10.2 Rotation with Constant Angular Acceleration

Kinematics of Rotational Motion

Using our intuition, we can begin to see how the rotational quantities $\theta ,$ $\omega ,$ $\alpha$, and t are related to one another. For example, we saw in the preceding section that if a flywheel has an angular acceleration in the same direction as its angular velocity vector, its angular velocity increases with time and its angular displacement also increases. On the contrary, if the angular acceleration is opposite to the angular velocity vector, its angular velocity decreases with time. We can describe these physical situations and many others with a consistent set of rotational kinematic equations under a constant angular acceleration. The method to investigate rotational motion in this way is called kinematics of rotational motion.

To begin, we note that if the system is rotating under a constant acceleration, then the average angular velocity follows a simple relation because the angular velocity is increasing linearly with time. The average angular velocity is just half the sum of the initial and final values:

From the definition of the average angular velocity, we can find an equation that relates the angular position, average angular velocity, and time:

Solving for $\theta$, we have

where we have set $t_0=0$. This equation can be very useful if we know the average angular velocity of the system. Then we could find the angular displacement over a given time period. Next, we find an equation relating $\omega$, $\alpha$, and t. To determine this equation, we start with the definition of angular acceleration:

We rearrange this to get $\alpha dt=d\omega$ and then we integrate both sides of this equation from initial values to final values, that is, from $t_0$ to t and ${\omega }_0\text{to}{\omega }_{\text{f}}$. In uniform rotational motion, the angular acceleration is constant so it can be pulled out of the integral, yielding two definite integrals:

Setting $t_0=0$, we have

We rearrange this to obtain

where ${\omega }_0$ is the initial angular velocity. Equation 10.11 is the rotational counterpart to the linear kinematics equation $v_{\text{f}}=v_0+at$. With Equation 10.11, we can find the angular velocity of an object at any specified time t given the initial angular velocity and the angular acceleration.

Let’s now do a similar treatment starting with the equation $\omega =\frac{d\theta }{dt}$. We rearrange it to obtain $\omega dt=d\theta$ and integrate both sides from initial to final values again, noting that the angular acceleration is constant and does not have a time dependence. However, this time, the angular velocity is not constant (in general), so we substitute in what we derived above:

where we have set $t_0=0$. Now we rearrange to obtain

Equation 10.12 is the rotational counterpart to the linear kinematics equation found in Motion Along a Straight Line for position as a function of time. This equation gives us the angular position of a rotating rigid body at any time t given the initial conditions (initial angular position and initial angular velocity) and the angular acceleration.

We can find an equation that is independent of time by solving for t in Equation 10.11 and substituting into Equation 10.12. Equation 10.12 becomes

or

Equation 10.10 through Equation 10.13 describe fixed-axis rotation for constant acceleration and are summarized in Table 10.1.

Applying the Equations for Rotational Motion

Now we can apply the key kinematic relations for rotational motion to some simple examples to get a feel for how the equations can be applied to everyday situations.

Example 10.4

Calculating the Acceleration of a Fishing Reel

A deep-sea fisherman hooks a big fish that swims away from the boat, pulling the fishing line from his fishing reel. The whole system is initially at rest, and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of $110{\text{rad/s}}^2$ for 2.00 s (Figure 10.11).

(a) What is the final angular velocity of the reel after 2 s?

(b) How many revolutions does the reel make?

Figure 10.11 Fishing line coming off a rotating reel moves linearly.

Strategy

Identify the knowns and compare with the kinematic equations for constant acceleration. Look for the appropriate equation that can be solved for the unknown, using the knowns given in the problem description.

Solution

1. We are given $\alpha$ and t and want to determine $\omega$. The most straightforward equation to use is ${\omega }_{\text{f}}={\omega }_0+\alpha t$, since all terms are known besides the unknown variable we are looking for. We are given that ${\omega }_0=0$ (it starts from rest), so
   $${\omega }_{\text{f}}=0+(110{\text{rad/s}}^2)(2.00\text{s})=220\text{rad/s}.$$
2. We are asked to find the number of revolutions. Because $1\text{rev}=2\pi \text{rad}$, we can find the number of revolutions by finding $\theta$ in radians. We are given $\alpha$ and t, and we know ${\omega }_0$ is zero, so we can obtain $\theta$ by using
   $$\begin{array}{cc}\hfill {\theta }_{\text{f}}& ={\theta }_{\text{i}}+{\omega }_{\text{i}}t+\frac{1}{2}\alpha t^2\hfill \\ & =0+0+\left(0.500\right)\left(110{\text{rad/s}}^2\right){\left(2.00\text{s}\right)}^2=220\text{rad}\text{.}\hfill \end{array}$$
   Converting radians to revolutions gives
   $$\text{Number of rev}=(220\text{rad})\frac{1\text{rev}}{2\pi \text{rad}}=35.0\text{rev}\text{.}$$

Significance

This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. The answers to the questions are realistic. After unwinding for two seconds, the reel is found to spin at 220 rad/s, which is 2100 rpm. (No wonder reels sometimes make high-pitched sounds.)

In the preceding example, we considered a fishing reel with a positive angular acceleration. Now let us consider what happens with a negative angular acceleration.

Example 10.6

Angular Acceleration of a Propeller

Figure 10.12 shows a graph of the angular velocity of a propeller on an aircraft as a function of time. Its angular velocity starts at 30 rad/s and drops linearly to 0 rad/s over the course of 5 seconds. (a) Find the angular acceleration of the object and verify the result using the kinematic equations. (b) Find the angle through which the propeller rotates during these 5 seconds and verify your result using the kinematic equations.

Figure 10.12 A graph of the angular velocity of a propeller versus time.

Strategy

1. Since the angular velocity varies linearly with time, we know that the angular acceleration is constant and does not depend on the time variable. The angular acceleration is the slope of the angular velocity vs. time graph, $\alpha =\frac{d\omega }{dt}$. To calculate the slope, we read directly from Figure 10.12, and see that ${\omega }_0=30\text{rad/s}$ at $t=0\text{s}$ and ${\omega }_{\text{f}}=0\text{rad/s}$ at $t=5\text{s}$. Then, we can verify the result using $\omega ={\omega }_0+\alpha t$.
2. We use the equation $\omega =\frac{d\theta }{dt};$ since the time derivative of the angle is the angular velocity, we can find the angular displacement by integrating the angular velocity, which from the figure means taking the area under the angular velocity graph. In other words:
   $${\displaystyle \underset{{\theta }_0}{\overset{{\theta }_{\text{f}}}{\int }}d\theta ={\theta }_{\text{f}}-{\theta }_0=}{\displaystyle \underset{t_0}{\overset{t_{\text{f}}}{\int }}\omega (t)dt}.$$
   Then we use the kinematic equations for constant acceleration to verify the result.

Solution

1. Calculating the slope, we get
   $$\alpha =\frac{\omega -{\omega }_0}{t-t_0}=\frac{(0-30.0)\text{rad/s}}{(5.0-0)\text{s}}=\mathrm{-6.0}{\text{rad/s}}^2.$$
   We see that this is exactly Equation 10.11 with a little rearranging of terms.
2. We can find the area under the curve by calculating the area of the right triangle, as shown in Figure 10.13.
   

   

   Figure 10.13 The area under the curve is the area of the right triangle.
   $$\begin{array}{ccc}\hfill \text{Δ}\theta & =\hfill & \text{area}\left(\text{triangle}\right);\hfill \\ \hfill \text{Δ}\theta & =\hfill & \frac{1}{2}(30\text{rad/s})(5\text{s})=75\text{rad}\text{.}\hfill \end{array}$$
   We verify the solution using Equation 10.12:
   $${\theta }_{\text{f}}={\theta }_0+{\omega }_{0}t+\frac{1}{2}\alpha t^2.$$
   Setting ${\theta }_0=0$, we have
   $${\theta }_f=(30.0\text{rad}\text{/}\text{s})(5.0\text{s})+\frac{1}{2}{(\mathrm{-6.0}\text{rad}\text{/}{\text{s}}^2)(5.0\text{rad}\text{/}\text{s})}^2=150.0-75.0=75.0\text{rad}.$$
   This verifies the solution found from finding the area under the curve.

Significance

We see from part (b) that there are alternative approaches to analyzing fixed-axis rotation with constant acceleration. We started with a graphical approach and verified the solution using the rotational kinematic equations. Since $\alpha =\frac{d\omega }{dt}$, we could do the same graphical analysis on an angular acceleration-vs.-time curve. The area under an $\alpha \text{-vs.-}t$ curve gives us the change in angular velocity. Since the angular acceleration is constant in this section, this is a straightforward exercise.