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{excerpt:hidden=true}Changes in Angular Velocity when the Moment of Inertia is changed, but no torque applied.{excerpt}


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|Original artwork and photo by http://commons.wikimedia.org/wiki/User:Laretrotienda 
Photo courtesy [Wikimedia Commons|http://commons.wikimedia.org]|

You can change the orientation of a spacecraft by rotating an internal flywheel in the direction _opposite_ the direction you want the spacecraft to move. Conservation of angular momentum will guarantee that the ship will rotate in space, without the use of any propellants or changes in the trajectory of the ship. This idea has been used in science fiction for a long time (See Robert A, Heinlein's 1947 novel *Rocket Ship Galileo*, or his 1959 novel *Starship Troopers*), and it is being used today on board unmanned satellites. There are several companies manufacturing such Attitude Control Devices. How do they work? 

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h3. Part A

Assume for simplicity that we can model the Spacecraft as a rod of length *D* and mass *M*. The Flywheel is a ring of radius *r* and mass *m*. We ignore the masses of the spokes, hub, and motor that effects the rotation of the wheel and Spacecraft relative to each other. 

h4. Solution





{toggle-cloak:id=sysa1} *System:*  {cloak:id=sysa1}The Spacecraft and the Flywheel are each treated as a [rigid body].{cloak}

{toggle-cloak:id=inta1} *Interactions:* {cloak:id=inta1}External influences -- none. The ship is in space and the only interactions are internal.{cloak}

{toggle-cloak:id=moda1} *Model:*  {cloak:id=moda1}[Single-Axis Rotation of a Rigid Body]{cloak}

{toggle-cloak:id=appa1} *Approach:*  

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{toggle-cloak:id=diaga1} {color:red} *Diagrammatic Representation* {color}

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As stated above, the Spacecraft is treated as a Rod of length *D* and mass *M*. The Flywheel is a ring of radius *r* and mass *m*. They are connected by an axis that runs through the Center of Mass of each.

|!Rotating Spacecraft 01.PNG!|

The axis of rotation is perpendicular to the page.

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{toggle-cloak:id=matha1} {color:red} *Mathematical Representation* {color}

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The Moment of Inertia of a rod of lenngth *D* and mass *M* about its center is:

{latex}\begin{large}\[ I_{\rm rod} = \frac{1}{12}M D^{2} \] \end{large}{latex}

The Moment of Inertia of a ring of radius *r* and mass *m* is:

{latex}\begin{large}\[ I_{\rm ring} =  m r^{2} \] \end{large}{latex}

Conservation of Angular Momentum means that:

{latex}\begin{large}\[ \sum L = L_{\rm rod} + L_{\rm ring} = 0 \] \end{large}{latex}


The Angular Momentum *L* is explicitly given by 


{latex}\begin{large}\[ L = I_{\rm rod}\omega_{\rm rod} + I_{\rm ring}\omega_{\rm ring} = 0 \] \end{large}{latex}

so the relationship between the angular velocities is: 

{latex}\begin{large}\[ \omega_{\rm ring} = - {\omega_{\rm rod}}\frac{I_{\rm rod}}{I_{\rm ring}} \] \end{large}{latex}
.
{latex}\begin{large}\[ \omega_{\rm ring} = - {\omega_{\rm rod}}\frac{MD^2}{12mr^2} \] \end{large}{latex}

Since *M* is probably orders of magnitude larger than *m*, and *D* is probably orders of magnitude larger than *r*, the angular velocity of the spacecraft is going to be very small relative to the angular velocity of the flywheel, so the flywheel will have to spin very rapidly to be useful. 

{info:title=Question}How can you decrease the needed angular velocity of the flywheel in order to produce the same angular velocity in the spacecraft?{info}  



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h3. Part B 


h4. Solution



What is the Kinetic Energy of the Spacecraft and the Flywheel? We ignore any velocity the combined system may have due to travelling at some velocity, and consider only the rotational Kinetic Energy.



From our expressions for Rotational Energy, we know that this energy is given by


{latex}\begin{large}\[E_{\rm total} = \sum \frac{1}{2} I \omega^2 \]\end{large}{latex}

\\
{latex}\begin{large}\[E_{\rm total} = \frac{1}{2}I_{\rm rod}{\omega_{\rm rod}}^2 + \frac{1}{2}I_{\rm ring}{\omega_{\rm ring}}^2 \]\end{large}{latex}

Using the expressions for the Moments of Inertia and the relationships between the angular velocities,we get:

{latex}\begin{large}\[ E_{\rm total} = \frac{1}{2}mr^2{\omega_{\rm ring}}^2(1 + \frac{12 m r^2}{M D^2}) \]\end{large}{latex}

{warning} TheThere aboveis equationZERO isKinetic NOT correct.{warning}

Clearly there is an error, since this does not agree with our previous result.


The error is the assumption that energy is conserved. In drawing his or her arms inwards, the skater must exert force to bring them closer tto the axis and rotate faster. This force acting over a distance performs work and increases the system energy. The change in energy is given by the difference between the above expressions

{latex}\begin{large}\[\Delta E = E_{\rm f} - E_{\rm i} \]\end{large}{latex}

Substituting in from the equations in Part A, one finds, after a little algebra:

{latex}\begin{large}\[\Delta E = E_{\rm i}(\frac{I_{\rm i}}{I_{\rm f}} - 1) \]\end{large}{latex}Energy before the Flywheel starts rotating and ZERO kinetic energy after it has stopped. But while the Flywheel (and the spacecraft) are rotating there very clearly is a nonzero kinetic energy. This is true even though there are no external forces or torques applied. So the kinetic energy of a closed system can be changed purely by internal interactions between the components of a system.{warning}









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