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The Figure Skater

  

Figure: A contrived central-force problem. The ball swings around (without friction, of course) on the end of a string fixed at the origin  O. The central force in the string cannot generate any torque about  O, so the angular momentum   L_O = m v r  about  O  must remain constant. As the string is pulled in slowly, the radius  r  gets shorter so the momentum   $p = m v = m r \omega$  has to increase to compensate.

Again, so what? Well, there are numerous examples of central forces in which angular momentum conservation is used to make sense of otherwise counterintuitive phenomena. For instance, consider the classic image of the figure skater doing a pirouette: she starts spinning with hands and feet as far extended as possible, then pulls them in as close to her body. As a result, even though no torques were applied, she spins much faster. Why? I can't draw a good figure skater, so I will resort to a cruder example [shown in Fig. 11.5] that has the same qualitative features: imagine a ball (mass  m) on the end of a string that emerges through a hole in an axle which is held rigidly fixed. The ball is swinging around in a circle in the end of the string. For an initial radius  r  and an initial velocity   $v = r \omega$,  the initial momentum is   $m r \omega$  and the angular momentum about  O  is   $L_O = m v r = m r^2 \omega$. Now suppose we pull in the string until   $r' = {1\over2} r$. To keep the same  L_O  the momentum (and therefore the velocity) must increase by a factor of 2, which means that the angular velocity   $\omega\,' = 4 \omega$  since the ball is now moving at twice the speed but has only half as far to go around the circumference of the circle. The period of the "orbit" has thus decreased by a factor of four!

Returning to our more æsthetic example of the figure skater, if she is able to pull in all her mass a factor of 2 closer to her centre (on average) then she will spin 4 times more rapidly in the sense of revolutions per second or "Hertz" (Hz).