Chapter 12, Problem 57PQ

(a)

To determine

The values of $b$ and ${\omega }_0$.

Answer to Problem 57PQ

The values of $b$ and ${\omega }_0$ are $0.277{\text{s}}^{-1}$ and $4.00\text{rad/s}$ respectively.

Explanation of Solution

Write the equation of angular speed of the wheel.

  $\frac{d\theta }{dt}={\omega }_0{e}^{-bt}$                                                                                            (I)

Here, $d\theta /dt$ is the rate of change of angular displacement, ${\omega }_0$ is the initial angular velocity of the wheel, $b$ is the constant, and $t$ is the time interval.

Write the expression for angular velocity.

  $\omega =\frac{d\theta }{dt}$                                                                                                  (II)

Here, $\omega$ is the angular speed of the wheel.

Compare equation (I) and (II).

  $\omega ={\omega }_0{e}^{-bt}$                                                                                               (III)

Conclusion:

Substitute $0\text{s}$ for $t$ and $4.00\text{rad/s}$ for $\omega$ in the equation (III) to find ${\omega }_0$.

  $\begin{array}{c}4.00\text{rad/s}={\omega }_0{e}^{-b\left(0\text{s}\right)}\\ 4.00\text{rad/s}={\omega }_0{e}^0\\ 4.00\text{rad/s}={\omega }_0\end{array}$

Substitute $5.00\text{s}$ for $t$, $4.00\text{rad/s}$ for ${\omega }_0$, and $1.00\text{rad/s}$ for $\omega$ in the equation (III) to find $b$.

  $\begin{array}{c}1.00\text{rad/s}=\left(4.00\text{rad/s}\right)e^{-b\left(5.00\text{s}\right)}\\ e^{-b\left(5.00\text{s}\right)}=\frac{1.00\text{rad/s}}{4.00\text{rad/s}}\\ e^{-b\left(5.00\text{s}\right)}=0.25\end{array}$

Take natural logarithms on both the side of the equation.

  $\begin{array}{c}\ln \left(e^{-b\left(5.00\text{s}\right)}\right)=\ln \left(0.25\right)\\ -b\left(5.00\text{s}\right)=-1.386\\ b\left(5.00\text{s}\right)=1.386\\ b=0.277{\text{s}}^{-1}\end{array}$

Therefore, the values of $b$ and ${\omega }_0$ are $0.277{\text{s}}^{-1}$ and $4.00\text{rad/s}$.

(b)

To determine

The magnitude of the angular acceleration of the disk at $t=5.00\text{s}$.

Answer to Problem 57PQ

The magnitude of the angular acceleration of the disk at $t=5.00\text{s}$ is $0.277{\text{rad/s}}^2$.

Explanation of Solution

Write the expression for angular acceleration.

  $\alpha =\frac{d\omega }{dt}$                                                                                          (IV)

Here, $\alpha$ is the angular acceleration and $d\omega /dt$ is the rate of change of angular velocity.

Substitute equation (III) in the equation (IV).

  $\begin{array}{c}\alpha =\frac{d\left({\omega }_0{e}^{-bt}\right)}{dt}\\ ={\omega }_0\frac{d{e}^{-bt}}{dt}\\ ={\omega }_0\left(-b\right)e^{-bt}\end{array}$                                                                                    (V)

Conclusion:

Substitute $5.00\text{s}$ for $t$, $4.00\text{rad/s}$ for ${\omega }_0$, and $0.277{\text{s}}^{-1}$ for $b$ in the equation (V) to find $\alpha$.

  $\begin{array}{c}\alpha =\left(4.00\text{rad/s}\right)\left(-0.277{\text{s}}^{-1}\right)e^{-\left(0.277{\text{s}}^{-1}\right)\left(5.00\text{s}\right)}\\ =\left(-1.108{\text{rad/s}}^2\right)e^{-1.385}\\ =\left(-1.108{\text{rad/s}}^2\right)\left(0.250\right)\\ =-0.277{\text{rad/s}}^2\end{array}$

Write the magnitude of the angular acceleration.

  $\left|\alpha \right|=0.277{\text{rad/s}}^2$

Therefore, the magnitude of the angular acceleration of the disk at $t=5.00\text{s}$ is $0.277{\text{rad/s}}^2$.

(c)

To determine

The revolution of the disk during the time interval $t=0\text{ to }t=5.00\text{s}$.

Answer to Problem 57PQ

The revolution of the disk during the time interval $t=0\text{ to }t=5.00\text{s}$ is $1.72\text{rev}$.

Explanation of Solution

Rewrite the expression (II) for angular displacement.

  $\Delta \theta =\omega dt$                                                                                    (VI)

Here, $\Delta \theta$ is the change in angular displacement.

Integrating the equation (VI) using the limits.

  $\Delta \theta ={\displaystyle \underset{0}{\overset{t}{\int }}\omega dt}$

Substitute equation (III) in the above equation.

  $\begin{array}{c}\Delta \theta ={\displaystyle \underset{0}{\overset{t}{\int }}{\omega }_0{e}^{-bt}dt}\\ =\frac{{\omega }_0}{-b}\left[e^{-bt}-e^0\right]\\ =\frac{{\omega }_0}{-b}\left[e^{-bt}-1\right]\\ =\frac{{\omega }_0}{b}\left[1-e^{-bt}\right]\end{array}$                                                                        (VII)

Conclusion:

Substitute $5.00\text{s}$ for $t$, $4.00\text{rad/s}$ for ${\omega }_0$, and $0.277{\text{s}}^{-1}$ for $b$ in the equation (VII) to find $\Delta \theta$.

  $\begin{array}{c}\Delta \theta =\frac{4.00\text{rad/s}}{0.277{\text{s}}^{-1}}\left[1-e^{-\left(0.277{\text{s}}^{-1}\right)\left(5.00\text{s}\right)}\right]\\ =14.4\text{rad}\left[1-e^{-1.385}\right]\\ =14.4\text{rad}\left(0.75\right)\\ =10.8\text{rad}\end{array}$

Convert the radians to revolutions.

  $\begin{array}{c}\Delta \theta =10.8\text{rad}\left(\frac{1\text{rev}}{2\pi \text{rad}}\right)\\ =1.72\text{rev}\end{array}$

Therefore, the revolution of the disk during the time interval $t=0\text{ to }t=5.00\text{s}$ is $1.72\text{rev}$.