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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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|!Yun_2009_National_sit_spin.jpg!|
|Yun Yea-Ji (KOR) at the 2009 South Korean Figure Skating Championships 
Photo courtesy [Wikimedia Commons|http://commons.wikimedia.org]|

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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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}\[ I = \sum {m r^{2}} \] \end{large}{latex}

 by 


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

 

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

 

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

 


{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}

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.

\begin{large}\[

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h3. Part B 


h4. Solution

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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.


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} \[ E_{\rm f)} = \frac{1}{2} II_{\rm f}{\omega_{\rm f} (}^2 \]\end{large}

\begin{large}\[ {\omega_{\rm f})^2  \] \end{large}{latex}



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{live-template:RELATE license}^2 = {\omega_{\rm i}}^2 \frac{d_{\rm i}}{d_{\rm f}} \]\end{large}

The above equation is NOT correct.

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

\begin{large}\[\Delta E = E_{\rm i}(\frac{I_{\rm i}}{I_{\rm f}} - 1) \]\end{large}