Example

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)).

(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
   $$\begin{array}{cc}\hfill {\theta }_{\text{f}}& ={\theta }_{\text{i}}+{\omega }_{\text{i}}t+\frac{1}{2}\alpha {t}^{2}\hfill \\ & =0+0+(0.500)(110{\,\text{rad/s}}^{2}){(2.00\,\text{s})}^{2}=220\,\text{rad}\text{.}\hfill \end{array}$$

   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.)

Example

Calculating the Duration When the Fishing Reel Slows Down and Stops

Now the fisherman applies a brake to the spinning reel, achieving an angular acceleration of $$ -300{\,\text{rad/s}}^{2}$$. How long does it take the reel to come to a stop?

Strategy

We are asked to find the time t for the reel to come to a stop. The initial and final conditions are different from those in the previous problem, which involved the same fishing reel. Now we see that the initial angular velocity is $$ {\omega }_{0}=220\,\text{rad}\text{/}\text{s} $$ and the final angular velocity $$ \omega $$ is zero. The angular acceleration is given as $$ \alpha =-300\,{\text{rad/s}}^{2}. $$ Examining the available equations, we see all quantities but t are known in $$ {\omega }_{\text{f}}={\omega }_{0}+\alpha t$$, making it easiest to use this equation.

Solution

The equation states

We solve the equation algebraically for t and then substitute the known values as usual, yielding

Significance

Note that care must be taken with the signs that indicate the directions of various quantities. Also, note that the time to stop the reel is fairly small because the acceleration is rather large. Fishing lines sometimes snap because of the accelerations involved, and fishermen often let the fish swim for a while before applying brakes on the reel. A tired fish is slower, requiring a smaller acceleration.

Example

Angular Acceleration of a Propeller(Figure) 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), 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:
   $$\underset{{\theta }_{0}}{\overset{{\theta }_{\text{f}}}{\int }}d\theta ={\theta }_{\text{f}}-{\theta }_{0}=\underset{{t}_{0}}{\overset{{t}_{\text{f}}}{\int }}\omega (t)dt.$$

   Then we use the kinematic equations for constant acceleration to verify the result.

Solution

1. Calculating the slope, we get
   $$\alpha =\frac{\omega -{\omega }_{0}}{t-{t}_{0}}=\frac{(0-30.0)\,\text{rad/s}}{(5.0-0)\,\text{s}}=-6.0\,{\text{rad/s}}^{2}.$$

   We see that this is exactly (Figure) 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).
   

   Figure 10.13 The area under the curve is the area of the right triangle.
   $$\begin{array}{ccc}\hfill \text{Δ}\theta & =\hfill & \text{area}(\text{triangle});\hfill \\ \hfill \text{Δ}\theta & =\hfill & \frac{1}{2}(30\,\text{rad/s})(5\,\text{s})=75\,\text{rad}\text{.}\hfill \end{array}$$

   We verify the solution using (Figure):

   $${\theta }_{\text{f}}={\theta }_{0}+{\omega }_{0}t+\frac{1}{2}\alpha {t}^{2}.$$

   Setting $$ {\theta }_{0}=0$$, we have

   $${\theta }_{0}=(30.0\,\text{rad}\text{/}\text{s})(5.0\,\text{s})+\frac{1}{2}{(-6.0\,\text{rad}\text{/}{\text{s}}^{2})(5.0\,\text{rad}\text{/}\text{s})}^{2}=150.0-75.0=75.0\,\text{rad}.$$

   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.