Angular Acceleration

Site: Saylor Academy
Course: PHYS101: Introduction to Mechanics
Book: Angular Acceleration
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Date: Thursday, June 13, 2024, 7:43 PM

Description

We define angular acceleration as the change in angular velocity with respect to time. The equation is  \alpha = \frac{\Delta \omega}{\Delta t} , where  \alpha represents angular acceleration.

As you read, pay attention to Example 10.1, which shows how to calculate the angular acceleration of a bike wheel. In the first part of the problem, we calculate the angular acceleration of the wheel given the change in angular velocity and time. In the second part of the problem, we calculate the time needed to stop an already spinning wheel given angular deceleration as initial velocity, using the same angular acceleration equation. See a diagram of a rotating object showing the relationship between linear and angular velocity in Figure 10.3.

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

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 \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

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 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.

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 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).

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.

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
Creative Commons License This work is licensed under a Creative Commons Attribution 4.0 License.

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.


Solution

The magnitude of angular acceleration is \alpha and its most common units are \mathrm{rad} / \mathrm{s}^{2}. The direction of angular acceleration along a fixed axis is denoted by \mathrm{a}+ or \mathrm{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.