Chapter 9, Problem 38P

(a)

To determine

To Find: The average angular acceleration during $4.5\text{-s}$ spin-up and again during $\text{0}\text{.24-s}$ spin- down.

Answer to Problem 38P

  ${\alpha }_{avg1}=30{\text{rad/s}}^{\text{2}}$

  ${\alpha }_{avg2}=-1.2\times {10}^2{\text{rad/s}}^{\text{2}}$

Explanation of Solution

Given:

Initial angular speed of the fly wheel, ${\omega }_i=0$

Final angular speed of the fly wheel

  $\begin{array}{l}{\omega }_f=4.5\text{rev/s}\\ \text{=}\left(4.5\text{rev/s}\right)\times \left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\\ =9\pi \text{rad/s}\end{array}$

Time taken to pull the rope, $\Delta t=0.95\text{s}$

Time taken during spin up, $\Delta t_{up}=4.5\text{s}$

Time taken during spin down, $\Delta t_{down}=0.24\text{s}$

Formula used:

  $\begin{array}{l}{\alpha }_{avg}=\frac{\Delta \omega }{\Delta t}\\ {\alpha }_{avg}=\frac{{\omega }_f-{\omega }_i}{\Delta t}\end{array}$

  ${\alpha }_{avg}$ = average acceleration

  ${\omega }_f$ = final angular velocity

  ${\omega }_i$ = initial angular velocity

Calculation:

Average angular acceleration for spin up is:

  $\begin{array}{c}{\alpha }_{avg1}=\frac{\left(9\pi -0\right)\text{rad/s}}{0.95\text{s}}\\ {\alpha }_{avg1}=29.76{\text{rad/s}}^{\text{2}}\\ \approx 30{\text{rad/s}}^{\text{2}}\end{array}$

Average angular acceleration for spin down is:

  $\begin{array}{c}{\alpha }_{avg2}=\frac{\left(0-9\pi \right)\text{rad/s}}{0.24\text{s}}\\ {\alpha }_{avg2}=-117.8{\text{rad/s}}^{\text{2}}\\ \approx -1.2\times {10}^2{\text{rad/s}}^{\text{2}}\end{array}$

Conclusion:

Spin down acceleration $=-1.2\times {10}^2{\text{rad/s}}^2$

Spin up acceleration $=30{\text{rad/s}}^{\text{2}}$

(b)

To determine

To find: Maximum angular speed reached by fly wheel.

Answer to Problem 38P

  ${\omega }_{\max }\text{=28}\text{rad/s}$

Explanation of Solution

Given:

Initial angular speed of the fly wheel, ${\omega }_i=0$

Final angular speed of the fly wheel

  $\begin{array}{l}{\omega }_f=4.5\text{rev/s}\\ \text{=}\left(4.5\text{rev/s}\right)\times \left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\\ =9\pi \text{rad/s}\end{array}$

Formula used:

  ${\omega }_{\max }={\omega }_f$

Where,

  ${\omega }_{\max }$ = maximum angular velocity

  ${\omega }_f$ = final angular velocity

Calculation:

Substitute the given value in the equation,

  $\begin{array}{c}{\omega }_{\max }={\omega }_f\\ =9\pi \text{rad/s}\\ \approx \text{28}\text{rad/s}\end{array}$

Conclusion:

  ${\omega }_{\max }\text{=28}\text{rad/s}$ is maximum angular speed reached by fly wheel.

(c)

To determine

To find: The ratio of the number of revolutions made during spin-up to the number made during spin down.

Answer to Problem 38P

  $\frac{\Delta {\theta }_{up}}{\Delta {\theta }_{down}}\approx 4$

Explanation of Solution

Given:

Time taken to pull the rope, $\Delta t=0.95\text{s}$

Time taken during spin up, $\Delta t_{up}=4.5\text{s}$

Time taken during spin down, $\Delta t_{down}=0.24\text{s}$

Angular speed at spin up $=0$

Angular speed at spin down, ${\omega }_{\max }=28\text{rad/s}$

Angular acceleration for spin up, ${\alpha }_{up}=30{\text{rad/s}}^{\text{2}}$

Angular acceleration for spin up, ${\alpha }_{down}=-1.2\times {10}^2{\text{rad/s}}^{\text{2}}$

Formula used:

Number of revolutions can be obtained by:

  $\Delta \theta =\omega t+\frac{1}{2}\alpha t^2$

Where,

  $\theta$ is the number of revolutions.

  $\Delta t$ is the time taken.

  $\omega$ is the angular speed

  $\alpha$ is angular acceleration

Calculation:

For spin-up revolution:

  $\begin{array}{l}\Delta {\theta }_{up}=\omega t_{up}+\frac{1}{2}{\alpha }_{up}{t}_{up}^2\\ \Delta {\theta }_{up}=0+\frac{1}{2}\times 30\times {0.95}^2\\ \Delta {\theta }_{up}=\mathrm{13.5373.....}\left(1\right)\end{array}$

For spin down revolution:

  $\begin{array}{l}\Delta {\theta }_{down}=\omega t_{down}+\frac{1}{2}{\alpha }_{down}{t}_{down}^2\\ \Delta {\theta }_{down}=\left(28\times 0.24\right)+\frac{1}{2}\times \left(-1.2\times {10}^2\right){\left(0.24\right)}^2\\ \Delta {\theta }_{down}=6.72-3.456\\ \Delta {\theta }_{down}=3.264\mathrm{...}\left(2\right)\end{array}$

Divide equation (1) by equation (2)

  $\begin{array}{l}\frac{\Delta {\theta }_{up}}{\Delta {\theta }_{down}}=\frac{13.5373}{3.264}\\ \frac{\Delta {\theta }_{up}}{\Delta {\theta }_{down}}=4.147\\ \frac{\Delta {\theta }_{up}}{\Delta {\theta }_{down}}\approx 4\end{array}$

Conclusion:

Ratio of the number of revolutions made during spin-up to the number made during spin down is $4$ .