Chapter 5, Problem 53P

(a)

To determine

The time taken by the rotor to spin before coming to rest.

Answer to Problem 53P

The time taken by the rotor to spin before coming to rest is $\underset{\_}{\text{12}\text{.5}\times {\text{10}}^5\text{s}}$.

Explanation of Solution

Write the relation between angular speed and angular acceleration.

    ${\omega }_f-{\omega }_i=\alpha \Delta t$

Here, ${\omega }_f$ is the final angular speed, ${\omega }_i$ is the initial angular speed, $\alpha$ is the angular acceleration, and $\Delta t$ is the time taken by the rotor to spin.

Since the final angular speed of the rotor is zero and the rotor is slowing down at an angular acceleration of $-\alpha$ the above equation is reduced to

    $0-{\omega }_i=-\alpha \Delta t$

Rearrange the above equation to find $\Delta t$.

    $\Delta t=\frac{{\omega }_i}{\alpha }$        (I)

Conclusion:

Substitute $5\times {10}^5\text{rad}/\text{s}$ for ${\omega }_i$, and $0.4\text{rad}/{\text{s}}^2$ for $\alpha$ in equation (I), to find $\Delta t$.

    $\begin{array}{c}\Delta t=\frac{5\times {10}^5\text{rad}/\text{s}}{\left(0.4\text{rad}/{\text{s}}^2\right)}\\ =1250000\text{s}\\ \text{=12}\text{.5}\times {\text{10}}^5\text{s}\end{array}$

Therefore, the time taken by the rotor to spin before coming to rest is $\underset{\_}{\text{12}\text{.5}\times {\text{10}}^5\text{s}}$.

(b)

To determine

The angular displacement of the rotor during the time interval $\Delta t$.

Answer to Problem 53P

The angular displacement of the rotor during the time interval $\Delta t$ is $\underset{\_}{\text{5}\times {\text{10}}^{10}\text{rev}}$.

Explanation of Solution

Write the expression for angular displacement $\Delta \theta$.

    $\Delta \theta ={\omega }_i\Delta t+\frac{1}{2}\alpha {\left(\Delta t\right)}^2$        (II)

Here, $\Delta \theta$ is the angular displacement, ${\omega }_i$ is the initial angular speed, $\Delta t$ is the time taken to rotate, and $\alpha$ is the angular acceleration.

Write the expression for number of revolution of the rotor.

    $\text{The number of revolution =}\frac{\Delta \theta }{2\pi \text{rad}/\text{s}}$        (III)

Conclusion:

Substitute $\text{12}\text{.5}\times {\text{10}}^5\text{s}$ for $\Delta t$, $5\times {10}^5\text{rad}/\text{s}$ for ${\omega }_i$, and $-0.4\text{rad}/{\text{s}}^2$ for $\alpha$ in equation (II), to find $\Delta \theta$.

    $\begin{array}{c}\Delta \theta =\left(5\times {10}^5\text{rad}/\text{s}\right)\left(\text{12}\text{.5}\times {\text{10}}^5\text{s}\right)+\frac{1}{2}\left(0.4\text{rad}/{\text{s}}^2\right){\left(\text{12}\text{.5}\times {\text{10}}^5\text{s}\right)}^2\\ =6.25\times {10}^{11}\text{rad}-\text{3}\text{.125}\times {\text{10}}^{11}\text{rad}\\ \text{=3}\text{.125}\times {\text{10}}^{11}\text{rad}\end{array}$

Substitute $\text{3}\text{.125}\times {\text{10}}^{11}\text{rad}$ for $\Delta \theta$ in equation (III), to find the number of revolutions.

    $\begin{array}{c}\text{The number of revolution =}\frac{\text{3}\text{.125}\times {\text{10}}^{11}\text{rad}}{2\left(3.14\right)\text{rad}/\text{rev}}\\ =4.97\times {10}^{10}\text{rev}\\ \approx \text{5}\times {\text{10}}^{10}\text{rev}\end{array}$

Therefore, the angular displacement of the rotor during the time interval $\Delta t$ is $\underset{\_}{\text{5}\times {\text{10}}^{10}\text{rev}}$.