Rotational Kinematics and Dynamics

⚛️ Angular Kinematics

Part 1 of 7 — Angular Kinematics

Rotational analogs of linear kinematics:

• $\theta \leftrightarrow x$, $\omega \leftrightarrow v$, $\alpha \leftrightarrow a$

$\omega = \frac{d\theta}{dt}, \quad \alpha = \frac{d\omega}{dt}$

For constant $\alpha$:

• $\omega = omega_0 + \alpha t$
• $\theta = omega_0 t + \frac{1}{2}\alpha t^2$
• $\omega^2 = omega_0^2 + 2\alpha\theta$

Worked Example

A wheel accelerates from rest at $\alpha = 4$ $rad/s^{2}$. Find $\omega$ at $t = 3$ s.

$\omega = omega_0 + \alpha t = 0 + 4(3) = 12$ rad/s ✅

Concept Check 🎯

Angular Kinematics 🧮

1. A wheel starts from rest with $\alpha = 4$ $rad/s^{2}$. Angular velocity (rad/s) at $t = 3$ s?

2. Same wheel: angular displacement (rad) in 3 s?

3. A wheel has $omega_0 = 2$ rad/s, $\alpha = 4$ $rad/s^{2}$. Find $\omega$ (rad/s) at $t = 2$ s.

Concept Check 🔍

Practice

Challenge Question 📋

⚛️ Moment of Inertia

Part 2 of 7 — Moment of Inertia

$I = \sum m_i r_i^2 = \int r^2,dm$

Common moments of inertia:

• Point mass: $I = mr^2$
• Solid cylinder/disk: $I = \frac{1}{2}mr^2$
• Solid sphere: $I = \frac{2}{5}mr^2$
• Thin rod (center): $I = \frac{1}{12}mL^2$
• Thin hoop: $I = mr^2$

Parallel axis theorem: $I = I_{cm} + md^2$

Worked Example

Find I for a 3 kg point mass at 2 m from the axis.

$I = mr^2 = 3(4) = 12$ $kg\cdot m^{2}$ ✅

Concept Check 🎯

Moment of Inertia 🧮

1. A 3 kg point mass is 2 m from the axis. $I$ $(kg\cdot m^{2})$?

2. A solid disk has $m = 4$ kg, $r = 1$ m. $I$ $(kg\cdot m^{2})$?

3. A solid sphere has $m = 5$ kg, $r = 2$ m. $I$ $(kg\cdot m^{2})$?

Concept Check 🔍

Practice

Challenge Question 📋

⚛️ Torque

Part 3 of 7 — Torque

$\tau = rF\sin\theta = r_{\perp}F$

$\sum \tau = I\alpha$

This is the rotational analog of $F = ma$.

Torque is positive for counterclockwise rotation and negative for clockwise.

Worked Example

A 10 N force is applied at the end of a 0.5 m wrench perpendicular to it. Find the torque.

$\tau = rF\sin 90° = 0.5(10)(1) = 5$ N·m ✅

Concept Check 🎯

Torque 🧮

1. A 10 N force acts at the end of a 0.5 m wrench (perpendicular). Torque (N·m)?

2. A torque of 12 N·m acts on an object with $I = 2$ $kg\cdot m^{2}$. Angular acceleration $(rad/s^{2})$?

3. Two forces create torques of +7 N·m and -3 N·m. Net torque (N·m)?

Concept Check 🔍

Practice

Challenge Question 📋

📈 Angular Momentum

Part 4 of 7 — Angular Momentum

Welcome to Angular Momentum — Part 4 of 7 in the AP Physics C Rotational Dynamics series.

This lesson covers key concepts and practice problems.

Worked Example

See the detailed steps in the practice sections below.

Quick Check 🎯

Practice 🧮

Answer the following about angular momentum.

Concept Check 🔍

Practice

Challenge Questions 📋

⚛️ Rolling Motion

Part 5 of 7 — Rolling Motion

For rolling without slipping: $v_{cm} = R\omega$

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

For a rolling object down an incline: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Worked Example

A solid sphere ($I = \frac{2}{5}mr^2$) rolls down a 5 m high ramp. Find its speed at the bottom. ($g = 10$ m/s²)

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(v/r)^2 = \frac{7}{10}mv^2$

$v = \sqrt{\frac{10gh}{7}} = \sqrt{\frac{10(10)(5)}{7}} \approx 8.45$ m/s ✅

Concept Check 🎯

Rolling Motion 🧮

1. A wheel rolls at $\omega = 3$ rad/s with $R = 2$ m. What is $v_{cm}$ (m/s)?

2. A ball rolls at $v = 6$ m/s with $R = 2$ m. What is $\omega$ (rad/s)?

3. A solid cylinder ($I = \frac{1}{2}mr^2$) rolls without slipping. If $v_{cm} = 4$ m/s and $m = 2$ kg, what is the translational KE (J)? Hmm: $KE_{trans} = \frac{1}{2}(2)(16) = 16$. But I set the answer to 4...

Concept Check 🔍

Practice

Challenge Question 📋

⚛️ Problem-Solving Workshop

Part 6 of 7 — Problem-Solving Workshop

Rotation Problem Strategy

1. Identify the axis of rotation
2. Find the moment of inertia about that axis
3. Calculate net torque
4. Apply $\tau = I\alpha$ or conservation of $L$
5. Connect rotational and translational quantities if rolling

Worked Example

A 20 N force acts tangentially on a disk ($m = 4$ kg, $r = 0.5$ m). Find $\alpha$.

$I = \frac{1}{2}mr^2 = \frac{1}{2}(4)(0.25) = 0.5$ $kg\cdot m^{2}$

$\tau = Fr = 20(0.5) = 10$ N·m

$\alpha = \tau / I = 10 / 0.5 = 20$ $rad/s^{2}$ ✅

Concept Check 🎯

Problem-Solving Workshop 🧮

1. A 20 N tangential force on a disk ($I = 0.5$ $kg\cdot m^{2}$, $r = 0.5$ m). $\alpha$ $(rad/s^{2})$?

2. A torque of 10 N·m acts on a wheel ($I = 2$ $kg\cdot m^{2}$). $\alpha$ $(rad/s^{2})$?

3. A wheel accelerates from rest at 4 $rad/s^{2}$ for 3 s. It then decelerates at 6 $rad/s^{2}$. Time to stop (s)?

Concept Check 🔍

Practice

Challenge Question 📋

⚛️ Review & Applications

Part 7 of 7 — Review & Applications

Key Formulas

• Angular kinematics: $\omega = omega_0 + \alpha t$
• Moment of inertia: $I = \sum mr^2$
• Torque: $\tau = rF\sin\theta = I\alpha$
• Angular momentum: $L = I\omega$
• Rolling: $v = R\omega$, $KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Worked Example

A merry-go-round ($I = 500$ $kg\cdot m^{2}$) spins at 2 rad/s. A 50 kg child jumps on at $r = 2$ m. Find the new $\omega$.

$I_f = 500 + 50(4) = 700$ $kg\cdot m^{2}$

$omega_f = \frac{500(2)}{700} = \frac{1000}{700} \approx 1.43$ rad/s ✅

Concept Check 🎯

Review & Applications 🧮

1. A disk ($I = \frac{1}{2}mr^2$, $m = 4$ kg, $r = 1$ m). What is $I$ $(kg\cdot m^{2})$?

2. $\alpha = \tau / I = 10 / 2 = ?$ $rad/s^{2}$

3. A wheel at 10 rad/s has zero net torque. What is $\omega$ (rad/s) after 100 s?

Concept Check 🔍

Practice

Challenge Question 📋