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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.

\\


{card:Part A}

{card:label=Part B}


h3. Part B 

Consider the same basic situation as in Part A, but now suppose that the wall has friction.  If the coefficient of static friction between the floor and the ladder is only 0.15, what is the minimum coefficient of static friction needed between the wall and the ladder to prevent the ladder from slipping?  

h4. Solution

{toggle-cloak:id=sysb} *System:* {cloak:id=sysb} The ladder as a [rigid body].{cloak}

{toggle-cloak:id=intb} *Interactions:* {cloak:id=intb}External influences from the earth (gravity) the wall (normal force and friction) and the floor (normal force and frictional force).{cloak}

{toggle-cloak:id=modb} *Model:*  {cloak:id=modb}[Single-Axis Rotation of a Rigid Body] and [Point Particle Dynamics].{cloak}

{toggle-cloak:id=appb} *Approach:*  
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{toggle-cloak:id=prelim} {color:red} *An Important Assumption* {color}

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The minimum coefficient of friction will occur when both friction forces are maxima.  Thus, we have the important relationships:

{latex}\begin{large}\[ |F_{f,x}| = \mu_{s,f} |F_{f,y}| \]
\[ |F_{w,y}| = \mu_{s,w} |F_{w,x}| \]\end{large}{latex}
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{toggle-cloak:id=diagb} {color:red} *Diagrammatic Representation* {color}

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Because we know the floor's static friction coefficient, we can streamline the solution by switching our choice of axis.

!ladder2.jpg!

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

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The resulting equations are:

{latex}\begin{large}\[ \sum F_{x} = F_{w,x} - \mu_{s,f}F_{f,y} = 0 \]
\[ \sum F_{y} = - mg + F_{f,y} + \mu_{s,w}F_{w,x}= 0 \]
\[ \sum \tau = - mg (L/2) \cos\theta + F_{f,y} L\cos\theta - \mu_{s,f}F_{f,y} L \sin\theta \]\end{large}{latex}

{note}Note that the y-component of the floor force is the normal force from the floor on the ladder, and the x-component of the wall force is the normal force from the wall on the ladder.  Since we are assuming static friction is maximized, we have rewritten _F_~f,x~ and _F_~w,y~ as the coefficient of friction times the relevant normal force.{note}

The only equation that can be solved immediately is torque balance, which gives:

{latex}\begin{large}\[ F_{f,y} = \frac{mg}{2\left(1-\mu_{s,f}\tan\theta\right)} = \mbox{125 N} \]\end{large}{latex}

We can then substitute this result into the equation of x-force balance to find:

{latex}\begin{large}\[ F_{w,x} = \mu_{s,f}F_{f,y} = \mbox{18.8 N} \]\end{large}{latex}

Finally, we can find the required coefficient of static friction to be:

{latex}\begin{large}\[ \mu_{s,w} = \frac{mg-F_{f,y}}{F_{w,x}} = 1.19 \] \end{large}{latex}

{tip}We found in Part A that the ladder is stable as long as the coefficient of friction with the ground is 0.19 or greater, even if there is no friction at all from the wall.  Does it make sense that in Part B the wall's friction coefficient has to be so much larger than 0.19 if the wall is only partly responsible for keeping the ladder from sliding?{tip}

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