College Physics for AP® Courses

# 10.1Angular Acceleration

College Physics for AP® Courses10.1 Angular Acceleration

### Learning Objectives

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

• Describe uniform circular motion.
• Explain nonuniform 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 $ωω size 12{ω} {}$ was defined as the time rate of change of angle $θθ size 12{θ} {}$:

$ω=ΔθΔt,ω=ΔθΔt, size 12{ω= { {Δθ} over {Δt} } ","} {}$
10.1

where $θθ size 12{θ} {}$ is the angle of rotation as seen in Figure 10.3. The relationship between angular velocity $ωω size 12{ω} {}$ and linear velocity $vv size 12{v} {}$ was also defined in Rotation Angle and Angular Velocity as

$v = rω v = rω size 12{v=rω} {}$
10.2

or

$ω = v r , ω = v r , size 12{ω= { {v} over {r} } } {}$
10.3

where $rr size 12{r} {}$ 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

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 $ωω size 12{ω} {}$ changes. The faster the change occurs, the greater the angular acceleration. Angular acceleration $αα size 12{α} {}$ is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows:

$α=ΔωΔt,α=ΔωΔt, size 12{α= { {Δω} over {Δt} } ","} {}$
10.4

where $ΔωΔω size 12{Δω} {}$ is the change in angular velocity and $ΔtΔt size 12{Δt} {}$ is the change in time. The units of angular acceleration are $rad/s/srad/s/s size 12{ left ("rad/s" right )"/s"} {}$, or $rad/s2rad/s2 size 12{"rad/s" rSup { size 8{2} } } {}$. If $ωω size 12{ω} {}$ increases, then $αα size 12{α} {}$ is positive. If $ωω size 12{ω} {}$ decreases, then $αα size 12{α} {}$ is negative.

### Example 10.1Calculating 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 frequency of 250 rpm in 5.00 s. (a) Calculate the angular acceleration in $rad/s2rad/s2 size 12{"rad/s" rSup { size 8{2} } } {}$. (b) If she now slams on the brakes, causing an angular acceleration of $–87.3 rad/s2–87.3 rad/s2 size 12{"-87" "." 3"rad/s" rSup { size 8{2} } } {}$, how long does it take the wheel to stop?

Strategy for (a)

The angular acceleration can be found directly from its definition in $α=ΔωΔtα=ΔωΔt size 12{α= { {Δω} over {Δt} } } {}$ because the final frequency and time are given. We see that $ΔωΔω size 12{Δω} {}$ is the product of 2π rad/rev and is 250 rpm, and $ΔtΔt size 12{Δt} {}$ is 5.00 s.

Solution for (a)

First, calculate the change in angular velocity:

Δω = 250 rev min ⋅ 2π rad rev ⋅ 1 min 60 sec = 26.2 rad s . Δω = 250 rev min ⋅ 2π rad rev ⋅ 1 min 60 sec = 26.2 rad s . alignl { stack { size 12{Δω="250" { {"rev"} over {"min"} } cdot { {2π" rad"} over {"60" "." "0 s"} } } {} # size 12{ {}="26" "." 2 { {"rad"} over {"s"} } } {} } } {}

Entering this quantity into the expression for $αα size 12{α} {}$, we get

10.5

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 $ΔtΔt size 12{Δt} {}$, yielding

$Δt=Δωα .Δt=Δωα . size 12{Δt= { {Δω} over {α} } "."} {}$
10.6

Solution for (b)

Here the angular velocity decreases from $26.2 rad/s26.2 rad/s size 12{"26" "." 2"rad/s"} {}$ to zero, so that $ΔωΔω size 12{Δω} {}$ is $–26.2 rad/s–26.2 rad/s$, and $αα size 12{α} {}$ is given to be $–87.3rad/s2–87.3rad/s2 size 12{"-87" "." 3"rad/s" rSup { size 8{2} } } {}$. 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. alignl { stack { size 12{Δt= { { - "26" "." 2"rad/s"} over { - "87" "." 3"rad/s" rSup { size 8{2} } } } } {} # =0 "." "300""s" "." {} } } {}
10.7

Discussion

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 $atat size 12{a rSub { size 8{t} } } {}$.

Figure 10.4 In circular motion, linear acceleration $aa size 12{a} {}$, occurs as the magnitude of the velocity changes: $aa size 12{a} {}$ is tangent to the motion. In the context of circular motion, linear acceleration is also called tangential acceleration $atat size 12{a rSub { size 8{t} } } {}$.

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, $acac size 12{a rSub { size 8{t} } } {}$, 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, $atat size 12{a rSub { size 8{t} } } {}$ and $acac size 12{a rSub { size 8{t} } } {}$ are perpendicular and independent of one another. Tangential acceleration $atat size 12{a rSub { size 8{t} } } {}$ is directly related to the angular acceleration $αα size 12{α} {}$ and is linked to an increase or decrease in the velocity, but not its direction.

Figure 10.5 Centripetal acceleration $acac size 12{a rSub { size 8{t} } } {}$ 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 $atat size 12{a rSub { size 8{t} } } {}$ and angular acceleration $αα size 12{α} {}$. 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

$at= ΔvΔt .at= ΔvΔt . size 12{a rSub { size 8{t} } = { {Δv} over {Δt} } "."} {}$
10.8

For circular motion, note that $v=rωv=rω size 12{v=rω} {}$, so that

$at=ΔrωΔt .at=ΔrωΔt . size 12{a rSub { size 8{t} } = { {Δ left (rω right )} over {Δt} } "."} {}$
10.9

The radius $rr size 12{r} {}$ is constant for circular motion, and so $Δ(rω)=r(Δω)Δ(rω)=r(Δω) size 12{Δ $$rω$$ =r $$Δω$$ } {}$. Thus,

$at= r ΔωΔt .at= r ΔωΔt . size 12{a rSub { size 8{t} } =r { {Δω} over {Δt} } "."} {}$
10.10

By definition, $α=ΔωΔtα=ΔωΔt size 12{α= { {Δω} over {Δt} } } {}$. Thus,

$a t = rα , a t = rα , size 12{a rSub { size 8{t} } =rα} {}$
10.11

or

$α = a t r . α = a t r . size 12{α= { {a rSub { size 8{t} } } over {r} } } {}$
10.12

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 $αα size 12{α} {}$.

### Example 10.2Calculating 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.)

Figure 10.6 The linear acceleration of a motorcycle is accompanied by an angular acceleration of its wheels.

Strategy

We are given information about the linear velocities of the motorcycle. Thus, we can find its linear acceleration $atat size 12{a rSub { size 8{t} } } {}$. Then, the expression $α=atrα=atr size 12{a rSub { size 8{t} } =rα,α= { {a rSub { size 8{t} } } over {r} } } {}$ can be used to find the angular acceleration.

Solution

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. alignl { stack { size 12{a rSub { size 8{t} } = { {Δv} over {Δt} } } {} # = { {"30" "." 0" m/s"} over {4 "." "20 s"} } {} # =7 "." "14"" m/s" rSup { size 8{2} "."} {} } } {}
10.13

We also know the radius of the wheels. Entering the values for $atat size 12{a rSub { size 8{t} } } {}$ and $rr size 12{r} {}$ into $α=atr α=atr size 12{a rSub { size 8{t} } =rα,α= { {a rSub { size 8{t} } } over {r} } } {}$, 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 . alignl { stack { size 12{α= { {a rSub { size 8{t} } } over {r} } } {} # `= { {7 "." "14"" m/s" rSup { size 8{2} } } over {0 "." "320 m"} } {} # " "="22" "." "3 rad/s" rSup { size 8{2} } {} } } {}
10.14

Discussion

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

So far, we have defined three rotational quantities— , and $αα size 12{α} {}$. These quantities are analogous to the translational quantities , and $aa size 12{a} {}$. Table 10.1 displays rotational quantities, the analogous translational quantities, and the relationships between them.

Rotational Translational Relationship
$θ θ size 12{θ} {}$ $x x size 12{x} {}$ $θ = x r θ = x r size 12{θ= { {x} over {r} } } {}$
$ω ω size 12{ω} {}$ $v v size 12{v} {}$ $ω = v r ω = v r size 12{ω= { {v} over {r} } } {}$
$α α size 12{α} {}$ $a a size 12{a} {}$ $α = a t r α = a t r size 12{α= { {a rSub { size 8{t} } } over {r} } } {}$
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.

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

#### Solution

The magnitude of angular acceleration is $αα size 12{α} {}$ and its most common units are $rad/s2rad/s2 size 12{"rad/s" rSup { size 8{2} } } {}$. The direction of angular acceleration along a fixed axis is denoted by a + or a – sign, just as the direction of linear acceleration in one dimension is denoted by a + or a – sign. For example, consider a gymnast doing a forward flip. Her angular momentum would be parallel to the mat and to her left. The magnitude of her angular acceleration would be proportional to her angular velocity (spin rate) and her moment of inertia about her spin axis.

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.