Chapter 8, Problem 99QAP

To determine

The change of the angular speed station with the first wave of evacuees

Answer to Problem 99QAP

The change of the angular speed station with the first wave of evacuees = $1.6\times {10}^{-3}\pi$ rad/s

Explanation of Solution

Given info:

Mass of the station = $1.0*{10}^{10}$ kg

Radius of the station = $1.8$ km

Rotation rate of the station = $2\pi$ rad/min

Mass of an escape pod = $4.0*{10}^5$ kg

Formula used:

  $L_1=I_1{\omega }_1$

  $L_1=$ Initial angular momentum

  $I_1=$ Initial moment of inertia

  ${\omega }_1=$ Initial angular frequency

  $L_2=I_2{\omega }_2$

  $L_2=$ New angular momentum

  $I_2=$ New moment of inertia

  ${\omega }_2=$ New angular frequency

  $I_1=\frac{1}{2}{m}_1{r}_1{}^2$

  $I_1=$ Initial moment of inertia

  $m_1=$ Mass of the station

  $r_1=$ Radius of the station

  $I_2=\frac{1}{2}{m}_1{r}_1{}^2-m_2{r}_1{}^2$

  $I_2=$ Moment of inertia after the first wave of evacuees

  $m_1=$ Mass of the station

  $r_1=$ Radius of the station

  $m_2=$ Mass if the first wave of escape pods

  $\Delta \omega ={\omega }_2-{\omega }_1$

  $\Delta \omega =$ change in the angular speed

  ${\omega }_1=$ Initial angular speed

  ${\omega }_2=$ New angular speed

Calculation:

Initial moment of inertia,

  $I_1=\frac{1}{2}{m}_1{r}_1{}^2=\frac{1}{2}\times 1.0\times {10}^{10}kg\times {\left(1.8\times {10}^{3}m\right)}^2$

New moment of inertia,

  $I_2=\frac{1}{2}\times m_1{r}_1{}^2-m_2{r}_1{}^2=\left[\frac{1}{2}\times 1.0\times {10}^{10}kg\times {\left(1.8\times {10}^{3}m\right)}^2\right]-\left[4.0\times {10}^5\times {10}^{3}kg\times {\left(1.8\times {10}^{3}m\right)}^2\right]$

Applying conservation of angular momentum,

  $L_1=L_2$

  $I_1{\omega }_1=I_2{\omega }_2$

  $\frac{1}{2}\times 1.0\times {10}^{10}kg\times {\left(1.8\times {10}^{3}m\right)}^2\times \frac{2\pi }{60}rad/s=\left[\frac{1}{2}\times 1.0\times {10}^{10}kg\times {\left(1.8\times {10}^{3}m\right)}^2\right]-\left[4.0\times {10}^5\times {10}^{3}kg\times {\left(1.8\times {10}^{3}m\right)}^2\right]\times {\omega }_2$

  ${\omega }_2=\frac{{10}^9\pi }{3\times 9.2\times {10}^9}$

  ${\omega }_2=3.6\times {10}^{-2}\pi rad/s$

Change in angular speed,

  $\Delta \omega ={\omega }_2-{\omega }_1=(3.6\times {10}^{-2}\pi rad/s)-(2\pi /60rad/s)=2.67\times {10}^{-3}rad/s$

Conclusion:

The change of the angular speed station with the first wave of evacuees = $1.6\times {10}^{-3}\pi$ rad/s