# Angular Acceleration

## 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 $\omega$ was defined as the time rate of change of angle $\theta$

$\omega=\dfrac{\Delta \theta}{\Delta t},$

where $\theta$ is the angle of rotation as seen in Figure 10.3. The relationship between angular velocity $\omega$ and linear velocity $v$ was also defined in Rotation Angle and Angular Velocity. as

$v=r \omega$

or

$\omega=\dfrac{v}{r},$

where $r$ 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 $\omega$ changes. The faster the change occurs, the greater the angular acceleration. Angular acceleration $\alpha$ is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows:

$\alpha=\dfrac{\Delta \omega}{\Delta t},$

where $\Delta \omega$ is the change in angular velocity and $\Delta t$ is the change in time. The units of angular acceleration are $(\mathrm{rad} / \mathrm{s}) / \mathrm{s}$, or $\mathrm{rad} / \mathrm{s}^{2}$. If $\omega$ increases, then $\alpha$ is positive. If $\omega$ decreases, then $\alpha$ 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 $250 \mathrm{rpm}$ in $5.00$ s. (a) Calculate the angular acceleration in $\mathrm{rad} / \mathrm{s}^{2}$. (b) If she now slams on the brakes, causing an angular acceleration of $-87.3 \mathrm{rad} / \mathrm{s}^{2}$, how long does it take the wheel to stop?

#### Strategy for (a)

The angular acceleration can be found directly from its definition in $\alpha=\dfrac{\Delta \omega}{\Delta t}$ because the final angular velocity and time are given. We see that $\Delta \omega$ is $250 \mathrm{rpm}$ and $\Delta t$ is $5.00 \mathrm{~s}$.

#### Solution for (a)

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

\begin{aligned} \alpha &=\dfrac{\Delta \omega}{\Delta t} \\ &=\dfrac{250 \mathrm{rpm}}{5.00 \mathrm{~s}}. \end{aligned}

Because $\Delta \omega$ is in revolutions per minute (rpm) and we want the standard units of $\mathrm{rad} / \mathrm{s}^{2}$ for angular acceleration, we need to convert $\Delta \omega$ from rpm to $\mathrm{rad} / \mathrm{s}.$:

\begin{aligned} \Delta \omega &=250 \dfrac{\mathrm{rev}}{\mathrm{min}} \cdot \dfrac{2 \pi \mathrm{rad}}{\mathrm{rev}} \cdot \dfrac{1 \mathrm{~min}}{60 \mathrm{~s}} \\ &=26.2 \, \dfrac{\mathrm{rad}}{\mathrm{s}}.\end{aligned}

#### 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 $\Delta t$, yielding

$\Delta t=\dfrac{\Delta \omega}{\alpha}.$

#### Solution for (b)

Here the angular velocity decreases from $26.2 \, \mathrm{rad} / \mathrm{s}$ (250 rpm) to zero, so that $\Delta \omega$ is $-26.2 \, \mathrm{rad} / \mathrm{s}$, and $\alpha$ is given to be $-87.3 \, \mathrm{rad} / \mathrm{s}^{2}$. Thus,

\begin{aligned} \Delta t &=\dfrac{-26.2 \, \mathrm{rad} / \mathrm{s}}{-87.3 \, \mathrm{rad} / \mathrm{s}^{2}} \\ &=0.300 \mathrm{~s}.\end{aligned}

#### Discussion

Note that the angular acceleration as the girl spins the wheel is small and positive; it takes $5 \mathrm{~s}$ 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 at

Figure 10.4 In circular motion, linear acceleration a, occurs as the magnitude of the velocity changes: a is tangent to the motion. In the context of circular motion, linear acceleration is also called tangential acceleration at.

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, $a_{\mathrm{c}}$, 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, $a_{\mathrm{t}}$ and $a_{\mathrm{c}}$ are perpendicular and independent of one another. Tangential acceleration $a_{\mathrm{t}}$ is directly related to the angular acceleration $\alpha$ and is linked to an increase or decrease in the velocity, but not its direction.

Figure 10.5 Centripetal acceleration $a_{\mathrm{c}}$ 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 $a_{\mathrm{t}}$ and angular acceleration $\alpha$. 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

$a_{\mathrm{t}}=\dfrac{\Delta v}{\Delta t}.$

For circular motion, note that $v=r \omega$, so that

$a_{\mathrm{t}}=\dfrac{\Delta(r \omega)}{\Delta t}.$

The radius $r$ is constant for circular motion, and so $\Delta(r \omega)=r(\Delta \omega)$. Thus,

$a_{\mathrm{t}}=r \dfrac{\Delta \omega}{\Delta t}.$

By definition, $\alpha=\dfrac{\Delta \omega}{\Delta t}.$ Thus,

$a_{\mathrm{t}}=r \alpha$

or

$\alpha=\dfrac{a_{\mathrm{t}}}{r}.$

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 $\alpha$.

#### 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).

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 $a_{\mathrm{t}}$. Then, the expression $\alpha=\dfrac{a_{\mathrm{t}}}{r}$ can be used to find the angular acceleration.

#### Solution

The linear acceleration is

\begin{aligned} a_{\mathrm{t}} &=\dfrac{\Delta v}{\Delta t} \\ &=\dfrac{30.0 \mathrm{~m} / \mathrm{s}}{4.20 \mathrm{~s}} \\ &=7.14 \mathrm{~m} / \mathrm{s}^{2}.\end{aligned}

We also know the radius of the wheels. Entering the values for $a_{\mathrm{t}}$ and $r$ into $\alpha=\dfrac{a_{\mathrm{t}}}{r}$, we get

\begin{aligned} \alpha &=\dfrac{a_{\mathrm{t}}}{r} \\ &=\dfrac{7.14 \mathrm{~m} / \mathrm{s}^{2}}{0.320 \mathrm{~m}} \\ &=22.3 \, \mathrm{rad} / \mathrm{s}^{2}.\end{aligned}

#### Discussion

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

So far, we have defined three rotational quantities $-\theta, \omega$, and $\alpha.$ These quantities are analogous to the translational quantities $x, v$, and $a.$ Table 10.1 displays rotational quantities, the analogous translational quantities, and the relationships between them.

Rotational Translational Relationship
$\theta$ $x$ $\theta=\dfrac{x}{r}$
$\omega$ $v$ $\omega=\dfrac{v}{r}$
$\alpha$ $a$ $\alpha=\dfrac{a_{t}}{r}$

Table 10.1
Rotational and Translational Quantities

Source: Rice University, https://openstax.org/books/college-physics/pages/10-1-angular-acceleration