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Original artwork and photo by http://commons.wikimedia.org/wiki/User:Laretrotienda
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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?
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.
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
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.
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The axis of rotation is perpendicular to the page.
Mathematical Representation
The Moment of Inertia of a rod of lenngth D and mass M about its center is:
\begin
[ I_
= \frac
M D^
] \end
The Moment of Inertia of a ring of radius r and mass m is:
\begin
[ I_
= m r^
] \end
Conservation of Angular Momentum means that:
\begin
[ \sum L = L_
+ L_
= 0 ] \end
The Angular Momentum L is explicitly given by
\begin
[ L = I_
\omega_
+ I_
\omega_
= 0 ] \end
so the relationship between the angular velocities is:
\begin
[ \omega_
= - {\omega_{\rm rod}}\frac{I_{\rm rod}}{I_{\rm ring}} ] \end
.
\begin
[ \omega_
= - {\omega_{\rm rod}}\frac
] \end
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.
Question
How can you decrease the needed angular velocity of the flywheel in order to produce the same angular velocity in the spacecraft?
Answer
You can increase the mass of the flywheel, or you can increase its moment of inertia
Part B
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
\begin
[E_
= \sum \frac
I \omega^2 ]\end
\begin
[E_
= \frac
I_
{\omega_{\rm rod}}^2 + \frac
I_
{\omega_{\rm ring}}^2 ]\end
Using the expressions for the Moments of Inertia and the relationships between the angular velocities,we get:
\begin
[ E_
= \frac
mr^2{\omega_{\rm ring}}^2(1 + \frac
) ]\end
There is ZERO Kinetic 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.
For another example of a closed system with no external torques or forces, in which the Kinetic Energy changes due to interactions between the components of the system, see the worked example called Twirling Skater