Chapter 5, Problem 48P

(a)

To determine

The angle does the wheel rotate during the next $1.0\text{s}$ time interval.

Answer to Problem 48P

The angle does the wheel rotate during the next $1.0\text{s}$ time interval is $270°$.

Explanation of Solution

Write the expression for angular displacement from the kinematics equation.

  $\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$        (I)

Here, $\theta$ is the angular displacement, ${\omega }_0$ is the initial angular velocity, $t$ is the time interval, and $\alpha$ is the angular acceleration of the wheel.

Substitute $0$ for ${\omega }_0$ and rewrite for $\alpha$.

  $\begin{array}{c}\theta =\left(0\right)t+\frac{1}{2}\alpha t^2\\ \theta =\frac{1}{2}\alpha t^2\\ \alpha =\frac{2\theta }{t^2}\end{array}$        (II)

Write the expression for angular velocity of the wheel from the kinematics equation.

  $\omega =\alpha t$        (III)

Here, $\omega$ is the final angular velocity of the wheel.

The angular displacement for the end of the time period $1.0\text{s}$.

  ${\theta }_1=\omega t+\frac{1}{2}\alpha t^2$        (IV)

Here, ${\theta }_1$ is the angle does the wheel rotate during the next $1.0\text{s}$ time interval.

Conclusion:

Substitute $90.0°$ for $\theta$ and $1.0\text{s}$ for $t$ in equation (II) to find $\alpha$.

  $\begin{array}{c}\alpha =\frac{2\left(90.0\text{degree}\right)\left(\frac{\pi /180}{1\text{degree}}\right)}{{\left(1.0\text{s}\right)}^2}\\ =\frac{2\left(\frac{2\pi }{4}\right)}{{\left(1.0\text{s}\right)}^2}\\ =\pi {\text{rad/s}}^2\end{array}$

Substitute $\pi {\text{rad/s}}^2$ for $\alpha$ and $1.0\text{s}$ for $t$ in equation (III) to find $\omega$.

  $\begin{array}{c}\omega =\left(\pi {\text{rad/s}}^2\right)\left(1.0\text{s}\right)\\ =\pi \text{rad/s}\end{array}$

Substitute $\pi {\text{rad/s}}^2$ for $\alpha$, $\pi \text{rad/s}$ for $\omega$, and $1.0\text{s}$ for $t$ in equation (IV) to find ${\theta }_1$.

  $\begin{array}{c}{\theta }_1=\left(\pi \text{rad/s}\right)\left(1.0\text{s}\right)+\frac{1}{2}\left(\pi {\text{rad/s}}^2\right){\left(1.0\text{s}\right)}^2\\ =\frac{3\pi }{2}\text{rad}\left(\frac{180/\pi }{1\text{rad}}\right)\\ =270°\end{array}$

Therefore, the angle does the wheel rotate during the next $1.0\text{s}$ time interval is $270°$.

(b)

To determine

The angle does the wheel rotate during the third $1.0\text{s}$ time interval.

Answer to Problem 48P

The angle does the wheel rotate during the third $1.0\text{s}$ time interval is $450°$.

Explanation of Solution

Write the expression for angular velocity of the wheel from the kinematics equation.

  ${\omega }^{\prime }=\omega +\alpha t$        (V)

Here, ${\omega }^{\prime }$ is the new angular velocity of the wheel.

The angular displacement for the end of the third time period $1.0\text{s}$.

  ${\theta }_2={\omega }^{\prime }t+\frac{1}{2}\alpha t^2$        (VI)

Here, ${\theta }_2$ is the angle does the wheel rotate during the third $1.0\text{s}$ time interval.

Conclusion:

Substitute $\pi \text{rad/s}$ for $\omega$, $\pi {\text{rad/s}}^2$ for $\alpha$ and $1.0\text{s}$ for $t$ in equation (V) to find ${\omega }^{\prime }$.

  $\begin{array}{c}\omega =\pi \text{rad/s}+\left(\pi {\text{rad/s}}^2\right)\left(1.0\text{s}\right)\\ =\pi \text{rad/s}+\pi \text{rad/s}\\ \text{=2}\pi \text{rad/s}\end{array}$

Substitute $2\pi \text{rad/s}$ for ${\omega }^{\prime }$, $\pi {\text{rad/s}}^2$ for $\alpha$ and $1.0\text{s}$ for $t$ in equation (VI) to find ${\omega }^{\prime }$.

  $\begin{array}{c}{\theta }_2=\left(2\pi \text{rad/s}\right)\left(1.0\text{s}\right)+\frac{1}{2}\left(\pi {\text{rad/s}}^2\right){\left(1.0\text{s}\right)}^2\\ =2\pi \text{rad}+\frac{\pi }{2}\text{rad}\\ \text{=}\frac{5\pi }{2}\text{rad}\left(\frac{180/\pi }{1\text{rad}}\right)\\ =450°\end{array}$

Therefore, the angle does the wheel rotate during the third $1.0\text{s}$ time interval is $450°$.