## Section9.4Rotational Acceleration

Just as angular velocity $\omega$ is the rate of change of rotation angle $\theta\text{,}$ angular or rotational acceleration is the rate of change of angular velocity $\omega\text{.}$ We denote angular acceleration by the Greek letter $\alpha$ (alpha). Similar to the angular velocity, we have average angular acceleration over some period of time and an instantaneous angular velocity at tha particular instant. Also, it is important to note that angular acceleration is also a vector. In the following we will be working in a fixed-axis rotation and use simple notation of using only velocity and acceleration components along the axis of rotation, when we mean their vectors.

### Subsection9.4.1Average Angular Acceleration

By dividing the change in angular velocity by the time interval we get the average angular acceleration for that time interval. Consider an interval from $t_i$ to $t_f\text{.}$ Let the instantaneous angular velocity at these instants be $\omega_i$ and $\omega_f \text{.}$ Then, the average angular acceleration, $\alpha_{\text{av}}$ will be

\begin{equation} \alpha_{\text{av}} = \dfrac{\omega_f - \omega_i }{t_f - t_i} \equiv \dfrac{\Delta \omega }{\Delta t}.\tag{9.4.1} \end{equation}

This is component of angular acceleration vector along the axis of rotation. Since, we will work with only fixed-axis rotation in this chapter, we need only this component.

### Subsection9.4.2Instantaneous Angular Acceleration

Often we want angular acceleration at a particular instant. What we mean by this is the average angular acceleration between that instant, say $t$ and an instant very close to $t\text{,}$ $t + \Delta t \text{.}$

Denoting the instantaneous angular acceleration by symbol $\alpha \text{,}$ we write this special average quantity formally as

\begin{equation} \alpha = \dfrac{\omega\left(\text{at } (t+\Delta t)\right) - \omega\left(\text{at } t\right)}{\Delta t} = \frac{\Delta \omega}{\Delta t}.\label{eq-instantaneous-angular-acceleration-limit}\tag{9.4.2} \end{equation}

Since $\alpha$ is a ratio of a change in $\omega$ to a change in $t$ it can be computed by the slope of $\omega$ versus $t$ plot. The “as close as possible” is more precisely implemented in Calculus by the concept of derivative.

The angular acceleration is a vector quantity. Therefore, it has a direction and a magnitude. The direction for a fixed-axis rotation wil either be in the direction of the axis of rotation or in the opposite direction as illustrated in Figure 9.4.1.

To tell the direction of angular acceleration, you will need to look at the direction of angular velocity and whether the motion is slowing or speeding up. The angular velocity is in the direction of the axis if the object is rotating counterclockwise otherwise it is in the opposite direction. The angular acceleration will then be in the direction of the angular velocity or opposite to it depending upon whether the rotation is speeding up or slowing down.

### Subsection9.4.3Instantaneous Angular Acceleration - Graphically

The rate of change of $\omega$ at an instant can be computed by the slope of the tangent to $\omega$ versus $t$ plot - similar to the way we defined the rate of chang of $x \text{.}$

If angular velocity is increasing or decreasing at a constant rate, we say the object has uniform angular acceleration. An example is shown in Figure 9.4.2.

Since $\omega$ versus $t$ plot is a straight line, angular velocity changes uniformly wth time with a constant acceleration. From slope, we get angular acceleration, $\alpha\text{.}$

\begin{equation*} \alpha = \alpha_\text{av} = \dfrac{\omega_2 - \omega_1}{t_2 - t_1}. \end{equation*}

On the other hand, if the angular velocity is increasing or decreasing at a non-constant rate, $\omega$ versus $t$ plot will not be a straight line. An example is shown in Figure 9.4.3. In this case, slope of tangent to the curve at the instant of interest gives acceleration at that instant.

The figure shows tangent lines for instants $t_1$ and $t_2\text{.}$ The slope of the tangent lines $L_1$ and $L_2$ are then the instantaneous angular accelerations at these instants.

Figure 9.4.5 shows the angular velocity of a rotating fan at different times. We wish to study this plot and determine the following aspects of the rotation.

(a) Is the rotation at (i) $t = 5\text{ sec} \text{,}$ (ii) $t = 15\text{ sec} \text{,}$ (iii) $t = 25\text{ sec} \text{,}$ (iv) $t = 35\text{ sec} \text{,}$ (v) $t = 45\text{ sec} \text{,}$ (vi) $t = 55\text{ sec} \text{,}$ clockwise or a counterclockwise or no rotation? How can you tell?

(b) What are the values of instantaneous angular accelerations at (i) $t = 5\text{ sec} \text{,}$ (ii) $t = 15\text{ sec} \text{,}$ (iii) $t = 25\text{ sec} \text{,}$ (iv) $t = 35\text{ sec} \text{,}$ (v) $t = 45\text{ sec} \text{,}$ (vi) $t = 55\text{ sec} \text{?}$

(c) What are the average accelerations during the following intervals (i) $t = 0$ to $t = 20\text{ sec} \text{,}$ (ii) $t = 20\text{ sec}$ to $t = 30\text{ sec} \text{,}$ (iii) $t = 30\text{ sec}$ to $t = 40\text{ sec} \text{,}$ (iv) $t = 40\text{ sec}$ to $t = 50\text{ sec} \text{,}$ (v) $t = 50\text{ sec}$ to $t = 60\text{ sec} \text{,}$ and (vi) $t = 0$ to $t = 60\text{ sec} \text{.}$

Hint 1 (a)

Look at the sign of $\omega\text{.}$

Hint 2 (b) and (c)

Use slope

(a) Counterclockwise: (i), (ii), (iii); No rotation at (iv), and clockwise at (v), and (vi).

(b) (i) $1\text{ rad/sec}^2\text{,}$ (ii) $1\text{ rad/sec}^2\text{,}$ (iii) $0\text{,}$ (iv) $-4\text{ rad/sec}^2$ , (v) $1\text{ rad/sec}^2$ , and (vi) $0\text{.}$

(c) (i) $1\text{ rad/sec}^2\text{,}$ (ii)$0\text{,}$ (iii) $-4\text{ rad/sec}^2$ , (iv) $1\text{ rad/sec}^2$ , (v) $0\text{,}$ and (vi) $-\dfrac{1}{6}\,\text{ rad/sec}^2\text{.}$

Solution 1 (a)

(a) From the definition of angular velocity, we know that a clockwise rotation will have a negative value of $\omega$ and a counterclockwise rotation a positive value. Therefore, we can answer this question immediately from reading off the sign of $\omega \text{.}$

Counterclockwise at (i) $t = 5\text{ sec} \text{,}$ (ii) $t = 15\text{ sec} \text{,}$ and (iii) $t = 25\text{ sec} \text{,}$

no rotation at (iv) $t = 35\text{ sec} \text{,}$ and

Clockwise at (v) $t = 45\text{ sec} \text{,}$ and (vi) $t = 55\text{ sec} \text{.}$

Solution 2 (b)

(b) Since the segments are all linear, we do not need to draw any tangents; we just use the slope of the line upon which the point of interest falls.

(i) The slope here is

\begin{equation*} \alpha = \dfrac{20-0}{20-0} = 1\text{ rad/sec}^2. \end{equation*}

(ii) We do not need to do any work. This is same as (i).

(iii) The angular velocity is not changing here. Therefore, the slope here is zero. Hence, $\alpha = 0\text{.}$

(iv) the slope here is

\begin{equation*} \alpha = \dfrac{-20-20}{40-30} = -4\text{ rad/sec}^2. \end{equation*}

(v) The slope here is

\begin{equation*} \alpha = \dfrac{-10-(-20)}{50-40} = 1\text{ rad/sec}^2. \end{equation*}

You could have just used the parallel nature of this line to the line for (i) and guessed the answer.

(vi) The angular velocity is not changing here. Hence, $\alpha = 0\text{.}$

Solution 3 (c)

(c) We just use the values of angular velocity and times at the end points of the segment and calculate the slopes. The slopes here will give us the average angular accelerations, and not the instantaneous angular accelerations.

(i) The slope here is

\begin{equation*} \alpha_{\text{av}} = \dfrac{20-0}{20-0} = 1\text{ rad/sec}^2. \end{equation*}

(ii) The angular velocity at the two ends is same. Therefore, the slope here is zero. Hence, $\alpha_{\text{av}} = 0\text{.}$

(iii) the slope here is

\begin{equation*} \alpha_{\text{av}} = \dfrac{-20-20}{40-30} = -4\text{ rad/sec}^2. \end{equation*}

(iv) The slope here is

\begin{equation*} \alpha_{\text{av}} = \dfrac{-10-(-20)}{50-40} = 1\text{ rad/sec}^2. \end{equation*}

(v) The final angular velocity is same as initial angular velocity, here. Hence, $\alpha_{\text{av}} = 0\text{.}$

(vi) The slope here is

\begin{equation*} \alpha_{\text{av}} = \dfrac{-10-0}{60-0} = -\dfrac{1}{6}\,\text{ rad/sec}^2. \end{equation*}

### Subsection9.4.4Angular Acceleration Vector

The rate of change of the angular velocity vector is the angular acceleration vector. Since a change in angular velocity vector can occur by a change in its magnitude or its direction, we need to deal with the change as vectors.

The average acceleration vector during an interval $t$ to $t + \Delta t$ will be given by

\begin{equation} \vec \alpha_{\text{av}} = \dfrac{\Delta\vec\omega}{\Delta t}.\label{eq-angular-acceleration-vector-average}\tag{9.4.3} \end{equation}

Of course, the instantaneous rate of change of the angular velocity is captured, as usual, by the average angular acceleration between two instants arbitrarily close to the instant of interest, e.g., between $t$ and another instant that is arbitrarily close to $t\text{.}$ We have learned how to handle this by the method of slope of the tangent to plots of components of $\vec\omega$ vs $t\text{,}$ which we will nit repeat here.

### Subsection9.4.5(Calculus) Instantaneous Angular Acceleration - Analytically

Analytically, the component of instantaneous angular acceleration along the axis of rotation is the derivative of the anglular velocity component with respect to time.

\begin{equation} \alpha = \dfrac{d\omega}{dt},\tag{9.4.4} \end{equation}

where $d\omega$ is positive for increasing counterclockwise rotation and negative for increasing clockwise rotation. Make sure you understand how the sign of $\alpha$ is related to what is happening with the rotation as stated in the following paragraph.

If $\omega \gt 0$ (i.e., rotating counterclockwise) and rotation is speeding up, then $\alpha \gt 0\text{.}$ If $\omega \lt 0$ (i.e., rotating clockwise) and rotation is speeding up, then $\alpha \lt 0\text{.}$ Now, if $\omega \gt 0$ (i.e., rotating counterclockwise) and rotation is slowing down, then $\alpha \lt 0\text{,}$ and if $\omega \lt 0$ (i.e., rotating clockwise) and rotation is slowing down, then $\alpha \gt 0\text{.}$

### Subsection9.4.6(Calculus) Angular Acceleration Vector

By taking $\Delta t \rightarrow 0$ in Eq. (9.4.3), we get the instantaneous angular acceleration as derivative of angular velocity vector.

\begin{equation} \vec \alpha = \dfrac{d\vec\omega}{d t}.\tag{9.4.5} \end{equation}