Answer:
The angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s
Explanation:
Given;
moment of inertia of a skater with arms out, [tex]I_{arms \ out}[/tex] = 3.1 kg.m²
moment of inertia of a skater with arms in, [tex]I_{arms \ in}[/tex] = 0.9 kg.m²
inward angular speed, [tex]\omega _{in}[/tex] = 4 rev/s
The angular momentum of the skater when her arms are out and one leg extended is equal to her angular momentum when her arms and legs are in.
[tex]L_{out} = L_{in}[/tex]
[tex]I_{out} \omega_{out} = I_{in} \omega_{in}\\\\\omega_{out} = \frac{ I_{in} \omega_{in} }{I_{out} } \\\\\omega_{out} = \frac{0.9*4}{3.1} \\\\\omega_{out} = 1.161 \ rev/s[/tex]
Therefore, the angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s
