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Consider an ice skater performing a spin. The ice is very nearly a frictionless surface 

A skater spinning around has constant angular momentum, but can change his or her Moment of Interia  by changing body position. What happens to their rate of rotation when they do so?


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{card:label=Part A}

h3. Part A

Assume that the skater originally is standing straight up while spinning, with arms extended. He or she then draws his or her arms inwards tightly to the sides. How does the angular velocity change?

h4. Solution





{toggle-cloak:id=sysa1} *System:*  {cloak:id=sysa1}The skater is treated as a [rigid body] in two configurations -- one with the arms extended, the other with arms held close.{cloak}

{toggle-cloak:id=inta1} *Interactions:* {cloak:id=inta1}External influences -- none. The skater is moving on a frictionless surface.{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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Consider the skater intially as a massless vertical pole with the arms modeled as massless rods of length {*}L{~}i{~}{*} with a point mass *m* at each end. 

|!Skater Initial 02.PNG!|

After contracting the skater's arms, the two masses are each a distance {*}L{~}f{~}{*} from the body

|!Skater Final 02.PNG!|


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

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The definition of the Moment of Inertia is:

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

So that if we calculate the initial Moment of Inertia about the vertical pole that is the skater's torso, we get:

{latex}

\begin{large}\[ I_{\rm iring} = 2 m d_{\rm ir^{2} \] \end{large}{latex}

For the "final" configuration the Moment

 of Inertia becomes:

{latex}\begin{large}\[ I \sum L = L_{\rm frod} = 2 m d+ L_{\rm fring} = 0 \] \end{large}{latex}


The Angular Momentum *L* has a magnitude given by 


{latex}

\begin{large}\[ L = I \omega \] \end{large}{latex}

so the _initial_ angular momentum is 

{latex}\begin{large\rm rod}\[ Lomega_{\rm irod} =+ I_{\rm iring} \omega_{\rm iring} = 2 m d_{\rm i} \omega_{\rm i} 0 \] \end{large}{latex}

and the _final_ angular momentum is 

{latex}

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

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


{latex}\begin{large}\[  drod}}\frac{I_{\rm i} \omega_{\rm i} =  drod}}{I_{\rm f} \omega_{\rm fring}} \] \end{large}

{latex}

or

{latex}\begin{large} \[ \omega_{\rm fring} = - {\omega_{\rm irod}} \frac{d_{\rm i}}{d_{\rm f}}MD^2}{12mr^2} \] \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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{card:Part A}

{card:label=Part B}


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

How can you decrease the needed angular velocity of the flywheel in order to produce the same angular velocity in the spacecraft?

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

{latex}

\begin{large}\[E_{\rm ftotal} = \frac{1}{2}I_{\rm frod}{\omega_{\rm frod}}^2 + \]\endfrac{large1}{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}\[ {\omega2}I_{\rm fring}}^2 = {\omega_{\rm iring}}^2 \frac{d_{\rm i}}{d_{\rm f}} \]\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}\[\Delta E = E_{\rm ftotal} -= E_{\rm i} \]\end{large}{latex}

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

{latex}\begin{large}\[\Delta E = E\frac{1}{2}mr^2{\omega_{\rm iring}}^2(1 + \frac{I_{\rm i}}{I_{\rm f}} - 112 m r^2}{M D^2}) \]\end{large}{latex}








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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.