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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}

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

For the "final" configuration the Moment of Inertia becomes:

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

What if we looked at this from the point of view of Conservation of Energy?

Sample Samnple Sample Sample






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