivalent of linear momentum:

$L = I\omega$

where:

• $L$ = angular momentum (kg·m²/s)
• $I$ = moment of inertia (kg·m²)
• $\omega$ = angular velocity (rad/s)

Analogy: $p = mv$ for linear motion

For Point Particle

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

Magnitude: $L = mvr\sin\theta$

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Conservation of Angular Momentum

In an isolated system (no external torques):

$L_{initial} = L_{final}$

$I_i \omega_i = I_f \omega_f$

💡 Fundamental Law: Like linear momentum, angular momentum is conserved when no external torques act on the system.

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Applications of Conservation

Figure Skating Spins

• Arms out: large $I$, small $\omega$
• Arms in: small $I$, large $\omega$
• $I_i \omega_i = I_f \omega_f$

If $I$ decreases by factor of 3, then $\omega$ increases by factor of 3!

Diving

• Tucked position: small $I$ → fast rotation
• Layout position: large $I$ → slow rotation
• Controls rotation rate mid-air

Planetary Orbits

• Closer to sun: smaller $r$, larger $v$
• Farther from sun: larger $r$, smaller $v$
• Kepler's 2nd Law (equal areas in equal times)

Gyroscopes

• Large angular momentum resists changes in orientation
• Used in navigation, stabilization

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Relationship: Torque and Angular Momentum

$\tau_{net} = \frac{dL}{dt}$

Analogy: $F = \frac{dp}{dt}$ (Newton's 2nd Law)

If no external torque: $\frac{dL}{dt} = 0$ → $L$ is constant ✓

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Problem-Solving Strategy

For Rotational Dynamics:

1. Identify moment of inertia $I$ (use formula or calculate)
2. Find net torque: $\tau_{net} = \sum \tau_i$
3. Apply: $\tau_{net} = I\alpha$
4. Solve for unknown ($\tau$, $I$, or $\alpha$)
5. Use kinematics if needed (to find $\omega$, $\theta$, etc.)

For Angular Momentum Conservation:

1. Check: Is system isolated? (No external torques?)
2. Write initial state: $L_i = I_i \omega_i$
3. Write final state: $L_f = I_f \omega_f$
4. Set equal: $I_i \omega_i = I_f \omega_f$
5. Solve for unknown

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⚠️ Common Mistakes

Mistake 1: Forgetting $I$ Depends on Axis

Same object, different axis → different $I$!

Mistake 2: Using Wrong $I$ Formula

Check axis location and object shape carefully!

Mistake 3: Confusing $L = I\omega$ with $p = mv$

These are analogous but not the same!

• Angular momentum has different units (kg·m²/s)
• Depends on axis choice

Mistake 4: Assuming $I$ is Constant

In conservation problems, $I$ often changes (figure skater pulling arms in)!

Mistake 5: Forgetting Rolling Kinetic Energy

Rolling objects have BOTH translational and rotational KE!

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Energy in Rotational Motion

Work Done by Torque

$W = \tau \theta$

(when $\tau$ is constant)

Analogy: $W = Fd$ for linear motion

Work-Energy Theorem (Rotational)

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

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Rolling Motion

For object rolling without slipping down incline:

Energy conservation: $mgh = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2$

Using $v_{cm} = r\omega$ and $I = \beta mr^2$ (where $\beta$ is shape factor):

$gh = \frac{1}{2}v_{cm}^2\left(1 + \beta\right)$

$v_{cm} = \sqrt{\frac{2gh}{1 + \beta}}$

Different objects, different speeds:

• Hollow cylinder ($\beta = 1$): slowest
• Solid cylinder ($\beta = 1/2$): medium
• Solid sphere ($\beta = 2/5$): fastest

All faster than sliding (no friction) box!

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Comparing Linear and Rotational

| Linear | Rotational | Relationship | |--------|-----------|--------------| | Mass $m$ | Moment of inertia $I$ | $I = \sum mr^2$ | | Velocity $v$ | Angular velocity $\omega$ | $v = r\omega$ | | Acceleration $a$ | Angular acceleration $\alpha$ | $a_t = r\alpha$ | | Force $F$ | Torque $\tau$ | $\tau = rF\sin\theta$ | | $F = ma$ | $\tau = I\alpha$ | Newton's 2nd Law | | Momentum $p = mv$ | Angular momentum $L = I\omega$ | $L = r \times p$ | | $KE = \frac{1}{2}mv^2$ | $KE_{rot} = \frac{1}{2}I\omega^2$ | Kinetic energy | | $p$ conserved | $L$ conserved | No external force/torque |

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Real-World Applications

Spinning Top/Gyroscope

• Large $L$ → resists tipping
• Precession when external torque applied
• Used in navigation systems

Bicycle Wheels

• Spinning wheels have angular momentum
• Harder to tip over when moving
• Gyroscopic stability

Earth's Rotation

• Huge moment of inertia
• Angular momentum conserved
• Day length nearly constant

Tornadoes and Hurricanes

• Air spiraling inward
• $r$ decreases → $\omega$ increases
• Conservation of angular momentum

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Key Formulas Summary

| Concept | Formula | Units | |---------|---------|-------| | Moment of inertia | $I = \sum mr^2$ | kg·m² | | Rotational 2nd Law | $\tau = I\alpha$ | N·m | | Rotational KE | $KE_{rot} = \frac{1}{2}I\omega^2$ | J | | Angular momentum | $L = I\omega$ | kg·m²/s | | Conservation | $I_i\omega_i = I_f\omega_f$ | (isolated) | | Torque-momentum | $\tau = \frac{dL}{dt}$ | N·m | | Work by torque | $W = \tau\theta$ | J |

Common $I$ values:

• Point mass: $mr^2$
• Disk: $\frac{1}{2}mr^2$
• Sphere: $\frac{2}{5}mr^2$
• Rod (center): $\frac{1}{12}mL^2$
• Rod (end): $\frac{1}{3}mL^2$