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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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|!800px-TZ_Rocketmod.jpg!|
|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 _initial_relationship between the angular momentumvelocities is: 

{latex}\begin{large}\[ L\omega_{\rm iring} = I_{\rm i}- \omega_{\rm i} = 2 m drod)\frac{I_{\rm i} \omegarod}}{I_{\rm iring}} \] \end{large}{latex}

and the _final_ angular momentum is 

{latex}\begin{large}\[ L_{\rm f} = I_{\rm f} \omega_{\rm f} = 2 m d_{\rm f} \omega_{\rm f} \] \end{large}{latex}

Since the Angular Momentum is unchanged, the initial and final expressions should be equal. This means that 


{latex}\begin{large}\[  d_{\rm i} \omega_{\rm i} =  d_{\rm f} \omega_{\rm f} \] \end{large}{latex}

or

{latex}\begin{large} \[\omega_{\rm f} = \omega_{\rm i} \frac{d_{\rm i}}{d_{\rm f}} \] \end{large}{latex}

After drawing in his or her arms, the skater is spinning much more rapidly, without the application of any external forces or torques.

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h3. Part B 


h4. Solution



What if we address this from the standpoint of Energy? If the Energy is conserved (since there is no external work done on the system), then we ought to be able to derive the angular velocity from Conservation of Energy.

{warning} This assumption really is not correct -- Energy is NOT conserved in this case, despite the absence of outside forces. But we'll proceed to see where this assuimption leads.{warning}

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


{latex}\begin{large}\[E_{\rm i} = \frac{1}{2}I_{\rm i}{\omega_{\rm i}}^2 \]\end{large}{latex}

{latex}\begin{large}\[E_{\rm f} = \frac{1}{2}I_{\rm f}{\omega_{\rm f}}^2 \]\end{large}{latex}

If the energy is the same at the end as at the beginning, then we can equate these expressions. After some algebra, we get:

{latex}\begin{large}\[ {\omega_{\rm f}}^2 = {\omega_{\rm i}}^2 \frac{d_{\rm i}}{d_{\rm f}} \]\end{large}{latex}

{warning} The above equation is 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}








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