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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

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.

The axis of rotation is perpendicular to the page.

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

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

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

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

Conservation of Angular Momentum means that:

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

The Angular Momentum L is explicitly given by

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

so the relationship between the angular velocities is:

\begin{large}\[ \omega_{\rm ring} = - {\omega_{\rm rod}}\frac{I_{\rm rod}}{I_{\rm ring}} \] \end{large}

.

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

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.

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

\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}

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

\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}

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