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!

Common Moments of Inertia

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

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!

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

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 ($v_{cm} = r\omega$): $KE_{total} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\left(\frac{v_{cm}}{r}\right)^2$

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: $L = mvr\sin\theta$

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.

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

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 ✓

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

⚠️ 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!

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)

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

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!

Comparing Linear and Rotational

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

Key Formulas Summary

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): $\frac{1}{3}mL^2$

(a) Find moment of inertia

Step 1: Use formula for solid disk

For a solid disk rotating about its center:

$I = \frac{1}{2}mr^2$

$I = \frac{1}{2}(2)(0.4)^2$

$I = \frac{1}{2}(2)(0.16)$

$I = 0.16 \text{ kg·m}^2$

Answer (a): Moment of inertia = 0.16 kg·m²

(b) Find angular acceleration

Step 2: Apply Newton's 2nd Law for rotation

$\tau = I\alpha$

$\alpha = \frac{\tau}{I} = \frac{3}{0.16}$

$\alpha = 18.75 \text{ rad/s}^2$

Answer (b): Angular acceleration = 18.75 rad/s²

(c) Find angular velocity after 5 seconds

Step 3: Use rotational kinematics

$\omega_f = \omega_i + \alpha t$

$\omega_f = 0 + (18.75)(5)$

$\omega_f = 93.75 \text{ rad/s}$

Answer (c): Angular velocity after 5 s = 93.75 rad/s (about 94 rad/s)

Check:

• In revolutions: $\frac{93.75}{2\pi} \approx 14.9$ rev/s
• Tangential speed at rim: $v = r\omega = 0.4(93.75) = 37.5$ m/s ✓