Physics of a Fouetté Turn

In ballet, there are several types of turns a dancer can do. One of the most difficult types is a Fouetté. The video below describes the motions involved in such a turn.

As the dancer turns, she brings her leg inward which decreases her moment of inertia. This in turn increases her angular velocity. I decided to calculate what the dancer's velocity would be right after she brings her leg in. To do this, I timed how long it took the dancer to turn with her leg extended. She completed one turn in 0.8 seconds, which is about 2.5*pi rad/s. Additionally, I considered the dancer with her leg close to the body as a uniform cylinder, giving a moment of inertia as I=1/2MR^2 (Even though the dancer does not bring her leg all the way in, I did this to simplify my calculations). I considered her leg to be a uniform rod rotating about its end.

Here are my knowns:

Radius of body= 0.1524 meters

Mass of person= 60 kg

Mass of leg= 6 kg

Length of leg= 0.762 meters

w2=7.85 rad/s

To solve for the angular velocity of the dancer with her leg brought in:

I1w1=I2w2

(1/2)MR^2(w1) = (1/2MR^2 + 1/3mL^2)(w2)

1/2(60)(0.1524^2)(w1)= (1/2(60)(0.1524^2)+1/3(6)(0.762^2))(7.85)

w1= 20.9 rad/s

In linear terms, this is about 3.2 m/s which is pretty fast.

An observation I made was that when a dancer does a fouetté turn en pointe, they spin faster. This is because the dancer decreases the radius that they are spinning on. This is turn decreases their moment of inertia, which would later increase their angular velocity. This is a video of fouetté turns en pointe:

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