Rotational kinematics and dynamics — AP Physics C: Mechanics Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Angular displacement, velocity, and acceleration, constant angular acceleration kinematics, torque, rotational inertia, parallel axis theorem, and Newton’s second law for fixed-axis rotation, including relations between rotational and linear quantities.

You should already know: Linear kinematics for constant acceleration. Newton’s second law for translational motion. Integration of kinematic functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Rotational kinematics and dynamics?

Rotational kinematics and dynamics is the study of fixed-axis rotation of rigid bodies, describing how rotation occurs and what causes changes in rotational speed. This topic is the core of Unit 5 Rotation, which accounts for 14-20% of the total score on the AP Physics C: Mechanics exam, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with translational motion to form multi-part problems. By convention, counterclockwise rotation is defined as positive, with all angular quantities measured in radians for calculations. Rotational kinematics describes the relationships between angular displacement, angular velocity, and angular acceleration, independent of what causes the rotation, while rotational dynamics connects these motion quantities to torque, the rotational analog of force. This topic builds a direct analogy between translational and rotational mechanics that simplifies learning the rules of rigid body motion.

2. Rotational Kinematics

Rotational kinematics is the description of rotational motion without reference to its causes, analogous to translational kinematics for linear motion. The core quantities are:

• Angular displacement $\Delta \theta$: change in angle of a rigid body about the rotation axis, measured in radians
• Angular velocity $\omega =\frac{d\theta }{dt}$: rate of change of angular displacement, units rad/s
• Angular acceleration $\alpha =\frac{d\omega }{dt}=\frac{d^2\theta }{d{t}^2}$: rate of change of angular velocity, units rad/s²

For constant angular acceleration, we derive kinematic equations directly analogous to the constant linear acceleration equations, by swapping $x\to \theta$, $v\to \omega$, $a\to \alpha$: $$ \begin{align*} \omega &= \omega_0 + \alpha t \ \theta &= \omega_0 t + \frac{1}{2}\alpha t^2 \ \omega^2 &= \omega_0^2 + 2\alpha \Delta\theta \end{align*} $$

For any point on the rotating rigid body at distance $r$ from the axis, we relate angular quantities to linear tangential quantities: $s=r\theta$, $v=r\omega$, $a_t=r\alpha$. All points also have a centripetal (radial) acceleration $a_c=r{\omega }^2$ directed toward the rotation axis.

Worked Example

A blender blade starts from rest and accelerates uniformly to 1800 rpm in 1.5 seconds. Find (a) the angular acceleration of the blade, (b) the total number of rotations the blade makes while accelerating, and (c) the tangential acceleration of a point 0.04 m from the rotation axis.

1. Convert final angular speed to radians per second: ${\omega }_f=1800\frac{\text{rev}}{\text{min}}\times \frac{2\pi \text{rad}}{\text{rev}}\times \frac{1\text{min}}{60\text{s}}=60\pi \text{rad/s}$.
2. Calculate angular acceleration: $\alpha =\frac{{\omega }_f-{\omega }_0}{t}=\frac{60\pi -0}{1.5}=40\pi \approx 125.7{\text{rad/s}}^2$.
3. Find total angular displacement: $\Delta \theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2=0+0.5(40\pi )(1.5{)}^2=45\pi \text{rad}$. Convert to rotations: $\frac{45\pi }{2\pi }=22.5\text{rotations}$.
4. Calculate tangential acceleration: $a_t=r\alpha =0.04(40\pi )\approx 5.03{\text{m/s}}^2$.

Exam tip: Always convert angular speed from rpm or degrees to radians at the start of every calculation; all standard rotational kinematics formulas only work with radians, and unit errors will lead to incorrect numerical answers on the AP exam.

3. Torque and Rotational Inertia

Torque is the rotational analog of force: it is the quantity that causes changes in rotational motion, just as force causes changes in linear motion. For a force $F$ applied at a distance $r$ from the rotation axis, the magnitude of torque is: $$\tau =rF\sin \theta =r{F}_{\perp }=r_{\perp }F$$ where $\theta$ is the angle between the position vector $\vec{r}$ (from the axis to the point of application) and $\vec{F}$. $F_{\perp }$ is the component of force perpendicular to $r$, and $r_{\perp }$ is the perpendicular lever arm from the axis to the line of action of the force. By convention, counterclockwise torque is positive, and clockwise torque is negative.

Rotational inertia (or moment of inertia) is the rotational analog of mass, describing how much torque is needed to produce a given angular acceleration. For a system of discrete masses, $I=\sum m_i{r}_i^2$, where $r_i$ is the distance of mass $m_i$ from the axis. For a continuous rigid body, $I=\int r^{2}dm$. The parallel axis theorem lets you calculate $I$ for any axis parallel to the axis through the center of mass (cm): $$I=I_{cm}+M{d}^2$$ where $M$ is total mass of the object, and $d$ is the distance between the two axes.

Worked Example

A uniform solid square plate of mass $M=3\text{kg}$ and side length $L=0.8\text{m}$ has $I_{cm}=\frac{1}{6}M{L}^2$ for rotation about an axis parallel to one edge through the center. Find the rotational inertia for rotation about an axis along one edge of the plate.

1. Locate the center of mass of the uniform square: it is at the geometric center, so the distance from the center axis to the edge axis is $d=\frac{L}{2}$.
2. Apply the parallel axis theorem: $I=I_{cm}+M{d}^2=\frac{1}{6}M{L}^2+M{\left(\frac{L}{2}\right)}^2$.
3. Simplify: $I=M{L}^2\left(\frac{1}{6}+\frac{1}{4}\right)=\frac{5}{12}M{L}^2$.
4. Substitute values: $I=\frac{5}{12}(3\text{kg})(0.8\text{m}{)}^2=\frac{5}{12}(3)(0.64)=0.8{\text{kgcdotpm}}^2$.

Exam tip: Always confirm that $d$ in the parallel axis theorem is measured from the center of mass axis, not from the nearest edge of the object; this is one of the most common student mistakes on AP exam problems.

4. Newton's Second Law for Rotation

Newton's second law for rotation connects net torque to angular acceleration, analogous to Newton's second law for translation $\sum \vec{F}=m\vec{a}$. For rotation about a fixed axis, the law states: $$\sum {\tau }_{\text{axis}}=I_{\text{axis}}\alpha$$ This means the net torque on a rigid body about the rotation axis equals the product of the rotational inertia about that axis and the angular acceleration. For problems involving both translation and rotation (e.g., pulley systems with massive pulleys, rolling motion), this law is used alongside Newton's second law for translation, with the relation $a=r\alpha$ connecting linear acceleration of a point on the rigid body to angular acceleration when there is no slipping.

Intuition: Just as a larger net force gives a larger linear acceleration for a given mass, a larger net torque gives a larger angular acceleration for a given rotational inertia. Rotational inertia acts as "rotational mass": higher $I$ means more resistance to angular acceleration, just like higher mass means more resistance to linear acceleration.

Worked Example

A massive pulley is a solid disk of mass $M=2\text{kg}$ and radius $R=0.1\text{m}$, rotating about a fixed axis through its center. A string wrapped around the pulley supports a hanging block of mass $m=1\text{kg}$. The string does not slip on the pulley. Find the angular acceleration of the pulley.

1. Write Newton's second law for the hanging block: $mg-T=ma$, where $T$ is tension in the string, $a$ is the linear acceleration of the block.
2. Write Newton's second law for the pulley: net torque from tension is $\tau =TR=I\alpha$. Rotational inertia of a solid disk is $I=\frac{1}{2}M{R}^2$.
3. Relate $a$ and $\alpha$ for no slipping: $a=R\alpha$.
4. Substitute $I$ into the torque equation: $TR=\left(\frac{1}{2}M{R}^2\right)\alpha ⟹T=\frac{1}{2}MR\alpha =\frac{1}{2}Ma$ (since $a=R\alpha$).
5. Substitute $T$ into the block equation: $mg-\frac{1}{2}Ma=ma⟹mg=a\left(m+\frac{M}{2}\right)⟹a=\frac{mg}{m+M/2}=\frac{(1)(9.8)}{1+1}=4.9{\text{m/s}}^2$.
6. Solve for $\alpha$: $\alpha =\frac{a}{R}=\frac{4.9}{0.1}=49{\text{rad/s}}^2$.

Exam tip: Always use $I$ for the actual rotation axis when applying $\sum \tau =I\alpha$, not $I_{cm}$; if the object rotates about an axis that is not through the center of mass, apply the parallel axis theorem first to get the correct $I$.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using degrees or revolutions instead of radians when calculating tangential acceleration, arc length, or any derived rotational quantity. Why: Many problems give angular speed in revolutions per minute, so students forget to convert to radians, the required unit for all standard rotational formulas. Correct move: Convert all angular units to radians or radians per second at the start of every problem.
• Wrong move: Forgetting that a point on a rotating object with non-zero angular acceleration still has centripetal acceleration. Why: Students focus on tangential acceleration from angular acceleration and forget that any point in circular motion requires centripetal acceleration to stay on its path. Correct move: When asked for total acceleration, always calculate both $a_t=r\alpha$ and $a_c=r{\omega }^2$, then combine them with the Pythagorean theorem.
• Wrong move: Calculating torque from the weight of a uniform rigid body about an end axis using $r=L$ (full length) instead of $r=L/2$. Why: Students incorrectly assume weight acts at the end of the object instead of at the center of mass. Correct move: For any uniform rigid body, weight always acts at the center of mass, so use the distance from the axis to the center of mass to calculate torque from weight.
• Wrong move: Forgetting to add the $M{d}^2$ term when using the parallel axis theorem, or using $d$ as distance from the nearest end of the object. Why: Students memorize the formula but misinterpret what $d$ measures. Correct move: Explicitly locate the center of mass, measure $d$ from the center of mass to the new axis, then add $M{d}^2$ to $I_{cm}$.
• Wrong move: Neglecting the torque from tension in problems with massive pulleys, treating tension as the same on both sides of the pulley. Why: Students are used to massless pulleys where tension is equal, but this does not hold for massive pulleys with angular acceleration. Correct move: For massive pulleys, write separate torque equations that account for different tensions on each side of the pulley.

6. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (Multiple Choice)

A turntable accelerates uniformly from rest to an angular speed of $3\pi$ rad/s in 2 seconds. What is the magnitude of the total acceleration of a point 0.1 m from the center of the turntable at $t=2$ s? A) $0.15\pi$ m/s² B) $0.9{\pi }^2$ m/s² C) $0.15\pi \sqrt{1+36{\pi }^2}$ m/s² D) $1.5\pi$ m/s²

Worked Solution: First, calculate angular acceleration: $\alpha =({\omega }_f-{\omega }_i)/t=(3\pi -0)/2=1.5\pi$ rad/s². Next, tangential acceleration is $a_t=r\alpha =0.1(1.5\pi )=0.15\pi$ m/s². Centripetal acceleration at $t=2$ s is $a_c=r{\omega }_f^2=0.1(3\pi {)}^2=0.9{\pi }^2=0.15\pi (6\pi )$ m/s². Since $a_t$ and $a_c$ are perpendicular, total acceleration is the hypotenuse: $a=\sqrt{a_t^2+a_c^2}=0.15\pi \sqrt{1+(6\pi {)}^2}=0.15\pi \sqrt{1+36{\pi }^2}$. The correct answer is C.

Question 2 (Free Response)

A uniform solid sphere of mass $M=1.2$ kg and radius $R=0.2$ m rotates about a fixed axis tangent to its surface. The moment of inertia of a solid sphere about its center of mass is $I_{cm}=\frac{2}{5}M{R}^2$. (a) Calculate the moment of inertia of the sphere about the tangent axis. (b) A constant tangential force of 3 N is applied to the surface of the sphere, in the direction of rotation. What is the angular acceleration of the sphere? (c) If the sphere starts from rest, what is the angular speed after 4 seconds, and what is the magnitude of the total acceleration of the point where the force is applied?

Worked Solution: (a) Use the parallel axis theorem: the distance from center of mass to the tangent axis is $d=R$, so $I=I_{cm}+M{d}^2=\frac{2}{5}M{R}^2+M{R}^2=\frac{7}{5}M{R}^2$. Substitute values: $I=1.4(1.2)(0.2{)}^2=0.0672$ kg·m². (b) Net torque from the tangential force is $\tau =FR=(3\text{ N})(0.2\text{ m})=0.6$ N·m. By Newton's second law for rotation: $\alpha =\tau /I=0.6/0.0672\approx 8.93$ rad/s². (c) For constant angular acceleration starting from rest: $\omega =\alpha t=(8.93)(4)\approx 35.7$ rad/s. For the point of application: $a_t=R\alpha =0.2(8.93)\approx 1.79$ m/s², $a_c=R{\omega }^2=0.2(35.7{)}^2\approx 255$ m/s². Total acceleration: $a=\sqrt{(1.79{)}^2+(255{)}^2}\approx 255$ m/s² (centripetal acceleration dominates here).

Question 3 (Application / Real-World Style)

A bicycle wheel can be approximated as a thin hoop of mass 1.0 kg and radius 0.35 m. A cyclist applies a constant tangential force of 15 N to the pedal crank, which is 0.17 m long. The bicycle's gear ratio means the wheel receives 4 times the torque applied to the crank (the wheel rotates 4 radians for every 1 radian rotated by the crank). Neglecting friction and rotational inertia of all other components, calculate the angular acceleration of the wheel.

Worked Solution: First, calculate torque applied to the crank: ${\tau }_{\text{crank}}=r_{\text{crank}}F=(0.17\text{ m})(15\text{ N})=2.55$ N·m. Torque on the wheel is ${\tau }_{\text{wheel}}=4{\tau }_{\text{crank}}=4(2.55)=10.2$ N·m. Rotational inertia of a thin hoop is $I=M{R}^2=(1.0\text{ kg})(0.35\text{ m}{)}^2=0.1225$ kg·m². By Newton's second law for rotation: $\alpha ={\tau }_{\text{wheel}}/I=10.2/0.1225\approx 83$ rad/s². This means the wheel's angular speed increases by ~13 revolutions per second every second, which is consistent with a cyclist accelerating aggressively from a stop.

7. Quick Reference Cheatsheet

8. What's Next

Rotational kinematics and dynamics is the foundational topic for all of Unit 5 Rotation. Next, you will apply these core relationships to rotational kinetic energy, angular momentum, and rolling without slipping motion, which make up the remaining parts of the unit. Without mastering the relationships between torque, rotational inertia, and angular acceleration, as well as the kinematic rules for fixed-axis rotation, you will not be able to solve multi-part problems involving rolling motion or conserved angular momentum, which are frequent high-weight FRQ topics on the AP exam. This topic also unifies translational and rotational Newtonian mechanics to create a complete description of rigid body motion that you will use for all remaining AP Physics C mechanics topics.

• Rotational kinetic energy
• Angular momentum and conservation
• Rolling without slipping