A roller coaster car doing a loop-the-loop slows down as it goes up one side, speeds up as it comes back down the other. The ball in a roulette wheel gradually slows until it stops. Circular motion with a changing speed is called nonuniform circular motion. As you’ll see, nonuniform circular motion is analogous to accelerated linear motion.

FIGURE 4.29 shows a point speeding up as it moves around a circle. This might be a car speeding up around a curve or simply a point on a solid object that is rotating faster and faster. The key feature of the motion is a changing angular velocity. For linear motion, we defined acceleration as $a_x=d{v}_x\text{/}dt$. By analogy, let’s define the angular acceleration $\alpha$ (Greek alpha) of a rotating object, or a point on the object, to be

FIGURE 4.29 Circular motion with a changing angular velocity.

Angular acceleration is the rate at which the angular velocity $\omega$ changes, just as linear acceleration is the rate at which the linear velocity $v_x$ changes. The units of angular acceleration are ${\text{rad/s}}^2$.

For linear acceleration, you learned that $a_x$ and $v_x$ have the same sign when an object is speeding up, opposite signs when it is slowing down. The same rule applies to circular and rotational motion: $\omega$ and $\alpha$ have the same sign when the rotation is speeding up, opposite signs if it is slowing down. These ideas are illustrated in FIGURE 4.30.

FIGURE 4.30 The signs of angular velocity and acceleration. The rotation is speeding up if $\omega$ and $\alpha$ have the same sign, slowing down if they have opposite signs.

Angular position, angular velocity, and angular acceleration are defined exactly the same as linear position, velocity, and acceleration—simply starting with an angular rather than a linear measurement of position. Consequently, the graphical interpretation and the kinematic equations of circular/rotational motion with constant angular acceleration are exactly the same as for linear motion with constant acceleration. This is shown in the constant angular acceleration model below. All the problem-solving techniques you learned in Chapter 2 for linear motion carry over to circular and rotational motion.

MODEL 4.3 Constant angular acceleration

For motion with constant angular acceleration $\alpha$.

• Applies to particles with circular trajectories and to rotating solid objects.

• Mathematically: The graphs and equations for this circular/rotational motion are analogous to linear motion with constant acceleration.

  • Analogs: $s\to \theta v_s\to {\omega }_{}{a}_s\to \alpha$

Rotational kinematics	Linear kinematics
$\begin{array}{l}{\omega }_{\text{f}}={\omega }_i+\alpha \text{ Δ}t\hfill \\ {\theta }_{\text{f}}={\theta }_i+{\omega }_i\text{ Δ}t+\frac{1}{2}\alpha {\left(\text{Δ}t\right)}^2\hfill \\ {\omega }_{\text{f}}{}^2={\omega }_{\text{i}}{}^2+2\alpha \text{ Δ}\theta \hfill \end{array}$	$\begin{array}{l}{v}_{\text{f}s}=v_{\text{i}s}+a_s\text{ Δ}t\hfill \\ s_{\text{f}}=s_{\text{i}}+v_{\text{i}s}\text{ Δ}t+\frac{1}{2}{a}_s{\left(\text{Δ}t\right)}^2\hfill \\ v_{\text{f}s}{}^2=v_{\text{i}s}{}^2+2a_s\text{ Δ}s\hfill \end{array}$

Full Alternative Text

EXAMPLE 4.13 A rotating wheel

Figure 4.31a is a graph of angular velocity versus time for a rotating wheel. Describe the motion and draw a graph of angular acceleration versus time.

Figure 4.31 $\omega$-versus-$t$ graph and the corresponding $\alpha$-versus-$t$ graph for a rotating wheel.

Figure 4.31 Full Alternative Text

SOLVE This is a wheel that starts from rest, gradually speeds up counterclockwise until reaching top speed at $t_1$, maintains a constant angular velocity until $t_2$, then gradually slows down until stopping at $t_3$. The motion is always ccw because $\omega$ is always positive. The angular acceleration graph of FIGURE 4.31b is based on the fact that $\alpha$ is the slope of the $\omega \text{-}\text{versus}\text{-}t$ graph.

Conversely, the initial linear increase of $\omega$ can be seen as the increasing area under the $\alpha \text{-}\text{versus}\text{-}t$ graph as $t$ increases from $0$ to $t_1$. The angular velocity doesn’t change from $t_1$ to $t_2$ when the area under the $\alpha \text{-}\text{versus}\text{-}t$ is zero.