Yun Yea-Ji (KOR) at the 2009 South Korean Figure Skating Championships |
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?
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?
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
Consider the skater intially as a massless vertical pole with the arms modeled as massless rods of length Li with a point mass m at each end.
After contracting the skater's arms, the two masses are each a distance Lf from the body
Mathematical Representation
The definition of the Moment of Inertia is:
\begin
[ I = \sum {m r^{2}} ] \end
So that if we calculate the initial Moment of Inertia about the vertical pole that is the skater's torso, we get:
\begin
[ I_
= 2 m d_
] \end
For the "final" configuration the Moment of Inertia becomes:
\begin
[ I_
= 2 m d_
] \end
The Angular Momentum L has a magnitude given by
\begin
[ L = I \omega ] \end
so the initial angular momentum is
\begin
[ L_
= I_
\omega_
= 2 m d_
\omega_
] \end
and the final angular momentum is
\begin
[ L_
= I_
\omega_
= 2 m d_
\omega_
] \end
Since the Angular Momentum is unchanged, the initial and final expressions should be equal. This means that
\begin
[ d_
\omega_
= d_
\omega_
] \end
or
\begin
[\omega_
= \omega_
\frac{d_{\rm i}}{d_{\rm f}} ] \end
After drawing in his or her arms, the skater is spinning much more rapidly, without the application of any external forces or torques.
Part B
What if we looked at this fro the point of view of Conservation of Energy?
\begin
[ K_
+ K_
+ U_
= K_
+ K_
+ U_
]\end
\begin
[ h = - x \sin\theta]\end
\begin
[ \omega_
R = v_
]\end
\begin
[ v_
= \sqrt{\frac
{1+\frac
{\displaystyle mR^{2}}}} ] \end
[ F_
= I\alpha /R ]
[ W = - \frac
{R^{2}} \frac
{1+\frac
{\displaystyle mR^
}} ]