The rotational speed of the trapeze artist is 32.372 revolutions per second.
In this question we should apply the principle of angular momentum conservation to determine the final angular speed of the trapeze artist, since there are no external forces acting on the trapeze artist:
[tex]I_{A}\cdot \omega_{A} = I_{B}\cdot \omega_{B}[/tex] (1)
Where:
[tex]I_{A}[/tex], [tex]I_{B}[/tex] - Initial and final moments of the trapeze artist, in kilograms-square meter.
[tex]\omega_{A}[/tex], [tex]\omega_{B}[/tex] - Initial and final angular speeds, in radians per second.
If we know that [tex]I_{A} = 17.9\,kg\cdot m^{2}[/tex], [tex]I_{B} = 33.1\,kg\cdot m^{2}[/tex] and [tex]\omega_{A} = 6.27\,\frac{rad}{s}[/tex], then the final angular speed is:
[tex]\omega_{B} = \frac{I_{A}}{I_{B}}\times \omega_{A}[/tex]
[tex]\omega_{B} = \frac{17.9\,kg\cdot m^{3}}{33.1\,kg\cdot m^{2}}\times 6.27\,\frac{rad}{s}[/tex]
[tex]\omega_{B} = 3.390\,\frac{rad}{s}[/tex] ([tex]32.372\,rpm[/tex])
The rotational speed of the trapeze artist is 32.372 revolutions per second.
We kindly invite to check this question on conservation of rotational momentum: https://brainly.com/question/13183618
