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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 Inertia 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

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.

After contracting the skater's arms, the two masses are each a distance L_f from the body

The definition of the Moment of Inertia is:

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

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

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

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

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

The Angular Momentum L has a magnitude given by

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

so the initial angular momentum is

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

and the final angular momentum is

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

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

\begin{large}\[  d_{\rm i} \omega_{\rm i} =  d_{\rm f} \omega_{\rm f} \] \end{large}

or

\begin{large} \[\omega_{\rm f} = \omega_{\rm i} \frac{d_{\rm i}}{d_{\rm f}} \] \end{large}

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

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.

From our expressions for Rotational Energy, we know that the energy is given by

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

\begin{large}\[E_{\rm f} = \frac{1}{2}I_{\rm f}{\omega_{\rm f}}^2 \]\end{large}

If the energy is the same at the end as at the beginning, then we can equate these expressions. After some algebra, we get:

\begin{large}\[ {\omega_{\rm f}}^2 = {\omega_{\rm i}}^2 \frac{d_{\rm i}}{d_{\rm f}} \]\end{large}

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

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

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

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

If it is not clear how the Kinetic Energy can change within a closed system with no external torques or forces, see the example Rotating a Space Ship

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