Rotational kinematics and dynamics

AP Physics C: Mechanics· AP Physics C: Mechanics CED — Rotation· 14 min read

1. What is Rotational Kinematics and Dynamics?★★☆☆☆⏱ 3 min

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, accounting for 14-20% of the total score on the AP Physics C: Mechanics exam, and it appears in both multiple-choice and free-response 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 motion quantities independent of what causes rotation, while rotational dynamics connects these quantities to torque, the rotational analog of force. This topic builds a direct analogy between translational and rotational mechanics that simplifies learning rigid body motion.

2. Rotational Kinematics★★☆☆☆⏱ 4 min

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

Angular displacement $Δ θ$ : change in angle of a rigid body about the rotation axis, measured in radians
Angular velocity $ω = \frac{d θ}{d t}$ : rate of change of angular displacement, units rad/s
Angular acceleration $α = \frac{d ω}{d t} = \frac{d ^{2} θ}{d t ^{2}}$ : 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 $x \to θ$ , $v \to ω$ , $a \to α$ :

\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, and all points also have a centripetal (radial) acceleration directed toward the rotation axis:

Arc length: $s = r θ$
Tangential speed: $v = r ω$
Tangential acceleration: $a_{t} = r α$
Centripetal acceleration: $a_{c} = r ω^{2}$

📐 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 and convert to rotations:
$\Delta\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + 0.5(40\pi)(1.5)^2 = 45\pi \text{rad} \implies \frac{45\pi}{2\pi} = 22.5 \text{rotations}$
4
Calculate tangential acceleration:
$a_t = r\alpha = 0.04(40\pi) \approx 5.03 \text{m/s}^2$

3. Torque and Rotational Inertia★★★☆☆⏱ 3 min

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 = rF_\perp = r_\perp F

Where $θ$ is the angle between the position vector $r$ (from the axis to the point of application) and $F$ . $F_{⊥}$ is the component of force perpendicular to $r$ , and $r_{⊥}$ 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, $I = \int r^{2} d m$ . The parallel axis theorem lets you calculate $I$ for any axis parallel to the axis through the center of mass:

I = I_{cm} + Md^2

Where $M$ is the total mass of the object, and $d$ is the distance between the two parallel axes.

📐 Worked Example

A uniform solid square plate of mass $M = 3 kg$ and side length $L = 0.8 m$ has $I_{c m} = \frac{1}{6} M L^{2}$ 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 at its geometric center. The distance from the center axis to the edge axis is:
$d = \frac{L}{2}$
2
Apply the parallel axis theorem:
$I = I_{cm} + Md^2 = \frac{1}{6}ML^2 + M\left(\frac{L}{2}\right)^2$
3
Simplify the expression:
$I = ML^2\left(\frac{1}{6} + \frac{1}{4}\right) = \frac{5}{12}ML^2$
4
Substitute numerical values:
$I = \frac{5}{12}(3 \text{kg})(0.8 \text{m})^2 = 0.8 \text{kg·m}^2$

4. Newton's Second Law for Rotation★★★☆☆⏱ 4 min

Newton's second law for rotation connects net torque to angular acceleration, analogous to Newton's second law for translation $\sum F = m a$ . 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 $a = r α$ connecting linear acceleration of a point on the rigid body to angular acceleration when there is no slipping.

📐 Worked Example

A massive pulley is a solid disk of mass $M = 2 kg$ and radius $R = 0.1 m$ , rotating about a fixed axis through its center. A string wrapped around the pulley supports a hanging block of mass $m = 1 kg$ . 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 ( $T$ = tension, $a$ = linear acceleration):
$mg - T = ma$
2
Write Newton's second law for the pulley. Rotational inertia of a solid disk is $I = \frac{1}{2} M R^{2}$ :
$TR = I\alpha$
3
Relate linear and angular acceleration for no slipping:
$a = R\alpha$
4
Substitute $I$ and $a = R α$ into the torque equation:
$TR = \left(\frac{1}{2}MR^2\right)\alpha \implies T = \frac{1}{2}Ma$
5
Substitute $T$ into the block equation and solve for $a$ :
$mg - \frac{1}{2}Ma = ma \implies a = \frac{mg}{m + M/2} = \frac{(1)(9.8)}{1 + 1} = 4.9 \text{m/s}^2$
6
Solve for angular acceleration:
$\alpha = \frac{a}{R} = \frac{4.9}{0.1} = 49 \text{rad/s}^2$

5. Common Pitfalls

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 $a_{t} = r α$ and $a_{c} = r ω^{2}$ , 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 $I_{c m}$ .

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. Quick Reference Cheatsheet

Category	Formula	Notes
Angular-Linear Relation	$s = r θ$ , $v = r ω$ , $a_{t} = r α$ , $a_{c} = r ω^{2}$	$θ, ω, α$ must be in radians
Constant $α$ Kinematics	$ω = ω_{0} + α t$ ; $θ = ω_{0} t + \frac{1}{2} α t^{2}$ ; $ω^{2} = ω_{0}^{2} + 2 α Δ θ$	Direct analog to linear constant acceleration kinematics
Torque Magnitude	$τ = r F sin θ = r F_{⊥} = r_{⊥} F$	Counterclockwise torque = positive by convention
Discrete Rotational Inertia	$I = \sum m_{i} r_{i}^{2}$	$r_{i}$ = distance from rotation axis
Parallel Axis Theorem	$I = I_{c m} + M d^{2}$	$d$ = distance between parallel axes
Newton's 2nd Law (Rotation)	$\sum τ_{axis} = I_{axis} α$	Use $I$ for the actual rotation axis
Common $I_{c m}$ Values	Rod (cm): $\frac{1}{12} M L^{2}$ ; Solid disk: $\frac{1}{2} M R^{2}$ ; Solid sphere: $\frac{2}{5} M R^{2}$ ; Thin hoop: $M R^{2}$	All for axis through center of mass
Total Acceleration	$a_{total} = a_{t}^{2} + a_{c}^{2}$	$a_{t}$ tangential, $a_{c}$ radial; components are perpendicular

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Constant angular acceleration problem
2022 · FRQ
Massive pulley system problem

Going deeper

unit_overviewUnit 5 Rotation OverviewParent unit overview for this topic

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 free response questions on the AP exam.