A disk has θ(t) = 4t^2 (rad). Calculate its angular acceleration. [2 points]

United StatesPhysics C: MechanicsSyllabus dot point

How do angular position, velocity and acceleration describe rotation, and how does calculus link them just as it does for linear motion?

Topic 5.1 Rotational Kinematics: define angular position, velocity and acceleration as derivatives, apply the constant-angular-acceleration equations, and use integration for variable angular acceleration.

A focused answer to AP Physics C: Mechanics Topic 5.1, covering angular position, velocity and acceleration as time derivatives, the constant-angular-acceleration equations as analogues of the linear ones, integration for variable angular acceleration, and the sign convention for rotation, with calculus-based worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Angular position $\theta$ (rad) has angular velocity $\omega = \dfrac{d\theta}{dt}$ (rad/s) and angular acceleration $\alpha = \dfrac{d\omega}{dt} = \dfrac{d^2\theta}{dt^2}$ (rad/s squared) as its first and second time derivatives. To reverse, integrate: $\omega = \omega_0 + \int\alpha\,dt$ and $\theta = \theta_0 + \int\omega\,dt$ . For constant $\alpha$ , the equations parallel the linear ones: $\omega = \omega_0 + \alpha t$ , $\;\theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha t^2$ , $\;\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$ . A consistent sign convention (counterclockwise positive) keeps directions straight.

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What this topic is asking
Angular quantities as derivatives
Recovering rotation by integration
The constant-angular-acceleration equations
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What this topic is asking

The College Board (Topic 5.1) wants you to define angular position, velocity and acceleration as time derivatives, to apply the constant-angular-acceleration equations as direct analogues of the linear kinematic equations, and to integrate when the angular acceleration varies. Rotational kinematics mirrors the linear kinematics of Unit 1 exactly, with angles replacing distances, so the calculus you already know transfers directly.

Angular quantities as derivatives

These are the rotational counterparts of position, velocity and acceleration, and the calculus relationships are identical: differentiate the angle to get the angular velocity, differentiate again for the angular acceleration. If you are handed $\theta(t)$ , you find $\omega$ and $\alpha$ by differentiating, exactly as in linear kinematics. Radians are the natural unit because they make the link to linear quantities (arc length $s = r\theta$ ) clean, which the next topic exploits.

Recovering rotation by integration

If the angular acceleration is given as a function of time, integrate to build up the motion, pinning down each constant with an initial condition:

\omega(t) = \omega_0 + \int_0^t \alpha\,dt', \qquad \theta(t) = \theta_0 + \int_0^t \omega\,dt'.

Graphically, the area under an $\alpha$ - $t$ graph is the change in angular velocity, and the area under an $\omega$ - $t$ graph is the angular displacement, the rotational versions of the slope-and-area rules from Unit 1. This integration approach handles a flywheel whose drive torque (and so $\alpha$ ) changes with time.

The constant-angular-acceleration equations

When the angular acceleration is constant, integrating gives three equations that are exact analogues of the linear kinematic equations:

\omega = \omega_0 + \alpha t, \qquad \theta = \theta_0 + \omega_0 t + \tfrac{1}{2}\alpha t^2, \qquad \omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0).

The correspondence is term for term: $\theta \leftrightarrow x$ , $\omega \leftrightarrow v$ , $\alpha \leftrightarrow a$ . Every technique from linear kinematics, choosing the equation that omits the unknown you do not want, splitting multi-phase motion into segments, applies unchanged. These equations are valid only while $\alpha$ is constant; for varying $\alpha$ you return to integrating directly.

A wheel slowing under constant angular acceleration

A wheel spinning at $20$ rad/s is brought to rest in $40$ rad of rotation by a constant braking angular acceleration. Find the angular acceleration and the time to stop.

step 1 Choose the equation without time for $\alpha$

Use $\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$ with $\omega = 0$ , $\omega_0 = 20$ rad/s, $\theta - \theta_0 = 40$ rad.

step 2 Solve for the angular acceleration

$0 = 20^2 + 2\alpha(40) \Rightarrow 0 = 400 + 80\alpha \Rightarrow \alpha = -\dfrac{400}{80} = -5.0$ rad/s squared.

step 3 Find the time with the velocity equation

$\omega = \omega_0 + \alpha t \Rightarrow 0 = 20 + (-5.0)t \Rightarrow t = 4.0$ s.

step 4 Sanity check with the angle

$\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 = 20(4.0) + \tfrac{1}{2}(-5.0)(4.0)^2 = 80 - 40 = 40$ rad. Consistent.

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Q1. A turntable accelerates from rest at $2.0$ rad/s squared. Calculate its angular velocity after $5.0$ s. [2 points]

Cue. $\omega = \omega_0 + \alpha t = 0 + (2.0)(5.0) = 10$ rad/s.

Q2. A disk has $\theta(t) = 4t^2$ (rad). Calculate its angular acceleration. [2 points]

Cue. $\omega = 8t$ , $\alpha = d\omega/dt = 8$ rad/s squared (constant).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)5 marksSection II (FRQ, calculus). A wheel's angular position is

\theta(t) = 2.0t^3 - 3.0t^2

(rad,

t

in s). (a) Derive expressions for the angular velocity and angular acceleration. (b) Determine the angular velocity at

t = 2.0

s. (c) Determine the times at which the wheel is momentarily at rest. (d) A separate flywheel starts from rest and has constant angular acceleration

4.0

rad/s squared; determine its angular velocity and the angle turned after

3.0

Show worked answer →

A 5-point rotational-kinematics FRQ.

(a) Derivatives (1 point): $\omega(t) = \dfrac{d\theta}{dt} = 6.0t^2 - 6.0t$ ; $\alpha(t) = \dfrac{d\omega}{dt} = 12t - 6.0$ .
(b) Angular velocity at $t = 2.0$ s (1 point): $\omega = 6.0(2.0)^2 - 6.0(2.0) = 24 - 12 = 12$ rad/s.
(c) At rest (1 point): $\omega = 0$ when $6.0t(t - 1) = 0$ , so $t = 0$ and $t = 1.0$ s.
(d) Constant angular acceleration (2 points): $\omega = \omega_0 + \alpha t = 0 + (4.0)(3.0) = 12$ rad/s; $\theta = \tfrac{1}{2}\alpha t^2 = \tfrac{1}{2}(4.0)(3.0)^2 = 18$ rad.

Markers reward differentiating $\theta(t)$ and using the constant-angular-acceleration equations as analogues of the linear ones.

AP 2021 (style)1 marksSection I (multiple choice). A disk rotating at

10

rad/s slows uniformly and stops in

5.0

s. What is the magnitude of its angular acceleration? (A)

0.50

rad/s squared (B)

2.0

rad/s squared (C)

50

rad/s squared (D)

15

rad/s squared. Justify your reasoning.

Show worked answer →

A 1-point rotational MCQ. The answer is (B).

Angular acceleration is the rate of change of angular velocity: $\alpha = \dfrac{\Delta\omega}{\Delta t} = \dfrac{0 - 10}{5.0} = -2.0$ rad/s squared, magnitude $2.0$ rad/s squared. This is the rotational analogue of $a = \Delta v/\Delta t$ . The trap (A) inverts the ratio.

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Sources & how we know this

AP Physics C: Mechanics Course and Exam Description — College Board (2024)