A space station has the form of a hoop of radius $\mathit{R}\mathbf{=}\mathbf{14}\mathbf{\text{m}}$, with mass $\mathit{M}\mathbf{=}\mathbf{625}\mathbf{\text{0 kg}}$(Figure 11.91). Initially its center of mass is not moving, but it is spinning with angular speed ${\mathit{\omega }}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0013}\mathbf{}\mathbf{\text{rad/s}}$Then a small package of mass $\mathit{m}\mathbf{=}\mathbf{6}\mathbf{\text{kg}}$is thrown by a spring-loaded gun toward a nearby spacecraft as shown; the package has a speed$\mathit{v}\mathbf{=}\mathbf{40}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$ after launch. Calculate the center-of-mass velocity of the space station and its rational speed after the launch.

The general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant.

Apply conservation of linear momentum to solve for the velocity of the center of mass of the space station along $x$and $y$directions. Also apply conservation of angular momentum to solve for the angular speed of rotation of the hoop.

Assume the space station and package as a system.

Thus, the net external force acting the system is zero, so the momentum of the system is conserved.

$\begin{array}{rcl}{\overrightarrow{F}}_{net}& =& \frac{d\overrightarrow{p}}{dt}\\ 0& =& \frac{d\overrightarrow{p}}{dt}\\ & & \end{array}$

Thus, final linear momentum of the system is equal to the linear momentum of the system.

Hence, ${\overrightarrow{p}}_f={\overrightarrow{p}}_i$

Let $\overrightarrow{V}$be the final speed of the centre of mass of the space station.

So, the final linear momentum of the system is given by,

${\overrightarrow{p}}_f=M\overrightarrow{V}+m\overrightarrow{v}$

Here, $M$is the mass of the space-station, $m$is the mass of the package, $\overrightarrow{v}$is the velocity of the package, and $\overrightarrow{V}$ is the velocity of the space station.

The $x$component of the speed of the small package is,

$v_x=v\cos \theta$

The $y$component of the speed of the small package is,

$v_y=v\cos \theta$

The $z$component of the speed of the small package is,

$v_z=0$

Now, the final momentum of the system can be written as,

Initial linear momentum of the system is zero, because the centre of mass of the space station is at rest.

Thus,${\overrightarrow{p}}_i=0$

As thefinal linear momentum of the system is equal to the linear momentum of the system.

Therefore, the components of the centre of mass velocity of mass velocity of the space station are,

$\begin{array}{rcl}{V}_x& =& -\frac{m}{M}v\cos \theta \\ V_y& =& -\frac{m}{M}vsn\theta \\ V_z& =& 0\\ & & \end{array}$

Thus, the magnitude of centre of mass velocity of space-station is,

$\begin{array}{rcl}V& =& \sqrt{{\left(-\frac{m}{M}v\cos \theta \right)}^2+{\left(-\frac{m}{M}v\sin \theta \right)}^2+0}\\ & =& \sqrt{{\left(\frac{m}{M}v\right)}^2{\cos }^2\theta +{\left(\frac{m}{M}v\right)}^2{\sin }^2\theta }\\ & =& \left(\frac{m}{M}v\right)\sqrt{{\cos }^2\theta +{\sin }^2\theta }\\ & =& \frac{m}{M}v\\ & & \end{array}$

Substitute $6\text{kg}$ for $m$, $6250\text{kg}$for $M$, $40\mathrm{m}/\mathrm{s}$ for $v$ in the above equation.

$\begin{array}{rcl}V& =& \left(\frac{6\text{kg}}{6250\text{kg}}\right)\left(40\mathrm{m}/\mathrm{s}\right)\\ & =& 0.0384\mathrm{m}/\mathrm{s}\end{array}$

The net external torque acting on the system is zero, so the angular momentum of the system is conserved.

$\begin{array}{rcl}{\overrightarrow{\tau }}_{net}& =& \frac{d\overrightarrow{L}}{dt}\\ 0& =& \frac{d\overrightarrow{L}}{dt}\\ & & \end{array}$

Thus, final angular momentum of the system is equal to initial angular momentum of the system.

${\overrightarrow{L}}_f={\overrightarrow{L}}_i$

Let $\hat{i},\hat{j}$ and $\hat{k}$are the unit vector along the $x,y$ and $z$ axis respectively.

The moment of inertia of the space station (hoop) of mass $M$ and radius $R$ is,

$I=M{R}^2$

The initial angular momentum of the system is given by,

${\overrightarrow{L}}_i=I{\omega }_0\hat{k}$

The final angular momentum of the system is given by,

Applying conservation of momentum to the space-station package system yields,

$\begin{array}{rcl}{\overrightarrow{L}}_f& =& \overrightarrow{L_i}\\ I\omega \hat{k}+Rmv\sin \theta \hat{k}& =& I{\omega }_0\hat{k}\\ I\omega +Rmv\sin \theta & =& I{\omega }_0\\ I\omega & =& I{\omega }_0-Rmv\sin \theta \\ & & \end{array}$

Therefore, the final rotational speed of the space station after the launch is,

\(\begin{aligned}{}\omega = {\omega _0} - \frac{{Rmv\sin \theta }}{I}\\ = {\omega _0} - \frac{{Rmv\sin \theta }}{{M{R^2}}}\\ = {\omega _0} - \frac{{mv\sin \theta }}{{MR}}\end{aligned}\)

Here,\(\frac{{mv\sin \theta }}{M}\)is\({V_y}\), so the above equation will be,

\(\omega = {\omega _0} - \frac{{{V_y}}}{R}\)

Substitute\(\frac{V}{{\sqrt 2 }}\)for\({V_y}\)

Thus, the final expression for the rational speed is,

\(\omega = {\omega _o} - \frac{V}{{\sqrt 2 R}}\)

Substitute \(0.0013{\rm{ }} {{{\rm{rad}}} \mathord{\left/ {\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}}\)for\({\omega _0}\),\(0.0384{\rm{ }}{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\)for\(V\), \(6{\rm{ kg}}\)for\(m\), \(6250{\rm{ kg}}\)and\(14{\rm{ m}}\)for \(R\)in the above equation.

\(\begin{aligned}{}\omega = 0.0013{\rm{ }}{{{\rm{rad}}} \mathord{\left/

{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.} {\rm{s}}} - \frac{{0.0384{\rm{ }}{{\rm{m}} \mathord{\left/

{\vphantom {{\rm{m}} {\rm{s}}}} \right.

\kern-\nulldelimiterspace} {\rm{s}}}}}{{\sqrt 2 \left( {14{\rm{ m}}} \right)}}\\ = 0.0013{\rm{ }}{{{\rm{rad}}} \mathord{\left/

{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.

\kern-\nulldelimiterspace} {\rm{s}}} - 0.001939{\rm{ }}{{{\rm{rad}}} \mathord{\left/

{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.

\kern-\nulldelimiterspace} {\rm{s}}}\\ = - 6.39 \times {10^{ - 4}}{\rm{ }}{{{\rm{rad}}} \mathord{\left/

{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.

\kern-\nulldelimiterspace} {\rm{s}}}\end{aligned}\)

Hence, the final rotational velocity of space station is\( - 6.39 \times {10^{ - 4}}{\rm{ }}{{{\rm{rad}}} \mathord{\left/

{\vphantom {{{\rm{rad}}} {\rm{s}}}} \right.

\kern-\nulldelimiterspace} {\rm{s}}}\)and the station rotates clockwise.