Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo
College Physics for AP® Courses 2e

10.1 Angular Acceleration

College Physics for AP® Courses 2e10.1 Angular Acceleration

Learning Objectives

By the end of this section, you will be able to:

  • Describe uniform circular motion.
  • Explain non-uniform circular motion.
  • Calculate angular acceleration of an object.
  • Observe the link between linear and angular acceleration.

Uniform Circular Motion and Gravitation discussed only uniform circular motion, which is motion in a circle at constant speed and, hence, constant angular velocity. Recall that angular velocity ωω was defined as the time rate of change of angle θθ:


where θθ is the angle of rotation as seen in Figure 10.3. The relationship between angular velocity ωω and linear velocity vv was also defined in Rotation Angle and Angular Velocity as

v = v =


ω = v r , ω = v r ,

where rr is the radius of curvature, also seen in Figure 10.3. According to the sign convention, the counter clockwise direction is considered as positive direction and clockwise direction as negative

The given figure shows counterclockwise circular motion with a horizontal line, depicting radius r, drawn from the center of the circle to the right side on its circumference and another line is drawn in such a manner that it makes an acute angle delta theta with the horizontal line. Tangential velocity vectors are indicated at the end of the two lines. At the bottom right side of the figure, the formula for angular velocity is given as v upon r.
Figure 10.3 This figure shows uniform circular motion and some of its defined quantities.

Angular velocity is not constant when a skater pulls in her arms, when a child starts up a merry-go-round from rest, or when a computer’s hard disk slows to a halt when switched off. In all these cases, there is an angular acceleration, in which ωω changes. The faster the change occurs, the greater the angular acceleration. Angular acceleration αα is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows:


where ΔωΔω is the change in angular velocity and ΔtΔt is the change in time. The units of angular acceleration are rad/s/srad/s/s, or rad/s2rad/s2. If ωω increases, then αα is positive. If ωω decreases, then αα is negative.

Example 10.1

Calculating the Angular Acceleration and Deceleration of a Bike Wheel

Suppose a teenager puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the angular acceleration in rad/s2rad/s2. (b) If she now slams on the brakes, causing an angular acceleration of 87.3 rad/s287.3 rad/s2, how long does it take the wheel to stop?

Strategy for (a)

The angular acceleration can be found directly from its definition in α=ΔωΔtα=ΔωΔt because the final angular velocity and time are given. We see that ΔωΔω is 250 rpm and ΔtΔt is 5.00 s.

Solution for (a)

Entering known information into the definition of angular acceleration, we get

α = ΔωΔt = 250 rpm5.00 s.α = ΔωΔt = 250 rpm5.00 s.

Because ΔωΔω is in revolutions per minute (rpm) and we want the standard units of rad/s2rad/s2 for angular acceleration, we need to convert ΔωΔω from rpm to rad/s:

Δω = 250 rev min 2π rad rev 1 min 60 s = 26.2 rad s . Δω = 250 rev min 2π rad rev 1 min 60 s = 26.2 rad s .

Entering this quantity into the expression for αα, we get

α = Δω Δt = 26.2 rad/s 5.00 s = 5.24  rad/s 2 . α = Δω Δt = 26.2 rad/s 5.00 s = 5.24  rad/s 2 .

Strategy for (b)

In this part, we know the angular acceleration and the initial angular velocity. We can find the stoppage time by using the definition of angular acceleration and solving for ΔtΔt, yielding

Δt=Δωα .Δt=Δωα .

Solution for (b)

Here the angular velocity decreases from 26.2 rad/s26.2 rad/s (250 rpm) to zero, so that ΔωΔω is 26.2 rad/s26.2 rad/s, and αα is given to be 87.3rad/s287.3rad/s2. Thus,

Δt = 26.2 rad/s 87.3 rad/s 2 = 0.300 s. Δt = 26.2 rad/s 87.3 rad/s 2 = 0.300 s.


Note that the angular acceleration as the girl spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero. In both cases, the relationships are analogous to what happens with linear motion. For example, there is a large deceleration when you crash into a brick wall—the velocity change is large in a short time interval.

If the bicycle in the preceding example had been on its wheels instead of upside-down, it would first have accelerated along the ground and then come to a stop. This connection between circular motion and linear motion needs to be explored. For example, it would be useful to know how linear and angular acceleration are related. In circular motion, linear acceleration is tangent to the circle at the point of interest, as seen in Figure 10.4. Thus, linear acceleration is called tangential acceleration atat.

In the figure, a semicircle is drawn, with its radius r, shown here as a line segment. The anti-clockwise motion of the circle is shown with an arrow on the path of the circle. Tangential velocity vector, v, of the point, which is on the meeting point of radius with the circle, is shown as a green arrow and the linear acceleration, a-t is shown as a yellow arrow in the same direction along v.
Figure 10.4 In circular motion, linear acceleration aa, occurs as the magnitude of the velocity changes: aa is tangent to the motion. In the context of circular motion, linear acceleration is also called tangential acceleration atat.

Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction. We know from Uniform Circular Motion and Gravitation that in circular motion centripetal acceleration, acac, refers to changes in the direction of the velocity but not its magnitude. An object undergoing circular motion experiences centripetal acceleration, as seen in Figure 10.5. Thus, atat and acac are perpendicular and independent of one another. Tangential acceleration atat is directly related to the angular acceleration αα and is linked to an increase or decrease in the velocity, but not its direction.

In the figure, a semicircle is drawn, with its radius r, shown here as a line segment. The anti-clockwise motion of the circle is shown with an arrow on the path of the circle. Tangential velocity vector, v, of the point, which is on the meeting point of radius with the circle, is shown as a green arrow and the linear acceleration, a sub t is shown as a yellow arrow in the same direction along v. The centripetal acceleration, a sub c, is also shown as a yellow arrow drawn perpendicular to a sub t, toward the direction of the center of the circle. A label in the figures states a sub t affects magnitude and a sub c affects direction.
Figure 10.5 Centripetal acceleration acac occurs as the direction of velocity changes; it is perpendicular to the circular motion. Centripetal and tangential acceleration are thus perpendicular to each other.

Now we can find the exact relationship between linear acceleration atat and angular acceleration αα. Because linear acceleration is proportional to a change in the magnitude of the velocity, it is defined (as it was in One-Dimensional Kinematics) to be

at= ΔvΔt .at= ΔvΔt .

For circular motion, note that v=v=, so that

at=ΔΔt .at=ΔΔt .

The radius rr is constant for circular motion, and so Δ()=r(Δω)Δ()=r(Δω). Thus,

at= r ΔωΔt .at= r ΔωΔt .

By definition, α=ΔωΔtα=ΔωΔt. Thus,

a t = , a t = ,


α = a t r . α = a t r .

These equations mean that linear acceleration and angular acceleration are directly proportional. The greater the angular acceleration is, the larger the linear (tangential) acceleration is, and vice versa. For example, the greater the angular acceleration of a car’s drive wheels, the greater the acceleration of the car. The radius also matters. For example, the smaller a wheel, the smaller its linear acceleration for a given angular acceleration αα.

Example 10.2

Calculating the Angular Acceleration of a Motorcycle Wheel

A powerful motorcycle can accelerate from 0 to 30.0 m/s (about 108 km/h) in 4.20 s. What is the angular acceleration of its 0.320-m-radius wheels? (See Figure 10.6.)

The figure shows the right side view of a man riding a motorcycle hence, depicting linear acceleration a of the motorcycle pointing toward the front of the bike as a horizontal arrow and the angular acceleration alpha of its wheels, shown here as curved arrows along the front of both the wheels pointing downward.
Figure 10.6 The linear acceleration of a motorcycle is accompanied by an angular acceleration of its wheels.


We are given information about the linear velocities of the motorcycle. Thus, we can find its linear acceleration atat. Then, the expression α=atrα=atr can be used to find the angular acceleration.


The linear acceleration is
at = ΔvΔt = 30.0 m/s4.20 s = 7.14 m/s2. at = ΔvΔt = 30.0 m/s4.20 s = 7.14 m/s2.

We also know the radius of the wheels. Entering the values for atat and rr into α=atr α=atr, we get

α = a t r = 7.14 m/s 2 0.320 m = 22.3 rad/s 2 . α = a t r = 7.14 m/s 2 0.320 m = 22.3 rad/s 2 .


Units of radians are dimensionless and appear in any relationship between angular and linear quantities.

So far, we have defined three rotational quantities— θωθω, and αα. These quantities are analogous to the translational quantities xvxv, and aa. Table 10.1 displays rotational quantities, the analogous translational quantities, and the relationships between them.

Rotational Translational Relationship
θ θ x x θ = x r θ = x r
ω ω v v ω = v r ω = v r
α α a a α = a t r α = a t r
Table 10.1 Rotational and Translational Quantities

Making Connections: Take-Home Experiment

Sit down with your feet on the ground on a chair that rotates. Lift one of your legs such that it is unbent (straightened out). Using the other leg, begin to rotate yourself by pushing on the ground. Stop using your leg to push the ground but allow the chair to rotate. From the origin where you began, sketch the angle, angular velocity, and angular acceleration of your leg as a function of time in the form of three separate graphs. Estimate the magnitudes of these quantities.

Check Your Understanding

Angular acceleration is a vector, having both magnitude and direction. How do we denote its magnitude and direction? Illustrate with an example.

Ladybug Revolution

Join the ladybug in an exploration of rotational motion. Rotate the merry-go-round to change its angle, or choose a constant angular velocity or angular acceleration. Explore how circular motion relates to the bug's x,y position, velocity, and acceleration using vectors or graphs.

Click to view content.

Order a print copy

As an Amazon Associate we earn from qualifying purchases.


This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at
Citation information

© Jan 19, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.