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:

$$\stackrel{-}{\omega }=\frac{{\omega }_0+{\omega }_{\text{f}}}{2}.$$

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

$$\stackrel{-}{\omega }=\frac{\text{Δ}\theta }{\text{Δ}t}.$$

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:

$$\alpha =\frac{d\omega }{dt}.$$

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:

$$\alpha {\displaystyle \underset{t_0}{\overset{t}{\int }}d{t}^{\prime }}={\displaystyle \underset{{\omega }_0}{\overset{{\omega }_{\text{f}}}{\int }}d\omega }.$$

Setting $t_0=0$, we have

$$\alpha t={\omega }_f-{\omega }_0.$$

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:

$$\begin{array}{c}{\displaystyle \underset{t_0}{\overset{t_f}{\int }}({\omega }_0+\alpha t^{\prime })d{t}^{\prime }={\displaystyle \underset{{\theta }_0}{\overset{{\theta }_{\text{f}}}{\int }}d\theta };}\hfill \\ {\displaystyle \underset{t_0}{\overset{t}{\int }}{\omega }_{0}dt+{\displaystyle \underset{t_0}{\overset{t}{\int }}\alpha t^{\prime }d{t}^{\prime }=}{\displaystyle \underset{{\theta }_0}{\overset{{\theta }_{\text{f}}}{\int }}d\theta }={\left[{\omega }_0{t}^{\prime }+\alpha \left(\frac{{(t^{\prime })}^2}{2}\right)\right]}_{t_0}^t={\omega }_{0}t+\alpha \left(\frac{t^2}{2}\right)={\theta }_{\text{f}}-{\theta }_0,}\hfill \end{array}$$

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

$$\begin{array}{ccc}\hfill {\theta }_{\text{f}}& =\hfill & {\theta }_0+{\omega }_0\left(\frac{{\omega }_{\text{f}}-{\omega }_0}{\alpha }\right)+\frac{1}{2}\alpha {\left(\frac{{\omega }_{\text{f}}-{\omega }_0}{\alpha }\right)}^2\hfill \\ & =\hfill & {\theta }_0+\frac{{\omega }_0{\omega }_{\text{f}}}{\alpha }-\frac{{\omega }_0^2}{\alpha }+\frac{1}{2}\frac{{\omega }_{\text{f}}^2}{\alpha }-\frac{{\omega }_0{\omega }_{\text{f}}}{\alpha }+\frac{1}{2}\frac{{\omega }_0^2}{\alpha }\hfill \\ & =\hfill & {\theta }_0+\frac{1}{2}\frac{{\omega }_{\text{f}}^2}{\alpha }-\frac{1}{2}\frac{{\omega }_0^2}{\alpha },\hfill \\ {\theta }_{\text{f}}-{\theta }_0\hfill & =\hfill & \frac{{\omega }_{\text{f}}^2-{\omega }_0^2}{2\alpha }\hfill \end{array}$$

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.

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.

Summary

• The kinematics of rotational motion describes the relationships among rotation angle (angular position), angular velocity, angular acceleration, and time.
• For a constant angular acceleration, the angular velocity varies linearly. Therefore, the average angular velocity is 1/2 the initial plus final angular velocity over a given time period:

  $$\stackrel{-}{\omega }=\frac{{\omega }_0+{\omega }_{\text{f}}}{2}.$$

• We used a graphical analysis to find solutions to fixed-axis rotation with constant angular acceleration. From the relation $\omega =\frac{d\theta }{dt}$, we found that the area under an angular velocity-vs.-time curve gives the angular displacement, ${\theta }_{\text{f}}-{\theta }_0=\text{Δ}\theta ={\displaystyle \underset{t_0}{\overset{t}{\int }}\omega (t)dt}$. The results of the graphical analysis were verified using the kinematic equations for constant angular acceleration. Similarly, since $\alpha =\frac{d\omega }{dt}$, the area under an angular acceleration-vs.-time graph gives the change in angular velocity: ${\omega }_f-{\omega }_0=\text{Δ}\omega ={\displaystyle \underset{t_0}{\overset{t}{\int }}\alpha (t)dt}$.

Conceptual Questions

Problems

The angular velocity vs. time for a fan on a hovercraft is shown below. (a) What is the angle through which the fan blades rotate in the first 8 seconds? (b) Verify your result using the kinematic equations.