# 10.1 Rotational variables  (Page 5/11)

 Page 5 / 11

Now let’s apply this problem-solving strategy to a few specific examples.

## A spinning bicycle wheel

A bicycle mechanic mounts a bicycle on the repair stand and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00 s. (a) Calculate the average angular acceleration in ${\text{rad/s}}^{2}$ . (b) If she now hits the brakes, causing an angular acceleration of −87.3 ${\text{rad/s}}^{2}$ , how long does it take the wheel to stop?

## Strategy

The average angular acceleration can be found directly from its definition $\stackrel{–}{\alpha }=\frac{\text{Δ}\omega }{\text{Δ}t}$ because the final angular velocity and time are given. We see that $\text{Δ}\omega ={\omega }_{\text{final}}-{\omega }_{\text{initial}}=250\phantom{\rule{0.2em}{0ex}}\text{rev/min}$ and $\text{Δ}t$ is 5.00 s. For part (b), we know the angular acceleration and the initial angular velocity. We can find the stopping time by using the definition of average angular acceleration and solving for $\text{Δ}t$ , yielding

$\text{Δ}t=\frac{\text{Δ}\omega }{\alpha }.$

## Solution

1. Entering known information into the definition of angular acceleration, we get
$\stackrel{–}{\alpha }=\frac{\text{Δ}\omega }{\text{Δ}t}=\frac{250\phantom{\rule{0.2em}{0ex}}\text{rpm}}{5.00\phantom{\rule{0.2em}{0ex}}\text{s}}.$

Because $\text{Δ}\omega$ is in revolutions per minute (rpm) and we want the standard units of ${\text{rad/s}}^{2}$ for angular acceleration, we need to convert from rpm to rad/s:
$\text{Δ}\omega =250\frac{\text{rev}}{\text{min}}·\frac{2\pi \phantom{\rule{0.2em}{0ex}}\text{rad}}{\text{rev}}·\frac{1\phantom{\rule{0.2em}{0ex}}\text{min}}{60\phantom{\rule{0.2em}{0ex}}\text{s}}=26.2\frac{\text{rad}}{\text{s}}.$

Entering this quantity into the expression for $\alpha$ , we get
$\alpha =\frac{\text{Δ}\omega }{\text{Δ}t}=\frac{26.2\phantom{\rule{0.2em}{0ex}}\text{rad/s}}{5.00\phantom{\rule{0.2em}{0ex}}\text{s}}=5.24{\phantom{\rule{0.2em}{0ex}}\text{rad/s}}^{2}.$
2. Here the angular velocity decreases from 26.2 rad/s (250 rpm) to zero, so that $\text{Δ}\omega$ is −26.2 rad/s, and $\alpha$ is given to be –87.3 ${\text{rad/s}}^{2}$ . Thus,
$\text{Δ}t=\frac{-26.2\phantom{\rule{0.2em}{0ex}}\text{rad/s}}{-87.3{\phantom{\rule{0.2em}{0ex}}\text{rad/s}}^{2}}=0.300\phantom{\rule{0.2em}{0ex}}\text{s}.$

## Significance

Note that the angular acceleration as the mechanic spins the wheel is small and positive; it takes 5 s to produce an appreciable angular velocity. When she hits the brake, the angular acceleration is large and negative. The angular velocity quickly goes to zero.

Check Your Understanding The fan blades on a turbofan jet engine (shown below) accelerate from rest up to a rotation rate of 40.0 rev/s in 20 s. The increase in angular velocity of the fan is constant in time. (The GE90-110B1 turbofan engine mounted on a Boeing 777, as shown, is currently the largest turbofan engine in the world, capable of thrusts of 330–510 kN.)

(a) What is the average angular acceleration?

(b) What is the instantaneous angular acceleration at any time during the first 20 s?

a. $40.0\phantom{\rule{0.2em}{0ex}}\text{rev}\text{/}\text{s}=2\pi \left(40.0\right)\phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s}$ , $\stackrel{–}{\alpha }=\frac{\text{Δ}\omega }{\text{Δ}t}=\frac{2\pi \left(40.0\right)-0\phantom{\rule{0.2em}{0ex}}\text{rad/s}}{20.0\phantom{\rule{0.2em}{0ex}}\text{s}}=2\pi \left(2.0\right)=4.0\pi \phantom{\rule{0.2em}{0ex}}\text{rad/}{\text{s}}^{2}$ ; b. Since the angular velocity increases linearly, there has to be a constant acceleration throughout the indicated time. Therefore, the instantaneous angular acceleration at any time is the solution to $4.0\pi \phantom{\rule{0.2em}{0ex}}\text{rad/}{\text{s}}^{2}$ .

## Wind turbine

A wind turbine ( [link] ) in a wind farm is being shut down for maintenance. It takes 30 s for the turbine to go from its operating angular velocity to a complete stop in which the angular velocity function is $\omega \left(t\right)=\left[{\left(t{\text{s}}^{-1}-30.0\right)}^{2}\text{/}100.0\right]\text{rad}\text{/}\text{s}$ . If the turbine is rotating counterclockwise looking into the page, (a) what are the directions of the angular velocity and acceleration vectors? (b) What is the average angular acceleration? (c) What is the instantaneous angular acceleration at $t=0.0,15.0,30.0\phantom{\rule{0.2em}{0ex}}\text{s}?$

## Strategy

1. We are given the rotational sense of the turbine, which is counterclockwise in the plane of the page. Using the right hand rule ( [link] ), we can establish the directions of the angular velocity and acceleration vectors.
2. We calculate the initial and final angular velocities to get the average angular acceleration. We establish the sign of the angular acceleration from the results in (a).
3. We are given the functional form of the angular velocity, so we can find the functional form of the angular acceleration function by taking its derivative with respect to time.

## Solution

1. Since the turbine is rotating counterclockwise, angular velocity $\stackrel{\to }{\omega }$ points out of the page. But since the angular velocity is decreasing, the angular acceleration $\stackrel{\to }{\alpha }$ points into the page, in the opposite sense to the angular velocity.
2. The initial angular velocity of the turbine, setting $t=0,\phantom{\rule{0.2em}{0ex}}\text{is}\phantom{\rule{0.2em}{0ex}}\omega =9.0\phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s}$ . The final angular velocity is zero, so the average angular acceleration is
$\stackrel{–}{\alpha }=\frac{\text{Δ}\omega }{\text{Δ}t}=\frac{\omega -{\omega }_{0}}{t-{t}_{0}}=\frac{0-9.0\phantom{\rule{0.2em}{0ex}}\text{rad/s}}{30.0-0\phantom{\rule{0.2em}{0ex}}\text{s}}=-0.3\phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}{\text{s}}^{2}.$
3. Taking the derivative of the angular velocity with respect to time gives $\alpha =\frac{d\omega }{dt}=\left(t-30.0\right)\text{/}50.0\phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}{\text{s}}^{2}$
$\alpha \left(0.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=-0.6\phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}{\text{s}}^{2},\alpha \left(15.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=-0.3\phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}{\text{s}}^{2},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\alpha \left(30.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=0\phantom{\rule{0.2em}{0ex}}\text{rad/s}.$

## Significance

We found from the calculations in (a) and (b) that the angular acceleration $\alpha$ and the average angular acceleration $\stackrel{–}{\alpha }$ are negative. The turbine has an angular acceleration in the opposite sense to its angular velocity.

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